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engineering/linear-systems-and-signals-3rd-edition-bp-lathi/060_4.8 FREQUENCY RESPONSE OF AN LTIC SYSTEM.md

@@ -1,266 +0,0 @@
-## **4.8 FREQUENCY [RESPONSE OF AN](#page-10-0) LTIC SYSTEM**
-
-Filtering is an important area of signal processing. Filtering characteristics of a system are indicated by its response to sinusoids of various frequencies varying from 0 to ∞. Such characteristics are called the frequency response of the system. In this section, we shall find the frequency response of LTIC systems.
-
-In Sec. 2.4-4 we showed that an LTIC system response to an everlasting exponential input *x*(*t*) = *est* is also an everlasting exponential *H*(*s*)*est*. As before, we use an arrow directed from the input to the output to represent an input–output pair:
-
-$$
-e^{st} \Longrightarrow H(s)e^{st} \tag{4.40}
-$$
-
-Setting *s* = *j*ω in this relationship yields
-
-$$
-e^{j\omega t} \Longrightarrow H(j\omega)e^{j\omega t} \tag{4.41}
-$$
-
-Noting that cosω*t* is the real part of *ej*ω*<sup>t</sup>* , use of Eq. (2.31) yields
-
-$$
-\cos \omega t \Longrightarrow \text{Re}[H(j\omega)e^{j\omega t}] \tag{4.42}
-$$
-
-We can express *H*(*j*ω) in the polar form as
-
-$$
-H(j\omega) = |H(j\omega)|e^{j\angle H(j\omega)}
-$$
-
-With this result, Eq. (4.42) becomes
-
-$$
-\cos \omega t \Longrightarrow |H(j\omega)| \cos [\omega t + \angle H(j\omega)]
-$$
-
-In other words, the system response *y*(*t*) to a sinusoidal input cosω*t* is given by
-
-$$
-y(t) = |H(j\omega)| \cos[\omega t + \angle H(j\omega)]
-$$
-
-Using a similar argument, we can show that the system response to a sinusoid cos(ω*t* +θ ) is
-
-$$
-y(t) = |H(j\omega)|\cos[\omega t + \theta + \angle H(j\omega)]\tag{4.43}
-$$
-
-This result is valid only for BIBO-stable systems. The frequency response is meaningless for BIBO-unstable systems. This follows from the fact that the frequency response in Eq. (4.41) is obtained by setting *s* = *j*ω in Eq. (4.40). But, as shown in Sec. 2.4-4 [Eqs. (2.38) and (2.39)], Eq. (4.40) applies only for the values of *s* for which *H*(*s*) exists. For BIBO-unstable systems, the ROC for *H*(*s*) does not include the ω axis where *s* = *j*ω [see Eq. (4.10)]. This means that *H*(*s*) when *s* = *j*ω is meaningless for BIBO-unstable systems.†
-
-Equation (4.43) shows that for a sinusoidal input of radian frequency ω, the system response is also a sinusoid of the same frequency ω. *The amplitude of the output sinusoid is* |*H*(*j*ω)| *times the input amplitude, and the phase of the output sinusoid is shifted by H*(*j*ω) *with respect to the input phase* (see later Fig. 4.38 in Ex. 4.27). For instance, a certain system with |*H*(*j*10)| = 3 and *H*(*j*10) = −30◦ amplifies a sinusoid of frequency ω = 10 by a factor of 3 and delays its phase by 30◦. The system response to an input 5cos(10*t* + 50◦) is 3 × 5 cos(10*t* + 50◦ − 30◦) = 15 cos(10*t* +20◦).
-
-Clearly |*H*(*j*ω)| is the amplitude *gain* of the system, and a plot of |*H*(*j*ω)| versus ω shows the amplitude gain as a function of frequency ω. We shall call |*H*(*j*ω)| the *amplitude response*. It also goes under the name *magnitude response*. ‡ Similarly, *H*(*j*ω) is the *phase response*, and a plot of *H*(*j*ω) versus ω shows how the system modifies or changes the phase of the input sinusoid. Plots of the magnitude response |*H*(*j*ω)| and phase response *H*(*j*ω) show at a glance how a system responds to sinusoids of various frequencies. Observe that *H*(*j*ω) has the information of |*H*(*j*ω)| and *H*(*j*ω) and is therefore termed the *frequency response* of the system. Clearly, the frequency response of a system represents its filtering characteristics.
-
-### **EXAMPLE 4.27 Frequency Response**
-
-Find the frequency response (amplitude and phase responses) of a system whose transfer function is
-
-$$
-H(s) = \frac{s+0.1}{s+5}
-$$
-
-Also, find the system response *y*(*t*) if the input *x*(*t*) is
-
-**(a)** cos 2*t*
-
-**(b)** cos(10*t* −50◦)
-
-In this case,
-
-$$
-H(j\omega) = \frac{j\omega + 0.1}{j\omega + 5}
-$$
-
-<sup>†</sup> This may also be argued as follows. For BIBO-unstable systems, the zero-input response contains nondecaying natural mode terms of the form cosω0*t* or *eat* cosω0*t* (*a* > 0). Hence, the response of such a system to a sinusoid cosω*t* will contain not just the sinusoid of frequency ω, but also nondecaying natural modes, rendering the concept of frequency response meaningless.
-
-<sup>‡</sup> Strictly speaking, <sup>|</sup>*H*(ω)<sup>|</sup> is magnitude response. There is a fine distinction between amplitude and magnitude. Amplitude *A* can be positive and negative. In contrast, the magnitude |*A*| is always nonnegative. We refrain from relying on this useful distinction between amplitude and magnitude in the interest of avoiding proliferation of essentially similar entities. This is also why we shall use the "amplitude" (instead of "magnitude") spectrum for |*H*(ω)|.
-
-### 414 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS
-
-Therefore,
-
-$$
-|H(j\omega)| = \frac{\sqrt{\omega^2 + 0.01}}{\sqrt{\omega^2 + 25}} \quad \text{and} \quad \angle H(j\omega) = \tan^{-1}\left(\frac{\omega}{0.1}\right) - \tan^{-1}\left(\frac{\omega}{5}\right)
-$$
-
-Both the amplitude and the phase response are depicted in Fig. 4.38a as functions of ω. These plots furnish the complete information about the frequency response of the system to sinusoidal inputs.
-
-**(a)** For the input *x*(*t*) = cos 2*t*, ω = 2, and
-
-$$
-|H(j2)| = \frac{\sqrt{(2)^2 + 0.01}}{\sqrt{(2)^2 + 25}} = 0.372
-$$
-  
-\n
-$$
-\angle H(j2) = \tan^{-1}\left(\frac{2}{0.1}\right) - \tan^{-1}\left(\frac{2}{5}\right) = 87.1^{\circ} - 21.8^{\circ} = 65.3^{\circ}
-$$
-
-**Figure 4.38** Responses for the system of Ex. 4.27.
-
-We also could have read these values directly from the frequency response plots in Fig. 4.38a corresponding to ω = 2. This result means that for a sinusoidal input with frequency ω = 2, the amplitude gain of the system is 0.372, and the phase shift is 65.3◦. In other words, the output amplitude is 0.372 times the input amplitude, and the phase of the output is shifted with respect to that of the input by 65.3◦. Therefore, the system response to the input cos 2*t* is
-
-$$
-y(t) = 0.372 \cos(2t + 65.3^{\circ})
-$$
-
-The input cos 2*t* and the corresponding system response 0.372cos(2*t* + 65.3◦) are illustrated in Fig. 4.38b.
-
-**(b)** For the input cos(10*t* − 50◦), instead of computing the values |*H*(*j*ω)| and *H*(*j*ω) as in part (a), we shall read them directly from the frequency response plots in Fig. 4.38a corresponding to ω = 10. These are
-
-$$
-|H(j10)| = 0.894
-$$
- and  $\angle H(j10) = 26^{\circ}$ 
-
-Therefore, for a sinusoidal input of frequency ω = 10, the output sinusoid amplitude is 0.894 times the input amplitude, and the output sinusoid is shifted with respect to the input sinusoid by 26◦. Therefore, the system response *y*(*t*) to an input cos(10*t* −50◦) is
-
-$$
-y(t) = 0.894 \cos (10t - 50^\circ + 26^\circ) = 0.894 \cos (10t - 24^\circ)
-$$
-
-If the input were sin(10*t* − 50◦), the response would be 0.894 sin(10*t* − 50◦ + 26◦) = 0.894 sin(10*t* −24◦).
-
-The frequency response plots in Fig. 4.38a show that the system has highpass filtering characteristics; it responds well to sinusoids of higher frequencies (ω well above 5), and suppresses sinusoids of lower frequencies (ω well below 5).
-
-### PLOTTING FREQUENCY RESPONSE WITH MATLAB
-
-It is simple to use MATLAB to create magnitude and phase response plots. Here, we consider two methods. In the first method, we use an anonymous function to define the transfer function *H*(*s*) and then obtain the frequency response plots by substituting *j*ω for *s*.
-
-```
->> H = @(s) (s+0.1)./(s+5); omega = 0:.01:20;
->> subplot(1,2,1); plot(omega,abs(H(1j*omega)),'k-');
->> subplot(1,2,2); plot(omega,angle(H(1j*omega))*180/pi,'k-');
-```
-
-In the second method, we define vectors that contain the numerator and denominator coefficients of *H*(*s*) and then use the freqs command to compute frequency response.
-
-```
->> B = [1 0.1]; A = [1 5]; H = freqs(B,A,omega); omega = 0:.01:20;
->> subplot(1,2,1); plot(omega,abs(H),'k-');
->> subplot(1,2,2); plot(omega,angle(H)*180/pi,'k-');
-```
-
-Both approaches generate plots that match Fig. 4.38a.
-
-### **EXAMPLE 4.28 Frequency Responses of Delay, Differentiator, and Integrator Systems**
-
-Find and sketch the frequency responses (magnitude and phase) for **(a)** an ideal delay of *T* seconds, **(b)** an ideal differentiator, and **(c)** an ideal integrator.
-
-**(a) Ideal delay of** *T* **seconds.** The transfer function of an ideal delay is [see Eq. (4.30)]
-
-$$
-H(s) = e^{-sT}
-$$
-
-Therefore,
-
-$$
-H(j\omega) = e^{-j\omega T}
-$$
-
-Consequently,
-
-$$
-|H(j\omega)| = 1 \quad \text{and} \quad \angle H(j\omega) = -\omega T
-$$
-
-These amplitude and phase responses are shown in Fig. 4.39a. The amplitude response is constant (unity) for all frequencies. The phase shift increases linearly with frequency with a slope of −*T*. This result can be explained physically by recognizing that if a sinusoid cosω*t* is passed through an ideal delay of *T* seconds, the output is cosω(*t* − *T*). The output sinusoid amplitude is the same as that of the input for all values of ω. Therefore, the amplitude response (gain) is unity for all frequencies. Moreover, the output cosω(*t* − *T*) = cos(ω*t* − ω*T*) has a phase shift −ω*T* with respect to the input cosω*t*. Therefore, the phase response is linearly proportional to the frequency ω with a slope −*T*.
-
-**(b) An ideal differentiator.** The transfer function of an ideal differentiator is [see Eq. (4.31)]
-
-*H*(*s*) = *s*
-
-Therefore,
-
-*<sup>H</sup>*(*j*ω) <sup>=</sup> *<sup>j</sup>*<sup>ω</sup> <sup>=</sup> <sup>ω</sup>*ej*π/<sup>2</sup>
-
-Consequently,
-
-$$
-|H(j\omega)| = \omega
-$$
- and  $\angle H(j\omega) = \frac{\pi}{2}$ 
-
-These amplitude and phase responses are depicted in Fig. 4.39b. The amplitude response increases linearly with frequency, and phase response is constant (π/2) for all frequencies. This result can be explained physically by recognizing that if a sinusoid cosω*t* is passed through an ideal differentiator, the output is −ωsin ω*t* = ωcos[ω*t* + (π/2)]. Therefore, the output sinusoid amplitude is ω times the input amplitude; that is, the amplitude response (gain) increases linearly with frequency ω. Moreover, the output sinusoid undergoes a phase shift π/2 with respect to the input cosω*t*. Therefore, the phase response is constant (π/2) with frequency.
-
-**Figure 4.39** Frequency response of an ideal **(a)** delay, **(b)** differentiator, and **(c)** integrator.
-
-In an ideal differentiator, the amplitude response (gain) is proportional to frequency [|*H*(*j*ω)| = ω] so that the higher-frequency components are enhanced (see Fig. 4.39b). All practical signals are contaminated with noise, which, by its nature, is a broadband (rapidly varying) signal containing components of very high frequencies. A differentiator can increase the noise disproportionately to the point of drowning out the desired signal. This is why ideal differentiators are avoided in practice.
-
-**(c) An ideal integrator.** The transfer function of an ideal integrator is [see Eq. (4.32)]
-
-$$
-H(s) = \frac{1}{s}
-$$
-
-Therefore,
-
-$$
-H(j\omega) = \frac{1}{j\omega} = \frac{-j}{\omega} = \frac{1}{\omega}e^{-j\pi/2}
-$$
-
-Consequently,
-
-$$
-|H(j\omega)| = \frac{1}{\omega} \quad \text{and} \quad \angle H(j\omega) = -\frac{\pi}{2}
-$$
-
-These amplitude and phase responses are illustrated in Fig. 4.39c. The amplitude response is inversely proportional to frequency, and the phase shift is constant (−π/2) with frequency. This result can be explained physically by recognizing that if a sinusoid cosω*t* is passed through an ideal integrator, the output is (1/ω)sin ω*t* = (1/ω) cos[ω*t* −(π/2)]. Therefore, the amplitude response is inversely proportional to ω, and the phase response is constant (−π/2)
-
-### <span id="page-437-0"></span>418 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS
-
-with frequency.† Because its gain is 1/ω, the ideal integrator suppresses higher-frequency components but enhances lower-frequency components with ω < 1. Consequently, noise signals (if they do not contain an appreciable amount of very-low-frequency components) are suppressed (smoothed out) by an integrator.
-
-### **DR ILL 4.15 Sinusoidal Response of an LTIC System**
-
-Find the response of an LTIC system specified by
-
-$$
-\frac{d^2y(t)}{dt^2} + 3\frac{dy(t)}{dt} + 2y(t) = \frac{dx(t)}{dt} + 5x(t)
-$$
-
-if the input is a sinusoid 20 sin(3*t* +35◦).
-
-### **ANSWER**
-
-10.23 sin(3*t* −61.91◦)
-
-### **[4.8-1 Steady-State Response to Causal Sinusoidal Inputs](#page-10-0)**
-
-So far we have discussed the LTIC system response to everlasting sinusoidal inputs (starting at *t* = −∞). In practice, we are more interested in causal sinusoidal inputs (sinusoids starting at *t* = 0). Consider the input *ej*ω*<sup>t</sup> u*(*t*), which starts at *t* = 0 rather than at *t* = −∞. In this case *X*(*s*) = 1/(*s* + *j*ω). Moreover, according to Eq. (4.27), *H*(*s*) = *P*(*s*)/*Q*(*s*), where *Q*(*s*) is the
-
-<sup>†</sup> A puzzling aspect of this result is that in deriving the transfer function of the integrator in Eq. (4.32), we have assumed that the input starts at *t* = 0. In contrast, in deriving its frequency response, we assume that the everlasting exponential input *ej*ω*<sup>t</sup>* starts at *<sup>t</sup>* = −∞. There appears to be a fundamental contradiction between the everlasting input, which starts at *t* = −∞, and the integrator, which opens its gates only at *t* = 0. Of what use is everlasting input, since the integrator starts integrating at *t* = 0? The answer is that the integrator gates are always open, and integration begins whenever the input starts. We restricted the input to start at *t* = 0 in deriving Eq. (4.32) because we were finding the transfer function using the unilateral transform, where the inputs begin at *t* = 0. So the integrator starting to integrate at *t* = 0 is restricted because of the limitations of the unilateral transform method, not because of the limitations of the integrator itself. If we were to find the integrator transfer function using Eq. (2.40), where there is no such restriction on the input, we would still find the transfer function of an integrator as 1/*s*. Similarly, even if we were to use the bilateral Laplace transform, where *t* starts at −∞, we would find the transfer function of an integrator to be 1/*s*. The transfer function of a system is the property of the system and does not depend on the method used to find it.
-
-<span id="page-438-0"></span>characteristic polynomial given by *Q*(*s*) = (*s*−λ1)(*s*−λ2)··· (*s*−λ*N*). † Hence,
-
-$$
-Y(s) = X(s)H(s) = \frac{P(s)}{(s - \lambda_1)(s - \lambda_2) \cdots (s - \lambda_N)(s - j\omega)}
-$$
-
-In the partial fraction expansion of the right-hand side, let the coefficients corresponding to the *N* terms (*s* − λ1), (*s* − λ2), ... , (*s* − λ*N*) be *k*1, *k*2, ... , *kN*. The coefficient corresponding to the last term (*s*−*j*ω) is *P*(*s*)/*Q*(*s*)|*<sup>s</sup>*=*j*<sup>ω</sup> = *H*(*j*ω). Hence,
-
-$$
-Y(s) = \sum_{i=1}^{n} \frac{k_i}{s - \lambda_i} + \frac{H(j\omega)}{s - j\omega}
-$$
-
-and
-
-$$
-y(t) = \underbrace{\sum_{i=1}^{n} k_i e^{\lambda_i t} u(t)}_{\text{transient component } y_{\text{tr}}(t)} + \underbrace{H(j\omega)e^{j\omega t} u(t)}_{\text{steady-state component } y_{\text{ss}}(t)}
-$$
-
-For an asymptotically stable system, the characteristic mode terms *e*<sup>λ</sup>*it* decay with time, and, therefore, constitute the so-called *transient* component of the response. The last term *H*(*j*ω)*ej*ω*<sup>t</sup>* persists forever, and is the *steady-state* component of the response given by
-
-$$
-y_{ss}(t) = H(j\omega)e^{j\omega t}u(t)
-$$
-
-This result also explains why an everlasting exponential input *ej*ω*<sup>t</sup>* results in the total response *H*(*j*ω)*ej*ω*<sup>t</sup>* for BIBO systems. Because the input started at *t* = −∞, at any finite time the decaying transient component has long vanished, leaving only the steady-state component. Hence, the total response appears to be *H*(*j*ω)*ej*ω*<sup>t</sup>* .
-
-From the argument that led to Eq. (4.43), it follows that for a causal sinusoidal input cosω*t*, the steady-state response *yss*(*t*) is given by
-
-$$
-y_{ss}(t) = |H(j\omega)| \cos[\omega t + \angle H(j\omega)]u(t)
-$$
-
-In summary, |*H*(*j*ω)| cos[ω*t* + *H*(*j*ω)] is the total response to everlasting sinusoid cosω*t*. In contrast, it is the steady-state response to the same input applied at *t* = 0.

engineering/linear-systems-and-signals-3rd-edition-bp-lathi/061_4.9 BODE PLOTS.md

@@ -1,417 +0,0 @@
-## **[4.9 BODE](#page-10-0) PLOTS**
-
-Sketching frequency response plots (|*H*(*j*ω)| and *H*(*j*ω) versus ω) is considerably facilitated by the use of logarithmic scales. The amplitude and phase response plots as a function of ω on a logarithmic scale are known as *Bode plots*. By using the asymptotic behavior of the amplitude and the phase responses, we can sketch these plots with remarkable ease, even for higher-order transfer functions.
-
-<sup>†</sup> For simplicity, we have assumed nonrepeating characteristic roots. The procedure is readily modified for repeated roots, and the same conclusion results.
-
-#### 420 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS
-
-Let us consider a system with the transfer function
-
-$$
-H(s) = \frac{K(s+a_1)(s+a_2)}{s(s+b_1)(s^2+b_2s+b_3)}
-$$
-\n(4.44)
-
-where the second-order factor (*s*<sup>2</sup> + *b*2*s* + *b*3) is assumed to have complex conjugate roots.† We shall rearrange Eq. (4.44) in the form
-
-$$
-H(s) = \frac{Ka_1a_2}{b_1b_3} \frac{\left(\frac{s}{a_1} + 1\right)\left(\frac{s}{a_2} + 1\right)}{s\left(\frac{s}{b_1} + 1\right)\left(\frac{s^2}{b_3} + \frac{b_2}{b_3} s + 1\right)}
-$$
-
-and
-
-$$
-H(j\omega) = \frac{Ka_1a_2}{b_1b_3} \frac{\left(1 + \frac{j\omega}{a_1}\right)\left(1 + \frac{j\omega}{a_2}\right)}{j\omega\left(1 + \frac{j\omega}{b_1}\right)\left[1 + j\frac{b_2\omega}{b_3} + \frac{(j\omega)^2}{b_3}\right]}
-$$
-
-This equation shows that *H*(*j*ω) is a complex function of ω. The amplitude response |*H*(*j*ω)| and the phase response *H*(*j*ω) are given by
-
-$$
-|H(j\omega)| = \left| \frac{Ka_1a_2}{b_1b_3} \right| \frac{\left| 1 + \frac{j\omega}{a_1} \right| \left| 1 + \frac{j\omega}{a_2} \right|}{|j\omega| \left| 1 + \frac{j\omega}{b_1} \right| \left| 1 + j\frac{b_2\omega}{b_3} + \frac{(j\omega)^2}{b_3} \right|}
-$$
-(4.45)
-
-and
-
-$$
-\angle H(j\omega) = \angle \left(\frac{Ka_1a_2}{b_1b_3}\right) + \angle \left(1 + \frac{j\omega}{a_1}\right) + \angle \left(1 + \frac{j\omega}{a_2}\right)
-$$
-$$
-- \angle j\omega - \angle \left(1 + \frac{j\omega}{b_1}\right) - \angle \left[1 + \frac{jb_2\omega}{b_3} + \frac{(j\omega)^2}{b_3}\right]
-$$
-(4.46)
-
-From Eq. (4.46) we see that the phase function consists of the addition of terms of four kinds: (i) the phase of a constant, (ii) the phase of *j*ω, which is 90◦ for all values of ω, (iii) the phase for the first-order term of the form 1+*j*ω/*a*, and (iv) the phase of the second-order term
-
-$$
-\[1+\frac{jb_2\omega}{b_3}+\frac{(j\omega)^2}{b_3}\]
-$$
-
-We can plot these basic phase functions for ω in the range 0 to ∞ and then, using these plots, we can construct the phase function of any transfer function by properly adding these basic responses. Note that if a particular term is in the numerator, its phase is added, but if the term is in the
-
-<sup>†</sup> Coefficients *a*1, *a*<sup>2</sup> and *b*1, *b*2, *b*<sup>3</sup> used in this section are not to be confused with those used in the representation of *N*th-order LTIC system equations given earlier [Eqs. (2.1) or (4.26)].
-
-denominator, its phase is subtracted. This makes it easy to plot the phase function *H*(*j*ω) as a function of ω. Computation of |*H*(*j*ω)|, unlike that of the phase function, however, involves the multiplication and division of various terms. This is a formidable task, especially when we have to plot this function for the entire range of ω (0 to ∞).
-
-We know that a log operation converts multiplication and division to addition and subtraction. So, instead of plotting |*H*(*j*ω)|, why not plot log |*H*(*j*ω)| to simplify our task? We can take advantage of the fact that logarithmic units are desirable in several applications, where the variables considered have a very large range of variation. This is particularly true in frequency response plots, where we may have to plot frequency response over a range from a very low frequency, near 0, to a very high frequency, in the range of 1010 or higher. A plot on a linear scale of frequencies for such a large range will bury much of the useful information at lower frequencies. Also, the amplitude response may have a very large dynamic range from a low of 10−<sup>6</sup> to a high of 10<sup>6</sup> . A linear plot would be unsuitable for such a situation. Therefore, logarithmic plots not only simplify our task of plotting, but, fortunately, they are also desirable in this situation.
-
-There is another important reason for using logarithmic scale. The Weber–Fechner law (first observed by Weber in 1834) states that human senses (sight, touch, hearing, etc.) generally respond in a logarithmic way. For instance, when we hear sound at two different power levels, we judge one sound twice as loud when the ratio of the two sound powers is 10. Human senses respond to equal ratios of power, not equal increments in power [10]. This is clearly a logarithmic response.†
-
-The logarithmic unit is the *decibel* and is equal to 20 times the logarithm of the quantity (log to the base 10). Therefore, 20log10 |*H*(*j*ω)| is simply the log amplitude in decibels (dB).‡ Thus, instead of plotting |*H*(*j*ω)|, we shall plot 20log10 |*H*(*j*ω)| as a function of ω. These plots (log amplitude and phase) are called *Bode plots*. For the transfer function in Eq. (4.45), the *log amplitude* is
-
-$$
-20\log|H(j\omega)| = 20\log\left|\frac{Ka_1a_2}{b_1b_3}\right| + 20\log\left|1 + \frac{j\omega}{a_1}\right| + 20\log\left|1 + \frac{j\omega}{a_2}\right| - 20\log|j\omega|
-$$
-$$
--20\log\left|1 + \frac{j\omega}{b_1}\right| - 20\log\left|1 + \frac{jb_2\omega}{b_3} + \frac{(j\omega)^2}{b_3}\right| \tag{4.47}
-$$
-
-The term 20log(*Ka*1*a*2/*b*1*b*3) is a constant. We observe that the log amplitude is a sum of four basic terms corresponding to a constant, a pole or zero at the origin (20log|*j*ω|), a first-order pole or zero (20log|1+*j*ω/*a*|), and complex-conjugate poles or zeros (20log|1+*j*ω*b*2/*b*<sup>3</sup> +(*j*ω)<sup>2</sup>/*b*3|).
-
-<sup>†</sup> Observe that the frequencies of musical notes are spaced logarithmically (not linearly). The octave is a ratio of 2. The frequencies of the same note in the successive octaves have a ratio of 2. On the Western musical scale, there are 12 distinct notes in each octave. The frequency of each note is about 6% higher than the frequency of the preceding note. Thus, the successive notes are separated not by some constant frequency, but by constant ratio of 1.06.
-
-<sup>‡</sup> Originally, the unit *bel* (after the inventor of telephone, Alexander Graham Bell) was introduced to represent power ratio as log10 *P*2/*P*<sup>1</sup> bels. A tenth of this unit is a decibel, as in 10 log10 *P*2/*P*<sup>1</sup> decibels. Since the power ratio of two signals is proportional to the amplitude ratio squared, or |*H*(*j*ω)| 2, we have 10 log10 *P*2/*P*<sup>1</sup> = 10 log10 |*H*(*j*ω)| <sup>2</sup> = 20 log10 |*H*(*j*ω)| dB.
-
-### <span id="page-441-0"></span>422 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS
-
-We can sketch these four basic terms as functions of ω and use them to construct the log-amplitude plot of any desired transfer function. Let us discuss each of the terms.
-
-## **[4.9-1 Constant](#page-10-0)** *Ka***1***a***2***/b***1***b***<sup>3</sup>**
-
-The log amplitude of the constant *Ka*1*a*2/*b*1*b*<sup>2</sup> term is also a constant, 20log|*Ka*1*a*2/*b*1*b*3|. The phase contribution from this term is zero for positive value and π for negative value of the constant (complex constants can have different phases).
-
-### **[4.9-2 Pole \(or Zero\) at the Origin](#page-10-0)**
-
-LOG MAGNITUDE
-
-A pole at the origin gives rise to the term −20log|*j*ω|, which can be expressed as
-
-$$
--20\log|j\omega| = -20\log\omega
-$$
-
-This function can be plotted as a function of ω. However, we can effect further simplification by using the logarithmic scale for the variable ω itself. Let us define a new variable *u* such that
-
-*u* = logω
-
-Hence,
-
-$$
--20\log\omega = -20u
-$$
-
-The log-amplitude function −20*u* is plotted as a function of *u* in Fig. 4.40a. This is a straight line with a slope of −20. It crosses the *u* axis at *u* = 0. The ω-scale (*u* = logω) also appears in Fig. 4.40a. Semilog graphs can be conveniently used for plotting, and we can directly plot ω on semilog paper. A ratio of 10 is a *decade*, and a ratio of 2 is known as an *octave*. Furthermore, a decade along the ω scale is equivalent to 1 unit along the *u* scale. We can also show that a ratio of 2 (an octave) along the ω scale equals to 0.3010 (which is log10 2) along the *u* scale.†
-
-$$
-u_2 - u_1 = \log_{10} \omega_2 - \log_{10} \omega_1 = \log_{10} (\omega_2/\omega_1)
-$$
-
-Thus, if
-
-(ω2/ω1) = 10 (which is a decade)
-
-then
-
-*u*<sup>2</sup> −*u*<sup>1</sup> = log10 10 = 1
-
-and if
-
-(ω2/ω1) = 2 (which is an octave)
-
-then
-
-$$
-u_2 - u_1 = \log_{10} 2 = 0.3010
-$$
-
-<sup>†</sup> This point can be shown as follows. Let ω<sup>1</sup> and ω<sup>2</sup> along the ω scale correspond to *u*<sup>1</sup> and *u*<sup>2</sup> along the *u* scale so that logω<sup>1</sup> = *u*<sup>1</sup> and logω<sup>2</sup> = *u*2. Then
-
-**Figure 4.40 (a)** Amplitude and **(b)** phase responses of a pole or a zero at the origin.
-
-Note that equal increments in *u* are equivalent to equal ratios on the ω scale. Thus, 1 unit along the *u* scale is the same as one decade along the ω scale. This means that the amplitude plot has a slope of −20 dB/decade or −20(0.3010) = −6.02 dB/octave (commonly stated as −6 dB/octave). Moreover, the amplitude plot crosses the ω axis at ω = 1, since *u* = log10ω = 0 when ω = 1.
-
-For the case of a zero at the origin, the log-amplitude term is 20 log ω. This is a straight line passing through ω = 1 and having a slope of 20 dB/decade (or 6 dB/octave). This plot is a mirror image about the ω axis of the plot for a pole at the origin and is shown dashed in Fig. 4.40a.
-
-### PHASE
-
-The phase function corresponding to the pole at the origin is − *j*ω [see Eq. (4.46)]. Thus,
-
-$$
-\angle H(j\omega) = -\angle j\omega = -90^{\circ}
-$$
-
-The phase is constant (−90◦) for all values of ω, as depicted in Fig. 4.40b. For a zero at the origin, the phase is *j*ω = 90◦. This is a mirror image of the phase plot for a pole at the origin and is shown dashed in Fig. 4.40b.
-
-### <span id="page-443-0"></span>**[4.9-3 First-Order Pole \(or Zero\)](#page-10-0)**
-
-### THE LOG MAGNITUDE
-
-The log amplitude of a first-order pole at −*a* is −20log|1+*j*ω/*a*|. Let us investigate the asymptotic behavior of this function for extreme values of ω (ω *a* and ω *a*).
-
-**(a)** For ω *a*,
-
-$$
--20\log\left|1+\frac{j\omega}{a}\right|\approx-20\log 1=0
-$$
-
-Hence, the log-amplitude function → 0 asymptotically for ω *a* (Fig. 4.41a).
-
-**(a)** For the other extreme case, where ω *a*,
-
-$$
--20\log\left|1+\frac{j\omega}{a}\right| \approx -20\log\left(\frac{\omega}{a}\right) = -20\log\omega + 20\log a = -20u + 20\log a
-$$
-
-This represents a straight line (when plotted as a function of *u*, the log of ω) with a slope of −20 dB/decade (or −6 dB/octave). When ω = *a*, the log amplitude is zero. Hence, this line crosses the ω axis at ω = *a*, as illustrated in Fig. 4.41a. Note that the asymptotes in (a) and (b) meet at ω = *a*.
-
-The exact log amplitude for this pole is
-
-$$
--20\log\left|1+\frac{j\omega}{a}\right| = -20\log\left(1+\frac{\omega^2}{a^2}\right)^{1/2} = -10\log\left(1+\frac{\omega^2}{a^2}\right)
-$$
-
-This exact log magnitude function also appears in Fig. 4.41a. Observe that the actual and the asymptotic plots are very close. A maximum error of 3 dB occurs at ω = *a*. This frequency is known as the *corner frequency* or *break frequency*. The error everywhere else is less than 3 dB. A plot of the error as a function of ω is shown in Fig. 4.42a. This figure shows that the error at 1 octave above or below the corner frequency is 1 dB and the error at 2 octaves above or below the corner frequency is 0.3 dB. The actual plot can be obtained by adding the error to the asymptotic plot.
-
-The amplitude response for a zero at −*a* (shown dotted in Fig. 4.41a) is identical to that of the pole at −*a* with a sign change and therefore is the mirror image (about the 0 dB line) of the amplitude plot for a pole at −*a*.
-
-### PHASE
-
-The phase for the first-order pole at −*a* is
-
-$$
-\angle H(j\omega) = -\angle \left(1 + \frac{j\omega}{a}\right) = -\tan^{-1}\left(\frac{\omega}{a}\right)
-$$
-
-Let us investigate the asymptotic behavior of this function. For ω *a*,
-
-$$
--\tan^{-1}\left(\frac{\omega}{a}\right) \approx 0
-$$
-
-and, for ω *a*,
-
-$$
--\tan^{-1}\left(\frac{\omega}{a}\right) \approx -90^{\circ}
-$$
-
-**Figure 4.41 (a)** Amplitude and **(b)** phase responses of a first-order pole or zero at *s* = −*a*.
-
-The actual plot along with the asymptotes is depicted in Fig. 4.41b. In this case, we use a three-line segment asymptotic plot for greater accuracy. The asymptotes are a phase angle of 0◦ for ω ≤ *a*/10, a phase angle of −90◦ for ω ≥ 10*a*, and a straight line with a slope −45◦/decade connecting these two asymptotes (from ω = *a*/10 to 10*a*) crossing the ω axis at ω = *a*/10. It can be seen from Fig. 4.41b that the asymptotes are very close to the curve and the maximum error is 5.7◦. Figure 4.42b plots the error as a function of ω; the actual plot can be obtained by adding the error to the asymptotic plot.
-
-<span id="page-445-0"></span>**Figure 4.42** Errors in asymptotic approximation of a first-order pole at *s* = −*a*.
-
-The phase for a zero at −*a* (shown dotted in Fig. 4.41b) is identical to that of the pole at −*a* with a sign change, and therefore is the mirror image (about the 0◦ line) of the phase plot for a pole at −*a*.
-
-### **[4.9-4 Second-Order Pole \(or Zero\)](#page-10-0)**
-
-Let us consider the second-order pole in Eq. (4.44). The denominator term is *s*<sup>2</sup> + *b*2*s* + *b*3. We shall introduce the often-used standard form *s*<sup>2</sup> + 2ζω*ns* + ω<sup>2</sup> *<sup>n</sup>* instead of *s*<sup>2</sup> + *b*2*s* + *b*3. With this form, the log amplitude function for the second-order term in Eq. (4.47) becomes
-
-$$
--20\log\left|1+2j\zeta\frac{\omega}{\omega_n}+\left(\frac{j\omega}{\omega_n}\right)^2\right|
-$$
-
-and the phase function is
-
-$$
--\angle \left[1+2j\zeta \frac{\omega}{\omega_n}+\left(\frac{j\omega}{\omega_n}\right)^2\right]
-$$
-\n(4.48)
-
-### THE LOG MAGNITUDE
-
-The log amplitude is given by
-
-log amplitude = 
-$$
--20 \log \left| 1 + 2j\zeta \left( \frac{\omega}{\omega_n} \right) + \left( \frac{j\omega}{\omega_n} \right)^2 \right|
-$$
- (4.49)
-
-For ω ω*n*, the log amplitude becomes
-
-$$
-log amplitude \approx -20 log 1 = 0
-$$
-
-For ω ω*n*, the log amplitude is
-
-$$
-\log amplitude \approx -20 \log \left| \left( -\frac{\omega}{\omega_n} \right)^2 \right| = -40 \log \left( \frac{\omega}{\omega_n} \right)
-$$
-  
-= -40 log  $\omega$  - 40 log  $\omega_n$  = -40u - 40 log  $\omega_n$  (4.50)
-
-The two asymptotes are zero for ω<ω*<sup>n</sup>* and −40*u*−40logω*<sup>n</sup>* for ω>ω*n*. The second asymptote is a straight line with a slope of −40 dB/decade (or −12 dB/octave) when plotted against the log ω scale. It begins at ω = ω*<sup>n</sup>* [see Eq. (4.50)]. The asymptotes are depicted in Fig. 4.43a. The exact log amplitude is given by [see Eq. (4.49)]
-
-log amplitude = 
-$$
--20 \log \left\{ \left[ 1 - \left( \frac{\omega}{\omega_n} \right)^2 \right]^2 + 4 \zeta^2 \left( \frac{\omega}{\omega_n} \right)^2 \right\}^{1/2}
-$$
- (4.51)
-
-The log amplitude in this case involves a parameter ζ , resulting in a different plot for each value of ζ . For complex-conjugate poles,† ζ < 1. Hence, we must sketch a family of curves for a number of values of ζ in the range 0 to 1. This is illustrated in Fig. 4.43a. The error between the actual plot and the asymptotes is shown in Fig. 4.44. The actual plot can be obtained by adding the error to the asymptotic plot.
-
-For second-order zeros (complex-conjugate zeros), the plots are mirror images (about the 0 dB line) of the plots depicted in Fig. 4.43a. Note the resonance phenomenon of the complex-conjugate poles. This phenomenon is barely noticeable for ζ > 0.707 but becomes pronounced as ζ → 0.
-
-### PHASE
-
-The phase function for second-order poles, as apparent in Eq. (4.48), is
-
-$$
-\angle H(j\omega) = -\tan^{-1}\left[\frac{2\zeta\left(\frac{\omega}{\omega_n}\right)}{1 - \left(\frac{\omega}{\omega_n}\right)^2}\right]
-$$
-(4.52)
-
-For ω ω*n*,
-
-$$
-\angle H(j\omega) \approx 0
-$$
-
-<sup>†</sup> For <sup>ζ</sup> <sup>≥</sup> 1, the two poles in the second-order factor are no longer complex but real, and each of these two real poles can be dealt with as a separate first-order factor.
-
-**Figure 4.43** Amplitude and phase response of a second-order pole.
-
-For ω ω*n*,
-
-$$
-\angle H(j\omega) \simeq -180^\circ
-$$
-
-Hence, the phase → −180◦ as ω → ∞. As in the case of amplitude, we also have a family of phase plots for various values of ζ , as illustrated in Fig. 4.43b. A convenient asymptote for the phase of complex-conjugate poles is a step function that is 0◦ for ω<ω*<sup>n</sup>* and −180◦ for ω>ω*n*.
-
-**Figure 4.44** Errors in the asymptotic approximation of a second-order pole.
-
-Error plots for such an asymptote are shown in Fig. 4.44 for various values of ζ . The exact phase is the asymptotic value plus the error.
-
-For complex-conjugate zeros, the amplitude and phase plots are mirror images of those for complex conjugate-poles.
-
-We shall demonstrate the application of these techniques with two examples.
-
-### **EXAMPLE 4.29 Bode Plots for Second-Order Transfer Function with Real Roots**
-
-Sketch Bode plots for the transfer function
-
-$$
-H(s) = \frac{20s(s+100)}{(s+2)(s+10)}
-$$
-
-### MAGNITUDE PLOT
-
-First, we write the transfer function in normalized form
-
-$$
-H(s) = \frac{20 \times 100}{2 \times 10} \frac{s \left(1 + \frac{s}{100}\right)}{\left(1 + \frac{s}{2}\right)\left(1 + \frac{s}{10}\right)} = 100 \frac{s \left(1 + \frac{s}{100}\right)}{\left(1 + \frac{s}{2}\right)\left(1 + \frac{s}{10}\right)}
-$$
-
-Here, the constant term is 100; that is, 40 dB (20 log 100 = 40). This term can be added to the plot by simply relabeling the horizontal axis (from which the asymptotes begin) as the 40 dB line (see Fig. 4.45a). Such a step implies shifting the horizontal axis upward by 40 dB. This is precisely what is desired.
-
-In addition, we have two first-order poles at −2 and −10, one zero at the origin, and one zero at −100.
-
-**Step 1.** For each of these terms, we draw an asymptotic plot as follows (shown in Fig. 4.45a by dashed lines):
-
-- **(a)** For the zero at the origin, draw a straight line with a slope of 20 dB/decade passing through ω = 1.
-- **(b)** For the pole at −2, draw a straight line with a slope of −20 dB/decade (for ω > 2) beginning at the corner frequency ω = 2.
-- **(c)** For the pole at −10, draw a straight line with a slope of −20 dB/decade beginning at the corner frequency ω = 10.
-- **(d)** For the zero at −100, draw a straight line with a slope of 20 dB/decade beginning at the corner frequency ω = 100.
-- **Step 2.** Add all the asymptotes, as depicted in Fig. 4.45a by solid line segments.
-- **Step 3.** Apply the following corrections (see Fig. 4.42a):
-  - **(a)** The correction at ω = 1 because of the corner frequency at ω = 2 is −1 dB. The correction at ω = 1 because of the corner frequencies at ω = 10 and ω = 100 is quite small (see Fig. 4.42a) and may be ignored. Hence, the net correction at ω = 1 is −1 dB.
-
-**Figure 4.45 (a)** Amplitude and **(b)** phase responses of the second-order system.
-
-- **(b)** The correction at ω = 2 because of the corner frequency at ω = 2 is −3 dB, and the correction because of the corner frequency at ω = 10 is −0.17 dB. The correction because of the corner frequency ω = 100 can be safely ignored. Hence the net correction at ω = 2 is −3.17 dB.
-- **(c)** The correction at ω = 10 because of the corner frequency at ω = 10 is −3 dB, and the correction because of the corner frequency at ω = 2 is −0.17 dB. The correction because of ω = 100 can be ignored. Hence the net correction at ω = 10 is −3.17 dB.
-
-- **(d)** The correction at ω = 100 because of the corner frequency at ω = 100 is 3 dB, and the corrections because of the other corner frequencies may be ignored.
-- **(e)** In addition to the corrections at corner frequencies, we may consider corrections at intermediate points for more accurate plots. For instance, the corrections at ω = 4 because of corner frequencies at ω = 2 and 10 are −1 and about −0.65, totaling −1.65 dB. In the same way, the corrections at ω = 5 because of corner frequencies at ω = 2 and 10 are −0.65 and −1, totaling −1.65 dB.
-
-With these corrections, the resulting amplitude plot is illustrated in Fig. 4.45a.
-
-### PHASE PLOT
-
-We draw the asymptotes corresponding to each of the four factors:
-
-- **(a)** The zero at the origin causes a 90◦ phase shift.
-- **(b)** The pole at *s* = −2 has an asymptote with a zero value for −∞ <ω< 0.2 and a slope of −45◦/decade beginning at ω = 0.2 and going up to ω = 20. The asymptotic value for ω > 20 is −90◦.
-- **(c)** The pole at *s* = −10 has an asymptote with a zero value for −∞ <ω< 1 and a slope of −45◦/decade beginning at ω = 1 and going up to ω = 100. The asymptotic value for ω > 100 is −90◦.
-- **(d)** The zero at *s* = −100 has an asymptote with a zero value for −∞ <ω< 10 and a slope of 45◦/decade beginning at ω = 10 and going up to ω = 1000. The asymptotic value for ω > 1000 is 90◦. All the asymptotes are added, as shown in Fig. 4.45b. The appropriate corrections are applied from Fig. 4.42b, and the exact phase plot is depicted in Fig. 4.45b.
-
-### **EXAMPLE 4.30 Bode Plots for Second-Order Transfer Function with Complex Poles**
-
-Sketch the amplitude and phase response (Bode plots) for the transfer function
-
-$$
-H(s) = \frac{10(s+100)}{s^2 + 2s + 100} = 10 \frac{1 + \frac{s}{100}}{1 + \frac{s}{50} + \frac{s^2}{100}}
-$$
-
-### MAGNITUDE PLOT
-
-Here, the constant term is 10: that is, 20 dB(20 log 10 = 20). To add this term, we simply label the horizontal axis (from which the asymptotes begin) as the 20 dB line, as before (see Fig. 4.46a).
-
-**Figure 4.46 (a)** Amplitude and **(b)** phase responses of the second-order system.
-
-In addition, we have a real zero at *s* = −100 and a pair of complex conjugate poles. When we express the second-order factor in standard form,
-
-$$
-s^2 + 2s + 100 = s^2 + 2\zeta \omega_n s + \omega_n^2
-$$
-
-### 434 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS
-
-we have
-
-$$
-\omega_n = 10 \quad \text{and} \quad \zeta = 0.1
-$$
-
-**Step 1.** Draw an asymptote of −40 dB/decade (−12 dB/octave) starting at ω = 10 for the complex conjugate poles, and draw another asymptote of 20 dB/decade starting at ω = 100 for the (real) zero.
-
-**Step 2.** Add both asymptotes.
-
-**Step 3.** Apply the correction at ω = 100, where the correction because of the corner frequency ω = 100 is 3 dB. The correction because of the corner frequency ω = 10, as seen from Fig. 4.44a for ζ = 0.1, can be safely ignored. Next, the correction at ω = 10 because of the corner frequency ω = 10 is 13.90 dB (see Fig. 4.44a for ζ = 0.1). The correction because of the real zero at −100 can be safely ignored at ω = 10. We may find corrections at a few more points. The resulting plot is illustrated in Fig. 4.46a.
-
-### PHASE PLOT
-
-The asymptote for the complex conjugate poles is a step function with a jump of −180◦ at ω = 10. The asymptote for the zero at *s* = −100 is zero for ω ≤ 10 and is a straight line with a slope of 45◦/decade, starting at ω = 10 and going to ω = 1000. For ω ≥ 1000, the asymptote is 90◦. The two asymptotes add to give the sawtooth shown in Fig. 4.46b. We now apply the corrections from Figs. 4.42b and 4.44b to obtain the exact plot.
-
-**Figure 4.47** MATLAB-generated Bode plots for Ex. 4.30.
-
-### <span id="page-454-0"></span>BODE PLOTS WITH MATLAB
-
-Bode plots make it relatively simple to hand-draw straight-line approximations to a system's magnitude and frequency responses. To produce exact Bode plots, we turn to MATLAB and its bode command.
-
->> bode(tf([10 1000],[1 2 100]),'k-');
-
-The resulting MATLAB plots, shown in Fig. 4.47, match the plots shown in Fig. 4.46.
-
-**Comment.** These two examples demonstrate that actual frequency response plots are very close to asymptotic plots, which are so easy to construct. Thus, by mere inspection of *H*(*s*) and its poles and zeros, one can rapidly construct a mental image of the frequency response of a system. This is the principal virtue of Bode plots.
-
-### POLES AND ZEROS IN THE RIGHT HALF-PLANE
-
-In our discussion so far, we have assumed the poles and zeros of the transfer function to be in the left half-plane. What if some of the poles and/or zeros of *H*(*s*) lie in the RHP? If there is a pole in the RHP, the system is unstable. Such systems are useless for any signal-processing application. For this reason, we shall consider only the case of the RHP zero. The term corresponding to RHP zero at *s* = *a* is (*s*/*a*) −1, and the corresponding frequency response is (*j*ω/*a*) −1. The amplitude response is
-
-$$
-\left|\frac{j\omega}{a} - 1\right| = \left(\frac{\omega^2}{a^2} + 1\right)^{1/2}
-$$
-
-This shows that the amplitude response of an RHP zero at *s* = *a* is identical to that of an LHP zero or *s* = −*a*. Therefore, the log amplitude plots remain unchanged whether the zeros are in the LHP or the RHP. However, the phase corresponding to the RHP zero at *s* = *a* is
-
-$$
-\angle \left(\frac{j\omega}{a} - 1\right) = \angle -\left(1 - \frac{j\omega}{a}\right) = \pi + \tan^{-1}\left(\frac{-\omega}{a}\right) = \pi - \tan^{-1}\left(\frac{\omega}{a}\right)
-$$
-
-whereas the phase corresponding to the LHP zero at *s* = −*a* is tan−<sup>1</sup>(ω/*a*).
-
-The complex-conjugate zeros in the RHP give rise to a term *s*<sup>2</sup>−2ζω*ns*+ω<sup>2</sup> *<sup>n</sup>*, which is identical to the term *s*<sup>2</sup> +2ζω*ns*+ω<sup>2</sup> *<sup>n</sup>* with a sign change in ζ . Hence, from Eqs. (4.51) and (4.52), it follows that the amplitudes are identical, but the phases are of opposite signs for the two terms.
-
-Systems whose poles and zeros are restricted to the LHP are classified as *minimum phase* systems. Minimum phase systems are particularly desirable because the system *and its inverse* are both stable.
-
-### **[4.9-5 The Transfer Function from the Frequency Response](#page-10-0)**
-
-In the preceding section we were given the transfer function of a system. From a knowledge of the transfer function, we developed techniques for determining the system response to sinusoidal inputs. We can also reverse the procedure to determine the transfer function of a minimum phase
-
-### <span id="page-455-0"></span>436 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS
-
-system from the system's response to sinusoids. This application has significant practical utility. If we are given a system in a black box with only the input and output terminals available, the transfer function has to be determined by experimental measurements at the input and output terminals. The frequency response to sinusoidal inputs is one of the possibilities that is very attractive because the measurements involved are so simple. One needs only to apply a sinusoidal signal at the input and observe the output. We find the amplitude gain |*H*(*j*ω)| and the output phase shift *H*(*j*ω) (with respect to the input sinusoid) for various values of ω over the entire range from 0 to ∞. This information yields the frequency response plots (Bode plots) when plotted against log ω. From these plots we determine the appropriate asymptotes by taking advantage of the fact that the slopes of all asymptotes must be multiples of ±20 dB/decade if the transfer function is a rational function (function that is a ratio of two polynomials in *s*). From the asymptotes, the corner frequencies are obtained. Corner frequencies determine the poles and zeros of the transfer function. Because of the ambiguity about the location of zeros since LHP and RHP zeros (zeros at *s* = ±*a*) have identical magnitudes, this procedure works only for minimum phase systems.

engineering/linear-systems-and-signals-3rd-edition-bp-lathi/062_4.10 FILTER DESIGN BY PLACEMENT OF POLES AND ZEROS OF H(s).md

@@ -1,164 +0,0 @@
-## **4.10 FILTER DESIGN BY [PLACEMENT OF](#page-10-0) POLES AND ZEROS OF** *H(s)*
-
-In this section we explore the strong dependence of frequency response on the location of poles and zeros of *H*(*s*). This dependence points to a simple intuitive procedure to filter design.
-
-### **[4.10-1 Dependence of Frequency Response on Poles](#page-10-0) and Zeros of** *H(s)*
-
-Frequency response of a system is basically the information about the filtering capability of the system. A system transfer function can be expressed as
-
-$$
-H(s) = \frac{P(s)}{Q(s)} = b_0 \frac{(s - z_1)(s - z_2) \cdots (s - z_N)}{(s - \lambda_1)(s - \lambda_2) \cdots (s - \lambda_N)}
-$$
-
-where *z*1, *z*2, ... , *zN* are λ1, λ2, ... , λ*<sup>N</sup>* are the poles of *H*(*s*). Now the value of the transfer function *H*(*s*) at some frequency *s* = *p* is
-
-$$
-H(s)|_{s=p} = b_0 \frac{(p-z_1)(p-z_2)\cdots(p-z_N)}{(p-\lambda_1)(p-\lambda_2)\cdots(p-\lambda_N)}
-$$
-(4.53)
-
-This equation consists of factors of the form *p*−*zi* and *p*−λ*i*. The factor *p*−*zi* is a complex number represented by a vector drawn from point *z* to the point *p* in the complex plane, as illustrated in Fig. 4.48a. The length of this line segment is |*p* − *zi*|, the magnitude of *p* − *zi*. The angle of this directed line segment (with the horizontal axis) is (*p* − *zi*). To compute *H*(*s*) at *s* = *p*, we draw line segments from all poles and zeros of *H*(*s*) to the point *p*, as shown in Fig. 4.48b. The vector connecting a zero *zi* to the point *p* is *p* − *zi*. Let the length of this vector be *ri*, and let its angle with the horizontal axis be φ*i*. Then *p*−*zi* = *riej*<sup>φ</sup>*<sup>i</sup>* . Similarly, the vector connecting a pole λ*<sup>i</sup>* to the point *p* is *p* − λ*<sup>i</sup>* = *diej*<sup>θ</sup>*<sup>i</sup>* , where *di* and θ*<sup>i</sup>* are the length and the angle (with the horizontal axis),
-
-**Figure 4.48** Vector representations of **(a)** complex numbers and **(b)** factors of *H*(*s*).
-
-respectively, of the vector *p*−λ*i*. Now from Eq. (4.53) it follows that
-
-$$
-H(s)|_{s=p} = b_0 \frac{(r_1 e^{j\phi_1})(r_2 e^{j\phi_2}) \cdots (r_N e^{j\phi_N})}{(d_1 e^{j\theta_1})(d_2 e^{j\theta_2}) \cdots (d_N e^{j\theta_N})}
-$$
-  
-=  $b_0 \frac{r_1 r_2 \cdots r_N}{d_1 d_2 \cdots d_N} e^{j[(\phi_1 + \phi_2 + \cdots + \phi_N) - (\theta_1 + \theta_2 + \cdots + \theta_N)]}$ 
-
-Therefore
-
-$$
-|H(s)|_{s=p} = b_0 \frac{r_1 r_2 \cdots r_N}{d_1 d_2 \cdots d_N} = b_0 \frac{\text{product of distances of zeros to } p}{\text{product of distances of poles to } p}
-$$
-(4.54)
-
-and
-
-$$
-\angle H(s)|_{s=p} = (\phi_1 + \phi_2 + \dots + \phi_N) - (\theta_1 + \theta_2 + \dots + \theta_N)
-$$
-  
-= sum of angles of zeros to  $p$  – sum of angles of poles to  $p$  (4.55)
-
-Here, we have assumed positive *b*0. If *b*<sup>0</sup> is negative, there is an additional phase π. Using this procedure, we can determine *H*(*s*) for any value of *s*. To compute the frequency response *H*(*j*ω), we use *s* = *j*ω (a point on the imaginary axis), connect all poles and zeros to the point *j*ω, and determine |*H*(*j*ω)| and *H*(*j*ω) from Eqs. (4.54) and (4.55). We repeat this procedure for all values of ω from 0 to ∞ to obtain the frequency response.
-
-### GAIN ENHANCEMENT BY A POLE
-
-To understand the effect of poles and zeros on the frequency response, consider a hypothetical case of a single pole −α + *j*ω0, as depicted in Fig. 4.49a. To find the amplitude response |*H*(*j*ω)| for a certain value of ω, we connect the pole to the point *j*ω (Fig. 4.49a). If the length of this line is *d*, then |*H*(*j*ω)| is proportional to 1/*d*,
-
-$$
-|H(j\omega)| = \frac{K}{d} \tag{4.56}
-$$
-
-where the exact value of constant *K* is not important at this point. As ω increases from zero, *d* decreases progressively until ω reaches the value ω0. As ω increases beyond ω0, *d* increases
-
-**Figure 4.49** The role of poles and zeros in determining the frequency response of an LTIC system.
-
-progressively. Therefore, according to Eq. (4.56), the amplitude response |*H*(*j*ω)| increases from ω = 0 until ω = ω0, and it decreases continuously as ω increases beyond ω0, as illustrated in Fig. 4.49b. Therefore, a pole at −α + *j*ω<sup>0</sup> results in a frequency-selective behavior that enhances the gain at the frequency ω<sup>0</sup> (resonance). Moreover, as the pole moves closer to the imaginary axis (as α is reduced), this enhancement (resonance) becomes more pronounced. This is because α, the distance between the pole and *j*ω<sup>0</sup> (*d* corresponding to *j*ω0), becomes smaller, which increases the gain *K*/*d*. In the extreme case, when α = 0 (pole on the imaginary axis), the gain at ω<sup>0</sup> goes to infinity. Repeated poles further enhance the frequency-selective effect. To summarize, we can enhance a gain at a frequency ω<sup>0</sup> by placing a pole opposite the point *j*ω0. The closer the pole is to *j*ω0, the higher is the gain at ω0, and the gain variation is more rapid (more frequency selective) in the vicinity of frequency ω0. Note that a pole must be placed in the LHP for stability.
-
-Here we have considered the effect of a single complex pole on the system gain. For a real system, a complex pole −α + *j*ω<sup>0</sup> must accompany its conjugate −α − *j*ω0. We can readily show that the presence of the conjugate pole does not appreciably change the frequency-selective <span id="page-458-0"></span>behavior in the vicinity of ω0. This is because the gain in this case is *K*/*dd* , where *d* is the distance of a point *j*ω from the conjugate pole −α − *j*ω0. Because the conjugate pole is far from *j*ω0, there is no dramatic change in the length *d* as ω varies in the vicinity of ω0. There is a gradual increase in the value of *d* as ω increases, which leaves the frequency-selective behavior as it was originally, with only minor changes.
-
-### GAIN SUPPRESSION BY A ZERO
-
-Using the same argument, we observe that zeros at −α ± *j*ω<sup>0</sup> (Fig. 4.49d) will have exactly the opposite effect of suppressing the gain in the vicinity of ω0, as shown in Fig. 4.49e). A zero on the imaginary axis at *j*ω<sup>0</sup> will totally suppress the gain (zero gain) at frequency ω0. Repeated zeros will further enhance the effect. Also, a closely placed pair of a pole and a zero (dipole) tend to cancel out each other's influence on the frequency response. Clearly, a proper placement of poles and zeros can yield a variety of frequency-selective behavior. We can use these observations to design lowpass, highpass, bandpass, and bandstop (or notch) filters.
-
-Phase response can also be computed graphically. In Fig. 4.49a, angles formed by the complex conjugate poles −α±*j*ω<sup>0</sup> at ω =0 (the origin) are equal and opposite. As ω increases from 0 up, the angle θ<sup>1</sup> (due to the pole −α +*j*ω0), which has a negative value at ω = 0, is reduced in magnitude; the angle θ<sup>2</sup> because of the pole −α − *j*ω0, which has a positive value at ω = 0, increases in magnitude. As a result, θ<sup>1</sup> + θ2, the sum of the two angles, increases continuously, approaching a value π as ω → ∞. The resulting phase response *H*(*j*ω) = −(θ<sup>1</sup> +θ2) is illustrated in Fig. 4.49c. Similar arguments apply to zeros at −α ± *j*ω0. The resulting phase response *H*(*j*ω) = (φ<sup>1</sup> + φ2) is depicted in Fig. 4.49f.
-
-We now focus on simple filters, using the intuitive insights gained in this discussion. The discussion is essentially qualitative.
-
-### **[4.10-2 Lowpass Filters](#page-10-0)**
-
-A typical lowpass filter has a maximum gain at ω = 0. Because a pole enhances the gain at frequencies in its vicinity, we need to place a pole (or poles) on the real axis opposite the origin (*j*ω = 0), as shown in Fig. 4.50a. The transfer function of this system is
-
-$$
-H(s) = \frac{\omega_c}{s + \omega_c}
-$$
-
-We have chosen the numerator of *H*(*s*) to be ω*<sup>c</sup>* to normalize the dc gain *H*(0) to unity. If *d* is the distance from the pole −ω*<sup>c</sup>* to a point *j*ω (Fig. 4.50a), then
-
-$$
-|H(j\omega)| = \frac{\omega_c}{d}
-$$
-
-with *H*(0) = 1. As ω increases, *d* increases and |*H*(*j*ω)| decreases monotonically with ω, as illustrated in Fig. 4.50d with label *N* = 1. This is clearly a lowpass filter with gain enhanced in the vicinity of ω = 0.
-
-### WALL OF POLES
-
-An ideal lowpass filter characteristic (shaded in Fig. 4.50d) has a constant gain of unity up to frequency ω*c*. Then the gain drops suddenly to 0 for ω>ω*c*. To achieve the ideal lowpass
-
-**Figure 4.50** Pole-zero configuration and the amplitude response of a lowpass (Butterworth) filter.
-
-characteristic, we need enhanced gain over the entire frequency band from 0 to ω*c*. We know that to enhance a gain at any frequency ω, we need to place a pole opposite ω. To achieve an enhanced gain for all frequencies over the band (0 to ω*c*), we need to place a pole opposite every frequency in this band. In other words, we need a *continuous wall of poles* facing the imaginary axis opposite the frequency band 0 to ω*<sup>c</sup>* (and from 0 to −ω*<sup>c</sup>* for conjugate poles), as depicted in Fig. 4.50b. At this point, the optimum shape of this wall is not obvious because our arguments are qualitative and intuitive. Yet, it is certain that to have enhanced gain (constant gain) at every frequency over this range, we need an infinite number of poles on this wall. We can show that for a maximally flat† response over the frequency range (0 to ω*c*), the wall is a semicircle with an infinite number of poles uniformly distributed along the wall [11]. In practice, we compromise by using a finite number (*N*) of poles with less-than-ideal characteristics. Figure 4.50c shows the pole configuration for a fifth-order (*N* = 5) filter. The amplitude response for various values of *N* is illustrated in Fig. 4.50d. As *N* → ∞, the filter response approaches the ideal. This family of filters is known as the *Butterworth* filters. There are also other families. In *Chebyshev* filters, the wall shape is a semiellipse rather than a semicircle. The characteristics of a Chebyshev filter are inferior to those of Butterworth over the passband (0,ω*c*), where the characteristics show a rippling effect
-
-<sup>†</sup> Maximally flat amplitude response means the first 2*<sup>N</sup>* <sup>−</sup> 1 derivatives of <sup>|</sup>*H*(*j*ω)<sup>|</sup> with respect to <sup>ω</sup> are zero at ω = 0.
-
-<span id="page-460-0"></span>instead of the maximally flat response of Butterworth. But in the stopband (ω>ω*c*), Chebyshev behavior is superior in the sense that Chebyshev filter gain drops faster than that of the Butterworth.
-
-### **[4.10-3 Bandpass Filters](#page-10-0)**
-
-The shaded characteristic in Fig. 4.51b shows the ideal bandpass filter gain. In the bandpass filter, the gain is enhanced over the entire passband. Our earlier discussion indicates that this can be realized by a wall of poles opposite the imaginary axis in front of the passband centered at ω0. (There is also a wall of conjugate poles opposite −ω0.) Ideally, an infinite number of poles is required. In practice, we compromise by using a finite number of poles and accepting less-than-ideal characteristics (Fig. 4.51).
-
-### **[4.10-4 Notch \(Bandstop\) Filters](#page-10-0)**
-
-An ideal notch filter amplitude response (shaded in Fig. 4.52b) is a complement of the amplitude response of an ideal bandpass filter. Its gain is zero over a small band centered at some frequency ω<sup>0</sup> and is unity over the remaining frequencies. Realization of such a characteristic requires an infinite number of poles and zeros. Let us consider a practical second-order notch filter to obtain zero gain at a frequency ω = ω0. For this purpose, we must have zeros at ±*j*ω0. The requirement of unity gain at ω = ∞ requires the number of poles to be equal to the number of zeros (*M* = *N*). This ensures that for very large values of ω, the product of the distances of poles from ω will be equal to the product of the distances of zeros from ω. Moreover, unity gain at ω = 0 requires a pole and the corresponding zero to be equidistant from the origin. For example, if we use two (complex-conjugate) zeros, we must have two poles; the distance from the origin of the poles and of the zeros should be the same. This requirement can be met by placing the two conjugate poles on the semicircle of radius ω0, as depicted in Fig. 4.52a. The poles can be anywhere on the semicircle to satisfy the equidistance condition. Let the two conjugate poles be at angles ±θ with respect to the negative real axis. Recall that a pole and a zero in the same vicinity tend to cancel out
-
-**Figure 4.51 (a)** Pole-zero configuration and **(b)** the amplitude response of a bandpass filter.
-
-**Figure 4.52 (a)** Pole-zero configuration and **(b)** the amplitude response of a bandstop (notch) filter.
-
-each other's influences. Therefore, placing poles closer to zeros (selecting θ closer to π/2) results in a rapid recovery of the gain from value 0 to 1 as we move away from ω<sup>0</sup> in either direction. Figure 4.52b shows the gain |*H*(*j*ω)| for three different values of θ.
-
-### **EXAMPLE 4.31 Notch Filter Design**
-
-Design a second-order notch filter to suppress 60 Hz hum in a radio receiver.
-
-We use the poles and zeros in Fig. 4.52a with ω<sup>0</sup> = 120π. The zeros are at *s* = ±*j*ω0. The two poles are at −ω<sup>0</sup> cos θ ±*j*ω<sup>0</sup> sin θ. The filter transfer function is (with ω<sup>0</sup> = 120π)
-
-$$
-H(s) = \frac{(s - j\omega_0)(s + j\omega_0)}{(s + \omega_0 \cos \theta + j\omega_0 \sin \theta)(s + \omega_0 \cos \theta - j\omega_0 \sin \theta)}
-$$
-  
-= 
-$$
-\frac{s^2 + \omega_0^2}{s^2 + (2\omega_0 \cos \theta)s + \omega_0^2} = \frac{s^2 + 142122.3}{s^2 + (753.98 \cos \theta)s + 142122.3}
-$$
-
-and
-
-$$
-|H(j\omega)| = \frac{-\omega^2 + 142122.3}{\sqrt{(-\omega^2 + 142122.3)^2 + (753.98\omega\cos\theta)^2}}
-$$
-
-The closer the poles are to the zeros (the closer θ is to π/2), the faster the gain recovery from 0 to 1 on either side of ω<sup>0</sup> = 120π. Figure 4.52b shows the amplitude response for three different values of θ. This example is a case of very simple design. To achieve zero gain over a band, we need an infinite number of poles as well as an infinite number of zeros.
-
-MATLAB easily computes and plots the magnitude response curves of Fig. 4.52b. To illustrate, let us plot the magnitude response using θ = 60◦ over a frequency range of 0 ≤ *f* ≤ 150 Hz. The result, shown in Fig. 4.53, matches the θ = 60◦ case of Fig. 4.52b.
-
->> f = (0:.01:150); omega0 = 2\*pi\*60; theta = 60\*pi/180;
-
-- >> H = @(s) (s.^2+omega0^2)./(s.^2+2\*omega0\*cos(theta)\*s+omega0^2);
-- >> plot(f,abs(H(1j\*2\*pi\*f)),'k-');
-- >> xlabel('f [Hz]'); ylabel('|H(j2\pi f)|');
-
-## **DR ILL 4.16 Magnitude Response from Pole-Zero Plots**
-
-Use the qualitative method of sketching the frequency response to show that the system with the pole-zero configuration in Fig. 4.54a is a highpass filter and the configuration in Fig. 4.54b is a bandpass filter.
-
-### <span id="page-463-0"></span>**[4.10-5 Practical Filters and Their Specifications](#page-10-0)**
-
-For ideal filters, everything is black and white; the gains are either zero or unity over certain bands. As we saw earlier, real life does not permit such a worldview. Things have to be gray or shades of gray. In practice, we can realize a variety of filter characteristics that can only approach ideal characteristics.
-
-An ideal filter has a passband (unity gain) and a stopband (zero gain) with a sudden transition from the passband to the stopband. There is no transition band. For practical (or realizable) filters, on the other hand, the transition from the passband to the stopband (or vice versa) is gradual and takes place over a finite band of frequencies. Moreover, for realizable filters, the gain cannot be zero over a finite band (Paley–Wiener condition). As a result, there can be no true stopband for practical filters. We therefore define a *stopband* to be a band over which the gain is below some small number *Gs*, as illustrated in Fig. 4.55. Similarly, we define a *passband* to be a band over which the gain is between 1 and some number *Gp* (*Gp* < 1), as shown in Fig. 4.55. We have selected the passband gain of unity for convenience. It could be any constant. Usually the gains are specified in terms of decibels. This is simply 20 times the log (to base 10) of the gain. Thus,
-
-$$
-G(\text{dB}) = 20\log_{10} G
-$$
-
-A gain of unity is 0 dB and a gain of <sup>√</sup> 2 is 3.01 dB, usually approximated by 3 dB. Sometimes the specification may be in terms of attenuation, which is the negative of the gain in dB. Thus, a gain of 1/ <sup>√</sup>2, that is, 0.707, is <sup>−</sup>3 dB, but is an attenuation of 3 dB.
-
-**Figure 4.55** Passband, stopband, and transition band in filters of various types.
-
-<span id="page-464-0"></span>In a typical design procedure, *Gp* (*minimum passband gain*) and *Gs* (*maximum stopband gain*) are specified. Figure 4.55 shows the passband, the stopband, and the transition band for typical lowpass, bandpass, highpass, and bandstop filters. Fortunately, the highpass, bandpass, and bandstop filters can be obtained from a basic lowpass filter by simple frequency transformations. For example, replacing *s* with ω*c*/*s* in the lowpass filter transfer function results in a highpass filter. Similarly, other frequency transformations yield the bandpass and bandstop filters. Hence, it is necessary to develop a design procedure only for a basic lowpass filter. Then, by using appropriate transformations, we can design filters of other types. The design procedures are beyond our scope here and will not be discussed. The interested reader is referred to [1].

engineering/linear-systems-and-signals-3rd-edition-bp-lathi/063_4.11 THE BILATERAL LAPLACE TRANSFORM.md

@@ -1,397 +0,0 @@
-## **4.11 THE BILATERAL LAPLACE [TRANSFORM](#page-10-0)**
-
-Situations involving noncausal signals and/or systems cannot be handled by the (unilateral) Laplace transform discussed so far. These cases can be analyzed by the *bilateral* (or *two-sided*) Laplace transform defined by
-
-$$
-X(s) = \int_{-\infty}^{\infty} x(t)e^{-st} dt
-$$
-
-and *x*(*t*) can be obtained from *X*(*s*) by the inverse transformation
-
-$$
-x(t) = \frac{1}{2\pi j} \int_{c-j\infty}^{c+j\infty} X(s)e^{st} ds
-$$
-
-Observe that the unilateral Laplace transform discussed so far is a special case of the bilateral Laplace transform, where the signals are restricted to the causal type. Basically, the two transforms are the same. For this reason we use the same notation for the bilateral Laplace transform.
-
-Earlier we showed that the Laplace transforms of *e*−*atu*(*t*) and of −*e*−*atu*(−*t*) are identical. The only difference is in their regions of convergence (ROC). The ROC for the former is Re*s* > −*a*; that for the latter is Re *s* < −*a*, as illustrated in Fig. 4.1. Clearly, the inverse Laplace transform of *X*(*s*) is not unique unless the ROC is specified. If we restrict all our signals to the causal type, however, this ambiguity does not arise. The inverse transform of 1/(*s*+*a*) is *e*−*atu*(*t*). Thus, in the unilateral Laplace transform, we can ignore the ROC in determining the inverse transform of *X*(*s*).
-
-We now show that any bilateral transform can be expressed in terms of two unilateral transforms. It is, therefore, possible to evaluate bilateral transforms from a table of unilateral transforms.
-
-Consider the function *x*(*t*) appearing in Fig. 4.56a. We separate *x*(*t*) into two components, *x*1(*t*) and *x*2(*t*), representing the positive time (*causal*) component and the negative time (*anticausal*) component of *x*(*t*), respectively (Figs. 4.56b and 4.56c):
-
-$$
-x_1(t) = x(t)u(t)
-$$
- and  $x_2(t) = x(t)u(-t)$ 
-
-**Figure 4.56** Expressing a signal as a sum of causal and anticausal components.
-
-The bilateral Laplace transform of *x*(*t*) is given by
-
-$$
-X(s) = \int_{-\infty}^{\infty} x(t)e^{-st} dt
-$$
-  
-= 
-$$
-\int_{-\infty}^{0^-} x_2(t)e^{-st} dt + \int_{0^-}^{\infty} x_1(t)e^{-st} dt
-$$
-  
-= 
-$$
-X_2(s) + X_1(s)
-$$
- (4.57)
-
-where *X*1(*s*) is the Laplace transform of the causal component *x*1(*t*), and *X*2(*s*) is the Laplace transform of the anticausal component *x*2(*t*). Consider *X*2(*s*), given by
-
-$$
-X_2(s) = \int_{-\infty}^{0^-} x_2(t)e^{-st} dt = \int_{0^+}^{\infty} x_2(-t)e^{st} dt
-$$
-
-Therefore,
-
-$$
-X_2(-s) = \int_{0^+}^{\infty} x_2(-t)e^{-st} dt
-$$
-
-If *x*(*t*) has any impulse or its derivative(s) at the origin, they are included in *x*1(*t*). Consequently, *x*2(*t*) = 0 at the origin; that is, *x*2(0) = 0. Hence, the lower limit on the integration in the preceding equation can be taken as 0<sup>−</sup> instead of 0+. Therefore,
-
-$$
-X_2(-s) = \int_{0^-}^{\infty} x_2(-t)e^{-st} dt
-$$
-
-Because *x*2(−*t*) is causal (Fig. 4.56d), *X*2(−*s*) can be found from the unilateral transform table. Changing the sign of *s* in *X*2(−*s*) yields *X*2(*s*).
-
-To summarize, the bilateral transform *X*(*s*) in Eq. (4.57) can be computed from the unilateral transforms in two steps:
-
-- 1. Split *x*(*t*) into its causal and anticausal components, *x*1(*t*) and *x*2(*t*), respectively.
-- 2. Since the signals *x*1(*t*) and *x*2(−*t*) are both causal, take the (unilateral) Laplace transform of *x*1(*t*) and add to it the (unilateral) Laplace transform of *x*2(−*t*), with *s* replaced by −*s*. This procedure gives the (bilateral) Laplace transform of *x*(*t*).
-
-Since *x*1(*t*) and *x*2(−*t*) are both causal, *X*1(*s*) and *X*2(−*s*) are both unilateral Laplace transforms. Let σ*<sup>c</sup>*<sup>1</sup> and σ*<sup>c</sup>*<sup>2</sup> be the abscissas of convergence of *X*1(*s*) and *X*2(−*s*), respectively. This statement implies that *X*1(*s*) exists for all *s* with Re *s* > σ*c*1, and *X*2(−*s*) exists for all *s* with Re*s* > σ*c*2. Therefore, *X*2(*s*) exists for all *s* with Re *s* < −σ*c*2. † Therefore, *X*(*s*) = *X*1(*s*) + *X*2(*s*) exists for all *s* such that
-
-$$
-\sigma_{c1} < \text{Re}\,s < -\sigma_{c2}
-$$
-
-The regions of convergence of *X*1(*s*), *X*2(*s*), and *X*(*s*) are shown in Fig. 4.57. Because *X*(*s*) is finite for all values of *s* lying in the strip of convergence (σ*<sup>c</sup>*<sup>1</sup> < Re *s* < −σ*c*2), poles of *X*(*s*) must lie outside this strip. The poles of *X*(*s*) arising from the causal component *x*1(*t*) lie to the left of the *strip* (region) *of convergence*, and those arising from its anticausal component *x*2(*t*) lie to its right (see Fig. 4.57). This fact is of crucial importance in finding the inverse bilateral transform.
-
-This result can be generalized to left-sided and right-sided signals. We define a signal *x*(*t*) as a *right-sided* signal if *x*(*t*) = 0 for *t* < *T*<sup>1</sup> for some finite positive or negative number *T*1. A causal signal is always a right-sided signal, but the converse is not necessarily true. A signal is said to *left-sided* if it is zero for *t* > *T*<sup>2</sup> for some finite, positive, or negative number *T*2. An anticausal signal is always a left-sided signal, but the converse is not necessarily true. A *two-sided* signal is of infinite duration on both positive and negative sides of *t* and is neither right-sided nor left-sided.
-
-We can show that the conclusions for ROC for causal signals also hold for right-sided signals, and those for anticausal signals hold for left-sided signals. In other words, if *x*(*t*) is causal or
-
-<sup>†</sup> For instance, if *<sup>x</sup>*(*t*) exists for all *<sup>t</sup>* <sup>&</sup>gt; 10, then *<sup>x</sup>*(−*t*), its time-inverted form, exists for *<sup>t</sup>* <sup>&</sup>lt; <sup>−</sup>10.
-
-**Figure 4.57** Regions of convergence for causal, anticausal, and combined signals.
-
-right-sided, the poles of *X*(*s*) lie to the left of the ROC, and if *x*(*t*) is anticausal or left-sided, the poles of *X*(*s*) lie to the right of the ROC.
-
-To prove this generalization, we observe that a right-sided signal can be expressed as *x*(*t*) + *xf*(*t*), where *x*(*t*) is a causal signal and *xf*(*t*) is some finite-duration signal. The ROC of any finite-duration signal is the entire *s*-plane (no finite poles). Hence, the ROC of the right-sided signal *x*(*t*) + *xf*(*t*) is the region common to the ROCs of *x*(*t*) and *xf*(*t*), which is same as the ROC for *x*(*t*). This proves the generalization for right-sided signals. We can use a similar argument to generalize the result for left-sided signals. Let us find the bilateral Laplace transform of
-
-$$
-x(t) = e^{bt}u(-t) + e^{at}u(t)
-$$
-\n(4.58)
-
-We already know the Laplace transform of the causal component
-
-$$
-e^{at}u(t) \Longleftrightarrow \frac{1}{s-a} \qquad \text{Re}\,s > a \tag{4.59}
-$$
-
-For the anticausal component, *x*2(*t*) = *ebtu*(−*t*), we have
-
-$$
-x_2(-t) = e^{-bt}u(t) \Longleftrightarrow \frac{1}{s+b} \qquad \text{Re}\, s > -b
-$$
-
-so that
-
-$$
-X_2(s) = \frac{1}{-s+b} = \frac{-1}{s-b} \qquad \text{Re}\, s < b
-$$
-
-Therefore,
-
-$$
-e^{bt}u(-t) \Longleftrightarrow \frac{-1}{s-b} \qquad \text{Re}\,s < b \tag{4.60}
-$$
-
-and the Laplace transform of *x*(*t*) in Eq. (4.58) is
-
-$$
-X(s) = -\frac{1}{s-b} + \frac{1}{s-a}
-$$
- Res > a and Res < b  
-= 
-$$
-\frac{a-b}{(s-b)(s-a)}
-$$
- Res < b  
-(4.61)
-
-Figure 4.58 shows *x*(*t*) and the ROC of *X*(*s*) for various values of *a* and *b*. Equation (4.61) indicates that the ROC of *X*(*s*) does not exist if *a* > *b*, which is precisely the case in Fig. 4.58f. Observe that the poles of *X*(*s*) are outside (on the edges) of the ROC. The poles of *X*(*s*) because of the anticausal component of *x*(*t*) lie to the right of the ROC, and those due to the causal component of *x*(*t*) lie to its left.
-
-When *X*(*s*) is expressed as a sum of several terms, the ROC for *X*(*s*) is the intersection of (region common to) the ROCs of all the terms. In general, if *<sup>x</sup>*(*t*) <sup>=</sup> %*<sup>k</sup> <sup>i</sup>*=<sup>1</sup> *xi*(*t*), then the ROC for *X*(*s*) is the intersection of the ROCs (region common to all ROCs) for the transforms *X*1(*s*), *X*2(*s*), ... , *Xk*(*s*).
-
-**Figure 4.58** Various two exponential signals and their regions of convergence.
-
-### **EXAMPLE 4.32 Inverse Bilateral Laplace Transform**
-
-Find the inverse bilateral Laplace transform of
-
-$$
-X(s) = \frac{-3}{(s+2)(s-1)}
-$$
-
-if the ROC is **(a)** −2 < Re *s* < 1, **(b)** Re *s* > 1, and **(c)** Re *s* < −2.
-
-**(a)**
-
-$$
-X(s) = \frac{1}{s+2} - \frac{1}{s-1}
-$$
-
-Now, *X*(*s*) has poles at −2 and 1. The strip of convergence is −2 < Re *s* < 1. The pole at −2, being to the left of the strip of convergence, corresponds to a causal signal. The pole at 1, being to the right of the strip of convergence, corresponds to an anticausal signal. Equations (4.59) and (4.60) yield
-
-$$
-x(t) = e^{-2t}u(t) + e^t u(-t)
-$$
-
-**(b)** Both poles lie to the left of the ROC, so both poles correspond to causal signals. Therefore,
-
-)*u*(*t*)
-
-*<sup>x</sup>*(*t*) <sup>=</sup> (*e*−2*<sup>t</sup>* <sup>−</sup>*et*
-
-**Figure 4.59** Three possible inverse transforms of −3/((*s* +2)(*s*−1)).
-
-<span id="page-470-0"></span>**(c)** Both poles lie to the right of the region of convergence, so both poles correspond to anticausal signals, and
-
-$$
-x(t) = (-e^{-2t} + e^t)u(-t)
-$$
-
-Figure 4.59 shows the three inverse transforms corresponding to the same *X*(*s*) but with different regions of convergence.
-
-### **[4.11-1 Properties of the Bilateral Laplace Transform](#page-11-0)**
-
-Properties of the bilateral Laplace transform are similar to those of the unilateral transform. We shall merely state the properties here without proofs. Let the ROC of *X*(*s*) be *a* < Re *s* < *b*. Similarly, let the ROC of *Xi*(*s*) be *ai* < Re *s* < *bi* for (*i* = 1, 2).
-
-LINEARITY
-
-$$
-a_1x_1(t) + a_2x_2(t) \Longleftrightarrow a_1X_1(s) + a_2X_2(s)
-$$
-
-The ROC for *a*1*X*1(*s*) + *a*2*X*2(*s*) is the region common to (intersection of) the ROCs for *X*1(*s*) and *X*2(*s*).
-
-TIME SHIFT
-
-$$
-x(t-T) \Longleftrightarrow X(s)e^{-sT}
-$$
-
-The ROC for *X*(*s*)*e*−*sT* is identical to the ROC for *X*(*s*).
-
-FREQUENCY SHIFT
-
-$$
-x(t)e^{s_0t} \Longleftrightarrow X(s-s_0)
-$$
-
-The ROC for *X*(*s*−*s*0) is *a*+*c* < Re *s* < *b*+*c*, where *c* = Re *s*0.
-
-TIME DIFFERENTIATION
-
-$$
-\frac{dx(t)}{dt} \Longleftrightarrow sX(s)
-$$
-
-The ROC for *sX*(*s*) contains the ROC for *X*(*s*) and may be larger than that of *X*(*s*) under certain conditions [e.g., if *X*(*s*) has a first-order pole at *s* = 0, it is canceled by the factor *s* in *sX*(*s*)].
-
-TIME INTEGRATION
-
-$$
-\int_{-\infty}^{t} x(\tau) d\tau \Longleftrightarrow X(s)/s
-$$
-
-The ROC for *sX*(*s*) is max (*a*, 0) < Re *s* < *b*.
-
-<span id="page-471-0"></span>TIME SCALING
-
-$$
-x(\beta t) \Longleftrightarrow \frac{1}{|\beta|}X\left(\frac{s}{\beta}\right)
-$$
-
-The ROC for *X*(*s*/β) is β*a* < Re *s* < β*b*. For β > 1, *x*(β*t*) represents time compression and the corresponding ROC expands by factor β. For 0 >β> 1, *x*(β*t*) represents time expansion and the corresponding ROC is compressed by factor β.
-
-TIME CONVOLUTION
-
-$$
-x_1(t) * x_2(t) \Longleftrightarrow X_1(s)X_2(s)
-$$
-
-The ROC for *X*1(*s*)*X*2(*s*) is the region common to (intersection of ) the ROCs for *X*1(*s*) and *X*2(*s*).
-
-FREQUENCY CONVOLUTION
-
-$$
-x_1(t)x_2(t) \Longleftrightarrow \frac{1}{2\pi j}\int_{c-j\infty}^{c+j\infty} X_1(w)X_2(s-w) dw
-$$
-
-The ROC for *X*1(*s*) ∗ *X*2(*s*) is *a*<sup>1</sup> +*a*<sup>2</sup> < Re *s* < *b*<sup>1</sup> +*b*2.
-
-TIME REVERSAL
-
-$$
-x(-t) \Longleftrightarrow X(-s)
-$$
-
-The ROC for *X*(−*s*) is −*b* < Re *s* < −*a*.
-
-### **[4.11-2 Using the Bilateral Transform for Linear System Analysis](#page-11-0)**
-
-Since the bilateral Laplace transform can handle noncausal signals, we can analyze noncausal LTIC systems using the bilateral Laplace transform. We have shown that the (zero-state) output *y*(*t*) is given by
-
-$$
-y(t) = \mathcal{L}^{-1}[X(s)H(s)]
-$$
-
-This expression is valid only if *X*(*s*)*H*(*s*) exists. The ROC of *X*(*s*)*H*(*s*) is the region in which both *X*(*s*) and *H*(*s*) exist. In other words, the ROC of *X*(*s*)*H*(*s*) is the region common to the regions of convergence of both *X*(*s*) and *H*(*s*). These ideas are clarified in the following examples.
-
-Find the current *y*(*t*) for the *RC* circuit in Fig. 4.60a if the voltage *x*(*t*) is
-
-$$
-x(t) = e^t u(t) + e^{2t} u(-t)
-$$
-
-**Figure 4.60** Response of a circuit to a noncausal input.
-
-The transfer function *H*(*s*) of the circuit is given by
-
-$$
-H(s) = \frac{s}{s+1} \qquad \text{Re}\, s > -1
-$$
-
-Because *h*(*t*) is a causal function, the ROC of *H*(*s*) is Re *s* > −1. Next, the bilateral Laplace transform of *x*(*t*) is given by
-
-$$
-X(s) = \frac{1}{s-1} - \frac{1}{s-2} = \frac{-1}{(s-1)(s-2)} \qquad 1 < \text{Re } s < 2
-$$
-
-The response *y*(*t*) is the inverse transform of *X*(*s*)*H*(*s*):
-
-$$
-y(t) = \mathcal{L}^{-1} \left[ \frac{-s}{(s+1)(s-1)(s-2)} \right] = \mathcal{L}^{-1} \left[ \frac{1}{6} \frac{1}{s+1} + \frac{1}{2} \frac{1}{s-1} - \frac{2}{3} \frac{1}{s-2} \right]
-$$
-
-The ROC of *X*(*s*)*H*(*s*) is that ROC common to both *X*(*s*) and *H*(*s*). This is 1 < Re *s* < 2. The poles *s* = ±1 lie to the left of the ROC and, therefore, correspond to causal signals; the pole
-
-### 454 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS
-
-*s* = 2 lies to the right of the ROC and thus represents an anticausal signal. Hence,
-
-$$
-y(t) = \frac{1}{6}e^{-t}u(t) + \frac{1}{2}e^{t}u(t) + \frac{2}{3}e^{2t}u(-t)
-$$
-
-Figure 4.60c shows *y*(*t*). Note that in this example, if
-
-$$
-x(t) = e^{-4t}u(t) + e^{-2t}u(-t)
-$$
-
-then the ROC of *X*(*s*) is −4 < Re *s* < −2. Here no region of convergence exists for *X*(*s*)*H*(*s*). Hence, the response *y*(*t*) goes to infinity.
-
-### **EXAMPLE 4.34 Response of a Noncausal System**
-
-Find the response *y*(*t*) of a noncausal system with the transfer function
-
-$$
-H(s) = \frac{-1}{s-1} \qquad \text{Re}\, s < 1
-$$
-
-to the input *x*(*t*) = *e*−2*<sup>t</sup> u*(*t*).
-
-We have
-
-$$
-X(s) = \frac{1}{s+2} \qquad \text{Re}\, s > -2
-$$
-
-and
-
-$$
-Y(s) = X(s)H(s) = \frac{-1}{(s-1)(s+2)}
-$$
-
-The ROC of *X*(*s*)*H*(*s*) is the region −2 < Re *s* < 1. By partial fraction expansion,
-
-$$
-Y(s) = \frac{-1/3}{s-1} + \frac{1/3}{s+2} \qquad -2 < \text{Re}\, s < 1
-$$
-
-and
-
-$$
-y(t) = \frac{1}{3} [e^t u(-t) + e^{-2t} u(t)]
-$$
-
-Note that the pole of *H*(*s*) lies in the RHP at 1. Yet the system is not unstable. The pole(s) in the RHP may indicate instability or noncausality, depending on its location with respect to the region of convergence of *H*(*s*). For example, if *H*(*s*) = −1/(*s*−1) with Re *s* > 1, the system is causal and unstable, with *h*(*t*) = −*et u*(*t*). In contrast, if *H*(*s*) = −1/(*s* − 1) with Re *s* < 1, the system is noncausal and stable, with *h*(*t*) = *et u*(−*t*).
-
-### <span id="page-474-0"></span>**EXAMPLE 4.35 System Response to a Noncausal Input**
-
-Find the response *y*(*t*) of a system with the transfer function
-
-$$
-H(s) = \frac{1}{s+5} \qquad \text{Re}\, s > -5
-$$
-
-and the input
-
-$$
-x(t) = e^{-t}u(t) + e^{-2t}u(-t)
-$$
-
-The input *x*(*t*) is of the type depicted in Fig. 4.58f, and the region of convergence for *X*(*s*) does not exist. In this case, we must determine separately the system response to each of the two input components, *x*1(*t*) = *e*−*<sup>t</sup> u*(*t*) and *x*2(*t*) = *e*−2*<sup>t</sup> u*(−*t*).
-
-$$
-X_1(s) = \frac{1}{s+1} \qquad \text{Re}\, s > -1
-$$
-\n
-$$
-X_2(s) = \frac{-1}{s+2} \qquad \text{Re}\, s < -2
-$$
-
-If *y*1(*t*) and *y*2(*t*) are the system responses to *x*1(*t*) and *x*2(*t*), respectively, then
-
-$$
-Y_1(s) = \frac{1}{(s+1)(s+5)} = \frac{1/4}{s+1} - \frac{1/4}{s+5}
-$$
- Re  $s > -1$ 
-
-so that
-
-$$
-y_1(t) = \frac{1}{4}(e^{-t} - e^{-5t})u(t)
-$$
-
-and
-
-$$
-Y_2(s) = \frac{-1}{(s+2)(s+5)} = \frac{-1/3}{s+2} + \frac{1/3}{s+5} \qquad -5 < \text{Re}\,s < -2
-$$
-
-so that
-
-$$
-y_2(t) = \frac{1}{3} [e^{-2t}u(-t) + e^{-5t}u(t)]
-$$
-
-Therefore,
-
-$$
-y(t) = y_1(t) + y_2(t) = \frac{1}{3}e^{-2t}u(-t) + \left(\frac{1}{4}e^{-t} + \frac{1}{12}e^{-5t}\right)u(t)
-$$

engineering/linear-systems-and-signals-3rd-edition-bp-lathi/064_4.12 MATLAB - CONTINUOUS-TIME FILTERS.md

@@ -1,360 +0,0 @@
-## **[4.12 MATLAB: CONTINUOUS-TIME](#page-11-0) FILTERS**
-
-Continuous-time filters are essential to many if not most engineering systems, and MATLAB is an excellent assistant for filter design and analysis. Although a comprehensive treatment of continuous-time filter techniques is outside the scope of this book, quality filters can be designed and realized with minimal additional theory.
-
-<span id="page-475-0"></span>A simple yet practical example demonstrates basic filtering concepts. Telephone voice signals are often lowpass-filtered to eliminate frequencies above a cutoff of 3 kHz, or ω*<sup>c</sup>* = 3000(2π ) ≈ 18,850 rad/s. Filtering maintains satisfactory speech quality and reduces signal bandwidth, thereby increasing the phone company's call capacity. How, then, do we design and realize an acceptable 3 kHz lowpass filter?
-
-### **[4.12-1 Frequency Response and Polynomial Evaluation](#page-11-0)**
-
-Magnitude response plots help assess a filter's performance and quality. The magnitude response of an ideal filter is a brick-wall function with unity passband gain and perfect stopband attenuation. For a lowpass filter with cutoff frequency ω*c*, the ideal magnitude response is
-
-$$
-|H_{\text{ideal}}(j\omega)| = \begin{cases} 1 & |\omega| \le \omega_c \\ 0 & |\omega| > \omega_c \end{cases}
-$$
-
-Unfortunately, ideal filters cannot be implemented in practice. Realizable filters require compromises, although good designs will closely approximate the desired brick-wall response.
-
-A realizable LTIC system often has a rational transfer function that is represented in the *s*-domain as
-
-$$
-H(s) = \frac{Y(s)}{X(s)} = \frac{B(s)}{A(s)} = \frac{\sum_{k=0}^{M} b_{k+N-M} s^{M-k}}{\sum_{k=0}^{N} a_k s^{N-k}}
-$$
-
-Frequency response *H*(*j*ω) is obtained by letting *s* = *j*ω, where frequency ω is in radians per second.
-
-MATLAB is ideally suited to evaluate frequency response functions. Defining a length-(*N* + 1) coefficient vector **A** = [*a*0,*a*1,...,*aN*] and a length-(*M* + 1) coefficient vector **B** = [*bN*−*<sup>M</sup>*,*bN*−*M*+1,..., *bN*], program CH4MP1 computes *H*(*j*ω) for each frequency in the input vector *ω*.
-
-```
-function [H] = CH4MP1(B,A,omega);
-% CH4MP1.m : Chapter 4, MATLAB Program 1
-% Function M-file computes frequency response for LTIC system
-% INPUTS: B = vector of feedforward coefficients
-% A = vector of feedback coefficients
-% omega = vector of frequencies [rad/s].
-% OUTPUTS: H = frequency response
-```
-
-```
-H = polyval(B,j*omega)./polyval(A,j*omega);
-```
-
-The function polyval efficiently evaluates simple polynomials and makes the program nearly trivial. For example, when A is the vector of coefficients [*a*0,*a*1,...,*aN*], polyval (A,j\*omega) computes
-
-$$
-\sum_{k=0}^N a_k (j\omega)^{N-k}
-$$
-
-for each value of the frequency vector omega. It is also possible to compute frequency responses by using the signal-processing toolbox function freqs.
-
-### DESIGN AND EVALUATION OF A SIMPLE *RC* FILTER
-
-One of the simplest lowpass filters is realized by using an *RC* circuit, as shown in Fig. 4.61. This one-pole system has transfer function *HRC*(*s*) = (*RCs* + 1)−<sup>1</sup> and magnitude response |*HRC*(*j*ω)|=|(*j*ω*RC* + 1)−<sup>1</sup>| = 1/ 1+(*RC*ω)2. Independent of component values *R* and *C*, this circuit has many desirable characteristics, such as unity gain at ω = 0 and magnitude response that monotonically decreases to zero as ω → ∞.
-
-Components *R* and *C* are chosen to set the desired 3 kHz cutoff frequency. For many filter types, the cutoff frequency corresponds to the half-power point, or |*HRC*(*j*ω*c*)| = 1/ <sup>√</sup>2. Assign *C* a realistic capacitance of 1 nF, then the required resistance is computed by *R* = 1/ *C*<sup>2</sup>ω<sup>2</sup> *c* = 1/ (10−<sup>9</sup>)<sup>2</sup>(2π3000)2.
-
-```
->> omega_c = 2*pi*3000; C = 1e-9; R = 1/sqrt(C^2*omega_c^2)
-    R = 5.3052e+004
-```
-
-The root of this first-order *RC* filter is directly related to the cutoff frequency, λ = −1/*RC* = −18,850 = −ω*c*.
-
-To evaluate the *RC* filter performance, the magnitude response is plotted over the mostly audible frequency range (0 ≤ *f* ≤ 20 kHz).
-
-```
->> f = linspace(0,20000,200); Hmag_RC = abs(CH4MP1([1],[R*C 1],f*2*pi));
-```
-
-```
->> plot(f,abs(f*2*pi)<=omega_c,'k-',f,Hmag_RC,'k--');
-```
-
-```
->> axis([0 20000 -0.05 1.05]); xlabel('f [Hz]'); ylabel('|H(j2\pi f)|');
-```
-
-```
->> legend('Ideal','First-order RC','location','best');
-```
-
-The linspace(X1,X2,N) command generates an N-length vector of linearly spaced points between X1 and X2.
-
-As shown in Fig. 4.62, the first-order *RC* response is indeed lowpass with a half-power cutoff frequency equal to 3 kHz. It rather poorly approximates the desired brick-wall response: the passband is not very flat, and stopband attenuation increases very slowly to less than 20 dB at 20 kHz.
-
-**Figure 4.62** Magnitude response |*HRC*(*j*2π*f*)| of a first-order *RC* filter.
-
-**Figure 4.63** A cascaded *RC* filter.
-
-### A CASCADED *RC* FILTER AND POLYNOMIAL EXPANSION
-
-A first-order *RC* filter is destined for poor performance; one pole is simply insufficient to obtain good results. A cascade of *RC* circuits increases the number of poles and improves the filter response. To simplify the analysis and prevent loading between stages, we employ op-amp followers to buffer the output of each stage, as shown in Fig. 4.63. A cascade of *N* stages results in an *N*th-order filter with transfer function given by
-
-$$
-H_{\text{cascade}}(s) = [H_{RC}(s)]^N = (RCs + 1)^{-N}
-$$
-
-Upon choosing a cascade of 10 stages and *C* = 1 nF, a 3 kHz cutoff frequency is obtained by setting *<sup>R</sup>* <sup>=</sup> <sup>√</sup> <sup>21</sup>/<sup>10</sup> <sup>−</sup>1/(*C*ω*c*) <sup>=</sup> <sup>√</sup> 21/<sup>10</sup> −1/(6π(10)−<sup>6</sup>).
-
->> R = sqrt(2^(1/10)-1)/(C\*omega\_c) R = 1.4213e+004
-
-This cascaded filter has a 10th-order pole at λ = −1/*RC* and no finite zeros. To compute the magnitude response, polynomial coefficient vectors **A** and **B** are needed. Setting **B** = [1] ensures there are no finite zeros or, equivalently, that all zeros are at infinity. The poly command, which expands a vector of roots into a corresponding vector of polynomial coefficients, is used to obtain **A**.
-
-- <span id="page-478-0"></span>>> B = 1; A = poly(-1/(R\*C)\*ones(10,1));A = A/A(end);
-- >> Hmag\_cascade = abs(CH4MP1(B,A,f\*2\*pi));
-- >> plot(f,abs(f\*2\*pi)<=omega\_c,'k-',f,Hmag\_cascade,'k--');
-- >> axis([0 20000 -0.05 1.05]); xlabel('f [Hz]'); ylabel('|H(j2\pi f)|');
-- >> legend('Ideal','Tenth-order RC cascade','location','best');
-
-Notice that scaling a polynomial by a constant does not change its roots. Conversely, the roots of a polynomial specify a polynomial within a scale factor. The command A = A/A(end) properly scales the denominator polynomial to ensure unity gain at ω = 0.
-
-The magnitude response plot of the tenth-order *RC* cascade is shown in Fig. 4.64. Compared with the simple *RC* response of Fig. 4.62, the passband remains relatively unchanged, but stopband attenuation is greatly improved to over 60 dB at 20 kHz.
-
-**Figure 4.64** Magnitude response |*H*cascade(*j*2π*f*)| of a tenth-order *RC* cascade.
-
-## **[4.12-2 Butterworth Filters and the](#page-11-0)** Find **Command**
-
-The pole location of a first-order lowpass filter is necessarily fixed by the cutoff frequency. There is little reason, however, to place all the poles of a 10th-order filter at one location. Better pole placement will improve our filter's magnitude response. One strategy, discussed in Sec. 4.10, is to place a wall of poles opposite the passband frequencies. A semicircular wall of poles leads to the Butterworth family of filters, and a semi-elliptical shape leads to the Chebyshev family of filters. Butterworth filters are considered first.
-
-To begin, notice that a transfer function *H*(*s*) with real coefficients has a squared magnitude response given by |*H*(*j*ω)| <sup>2</sup> <sup>=</sup> *<sup>H</sup>*(*j*ω)*H*∗(*j*ω) <sup>=</sup> *<sup>H</sup>*(*j*ω)*H*(−*j*ω) <sup>=</sup> *<sup>H</sup>*(*s*)*H*(−*s*)|*<sup>s</sup>*=*j*ω. Thus, half the poles of |*H*(*j*ω)| <sup>2</sup> correspond to the filter *H*(*s*) and the other half correspond to *H*(−*s*). Filters that are both stable and causal require *H*(*s*) to include only left-half-plane poles.
-
-The squared magnitude response of a Butterworth filter is
-
-$$
-|H_{\text{BW}}(j\omega)|^2 = \frac{1}{1 + (j\omega/j\omega_c)^{2N}}
-$$
-
-This function has the same appealing characteristics as the first-order *RC* filter: a gain that is unity at ω = 0 and monotonically decreases to zero as ω → ∞. By construction, the half-power gain
-
-occurs at ω*c*. Perhaps most importantly, however, the first 2*N* − 1 derivatives of |*H*BW(*j*ω)| with respect to ω are zero at ω = 0. Put another way, the passband is constrained to be very flat for low frequencies. For this reason, Butterworth filters are sometimes called maximally flat filters.
-
-As discussed in Sec. B.7, the roots of minus 1 must lie equally spaced on a circle centered at the origin. Thus, the 2*N* poles of |*H*BW(*j*ω)| <sup>2</sup> naturally lie equally spaced on a circle of radius ω*<sup>c</sup>* centered at the origin. Figure 4.65 displays the 20 poles corresponding to the case *N* = 10 and ω*<sup>c</sup>* = 3000(2π ) rad/s. An *N*th-order Butterworth filter that is both causal and stable uses the *N* left-half-plane poles of |*H*BW(*j*ω)| 2.
-
-To design a 10th-order Butterworth filter, we first compute the 20 poles of |*H*BW(*j*ω)| 2:
-
->> N=10; poles = roots([(1j\*omega\_c)^(-2\*N),zeros(1,2\*N-1),1]);
-
-The find command is a powerful and useful function that returns the indices of a vector's nonzero elements. Combined with relational operators, the find command allows us to extract the 10 left-half-plane roots that correspond to the poles of our Butterworth filter.
-
->> BW\_poles = poles(find(real(poles)<0));
-
-To compute the magnitude response, these roots are converted to coefficient vector **A**.
-
-```
->> A = poly(BW_poles); A = A/A(end); Hmag_BW = abs(CH4MP1(B,A,f*2*pi));
-```
-
-```
->> plot(f,abs(f*2*pi)<=omega_c,'k-',f,Hmag_BW,'k--');
-```
-
-```
->> axis([0 20000 -0.05 1.05]); xlabel('f [Hz]'); ylabel('|H(j2\pi f)|');
-```
-
-```
->> legend('Ideal','Tenth-order Butterworth','location','best');
-```
-
-The magnitude response plot of the Butterworth filter is shown in Fig. 4.66. The Butterworth response closely approximates the brick-wall function and provides excellent filter characteristics: flat passband, rapid transition to the stopband, and excellent stopband attenuation (>40 dB at 5 kHz).
-
-<span id="page-480-0"></span>**Figure 4.66** Magnitude response |*H*BW(*j*2π*f*)| of a tenth-order Butterworth filter.
-
-### **[4.12-3 Using Cascaded Second-Order Sections](#page-11-0) for Butterworth Filter Realization**
-
-For our *RC* filters, realization preceded design. For our Butterworth filter, however, design has preceded realization. For our Butterworth filter to be useful, we must be able to implement it.
-
-Since the transfer function *H*BW(*s*) is known, the differential equation is also known. Therefore, it is possible to try to implement the design by using op-amp integrators, summers, and scalar multipliers. Unfortunately, this approach will not work well. To understand why, consider the denominator coefficients *a*<sup>0</sup> = 1.766×10−<sup>43</sup> and *a*<sup>10</sup> = 1. The smallest coefficient is 43 orders of magnitude smaller than the largest coefficient! It is practically impossible to accurately realize such a broad range in scale values. To understand this, skeptics should try to find realistic resistors such that *Rf* /*R* = 1.766×10−43. Additionally, small component variations will cause large changes in actual pole location.
-
-A better approach is to cascade five second-order sections, where each section implements one complex conjugate pair of poles. By pairing poles in complex conjugate pairs, each of the resulting second-order sections has real coefficients. With this approach, the smallest coefficients are only about nine orders of magnitude smaller than the largest coefficients. Furthermore, pole placement is typically less sensitive to component variations for cascaded structures.
-
-The Sallen–Key circuit shown in Fig. 4.67 provides a good way to realize a pair of complex-conjugate poles.† The transfer function of this circuit is
-
-$$
-H_{SK}(s) = \frac{\frac{1}{R_1R_2C_1C_2}}{s^2 + \left(\frac{1}{R_1C_1} + \frac{1}{R_2C_1}\right)s + \frac{1}{R_1R_2C_1C_2}} = \frac{\omega_0^2}{s^2 + \left(\frac{\omega_0}{Q}\right)s + \omega_0^2}
-$$
-
-Geometrically, ω<sup>0</sup> is the distance from the origin to the poles and *Q* = 1/2 cosψ, where ψ is the angle between the negative real axis and the pole. Termed the "quality factor" of a circuit, *Q*
-
-<sup>†</sup> A more general version of the Sallen–Key circuit has a resistor *Ra* from the negative terminal to ground and a resistor *Rb* between the negative terminal and the output. In Fig. 4.67, *Ra* = ∞ and *Rb* = 0.
-
-provides a measure of the peakedness of the response. High-*Q* filters have poles close to the ω axis, which boost the magnitude response near those frequencies.
-
-Although many ways exist to determine suitable component values, a simple method is to assign *R*<sup>1</sup> a realistic value and then let *R*<sup>2</sup> = *R*1, *C*<sup>1</sup> = 2*Q*/ω0*R*1, and *C*<sup>2</sup> = 1/2*Q*ω0*R*2. Butterworth poles are a distance ω*<sup>c</sup>* from the origin, so ω<sup>0</sup> = ω*c*. For our 10th-order Butterworth filter, the angles ψ are regularly spaced at 9, 27, 45, 63, and 81 degrees. MATLAB program CH4MP2 automates the task of computing component values and magnitude responses for each stage.
-
-```
-% CH4MP2.m : Chapter 4, MATLAB Program 2
-% Script M-file computes Sallen-Key component values and magnitude
-% responses for each of the five cascaded second-order filter sections.
-omega_0 = 3000*2*pi; % Filter cut-off frequency
-psi = [9 27 45 63 81]*pi/180; % Butterworth pole angles
-f = linspace(0,6000,200); % Frequency range for magnitude response calculations
-Hmag_SK = zeros(5,200); % Pre-allocate array for magnitude responses
-for stage = 1:5,
-    Q = 1/(2*cos(psi(stage))); % Compute Q for current stage
-    % Compute and display filter components to the screen:
-    disp(['Stage ',num2str(stage),...
-            ' (Q = ',num2str(Q),...
-            '): R1 = R2 = ',num2str(56000),...
-            ', C1 = ',num2str(2*Q/(omega_0*56000)),...
-            ', C2 = ',num2str(1/(2*Q*omega_0*56000))]);
-    B = omega_0^2; A = [1 omega_0/Q omega_0^2]; % Compute filter coefficients
-    Hmag_SK(stage,:) = abs(CH4MP1(B,A,2*pi*f)); % Compute magnitude response
-end
-plot(f,Hmag_SK,'k',f,prod(Hmag_SK),'k:')
-xlabel('f [Hz]'); ylabel('Magnitude Response')
-```
-
-The disp command displays a character string to the screen. Character strings must be enclosed in single quotation marks. The num2str command converts numbers to character strings and facilitates the formatted display of information. The prod command multiplies along the columns of a matrix; it computes the total magnitude response as the product of the magnitude responses of the five stages.
-
-Executing the program produces the following output:
-
-```
->> CH4MP2
-   Stage 1 (Q = 0.50623): R1 = R2 = 56000, C1 = 9.5916e-10, C2 = 9.3569e-10
-   Stage 2 (Q = 0.56116): R1 = R2 = 56000, C1 = 1.0632e-09, C2 = 8.441e-10
-   Stage 3 (Q = 0.70711): R1 = R2 = 56000, C1 = 1.3398e-09, C2 = 6.6988e-10
-   Stage 4 (Q = 1.1013): R1 = R2 = 56000, C1 = 2.0867e-09, C2 = 4.3009e-10
-   Stage 5 (Q = 3.1962): R1 = R2 = 56000, C1 = 6.0559e-09, C2 = 1.482e-10
-```
-
-<span id="page-482-0"></span>**Figure 4.68** Magnitude responses for Sallen–Key filter stages.
-
-Since all the component values are practical, this filter is possible to implement. Figure 4.68 displays the magnitude responses for all five stages (solid lines). The total response (dotted line) confirms a 10th-order Butterworth response. Stage 5, which has the largest *Q* and implements the pair of conjugate poles nearest the ω axis, is the most peaked response. Stage 1, which has the smallest *Q* and implements the pair of conjugate poles furthest from the ω axis, is the least peaked response. In practice, it is best to order high-*Q* stages last; this reduces the risk that the high gains will saturate the filter hardware.
-
-### **[4.12-4 Chebyshev Filters](#page-11-0)**
-
-Like an order-*N* Butterworth lowpass filter (LPF), an order-*N* Chebyshev LPF is an all-pole filter that possesses many desirable characteristics. Compared with an equal-order Butterworth filter, the Chebyshev filter achieves better stopband attenuation and reduced transition bandwidth by allowing an adjustable amount of ripple within the passband.
-
-The squared magnitude response of a Chebyshev filter is
-
-$$
-|H_{\rm C}(j\omega)|^2 = \frac{1}{1 + \epsilon^2 C_N^2(\omega/\omega_c)}
-$$
-
-where controls the passband ripple, *CN*(ω/ω*c*) is a degree-*N* Chebyshev polynomial, and ω*<sup>c</sup>* is the radian cutoff frequency. Several characteristics of Chebyshev LPFs are noteworthy:
-
-• An order-*N* Chebyshev LPF is equi-ripple in the passband (|ω| ≤ ω*c*), has a total of *N* maxima and minima over (0 ≤ ω ≤ ω*c*), and is monotonic decreasing in the stopband (|ω| > ω*c*).
-
-### 464 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS
-
-- In the passband, the maximum gain is 1 and the minimum gain is 1/ <sup>√</sup>1<sup>+</sup> 2. For odd-valued *N*, |*H*(*j*0)| = 1. For even-valued *N*, |*H*C(*j*0)| = 1/ <sup>√</sup>1<sup>+</sup> 2.
-- Ripple is controlled by setting <sup>=</sup> <sup>√</sup> 10*<sup>R</sup>*/<sup>10</sup> −1, where *R* is the allowable passband ripple expressed in decibels. Reducing adversely affects filter performance (see Prob. 4.12-10).
-- Unlike Butterworth filters, the cutoff frequency ω*<sup>c</sup>* rarely specifies the 3 dB point. For = 1, |*H*C(*j*ω*c*)| <sup>2</sup> = 1/(1+ <sup>2</sup>) = 0.5. The cutoff frequency ω*<sup>c</sup>* simply indicates the frequency after which |*H*C(*j*ω)| < 1/ √ 1+ 2.
-
-The Chebyshev polynomial *CN*(*x*) is defined as
-
-$$
-C_N(x) = \cos[N\cos^{-1}(x)] = \cosh[N\cosh^{-1}(x)]
-$$
-
-In this form, it is difficult to verify that *CN*(*x*) is a degree-*N* polynomial in *x*. A recursive form of *CN*(*x*) makes this fact more clear (see Prob. 4.12-13).
-
-$$
-C_N(x) = 2xC_{N-1}(x) - C_{N-2}(x)
-$$
-
-With *C*0(*x*) = 1 and *C*1(*x*) = *x*, the recursive form shows that any *CN* is a linear combination of degree-*N* polynomials and is therefore a degree-*N* polynomial itself. For *N* ≥ 2, MATLAB program CH4MP3 generates the (*N* +1) coefficients of Chebyshev polynomial *CN*(*x*).
-
-```
-function [C_N] = CH4MP3(N);
-% CH4MP3.m : Chapter 4, MATLAB Program 3
-% Function M-file computes Chebyshev polynomial coefficients
-% using the recursion relation C_N(x) = 2xC_{N-1}(x) - C_{N-2}(x)
-% INPUTS: N = degree of Chebyshev polynomial
-% OUTPUTS: C_N = vector of Chebyshev polynomial coefficients
-C_Nm2 = 1; C_Nm1 = [1 0]; % Initial polynomial coefficients:
-for t = 2:N;
-   C_N = 2*conv([1 0],C_Nm1)-[zeros(1,length(C_Nm1)-length(C_Nm2)+1),C_Nm2];
-   C_Nm2 = C_Nm1; C_Nm1 = C_N;
-```
-
-```
-end
-```
-
-```
-As examples, consider C2(x) = 2xC1(x) − C0(x) = 2x(x) − 1 = 2x2 − 1 and C3(x) = 2xC2(x) −
-C1(x) = 2x(2x2 −1)−x = 4x3 −3x. CH4MP3 easily confirms these cases.
-```
-
-```
->> CH4MP3(2)
-  ans = 2 0 -1
->> CH4MP3(3)
-  ans = 4 0 -3 0
-```
-
-Since *CN*(ω/ω*c*) is a degree-*N* polynomial, |*H*C(*j*ω)| <sup>2</sup> is an all-pole rational function with 2*N* finite poles. Similar to the Butterworth case, the *N* poles specifying a causal and stable Chebyshev filter can be found by selecting the *N* left-half-plane roots of 1+ <sup>2</sup>*C*<sup>2</sup> *<sup>N</sup>*[*s*/(*j*ω*c*)].
-
-Root locations and dc gain are sufficient to specify a Chebyshev filter for a given *N* and . To demonstrate, consider the design of an order-8 Chebyshev filter with cutoff frequency *fc* = 1 kHz and allowable passband ripple *R* = 1 dB. First, filter parameters are specified.
-
-```
->> omega_c = 2*pi*1000; R = 1; N = 8;
->> epsilon = sqrt(10^(R/10)-1);
-```
-
-The coefficients of *CN*[*s*/(*j*ω*c*)] are obtained with the help of CH4MP3, and then the coefficients of [1+ <sup>2</sup>*C*<sup>2</sup> *<sup>N</sup>*(*s*/(*j*ω*c*))] are computed by using convolution to perform polynomial multiplication.
-
-```
->> CN = CH4MP3(N).*((1/(1j*omega_c)).^[N:-1:0]);
->> CP = epsilon^2*conv(CN,CN); CP(end) = CP(end)+1;
-```
-
-Next, the polynomial roots are found, and the left-half-plane poles are retained and plotted.
-
-```
->> poles = roots(CP); i = find(real(poles)<0); C_poles = poles(i);
->> plot(real(C_poles),imag(C_poles),'kx'); axis equal;
->> axis(omega_c*[-1.1 1.1 -1.1 1.1]);
->> xlabel('Real'); ylabel('Imaginary');
-```
-
-As shown in Fig. 4.69, the roots of a Chebyshev filter lie on an ellipse† (see Prob. 4.12-14).
-
-**Figure 4.69** Pole-zero plot for an order-8 Chebyshev LPF with *fc* = 1 kHz and *R* = 1 dB.
-
-To compute the filter's magnitude response, the poles are expanded into a polynomial, the dc gain is set based on the even value of *N*, and CH4MP1 is used.
-
-```
->> A = poly(C_poles); B = A(end)/sqrt(1+epsilon^2);
-```
-
->> omega = linspace(0,2\*pi\*2000,2001); H\_C = CH4MP1(B,A,omega);
-
-```
->> plot(omega/2/pi,abs(H_C),'k'); axis([0 2000 0 1.1]);
-```
-
-```
->> xlabel('f [Hz]'); ylabel('|H_C(j2\pi f)|');
-```
-
-<sup>†</sup> E. A. Guillemin demonstrates a wonderful relationship between the Chebyshev ellipse and the Butterworth circle in his book *Synthesis of Passive Networks* (Wiley, New York, 1957).
-
-<span id="page-485-0"></span>**Figure 4.70** Magnitude responses for an order-8 Chebyshev LPF with *fc* = 1 kHz and *R* = 1 dB.
-
-As seen in Fig. 4.70, the magnitude response exhibits correct Chebyshev filter characteristics: passband ripples are equal in height and never exceed *R* = 1 dB; there are a total of *N* = 8 maxima and minima in the passband; and the gain rapidly and monotonically decreases after the cutoff frequency of *fc* = 1 kHz.
-
-For higher-order filters, polynomial rooting may not provide reliable results. Fortunately, Chebyshev roots can also be determined analytically. For
-
-$$
-\phi_k = \frac{2k+1}{2N}\pi \quad \text{and} \quad \xi = \frac{1}{N}\sinh^{-1}\left(\frac{1}{\epsilon}\right)
-$$
-
-the Chebyshev poles are
-
-$$
-p_k = \omega_c \sinh(\xi) \sin(\phi_k) + j\omega_c \cosh(\xi) \cos(\phi_k)
-$$
-
-Continuing the same example, the poles are recomputed and again plotted. The result is identical to Fig. 4.69.
-
-```
->> k = [1:N]; xi = 1/N*asinh(1/epsilon); phi = (k*2-1)/(2*N)*pi;
->> C_poles = omega_c*(-sinh(xi)*sin(phi)+1j*cosh(xi)*cos(phi));
->> plot(real(C_poles),imag(C_poles),'kx'); axis equal;
->> axis(omega_c*[-1.1 1.1 -1.1 1.1]);
->> xlabel('Real'); ylabel('Imaginary');
-```
-
-As in the case of high-order Butterworth filters, a cascade of second-order filter sections facilitates practical implementation of Chebyshev filters. Problems 4.12-5 and 4.12-8 use second-order Sallen–Key circuit stages to investigate such implementations.

engineering/linear-systems-and-signals-3rd-edition-bp-lathi/065_4.13 SUMMARY.md

@@ -1,65 +0,0 @@
-## **[4.13 SUMMARY](#page-11-0)**
-
-This chapter discusses analysis of LTIC (linear, time-invariant, continuous-time) systems by the Laplace transform, which transforms integro-differential equations of such systems into algebraic equations. Therefore solving these integro-differential equations reduces to solving algebraic equations. The Laplace transform method cannot be used for time-varying-parameter systems or for nonlinear systems in general.
-
-The transfer function *H*(*s*) of an LTIC system is the Laplace transform of its impulse response. It may also be defined as a ratio of the Laplace transform of the output to the Laplace transform of the input when all initial conditions are zero (system in zero state). If *X*(*s*) is the Laplace transform of the input *x*(*t*) and *Y*(*s*) is the Laplace transform of the corresponding output *y*(*t*) (when all initial conditions are zero), then *Y*(*s*) = *X*(*s*)*H*(*s*). For an LTIC system described by an *N*th-order differential equation *Q*(*D*)*y*(*t*) = *P*(*D*)*x*(*t*), the transfer function *H*(*s*) = *P*(*s*)/*Q*(*s*). Like the impulse response *h*(*t*), the transfer function *H*(*s*) is also an external description of the system.
-
-Electrical circuit analysis can also be carried out by using a transformed circuit method, in which all signals (voltages and currents) are represented by their Laplace transforms, all elements by their impedances (or admittances), and initial conditions by their equivalent sources (initial condition generators). In this method, a network can be analyzed as if it were a resistive circuit.
-
-Large systems can be depicted by suitably interconnected subsystems represented by blocks. Each subsystem, being a smaller system, can be readily analyzed and represented by its input–output relationship, such as its transfer function. Analysis of large systems can be carried out with the knowledge of input–output relationships of its subsystems and the nature of interconnection of various subsystems.
-
-LTIC systems can be realized by scalar multipliers, adders, and integrators. A given transfer function can be synthesized in many different ways, such as canonic, cascade, and parallel. Moreover, every realization has a transpose, which also has the same transfer function. In practice, all the building blocks (scalar multipliers, adders, and integrators) can be obtained from operational amplifiers.
-
-The system response to an everlasting exponential *est* is also an everlasting exponential *H*(*s*)*est*. Consequently, the system response to an everlasting exponential *ej*ω*<sup>t</sup>* is *H*(*j*ω) *ej*ω*<sup>t</sup>* . Hence, *H*(*j*ω) is the frequency response of the system. For a sinusoidal input of unit amplitude and having frequency ω, the system response is also a sinusoid of the same frequency (ω) with amplitude |*H*(*j*ω)|, and its phase is shifted by *H*(*j*ω) with respect to the input sinusoid. For this reason |*H*(*j*ω)| is called the amplitude response (gain) and *H*(*j*ω) is called the phase response of the system. Amplitude and phase response of a system indicate the filtering characteristics of the system. The general nature of the filtering characteristics of a system can be quickly determined from a knowledge of the location of poles and zeros of the system transfer function.
-
-Most of the input signals and practical systems are causal. Consequently we are required most of the time to deal with causal signals. When all signals must be causal, the Laplace transform analysis is greatly simplified; the region of convergence of a signal becomes irrelevant to the analysis process. This special case of the Laplace transform (which is restricted to causal signals) is called the unilateral Laplace transform. Much of the chapter deals with this variety of Laplace transform. Section 4.11 discusses the general Laplace transform (the bilateral Laplace transform), which can handle causal and noncausal signals and systems. In the bilateral transform, the inverse transform of *X*(*s*) is not unique but depends on the region of convergence of *X*(*s*). Thus, the region of convergence plays a very crucial role in the bilateral Laplace transform.
-
-### <span id="page-487-0"></span>468 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS
-
-### **[REFERENCES](#page-11-0)**
-
-- 1. Lathi, B. P. *Signal Processing and Linear Systems*, 1st ed. Oxford University Press, New York, 1998.
-- 2. Doetsch, G. *Introduction to the Theory and Applications of the Laplace Transformation with a Table of Laplace Transformations.* Springer-Verlag, New York, 1974.
-- 3. LePage, W. R. *Complex Variables and the Laplace Transforms for Engineers*. McGraw-Hill, New York, 1961.
-- 4. Durant, Will, and Ariel Durant. *The Age of Napoleon*, Part XI in *The Story of Civilization Series*. Simon & Schuster, New York, 1975.
-- 5. Bell, E. T. *Men of Mathematics*. Simon & Schuster, New York, 1937.
-- 6. Nahin, P. J. "Oliver Heaviside: Genius and Curmudgeon." *IEEE Spectrum*, vol. 20, pp. 63–69, July 1983.
-- 7. Berkey, D. *Calculus*, 2nd ed. Saunders, Philadelphia, 1988.
-- 8. Encyclopaedia Britannica. *Micropaedia IV*, 15th ed., p. 981, Chicago, 1982.
-- 9. Churchill, R. V. *Operational Mathematics*, 2nd ed. McGraw-Hill, New York, 1958.
-- 10. Truxal, J. G. *The Age of Electronic Messages*. McGraw-Hill, New York, 1990.
-- 11. Van Valkenberg, M. *Analog Filter Design*. Oxford University Press, New York, 1982.
-
-## **[PROBLEMS](#page-11-0)**
-
-- **4.1-1** By direct integration [Eq. (4.1)] find the Laplace transforms and the region of convergence of the following functions:
-  - (a) *u*(*t*)−*u*(*t* −1)
-  - (b) *te*−*<sup>t</sup> u*(*t*)
-  - (c) *t* cos ω0*t u*(*t*)
-  - (d) (*e*2*<sup>t</sup>* <sup>−</sup>2*e*−*<sup>t</sup>* )*u*(*t*)
-  - (e) cos ω1*t* cos ω2*t u*(*t*)
-  - (f) cosh(*at*)*u*(*t*)
-  - (g) sinh(*at*)*u*(*t*)
-  - (h) *<sup>e</sup>*−2*<sup>t</sup>* cos(5*<sup>t</sup>* <sup>+</sup>θ )*u*(*t*)
-- **4.1-2** By direct integration [Eq. (4.1)] find the Laplace transforms and the region of convergence of the following functions:
-
-(a) 
-$$
-e^{-2t}u(t-5) + \delta(t-1)
-$$
-
-(b) 
-$$
-\pi e^{3t} u(t+5) - \delta(2t)
-$$
-
-(c) 
-$$
-\sum_{k=0}^{\infty} \delta(t - kT), T > 0
-$$
-
-- **4.1-3** By direct integration find the Laplace transforms of the signals shown in Fig. P4.1-3.
-- **4.1-4** Find the inverse (unilateral) Laplace transforms of the following functions:
-
-(a) 
-$$

engineering/linear-systems-and-signals-3rd-edition-bp-lathi/066_REFERENCES.md

@@ -1,1159 +0,0 @@
-\frac{2s+5}{s^2+5s+6}
-$$
-  
-\n(b) 
-$$
-\frac{3s+5}{s^2+4s+13}
-$$
-  
-\n(c) 
-$$
-\frac{(s+1)^2}{s^2-s-6}
-$$
-  
-\n(d) 
-$$
-\frac{5}{s^2(s+2)}
-$$
-
-**Figure P4.1-3**
-
-(e) 
-$$
-\frac{2s+1}{(s+1)(s^2+2s+2)}
-$$
-  
-\n(f) 
-$$
-\frac{s+2}{s(s+1)^2}
-$$
-  
-\n(g) 
-$$
-\frac{1}{(s+1)(s+2)^4}
-$$
-  
-\n(h) 
-$$
-\frac{s+1}{s(s+2)^2(s^2+4s+5)}
-$$
-
-(i) 
-$$
-\frac{s(s+2)^2(s^2+4s+5)}{(s+1)^2(s^2+2s+5)}
-$$
-
-- **4.2-1** Suppose a CT signal *x*(*t*)=2[*u*(*t* −2)−*u*(*t* +1)] has a transform *X*(*s*).
-  - (a) If *<sup>Y</sup>*a(*s*) <sup>=</sup> *<sup>e</sup>*−5*<sup>s</sup> sX <sup>s</sup>* <sup>+</sup> <sup>1</sup> 2 , determine and sketch the corresponding signal *y*a(*t*).
-  - (b) If *<sup>Y</sup>*b(*s*) <sup>=</sup> <sup>2</sup>−*<sup>s</sup> sX*(*s* −2), determine and sketch the corresponding signal *y*b(*t*).
-- **4.2-2** Find the Laplace transforms of the following functions using only Table 4.1 and the time-shifting property (if needed) of the unilateral Laplace transform:
-  - (a) *u*(*t*) −*u*(*t* −1)
-  - (b) *<sup>e</sup>*−(*t*−τ )*u*(*<sup>t</sup>* <sup>−</sup>τ )
-  - (c) *e*−(*t*−τ )*u*(*t*)
-  - (d) *e*−*<sup>t</sup> u*(*t* −τ )
-  - (e) *te*−*<sup>t</sup> u*(*t* −τ )
-  - (f) sin[ω0(*t* −τ )]*u*(*t* −τ )
-  - (g) sin[ω0(*t* −τ )]*u*(*t*)
-  - (h) sin ω0*t u*(*t* −τ )
-  - (i) *t*sin(*t*)*u*(*t*)
-  - (j) (1−*t*) cos(*t* −1)*u*(*t* −1)
-- **4.2-3** Using only Table 4.1 and the time-shifting property, determine the Laplace transform of the signals in Fig. P4.1-3. [*Hint:* See Sec. 1.4 for discussion of expressing such signals analytically.]
-- **4.2-4** Prove the frequency-differentiation property, <sup>−</sup>*tx*(*t*) ⇐⇒ *<sup>d</sup> dsX*(*s*). This property holds for both the unilateral and bilateral Laplace transforms.
-- **4.2-5** Consider the signal *<sup>x</sup>*(*t*) <sup>=</sup> *te*−2(*t*−3) *u*(*t*−2).
-  - (a) Determine the *unilateral* Laplace transform *X*u(*s*) = *L*<sup>u</sup> {*x*(*t*)}.
-
-- (b) Determine the *bilateral* Laplace transform *X*(*s*) = *L*{*x*(*t*)}.
-- **4.2-6** Consider the signals *x*(*t*) and *y*(*t*), as shown in Fig. P4.2-6.
-  - (a) Using the definition, compute *X*(*s*), the bilateral Laplace transform of *x*(*t*).
-  - (b) Using Laplace transform properties, express *Y*(*s*), the bilateral Laplace transform of *y*(*t*), as a function of *X*(*s*), the bilateral Laplace transform of *x*(*t*). Simplify as much as possible without substituting your answer from part (a).
-- **4.2-7** Find the inverse Laplace transforms of the following functions:
-
-(a) 
-$$
-\frac{(2s+5)e^{-2s}}{s^2+5s+6}
-$$
-  
-(b) 
-$$
-\frac{se^{-3s}+2}{s^2+2s+2}
-$$
-
-(c) 
-$$
-\frac{e^{-(s-1)}+3}{2(2s+5)}
-$$
-
-$$
-\begin{array}{c}\n\text{(c)} \quad s^2 - 2s + 5 \\
-\text{(d)} \quad \frac{e^{-s} + e^{-2s} + 1}{s^2 + 3s + 2}\n\end{array}
-$$
-
-- **4.2-8** Using ROC σ > 0, determine the inverse Laplace transform of *<sup>X</sup>*(*s*) <sup>=</sup> *<sup>s</sup>*−<sup>1</sup> *<sup>d</sup> ds e*−2*<sup>s</sup> s* .
-- **4.2-9** The Laplace transform of a causal periodic signal can be determined from the knowledge of the Laplace transform of its first cycle (period).
-  - (a) If the Laplace transform of *x*(*t*) in Fig. P4.2-9a is *X*(*s*), then show that *G*(*s*), the Laplace transform of *g*(*t*) (Fig. P4.2-9b), is
-
-$$
-G(s) = \frac{X(s)}{1 - e^{-sT_0}} \qquad \text{Re}\, s > 0
-$$
-
-- (b) Use this result to find the Laplace transform of the signal *p*(*t*) illustrated in Fig. P4.2-9c.
-- **4.2-10** Starting only with the fact that δ(*t*) ⇐⇒ 1, build pairs 2 through 10b in Table 4.1, using various properties of the Laplace transform.
-
-#### **Figure P4.2-9**
-
-- **4.2-11** (a) Find the Laplace transform of the pulses in Fig. 4.2 by using only the time-different iation property, the time-shifting property, and the fact that δ(*t*) ⇐⇒ 1.
-  - (b) In Ex. 4.9, the Laplace transform of *x*(*t*) is found by finding the Laplace transform of *d*2*x*/*dt*2. Find the Laplace transform of *x*(*t*) in that example by finding the Laplace transform of *dx*/*dt* and using Table 4.1, if necessary.
-- **4.2-12** Determine the inverse unilateral Laplace transform of
-
-$$
-X(s) = \frac{1}{e^{s+3}} \frac{s^2}{(s+1)(s+2)}
-$$
-
-- **4.2-13** Since 13 is such a lucky number, determine the inverse Laplace transform of *X*(*s*) = 1/(*s* +1)<sup>13</sup> given region of convergence σ > −1. [*Hint:* What is the *n*th derivative of 1/(*s* +*a*)?]
-- **4.2-14** It is difficult to compute the Laplace transform *X*(*s*) of signal
-
-$$
-x(t) = \frac{1}{t}u(t)
-$$
-
-by using direct integration. Instead, properties provide a simpler method.
-
-- (a) Use Laplace transform properties to express the Laplace transform of *tx*(*t*) in terms of the unknown quantity *X*(*s*).
-- (b) Use the definition to determine the Laplace transform of *y*(*t*) = *tx*(*t*).
-
-- (c) Solve for *X*(*s*) by using the two pieces from **()**(a) and **()**(b). Simplify your answer.
-- **4.3-1** Use the Laplace transform to solve the following differential equations:
-  - (a) (*D*<sup>2</sup> + 3*D* + 2)*y*(*t*) = *Dx*(*t*) if *y*(0−) = *y*˙(0−) = 0 and *x*(*t*) = *u*(*t*)
-  - (b) (*D*<sup>2</sup> + 4*D* + 4)*y*(*t*) = (*D* + 1)*x*(*t*) if *y*(0−) = 2, *<sup>y</sup>*˙(0−) <sup>=</sup> 1 and *<sup>x</sup>*(*t*) <sup>=</sup> *<sup>e</sup>*−*<sup>t</sup> u*(*t*)
-  - (c) (*D*<sup>2</sup> +6*D*+25)*y*(*t*) = (*D*+2)*x*(*t*) if *y*(0−) = *y*˙(0−) = 1 and *x*(*t*) = 25*u*(*t*)
-- **4.3-2** Solve the differential equations in Prob. 4.3-1 using the Laplace transform. In each case determine the zero-input and zero-state components of the solution.
-- **4.3-3** Consider a causal LTIC system described by the differential equation
-
-$$
-2\dot{y}(t) + 6y(t) = \dot{x}(t) - 4x(t)
-$$
-
-- (a) Using transform-domain techniques, determine the ZIR *y*zir(*t*) if *y*(0−) = −3.
-- (b) Using transform-domain techniques, determine the ZSR *y*zsr(*t*) to the input *x*(*t*) = *e*δ(*t* −π ).
-- **4.3-4** Consider a causal LTIC system described by the differential equation
-
-$$
-\ddot{y}(t) + 3\dot{y}(t) + 2y(t) = 2\dot{x}(t) - x(t)
-$$
-
-- (a) Using transform-domain techniques, determine the ZIR *y*zir(*t*) if *y*˙(0−) = 2 and *y*(0−) = −3.
-- (b) Using transform-domain techniques, determine the ZSR *y*zsr(*t*) to the input *x*(*t*) = *u*(*t*).
-- **4.3-5** Solve the following simultaneous differential equations using the Laplace transform, assuming all initial conditions to be zero and the input *x*(*t*) = *u*(*t*):
-  - (a) (*D*+3)*y*1(*t*)−2*y*2(*t*) = *x*(*t*) −2*y*1(*t*) +(2*D*+4)*y*2(*t*) = 0
-  - (b) (*D*+2)*y*1(*t*)−(*D*+1)*y*2(*t*) = 0 −(*D*+1)*y*1(*t*) +(2*D*+1)*y*2(*t*) = *x*(*t*)
-
-Determine the transfer functions relating outputs *y*1(*t*) and *y*2(*t*) to the input *x*(*t*).
-
-- **4.3-6** Consider a causal LTIC system described by *y*˙(*t*) +2*y*(*t*) = ˙*x*(*t*).
-  - (a) Determine the transfer function *H*(*s*) for this system.
-  - (b) Using your result from part (a), determine the impulse response *h*(*t*) for this system.
-  - (c) Using Laplace transform techniques, determine the output *y*(*t*) if the input is *x*(*t*) = *e*−*<sup>t</sup> <sup>u</sup>*(*t*) and *<sup>y</sup>*(0−) <sup>=</sup> <sup>√</sup> 2.
-- **4.3-7** Repeat Prob. 4.3-6 for a causal LTIC system described by 3*y*(*t*) + ˙*y*(*t*)+ ˙*x*(*t*) = 0.
-- **4.3-8** For the circuit in Fig. P4.3-8, the switch is in the open position for a long time before *t* = 0, when it is closed instantaneously.
-  - (a) Write loop equations (in time domain) for *t* ≥ 0.
-  - (b) Solve for *y*1(*t*) and *y*2(*t*) by taking the Laplace transform of loop equations found in part (a).
-
-**4.3-9** For each of the systems described by the following differential equations, find the system transfer function:
-
-(a) 
-$$
-\frac{d^2y(t)}{dt^2} + 11\frac{dy(t)}{dt} + 24y(t) = 5\frac{dx(t)}{dt} + 3x(t)
-$$
-  
-\n(b) 
-$$
-\frac{d^3y(t)}{dt^3} + 6\frac{d^2y(t)}{dt^2} - 11\frac{dy(t)}{dt} + 6y(t)
-$$
-$$
-= 3\frac{d^2x(t)}{dt^2} + 7\frac{dx(t)}{dt} + 5x(t)
-$$
-  
-\n(c) 
-$$
-\frac{d^4y(t)}{dt^4} + 4\frac{dy(t)}{dt} = 3\frac{dx(t)}{dt} + 2x(t)
-$$
-  
-\n(d) 
-$$
-\frac{d^2y(t)}{dt^2} - y(t) = \frac{dx(t)}{dt} - x(t)
-$$
-
-**4.3-10** For each of the systems specified by the following transfer functions, find the differential equation relating the output *y*(*t*) to the input *x*(*t*), assuming that the systems are controllable and observable:
-
-(a) 
-$$
-H(s) = \frac{s+5}{s^2 + 3s + 8}
-$$
-  
-\n(b)  $H(s) = \frac{s^2 + 3s + 5}{s^3 + 8s^2 + 5s + 7}$   
-\n(c)  $H(s) = \frac{5s^2 + 7s + 2}{s^2 - 2s + 5}$ 
-
-**4.3-11** For a system with transfer function
-
-$$
-H(s) = \frac{2s+3}{s^2+2s+5}
-$$
-
-- (a) Find the (zero-state) response for inputs *x*1(*t*) = 10*u*(*t*) and *x*2(*t*) = *u*(*t* −5).
-- (b) For this system write the differential equation relating the output *y*(*t*) to the input *x*(*t*), assuming that the systems are controllable and observable.
-- **4.3-12** For a system with transfer function
-
-$$
-H(s) = \frac{s}{s^2 + 9}
-$$
-
-- (a) Find the (zero-state) response if the input *<sup>x</sup>*(*t*) <sup>=</sup> (1−*e*−*<sup>t</sup>* )*u*(*t*)
-- (b) For this system write the differential equation relating the output *y*(*t*) to the input *x*(*t*), assuming that the systems are controllable and observable.
-- **4.3-13** Consider a system with transfer function
-
-$$
-H(s) = \frac{s+5}{s^2 + 5s + 6}
-$$
-
-Find the (zero-state) response for the following inputs:
-
-- (a) *<sup>x</sup>*a(*t*) <sup>=</sup> *<sup>e</sup>*−3*<sup>t</sup> u*(*t*)
-- (b) *<sup>x</sup>*b(*t*) <sup>=</sup> *<sup>e</sup>*−4*<sup>t</sup> u*(*t*)
-- (c) *<sup>x</sup>*c(*t*) <sup>=</sup> *<sup>e</sup>*−4(*t*−5) *u*(*t* −5)
-- (d) *<sup>x</sup>*d(*t*) <sup>=</sup> *<sup>e</sup>*−4(*t*−5) *u*(*t*)
-- (e) *<sup>x</sup>*e(*t*) <sup>=</sup> *<sup>e</sup>*−4*<sup>t</sup> u*(*t* −5)
-
-Assuming that the system *H*(*s*) is controllable and observable,
-
-- (f) write the differential equation relating the output *y*(*t*) to the input *x*(*t*).
-- **4.3-14** An LTI system has a step response given by *<sup>s</sup>*(*t*) <sup>=</sup> *<sup>e</sup>*−*<sup>t</sup> <sup>u</sup>*(*t*) <sup>−</sup> *<sup>e</sup>*−2*<sup>t</sup> u*(*t*). Determine the output of this system *y*(*t*) given an input *x*(*t*) = δ(*t* − π )−cos( <sup>√</sup>3)*u*(*t*).
-- **4.3-15** For an LTIC system with zero initial conditions (system initially in zero state), if an input *x*(*t*) produces an output *y*(*t*), then using the Laplace transform, show the following:
-  - (a) The input *dx*/*dt* produces an output *dy*/*dt*.
-  - (b) The input \$ *<sup>t</sup>* <sup>0</sup> *x*(τ )*d*τ produces an output \$ *t* <sup>0</sup> *y*(τ )*d*τ . Hence, show that the unit step response of a system is an integral of the impulse response; that is, \$ *<sup>t</sup>* <sup>0</sup> *h*(τ )*d*τ .
-- **4.3-16** Discuss asymptotic and BIBO stabilities for the systems described by the following transfer functions, assuming that the systems are controllable and observable:
-
-(a) 
-$$
-\frac{(s+5)}{s^2+3s+2}
-$$
-
-(b) 
-$$
-\frac{s+5}{s^2(s+2)}
-$$
-
-$$
-(c) \frac{s(s+2)}{s+5}
-$$
-
-(d) 
-$$
-\frac{s+5}{s(s+2)}
-$$
-
-(e) 
-$$
-\frac{s+5}{s^2-2s+3}
-$$
-
-- **4.3-17** Repeat Prob. 4.3-16 for systems described by the following differential equations. Systems may be uncontrollable and/or unobservable.
-  - (a) (*D*<sup>2</sup> +3*D*+2)*y*(*t*) = (*D*+3)*x*(*t*)
-  - (b) (*D*<sup>2</sup> +3*D*+2)*y*(*t*) = (*D*+1)*x*(*t*)
-  - (c) (*D*<sup>2</sup> +*D*−2)*y*(*t*) = (*D*−1)*x*(*t*)
-  - (d) (*D*<sup>2</sup> −3*D*+2)*y*(*t*) = (*D*−1)*x*(*t*)
-- **4.4-1** The circuit shown in Fig. P4.4-1 has system function given by *<sup>H</sup>*(*s*) <sup>=</sup> <sup>1</sup> <sup>1</sup>+*RCs*. Let *<sup>R</sup>* <sup>=</sup> 2 and *C* = 3 and use Laplace transform techniques to solve the following.
-  - (a) Find the output *y*(*t*) given an initial capacitor voltage of *y*(0−) = 3 and an input *x*(*t*) = *u*(*t*).
-  - (b) Given an input *x*(*t*) = *u*(*t* − 3), determine the initial capacitor voltage *y*(0−) so that the output *y*(*t*) is 1 volt at *t* = 6 seconds.
-
-- **4.4-2** Consider the circuit shown in Fig. P4.4-2. Use Laplace transform techniques to solve the following.
-  - (a) Determine the standard-form, constantcoefficient differential equation description of this circuit.
-  - (b) Letting *R* = *C* = 1, determine the total response *<sup>y</sup>*(*t*) to input *<sup>x</sup>*(*t*) <sup>=</sup> <sup>3</sup>*e*−*<sup>t</sup> u*(*t*) and initial capacitor voltage of *vC*(0−) = 5.
-
-**4.4-3** Find the zero-state response *y*(*t*) of the network in Fig. P4.4-3 if the input voltage *<sup>x</sup>*(*t*) <sup>=</sup> *te*−*<sup>t</sup> u*(*t*). Find the transfer function relating the output *Y*(*s*) to the input *X*(*s*). From the transfer function, write the differential equation relating *y*(*t*) to *x*(*t*).
-
-### **Figure P4.4-3**
-
-**4.4-4** The switch in the circuit of Fig. P4.4-4 is closed for a long time and then opened instantaneously at *t* = 0. Find and sketch the current *y*(*t*).
-
-#### **Figure P4.4-4**
-
-**4.4-5** Find the current *y*(*t*) for the parallel resonant circuit in Fig. P4.4-5 if the input is: (a) *x*(*t*) = *A*cos ω0*t u*(*t*) (b) *x*(*t*) = *A*sin ω0*t u*(*t*) Assume all initial conditions to be zero and, in both cases, ω<sup>2</sup> <sup>0</sup> = 1/*LC*.
-
-**Figure P4.4-5**
-
-**Figure P4.4-6**
-
-- **4.4-6** Find the loop currents *y*1(*t*) and *y*2(*t*) for *t* ≥ 0 in the circuit of Fig. P4.4-6a for the input *x*(*t*) in Fig. P4.4-6b.
-- **4.4-7** For the network in Fig. P4.4-7, the switch is in a closed position for a long time before *t* = 0, when it is opened instantaneously. Find *y*1(*t*) and *vs*(*t*) for *t* ≥ 0.
-
-#### **Figure P4.4-7**
-
-- **4.4-8** Find the output voltage *v*0(*t*) for *t* ≥ 0 for the circuit in Fig. P4.4-8, if the input *x*(*t*) = 100*u*(*t*). The system is in the zero state initially.
-- **4.4-9** Find the output voltage *y*(*t*) for the network in Fig. P4.4-9 for the initial conditions *iL*(0) = 1 A and *vC*(0) = 3 V.
-- **4.4-10** For the network in Fig. P4.4-10, the switch is in position *a* for a long time and then is moved to position *b* instantaneously at *t* = 0. Determine the current *y*(*t*) for *t* > 0.
-
-**4.4-11** Consider the circuit of Fig. P4.4-11.
-
-- (a) Using transform-domain techniques, determine the system's standard-form transfer function *H*(*s*).
-- (b) Using transform-domain techniques and letting *R* = *L* = 1, determine the circuit's zero-state response *y*zsr(*t*) to the input *x*(*t*) = *e*−2*<sup>t</sup> u*(*t* −1).
-
-**Figure P4.4-10**
-
-(c) Using transform-domain techniques and letting *R* = 2*L* = 1, determine the circuit's zero-state response *y*zsr(*t*) to the input *x*(*t*) = *e*−2*<sup>t</sup> u*(*t* −1).
-
-**4.4-12** Show that the transfer function that relates the output voltage *y*(*t*) to the input voltage *x*(*t*) for the op-amp circuit in Fig. P4.4-12a is given by
-
-$$
-H(s) = \frac{Ka}{s+a} \quad \text{where}
-$$
-  
-$$
-K = 1 + \frac{R_b}{R_a} \quad \text{and} \quad a = \frac{1}{RC}
-$$
-
-and that the transfer function for the circuit in Fig. P4.4-12b is given by
-
-$$
-H(s) = \frac{Ks}{s+a}
-$$
-
-**4.4-13** For the second-order op-amp circuit in Fig. P4.4-13, show that the transfer function *H*(*s*) relating the output voltage *y*(*t*) to the input
-
-### **Figure P4.4-14**
-
-voltage *x*(*t*) is given by
-
-$$
-H(s) = \frac{-s}{s^2 + 8s + 12}
-$$
-
-- **4.4-14** Consider the op-amp circuit of Fig. P4.4-14.
-  - (a) Determine the standard-form transfer function *H*(*s*) of this system.
-  - (b) Determine the standard-form constant coefficient linear differential equation description of this circuit.
-
-- (c) Using transform-domain techniques, determine the circuit's zero-state response *y*zsr(*t*) to the input *<sup>x</sup>*(*t*) <sup>=</sup> *<sup>e</sup>*2*<sup>t</sup> u*(*t* +1).
-- (d) Using transform-domain techniques, determine the circuit's zero-input response *y*zir(*t*) if the *t* = 0<sup>−</sup> capacitor voltage (first op-amp output voltage) is 3 volts.
-- **4.4-15** We desire the op-amp circuit of Fig. P4.4-15 to behave as *y*˙(*t*)−1.5*y*(*t*) = −3*x*˙(*t*)+0.75*x*(*t*).
-  - (a) Determine resistors *R*1, *R*2, and *R*<sup>3</sup> so that the circuit's input–output behavior follows
-
-the desired differential equation of *y*˙(*t*) − 1.5*y*(*t*) = −3*x*˙(*t*)+0.75*x*(*t*).
-
-- (b) Using transform-domain techniques, determine the circuit's zero-input response *y*zir(*t*) if the *t* = 0 capacitor voltage (first op-amp output voltage) is 2 volts.
-- (c) Using transform-domain techniques, determine the impulse response *h*(*t*) of this circuit.
-- (d) Determine the circuit's zero-state response *y*zsr(*t*) to the input *x*(*t*) = *u*(*t* −2).
-- **4.4-16** We desire the op-amp circuit of Fig. P4.4-16 to behave as \$ \$ *<sup>y</sup>*(*t*)<sup>+</sup> <sup>2</sup> 5 \$ *<sup>y</sup>*(*t*)<sup>+</sup> <sup>1</sup> <sup>5</sup> *<sup>y</sup>*(*t*) <sup>=</sup> \$ \$ *<sup>x</sup>*(*t*)<sup>−</sup> \$ *x*(*t*)
-  - (a) Determine the resistors *R*1, *R*2, and *R*<sup>3</sup> to produce the desired behavior.
-  - (b) Using transform-domain techniques, determine the circuit's zero-input response *y*zir(*t*) if the *t* = 0 capacitor voltages (first two op-amp outputs) are each 1 volt.
-
-**4.4-17** (a) Using the initial and final value theorems, find the initial and final value of the zero-state response of a system with the transfer function
-
-$$
-H(s) = \frac{6s^2 + 3s + 10}{2s^2 + 6s + 5}
-$$
-
-and input *x*(*t*) = *u*(*t*).
-
-(b) Repeat part (a) for the input *<sup>x</sup>*(*t*) <sup>=</sup> *<sup>e</sup>*−*<sup>t</sup> u*(*t*).
-
-(c) Find y(0<sup>+</sup>) and y(
-$$
-\infty
-$$
-) if  $Y(s) = \frac{s^2 + 5s + 6}{s^2 + 3s + 2}$ 
-
-.
-
-- (d) Find *y*(0+) and *y*(∞) if *Y*(*s*) = *s*<sup>3</sup> +4*s*<sup>2</sup> +10*s*+7 *<sup>s</sup>*<sup>2</sup> <sup>+</sup>2*<sup>s</sup>* <sup>+</sup><sup>3</sup> .
-- **4.5-1** Consider two LTIC systems. The first has transfer function *<sup>H</sup>*1(*s*) <sup>=</sup> <sup>2</sup>*<sup>s</sup> <sup>s</sup>*+<sup>1</sup> , and the second has transfer function *<sup>H</sup>*2(*s*) <sup>=</sup> <sup>1</sup> *se*3(*s*−1) .
-
-**Figure P4.5-2**
-
-- (a) Determine the overall impulse response *h*s(*t*) if the two systems are connected in series.
-- (b) Determine the overall impulse response *h*p(*t*) if the two systems are connected in parallel.
-- **4.5-2** Figure P4.5-2a shows two resistive ladder segments. The transfer function of each segment (ratio of output to input voltage) is 1/2. Figure P4.5-2b shows these two segments connected in cascade.
-  - (a) Is the transfer function (ratio of output to input voltage) of this cascaded network (1/2)(1/2) = 1/4?
-  - (b) If your answer is affirmative, verify the answer by direct computation of the transfer function. Does this computation confirm the earlier value 1/4? If not, why?
-  - (c) Repeat the problem with *R*<sup>3</sup> = *R*<sup>4</sup> = 20 k. Does this result suggest the answer to the problem in part (b)?
-- **4.5-3** In communication channels, transmitted signal is propagated simultaneously by several paths of varying lengths. This causes the signal to reach the destination with varying time delays and varying gains. Such a system generally distorts the received signal. For error-free communication, it is necessary to undo this distortion as
-
-For simplicity, let us assume that a signal is propagated by two paths whose time delays differ by τ seconds. The channel over the intended path has a delay of *T* seconds and unity gain. The signal over the unintended path has a delay of *T* + τ seconds and gain *a*. Such a channel can be modeled, as shown in Fig. P4.5-3. Find the inverse system transfer function to correct the delay distortion and show that the inverse system can be realized by a feedback system. The inverse system should be causal to be realizable. [*Hint:* We want to correct only the distortion caused by the relative delay τ seconds. For distortionless transmission, the signal may be delayed. What is important is to maintain the shape of *x*(*t*). Thus, a received signal of the form *c x*(*t* −*T*) is considered to be distortionless.]
-
-**4.5-4** Discuss BIBO stability of the feedback systems depicted in Fig. P4.5-4. For the system in
-
-**Figure P4.5-4**
-
-Fig. P4.5-4b, consider three cases: **(a)** *K* = 10, **(b)** *K* = 50, and **(c)** *K* = 48.
-
-**4.6-1** Realize
-
-$$
-H(s) = \frac{s(s+2)}{(s+1)(s+3)(s+4)}
-$$
-
-by canonic direct, series, and parallel forms.
-
-- **4.6-2** Realize the transfer function in Prob. 4.6-1 by using the transposed form of the realizations found in Prob. 4.6-1.
-- **4.6-3** Repeat Prob. 4.6-1 for (a) *<sup>H</sup>*(*s*) <sup>=</sup> <sup>3</sup>*s*(*<sup>s</sup>* <sup>+</sup>2) (*s* +1)(*s*<sup>2</sup> +2*s*+2) (b) *<sup>H</sup>*(*s*) <sup>=</sup> <sup>2</sup>*<sup>s</sup>* <sup>−</sup><sup>4</sup> (*s* +2)(*s*<sup>2</sup> +4)
-- **4.6-4** Realize the transfer functions in Prob. 4.6-3 by using the transposed form of the realizations found in Prob. 4.6-3.
-- **4.6-5** Repeat Prob. 4.6-1 for
-
-$$
-H(s) = \frac{2s+3}{5s(s+2)^2(s+3)}
-$$
-
-- **4.6-6** Realize the transfer function in Prob. 4.6-5 by using the transposed form of the realizations found in Prob. 4.6-5.
-- **4.6-7** Repeat Prob. 4.6-1 for
-
-$$
-H(s) = \frac{s(s+1)(s+2)}{(s+5)(s+6)(s+8)}
-$$
-
-- **4.6-8** Realize the transfer function in Prob. 4.6-7 by using the transposed form of the realizations found in Prob. 4.6-7.
-- **4.6-9** Repeat Prob. 4.6-1 for
-
-$$
-H(s) = \frac{s^3}{(s+1)^2(s+2)(s+3)}
-$$
-
-- **4.6-10** Realize the transfer function in Prob. 4.6-9 by using the transposed form of the realizations found in Prob. 4.6-9.
-- **4.6-11** Repeat Prob. 4.6-1 for
-
-$$
-H(s) = \frac{s^3}{(s+1)(s^2+4s+13)}
-$$
-
-**4.6-12** Realize the transfer function in Prob. 4.6-11 by using the transposed form of the realizations found in Prob. 4.6-11.
-
-- **4.6-13** Draw a TDFII block realization of a causal LTIC system with transfer function *<sup>H</sup>*(*s*) <sup>=</sup> (*s*−2*j*)(*s*+2*j*) (*s*−*j*)(*s*+*j*)(*s*+2) . Give two reasons why TDFII tends to be a good structure.
-- **4.6-14** Consider a causal LTIC system with transfer function *<sup>H</sup>*(*s*) <sup>=</sup> (*s*−2*j*)(*s*+2*j*)(*s*−3*j*)(*s*+3*j*) <sup>9</sup>(*s*+1)(*s*+2)(*s*+1−*j*)(*s*+1+*j*) .
-  - (a) Realize *H*(*s*) using a single fourth-order real TDFII structure. Is this block realization unique? Explain.
-  - (b) Realize *H*(*s*) using a cascade of secondorder real DFII structures. Is this block realization unique? Explain.
-  - (c) Realize *H*(*s*) using a parallel connection of second-order real DFI structures. Is this block realization unique? Explain.
-- **4.6-15** In this problem we show how a pair of complex conjugate poles may be realized by using a cascade of two first-order transfer functions and feedback. Show that the transfer functions of the block diagrams in Figs. P4.6-15a and P4.6-15b are: (a)
-
-$$
-H_a(s) = \frac{1}{(s+a)^2 + b^2}
-$$
-  
-= 
-$$
-\frac{1}{s^2 + 2as + (a^2 + b^2)}
-$$
-
-(b)
-
-$$
-Hb(s) = \frac{s+a}{(s+a)^2 + b^2}
-$$
-  
-= 
-$$
-\frac{s+a}{s^2 + 2as + (a^2 + b^2)}
-$$
-
-Hence, show that the transfer function of the block diagram in Fig. P4.6-15c is (c)
-
-$$
-H_c(s) = \frac{As + B}{(s+a)^2 + b^2}
-$$
-  
-= 
-$$
-\frac{As + B}{s^2 + 2as + (a^2 + b^2)}
-$$
-
-**4.6-16** Show op-amp realizations of the following transfer functions:
-
-(a) 
-$$
-\frac{-10}{s+5}
-$$
-  
-\n(b)  $\frac{10}{s+5}$   
-\n(c)  $\frac{s+2}{s+5}$ 
-
-(c)
-
-*b*2
-
-**4.6-17** Show two different op-amp circuit realizations of the transfer function
-
-$$
-H(s) = \frac{s+2}{s+5} = 1 - \frac{3}{s+5}
-$$
-
-**4.6-18** Show an op-amp canonic direct realization of the transfer function
-
-$$
-H(s) = \frac{3s + 7}{s^2 + 4s + 10}
-$$
-
-**4.6-19** Show an op-amp canonic direct realization of the transfer function
-
-$$
-H(s) = \frac{s^2 + 5s + 2}{s^2 + 4s + 13}
-$$
-
-**4.6-20** Consider a system described by a constantcoefficient linear differential equation as *<sup>d</sup> dt y*(*t*)
-
-<sup>+</sup> <sup>2</sup>*y*(*t*) <sup>=</sup> *<sup>x</sup>*(*t*) <sup>−</sup> <sup>3</sup> *<sup>d</sup> dt x*(*t*). Draw an op-amp realization of this system if resistors and inductors are available but not capacitors. Would using inductors rather than capacitors in this circuit pose any problem? Explain.
-
-- **4.7-1** Feedback can be used to increase (or decrease) the system bandwidth. Consider the system in Fig. P4.7-1a with transfer function *G*(*s*) = ω*c*/(*s* +ω*c*).
-  - (a) Show that the 3 dB bandwidth of this system is ω*<sup>c</sup>* and the dc gain is unity; that is, |*H*(*j*0)| = 1.
-  - (b) To increase the bandwidth of this system, we use negative feedback with *H*(*s*) = 9, as depicted in Fig. P4.7-1b. Show that the 3 dB bandwidth of this system is 10ω*c*. What is the dc gain?
-
-- (c) To decrease the bandwidth of this system, we use positive feedback with *H*(*s*) = −0.9, as illustrated in Fig. P4.7-1c. Show that the 3 dB bandwidth of this system is ω*c*/10. What is the dc gain?
-- (d) The system gain at dc times its 3 dB bandwidth is the *gain-bandwidth product* of a system. Show that this product is the same for all the three systems in Fig. P4.7-1. This result shows that if we increase the bandwidth, the gain decreases and vice versa.
-- **4.8-1** Suppose an engineer builds a controllable, observable LTIC system with transfer function *<sup>H</sup>*(*s*) <sup>=</sup> *<sup>s</sup>*2+<sup>4</sup> <sup>2</sup>*s*2+4*s*+<sup>4</sup> .
-  - (a) By direct calculation, compute the magnitude response at frequencies ω = 0, 1, 2, 3, 5, 10, and ∞. Use these calculations to roughly sketch the magnitude response over 0 ≤ ω ≤ 10.
-  - (b) To test the system, the engineer connects a signal generator to the system in hopes to measure the magnitude response using a standard oscilloscope. What type of signal should the engineer input into the system to make the measurements? How should the engineer make the measurements? Provide sufficient detail to fully justify your answers.
-  - (c) Suppose the engineer accidentally constructs the system *<sup>H</sup>*−1(*s*)<sup>=</sup> <sup>1</sup> *<sup>H</sup>*(*s*) <sup>=</sup> <sup>2</sup>*s*2+4*s*+<sup>4</sup> *<sup>s</sup>*2+<sup>4</sup> . What impact will this mistake have on his tests?
-- **4.8-2** For an LTIC system described by the transfer function
-
-$$
-H(s) = \frac{s+2}{s^2 + 5s + 4}
-$$
-
-find the response to the following everlasting sinusoidal inputs:
-
-- (a) 5 cos(2*t* +30◦)
-- (b) 10 sin(2*t* +45◦)
-- (c) 10 cos(3*t* +40◦)
-
-Observe that these are everlasting sinusoids.
-
-**4.8-3** For an LTIC system described by the transfer function
-
-$$
-H(s) = \frac{s+3}{(s+2)^2}
-$$
-
-find the steady-state system response to the following inputs:
-
-- (a) 10*u*(*t*)
-- (b) cos(2*t* +60◦)*u*(*t*)
-- (c) sin(3*t* −45◦)*u*(*t*)
-- (d) *ej*3*<sup>t</sup> u*(*t*)
-- **4.8-4** For an allpass filter specified by the transfer function
-
-$$
-H(s) = \frac{-(s-10)}{s+10}
-$$
-
-find the system response to the following (everlasting) inputs:
-
-- (a) *ej*ω*<sup>t</sup>*
-- (b) cos(ω*t* +θ )
-- (c) cos *t*
-- (d) sin 2*t*
-- (e) cos 10*t*
-- (f) cos 100*t*
-
-Comment on the filter response.
-
-- **4.8-5** The pole-zero plot of a second-order system *H*(*s*) is shown in Fig. P4.8-5. The dc response of this system is minus 1, *H*(*j*0) = −1.
-  - (a) Letting *H*(*s*) = *k*(*s*<sup>2</sup> +*b*1*s*+*b*2)/(*s*<sup>2</sup> +*a*1*s*+ *a*2), determine the constants *k*, *b*1, *b*2, *a*1, and *a*2.
-  - (b) What is the output *y*(*t*) of this system in response to the input *x*(*t*) = 4 + cos(*t*/2 + π/3)?
-
-### **Figure P4.8-5**
-
-- **4.8-6** Consider a CT system described by (*D*+1)(*D*+ 2){*y*(*t*)} = *x*(*t* − 1). Notice that this differential equation is in terms of *x*(*t* −1), not *x*(*t*)!
-  - (a) Determine the output *y*(*t*) given input *x*(*t*) = 1.
-
-- (b) Determine the output *y*(*t*) given input *x*(*t*) = cos(*t*).
-- **4.8-7** An LTIC system has transfer function *<sup>H</sup>*(*s*) <sup>=</sup> <sup>4</sup>*<sup>s</sup> <sup>s</sup>*2+2*s*+<sup>37</sup> <sup>=</sup> <sup>4</sup>*<sup>s</sup>* (*s*+1+6*j*)(*s*+1−6*j*). Determine the steady-state output in response to input *<sup>x</sup>*(*t*) <sup>=</sup> <sup>1</sup> <sup>3</sup> *ej*(6*t*+π/3) *u*(6*t* +π/3).
-- **4.9-1** Suppose a real first-order lowpass system *H*(*s*) has unity gain in the passband, one finite pole at *s* = −2, and one finite zero at an unspecified location.
-  - (a) Determine the location of the system zero so that the filter achieves 40 dB of stopband attenuation. Sketch the corresponding straight-line Bode approximation of the system magnitude response.
-  - (b) Determine the location of the system zero so that the filter achieves 30 dB of stopband attenuation. Sketch the corresponding straight-line Bode approximation of the system magnitude response.
-- **4.9-2** Repeat Prob. 4.9-1 for a highpass rather than a lowpass system.
-- **4.9-3** Repeat Prob. 4.9-1 for a second-order system that has a pair of repeated poles and a pair of repeated zeros.
-- **4.9-4** Sketch Bode plots for the following transfer functions:
-
-(a) 
-$$
-\frac{s(s+100)}{(s+2)(s+20)}
-$$
-  
-(b) 
-$$
-\frac{(s+10)(s+20)}{s^2(s+100)}
-$$
-
-(c) 
-$$
-\frac{(s+10)(s+200)}{(s+20)^2(s+1000)}
-$$
-
-$$
-4.9-5 \quad \text{Repeat Prob. } 4.9-4 \text{ for}
-$$
-
-(a) 
-$$
-\frac{s^2}{(s+1)(s^2+4s+16)}
-$$
-  
-\n(b) 
-$$
-\frac{s}{(s+1)(s^2+14.14s+100)}
-$$
-  
-\n(c) 
-$$
-\frac{(s+10)}{s(s^2+14.14s+100)}
-$$
-
-- **4.9-6** Using the lowest order possible, determine a system function *H*(*s*) with real-valued roots that matches the frequency response in Fig. P4.9-6. Verify your answer with MATLAB.
-- **4.9-7** A graduate student recently implemented an analog phase lock loop (PLL) as part of his thesis. His PLL consists of four basic components: a phase/frequency detector, a charge pump, a loop filter, and a voltage-controlled oscillator. This problem considers only the loop filter, which is shown in Fig. P4.9-7a. The loop filter input is the current *x*(*t*), and the output is the voltage *y*(*t*).
-  - (a) Derive the loop filter's transfer function *H*(*s*). Express *H*(*s*) in standard form.
-  - (b) Figure P4.9-7b provides four possible frequency response plots, labeled A through D. Each log-log plot is drawn to the same scale, and line slopes are either 20 dB/decade, 0 dB/decade, or −20 dB/decade. Clearly identify which plot(s), if any, could represent the loop filter.
-  - (c) Holding the other components constant, what is the general effect of increasing the resistance *R* on the magnitude response for low-frequency inputs?
-
-**Figure P4.9-6**
-
-(d) Holding the other components constant, what is the general effect of increasing the resistance *R* on the magnitude response for high-frequency inputs?
-
-**Figure P4.9-7**
-
-- **4.10-1** A causal LTIC system *<sup>H</sup>*(*s*) <sup>=</sup> <sup>2</sup>(*s*−4*j*)(*s*+4*j*) (*s*+1+2*j*)(*s*+1−2*j*) has input *x*(*t*) = −1 + 2 cos(2*t*) − 3 sin(4*t* + π/3) + 4 cos(10*t*). Below, perform accurate calculations at ω = 0, ±2, ±4, and ±10.
-  - (a) Using the graphical method of Sec. 4.10-1, accurately sketch the magnitude response |*H*(*j*ω)| over −10 ≤ ω ≤ 10.
-  - (b) Using the graphical method of Sec. 4.10-1, accurately sketch the phase response *H*(*j*ω) over −10 ≤ ω ≤ 10.
-  - (c) Approximate the system output *y*(*t*) in response to the input *x*(*t*).
-- **4.10-2** The pole-zero plot of a second-order system *H*(*s*) is shown in Fig. P4.10-2. The dc response of this system is minus 2, *H*(*j*0) = −2.
-  - (a) Letting *<sup>H</sup>*(*s*) <sup>=</sup> *<sup>k</sup> <sup>s</sup>*2+*b*1*s*+*b*<sup>2</sup> *<sup>s</sup>*2+*a*1*s*+*a*<sup>2</sup> , determine the constants *k*, *b*1, *b*2, *a*1, and *a*2.
-
-- (b) Using the graphical method of Sec. 4.10-1, hand-sketch the magnitude response |*H*(*j*ω)| over −10 ≤ ω ≤ 10. Verify your sketch with MATLAB.
-- (c) Using the graphical method of Sec. 4.10-1, hand-sketch the phase response *H*(*j*ω) over −10 ≤ ω ≤ 10. Verify your sketch with MATLAB.
-- (d) What is the output *y*(*t*) in response to input *x*(*t*) = −3+cos(3*t*+π/3)−sin(4*t*−π/8)?
-
-**Figure P4.10-2**
-
-**4.10-3** Using the graphical method of Sec. 4.10-1, draw a rough sketch of the amplitude and phase responses of an LTIC system described by the transfer function
-
-$$
-H(s) = \frac{s^2 - 2s + 50}{s^2 + 2s + 50}
-$$
-  
-= 
-$$
-\frac{(s - 1 - j7)(s - 1 + j7)}{(s + 1 - j7)(s + 1 + j7)}
-$$
-
-What kind of filter is this?
-
-**4.10-4** Using the graphical method of Sec. 4.10-1, draw a rough sketch of the amplitude and phase responses of LTIC systems whose pole-zero plots are shown in Fig. P4.10-4.
-
-**Figure P4.10-4**
-
-- **4.10-5** A causal LTIC system *<sup>H</sup>*(*s*) <sup>=</sup> (*s*−3*j*)(*s*+3*j*) 3(*s*+2+*j*)(*s*+2−*j*) has input *x*(*t*) = cos(*t*) + sin(3*t* + π/3) + cos(100*t*).
-  - (a) Using the graphical method of Sec. 4.10-1, sketch the magnitude response |*H*(*j*ω)| over −10 ≤ ω ≤ 10.
-  - (b) Determine the system output *y*(*t*) in response to the input *x*(*t*).
-  - (c) Suppose we create a second causal system with transfer function *H*2(*s*) = *H*(−*s*). Sketch this system's pole/zero plot. What is the response *y*2(*t*) of system *H*2(*s*) to input *x*(*t*)?
-- **4.10-6** Design a second-order bandpass filter with center frequency ω = 10. The gain should be zero at ω = 0 and at ω = ∞. Select poles at −*a* ± *j*10. Leave your answer in terms of *a*. Explain the influence of *a* on the frequency response.
-- **4.10-7** The LTIC system described by *H*(*s*) = (*s*−1)/(*s* +1) has unity magnitude response |*H*(*j*ω)| = 1. Positive Pat claims that the output *y*(*t*) of this system is equal the input *x*(*t*), since the system is allpass. Cynical Cynthia doesn't think so. "This is *signals and systems* class," she complains. "It *has* to be more complicated!" Who is correct, Pat or Cynthia? Justify your answer.
-- **4.10-8** Two students, Amy and Jeff, disagree about an analog system function given by *H*1(*s*) = *s*. Sensible Jeff claims the system has a zero at *s* = 0. Rebellious Amy, however, notes that the system function can be rewritten as *H*1(*s*) = 1/*s*−<sup>1</sup> and claims that this implies a system pole
-
-at *s* = ∞. Who is correct? Why? What are the poles and zeros of the system *H*2(*s*) = 1/*s*?
-
-- **4.10-9** A rational transfer function *H*(*s*) is often used to represent an analog filter. Why must *H*(*s*) be strictly proper for lowpass and bandpass filters? Why must *H*(*s*) be proper for highpass and bandstop filters?
-- **4.10-10** For a given filter order *N*, why is the stopband attenuation rate of an all-pole lowpass filter better than filters with finite zeros?
-- **4.10-11** Is it possible, with real coefficients ([*k*,*b*1,*b*2,*a*1,*a*2] ∈ *R*), for a system
-
-$$
-H(s) = k \frac{s^2 + b_1 s + b_2}{s^2 + a_1 s + a_2}
-$$
-
-to function as a lowpass filter? Explain your answer.
-
-- **4.10-12** Nick recently built a simple second-order Butterworth lowpass filter for his home stereo. Although the system performs fairly well, Nick is an overachiever and hopes to improve the system performance. Unfortunately, Nick is lazy and doesn't want to design another filter. Thinking "Twice the filtering gives twice the performance," he suggests filtering the audio signal not once but twice with a cascade of two identical filters. His overworked, underpaid signals professor is skeptical and states, "If you are using *identical* filters, it makes no difference whether you filter once or twice!" Who is correct? Why?
-- **4.10-13** An LTIC system impulse response is given by *h*(*t*) = *u*(*t*)−*u*(*t* −1).
-  - (a) Determine the transfer function *H*(*s*). Using *H*(*s*), determine and plot the magnitude response |*H*(*j*ω)|. Which type of filter most accurately describes the behavior of this system: lowpass, highpass, bandpass, or bandstop?
-  - (b) What are the poles and zeros of *H*(*s*)? Explain your answer.
-  - (c) Can you determine the impulse response of the inverse system? If so, provide it. If not, suggest a method that could be used to approximate the impulse response of the inverse system.
-- **4.10-14** An ideal lowpass filter *H*LP(*s*) has magnitude response that is unity for low frequencies
-
-and zero for high frequencies. An ideal highpass filter *H*HP(*s*) has an opposite magnitude response: zero for low frequencies and unity for high frequencies. A student suggests a possible lowpass-to-highpass filter transformation: *H*HP(*s*) = 1 − *H*LP(*s*). In general, will this transformation work? Explain your answer.
-
-- **4.10-15** An LTIC system has a rational transfer function *H*(*s*). When appropriate, assume that all initial conditions are zero.
-  - (a) Is is possible for this system to output *y*(*t*) = sin(100π*t*)*u*(*t*) in response to an input *x*(*t*) = cos(100π*t*)*u*(*t*)? Explain.
-  - (b) Is is possible for this system to output *y*(*t*) = sin(100π*t*)*u*(*t*) in response to an input *x*(*t*) = sin(50π*t*)*u*(*t*)? Explain.
-  - (c) Is is possible for this system to output *y*(*t*) = sin(100π*t*) in response to an input *x*(*t*) = cos(100π*t*)? Explain.
-  - (d) Is is possible for this system to output *y*(*t*) = sin(100π*t*) in response to an input *x*(*t*) = sin(50π*t*)? Explain.
-- **4.11-1** Find the ROC, if it exists, of the (bilateral) Laplace transform of the following signals:
-  - (a) *etu*(*t*) (b) *e*−*tu*(*t*)
-
-(c) 
-$$
-\frac{1}{1+t^2}
-$$
-
-(d) 
-$$
-\frac{1}{1+e^t}
-$$
-  
-(e)  $e^{-kt^2}$ 
-
-- **4.11-2** Using the definition and direct integration, find the (bilateral) Laplace transform and the corresponding region of convergence for the following signals. If the Laplace transform does not exist, carefully explain why.
-  - (a) *<sup>x</sup>*a(*t*) <sup>=</sup> *<sup>e</sup>*(−1−*j*)*<sup>t</sup> u*(1−*t*)
-  - (b) *x*b(*t*) = *j* (*t*+1) *u*(−*t* −1)
-  - (c) *<sup>x</sup>*c(*t*) <sup>=</sup> *ej*π/3*u*(2−*t*) <sup>+</sup>*j*δ(*<sup>t</sup>* <sup>−</sup>5)
-  - (d) *x*d(*t*) = 1+1 = 2
-  - (e) *<sup>x</sup>*e(*t*) <sup>=</sup> <sup>3</sup>*u*(−*t*) <sup>+</sup>*e*−2*<sup>t</sup>* [*u*(*t*)−*u*(*t* −10)]
-  - (f) *<sup>x</sup>*f(*t*) <sup>=</sup> *et*−2*u*(1−*t*) <sup>+</sup>*e*−2*<sup>t</sup> u*(*t* +1)
-- **4.11-3** Determine the bilateral Laplace transform *X*(*s*) of the signal
-
-$$
-x(t) = \left[e^t u(-t)\right] * \left[t\cos(2t)u(t)\right]
-$$
-
-**4.11-4** A signal has bilateral Laplace transform *<sup>X</sup>*(*s*) <sup>=</sup> (*s*+1) (*s*−2)(*s*−3) but unknown region of convergence. What ROC results in the smallest maximum amplitude of *x*(*t*)? Justify your answer.
-
-- **4.11-5** Find the (bilateral) Laplace transform and the corresponding region of convergence for the following signals:
-  - (a) *e*−|*t*<sup>|</sup>
-  - (b) *e*−|*t*<sup>|</sup> cos *t*
-
-(c) 
-$$
-e^t u(t) + e^{2t} u(-t)
-$$
-  
-(d)  $e^{-tu(t)}$ 
-
-$$
-\begin{array}{cc}\n(e) & e^{-tu(-t)} \\
-(e) & e^{-tu(-t)}\n\end{array}
-$$
-
-- (f) cosω0*t u*(*t*) <sup>+</sup>*et <sup>u</sup>*(−*t*)
-- **4.11-6** Find the inverse (bilateral) Laplace transforms of the following functions:
-
-(a) 
-$$
-\frac{2s+5}{(s+2)(s+3)}
-$$
-  
-\n
-$$
--3 < \sigma < -2
-$$
-  
-\n(b) 
-$$
-\frac{2s-5}{(s-2)(s-3)}
-$$
-  
-\n
-$$
-2 < \sigma < 3
-$$
-  
-\n(c) 
-$$
-\frac{2s+3}{(s+1)(s+2)}
-$$
-  
-\n
-$$
-\sigma > -1
-$$
-  
-\n(d) 
-$$
-\frac{2s+3}{(s+1)(s+2)}
-$$
-  
-\n
-$$
-\sigma < -2
-$$
-
-(d) 
-$$
-\frac{2s+6}{(s+1)(s+2)} \quad \sigma < -2
-$$
-  
-(e) 
-$$
-\frac{3s^2 - 2s - 17}{(s+1)(s+3)(s-5)} \quad -1 < \sigma < 5
-$$
-
-**4.11-7** Find
-
-$$
-\mathcal{L}^{-1}\left[\frac{2s^2 - 2s - 6}{(s+1)(s-1)(s+2)}\right]
-$$
-
-if the ROC is
-
-- (a) Re *s* > 1
-- (b) Re *s* < −2
-- (c) −1 < Re*s* < 1
-- (d) −2 < Re*s* < −1
-- **4.11-8** For a causal LTIC system having a transfer function *H*(*s*) = 1/(*s*+1), find the output *y*(*t*) if the input *x*(*t*) is given by (a) *e*−|*t*|/<sup>2</sup> (b) *et <sup>u</sup>*(*t*)+*e*2*<sup>t</sup> u*(−*t*) (c) *<sup>e</sup>*−*t*/2*u*(*t*) <sup>+</sup>*e*−*t*/4*u*(−*t*) (d) *e*2*<sup>t</sup> <sup>u</sup>*(*t*)+*et u*(−*t*) (e) *<sup>e</sup>*−*t*/4*u*(*t*) <sup>+</sup>*e*−*t*/2*u*(−*t*)
-  - (f) *e*−3*<sup>t</sup> <sup>u</sup>*(*t*)+*e*−2*<sup>t</sup> u*(−*t*)
-- **4.11-9** The autocorrelation function *rxx*(*t*) of a signal *x*(*t*) is given by
-
-$$
-r_{xx}(t) = \int_{-\infty}^{\infty} x(\tau) x(\tau + t) d\tau
-$$
-
-Derive an expression for *Rxx*(*s*) = *L*(*rxx*(*t*)) in terms of *X*(*s*), where *X*(*s*) = *L*(*x*(*t*)).
-
-**4.11-10** Determine the inverse Laplace transform of
-
-$$
-X(s) = \frac{2}{s} + \frac{s}{2}
-$$
-
-given that the region of convergence is σ < 0.
-
-**4.11-11** An absolutely integrable signal *x*(*t*) has a pole at *s* = π. It is possible that other poles may be present. Recall that an absolutely integrable signal satisfies
-
-$$
-\int_{-\infty}^{\infty} |x(t)| dt < \infty
-$$
-
-- (a) Can *x*(*t*) be left-sided? Explain.
-- (b) Can *x*(*t*) be right-sided? Explain.
-- (c) Can *x*(*t*) be two-sided? Explain.
-- (d) Can *x*(*t*) be of finite duration? Explain.
-- **4.11-12** Using ROC σ < 0, determine the inverse Laplace transform of *<sup>X</sup>*(*s*) <sup>=</sup> *<sup>s</sup> <sup>d</sup> ds e*−2*<sup>s</sup> s* . [*Hint:* Use Laplace transform properties to avoid tedious calculus.]
-- **4.11-13** With the assistance of Laplace transform properties, determine the inverse bilateral Laplace transform *x*(*t*) of signal
-
-$$
-X(s) = \frac{2}{e^s} + \frac{1}{s} \left[ e^s \frac{4}{\frac{s}{3} + 2} \right]
-$$
-
-where the ROC is −6 < Re{*s*} < 0.
-
-**4.11-14** With the assistance of Laplace transform properties, determine the inverse bilateral Laplace transform *x*(*t*) of signal
-
-$$
-X(s) = \frac{d^7}{ds^7} \left[ \frac{e^{-4s}}{(s+2)(s+3)} \right]
-$$
-
-where the ROC is −3 < Re{*s*} < −2.
-
-**4.11-15** Using the definition, compute the bilateral Laplace transform, including the region of convergence (ROC), of the following complexvalued functions:
-
-(a) 
-$$
-x_1(t) = (j + e^{jt})u(t)
-$$
-
-- (b) *x*2(*t*) = *j* cosh(*t*)*u*(−*t*)
-- (c) *<sup>x</sup>*3(*t*) <sup>=</sup> *ej*( <sup>π</sup> 4 ) *u*(−*t* +1)+*j*δ(*t* −5)
-- (d) *x*4(*t*) = *j t u*(−*t*) +δ(*t* −π )
-
-**4.11-16** A bounded-amplitude signal *x*(*t*) has bilateral Laplace transform *X*(*s*) given by
-
-$$
-X(s) = \frac{s2^s}{(s-1)(s+1)}
-$$
-
-- (a) Determine the corresponding region of convergence.
-- (b) Determine the time-domain signal *x*(*t*).
-- **4.12-1** Express the polynomial *C*20(*x*) in standard form. That is, determine the coefficients % *ak* of *<sup>C</sup>*20(*x*)<sup>=</sup> <sup>20</sup> *<sup>k</sup>*=<sup>0</sup> *akx*<sup>20</sup>−*k*.
-- **4.12-2** Consider an LTIC system with
-
-$$
-H(s) = \frac{1}{s^3 + 4s^2 + 8s + 8} = \frac{1}{(s^2 + 2s + 4)(s + 2)}
-$$
-  
-= 
-$$
-\frac{1}{(s - 2e^{j2\pi/3})(s - 2e^{-j2\pi/3})(s + 2)}.
-$$
-
-- (a) Write MATLAB code that accurately plots the system magnitude response |*H*(*j*ω)| over −10 ≤ ω ≤ 10.
-- (b) Write MATLAB code that accurately plots the system phase response over −10 ≤ ω ≤ 10.
-- (c) Determine the max value *y*max of output *y*(*t*) in response to input *x*(*t*) = 2−sin(2*t*+π/3).
-- (d) Draw a parallel representation of this system using real DFI structures of order 2 or less. [*Hint:* Use MATLAB to perform a partial fraction expansion of *H*(*s*).]
-- **4.12-3** Consider the op-amp circuit of Fig. P4.12-3. Further, let *RC* = 1.
-  - (a) From Fig. P4.12-3, determine the (simplified, standard form, rational) transfer function *H*(*s*).
-  - (b) Use MATLAB to accurately plot |*H*(*j*ω)| over −10 ≤ ω ≤ 10.
-  - (c) Determine the output of this system in response to *x*(*t*) = cos(10*t*)−1.
-  - (d) The circuit of Fig. P4.12-3 contains two capacitors. Suppose one capacitor must be a 25% tolerance part, while the other must be a 10% tolerance part. If the goal is to preserve the original magnitude response, should you use the 25% tolerance capacitor with the first op-amp or the second op-amp? Justify your answer with appropriate MAT-LAB simulations.
-
-**Figure P4.12-3**
-
-- **4.12-4** Design an order-12 Butterworth lowpass filter with a cutoff frequency of ω*<sup>c</sup>* = 2π5000 by completing the following.
-  - (a) Locate and plot the filter's poles and zeros in the complex plane. Plot the corresponding magnitude response |*H*LP(*j*ω)| to verify proper design.
-  - (b) Setting all resistor values to 100,000, determine the capacitor values to implement the filter using a cascade of six second-order Sallen–Key circuit sections. The form of a Sallen–Key stage is shown in Fig. P4.12-4. On a single plot, plot the magnitude response of each section as well as the overall magnitude response. Identify the poles that correspond to each section's magnitude response curve. Are the capacitor values realistic?
-- **4.12-5** Rather than a Butterworth filter, repeat Prob. 4.12-4 for a Chebyshev LPF with *R* = 3 dB of passband ripple. Since each Sallen–Key stage is constrained to have unity gain at dc, an overall gain error of 1/ √ 1+ <sup>2</sup> is acceptable.
-- **4.12-6** An analog lowpass filter with cutoff frequency ω*<sup>c</sup>* can be transformed into a highpass filter
-
-with cutoff frequency ω*<sup>c</sup>* by using an *RC*–*CR* transformation rule: each resistor *Ri* is replaced by a capacitor *C <sup>i</sup>* = 1/*Ri*ω*<sup>c</sup>* and each capacitor *Ci* is replaced by a resistor *R <sup>i</sup>* = 1/*Ci*ω*c*.
-
-Use this rule to design an order-8 Butterworth highpass filter with ω*<sup>c</sup>* = 2π4000 by completing the following.
-
-- (a) Design an order-8 Butterworth lowpass filter with ω*<sup>c</sup>* = 2π4000 by using four second-order Sallen–Key circuit stages, the form of which is shown in Fig. P4.12-4. Give resistor and capacitor values for each stage. Choose the resistors so that the *RC*–*CR* transformation will result in 1 nF capacitors. At this point, are the component values realistic?
-- (b) Draw an *RC*–*CR* transformed Sallen–Key circuit stage. Determine the transfer function *H*(*s*) of the transformed stage in terms of the variables *R* <sup>1</sup>, *R* <sup>2</sup>, *C* <sup>1</sup>, and *C* 2.
-- (c) Transform the LPF designed in part (a) by using an *RC*–*CR* transformation. Give the resistor and capacitor values for each stage. Are the component values realistic? Using *H*(*s*) derived in part (b), plot the
-
-magnitude response of each section as well
-
-as the overall magnitude response. Does the overall response look like a highpass Butterworth filter?
-
-Plot the HPF system poles and zeros in the complex *s* plane. How do these locations compare with those of the Butterworth LPF?
-
-- **4.12-7** Repeat Prob. 4.12-6, using ω*<sup>c</sup>* = 2π1500 and an order-16 filter. That is, eight second-order stages need to be designed.
-- **4.12-8** Rather than a Butterworth filter, repeat Prob. 4.12-6 for a Chebyshev LPF with *R* = 3 dB of passband ripple. Since each transformed Sallen–Key stage is constrained to have unity gain at ω = ∞, an overall gain error of 1/ √ 1+ <sup>2</sup> is acceptable.
-- **4.12-9** The MATLAB signal-processing toolbox function butter helps design analog Butterworth filters. Use MATLAB help to learn how butter works. For each of the following cases, design the filter, plot the filter's poles and zeros in the complex *s* plane, and plot the decibel magnitude response 20log10 |*H*(*j*ω)|:
-  - (a) Design a sixth-order analog lowpass filter with ω*<sup>c</sup>* = 2π3500.
-  - (b) Design a sixth-order analog highpass filter with ω*<sup>c</sup>* = 2π3500.
-  - (c) Design a sixth-order analog bandpass filter with a passband between 2 and 4 kHz.
-  - (d) Design a sixth-order analog bandstop filter with a stopband between 2 and 4 kHz.
-- **4.12-10** The MATLAB signal-processing toolbox function cheby1 helps design analog Chebyshev
-
-type I filters. A Chebyshev type I filter has a passband ripple and a smooth stopband. Setting the passband ripple to *Rp* = 3 dB, repeat Prob. 4.12-9 using the cheby1 command. With all other parameters held constant, what is the general effect of reducing *Rp*, the allowable passband ripple?
-
-- **4.12-11** The MATLAB signal-processing toolbox function cheby2 helps design analog Chebyshev type II filters. A Chebyshev type II filter has a smooth passband and ripple in the stopband. Setting the stopband ripple *Rs* = 20 dB down, repeat Prob. 4.12-9 using the cheby2 command. With all other parameters held constant, what is the general effect of increasing *Rs*, the minimum stopband attenuation?
-- **4.12-12** The MATLAB signal-processing toolbox function ellip helps design analog elliptic filters. An elliptic filter has ripple in both the passband and the stopband. Setting the passband ripple to *Rp* = 3 dB and the stopband ripple *Rs* = 20 dB down, repeat Prob. 4.12-9 using the ellip command.
-- **4.12-13** Using the definition *CN*(*x*)=cosh(*<sup>N</sup>* cosh−1(*x*)), prove the recursive relation *CN*(*x*) = 2*xCN*−1(*x*) −*CN*−2(*x*).
-- **4.12-14** Prove that the poles of a Chebyshev filter, which are located at *pk* = ω*<sup>c</sup>* sinh(ξ )sin(φ*k*) + *j*ω*<sup>c</sup>* cosh(ξ ) cos(φ*k*), lie on an ellipse. [*Hint:* The equation of an ellipse in the *x*–*y* plane is (*x*/*a*) 2+ (*y*/*b*) <sup>2</sup> <sup>=</sup> 1, where constants *<sup>a</sup>* and *<sup>b</sup>* define the major and minor axes of the ellipse.]
-
-# **[DISCRETE-TIME](#page-11-0) SYSTEM ANALYSIS USING THE** *z***-TRANSFORM**
-
-The counterpart of the Laplace transform for discrete-time systems is the *z*-transform. The Laplace transform converts integro-differential equations into algebraic equations. In the same way, the *z*-transforms changes difference equations into algebraic equations, thereby simplifying the analysis of discrete-time systems. The *z*-transform method of analysis of discrete-time systems parallels the Laplace transform method of analysis of continuous-time systems, with some minor differences. In fact, we shall see that *the z-transform is the Laplace transform in disguise*.
-
-The behavior of discrete-time systems is similar to that of continuous-time systems (with some differences). The frequency-domain analysis of discrete-time systems is based on the fact (proved in Sec. 3.8-2) that the response of a linear, time-invariant, discrete-time (LTID) system to an everlasting exponential *z<sup>n</sup>* is the same exponential (within a multiplicative constant) given by *H*[*z*]*z<sup>n</sup>*. We then express an input *x*[*n*] as a sum of (everlasting) exponentials of the form *z<sup>n</sup>*. The system response to *x*[*n*] is then found as a sum of the system's responses to all these exponential components. The tool that allows us to represent an arbitrary input *x*[*n*] as a sum of (everlasting) exponentials of the form *z<sup>n</sup>* is the *z*-transform.
-
-## **5.1 THE** *z***[-TRANSFORM](#page-11-0)**
-
-We define *X*[*z*], the direct *z*-transform of *x*[*n*], as
-
-$$
-X[z] = \sum_{n = -\infty}^{\infty} x[n]z^{-n}
-$$
- (5.1)
-
-where *z* is a complex variable. The signal *x*[*n*], which is the inverse *z*-transform of *X*[*z*], can be obtained from *X*[*z*] by using the following inverse *z*-transformation:
-
-$$
-x[n] = \frac{1}{2\pi j} \oint X[z] z^{n-1} dz
-$$
-\n(5.2)
-
-The symbol 6 indicates an integration in counterclockwise direction around a closed path in the complex plane (see Fig. 5.1). We derive this *z*-transform pair later, in Ch. 9, as an extension of the discrete-time Fourier transform pair.
-
-<span id="page-507-0"></span>CHAPTER
-
-**5**
-
-As in the case of the Laplace transform, we need not worry about this integral at this point because inverse *z*-transforms of many signals of engineering interest can be found in a *z*-transform table. The direct and inverse *z*-transforms can be expressed symbolically as
-
-$$
-X[z] = \mathcal{Z}\{x[n]\} \qquad \text{and} \qquad x[n] = \mathcal{Z}^{-1}\{X[z]\}
-$$
-
-or simply as
-
-*x*[*n*] ⇐⇒ *X*[*z*]
-
-Note that
-
-*<sup>Z</sup>*<sup>−</sup><sup>1</sup> [*Z*{*x*[*n*]}] = *<sup>x</sup>*[*n*] and *<sup>Z</sup>*[*Z*<sup>−</sup><sup>1</sup> {*X*[*z*]}] = *X*[*z*]
-
-### LINEARITY OF THE *z*-TRANSFORM
-
-Like the Laplace transform, the *z*-transform is a linear operator. If
-
-*x*1[*n*] ⇐⇒ *X*1[*z*] and *x*2[*n*] ⇐⇒ *X*2[*z*]
-
-then
-
-*a*1*x*1[*n*] +*a*2*x*2[*n*] ⇐⇒ *a*1*X*1[*z*] +*a*2*X*2[*z*]
-
-The proof is trivial and follows from the definition of the *z*-transform. This result can be extended to finite sums.
-
-### THE UNILATERAL *z*-TRANSFORM
-
-For the same reasons discussed in Ch. 4, we find it convenient to consider the unilateral *z*-transform. As seen for the Laplace case, the bilateral transform has some complications because of the non-uniqueness of the inverse transform. In contrast, the unilateral transform has a unique inverse. This fact simplifies the analysis problem considerably, but at a price: the unilateral version can handle only causal signals and systems. Fortunately, most of the practical cases are causal. The more general *bilateral z-transform* is discussed later, in Sec. 5.8. In practice, the term *z-transform* generally means *the unilateral z-transform*.
-
-In a basic sense, there is no difference between the unilateral and the bilateral *z*-transform. The unilateral transform is the bilateral transform that deals with a subclass of signals starting at *n* = 0 (causal signals). Hence, the definition of the unilateral transform is the same as that of the bilateral [Eq. (5.1)], except that the limits of the sum are from 0 to ∞:
-
-$$
-X[z] = \sum_{n=0}^{\infty} x[n]z^{-n}
-$$
- (5.3)
-
-The expression for the inverse *z*-transform in Eq. (5.2) remains valid for the unilateral case also.
-
-## THE REGION OF CONVERGENCE (ROC) OF *X*[*z*]
-
-The sum in Eq. (5.1) [or Eq. (5.3)] defining the direct *z*-transform *X*[*z*] may not converge (exist) for all values of *z*. The values of *z* (the region in the complex plane) for which the sum in Eq. (5.1) converges (or exists) are called the *region of existence,* or more commonly the *region of convergence* (ROC), for *X*[*z*]. This concept will become clear in the following example.
-
-### **EXAMPLE 5.1 Bilateral** *z***-Transform of a Causal Exponential**
-
-Find the *z*-transform and the corresponding ROC for the signal γ *nu*[*n*].
-
-By definition,
-
-$$
-X[z] = \sum_{n=0}^{\infty} \gamma^n u[n] z^{-n}
-$$
-
-Since *u*[*n*] = 1 for all *n* ≥ 0,
-
-$$
-X[z] = \sum_{n=0}^{\infty} \left(\frac{\gamma}{z}\right)^n = 1 + \left(\frac{\gamma}{z}\right) + \left(\frac{\gamma}{z}\right)^2 + \left(\frac{\gamma}{z}\right)^3 + \dots + \dots
-$$
- (5.4)
-
-It is helpful to remember the geometric progression and its sum [see Sec. B.8-3]:
-
-$$
-1 + x + x2 + x3 + \dots = \frac{1}{1 - x} \quad \text{if} \quad |x| < 1
-$$
-
-Applying this relationship to Eq. (5.4) yields
-
-$$
-X[z] = \frac{1}{1 - \frac{\gamma}{z}} \qquad \left| \frac{\gamma}{z} \right| < 1
-$$
-\n
-$$
-= \frac{z}{z - \gamma} \qquad |z| > |\gamma| \tag{5.5}
-$$
-
-Observe that *X*[*z*] exists only for |*z*| > |γ |. For |*z*| < |γ |, the sum in Eq. (5.4) does not converge; it goes to infinity. Therefore, the ROC of *X*[*z*] is the shaded region outside the circle of radius |γ |, centered at the origin, in the *z*-plane, as depicted in Fig. 5.1b.
-
-**Figure 5.1** <sup>γ</sup> *nu*[*n*] and the region of convergence of its *<sup>z</sup>*-transform.
-
-Later in Eq. (5.52), we show that the *z*-transform of another signal, −γ *nu*[−(*n* + 1)], is also *z*/(*z* − γ ). However, the ROC in this case is |*z*| < |γ |. Clearly, the inverse *z*-transform of *z*/(*z* − γ ) is not unique. However, if we restrict the inverse transform to be causal, then the inverse transform is unique, namely, γ *nu*[*n*].
-
-The ROC is required for evaluating *x*[*n*] from *X*[*z*], according to Eq. (5.2). The integral in Eq. (5.2) is a contour integral, implying integration in a counterclockwise direction along a closed path centered at the origin and satisfying the condition |*z*| > |γ |. Thus, any circular path centered at the origin and with a radius greater than |γ | (Fig. 5.1b) will suffice. We can show that the integral in Eq. (5.2) along any such path (with a radius greater than |γ |) yields the same result, namely, *x*[*n*]. † Such integration in the complex plane requires a background in the theory of functions of complex variables. We can avoid this integration by compiling a table of *z*-transforms (Table 5.1), where *z*-transform pairs are tabulated for a variety of signals. To find the inverse *z*-transform of say, *z*/(*z* − γ ), instead of using the complex integration in Eq. (5.2), we consult the table and find the inverse *z*-transform of *z*/(*z*−γ ) as γ *nu*[*n*]. Because of the uniqueness property of the unilateral *z*-transform, there is only one inverse for each *X*[*z*]. Although the table given here is rather short, it comprises the functions of most practical interest.
-
-The situation of the *z*-transform regarding the uniqueness of the inverse transform is parallel to that of the Laplace transform. For the bilateral case, the inverse *z*-transform is not unique unless the ROC is specified. For the unilateral case, the inverse transform is unique; the region of convergence need not be specified to determine the inverse *z*-transform. For this reason, we shall ignore the ROC in the unilateral *z*-transform Table 5.1.
-
-## EXISTENCE OF THE *z*-TRANSFORM
-
-By definition,
-
-$$
-X[z] = \sum_{n=0}^{\infty} x[n]z^{-n} = \sum_{n=0}^{\infty} \frac{x[n]}{z^n}
-$$
-
-The existence of the *z*-transform is guaranteed if
-
-$$
-|X[z]| \leq \sum_{n=0}^{\infty} \frac{|x[n]|}{|z|^n} < \infty
-$$
-
-for some <sup>|</sup>*z*|. Any signal *<sup>x</sup>*[*n*] that grows no faster than an exponential signal *<sup>r</sup><sup>n</sup>* <sup>0</sup>, for some *r*0, satisfies this condition. Thus, if
-
-$$
-|x[n]| \le r_0^n \qquad \text{for some } r_0 \tag{5.6}
-$$
-
-<sup>†</sup> Indeed, the path need not even be circular. It can have any odd shape, as long as it encloses the pole(s) of *X*[*z*] and the path of integration is counterclockwise.
-
-| No. | x[n]                                               | X[z]                                                            |
-|-----|----------------------------------------------------|-----------------------------------------------------------------|
-| 1   | δ[n−k]                                             | z−k                                                             |
-| 2   | u[n]                                               | z<br>z−1                                                        |
-| 3   | nu[n]                                              | z<br>(z−1)2                                                     |
-| 4   | n2u[n]                                             | z(z+1)<br>(z−1)3                                                |
-| 5   | n3u[n]                                             | z(z2 +4z+1)<br>(z−1)4                                           |
-| 6   | γ nu[n]                                            | z<br>z−γ                                                        |
-| 7   | γ n−1u[n−1]                                        | 1<br>z−γ                                                        |
-| 8   | nγ nu[n]                                           | γ z<br>(z−γ )2                                                  |
-| 9   | n2γ<br>nu[n]                                       | γ z(z+γ )<br>(z−γ )3                                            |
-| 10  | n(n−1)(n−2)···(n−m+1)<br>γ nu[n]<br>γ mm!          | z<br>(z−γ )m+1                                                  |
-| 11a | n cos<br> γ  <br>βn u[n]                           | z(z−  γ   cos β)<br>z2 −(2 γ<br>2<br>  cos β)z+  γ              |
-| 11b | n sin<br> γ  <br>βn u[n]                           | z γ  sin β<br>z2 −(2 γ<br>2<br>  cos β)z+  γ                    |
-| 12a | n cos(βn+θ<br>r γ  <br>)u[n]                       | rz[z cos θ −  γ   cos(β −θ )]<br>z2 −(2 γ<br>2<br>  cos β)z+  γ |
-| 12b | n cos(βn+θ<br>γ =  γ  ejβ<br>r γ  <br>)u[n]        | (0.5rejθ )z<br>(0.5re−jθ )z<br>+<br>z−γ<br>z−γ ∗                |
-| 12c | n cos(βn+θ<br>r γ  <br>)u[n]                       | z(Az+B)<br>z2 +2az+  γ<br>2<br>                                 |
-|     | <br>A2 γ<br>2 +B2 −2AaB<br> <br>r =<br>2 −a2<br> γ |                                                                 |
-|     | β = cos−1 −a<br> γ                                 |                                                                 |
-|     | Aa−B<br>θ = tan−1<br><br>2 −a2<br>A<br> γ          |                                                                 |
-
-**TABLE 5.1** Select (Unilateral) *z*-Transform Pairs
-
-$$
-|X[z]| \le \sum_{n=0}^{\infty} \left(\frac{r_0}{|z|}\right)^n = \frac{1}{1 - \frac{r_0}{|z|}} \qquad |z| > r_0
-$$
-
-Therefore, *X*[*z*] exists for |*z*| > *r*0. Almost all practical signals satisfy Eq. (5.6) and are therefore *z*-transformable. Some signal models (e.g., γ *<sup>n</sup>*<sup>2</sup> ) grow faster than the exponential signal *r<sup>n</sup>* <sup>0</sup> (for any *r*0) and do not satisfy Eq. (5.6) and therefore are not *z*-transformable. Fortunately, such signals are of little practical or theoretical interest. Even such signals over a finite interval are *z*-transformable.
-
-then
-
-This geometric sum simplifies [see Sec. B.8-3] to
-
-$$
-X[z] = \frac{1}{1 - \frac{1}{z}} \qquad \left| \frac{1}{z} \right| < 1
-$$
-\n
-$$
-= \frac{z}{z - 1} \qquad |z| > 1
-$$
-
-Therefore,
-
-$$
-u[n] \Longleftrightarrow \frac{z}{z-1} \qquad |z| > 1
-$$
-
-**(c)** Recall that cos β*n* = (*ej*β*<sup>n</sup>* +*e*−*j*β*<sup>n</sup>*)/2. Moreover, according to Eq. (5.5),
-
-$$
-e^{\pm j\beta n}u[n] \Longleftrightarrow \frac{z}{z - e^{\pm j\beta}} \qquad |z| > |e^{\pm j\beta}| = 1
-$$
-
-Therefore,
-
-$$
-X[z] = \frac{1}{2} \left[ \frac{z}{z - e^{j\beta}} + \frac{z}{z - e^{-j\beta}} \right] = \frac{z(z - \cos \beta)}{z^2 - 2z \cos \beta + 1} \qquad |z| > 1
-$$
-
-**(d)** Here *x*[0] = *x*[1] = *x*[2] = *x*[3] = *x*[4] = 1 and *x*[5] = *x*[6]=···= 0. Therefore, according to Eq. (5.7),
-
-$$
-X[z] = 1 + \frac{1}{z} + \frac{1}{z^2} + \frac{1}{z^3} + \frac{1}{z^4} = \frac{z^4 + z^3 + z^2 + z + 1}{z^4}
-$$
- for all  $z \neq 0$ 
-
-We can also express this result in a more compact form by summing the geometric progression on the right-hand side of the foregoing equation. From the result in Sec. B.8-3 with *r* =1/*z*,*m*= 0, and *n* = 4, we obtain
-
-$$
-X[z] = \frac{\left(\frac{1}{z}\right)^5 - \left(\frac{1}{z}\right)^0}{\frac{1}{z} - 1} = \frac{z}{z - 1} (1 - z^{-5})
-$$
-
-### **DR ILL 5.1 Bilateral** *z***-Transform**
-
-- **(a)** Find the *z*-transform of a signal shown in Fig. 5.3.
-- **(b)** Use pair 12a (Table 5.1) to find the *z*-transform of *x*[*n*] = 20.65( <sup>√</sup>2)*<sup>n</sup>* cos[(π/4)*<sup>n</sup>* <sup>−</sup> 1.415]*u*[*n*].
-
-<span id="page-514-0"></span>
-
-### **[5.1-1 Inverse Transform by Partial Fraction Expansion and Tables](#page-11-0)**
-
-As in the Laplace transform, we shall avoid the integration in the complex plane required to find the inverse *z*-transform [Eq. (5.2)] by using the (unilateral) transform table (Table 5.1). Many of the transforms *X*[*z*] of practical interest are rational functions (ratio of polynomials in *z*), which can be expressed as a sum of partial fractions, whose inverse transforms can be readily found in a table of transform. The partial fraction method works because for every transformable *x*[*n*] defined for *n* ≥ 0, there is a corresponding unique *X*[*z*] defined for |*z*| > *r*<sup>0</sup> (where *r*<sup>0</sup> is some constant), and vice versa.
-
-### **EXAMPLE 5.3 Inverse** *z***-Transform by Partial Fraction Expansion**
-
-Find the inverse *z*-transforms of
-
-(a) 
-$$
-\frac{8z-19}{(z-2)(z-3)}
-$$
-  
-\n(b) 
-$$
-\frac{z(2z^2-11z+12)}{(z-1)(z-2)^3}
-$$
-  
-\n(c) 
-$$
-\frac{2z(3z+17)}{(z-1)(z^2-6z+25)}
-$$
-
-**(a)** Expanding *X*[*z*] into partial fractions yields
-
-$$
-X[z] = \frac{8z - 19}{(z - 2)(z - 3)} = \frac{3}{z - 2} + \frac{5}{z - 3}
-$$

engineering/linear-systems-and-signals-3rd-edition-bp-lathi/067_5 DISCRETE-TIME SYSTEM ANALYSIS USING THE z-TRANSFORM.md

@@ -1,304 +0,0 @@
-### 496 CHAPTER 5 DISCRETE-TIME SYSTEM ANALYSIS USING THE *Z*-TRANSFORM
-
-From Table 5.1, pair 7, we obtain
-
-$$
-x[n] = [3(2)^{n-1} + 5(3)^{n-1}]u[n-1]
-$$
-\n(5.8)
-
-If we expand rational *X*[*z*] into partial fractions directly, we shall always obtain an answer that is multiplied by *u*[*n* − 1] because of the nature of pair 7 in Table 5.1. This form is rather awkward as well as inconvenient. We prefer the form that contains *u*[*n*] rather than *u*[*n*−1]. A glance at Table 5.1 shows that the *z*-transform of every signal that is multiplied by *u*[*n*] has a factor *z* in the numerator. This observation suggests that we expand *X*[*z*] into *modified partial fractions*, where each term has a factor *z* in the numerator. This goal can be accomplished by expanding *X*[*z*]/*z* into partial fractions and then multiplying both sides by *z*. We shall demonstrate this procedure by reworking part (a). For this case,
-
-$$
-\frac{X[z]}{z} = \frac{8z - 19}{z(z - 2)(z - 3)} = \frac{(-19/6)}{z} + \frac{(3/2)}{z - 2} + \frac{(5/3)}{z - 3}
-$$
-
-Multiplying both sides by *z* yields
-
-$$
-X[z] = -\frac{19}{6} + \frac{3}{2} \left( \frac{z}{z-2} \right) + \frac{5}{3} \left( \frac{z}{z-3} \right)
-$$
-
-From pairs 1 and 6 in Table 5.1, it follows that
-
-$$
-x[n] = -\frac{19}{6}\delta[n] + \left[\frac{3}{2}(2)^n + \frac{5}{3}(3)^n\right]u[n] \tag{5.9}
-$$
-
-The reader can verify that this answer is equivalent to that in Eq. (5.8) by computing *x*[*n*] in both cases for *n* = 0, 1, 2, 3,..., and comparing the results. The form in Eq. (5.9) is more convenient than that in Eq. (5.8). For this reason, we shall always expand *X*[*z*]/*z* rather than *X*[*z*] into partial fractions and then multiply both sides by *z* to obtain modified partial fractions of *X*[*z*], which have a factor *z* in the numerator.
-
-**(b)**
-
-$$
-X[z] = \frac{z(2z^2 - 11z + 12)}{(z - 1)(z - 2)^3}
-$$
-
-and
-
-$$
-\frac{X[z]}{z} = \frac{2z^2 - 11z + 12}{(z - 1)(z - 2)^3} = \frac{k}{z - 1} + \frac{a_0}{(z - 2)^3} + \frac{a_1}{(z - 2)^2} + \frac{a_2}{(z - 2)}
-$$
-
-where
-
-$$
-k = \frac{2z^2 - 11z + 12}{(z - 1)(z - 2)^3}\Big|_{z=1} = -3
-$$
-  
-$$
-a_0 = \frac{2z^2 - 11z + 12}{(z - 1)(z - 2)^3}\Big|_{z=2} = -2
-$$
-
-Therefore,
-
-$$
-\frac{X[z]}{z} = \frac{2z^2 - 11z + 12}{(z - 1)(z - 2)^3} = \frac{-3}{z - 1} - \frac{2}{(z - 2)^3} + \frac{a_1}{(z - 2)^2} + \frac{a_2}{(z - 2)}
-$$
-(5.10)
-
-We can determine *a*<sup>1</sup> and *a*<sup>2</sup> by clearing fractions. Or we may use a shortcut. For example, to determine *a*2, we multiply both sides of Eq. (5.10) by *z* and let *z* → ∞. This yields
-
-$$
-0 = -3 - 0 + 0 + a_2 \implies a_2 = 3
-$$
-
-This result leaves only one unknown, *a*1, which is readily determined by letting *z* take any convenient value, say, *z* = 0, on both sides of Eq. (5.10). This produces
-
-$$
-\frac{12}{8} = 3 + \frac{1}{4} + \frac{a_1}{4} - \frac{3}{2}
-$$
-
-which yields *a*<sup>1</sup> = −1. Therefore,
-
-$$
-\frac{X[z]}{z} = \frac{-3}{z-1} - \frac{2}{(z-2)^3} - \frac{1}{(z-2)^2} + \frac{3}{z-2}
-$$
-
-and
-
-$$
-X[z] = -3\frac{z}{z-1} - 2\frac{z}{(z-2)^3} - \frac{z}{(z-2)^2} + 3\frac{z}{z-2}
-$$
-
-Now the use of Table 5.1, pairs 6 and 10, yields
-
-$$
-x[n] = \left[ -3 - 2\frac{n(n-1)}{8} (2)^n - \frac{n}{2} (2)^n + 3(2)^n \right] u[n]
-$$
-  
-= 
-$$
-- \left[ 3 + \frac{1}{4} (n^2 + n - 12) 2^n \right] u[n]
-$$
-
-**(c) Complex Poles.**
-
-$$
-X[z] = \frac{2z(3z+17)}{(z-1)(z^2-6z+25)} = \frac{2z(3z+17)}{(z-1)(z-3-j4)(z-3+j4)}
-$$
-
-The poles of *X*[*z*] are 1, 3 + *j*4, and 3 − *j*4. Whenever there are complex-conjugate poles, the problem can be worked out in two ways. In the first method we expand *X*[*z*] into (modified) first-order partial fractions. In the second method, rather than obtain one factor corresponding to each complex-conjugate pole, we obtain quadratic factors corresponding to each pair of complex-conjugate poles. This procedure is explained next.
-
-MENT-ORDER FACTORS
-
-\n
-$$
-\frac{X[z]}{z} = \frac{2(3z+17)}{(z-1)(z^2-6z+25)} = \frac{2(3z+17)}{(z-1)(z-3-j4)(z-3+j4)}
-$$
-
-We find the partial fraction of *X*[*z*]/*z* using the Heaviside "cover-up" method:
-
-$$
-\frac{X[z]}{z} = \frac{2}{z-1} + \frac{1.6e^{-j2.246}}{z-3-j4} + \frac{1.6e^{j2.246}}{z-3+j4}
-$$
-
-and
-
-$$
-X[z] = 2\frac{z}{z-1} + (1.6e^{-j2.246})\frac{z}{z-3-j4} + (1.6e^{j2.246})\frac{z}{z-3+j4}
-$$
-
-The inverse transform of the first term on the right-hand side is 2*u*[*n*]. The inverse transform of the remaining two terms (complex conjugate poles) can be obtained from pair 12b (Table 5.1) by identifying *r*/2 = 1.6, θ = −2.246 rad, γ = 3 + *j*4 = 5*ej*0.927, so that |γ | = 5, β = 0.927. Therefore,
-
-$$
-x[n] = [2 + 3.2(5)n \cos(0.927n - 2.246)]u[n]
-$$
-
-### METHOD OF QUADRATIC FACTORS
-
-$$
-\frac{X[z]}{z} = \frac{2(3z+17)}{(z-1)(z^2-6z+25)} = \frac{2}{z-1} + \frac{Az+B}{z^2-6z+25}
-$$
-
-Multiplying both sides by *z* and letting *z* → ∞, we find
-
-$$
-0 = 2 + A \Longrightarrow A = -2
-$$
-
-and
-
-$$
-\frac{2(3z+17)}{(z-1)(z^2-6z+25)} = \frac{2}{z-1} + \frac{-2z+B}{z^2-6z+25}
-$$
-
-To find *B*, we let *z* take any convenient value, say, *z* = 0. This step yields
-
-$$
-\frac{-34}{25} = -2 + \frac{B}{25} \Longrightarrow B = 16
-$$
-
-Therefore,
-
-$$
-\frac{X[z]}{z} = \frac{2}{z-1} + \frac{-2z+16}{z^2-6z+25}
-$$
-
-and
-
-$$
-X[z] = \frac{2z}{z-1} + \frac{z(-2z+16)}{z^2 - 6z + 25}
-$$
-
-We now use pair 12c, where we identify *A* = −2, *B* = 16, |γ | = 5, and *a* = −3. Therefore,
-
-$$
-r = \sqrt{\frac{100 + 256 - 192}{25 - 9}} = 3.2
-$$
-,  $\beta = \cos^{-1}\left(\frac{3}{5}\right) = 0.927$  rad
-
-and
-
-$$
-\theta = \tan^{-1}\left(\frac{-10}{-8}\right) = -2.246 \,\text{rad}
-$$
-
-so that
-
-$$
-x[n] = [2 + 3.2(5)n \cos (0.927n - 2.246)]u[n]
-$$
-
-### <span id="page-518-0"></span>**DR ILL 5.2 Inverse** *z***-Transform by Partial Fraction Expansion**
-
-Find the inverse *z*-transform of the following functions:
-
-(a) 
-$$
-\frac{z(2z-1)}{(z-1)(z+0.5)}
-$$
-  
-\n(b) 
-$$
-\frac{1}{(z-1)(z+0.5)}
-$$
-  
-\n(c) 
-$$
-\frac{9}{(z+2)(z-0.5)^2}
-$$
-  
-\n(d) 
-$$
-\frac{5z(z-1)}{z^2-1.6z+0.8}
-$$
-  
-\n[*Hint:*  $\sqrt{0.8} = 2/\sqrt{5}$ .]  
-\n**ANSWERS**  
-\n(a)  $\left[\frac{2}{3} + \frac{4}{3}(-0.5)^n\right]u[n]$   
-\n(b)  $-2\delta[n] + \left[\frac{2}{3} + \frac{4}{3}(-0.5)^n\right]u[n]$ 
-
-- **(c)** 18δ[*n*]−[0.72(−2)*<sup>n</sup>* +17.28(0.5)*<sup>n</sup>* −14.4*n*(0.5)*<sup>n</sup>*]*u*[*n*]
-- **(d)** <sup>5</sup> √5 2 √ 2 5 *n* cos(0.464*n*+0.464)*u*[*n*]
-
-## **5.1-2 Inverse** *z***[-Transform by Power Series Expansion](#page-11-0)**
-
-By definition,
-
-$$
-X[z] = \sum_{n=0}^{\infty} x[n]z^{-n}
-$$
-  
-=  $x[0] + \frac{x[1]}{z} + \frac{x[2]}{z^2} + \frac{x[3]}{z^3} + \cdots$   
-=  $x[0]z^0 + x[1]z^{-1} + x[2]z^{-2} + x[3]z^{-3} + \cdots$ 
-
-This result is a power series in *z*−1. Therefore, if we can expand *X*[*z*] into the power series in *z*−1, the coefficients of this power series can be identified as *x*[0], *x*[1], *x*[2], *x*[3], .... A rational *X*[*z*] can be expanded into a power series of *z*−<sup>1</sup> by dividing its numerator by the denominator. Consider, for example,
-
-$$
-X[z] = \frac{z^2(7z-2)}{(z-0.2)(z-0.5)(z-1)} = \frac{7z^3 - 2z^2}{z^3 - 1.7z^2 + 0.8z - 0.1}
-$$
-
-### 500 CHAPTER 5 DISCRETE-TIME SYSTEM ANALYSIS USING THE *Z*-TRANSFORM
-
-To obtain a series expansion in powers of *z*−1, we divide the numerator by the denominator as follows:
-
-$$
-z^{3}-1.7z^{2}+0.8z-0.1\overline{)7z^{3}-2z^{2}}
-$$
-  
-$$
-\underline{7z^{3}-11.9z^{2}+5.60z-0.7}
-$$
-  
-$$
-\underline{7z^{3}-11.9z^{2}+5.60z-0.7}
-$$
-  
-$$
-\underline{9.9z^{2}-5.60z+0.7}
-$$
-  
-$$
-\underline{9.9z^{2}-16.83z+7.92-0.99z^{-1}}
-$$
-  
-$$
-\underline{11.23z-7.22+0.99z^{-1}}
-$$
-  
-$$
-\underline{11.23z-19.09+8.98z^{-1}}
-$$
-  
-$$
-\underline{11.87-7.99z^{-1}}
-$$
-
-Thus,
-
-$$
-X[z] = \frac{z^2(7z-2)}{(z-0.2)(z-0.5)(z-1)} = 7 + 9.9z^{-1} + 11.23z^{-2} + 11.87z^{-3} + \cdots
-$$
-
-Therefore,
-
-$$
-x[0] = 7, x[1] = 9.9, x[2] = 11.23, x[3] = 11.87, \dots
-$$
-
-Although this procedure yields *x*[*n*] directly, it does not provide a closed-form solution. For this reason, it is not very useful unless we want to know only the first few terms of the sequence *x*[*n*].
-
-### **DR ILL 5.3 Inverse** *z***-Transform by Long Division**
-
-Using long division to find the power series in *z*−1, show that the inverse *z*-transform of *z*/(*z*−0.5) is (0.5)*nu*[*n*] or (2)−*nu*[*n*].
-
-## RELATIONSHIP BETWEEN *h*[*n*] AND *H*[*z*]
-
-For an LTID system, if *h*[*n*] is its unit impulse response, then from Eq. (3.39), where we defined *H*[*z*], the system transfer function, we write
-
-$$
-H[z] = \sum_{n=-\infty}^{\infty} h[n]z^{-n}
-$$
- (5.11)
-
-For causal systems, the limits on the sum are from *n* = 0 to ∞. This equation shows that the transfer function *H*[*z*] is the *z*-transform of the impulse response *h*[*n*] of an LTID system; that is,
-
-$$
-h[n] \Longleftrightarrow H[z]
-$$
-
-This important result relates the time-domain specification *h*[*n*] of a system to *H*[*z*], the frequency-domain specification of a system. The result is parallel to that for LTIC systems.
-
-### <span id="page-520-0"></span>**DR ILL 5.4 Impulse Response by Inverse** *z***-Transform**
-
-Redo Drill 3.14 by taking the inverse *z*-transform of *H*[*z*], as given by Eq. (3.41).

engineering/linear-systems-and-signals-3rd-edition-bp-lathi/068_5.2 SOME PROPERTIES OF THE z-TRANSFORM.md

@@ -1,924 +0,0 @@
-## **5.2 SOME [PROPERTIES OF THE](#page-11-0)** *z***-TRANSFORM**
-
-The *z*-transform properties are useful in the derivation of *z*-transforms of many functions and also in the solution of linear difference equations with constant coefficients. Here we consider a few important properties of the *z*-transform.
-
-In our discussion, the variable *n* appearing in signals, such as *x*[*n*] and *y*[*n*], may or may not stand for time. However, in most applications of our interest, *n* is proportional to time. For this reason, we shall loosely refer to the variable *n* as time.
-
-### **[5.2-1 Time-Shifting Properties](#page-11-0)**
-
-In the following discussion of the shift property, we deal with shifted signals *x*[*n*]*u*[*n*], *x*[*n*−*k*]*u*[*n*− *k*], *x*[*n*−*k*]*u*[*n*], and *x*[*n*+*k*]*u*[*n*]. Unless we physically understand the meaning of such shifts, our understanding of the shift property remains mechanical rather than intuitive or heuristic. For this reason, using a hypothetical signal *x*[*n*], we have illustrated various shifted signals for *k* = 1 in Fig. 5.4.
-
-## RIGHT SHIFT (DELAY) If
-
-*x*[*n*]*u*[*n*] ⇐⇒ *X*[*z*]
-
-then
-
-$$
-x[n-1]u[n-1] \Longleftrightarrow \frac{1}{z}X[z] \tag{5.12}
-$$
-
-In general,
-
-$$
-x[n-m]u[n-m] \Longleftrightarrow \frac{1}{z^m}X[z]
-$$
-\n(5.13)
-
-Moreover,
-
-$$
-x[n-1]u[n] \Longleftrightarrow \frac{1}{z}X[z] + x[-1]
-$$
-\n(5.14)
-
-Repeated application of this property yields
-
-$$
-x[n-2]u[n] \Longleftrightarrow \frac{1}{z} \left[ \frac{1}{z} X[z] + x[-1] \right] + x[-2] = \frac{1}{z^2} X[z] + \frac{1}{z} x[-1] + x[-2]
-$$
-
-In general, for integer value of *m*,
-
-$$
-x[n-m]u[n] \Longleftrightarrow z^{-m}X[z] + z^{-m}\sum_{n=1}^{m}x[-n]z^{n}
-$$
-\n(5.15)
-
-A look at Eqs. (5.12) and (5.14) shows that they are identical except for the extra term *x*[−1] in Eq. (5.14). We see from Figs. 5.4c and 5.4d that *x*[*n* − 1]*u*[*n*] is the same as *x*[*n* − 1]*u*[*n* − 1] plus *x*[−1]δ[*n*]. Hence, the difference between their transforms is *x*[−1].
-
-### 5.2 Some Properties of the *z*-Transform 503
-
-**Proof.** For the integer value of *m*,
-
-$$
-\mathcal{Z}{x[n-m]u[n-m]} = \sum_{n=0}^{\infty} x[n-m]u[n-m]z^{-n}
-$$
-
-Recall that *x*[*n* − *m*]*u*[*n* − *m*] = 0 for *n* < *m* so that the limits on the summation on the right-hand side can be taken from *n* = *m* to ∞. Therefore,
-
-$$
-\mathcal{Z}\{x[n-m]u[n-m]\} = \sum_{n=m}^{\infty} x[n-m]z^{-n}
-$$
-$$
-= \sum_{r=0}^{\infty} x[r]z^{-(r+m)}
-$$
-$$
-= \frac{1}{z^m} \sum_{r=0}^{\infty} x[r]z^{-r} = \frac{1}{z^m} X[z]
-$$
-
-To prove Eq. (5.15), we have
-
-$$
-\mathcal{Z}\{x[n-m]u[n]\} = \sum_{n=0}^{\infty} x[n-m]z^{-n} = \sum_{r=-m}^{\infty} x[r]z^{-(r+m)}
-$$
-$$
-= z^{-m} \left[ \sum_{r=-m}^{-1} x[r]z^{-r} + \sum_{r=0}^{\infty} x[r]z^{-r} \right]
-$$
-$$
-= z^{-m} \sum_{n=1}^{m} x[-n]z^{n} + z^{-m}X[z]
-$$
-
-LEFT SHIFT (ADVANCE) If
-
-$$
-x[n]u[n] \Longleftrightarrow X[z]
-$$
-
-then
-
-$$
-x[n+1]u[n] \Longleftrightarrow zX[z] - zx[0]
-$$
-
-Repeated application of this property yields
-
-$$
-x[n+2]u[n] \Longleftrightarrow z\{z(X[z] - zx[0]) - x[1]\} = z^2 X[z] - z^2 x[0] - zx[1]
-$$
-
-and for the integer value of *m*,
-
-$$
-x[n+m]u[n] \Longleftrightarrow z^m X[z] - z^m \sum_{n=0}^{m-1} x[n]z^{-n}
-$$
-\n(5.16)
-
-### 504 CHAPTER 5 DISCRETE-TIME SYSTEM ANALYSIS USING THE *Z*-TRANSFORM
-
-**Proof.** By definition,
-
-$$
-\mathcal{Z}\{x[n+m]u[n]\} = \sum_{n=0}^{\infty} x[n+m]z^{-n}
-$$
-$$
-= \sum_{r=m}^{\infty} x[r]z^{-(r-m)}
-$$
-$$
-= z^m \sum_{r=m}^{\infty} x[r]z^{-r}
-$$
-$$
-= z^m \left[ \sum_{r=0}^{\infty} x[r]z^{-r} - \sum_{r=0}^{m-1} x[r]z^{-r} \right]
-$$
-$$
-= z^m X[z] - z^m \sum_{r=0}^{m-1} x[r]z^{-r}
-$$
-
-### **EXAMPLE 5.4** *z***-Transform Using the Right-Shift Property**
-
-The signal *x*[*n*] can be expressed as a product of *n* and a gate pulse *u*[*n*] − *u*[*n* − 6]. Therefore,
-
-$$
-x[n] = n\{u[n] - u[n-6]\} = nu[n] - nu[n-6]
-$$
-
-We cannot find the *z*-transform of *nu*[*n* − 6] directly by using the right-shift property [Eq. (5.13)]. So we rearrange it in terms of (*n*−6)*u*[*n*−6] as follows:
-
-$$
-x[n] = nu[n] - (n - 6 + 6)u[n - 6]
-$$
-  
-=  $nu[n] - (n - 6)u[n - 6] - 6u[n - 6]$ 
-
-<span id="page-524-0"></span>We can now find the *z*-transform of the bracketed term by using the right-shift property [Eq. (5.13)]. Because *u*[*n*] ⇐⇒ *z*/(*z*−1),
-
-$$
-u[n-6] \Longleftrightarrow \frac{1}{z^6} \frac{z}{z-1} = \frac{1}{z^5(z-1)}
-$$
-
-Also, because *nu*[*n*] ⇐⇒ *z*/(*z*−1)2,
-
-$$
-(n-6)u[n-6] \Longleftrightarrow \frac{1}{z^6} \frac{z}{(z-1)^2} = \frac{1}{z^5(z-1)^2}
-$$
-
-Therefore,
-
-$$
-X[z] = \frac{z}{(z-1)^2} - \frac{1}{z^5(z-1)^2} - \frac{6}{z^5(z-1)} = \frac{z^6 - 6z + 5}{z^5(z-1)^2}
-$$
-
-### **DR ILL 5.5** *z***-Transform Using the Right-Shift Property**
-
-Using only the fact that *u*[*n*] ⇐⇒ *z*/(*z*−1) and the right-shift property [Eq. (5.13)], find the *z*-transforms of the signals in Figs. 5.2 and 5.3.
-
-### **ANSWERS**
-
-See Ex. 5.2d and Drill 5.1a.
-
-## **5.2-2** *z***[-Domain Scaling Property \(Multiplication by](#page-11-0)** *γ <sup>n</sup>***)**
-
-Scaling in the *z*-domain is equivalent to multiplying a time-domain signal by an exponential. That is, if
-
-$$
-x[n]u[n] \Longleftrightarrow X[z]
-$$
-
-then
-
-$$
-\gamma^n x[n]u[n] \Longleftrightarrow X\left[\frac{z}{\gamma}\right] \tag{5.17}
-$$
-
-**Proof.**
-
-$$
-\mathcal{Z}\{\gamma^{n}x[n]u[n]\} = \sum_{n=0}^{\infty} \gamma^{n}x[n]z^{-n} = \sum_{n=0}^{\infty} x[n]\left(\frac{z}{\gamma}\right)^{-n} = X\left[\frac{z}{\gamma}\right]
-$$
-
-### **DR ILL 5.6 Using the** *z***-Domain Scaling Property**
-
-Use Eq. (5.17) to derive pairs 6 and 8 in Table 5.1 from pairs 2 and 3, respectively.
-
-## <span id="page-525-0"></span>**5.2-3** *z***[-Domain Differentiation Property \(Multiplication by](#page-11-0)** *n***)**
-
-Multiplying a signal by *n* in the time domain produces differentiation in the *z*-domain. That is, if
-
-$$
-x[n]u[n] \Longleftrightarrow X[z]
-$$
-
-then
-
-$$
-nx[n]u[n] \Longleftrightarrow -z\frac{d}{dz}X[z] \tag{5.18}
-$$
-
-**Proof.**
-
-$$
--z\frac{d}{dz}X[z] = -z\frac{d}{dz}\sum_{n=0}^{\infty}x[n]z^{-n}
-$$
-$$
-= -z\sum_{n=0}^{\infty} -nx[n]z^{-n-1}
-$$
-$$
-= \sum_{n=0}^{\infty}nx[n]z^{-n} = \mathcal{Z}\{nx[n]u[n]\}
-$$
-
-### **DR ILL 5.7 Using the** *z***-Domain Differentiation Property**
-
-Use Eq. (5.18) to derive pairs 3 and 4 in Table 5.1 from pair 2. Similarly, derive pairs 8 and 9 from pair 6.
-
-### **[5.2-4 Time-Reversal Property](#page-11-0)**
-
-If
-
-$$
-x[n] \Longleftrightarrow X[z]
-$$
-
-then†
-
-$$
-x[-n] \Longleftrightarrow X[1/z]
-$$
-
-**Proof.**
-
-$$
-\mathcal{Z}{x[-n]} = \sum_{n=-\infty}^{\infty} x[-n]z^{-n}
-$$
-
-$$
-x^*[-n] \Longleftrightarrow X^*[1/z^*]
-$$
-
-<sup>†</sup> For complex signal *<sup>x</sup>*[*n*], the time-reversal property is modified as follows:
-
-<span id="page-526-0"></span>Changing the sign of the dummy variable *n* yields
-
-$$
-\mathcal{Z}{x[-n]} = \sum_{n=-\infty}^{\infty} x[n]z^n
-$$
-$$
-= \sum_{n=-\infty}^{\infty} x[n](1/z)^{-n}
-$$
-$$
-= X[1/z]
-$$
-
-The region of convergence is also inverted; that is, if the ROC of *x*[*n*] is |*z*| > |γ |, then the ROC of *x*[−*n*] is |*z*| < 1/|γ |.
-
-### **DR ILL 5.8 Using the Time-Reversal Property**
-
-Use the time-reversal property and pair 2 in Table 5.1 to show that *u*[−*n*] ⇐⇒ −1/(*z*−1) with the ROC |*z*| < 1.
-
-### **[5.2-5 Convolution Property](#page-11-0)**
-
-The time-convolution property states that if‡
-
-$$
-x_1[n] \Longleftrightarrow X_1[z]
-$$
- and  $x_2[n] \Longleftrightarrow X_2[z]$ ,
-
-then (*time convolution*)
-
-$$
-x_1[n] * x_2[n] \Longleftrightarrow X_1[z]X_2[z] \tag{5.19}
-$$
-
-**Proof.** This property applies to causal as well as noncausal sequences. We shall prove it for the more general case of noncausal sequences, where the convolution sum ranges from −∞ to ∞.
-
-We have
-
-$$
-\mathcal{Z}\{x_1[n] * x_2[n]\} = \mathcal{Z}\left[\sum_{m=-\infty}^{\infty} x_1[m]x_2[n-m]\right]
-$$
-$$
-= \sum_{n=-\infty}^{\infty} z^{-n} \sum_{m=-\infty}^{\infty} x_1[m]x_2[n-m]
-$$
-
-‡ There is also the frequency-convolution property, which states that
-
-$$
-x_1[n]x_2[n] \Longleftrightarrow \frac{1}{2\pi j} \oint X_1[u]X_2\left[\frac{z}{u}\right]u^{-1} du
-$$
-
-### 508 CHAPTER 5 DISCRETE-TIME SYSTEM ANALYSIS USING THE *Z*-TRANSFORM
-
-Interchanging the order of summation, we have
-
-$$
-\mathcal{Z}[x_1[n]*x_2[n]] = \sum_{m=-\infty}^{\infty} x_1[m] \sum_{n=-\infty}^{\infty} x_2[n-m]z^{-n}
-$$
-$$
-= \sum_{m=-\infty}^{\infty} x_1[m] \sum_{r=-\infty}^{\infty} x_2[r]z^{-(r+m)}
-$$
-$$
-= \sum_{m=-\infty}^{\infty} x_1[m]z^{-m} \sum_{r=-\infty}^{\infty} x_2[r]z^{-r}
-$$
-$$
-= X_1[z]X_2[z]
-$$
-
-### LTID SYSTEM RESPONSE
-
-It is interesting to apply the time-convolution property to the LTID input–output equation *y*[*n*] = *x*[*n*] ∗ *h*[*n*]. Since *h*[*n*] ⇐⇒ *H*[*z*], it follows from Eq. (5.19) that
-
-$$
-Y[z] = X[z]H[z] \tag{5.20}
-$$
-
-### **DR ILL 5.9 Using the Convolution Property**
-
-Use the time-convolution property and appropriate pairs in Table 5.1 to show that *u*[*n*] ∗ *u*[*n*−1] = *nu*[*n*].
-
-### INITIAL AND FINAL VALUES
-
-For a causal *x*[*n*], the initial value theorem states that
-
-$$
-x[0] = \lim_{z \to \infty} X[z]
-$$
-
-This result follows immediately from Eq. (5.7).
-
-If (*z*−1)*X*[*z*] has no poles outside the unit circle, then the final value theorem states that
-
-$$
-\lim_{N \to \infty} x[N] = \lim_{z \to 1} (z - 1)X[z]
-$$
-
-This can be shown from the fact that
-
-$$
-x[n] - x[n-1] \Longleftrightarrow \left\{1 - \frac{1}{z}\right\} X[z] = \frac{(z-1)X[z]}{z}
-$$
-
-and
-
-$$
-\frac{(z-1)X[z]}{z} = \sum_{n=-\infty}^{\infty} \{x[n] - x[n-1]\}z^{-n}
-$$
-
-$$
-\lim_{z \to 1} \frac{(z-1)X[z]}{z} = \lim_{z \to 1} (z-1)X[z] = \lim_{z \to 1} \lim_{N \to \infty} \sum_{n=-\infty}^{N} \{x[n] - x[n-1]\}z^{-n} = \lim_{N \to \infty} x[N]
-$$
-
-All these properties of the *z*-transform are listed in Table 5.2.
-
-| Operation             | x[n]               | X[z]                                                        |
-|-----------------------|--------------------|-------------------------------------------------------------|
-| Addition              | x1[n] +x2[n]       | X1[z] +X2[z]                                                |
-| Scalar multiplication | ax[n]              | aX[z]                                                       |
-| Right shifting        | x[n−m]u[n−m]       | 1<br>zm X[z]                                                |
-|                       | x[n−m]u[n]         | "m<br>1<br>1<br>n<br>zm X[z] +<br>x[−n]z<br>zm<br>n=1       |
-|                       | x[n−1]u[n]         | 1<br>X[z] +x[−1]<br>z                                       |
-|                       | x[n−2]u[n]         | 1<br>1<br>z2 X[z] +<br>x[−1] +x[−2]<br>z                    |
-|                       | x[n−3]u[n]         | 1<br>1<br>1<br>z3 X[z] +<br>z2 x[−1] +<br>x[−2] +x[−3]<br>z |
-| Left shifting         | x[n+m]u[n]         | m<br>"−1<br>−n<br>zmX[z]<br>−zm<br>x[n]z                    |
-|                       | x[n+1]u[n]         | n=0<br>zX[z] −zx[0]                                         |
-|                       | x[n+2]u[n]         | z2X[z]<br>−z2x[0]<br>−zx[1]                                 |
-|                       | x[n+3]u[n]         | z3X[z]<br>−z3x[0]<br>−z2x[1]<br>−zx[2]                      |
-| Multiplication by γ n | γ nx[n]u[n]        | z<br>!<br>X<br>γ                                            |
-| Multiplication by n   | nx[n]u[n]          | d<br>−z<br>X[z]<br>dz                                       |
-| Time reversal         | x[−n]              | X[1/z]                                                      |
-| Time convolution      | x1[n] ∗ x2[n]      | X1[z]X2[z]                                                  |
-| Initial value         | x[0]               | lim<br>X[z]<br>z→∞                                          |
-| Final value           | lim<br>x[N]<br>N→∞ | lim<br>(z−1)X[z]<br>Poles of (z −1)X[z]<br>z→1              |
-|                       |                    | inside the unit circle                                      |
-
-**TABLE 5.2** *z*-Transform Properties
-
-and
-
-## <span id="page-529-0"></span>**5.3** *z***-TRANSFORM [SOLUTION OF](#page-11-0) LINEAR DIFFERENCE EQUATIONS**
-
-The time-shifting (left-shift or right-shift) property has set the stage for solving linear difference equations with constant coefficients. As in the case of the Laplace transform with differential equations, the *z*-transform converts difference equations into algebraic equations that are readily solved to find the solution in the *z* domain. Taking the inverse *z*-transform of the *z*-domain solution yields the desired time-domain solution. The following examples demonstrate the procedure.
-
-### **EXAMPLE 5.5** *z***-Transform Solution of a Linear Difference Equation**
-
-Solve
-
-$$
-y[n+2] - 5y[n+1] + 6y[n] = 3x[n+1] + 5x[n]
-$$
-
-if the initial conditions are *y*[−1] = 11/6, *y*[−2] = 37/36, and the input *x*[*n*] = (2)−*nu*[*n*].
-
-As we shall see, difference equations can be solved by using the right-shift or the leftshift property. Because the difference equation here is in advance form, the use of the left-shift property in Eq. (5.16) may seem appropriate for its solution. Unfortunately, this left-shift property requires a knowledge of auxiliary conditions *y*[0], *y*[1], ... , *y*[*N* − 1] rather than of the initial conditions *y*[−1], *y*[−2], ... , *y*[−*n*], which are generally given. This difficulty can be overcome by expressing the difference equation in delay form (obtained by replacing *n* with *n*−2) and then using the right-shift property.† The resulting delay-form difference equation is
-
-$$
-y[n] - 5y[n-1] + 6y[n-2] = 3x[n-1] + 5x[n-2]
-$$
-\n(5.21)
-
-We now use the right-shift property to take the *z*-transform of this equation. But before proceeding, we must be clear about the meaning of a term like *y*[*n* − 1] here. Does it mean *y*[*n* − 1]*u*[*n* − 1] or *y*[*n* − 1]*u*[*n*]? In any equation, we must have some time reference *n* = 0, and every term is referenced from this instant. Hence, *y*[*n*−*k*] means *y*[*n*−*k*]*u*[*n*]. Remember also that although we are considering the situation for *n* ≥ 0, *y*[*n*] is present even before *n* = 0 (in the form of initial conditions). Now
-
-$$
-y[n]u[n] \Longleftrightarrow Y[z]
-$$
-  
-\n
-$$
-y[n-1]u[n] \Longleftrightarrow \frac{1}{z}Y[z] + y[-1] = \frac{1}{z}Y[z] + \frac{11}{6}
-$$
-  
-\n
-$$
-y[n-2]u[n] \Longleftrightarrow \frac{1}{z^2}Y[z] + \frac{1}{z}y[-1] + y[-2] = \frac{1}{z^2}Y[z] + \frac{11}{6z} + \frac{37}{36}
-$$
-
-Noting that for causal input *x*[*n*],
-
-$$
-x[-1] = x[-2] = \cdot \cdot \cdot = x[-n] = 0
-$$
-
-<sup>†</sup> Another approach is to find *<sup>y</sup>*[0], *<sup>y</sup>*[1], *<sup>y</sup>*[2], ... , *<sup>y</sup>*[*<sup>N</sup>* <sup>−</sup> <sup>1</sup>] from *<sup>y</sup>*[−1], *<sup>y</sup>*[−2], ... , *<sup>y</sup>*[−*n*] iteratively, as in Sec. 3.5-1, and then apply the left-shift property to the advance-form difference equation.
-
-We obtain
-
-$$
-x[n] = (2)^{-n}u[n] = (2^{-1})^{n}u[n] = (0.5)^{n}u[n] \Longleftrightarrow \frac{z}{z - 0.5}
-$$
-
-$$
-x[n-1]u[n] \Longleftrightarrow \frac{1}{z}X[z] + x[-1] = \frac{1}{z}\frac{z}{z-0.5} + 0 = \frac{1}{z-0.5}
-$$
-$$
-x[n-2]u[n] \Longleftrightarrow \frac{1}{z^2}X[z] + \frac{1}{z}x[-1] + x[-2] = \frac{1}{z^2}X[z] + 0 + 0 = \frac{1}{z(z-0.5)}
-$$
-
-In general,
-
-$$
-x[n - r]u[n] \Longleftrightarrow \frac{1}{z^r}X[z]
-$$
-
-Taking the *z*-transform of Eq. (5.21) and substituting the foregoing results, we obtain
-
-$$
-Y[z] - 5\left[\frac{1}{z}Y[z] + \frac{11}{6}\right] + 6\left[\frac{1}{z^2}Y[z] + \frac{11}{6z} + \frac{37}{36}\right] = \frac{3}{z - 0.5} + \frac{5}{z(z - 0.5)}
-$$
-
-or
-
-<sup>1</sup><sup>−</sup> <sup>5</sup> *z* + 6 *z*2 *<sup>Y</sup>*[*z*] − <sup>3</sup><sup>−</sup> <sup>11</sup> *z* <sup>=</sup> <sup>3</sup> *<sup>z</sup>*−0.5 <sup>+</sup> 5 *<sup>z</sup>*(*z*−0.5) (5.22)
-
-from which we obtain
-
-$$
-(z2 - 5z + 6)Y[z] = \frac{z(3z2 - 9.5z + 10.5)}{(z - 0.5)}
-$$
-
-so that
-
-$$
-Y[z] = \frac{z(3z^2 - 9.5z + 10.5)}{(z - 0.5)(z^2 - 5z + 6)}
-$$
-
-and
-
-$$
-\frac{Y[z]}{z} = \frac{3z^2 - 9.5z + 10.5}{(z - 0.5)(z - 2)(z - 3)} = \frac{(26/15)}{z - 0.5} - \frac{(7/3)}{z - 2} + \frac{(18/5)}{z - 3}
-$$
-
-Therefore,
-
-$$
-Y[z] = \frac{26}{15} \left(\frac{z}{z - 0.5}\right) - \frac{7}{3} \left(\frac{z}{z - 2}\right) + \frac{18}{5} \left(\frac{z}{z - 3}\right)
-$$
-$$
-y[n] = \left[\frac{26}{15} (0.5)^n - \frac{7}{3} (2)^n + \frac{18}{5} (3)^n\right] u[n] \tag{5.23}
-$$
-
-and
-
-This example demonstrates the ease with which linear difference equations with constant coefficients can be solved by the *z*-transform. This method is general: it can be used to solve a single difference equation or a set of simultaneous difference equations of any order as long as the equations are linear with constant coefficients.
-
-### **Comment.**
-
-Sometimes, instead of initial conditions *y*[−1], *y*[−2], ... , *y*[−*n*], auxiliary conditions *y*[0], *y*[1], ... , *y*[*N* − 1] are given to solve a difference equation. In this case, the equation can be solved by expressing it in the advance form and then using the left-shift property (see Drill 5.11).
-
-### **DR ILL 5.10** *z***-Transform Solution of a Linear Difference Equation**
-
-Solve the following equation if the initial conditions *y*[−1] = 2, *y*[−2] = 0, and the input *x*[*n*] = *u*[*n*]:
-
-$$
-y[n+2] - \frac{5}{6}y[n+1] + \frac{1}{6}y[n] = 5x[n+1] - x[n]
-$$
-
-### **ANSWER**
-
-*<sup>y</sup>*[*n*] = <sup>12</sup>−15 <sup>1</sup> 2 *n* <sup>+</sup> <sup>14</sup> 3 1 3 *<sup>n</sup> u*[*n*]
-
-## **DR ILL 5.11 Difference Equation Solution Using** *y*[**0**]**,** *y*[**1**]**,** ... **,** *y*[*N* −**1**]
-
-Solve the following equation if the auxiliary conditions are *y*[0] = 1, *y*[1] = 2, and the input *x*[*n*] = *u*[*n*]:
-
-*y*[*n*] +3*y*[*n*−1] +2*y*[*n*−2] = *x*[*n*−1] +3*x*[*n*−2]
-
-### **ANSWER**
-
-*<sup>y</sup>*[*n*] = <sup>2</sup> <sup>3</sup> <sup>+</sup>2(−1)*<sup>n</sup>* <sup>−</sup> <sup>5</sup> <sup>3</sup> (−2)*<sup>n</sup> u*[*n*]
-
-### ZERO-INPUT AND ZERO-STATE COMPONENTS
-
-In Ex. 5.5 we found the total solution of the difference equation. It is relatively easy to separate the solution into zero-input and zero-state components. All we have to do is to separate the response into terms arising from the input and terms arising from initial conditions (IC). We can separate the response in Eq. (5.22) as follows:
-
-$$
-\left(1 - \frac{5}{z} + \frac{6}{z^2}\right)Y[z] - \underbrace{\left(3 - \frac{11}{z}\right)}_{\text{IC terms}} = \underbrace{\frac{3}{z - 0.5} + \frac{5}{z(z - 0.5)}}_{\text{input terms}}
-$$
-
-Therefore,
-
-$$
-\left(1 - \frac{5}{z} + \frac{6}{z^2}\right)Y[z] = \underbrace{\left(3 - \frac{11}{z}\right)}_{\text{IC terms}} + \underbrace{\frac{(3z + 5)}{z(z - 0.5)}}_{\text{input terms}}
-$$
-
-Multiplying both sides by *z*<sup>2</sup> yields
-
-$$
-(z2 - 5z + 6)Y[z] = \underbrace{z(3z - 11)}_{\text{IC terms}} + \underbrace{\frac{z(3z + 5)}{z - 0.5}}_{\text{input terms}}
-$$
-
-and
-
-$$
-Y[z] = \underbrace{\frac{z(3z-11)}{z^2 - 5z + 6}}_{\text{zero-input response}} + \underbrace{\frac{z(3z+5)}{(z-0.5)(z^2 - 5z + 6)}}_{\text{zero-state response}}
-$$
-
-We expand both terms on the right-hand side into modified partial fractions to yield
-
-$$
-Y[z] = \underbrace{\left[5\left(\frac{z}{z-2}\right) - 2\left(\frac{z}{z-3}\right)\right]}_{\text{zero-input response}} + \underbrace{\left[\frac{26}{15}\left(\frac{z}{z-0.5}\right) - \frac{22}{3}\left(\frac{z}{z-2}\right) + \frac{28}{5}\left(\frac{z}{z-3}\right)\right]}_{\text{zero-state response}}
-$$
-
-and
-
-$$
-y[n] = \underbrace{(5(2)^n - 2(3)^n) u[n]}_{\text{zero-input response}} + \underbrace{\left(\frac{26}{15}(0.5)^n - \frac{22}{3}(2)^n + \frac{28}{5}(3)^n\right) u[n]}_{\text{zero-state response}}
-$$
-  
-= 
-$$
-\left[-\frac{7}{3}(2)^n + \frac{18}{5}(3)^n + \frac{26}{15}(0.5)^n\right] u[n]
-$$
-
-which agrees with the result in Eq. (5.23).
-
-### **DR ILL 5.12 Separating Zero-Input and Zero-State Responses**
-
-Solve
-
-$$
-y[n+2] - \frac{5}{6}y[n+1] + \frac{1}{6}y[n] = 5x[n+1] - x[n]
-$$
-
-if the initial conditions are *y*[−1] = 2, *y*[−2] = 0, and the input *x*[*n*] = *u*[*n*]. Separate the response into zero-input and zero-state responses.
-
-### **ANSWER**
-
-$$
-y[n] = \underbrace{\left(3\left(\frac{1}{2}\right)^n - \frac{4}{3}\left(\frac{1}{3}\right)^n\right)u[n]}_{\text{zero-input response}} + \underbrace{\left(12 - 18\left(\frac{1}{2}\right)^n + 6\left(\frac{1}{3}\right)^n\right)u[n]}_{\text{zero-state response}}
-$$
-\n
-$$
-= \left[12 - 15\left(\frac{1}{2}\right)^n + \frac{14}{3}\left(\frac{1}{3}\right)^n\right]u[n]
-$$
-
-### <span id="page-533-0"></span>**[5.3-1 Zero-State Response of LTID Systems: The Transfer Function](#page-11-0)**
-
-Consider an *N*th-order LTID system specified by the difference equation
-
-$$
-Q[E]y[n] = P[E]x[n]
-$$
-
-or
-
-$$
-(EN + a1EN-1 + \dots + aN-1E + aN)y[n]
-$$
-  
-= (b<sub>0</sub>E<sup>N</sup> + b<sub>1</sub>E<sup>N-1</sup> + \dots + b<sub>N-1</sub>E + b<sub>N</sub>)x[n]
-
-or
-
-$$
-y[n+N] + a_1y[n+N-1] + \cdots + a_{N-1}y[n+1] + a_Ny[n]
-$$
-  
-=  $b_0x[n+N] + \cdots + b_{N-1}x[n+1] + b_Nx[n]$  (5.24)
-
-We now derive the general expression for the zero-state response: that is, the system response to input *x*[*n*] when all the initial conditions *y*[−1] = *y*[−2]=···= *y*[−*N*] = 0 (zero state). The input *x*[*n*] is assumed to be causal so that *x*[−1] = *x*[−2]=···= *x*[−*N*] = 0.
-
-Equation (5.24) can be expressed in delay form as
-
-$$
-y[n] + a_1y[n-1] + \dots + a_Ny[n-N] = b_0x[n] + b_1x[n-1] + \dots + b_Nx[n-N]
-$$
- (5.25)
-
-Because *y*[−*r*] = *x*[−*r*] = 0 for *r* = 1, 2,...,*N*,
-
-$$
-y[n-m]u[n] \Longleftrightarrow \frac{1}{z^m}Y[z]
-$$
-  
-$$
-x[n-m]u[n] \Longleftrightarrow \frac{1}{z^m}X[z] \qquad m=1,2,\ldots,N
-$$
-
-Now the *z*-transform of Eq. (5.25) is given by
-
-$$
-\left(1+\frac{a_1}{z}+\frac{a_2}{z^2}+\cdots+\frac{a_N}{z^N}\right)Y[z] = \left(b_0+\frac{b_1}{z}+\frac{b_2}{z^2}+\cdots+\frac{b_N}{z^N}\right)X[z]
-$$
-
-Multiplication of both sides by *z<sup>N</sup>* yields
-
-$$
-(zN + a1zN-1 + \dots + aN-1z + aN)Y[z]
-$$
-  
-= (b<sub>0</sub>z<sup>N</sup> + b<sub>1</sub>z<sup>N-1</sup> + \dots + b<sub>N-1</sub>z + b<sub>N</sub>)X[z]
-
-Therefore,
-
-$$
-Y[z] = \left(\frac{b_0 z^N + b_1 z^{N-1} + \dots + b_{N-1} z + b_N}{z^N + a_1 z^{N-1} + \dots + a_{N-1} z + a_N}\right) X[z]
-$$
-  
-=  $\frac{P[z]}{Q[z]} X[z]$ 
-
-We have shown in Eq. (5.20) that *Y*[*z*] = *X*[*z*]*H*[*z*]. Hence, it follows that
-
-$$
-H[z] = \frac{P[z]}{Q[z]} = \frac{b_0 z^N + b_1 z^{N-1} + \dots + b_{N-1} z + b_N}{z^N + a_1 z^{N-1} + \dots + a_{N-1} z + a_N}
-$$
-(5.26)
-
-As in the case of LTIC systems, this result leads to an alternative definition of the LTID system transfer function as the ratio of *Y*[*z*] to *X*[*z*] (assuming all initial conditions zero).
-
-$$
-H[z] \equiv \frac{Y[z]}{X[z]} = \frac{\mathcal{Z}[zero-state response]}{\mathcal{Z}[input]}
-$$
-
-### ALTERNATE INTERPRETATION OF THE *z*-TRANSFORM
-
-So far we have treated the *z*-transform as a machine that converts linear difference equations into algebraic equations. There is no physical understanding of how this is accomplished or what it means. We now discuss more intuitive interpretation and meaning of the *z*-transform.
-
-In Ch. 3, Eq. (3.38), we showed that the LTID system response to an everlasting exponential *z<sup>n</sup>* is *H*[*z*]*z<sup>n</sup>*. If we could express every discrete-time signal as a linear combination of everlasting exponentials of the form *z<sup>n</sup>*, we could readily obtain the system response to any input. For example, if
-
-$$
-x[n] = \sum_{k=1}^{K} X[z_k] z_k^n
-$$
-\n(5.27)
-
-the response of an LTID system to this input is given by
-
-$$
-y[n] = \sum_{k=1}^{K} X[z_k]H[z_k]z_k^n
-$$
-
-Unfortunately, a very small class of signals can be expressed in the form of Eq. (5.27). However, we can express almost all signals of practical utility as a sum of everlasting exponentials over a continuum of values of *z*. This is precisely what the *z*-transform in Eq. (5.2) does.
-
-$$
-x[n] = \frac{1}{2\pi j} \oint X[z] z^{n-1} dz
-$$
-\n(5.28)
-
-Invoking the linearity property of the *z*-transform, we can find the system response *y*[*n*] to input *x*[*n*] in Eq. (5.28) as†
-
-$$
-y[n] = \frac{1}{2\pi j} \oint X[z]H[z]z^{n-1} dz = \mathcal{Z}^{-1}{X[z]H[z]}
-$$
-
-Clearly,
-
-*Y*[*z*] = *X*[*z*]*H*[*z*]
-
-This viewpoint of finding the response of LTID system is illustrated in Fig. 5.6a. Just as in continuous-time systems, we can model discrete-time systems in the transformed manner by
-
-<sup>†</sup> In computing *<sup>y</sup>*[*n*], the contour along which the integration is performed is modified to consider the ROC of *X*[*z*] as well as *H*[*z*]. We ignore this consideration in this intuitive discussion.
-
-**Figure 5.6** The transformed representation of an LTID system.
-
-representing all signals by their *z*-transforms and all system components (or elements) by their transfer functions, as shown in Fig. 5.6b.
-
-The result *Y*[*z*] = *H*[*z*]*X*[*z*] greatly facilitates derivation of the system response to a given input. We shall demonstrate this assertion by an example.
-
-### **EXAMPLE 5.6 Transfer Function to Find the Zero-State Response**
-
-Find the response *y*[*n*] of an LTID system described by the difference equation
-
-$$
-y[n+2] + y[n+1] + 0.16y[n] = x[n+1] + 0.32x[n]
-$$
-
-or
-
-$$
-(E2 + E + 0.16)y[n] = (E + 0.32)x[n]
-$$
-
-for the input *x*[*n*] = (−2)−*nu*[*n*] and with all the initial conditions zero (system in the zero state).
-
-From the difference equation, we find
-
-$$
-H[z] = \frac{P[z]}{Q[z]} = \frac{z + 0.32}{z^2 + z + 0.16}
-$$
-
-For the input *x*[*n*] = (−2)−*nu*[*n*]=[(−2)−<sup>1</sup>] *nu*(*n*) = (−0.5)*nu*[*n*],
-
-$$
-X[z] = \frac{z}{z+0.5}
-$$
-
-and
-
-$$
-Y[z] = X[z]H[z] = \frac{z(z+0.32)}{(z^2 + z + 0.16)(z+0.5)}
-$$
-
-Therefore,
-
-$$
-\frac{Y[z]}{z} = \frac{(z+0.32)}{(z^2+z+0.16)(z+0.5)} = \frac{(z+0.32)}{(z+0.2)(z+0.8)(z+0.5)}
-$$
-$$
-= \frac{2/3}{z+0.2} - \frac{8/3}{z+0.8} + \frac{2}{z+0.5}
-$$
-
-so that
-
-$$
-Y[z] = \frac{2}{3} \left( \frac{z}{z+0.2} \right) - \frac{8}{3} \left( \frac{z}{z+0.8} \right) + 2 \left( \frac{z}{z+0.5} \right)
-$$
-
-and
-
-$$
-y[n] = \left[\frac{2}{3}(-0.2)^n - \frac{8}{3}(-0.8)^n + 2(-0.5)^n\right]u[n]
-$$
-
-### **EXAMPLE 5.7 Transfer Function of a Unit Delay**
-
-Show that the transfer function of a unit delay is 1/*z*.
-
-If the input to the unit delay is *x*[*n*]*u*[*n*], then its output (Fig. 5.7) is given by
-
-*y*[*n*] = *x*[*n*−1]*u*[*n*−1]
-
-The *z*-transform of this equation yields [see Eq. (5.12)]
-
-$$
-Y[z] = \frac{1}{z}X[z] = H[z]X[z]
-$$
-
-It follows that the transfer function of the unit delay is
-
-$$
-H[z] = \frac{1}{z}
-$$
-
-*x*[*n*]*u*[*n*] *X*[*z*] *x*[*n* - 1]*u*[*n* - 1] *<sup>Y</sup>*[*z*] *<sup>X</sup>*[*z*] <sup>1</sup> *z* 1 *z* **Figure 5.7** Ideal unit delay and its transfer function.
-
-### <span id="page-537-0"></span>**DR ILL 5.13 Transfer Function to Find Zero-State Response and Difference Equation**
-
-A discrete-time system is described by the following transfer function:
-
-$$
-H[z] = \frac{z - 0.5}{(z + 0.5)(z - 1)}
-$$
-
-- **(a)** Find the system response to input *x*[*n*] = 3−(*n*+1) *u*[*n*] if all initial conditions are zero.
-- **(b)** Write the difference equation relating the output *y*[*n*] to input *x*[*n*] for this system.
-
-### **ANSWERS**
-
-- **(a)** *<sup>y</sup>*[*n*] = <sup>1</sup> 3 1 <sup>2</sup> <sup>−</sup>0.8(−0.5)*<sup>n</sup>* <sup>+</sup>0.3 <sup>1</sup> 3 *<sup>n</sup> u*[*n*]
-- **(b)** *y*[*n*+2] −0.5*y*[*n*+1] −0.5*y*[*n*] = *x*[*n*+1] −0.5*x*[*n*]
-
-### **[5.3-2 Stability](#page-11-0)**
-
-Equation (5.26) shows that the denominator of *H*[*z*] is *Q*[*z*], which is apparently identical to the characteristic polynomial *Q*[γ ] defined in Ch. 3. Does this mean that the denominator of *H*[*z*] is the characteristic polynomial of the system? This may or may not be the case: if *P*[*z*] and *Q*[*z*] in Eq. (5.26) have any common factors, they cancel out, and the effective denominator of *H*[*z*] is not necessarily equal to *Q*[*z*]. Recall also that the system transfer function *H*[*z*], like *h*[*n*], is defined in terms of measurements at the external terminals. Consequently, *H*[*z*] and *h*[*n*] are both external descriptions of the system. In contrast, the characteristic polynomial *Q*[*z*] is an internal description. Clearly, we can determine only external stability, that is, BIBO stability, from *H*[*z*]. If all the poles of *H*[*z*] are within the unit circle, all the terms in *h*[*n*] are decaying exponentials, and as shown in Sec. 3.9, *h*[*n*] is absolutely summable. Consequently, the system is BIBO-stable. Otherwise the system is BIBO-unstable.
-
-If *P*[*z*] and *Q*[*z*] do not have common factors, then the denominator of *H*[*z*] is identical to *Q*[*z*]. † The poles of *H*[*z*] are the characteristic roots of the system. We can now determine internal stability. The internal stability criterion in Sec. 3.9-2 can be restated in terms of the poles of *H*[*z*], as follows.
-
-- 1. An LTID system is asymptotically stable if and only if all the poles of its transfer function *H*[*z*] are within the unit circle. The poles may be repeated or simple.
-- 2. An LTID system is unstable if and only if either one or both of the following conditions exist: (i) at least one pole of *H*[*z*] is outside the unit circle; (ii) there are repeated poles of *H*[*z*] on the unit circle.
-
-<sup>†</sup> There is no way of determining whether any common factors in *<sup>P</sup>*[*z*] and *<sup>Q</sup>*[*z*] were canceled out. This is because in our derivation of *H*[*z*], we generally get the final result after the cancellations have been effected. When we use internal description of the system to derive *Q*[*z*], however, we find pure *Q*[*z*] unaffected by any common factor in *P*[*z*].
-
-<span id="page-538-0"></span>3. An LTID system is marginally stable if and only if there are no poles of *H*[*z*] outside the unit circle, and there are some simple poles on the unit circle.
-
-### **DR ILL 5.14 Transfer Function to Determine Stability**
-
-Show that an *accumulator* whose impulse response is *h*[*n*] = *u*[*n*] is marginally stable but BIBO-unstable.
-
-### **[5.3-3 Inverse Systems](#page-11-0)**
-
-If *H*[*z*] is the transfer function of a system *S*, then *Si*, its inverse system, has a transfer function *Hi*[*z*] given by
-
-$$
-H_i[z] = \frac{1}{H[z]}
-$$
-
-This follows from the fact the inverse system *S<sup>i</sup>* undoes the operation of *S*. Hence, if *H*[*z*] is placed in cascade with *Hi*[*z*], the transfer function of the composite system (identity system) is unity. For example, an *accumulator* whose transfer function is *H*[*z*] = *z*/(*z* − 1) and a *backward difference system* whose transfer function is *Hi*[*z*] = (*z*−1)/*z* are inverse of each other. Similarly if
-
-$$
-H[z] = \frac{z - 0.4}{z - 0.7}
-$$
-
-its inverse system transfer function is
-
-$$
-H_i[z] = \frac{z - 0.7}{z - 0.4}
-$$
-
-as required by the property *H*[*z*]*Hi*[*z*] = 1. Hence, it follows that
-
-$$
-h[n] * h_i[n] = \delta[n]
-$$
-
-### **DR ILL 5.15 Inverse Systems**
-
-Find the impulse responses of an accumulator and a first-order backward difference system. Show that the convolution of the two impulse responses yields δ[*n*].
-
-## **5.4 SYSTEM [REALIZATION](#page-11-0)**
-
-Because of the similarity between LTIC and LTID systems, conventions for block diagrams and rules of interconnection for LTID are identical to those for continuous-time (LTIC) systems. It is not necessary to rederive these relationships. We shall merely restate them to refresh the reader's memory.
-
-The block diagram representations of the basic operations, such as an adder, a scalar multiplier, unit delay, and pickoff points, is shown in Fig. 3.13. In our development, the unit delay, which is represented by a box marked D in Fig. 3.13, will be represented by its transfer function 1/*z*. All the signals will also be represented in terms of their *z*-transforms. Thus, the input and the output will be labeled *X*[*z*] and *Y*[*z*], respectively.
-
-When two systems with transfer functions *H*1[*z*] and *H*2[*z*] are connected in cascade (as in Fig. 4.18b), the transfer function of the composite system is *H*1[*z*]*H*2[*z*]. If the same two systems are connected in parallel (as in Fig. 4.18c), the transfer function of the composite
-
-**Figure 5.8** Realization of an *N*th-order causal LTID system transfer function by using **(a)** DFI, **(b)** canonic direct (DFII), and **(c)** the transpose form of DFII.
-
-system is *H*1[*z*] + *H*2[*z*]. For a feedback system (as in Fig. 4.18d), the transfer function is *G*[*z*]/(1+*G*[*z*]*H*[*z*]).
-
-We now consider a systematic method for realization (or simulation) of an arbitrary *N*th-order LTID transfer function. Since realization is basically a synthesis problem, there is no unique way of realizing a system. A given transfer function can be realized in many different ways. We present here the two forms of *direct realization*. Each of these forms can be executed in several other ways, such as cascade and parallel. Furthermore, a system can be realized by the transposed version of any known realization of that system. This artifice doubles the number of system realizations. A transfer function *H*[*z*] can be realized by using time delays along with adders and multipliers.
-
-We shall consider a realization of a general *N*th-order causal LTID system, whose transfer function is given by
-
-$$
-H[z] = \frac{b_0 z^N + b_1 z^{N-1} + \dots + b_{N-1} z + b_N}{z^N + a_1 z^{N-1} + \dots + a_{N-1} z + a_N}
-$$
-(5.29)
-
-This equation is identical to the transfer function of a general *N*th-order proper LTIC system given in Eq. (4.36). The only difference is that the variable *z* in the former is replaced by the variable *s* in the latter. Hence, the procedure for realizing an LTID transfer function is identical to that for the LTIC transfer function with the basic element 1/*s* (integrator) replaced by the element 1/*z* (unit delay). The reader is encouraged to follow the steps in Sec. 4.6 and rederive the results for the LTID transfer function in Eq. (5.29). Here we shall merely reproduce the realizations from Sec. 4.6 with integrators (1/*s*) replaced by unit delays (1/*z*).
-
-The direct form I (DFI) is shown in Fig. 5.8a, the canonic direct form (DFII) is shown in Fig. 5.8b and the transpose of canonic direct is shown in Fig. 5.8c. The DFII and its transpose are canonic because they require *N* delays, which is the minimum number needed to implement the *N*th-order LTID transfer function in Eq. (5.29). In contrast, the form DFI is a noncanonic because it generally requires 2*N* delays. The DFII realization in Fig. 5.8b is also called a *canonic direct* form.
-
-### **EXAMPLE 5.8 Canonical Realizations of Transfer Functions**
-
-Find the canonic direct and the transposed canonic direct realizations of the following transfer functions: **(a)** 2 *z*+5 , **(b)** 4*z*+28 *<sup>z</sup>*+<sup>1</sup> , **(c)** *<sup>z</sup> z*+7 , and **(d)** 4*z*+28 *z*<sup>2</sup> +6*z*+5 .
-
-All four of these transfer functions are special cases of *H*[*z*] in Eq. (5.29). **(a)**
-
-$$
-H[z] = \frac{2}{z+5}
-$$
-
-For this case, the transfer function is of the first order (*N* = 1); therefore, we need only one delay for its realization. The feedback and feedforward coefficients are
-
-$$
-a_1 = 5
-$$
- and  $b_0 = 0$ ,  $b_1 = 2$ 
-
-### 522 CHAPTER 5 DISCRETE-TIME SYSTEM ANALYSIS USING THE *Z*-TRANSFORM
-
-We use Fig. 5.8 as our model and reduce it to the case of *N* = 1. Figure 5.9a shows the canonic direct (DFII) form, and Fig. 5.9b its transpose. The two realizations are almost the same. The minor difference is that in the DFII form, the gain 2 is provided at the output, and in the transpose, the same gain is provided at the input.
-
-**Figure 5.9** Realization of transfer function 2/(*z*+5): **(a)** canonic direct form and **(b)** its transpose.
-
-In a similar way, we realize the remaining transfer functions. **(b)**
-
-$$
-H[z] = \frac{4z + 28}{z + 1}
-$$
-
-In this case also, the transfer function is of the first order (*N* = 1); therefore, we need only one delay for its realization. The feedback and feedforward coefficients are
-
-$$
-a_1 = 1
-$$
- and  $b_0 = 4$ ,  $b_1 = 28$ 
-
-Figure 5.10 illustrates the canonic direct and its transpose for this case.†
-
-**Figure 5.10** Realization of (4*z* +28)/(*z* +1): **(a)** canonic direct form and **(b)** its transpose.
-
-$$
-H[z] = \frac{4z + 28}{z+1} = 4 + \frac{24}{z+1}
-$$
-
-Hence, this transfer function can also be realized as two transfer functions in parallel.
-
-<sup>†</sup> Transfer functions with *<sup>N</sup>* <sup>=</sup> *<sup>M</sup>* may also be expressed as a sum of a constant and a strictly proper transfer function. For example,
-
-$$
-H[z] = \frac{z}{z+7}
-$$
-
-Here *N* = 1 and *b*<sup>0</sup> = 1,*b*<sup>1</sup> = 0 and *a*<sup>1</sup> = 7. Figure 5.11 shows the direct and the transposed realizations. Observe that the realizations are almost alike.
-
-**Figure 5.11** Realization of *z*/(*z* +7): **(a)** canonic direct form and **(b)** its transpose.

engineering/linear-systems-and-signals-3rd-edition-bp-lathi/069_5.3 z-TRANSFORM SOLUTION OF LINEAR DIFFERENCE EQUATIONS.md

@@ -1,55 +0,0 @@
-**(d)**
-
-$$
-H[z] = \frac{4z + 28}{z^2 + 6z + 5}
-$$
-
-This is a second-order system (*N* = 2) with *b*<sup>0</sup> = 0, *b*<sup>1</sup> = 4, *b*<sup>2</sup> = 28, *a*<sup>1</sup> = 6, *a*<sup>2</sup> = 5. Figure 5.12 shows the canonic direct and transposed canonic direct realizations.
-
-**Figure 5.12** Realization of (4*z* +28)/(*z*<sup>2</sup> +6*z*+5): **(a)** canonic direct form and **(b)** its transpose.
-
-**(c)**
-
-### **DR ILL 5.16 Realization of a Second-Order Transfer Function**
-
-Realize the transfer function
-
-$$
-H[z] = \frac{2z}{z^2 + 6z + 25}
-$$
-
-### REALIZATION OF FINITE IMPULSE RESPONSE (FIR) FILTERS
-
-So far we have been quite general in our development of realization techniques. They can be applied to infinite impulse response (IIR) or FIR filters. For FIR filters, the coefficients *ai* = 0 for all *i* = 0.† Hence, FIR filters can be readily implemented by means of the schemes developed so far by eliminating all branches with *ai* coefficients. The condition *ai* = 0 implies that all the poles of a FIR filter are at *z* = 0.
-
-### **EXAMPLE 5.9 Realization of an FIR Filter**
-
-Realize *H*[*z*] = (*z*<sup>3</sup> +4*z*<sup>2</sup> +5*z*+2)/*z*<sup>3</sup> using canonic direct and transposed forms.
-
-<sup>†</sup> This statement is true for all *<sup>i</sup>* = 0 because *<sup>a</sup>*<sup>0</sup> is assumed to be unity.
-
-For *H*[*z*], *b*<sup>0</sup> = 1, *b*<sup>1</sup> = 4, *b*<sup>2</sup> = 5, and *b*<sup>3</sup> = 2. Hence, we obtain the canonic direct realization, shown in Fig. 5.13a. We have shown the horizontal orientation because it is easier to see that this filter is basically a tapped delay line. That is why this structure is also known as a *tapped delay line* or *transversal filter*. Figure 5.13b shows the corresponding transposed implementation.
-
-## CASCADE AND PARALLEL REALIZATIONS, COMPLEX AND REPEATED POLES
-
-The considerations and observations for cascade and parallel realizations as well as complex and multiple poles are identical to those discussed for LTIC systems in Sec. 4.6-3.
-
-## **DR ILL 5.17 Cascade and Parallel Realizations of a Transfer Function**
-
-Find canonic direct realizations of the following transfer function by using the cascade and parallel forms. The specific cascade decomposition is as follows:
-
-$$
-H[z] = \frac{z+3}{z^2 + 7z + 10} = \left(\frac{z+3}{z+2}\right)\left(\frac{1}{z+5}\right)
-$$
-
-### DO ALL REALIZATIONS LEAD TO THE SAME PERFORMANCE?
-
-For a given transfer function, we have presented here several possible different realizations (DFI, canonic form DFII, and its transpose). There are also cascade and parallel versions, and there are many possible grouping of the factors in the numerator and the denominator of *H*[*z*], leading to different realizations. We can also use various combinations of these forms in implementing different subsections of a system. Moreover, the transpose of each version doubles the number. However, this discussion by no means exhausts all the possibilities. Transforming variables affords limitless potential realizations of the same transfer function.
-
-Theoretically, all these realizations are equivalent; that is, they lead to the same transfer function. This, however, is true only when we implement them with infinite precision. In practice, finite wordlength restriction causes each realization to behave differently in terms of sensitivity to parameter variation, stability, frequency response distortion error, and so on. These effects are serious for higher-order transfer functions, which require correspondingly higher numbers of delay elements. The finite wordlength errors that plague these implementations are coefficient quantization, overflow errors, and round-off errors. From a practical viewpoint, parallel and cascade forms using low-order filters minimize the effects of finite wordlength. Parallel and certain cascade forms are numerically less sensitive than the canonic direct form to small parameter variations in the system. In the canonic direct form structure with large *N*, a small change in a filter coefficient due to parameter quantization results in a large change in the location of the poles and the zeros of the system. Qualitatively, this difference can be explained by the fact that
-
-### <span id="page-545-0"></span>526 CHAPTER 5 DISCRETE-TIME SYSTEM ANALYSIS USING THE *Z*-TRANSFORM
-
-in a direct form (or its transpose), all the coefficients interact with each other, and a change in any coefficient will be magnified through its repeated influence from feedback and feedforward connections. In a parallel realization, in contrast, a change in a coefficient will affect only a localized segment; the case of a cascade realization is similar. For this reason, the most popular technique for minimizing finite wordlength effects is to design filters by using cascade or parallel forms employing low-order filters. In practice, high-order filters are realized by using multiple second-order sections in cascade, because second-order filters not only are easier to design but are less susceptible to coefficient quantization and round-off errors, and their implementations allow easier data word scaling to reduce the potential overflow effects of data word-size growth. A cascaded system using second-order building blocks usually requires fewer multiplications for a given filter frequency response [1].
-
-There are several ways to pair the poles and zeros of an *N*th-order *H*[*z*] into a cascade of second-order sections, and several ways to order the resulting sections. Quantizing error will be different for each combination. Although several papers published provide guidelines in predicting and minimizing finite wordlength errors, it is advisable to resort to computer simulation of the filter design. This way, one can vary filter hardware characteristic, such as coefficient wordlengths, accumulator register sizes, sequencing of cascaded sections, and input signal sets. Such an approach is both reliable and economical [1].

engineering/linear-systems-and-signals-3rd-edition-bp-lathi/070_5.5 FREQUENCY RESPONSE OF DISCRETE-TIME SYSTEMS.md

@@ -1,406 +0,0 @@
-## **5.5 FREQUENCY RESPONSE OF [DISCRETE-TIME](#page-11-0) SYSTEMS**
-
-For (asymptotically or BIBO-stable) continuous-time systems, we showed that the system response to an input *ej*ω*<sup>t</sup>* is *H*(*j*ω)*ej*ω*<sup>t</sup>* and that the response to an input cos ω*t* is |*H*(*j*ω)| cos[ω*t* + *H*(*j*ω)]. Similar results hold for discrete-time systems. We now show that for an (asymptotically or BIBO-stable) LTID system, the system response to an input *ej<sup>n</sup>* is *H*[*ej*]*ej<sup>n</sup>* and the response to an input cos *n* is |*H*[*ej*]| cos(*n*+ *H*[*ej*]).
-
-The proof is similar to the one used for continuous-time systems. In Sec. 3.8-2, we showed that an LTID system response to an (everlasting) exponential *zn* is also an (everlasting) exponential *H*[*z*]*z<sup>n</sup>*. This result is valid only for values of *z* for which *H*[*z*], as defined in Eq. (5.11), exists (converges). As usual, we represent this input–output relationship by a directed arrow notation as
-
-$$
-z^n \Longrightarrow H[z]z^n \tag{5.30}
-$$
-
-Setting *z* = *ej* in this relationship yields
-
-$$
-e^{i\Omega n} \Longrightarrow H[e^{i\Omega}]e^{i\Omega n} \tag{5.31}
-$$
-
-Noting that cos *n* is the real part of *ej<sup>n</sup>*, use of Eq. (3.34) yields
-
-$$
-\cos \Omega n \Longrightarrow \text{Re}\{H[e^{i\Omega}]e^{i\Omega n}\}\tag{5.32}
-$$
-
-Expressing *H*[*ej*] in the polar form
-
-$$
-H[e^{i\Omega}] = |H[e^{i\Omega}]|e^{i\angle H[e^{i\Omega}]}
-$$
-
-Eq. (5.32) can be expressed as
-
-$$
-\cos \Omega n \Longrightarrow |H[e^{i\Omega}]|\cos(\Omega n + \angle H[e^{i\Omega}])
-$$
-
-In other words, the system response *y*[*n*] to a sinusoidal input cos *n* is given by
-
-$$
-y[n] = |H[e^{j\Omega}]|\cos(\Omega n + \angle H[e^{j\Omega}])
-$$
-
-Following the same argument, the system response to a sinusoid cos(*n*+θ ) is
-
-$$
-y[n] = |H[e^{j\Omega}]\cos(\Omega n + \theta + \angle H[e^{j\Omega}])\tag{5.33}
-$$
-
-This result is valid only for BIBO-stable or asymptotically stable systems. The frequency response is meaningless for BIBO-unstable systems (which include marginally stable and asymptotically unstable systems). This follows from the fact that the frequency response in Eq. (5.31) is obtained by setting *z*=*ej* in Eq. (5.30). But, as shown in Sec. 3.8-2 [Eqs. (3.38) and (3.39)], the relationship of Eq. (5.30) applies only for values of *z* for which *H*[*z*] exists. For BIBO-unstable systems, the ROC for *H*[*z*] does not include the unit circle where *z* = *ej*. This means, for BIBO-unstable systems, that *H*[*z*] is meaningless when *z* = *ej*. †
-
-This important result shows that the response of an asymptotically or BIBO-stable LTID system to a discrete-time sinusoidal input of frequency is also a discrete-time sinusoid of the same frequency. *The amplitude of the output sinusoid is* |*H*[*ej*]| *times the input amplitude, and the phase of the output sinusoid is shifted by H*[*ej*] *with respect to the input phase*. Clearly, |*H*[*ej*]| is the amplitude gain, and a plot of |*H*[*ej*]| versus is the amplitude response of the discrete-time system. Similarly, *H*[*ej*] is the phase response of the system, and a plot of *H*[*ej*] versus shows how the system modifies or shifts the phase of the input sinusoid. Note that *H*[*ej*] incorporates the information of both amplitude and phase responses and therefore is called the *frequency responses* of the system.
-
-### STEADY-STATE RESPONSE TO CAUSAL SINUSOIDAL INPUT
-
-As in the case of continuous-time systems, we can show that the response of an LTID system to a causal sinusoidal input cos *n u*[*n*] is *y*[*n*] in Eq. (5.33), plus a natural component consisting of the characteristic modes (see Prob. 5.5-9). For a stable system, all the modes decay exponentially, and only the sinusoidal component in Eq. (5.33) persists. For this reason, this component is called the sinusoidal *steady-state* response of the system. Thus, *yss*[*n*], the steady-state response of a system to a causal sinusoidal input cos *n u*[*n*], is
-
-$$
-y_{ss}[n] = |H[e^{j\Omega}]\cos{(\Omega n + \angle H[e^{j\Omega}])}u[n]
-$$
-
-### SYSTEM RESPONSE TO SAMPLED CONTINUOUS-TIME SINUSOIDS
-
-So far we have considered the response of a discrete-time system to a discrete-time sinusoid cos *n* (or exponential *ej<sup>n</sup>*). In practice, the input may be a sampled continuous-time sinusoid cos ω*t* (or an exponential *ej*ω*<sup>t</sup>* ). When a sinusoid cos ω*t* is sampled with sampling interval *T*, the resulting
-
-<sup>†</sup> This may also be argued as follows. For BIBO-unstable systems, the zero-input response contains nondecaying natural mode terms of the form cos0*n* or γ *<sup>n</sup>* cos0*n* (γ > 1). Hence, the response of such a system to a sinusoid cos*n* will contain not just the sinusoid of frequency but also nondecaying natural modes, rendering the concept of frequency response meaningless. Alternately, we can argue that when *<sup>z</sup>*=*ej*, a BIBO-unstable system violates the dominance condition <sup>|</sup>γ*i*<sup>|</sup> <sup>&</sup>lt; <sup>|</sup>*ej*<sup>|</sup> for all *<sup>i</sup>*, where <sup>γ</sup>*<sup>i</sup>* represents *<sup>i</sup>*th characteristic root of the system.
-
-### 528 CHAPTER 5 DISCRETE-TIME SYSTEM ANALYSIS USING THE *Z*-TRANSFORM
-
-signal is a discrete-time sinusoid cos ω*nT*, obtained by setting *t* = *nT* in cosω*t*. Therefore, all the results developed in this section apply if we substitute ω*T* for :
-
-$$
-\Omega = \omega T \tag{5.34}
-$$
-
-### **EXAMPLE 5.10 Sinusoidal Response of a Difference Equation System**
-
-For a system specified by the equation
-
-$$
-y[n+1] - 0.8y[n] = x[n+1]
-$$
-
-find the system response to the inputs
-
-(a) 
-$$
-1^n = 1
-$$
-  
-(b)  $\cos\left[\frac{\pi}{6}n - 0.2\right]$ 
-
-**(c)** a sampled sinusoid cos 1500*t* with sampling interval *T* = 0.001
-
-The system equation can be expressed as
-
-$$
-(E - 0.8)y[n] = Ex[n]
-$$
-
-Therefore, the transfer function of the system is
-
-$$
-H[z] = \frac{z}{z - 0.8} = \frac{1}{1 - 0.8z^{-1}}
-$$
-
-The frequency response is
-
-$$
-H[e^{i\Omega}] = \frac{1}{1 - 0.8e^{-i\Omega}} = \frac{1}{(1 - 0.8\cos\Omega) + i0.8\sin\Omega}
-$$
-
-Therefore,
-
-$$
-|H[e^{i\Omega}]| = \frac{1}{\sqrt{(1 - 0.8 \cos \Omega)^2 + (0.8 \sin \Omega)^2}} = \frac{1}{\sqrt{1.64 - 1.6 \cos \Omega}}
-$$
-(5.35)
-
-and
-
-$$
-\angle H[e^{i\Omega}] = -\tan^{-1}\left[\frac{0.8\sin\Omega}{1 - 0.8\cos\Omega}\right]
-$$
-\n(5.36)
-
-The amplitude response |*H*[*ej*]| can also be obtained by observing that |*H*| <sup>2</sup> = *HH*∗. Since our system is real, we therefore see that
-
-$$
-|H[e^{i\Omega}]|^{2} = H[e^{i\Omega}]H^{*}[e^{i\Omega}] = H[e^{i\Omega}]H[e^{-i\Omega}]
-$$
-\n(5.37)
-
-Substituting for *H*[*ej*], it follows that
-
-$$
-|H[e^{i\Omega}]|^2 = \left(\frac{1}{1 - 0.8e^{-i\Omega}}\right)\left(\frac{1}{1 - 0.8e^{i\Omega}}\right) = \frac{1}{1.64 - 1.6\cos\Omega}
-$$
-
-which matches the result found earlier.
-
-Figure 5.14 shows plots of amplitude and phase response as functions of . We now compute the amplitude and the phase response for the various inputs.
-
-**Figure 5.14** Frequency response of the LTID system.
-
-**(a)** Since 1*<sup>n</sup>* = (*ej*)*<sup>n</sup>* with = 0, the amplitude response is *H*[*ej*<sup>0</sup>]. From Eq. (5.35) we obtain
-
-$$
-H[e^{i0}] = \frac{1}{\sqrt{1.64 - 1.6 \cos(0)}} = \frac{1}{\sqrt{0.04}} = 5 = 5 \angle 0
-$$
-
-Therefore,
-
-$$
-|H[e^{j0}]| = 5 \quad \text{and} \quad \angle H[e^{j0}] = 0
-$$
-
-These values also can be read directly from Figs. 5.14a and 5.14b, respectively, corresponding to = 0. Therefore, the system response to input 1 is
-
-$$
-y[n] = 5(1^n) = 5 \qquad \text{for all } n
-$$
-
-**(b)** For *x*[*n*] = cos[(π/6)*n*−0.2], = π/6. According to Eqs. (5.35) and (5.36),
-
-$$
-|H[e^{j\pi/6}]| = \frac{1}{\sqrt{1.64 - 1.6 \cos \frac{\pi}{6}}} = 1.983
-$$
-  
-$$
-\angle H[e^{j\pi/6}] = -\tan^{-1} \left[ \frac{0.8 \sin \frac{\pi}{6}}{1 - 0.8 \cos \frac{\pi}{6}} \right] = -0.916 \text{ rad}
-$$
-
-These values also can be read directly from Figs. 5.14a and 5.14b, respectively, corresponding to = π/6. Therefore,
-
-$$
-y[n] = 1.983 \cos\left(\frac{\pi}{6}n - 0.2 - 0.916\right) = 1.983 \cos\left(\frac{\pi}{6}n - 1.116\right)
-$$
-
-Figure 5.15 shows the input *x*[*n*] and the corresponding system response.
-
-**Figure 5.15** Sinusoidal input and the corresponding output of the LTID system.
-
-**(c)** A sinusoid cos 1500*t* sampled every *T* seconds (*t* = *nT*) results in a discrete-time sinusoid
-
-$$
-x[n] = \cos 1500nT
-$$
-
-For *T* = 0.001, the input is
-
-*x*[*n*] = cos(1.5*n*)
-
-In this case, = 1.5. According to Eqs. (5.35) and (5.36),
-
-$$
-|H[e^{j1.5}]| = \frac{1}{\sqrt{1.64 - 1.6 \cos(1.5)}} = 0.809
-$$
-  
-$$
-\angle H[e^{j1.5}] = -\tan^{-1} \left[ \frac{0.8 \sin(1.5)}{1 - 0.8 \cos(1.5)} \right] = -0.702 \text{ rad}
-$$
-
-These values also could be read directly from Fig. 5.14 corresponding to = 1.5. Therefore,
-
-$$
-y[n] = 0.809 \cos(1.5n - 0.702)
-$$
-
-### FREQUENCY RESPONSE PLOTS USING MATLAB
-
-MATLAB makes it easy to compute and plot magnitude and phase responses directly using a system's transfer function. As the following code demonstrates, there is no need to derive separate expressions for the magnitude and phase responses.
-
->> Omega = linspace(-pi,pi,400); H = @(z) z./(z-0.8);
-
->> subplot(1,2,1); plot(Omega,abs(H(exp(1j\*Omega))),'k'); axis tight;
-
->> xlabel('\Omega'); ylabel('|H[e^{j \Omega}]|');
-
->> subplot(1,2,2); plot(Omega,angle(H(exp(1j\*Omega))\*180/pi),'k'); axis tight;
-
-```
->> xlabel('\Omega'); ylabel('\angle H[e^{j \Omega}] [deg]');
-```
-
-The resulting plots, shown in Fig. 5.16, confirm the earlier results of Fig. 5.14.
-
-### <span id="page-551-0"></span>532 CHAPTER 5 DISCRETE-TIME SYSTEM ANALYSIS USING THE *Z*-TRANSFORM
-
-**Comment:** Figures 5.14 and 5.16 show amplitude and phase response plots as functions of . These plots as well as Eqs. (5.35) and (5.36) indicate that the frequency response of a discrete-time system is a continuous (rather than discrete) function of frequency . There is no contradiction here. This behavior is merely an indication of the fact that the frequency variable is continuous (takes on all possible values) and therefore the system response exists at every value of .
-
-### **DR ILL 5.18 Frequency Response of Difference Equation**
-
-For a system specified by the equation
-
-$$
-y[n+1] - 0.5y[n] = x[n]
-$$
-
-find the amplitude and the phase response. Find the system response to sinusoidal input cos[1000*t* −(π/3)] sampled every *T* = 0.5 ms.
-
-### **ANSWER**
-
-$$
-|H[e^{i\Omega}]| = \frac{1}{\sqrt{1.25 - \cos \Omega}}
-$$
-  
-\n
-$$
-\angle H[e^{i\Omega}] = -\tan^{-1} \left[ \frac{\sin \Omega}{\cos \Omega - 0.5} \right]
-$$
-  
-\n
-$$
-y[n] = 1.639 \cos \left( 0.5n - \frac{\pi}{3} - 0.904 \right) = 1.639 \cos (0.5n - 1.951)
-$$
-
-### **DR ILL 5.19 Frequency Response of an Ideal Delay System**
-
-Show that for an ideal delay (*H*[*z*] = 1/*z*), the amplitude response |*H*[*ej*]| = 1, and the phase response *H*[*ej*]=−. Thus, a pure time delay does not affect the amplitude gain of sinusoidal input, but it causes a phase shift (delay) of radians in a discrete sinusoid of frequency . Thus, for an ideal delay, the phase shift of the output sinusoid is proportional to the frequency of the input sinusoid (linear phase shift).
-
-### **[5.5-1 The Periodic Nature of Frequency Response](#page-11-0)**
-
-In Ex. 5.10 and Fig. 5.14, we saw that the frequency response *H*[*ej*] is a periodic function of . This is not a coincidence. Unlike continuous-time systems, all LTID systems have periodic frequency response. This is seen clearly from the nature of the expression of the frequency response of an LTID system. Because *e*±*j*2π*<sup>m</sup>* = 1 for all integer values of *m* [see Eq. (B.10)],
-
-$$
-H[e^{j\Omega}] = H[e^{j(\Omega + 2\pi m)}] \qquad m \text{ integer}
-$$
-
-Therefore, the frequency response *H*[*ej*] is a periodic function of with a period 2π. This is the mathematical explanation of the periodic behavior. The physical explanation that follows provides a much better insight into the periodic behavior.
-
-### NON-UNIQUENESS OF DISCRETE-TIME SINUSOID WAVEFORMS
-
-A continuous-time sinusoid cos ω*t* has a unique waveform for every real value of ω in the range 0 to ∞. Increasing ω results in a sinusoid of ever-increasing frequency. Such is not the case for the discrete-time sinusoid cos *n* because
-
-$$
-\cos[(\Omega \pm 2\pi m)n] = \cos \Omega n \qquad m \text{ integer}
-$$
-
-and
-
-$$
-e^{j(\Omega \pm 2\pi m)n} = e^{j\Omega n} \qquad m \text{ integer}
-$$
-
-This shows that the discrete-time sinusoids cos *n* (and exponentials *ej<sup>n</sup>*) separated by values of in integral multiples of 2π are identical. The reason for the periodic nature of the frequency response of an LTID system is now clear. Since the sinusoids (or exponentials) with frequencies separated by interval 2π are identical, the system response to such sinusoids is also identical and, hence, is periodic with period 2π.
-
-This discussion shows that the discrete-time sinusoid cos *n* has a unique waveform only for the values of in the range −π to π. This band is called the *fundamental band*. Every frequency , no matter how large, is identical to some frequency, *a*, in the fundamental band (−π ≤ *<sup>a</sup>* < π), where
-
-$$
-\Omega_a = \Omega - 2\pi m \qquad -\pi \le \Omega_a < \pi \quad \text{and} \quad m \text{ integer} \tag{5.38}
-$$
-
-The integer *m* can be positive or negative. We use Eq. (5.38) to plot the fundamental band frequency *<sup>a</sup>* versus the frequency of a sinusoid (Fig. 5.17a). The frequency *<sup>a</sup>* is modulo 2π value of .
-
-All these conclusions are also valid for exponential *ej<sup>n</sup>*.
-
-### ALL DISCRETE-TIME SIGNALS ARE INHERENTLY BANDLIMITED
-
-This discussion leads to the surprising conclusion that all discrete-time signals are inherently bandlimited, with frequencies lying in the range −π to π radians per sample. In terms of frequency *F* = /2π, where *F* is in cycles per sample, all frequencies *F* separated by an integer number are identical. For instance, all discrete-time sinusoids of frequencies 0.3, 1.3, 2.3, ... cycles per sample are identical. The fundamental range of frequencies is −0.5 to 0.5 cycles per sample.
-
-Any discrete-time sinusoid of frequency beyond the fundamental band, when plotted, appears and behaves, in every way, like a sinusoid having its frequency in the fundamental band. It is impossible to distinguish between the two signals. Thus, in a basic sense, discrete-time frequencies beyond || = π or |*F*| = 1/2 do not exist. Yet, in a "mathematical" sense, we must admit the existence of sinusoids of frequencies beyond = π. What does this mean?
-
-**Figure 5.17 (a)** Actual frequency versus **(b)** apparent frequency.
-
-### A MAN NAMED ROBERT
-
-To give an analogy, consider a fictitious person Mr. Robert Thompson. His mother calls him Robby; his acquaintances call him Bob, his close friends call him by his nickname, Shorty. Yet, Robert, Robby, Bob, and Shorty are one and the same person. However, we cannot say that only Mr. Robert Thompson exists, or only Robby exists, or only Shorty exists, or only Bob exists. All these four persons exist, although they are one and the same individual. In a same way, we cannot say that the frequency π/2 exists and frequency 5π/2 does not exist; they are both the same entity, called by different names.
-
-It is in this sense that we have to admit the existence of frequencies beyond the fundamental band. Indeed, mathematical expressions in the frequency domain automatically cater to this need by their built-in periodicity. As seen earlier, the very structure of the frequency response is 2π-periodic. We shall also see later, in Ch. 9, that discrete-time signal spectra are also 2π-periodic.
-
-Admitting the existence of frequencies beyond π also serves mathematical and computational convenience in digital signal-processing applications. Values of frequencies beyond π may also originate naturally in the process of sampling continuous-time sinusoids. Because there is no upper limit on the value of ω, there is no upper limit on the value of the resulting discrete-time frequency = ω*T* either.†
-
-The highest possible frequency is π and the lowest frequency is 0 (dc or constant). Clearly, the high frequencies are those in the vicinity of = (2*m* + 1)π and the low frequencies are those in the vicinity of = 2π*m* for all positive or negative integer values of *m*.
-
-<sup>†</sup> However, if goes beyond π, the resulting aliasing reduces the apparent frequency to *<sup>a</sup>* < π.
-
-### FURTHER REDUCTION IN THE FREQUENCY RANGE
-
-Because cos(−*n* + θ ) = cos(*n* − θ ), a frequency in the range −π to 0 is identical to the frequency (of the same magnitude) in the range 0 to π (but with a change in phase sign). Consequently the *apparent frequency* for a discrete-time sinusoid of any frequency is equal to some value in the range 0 to π. Thus, cos(8.7π*n* + θ ) = cos(0.7π*n* + θ ), and the apparent frequency is 0.7π. Similarly,
-
-$$
-\cos(9.6\pi n + \theta) = \cos(-0.4\pi n + \theta) = \cos(0.4\pi n - \theta)
-$$
-
-Hence, the frequency 9.6π is identical (in every respect) to frequency −0.4π, which, in turn, is equal (within the sign of its phase) to frequency 0.4π. In this case, the apparent frequency reduces to |*a*| = 0.4π. We can generalize the result to say that the apparent frequency of a discrete-time sinusoid is |*a*|, as found from Eq. (5.38), and if *<sup>a</sup>* <0, there is a phase reversal. Figure 5.17b plots versus the apparent frequency |*a*|. The shaded bands represent the ranges of for which there is a phase reversal, when represented in terms of |*a*|. For example, the apparent frequency for both the sinusoids cos(2.4π + θ ) and cos(3.6π + θ ) is |*a*| = 0.4π, as seen from Fig. 5.17b. But 2.4π is in a clear band and 3.6π is in a shaded band. Hence, these sinusoids appear as cos(0.4π +θ ) and cos(0.4π −θ ), respectively.
-
-Although every discrete-time sinusoid can be expressed as having frequency in the range from 0 to π, we generally use the frequency range from −π to π instead of 0 to π for two reasons. First, exponential representation of sinusoids with frequencies in the range 0 to π requires a frequency range −π to π. Second, even when we are using a trigonometric representation, we generally need the frequency range −π to π to have exact identity (without phase reversal) of a higher-frequency sinusoid.
-
-For certain practical advantages, in place of the range −π to π, we often use other contiguous ranges of width 2π. The range 0 to 2π, for instance, is used in many applications. It is left as an exercise for the reader to show that the frequencies in the range from π to 2π are identical to those in the range from −π to 0.
-
-### **EXAMPLE 5.11 Apparent Frequency**
-
-Express the following signals in terms of their apparent frequencies: **(a)** cos(0.5π*n* + θ ), **(b)** cos(1.6π*n*+θ ), **(c)** sin(1.6π*n*+θ ), **(d)** cos(2.3π*n*+θ ), and **(e)** cos(34.699*n*+θ ).
-
-**(a)** =0.5π is in the reduced range already. This is also apparent from Fig. 5.17a or 5.17b. Because *<sup>a</sup>* = 0.5π, there is no phase reversal, and the apparent sinusoid is cos(0.5π*n*+θ ).
-
-**(b)** We express 1.6π = −0.4π + 2π so that *<sup>a</sup>* = −0.4π and |*a*| = 0.4. Also, *<sup>a</sup>* is negative, implying sign change for the phase. Hence, the apparent sinusoid is cos(0.4π*n*−θ ). This fact is also apparent from Fig. 5.17b.
-
-**(c)** We first convert the sine form to cosine form as sin(1.6π*n*+θ ) = cos(1.6π*n* − (π/2) + θ ). In part **(b)**, we found *<sup>a</sup>* = −0.4π. Hence, the apparent sinusoid is cos(0.4π*n* + (π/2)−θ ) = −sin(0.4π*n*−θ ). In this case, both the phase and the amplitude change signs.
-
-**(d)** 2.3π = 0.3π +2π so that *<sup>a</sup>* = 0.3π. Hence, the apparent sinusoid is cos(0.3π*n*+θ ).
-
-<span id="page-555-0"></span>**(e)** We have 34.699 = −3+6(2π ). Hence, *<sup>a</sup>* = −3, and the apparent frequency |*a*| = 3 rad/sample. Because *<sup>a</sup>* is negative, there is a sign change of the phase. Hence, the apparent sinusoid is cos(3*n*−θ ).
-
-### **DR ILL 5.20 Apparent Frequency**
-
-Show that the sinusoids having frequencies of **(a)** 2π, **(b)** 3π, **(c)** 5π, **(d)** 3.2π, **(e)** 22.1327, and **(f)** π + 2 can be expressed, respectively, as sinusoids of frequencies **(a)** 0, **(b)** π, **(c)** π, **(d)** 0.8π, **(e)** 3, and **(f)** π −2. Show that in cases (d), (e), and (f), phase changes sign.
-
-### **[5.5-2 Aliasing and Sampling Rate](#page-11-0)**
-
-The non-uniqueness of discrete-time sinusoids and the periodic repetition of the same waveforms at intervals of 2π may seem innocuous, but in reality it leads to a serious problem for processing continuous-time signals by digital filters. A continuous-time sinusoid cosω*t* sampled every *T* seconds (*t* = *nT*) results in a discrete-time sinusoid cosω*nT*, which is cos*n* with = ω*T*. The discrete-time sinusoids cos*n* have unique waveforms only for the values of frequencies in the range <π or ω*T* < π. Therefore, samples of continuous-time sinusoids of two (or more) different frequencies can generate the same discrete-time signal, as shown in Fig. 5.18. *This phenomenon is known as aliasing because through sampling, two entirely different analog sinusoids take on the same "discrete-time" identity*. †
-
-Aliasing causes ambiguity in digital signal processing, which makes it impossible to determine the true frequency of the sampled signal. Consider, for instance, digitally processing
-
-**Figure 5.18** Demonstration of the aliasing effect.
-
-<sup>†</sup> Figure 5.18 shows samples of two sinusoids cos 12π*t* and cos 2π*t* taken every 0.2 second. The corresponding discrete-time frequencies ( = ω*T* = 0.2ω) are cos 2.4π and cos 0.4π. The apparent frequency of 2.4π is 0.4π, identical to the discrete-time frequency corresponding to the lower sinusoid. This shows that the samples of both these continuous-time sinusoids at 0.2-second intervals are identical, as verified from Fig. 5.18.
-
-a continuous-time signal that contains two distinct components of frequencies ω<sup>1</sup> and ω2. The samples of these components appear as discrete-time sinusoids of frequencies <sup>1</sup> = ω1*T* and <sup>2</sup> = ω2*T*. If <sup>1</sup> and <sup>2</sup> happen to differ by an integer multiple of 2π (if ω<sup>2</sup> − ω<sup>1</sup> = 2*k*π/*T*), the two frequencies will be read as the same (lower of the two) frequency by the digital processor.‡ As a result, the higher-frequency component ω<sup>2</sup> not only is lost for good (by losing its identity to ω1), but also it reincarnates as a component of frequency ω1, thus distorting the true amplitude of the original component of frequency ω1. Hence, the resulting processed signal will be distorted. Clearly, aliasing is highly undesirable and should be avoided. To avoid aliasing, the frequencies of the continuous-time sinusoids to be processed should be kept within the fundamental band ω*T* ≤ π or ω ≤ π/*T*. Under this condition the question of ambiguity or aliasing does not arise because any continuous-time sinusoid of frequency in this range has a unique waveform when it is sampled. Therefore, if ω*<sup>h</sup>* is the highest frequency to be processed, then, to avoid aliasing,
-
-$$
-\omega_h < \frac{\pi}{T}
-$$
-
-If *fh* is the highest frequency in hertz, *fh* = ω*h*/2π, and we avoid aliasing if
-
-$$
-f_h < \frac{1}{2T} \qquad \text{or} \qquad T < \frac{1}{2f_h} \tag{5.39}
-$$
-
-This shows that discrete-time signal processing places the limit on the highest frequency *fh* that can be processed for a given value of the sampling interval *T*. Fortunately, we can process a signal of any frequency (without aliasing) by choosing a suitably small value of *T*. Since the sampling frequency *fs* is the reciprocal of the sampling interval *T*, we can also express Eq. (5.39) as
-
-$$
-f_s = \frac{1}{T} > 2f_h
-$$
- or  $f_h < \frac{f_s}{2}$  (5.40)
-
-This result is a special case of the well-known *sampling theorem* (to be proved in Ch. 8). It states that for a discrete-time system to process a continuous-time sinusoid, the sampling rate must be greater than twice the frequency (in hertz) of the sinusoid. In short, *a sampled sinusoid must have a minimum of two samples per cycle*. † For sampling rates below this minimum value, the output signal will be aliased, which means it will be mistaken for a sinusoid of lower frequency.
-
-### ANTI-ALIASING FILTER
-
-If the sampling rate fails to satisfy Eq. (5.40), aliasing occurs, causing the frequencies beyond *fs*/2 Hz to masquerade as lower frequencies to corrupt the spectrum at frequencies below *fs*/2. To avoid such a corruption, a signal to be sampled is passed through an *anti-aliasing* filter of bandwidth *fs*/2 prior to sampling. This operation ensures the condition of Eq. (5.40). The drawback of such a filter is that we lose the spectral components of the signal beyond frequency *fs*/2, which is preferable to the aliasing corruption of the signal at frequencies below *fs*/2. Chapter 8 presents a detailed analysis of the aliasing problem.
-
-<sup>‡</sup> In the case shown in Fig. 5.18, <sup>ω</sup><sup>1</sup> <sup>=</sup> <sup>12</sup>π, <sup>ω</sup><sup>2</sup> <sup>=</sup> <sup>2</sup>π, and *<sup>T</sup>* <sup>=</sup> 0.2. Hence, <sup>ω</sup><sup>2</sup> <sup>−</sup> <sup>ω</sup><sup>1</sup> <sup>=</sup> <sup>10</sup>π*<sup>T</sup>* <sup>=</sup> <sup>2</sup>π, and the two frequencies are read as the same frequency = 0.4π by the digital processor.
-
-<sup>†</sup> Strictly speaking, we must have more than two samples per cycle.
-
-### <span id="page-557-0"></span>**EXAMPLE 5.12 Maximum Sampling Interval**
-
-Determine the maximum sampling interval *T* that can be used in a discrete-time oscillator that generates a sinusoid of 50 kHz.
-
-Here the highest significant frequency *fh* = 50 kHz. Therefore from Eq. (5.39),
-
-$$
-T < \frac{1}{2f_h} = 10\,\mu\,\mathrm{s}
-$$
-
-The sampling interval must be less than 10µs. The sampling frequency is *fs* = 1/*T* > 100 kHz.
-
-### **EXAMPLE 5.13 Maximum Frequency Without Aliasing**
-
-A discrete-time amplifier uses a sampling interval *T* = 25µs. What is the highest frequency of a signal that can be processed with this amplifier without aliasing?
-
-From Eq. (5.39)
-
-$$
-f_h < \frac{1}{2T} = 20 \, \text{kHz}
-$$

engineering/linear-systems-and-signals-3rd-edition-bp-lathi/071_5.6 FREQUENCY RESPONSE FROM POLE-ZERO LOCATIONS.md

@@ -1,170 +0,0 @@
-## **5.6 FREQUENCY [RESPONSE FROM](#page-11-0) POLE-ZERO LOCATIONS**
-
-The frequency responses (amplitude and phase responses) of a system are determined by pole-zero locations of the transfer function *H*[*z*]. Just as in continuous-time systems, it is possible to determine quickly the amplitude and the phase response and to obtain physical insight into the filter characteristics of a discrete-time system by using a graphical technique. The general *N*th-order transfer function *H*[*z*] in Eq. (5.26) can be expressed in factored form as
-
-$$
-H[z] = b_0 \frac{(z - z_1)(z - z_2) \cdots (z - z_N)}{(z - \gamma_1)(z - \gamma_2) \cdots (z - \gamma_N)}
-$$
-
-We can compute *H*[*z*] graphically by using the concepts discussed in Sec. 4.10. The directed line segment from *zi* to *z* in the complex plane (Fig. 5.19a) represents the complex number *z* − *zi*. The length of this segment is |*z*−*zi*| and its angle with the horizontal axis is (*z*−*zi*).
-
-To compute the frequency response *H*[*ej*] we evaluate *H*[*z*] at *z* = *ej*. But for *z* = *ej*, |*z*| = 1 and *z* = so that *z* = *ej* represents a point on the unit circle at an angle with the horizontal. We now connect all zeros (*z*1, *z*2,...,*zN*) and all poles (γ1, γ2, ... , γ*N*) to the point *ej*, as indicated in
-
-**Figure 5.19** Vector representations of **(a)** complex numbers and **(b)** factors of *H*[*z*].
-
-Fig. 5.19b. Let *r*1, *r*2, ... , *rN* be the lengths and φ1, φ2, ... , φ*<sup>N</sup>* be the angles, respectively, of the straight lines connecting *z*1, *z*2, ... , *zN* to the point *ej*. Similarly, let *d*1, *d*2, ... , *dN* be the lengths and θ1, θ2, ... , θ*<sup>N</sup>* be the angles, respectively, of the lines connecting γ1, γ2, ... , γ*<sup>N</sup>* to *ej*. Then
-
-$$
-H[e^{j\Omega}] = H[z]|_{z=e^{j\Omega}} = b_0 \frac{(r_1 e^{j\phi_1})(r_2 e^{j\phi_2}) \cdots (r_N e^{j\phi_N})}{(d_1 e^{j\theta_1})(d_2 e^{j\theta_2}) \cdots (d_N e^{j\theta_N})}
-$$
-  
-=  $b_0 \frac{r_1 r_2 \cdots r_N}{d_1 d_2 \cdots d_N} e^{j[(\phi_1 + \phi_2 + \cdots + \phi_N) - (\theta_1 + \theta_2 + \cdots + \theta_N)]}$ 
-
-Therefore (assuming *b*<sup>0</sup> > 0),
-
-$$
-|H[e^{j\Omega}]| = b_0 \frac{r_1 r_2 \cdots r_N}{d_1 d_2 \cdots d_N} = b_0 \frac{\text{product of the distances of zeros to } e^{j\Omega}}{\text{product of distances of poles to } e^{j\Omega}}
-$$
-(5.41)
-
-and
-
-$$
-\angle H[e^{i\Omega}] = (\phi_1 + \phi_2 + \dots + \phi_N) - (\theta_1 + \theta_2 + \dots + \theta_N)
-$$
-  
-= sum of zero angles to  $e^{i\Omega}$  - sum of pole angles to  $e^{i\Omega}$ 
-
-In this manner, we can compute the frequency response *H*[*ej*] for any value of by selecting the point on the unit circle at an angle . This point is *ej*. To compute the frequency response *H*[*ej*], we connect all poles and zeros to this point and use the foregoing equations to determine |*H*[*ej*]| and *H*[*ej*]. We repeat this procedure for all values of from 0 to π to obtain the frequency response.
-
-### CONTROLLING GAIN BY PLACEMENT OF POLES AND ZEROS
-
-The nature of the influence of pole and zero locations on the frequency response is similar to that observed in continuous-time systems, with minor differences. In place of the imaginary axis of the continuous-time systems, we have the unit circle in the discrete-time case. The nearer the pole (or zero) is to a point *ej* (on the unit circle) representing some frequency , the more influence that pole (or zero) wields on the amplitude response at that frequency because the length of the vector joining that pole (or zero) to the point *ej* is small. The proximity of a pole (or a zero) has a similar effect on the phase response. From Eq. (5.41), it is clear that to enhance the amplitude response at a frequency , we should place a pole as close as possible to the point *ej* (which is on the unit circle).† Similarly, to suppress the amplitude response at a frequency , we should place a zero as close as possible to the point *ej* on the unit circle. Placing repeated poles or zeros will further enhance their influence.
-
-Total suppression of signal transmission at any frequency can be achieved by placing a zero on the unit circle at a point corresponding to that frequency. This observation is used in the notch (bandstop) filter design.
-
-Placing a pole or a zero at the origin does not influence the amplitude response because the length of the vector connecting the origin to any point on the unit circle is unity. However, a pole (or a zero) at the origin adds angle − (or ) to *H*[*ej*]. Hence, the phase spectrum − (or ) is a linear function of frequency and therefore represents a pure time delay (or time advance) of *T* seconds (see Drill 5.19). Therefore, a pole (a zero) at the origin causes a time delay (or a time advance) of *T* seconds in the response. There is no change in the amplitude response.
-
-For a stable system, all the poles must be located inside the unit circle. The zeros may lie anywhere. Also, for a physically realizable system, *H*[*z*] must be a proper fraction, that is, *N* ≥ *M*. If, to achieve a certain amplitude response, we require *M* > *N*, we can still make the system realizable by placing a sufficient number of poles at the origin to make *N* = *M*. This will not change the amplitude response, but it will increase the time delay of the response.
-
-In general, a pole at a point has the opposite effect of a zero at that point. Placing a zero closer to a pole tends to cancel the effect of that pole on the frequency response.
-
-### LOWPASS FILTERS
-
-A lowpass filter generally has a maximum gain at or near = 0, which corresponds to point *ej*<sup>0</sup> = 1 on the unit circle. Clearly, placing a pole inside the unit circle near the point *z* = 1 (Fig. 5.20a) would result in a lowpass response.‡ The corresponding amplitude and phase response appear in Fig. 5.20a. For smaller values of , the point *ej* (a point on the unit circle at an angle ) is closer to the pole, and consequently the gain is higher. As increases, the distance of the point *ej* from the pole increases. Consequently the gain decreases, resulting in a lowpass characteristic. Placing a zero at the origin does not change the amplitude response but it does modify the phase response, as illustrated in Fig. 5.20b. Placing a zero at *z* = −1, however, changes both the amplitude and the phase response (Fig. 5.20c). The point *z* = −1 corresponds to frequency
-
-<sup>†</sup> The closest we can place a pole is on the unit circle at the point representing . This choice would lead to infinite gain, but should be avoided because it will render the system marginally stable (BIBO-unstable). The closer the point to the unit circle, the more sensitive the system gain to parameter variations.
-
-<sup>‡</sup> Placing the pole at *<sup>z</sup>* <sup>=</sup> 1 results in maximum (infinite) gain but renders the system BIBO-unstable, hence should be avoided.
-
-**Figure 5.20** Various pole-zero configurations and the corresponding frequency responses.
-
-### 542 CHAPTER 5 DISCRETE-TIME SYSTEM ANALYSIS USING THE *Z*-TRANSFORM
-
- = π (*z* = *ej* = *ej*<sup>π</sup> = −1). Consequently, the amplitude response now becomes more attenuated at higher frequencies, with a zero gain at = π. We can approach ideal lowpass characteristics by using more poles staggered near *z* = 1 (but within the unit circle). Figure 5.20d shows a third-order lowpass filter with three poles near *z* = 1 and a third-order zero at *z* = −1, with corresponding amplitude and phase response. For an ideal lowpass filter, we need an enhanced gain at every frequency in the band (0, *c*). This can be achieved by placing a continuous wall of poles (requiring an infinite number of poles) opposite this band.
-
-### HIGHPASS FILTERS
-
-A highpass filter has a small gain at lower frequencies and a high gain at higher frequencies. Such a characteristic can be realized by placing a pole or poles near *z* = −1 because we want the gain at = π to be the highest. Placing a zero at *z* = 1 further enhances suppression of gain at lower frequencies. Figure 5.20e shows a possible pole-zero configuration of the third-order highpass filter with corresponding amplitude and phase responses.
-
-In the following two examples, we shall realize analog filters by using digital processors and suitable interface devices (C/D and D/C), as shown in Fig. 3.2. At this point, we shall examine the design of a digital processor with the transfer function *H*[*z*] for the purpose of realizing bandpass and bandstop filters in the following examples.
-
-As Fig. 3.2 shows, the C/D device samples the continuous-time input *x*(*t*) to yield a discrete-time signal *x*[*n*], which serves as the input to *H*[*z*]. The output *y*[*n*] of *H*[*z*] is converted to a continuous-time signal *y*(*t*) by a D/C device. We also saw in Eq. (5.34) that a continuous-time sinusoid of frequency ω, when sampled, results in a discrete-time sinusoid = ω*T*.
-
-### **EXAMPLE 5.14 Bandpass Filter by Pole-Zero Placement**
-
-By trial and error, design a tuned (bandpass) analog filter with zero transmission at 0 Hz and also at the highest frequency *fh* = 500 Hz. The resonant frequency is required to be 125 Hz.
-
-Because *fh* = 500, we require *T* < 1/1000 [see Eq. (5.39)]. Let us select *T* = 10−3. † Recall that the analog frequencies ω correspond to digital frequencies = ω*T*. Hence, analog frequencies ω = 0 and 1000π correspond to = 0 and π, respectively. The gain is required to be zero at these frequencies. Hence, we need to place zeros at *ej* corresponding to = 0 and = π. For = 0, *z* = *ej* = 1; for = π, *ej* = −1. Hence, there must be zeros at *z* = ±1. Moreover, we need enhanced response at the resonant frequency ω = 250π, which corresponds to = π/4, which, in turn, corresponds to *z* = *ej* = *ej*π/4. Therefore, to enhance the frequency response at ω = 250π, we place a pole in the vicinity of *ej*π/4. Because this is a complex pole, we also need its conjugate near *e*−*j*π/4, as indicated in Fig. 5.21a. Let us choose these poles γ<sup>1</sup> and γ<sup>2</sup> as
-
-<sup>γ</sup><sup>1</sup> = |<sup>γ</sup> <sup>|</sup>*ej*π/<sup>4</sup> and <sup>γ</sup><sup>2</sup> = |<sup>γ</sup> <sup>|</sup>*e*−*j*π/<sup>4</sup>
-
-<sup>†</sup> Strictly speaking, we need *T* < 0.001. However, we shall show in Ch. 8 that if the input does not contain a finite amplitude component of 500 Hz, *T* = 0.001 is adequate. Generally, practical signals satisfy this condition.
-
-where |γ | < 1 for stability. The closer γ is to the unit circle, the more sharply peaked is the response around ω = 250π. We also have zeros at ±1. Hence,
-
-$$
-H[z] = K \frac{(z-1)(z+1)}{(z-|\gamma|e^{j\pi/4})(z-|\gamma|e^{-j\pi/4})} = K \frac{z^2 - 1}{z^2 - \sqrt{2}|\gamma|z+|\gamma|^2}
-$$
-
-For convenience, we shall choose *K* = 1. The amplitude response is given by
-
-$$
-|H[e^{i\Omega}]| = \frac{|e^{i2\Omega} - 1|}{|e^{i\Omega} - |\gamma|e^{i\pi/4}||e^{i\Omega} - |\gamma|e^{-i\pi/4}|}
-$$
-
-Now, by using Eq. (5.37), we obtain
-
-$$
-|H[e^{i\Omega}]|^2 = \frac{2(1 - \cos 2\Omega)}{\left[1 + |\gamma|^2 - 2|\gamma|\cos\left(\Omega - \frac{\pi}{4}\right)\right] \left[1 + |\gamma|^2 - 2|\gamma|\cos\left(\Omega + \frac{\pi}{4}\right)\right]}
-$$
-
-**Figure 5.21** Designing a bandpass filter.
-
-### 544 CHAPTER 5 DISCRETE-TIME SYSTEM ANALYSIS USING THE *Z*-TRANSFORM
-
-Figure 5.21b shows the amplitude response as a function of ω, as well as = ω*T* = 10−3ω for values of |γ | = 0.83, 0.96, and 1. As expected, the gain is zero at ω = 0 and at 500 Hz (ω = 1000π). The gain peaks at about 125 Hz (ω = 250π). The resonance (peaking) becomes pronounced as |γ | approaches 1. Figure 5.21c shows a canonical realization of this filter, which follows from the transfer function *H*[*z*].
-
-## MULTIPLE MAGNITUDE RESPONSE CURVES USING MATLAB
-
-By defining an anonymous function of the two variables *z* and γ in MATLAB, it is straightforward to duplicate the three magnitude response curves in Fig. 5.21b, corresponding to the cases γ = 0.83, 0.96, and 1.
-
-```
->> Omega = linspace(0,pi,400);
->> H = @(z,gamma_m) (z.^2-1)./(z.^2-sqrt(2)*gamma_m*z+gamma_m^2);
->> plot(Omega,abs(H(exp(1j*Omega),0.83)),...
->> Omega,abs(H(exp(1j*Omega),0.96)),...
->> Omega,abs(H(exp(1j*Omega),0.99)));
->> text(.27*pi,35,'|\gamma|=1');
->> text(.28*pi,25.5,'|\gamma|=0.96');
->> text(.35*pi,6.41,'|\gamma|=0.83');
->> set(gca,'xtick',0:pi/4:pi,'ytick',[0 6.41,25.5]);
->> axis([0 pi 0 40]); xlabel('\Omega'); ylabel('|H[e^{j \Omega}]|');
-```
-
-The result, shown in Fig. 5.22, confirms the earlier result of Fig. 5.21b. Phase response curves can be generated with minor modification to the MATLAB code.
-
-**Figure 5.22** MATLAB-generated magnitude response curves for Ex. 5.14.
-
-### **EXAMPLE 5.15 Bandstop Filter by Pole-Zero Placement**
-
-Design a second-order notch filter to have zero transmission at 250 Hz and a sharp recovery of gain to unity on both sides of 250 Hz. The highest significant frequency to be processed is *fh* = 400 Hz.
-
-In this case, *T* < 1/2*fh* = 1.25 × 10−3. Let us choose *T* = 10−3. For the frequency 250 Hz, = 2π(250)*T* = π/2. Thus, the frequency 250 Hz is represented by a point *ej* = *ej*π/<sup>2</sup> = *j* on the unit circle, as depicted in Fig. 5.23a. Since we need zero transmission at this frequency, we must place a zero at *z* = *ej*π/<sup>2</sup> = *j* and its conjugate at *z* = *e*−*j*π/<sup>2</sup> = −*j*. We also require a sharp recovery of gain on both sides of frequency 250 Hz. To accomplish this goal, we place two poles close to the two zeros, to cancel out the effect of the two zeros as we move away from the point *j* (corresponding to frequency 250 Hz). For this reason, let us use poles at ±*ja* with *a* < 1 for stability. The closer the poles are to zeros (the closer the *a* to 1), the faster is the gain recovery on either side of 250 Hz. The resulting transfer function is
-
-$$
-H[z] = K \frac{(z-j)(z+j)}{(z-ja)(z+ja)} = K \frac{z^2+1}{z^2+a^2}
-$$
-
-The dc gain (gain at = 0, or *z* = 1 ) of this filter is
-
-$$
-H[1] = K \frac{2}{1 + a^2}
-$$
-
-Because we require a dc gain of unity, we must select *K* = (1+*a*<sup>2</sup>)/2. The transfer function is therefore
-
-$$
-H[z] = \frac{(1+a^2)(z^2+1)}{2(z^2+a^2)}
-$$
-
-and according to Eq. (5.37),
-
-$$
-|H[e^{j\Omega}]|^{2} = \frac{(1+a^{2})^{2}}{4} \frac{(e^{j2\Omega}+1)(e^{-j2\Omega}+1)}{(e^{j2\Omega}+a^{2})(e^{-j2\Omega}+a^{2})}
-$$
-$$
-= \frac{(1+a^{2})^{2}(1+\cos 2\Omega)}{2(1+a^{4}+2a^{2}\cos 2\Omega)}
-$$
-
-Figure 5.23b shows |*H*[*ej*]| for values of *a* = 0.3, 0.6, and 0.95. Figure 5.23c shows a realization of this filter.
-
-### <span id="page-566-0"></span>**DR ILL 5.21 Highpass Filter by Pole-Zero Placement**
-
-Use the graphical argument to show that a filter with transfer function
-
-$$
-H[z] = \frac{z - 0.9}{z}
-$$
-
-acts like a highpass filter. Make a rough sketch of the amplitude response.

engineering/linear-systems-and-signals-3rd-edition-bp-lathi/072_5.7 DIGITAL PROCESSING OF ANALOG SIGNALS.md

@@ -1,233 +0,0 @@
-## **5.7 DIGITAL [PROCESSING OF](#page-11-0) ANALOG SIGNALS**
-
-An analog (meaning continuous-time) signal can be processed digitally by sampling the analog signal and processing the samples by a digital (meaning discrete-time) processor. The output of the processor is then converted back to analog signal, as shown in Fig. 5.24a. We saw some simple cases of such processing in Exs. 3.8, 3.9, 5.14, and 5.15. In this section, we shall derive a criterion for designing such a digital processor for a general LTIC system.
-
-Suppose that we wish to realize an equivalent of an analog system with transfer function *Ha*(*s*), shown in Fig. 5.24b. Let the digital processor transfer function in Fig. 5.24a that realizes this desired *Ha*(*s*) be *H*[*z*]. In other words, we wish to make the two systems in Fig. 5.24 equivalent (at least approximately).
-
-By "equivalence" we mean that for a given input *x*(*t*), the systems in Fig. 5.24 yield the same output *y*(*t*). Therefore, *y*(*nT*), the samples of the output in Fig. 5.24b, are identical to *y*[*n*], the output of *H*[*z*] in Fig. 5.24a.
-
-**Figure 5.24** Analog filter realization with a digital filter.
-
-### 548 CHAPTER 5 DISCRETE-TIME SYSTEM ANALYSIS USING THE *Z*-TRANSFORM
-
-For the sake of generality, we are assuming a noncausal system. The argument and the results are also valid for causal systems. The output *y*(*t*) of the system in Fig. 5.24b is
-
-$$
-y(t) = \int_{-\infty}^{\infty} x(\tau)h_a(t-\tau) d\tau = \lim_{\Delta \tau \to 0} \sum_{m=-\infty}^{\infty} x(m\Delta \tau)h_a(t-m\Delta \tau) \Delta \tau
-$$
-
-For our purpose, it is convenient to use the notation *T* for τ . Assuming *T* (the sampling interval) to be small enough, such a change of notation yields
-
-$$
-y(t) = T \sum_{m=-\infty}^{\infty} x(mT)h_a(t - mT)
-$$
-
-The response at the *n*th sampling instant is *y*(*nT*) obtained by setting *t* = *nT* in the equation is
-
-$$
-y(nT) = T \sum_{m=-\infty}^{\infty} x(mT)h_a[(n-m)T]
-$$
-\n(5.42)
-
-In Fig. 5.24a, the input to *H*[*z*] is *x*(*nT*) = *x*[*n*]. If *h*[*n*] is the unit impulse response of *H*[*z*], then *y*[*n*], the output of *H*[*z*], is given by
-
-$$
-y[n] = \sum_{m = -\infty}^{\infty} x[m]h[n-m]
-$$
-\n(5.43)
-
-If the two systems are to be equivalent, *y*(*nT*) in Eq. (5.42) must be equal to *y*[*n*] in Eq. (5.43). Therefore,
-
-$$
-h[n] = Th_a(nT) \tag{5.44}
-$$
-
-This is the time-domain criterion for equivalence of the two systems.† According to this criterion, *h*[*n*], the unit impulse response of *H*[*z*] in Fig. 5.24a, should be *T* times the samples of *ha*(*t*), the unit impulse response of the system in Fig. 5.24b. This is known as the *impulse invariance criterion* of filter design.
-
-Strictly speaking, this realization guarantees the output equivalence only at the sampling instants, that is, *y*(*nT*) = *y*[*n*], and that also requires the assumption that *T* → 0. Clearly, this criterion leads to an approximate realization of *Ha*(*s*). However, it can be shown that when the frequency response of |*Ha*(*j*ω)| is bandlimited, the realization is exact [2], provided the sampling rate is high enough to avoid any aliasing (*T* < 1/2*fh*).
-
-### REALIZATION OF RATIONAL *H*(*s*)
-
-If we wish to realize an analog filter with transfer function
-
-$$
-H_a(s) = \frac{c}{s - \lambda}
-$$
-
-<sup>†</sup> Because *<sup>T</sup>* is a constant, some authors ignore the factor *<sup>T</sup>*, which yields a simplified criterion *<sup>h</sup>*[*n*] = *ha*(*nT*). Ignoring *T* merely scales the amplitude response of the resulting filter.
-
-The impulse response *h*(*t*), given by the inverse Laplace transform of *Ha*(*s*), is
-
-$$
-h_a(t) = ce^{\lambda t}u(t)
-$$
-
-The corresponding digital filter unit impulse response *h*[*n*], per Eq. (5.44), is
-
-$$
-h[n] = Th_a(nT) = Tce^{n\lambda T}
-$$
-
-Figure 5.25 shows *ha*(*t*) and *h*[*n*]. The corresponding *H*[*z*], the *z*-transform of *h*[*n*], as found from Table 5.1, is
-
-$$
-H[z] = \frac{Tcz}{z - e^{\lambda T}}
-$$
-\n(5.45)
-
-**Figure 5.25** Impulse response for analog and digital systems in the impulse invariance method of filter design.
-
-| No. | Ha(s)            | ha(t)                         | h[n]                        | H[z]                                                            |
-|-----|------------------|-------------------------------|-----------------------------|-----------------------------------------------------------------|
-| 1   | K                | Kδ(t)                         | TKδ[n]                      | TK                                                              |
-| 2   | 1<br>s           | u(t)                          | Tu[n]                       | Tz<br>z−1                                                       |
-| 3   | 1<br>s2          | t                             | nT2                         | T2z<br>(z−1)2                                                   |
-| 4   | 1<br>s3          | 2<br>t<br>2                   | k2T3<br>2                   | T3z(z<br>+1)<br>2(z−1)3                                         |
-| 5   | 1<br>s −λ        | eλt                           | TeλnT                       | Tz<br>z−eλT                                                     |
-| 6   | 1<br>(s −λ)2     | teλt                          | nT2eλnT                     | T2zeλT<br>(z−eλT )2                                             |
-| 7   | As+B<br>s2+2as+c | Tre−at cos(bt+θ<br>)          | Tre−anT cos(bnT+θ<br>)      | Trz[z cos θ −e−aT cos(bT−θ<br>)]<br>z2−(2e−aT cos<br>bT)z+e−2aT |
-|     | r =              | <br>A2c+B2 −2ABa<br>,<br>c−a2 | b = √<br>θ = tan−1<br>c−a2, | Aa−B<br>√<br>c−a2<br>A                                          |
-
-**TABLE 5.3** Select Impulse-Invariance Pairs
-
-The procedure of finding *H*[*z*] can be systematized for any *N*th-order system. First we express an *N*th-order analog transfer function *Ha*(*s*) as a sum of partial fractions as‡
-
-$$
-H_a(s) = \sum_{i=1}^n \frac{c_i}{s - \lambda_i}
-$$
-
-Then the corresponding *H*[*z*] is given by
-
-$$
-H[z] = T \sum_{i=1}^{n} \frac{c_i z}{z - e^{\lambda_i T}}
-$$
-
-This transfer function can be readily realized, as explained in Sec. 5.4. Table 5.3 lists several pairs of *Ha*(*s*) and their corresponding *H*[*z*]. For instance, to realize a digital integrator, we examine its *Ha*(*s*) = 1/*s*. From Table 5.3, corresponding to *Ha*(*s*) = 1/*s* (pair 2), we find *H*[*z*] = *Tz*/(*z* − 1). This is exactly the result we obtained in Ex. 3.9 using another approach.
-
-Note that the frequency response *Ha*(*j*ω) of a practical analog filter cannot be bandlimited. Consequently, all these realizations are approximate.
-
-### CHOOSING THE SAMPLING INTERVAL *T*
-
-The impulse-invariance criterion (5.44) was derived under the assumption that *T* → 0. Such an assumption is neither practical nor necessary for satisfactory design. Avoiding of aliasing is the most important consideration for the choice of *T*. In Eq. (5.39), we showed that for a sampling interval *T* seconds, the highest frequency that can be sampled without aliasing is 1/2*T* Hz or π/*T* radians per second. This implies that *Ha*(*j*ω), the frequency response of the analog filter in Fig. 5.24b should not have spectral components beyond frequency π/*T* radians per second. In other words, to avoid aliasing, the frequency response of the system *Ha*(*s*) must be bandlimited to π/*T* radians per second. We shall see later in Ch. 7 that frequency response of a realizable LTIC system cannot be bandlimited; that is, the response generally exists for all frequencies up to ∞. Therefore, it is impossible to digitally realize an LTIC system exactly without aliasing. The saving grace is that the frequency response of every realizable LTIC system decays with frequency. This allows for a compromise in digitally realizing an LTIC system with an acceptable level of aliasing. The smaller the value of *T*, the smaller the aliasing, and the better the approximation. Since it is impossible to make |*Ha*(*j*ω)| zero, we are satisfied with making it negligible beyond the frequency π/*T*. As a rule of thumb [3], we choose *T* such that |*Ha*(*j*ω)| at the frequency ω = π/*T* is less than a certain fraction (often taken as 1%) of the peak value of |*Ha*(*j*ω)|. This ensures that aliasing is negligible. The peak |*Ha*(*j*ω)| usually occurs at ω = 0 for lowpass filters and at the band center frequency ω*<sup>c</sup>* for bandpass filters.
-
-<sup>‡</sup> Assuming *Ha*(*s*) has simple poles. For repeated poles, the form changes accordingly. Entry 6 in Table 5.3 is suitable for repeated poles.
-
-### **EXAMPLE 5.16 Butterworth Filter Design by the Impulse-Invariance Method**
-
-Design a digital filter to realize a first-order lowpass Butterworth filter with the transfer function
-
-$$
-H_a(s) = \frac{\omega_c}{s + \omega_c} \qquad \omega_c = 10^5 \tag{5.46}
-$$
-
-For this filter, we find the corresponding *H*[*z*] according to Eq. (5.45) (or pair 5 in Table 5.3) as
-
-$$
-H[z] = \frac{\omega_c T z}{z - e^{-\omega_c T}}
-$$
-\n(5.47)
-
-Next, we select the value of *T* by means of the criterion according to which the gain at ω = π/*T* drops to 1% of the maximum filter gain. However, this choice results in such a good design that aliasing is imperceptible. The resulting amplitude response is so close to the desired response that we can hardly notice the aliasing effect in our plot. For the sake of demonstrating the aliasing effect, we shall deliberately select a 10% criterion (instead of 1%). We have
-
-$$
-|H_a(j\omega)| = \left|\frac{\omega_c}{\sqrt{\omega^2 + \omega_c^2}}\right|
-$$
-
-In this case |*Ha*(*j*ω)|max = 1, which occurs at ω = 0. Use of 10% criterion leads to |*Ha*(π/*T*)| = 0.1. Observe that
-
-$$
-|H_a(j\omega)| \approx \frac{\omega_c}{\omega} \qquad \omega \gg \omega_c
-$$
-
-Hence,
-
-$$
-|H_a(\pi/T)| \approx \frac{\omega_c}{\pi/T} = 0.1 \quad \Longrightarrow \quad \pi/T = 10\omega_c = 10^6
-$$
-
-Thus, the 10% criterion yields *T* = 10−<sup>6</sup>π. The 1% criterion would have given *T* = 10−<sup>7</sup>π. Substitution of *T* = 10−<sup>6</sup>π in Eq. (5.47) yields
-
-$$
-H[z] = \frac{0.3142z}{z - 0.7304}
-$$
-\n(5.48)
-
-A canonical realization of this filter is shown in Fig. 5.26a.
-
-To find the frequency response of this digital filter, we rewrite *H*[*z*] as
-
-$$
-H[z] = \frac{0.3142}{1 - 0.7304z^{-1}}
-$$
-
-Therefore,
-
-$$
-H[e^{j\omega T}] = \frac{0.3142}{1 - 0.7304e^{-j\omega T}} = \frac{0.3142}{(1 - 0.7304 \cos \omega T) + j0.7304 \sin \omega T}
-$$
-
-**Figure 5.26** An example of filter design by the impulse-invariance method: **(a)** filter realization, **(b)** amplitude response, and **(c)** phase response.
-
-The corresponding magnitude response is
-
-$$
-|H[e^{j\omega T}]| = \frac{0.3142}{\sqrt{(1 - 0.7304 \cos \omega T)^2 + (0.7304 \sin \omega T)^2}}
-$$
-$$
-= \frac{0.3142}{\sqrt{1.533 - 1.4608 \cos \omega T}}
-$$
-(5.49)
-
-### 5.7 Digital Processing of Analog Signals 553
-
-and the phase response is
-
-$$
-\angle H[e^{j\omega T}] = -\tan^{-1}\left(\frac{0.7304 \sin \omega T}{1 - 0.7304 \cos \omega T}\right)
-$$
- (5.50)
-
-This frequency response differs from the desired response *Ha*(*j*ω) because aliasing causes frequencies above π/*T* to appear as frequencies below π/*T*. This generally results in increased gain for frequencies below π/*T*. For instance, the realized filter gain at ω = 0 is *H*[*ej*<sup>0</sup>] = *H*[1]. This value, as obtained from Eq. (5.48), is 1.1654 instead of the desired value 1. We can partly compensate for this distortion by multiplying *H*[*z*] or *H*[*ej*ω*<sup>T</sup>* ] by a normalizing constant *K* = *Ha*(0)/*H*[1] = 1/1.1654 = 0.858. This forces the resulting gain of *H*[*ej*ω*<sup>T</sup>* ] to be equal to 1 at ω = 0. The normalized *Hn*[*z*] = 0.858*H*[*z*] = 0.858(0.1π*z*/(*z*−0.7304)). The amplitude response in Eq. (5.49) is multiplied by *K* = 0.858 and plotted in Fig. 5.26b over the frequency range 0 ≤ ω ≤ π/*T* = 106. The multiplying constant *K* has no effect on the phase response in Eq. (5.50), which is shown in Fig. 5.26c.
-
-Also, the desired frequency response, according to Eq. (5.46) with ω*<sup>c</sup>* = 105, is
-
-$$
-H_a(j\omega) = \frac{\omega_c}{j\omega + \omega_c} = \frac{10^5}{j\omega + 10^5}
-$$
-
-Therefore,
-
-$$
-|H_a(j\omega)| = \frac{10^5}{\sqrt{\omega^2 + 10^{10}}} \quad \text{and} \quad \angle H_a(j\omega) = -\tan^{-1}\frac{\omega}{10^5}
-$$
-
-This desired amplitude and phase response are plotted (dotted) in Figs. 5.26b and 5.26c for comparison with realized digital filter response. Observe that the amplitude response behavior of the analog and the digital filter is very close over the range ω ≤ ω*<sup>c</sup>* = 105. However, for higher frequencies, there is considerable aliasing, especially in the phase spectrum. Had we used the 1% rule, the realized frequency response would have been closer over another decade of the frequency range.
-
-### IMPULSE INVARIANCE BY MATLAB
-
-We can readily use the MATLAB impinvar command to confirm our digital filter designed by the impulse-invariance method.
-
-```
->> omegac = 10^5; Ba = [omegac]; Aa = [1 omegac]; Fs = 10^6/pi;
->> [B,A] = impinvar(Ba,Aa,Fs)
-    B = 0.3142
-    A = 1.0000 -0.7304
-```
-
-This confirms our earlier result of Eq. (5.48) that the digital filter transfer function is
-
-$$
-H[z] = \frac{0.3142z}{z - 0.7304}
-$$
-
-### <span id="page-573-0"></span>**DR ILL 5.22 Filter Design by the Impulse-Invariance Method**
-
-Design a digital filter to realize an analog transfer function
-
-$$
-H_a(s) = \frac{20}{s+20}
-$$
-
-**ANSWER**
-
-*H*[*z*] = 20*Tz <sup>z</sup>*−*e*−20*<sup>T</sup>* with *<sup>T</sup>* <sup>=</sup> <sup>π</sup> 2000

engineering/linear-systems-and-signals-3rd-edition-bp-lathi/073_5.8 THE BILATERAL z-TRANSFORM.md

@@ -1,361 +0,0 @@
-## **5.8 THE BILATERAL** *z***[-TRANSFORM](#page-11-0)**
-
-Situations involving noncausal signals or systems cannot be handled by the (unilateral) *z*-transform discussed so far. Such cases can be analyzed by the *bilateral* (or two-sided) *z*-transform defined in Eq. (5.1) as
-
-$$
-X[z] = \sum_{n=-\infty}^{\infty} x[n]z^{-n}
-$$
-
-As in Eq. (5.2), the inverse *z*-transform is given by
-
-$$
-x[n] = \frac{1}{2\pi j} \oint X[z] z^{n-1} dz
-$$
-
-These equations define the bilateral *z*-transform. Earlier, we showed that
-
-$$
-\gamma^n u[n] \Longleftrightarrow \frac{z}{z - \gamma} \qquad |z| > |\gamma| \tag{5.51}
-$$
-
-In contrast, the *z*-transform of the signal −γ *nu*[−(*n*+1)], illustrated in Fig. 5.27a, is
-
-$$
-\mathcal{Z}\{-\gamma^n u[-(n+1)]\} = \sum_{-\infty}^{-1} -\gamma^n z^{-n} = \sum_{-\infty}^{-1} -\left(\frac{\gamma}{z}\right)^n
-$$
-$$
-= -\left[\frac{z}{\gamma} + \left(\frac{z}{\gamma}\right)^2 + \left(\frac{z}{\gamma}\right)^3 + \cdots\right]
-$$
-$$
-= 1 - \left[1 + \frac{z}{\gamma} + \left(\frac{z}{\gamma}\right)^2 + \left(\frac{z}{\gamma}\right)^3 + \cdots\right]
-$$
-$$
-= 1 - \frac{1}{1 - \frac{z}{\gamma}} \qquad \left|\frac{z}{\gamma}\right| < 1
-$$
-$$
-= \frac{z}{z - \gamma} \qquad |z| < |\gamma|
-$$
-
-**Figure 5.27 (a)** <sup>−</sup><sup>γ</sup> *nu*[−(*n*+1)] and **(b)** the region of convergence (ROC) of its *<sup>z</sup>*-transform.
-
-Therefore,
-
-$$
-\mathcal{Z}\{-\gamma^{n}u[-(n+1)]\} = \frac{z}{z-\gamma} \qquad |z| < |\gamma| \tag{5.52}
-$$
-
-A comparison of Eqs. (5.51) and (5.52) shows that the *z*-transform of γ *nu*[*n*] is identical to that of −γ *nu*[−(*n* + 1)]. The regions of convergence, however, are different. In the former case, *X*[*z*] converges for |*z*| > |γ |; in the latter, *X*[*z*] converges for |*z*| < |γ | (see Fig. 5.27b). Clearly, the inverse transform of *X*[*z*] is not unique unless the region of convergence is specified. If we add the restriction that all our signals be causal, however, this ambiguity does not arise. The inverse transform of *z*/(*z* − γ ) is γ *nu*[*n*] even without specifying the ROC. Thus, in the unilateral transform, we can ignore the ROC in determining the inverse *z*-transform of *X*[*z*].
-
-As in the case of the bilateral Laplace transform, if *<sup>x</sup>*[*n*] = %*<sup>k</sup> <sup>i</sup>*=<sup>1</sup> *xi*[*n*], then the ROC for *X*[*z*] is the intersection of the ROCs (region common to all ROCs) for the transforms *X*1[*z*],*X*2[*z*],...,*Xk*[*z*].
-
-The preceding results lead to the conclusion (similar to that for the Laplace transform) that if *z* = β is the largest magnitude pole for a causal sequence, its ROC is |*z*| > |β|. If *z* = α is the smallest magnitude nonzero pole for an anticausal sequence, its ROC is |*z*| < |α|.
-
-### REGION OF CONVERGENCE FOR LEFT-SIDED AND RIGHT-SIDED SEQUENCES
-
-Let us first consider a finite duration sequence *xf*[*n*], defined as a sequence that is nonzero for *N*<sup>1</sup> ≤ *n* ≤ *N*2, where both *N*<sup>1</sup> and *N*<sup>2</sup> are finite numbers and *N*<sup>2</sup> > *N*1. Also,
-
-$$
-X_f[z] = \sum_{n=N_1}^{N_2} x_f[n]z^{-n}
-$$
-
-For example, if *N*<sup>1</sup> = −2 and *N*<sup>2</sup> = 1, then
-
-$$
-X_f[z] = x_f[-2]z^2 + x_f[-1]z + x_f[0] + \frac{x_f[1]}{z}
-$$
-
-Assuming all the elements in *xf*[*n*] are finite, we observe that *Xf*[*z*] has two poles at *z* = ∞ because of terms *xf*[−2]*z*<sup>2</sup> +*xf*[−1]*z* and one pole at *z* = 0 because of term *xf*[1]/*z*. Thus, a finite-duration sequence could have poles at *z* = 0 and *z* = ∞. Observe that *Xf*[*z*] converges for all values of *z* except possibly *z* = 0 and *z* = ∞.
-
-This means that the ROC of a general signal *x*[*n*] + *xf*[*n*] is the same as the ROC of *x*[*n*] with the possible exception of *z* = 0 and *z* = ∞.
-
-A *right-sided* sequence is zero for *n* < *N*<sup>2</sup> < ∞ and a left-sided sequence is zero for *n* > *N*<sup>1</sup> > −∞. A causal sequence is always a right-sided sequence, but the converse is not necessarily true. An anticausal sequence is always a left-sided sequence, but the converse is not necessarily true. A *two-sided* sequence is of infinite duration and is neither right-sided nor left-sided.
-
-A right-sided sequence *xr*[*n*] can be expressed as *xr*[*n*] = *xc*[*n*] +*xf*[*n*], where *xc*[*n*] is a causal signal and *xf*[*n*] is a finite-duration signal. Therefore, the ROC for *xr*[*n*] is the same as the ROC for *xc*[*n*] except possibly *z* = ∞. If *z* = β is the largest magnitude pole for a right-sided sequence *xr*[*n*], its ROC is |β| < |*z*|≤∞. Similarly, a left-sided sequence can be expressed as *xl*[*n*] = *xa*[*n*]+*xf*[*n*], where *xa*[*n*] is an anticausal sequence and *xf*[*n*] is a finite-duration signal. Therefore, the ROC for *xl*[*n*] is the same as the ROC for *xa*[*n*] except possibly *z*=0. Thus, if*z*=α is the smallest magnitude nonzero pole for a left-sided sequence, its ROC is 0 ≤ |*z*| < |α|.
-
-### **EXAMPLE 5.17 Bilateral** *z***-Transform**
-
-Determine the bilateral *z*-transform of
-
-$$
-x[n] = \underbrace{(0.9)^n u[n]}_{x_1[n]} + \underbrace{(1.2)^n u[-(n+1)]}_{x_2[n]}
-$$
-
-From the results in Eqs. (5.51) and (5.52), we have
-
-$$
-X_1[z] = \frac{z}{z - 0.9} \qquad |z| > 0.9
-$$
-  
-$$
-X_2[z] = \frac{-z}{z - 1.2} \qquad |z| < 1.2
-$$
-
-The common region where both *X*1[*z*] and *X*2[*z*] converge is 0.9 < |*z*| < 1.2 (Fig. 5.28b). Hence,
-
-$$
-X[z] = X_1[z] + X_2[z]
-$$
-  
-=  $\frac{z}{z - 0.9} - \frac{z}{z - 1.2}$   
-=  $\frac{-0.3z}{(z - 0.9)(z - 1.2)}$  0.9 < |z| < 1.2
-
-The sequence *x*[*n*] and the ROC of *X*[*z*] are depicted in Fig. 5.28.
-
-### **EXAMPLE 5.18 Inverse Bilateral** *z***-Transform**
-
-Find the inverse bilateral *z*-transform of
-
-$$
-X[z] = \frac{-z(z+0.4)}{(z-0.8)(z-2)}
-$$
-
-if the ROC is **(a)** |*z*| > 2, **(b)** |*z*| < 0.8, and **(c)** 0.8 < |*z*| < 2.
-
-**(a)**
-
-$$
-\frac{X[z]}{z} = \frac{-(z+0.4)}{(z-0.8)(z-2)} = \frac{1}{z-0.8} - \frac{2}{z-2}
-$$
-
-and
-
-$$
-X[z] = \frac{z}{z - 0.8} - 2\frac{z}{z - 2}
-$$
-
-Since the ROC is |*z*| > 2, both terms correspond to causal sequences and
-
-$$
-x[n] = [(0.8)^n - 2(2)^n]u[n]
-$$
-
-This sequence appears in Fig. 5.29a.
-
-**(b)** In this case, |*z*| < 0.8, which is less than the magnitudes of both poles. Hence, both terms correspond to anticausal sequences, and
-
-$$
-x[n] = [-(0.8)^n + 2(2)^n]u[-(n+1)]
-$$
-
-This sequence appears in Fig. 5.29b.
-
-**(c)** In this case, 0.8 < |*z*| < 2; the part of *X*[*z*] corresponding to the pole at 0.8 is a causal sequence, and the part corresponding to the pole at 2 is an anticausal sequence:
-
-$$
-x[n] = (0.8)^n u[n] + 2(2)^n u[-(n+1)]
-$$
-
-This sequence appears in Fig. 5.29c.
-
-### <span id="page-578-0"></span>**DR ILL 5.23 Inverse Bilateral** *z***-Transform**
-
-Find the inverse bilateral *z*-transform of
-
-$$
-X[z] = \frac{z}{z^2 + \frac{5}{6}z + \frac{1}{6}} \qquad \frac{1}{2} > |z| > \frac{1}{3}
-$$
-
-### **ANSWER**
-
- − 1 3 *n u*[*n*] +6 − 1 2 *n u*[−(*n*+1)]
-
-INVERSE TRANSFORM BY EXPANSION OF *X*[*z*] IN POWER SERIES OF *z* We have
-
-$$
-X[z] = \sum_{n} x[n]z^{-n}
-$$
-
-For an anticausal sequence, which exists only for *n* ≤ −1, this equation becomes
-
-$$
-X[z] = x[-1]z + x[-2]z^{2} + x[-3]z^{3} + \cdots
-$$
-
-We can find the inverse *z*-transform of *X*[*z*] by dividing the numerator polynomial by the denominator polynomial, both in ascending powers of *z*, to obtain a polynomial in ascending powers of *z*. Thus, to find the inverse transform of *z*/(*z* − 0.5) (when the ROC is |*z*| < 0.5), we divide *z* by −0.5+*z* to obtain −2*z*−4*z*2−8*z*3−· · ·. Hence, *x*[−1]=−2, *x*[−2]=−4, *x*[−3]=−8, and so on.
-
-## **[5.8-1 Properties of the Bilateral](#page-11-0)** *z***-Transform**
-
-Properties of the bilateral *z*-transform are similar to those of the unilateral transform. We shall merely state the properties here, without proofs, for *xi*[*n*] ⇐⇒ *Xi*[*z*].
-
-LINEARITY
-
-$$
-a_1x_1[n]+a_2x_2[n] \Longleftrightarrow a_1X_1[z]+a_2X_2[z]
-$$
-
-The ROC for *a*1*X*1[*z*] + *a*2*X*2[*z*] is the region common to (intersection of) the ROCs for *X*1[*z*] and *X*2[*z*].
-
-SHIFT
-
-*x*[*n*−*m*] ⇐⇒ 1 *zm <sup>X</sup>*[*z*] *<sup>m</sup>* is positive or negative integer
-
-The ROC for *X*[*z*]/*zm* is the ROC for *X*[*z*] except for the addition or deletion of *z* = 0 or *z* = ∞ caused by the factor 1/*z<sup>m</sup>*.
-
-<span id="page-579-0"></span>CONVOLUTION
-
-$$
-x_1[n] * x_2[n] \Longleftrightarrow X_1[z]X_2[z]
-$$
-
-The ROC for *X*1[*z*]*X*2[*z*] is the region common to (intersection of) the ROCs for *X*1[*z*] and *X*2[*z*].
-
-MULTIPLICATION BY γ *<sup>n</sup>*
-
-$$
-\gamma^n x[n] \Longleftrightarrow X\left[\frac{z}{\gamma}\right]
-$$
-
-If the ROC for *X*[*z*] is |γ1| < |*z*| < |γ2|, then the ROC for *X*[*z*/γ ] is |γ γ1| < |*z*| < |γ γ2|, indicating that the ROC is scaled by the factor |γ |.
-
-MULTIPLICATION BY *n*
-
-$$
-nx[n]u[n] \Longleftrightarrow -z\frac{d}{dz}X[z]
-$$
-
-The ROC for −*z*(*dX*/*dz*) is the same as the ROC for *X*[*z*].
-
-TIME REVERSAL
-
-$$
-x[-n] \Longleftrightarrow X[1/z]
-$$
-
-If the ROC for *X*[*z*] is |γ1| < |*z*| < |γ2|, then the ROC for *X*[1/*z*] is 1/|γ1| > |*z*| > |1/γ2|.
-
-COMPLEX CONJUGATION
-
-*x*∗[*n*] ⇐⇒ *X*∗[*z* ∗]
-
-The ROC for *X*∗[*z*∗] is the same as the ROC for *X*[*z*].
-
-## **5.8-2 Using the Bilateral** *z***[-Transform for Analysis of LTID Systems](#page-11-0)**
-
-Because the bilateral *z*-transform can handle noncausal signals, we can use this transform to analyze noncausal linear systems. The zero-state response *y*[*n*] is given by
-
-$$
-y[n] = \mathcal{Z}^{-1}\{X[z]H[z]\}
-$$
-
-provided *X*[*z*]*H*[*z*] exists. The ROC of *X*[*z*]*H*[*z*] is the region in which both *X*[*z*] and *H*[*z*] exist, which means that the region is the common part of the ROC of both *X*[*z*] and *H*[*z*].
-
-### **EXAMPLE 5.19 Zero-State Response by Bilateral** *z***-Transform**
-
-For a causal system specified by the transfer function
-
-$$
-H[z] = \frac{z}{z - 0.5}
-$$
-
-find the zero-state response to input
-
-$$
-x[n] = (0.8)^n u[n] + 2(2)^n u[-(n+1)]
-$$
-  
-$$
-X[z] = \frac{z}{z - 0.8} - \frac{2z}{z - 2} = \frac{-z(z + 0.4)}{(z - 0.8)(z - 2)}
-$$
-
-The ROC corresponding to the causal term is |*z*| > 0.8, and that corresponding to the anticausal term is |*z*| < 2. Hence, the ROC for *X*[*z*] is the common region, given by 0.8 < |*z*| < 2. Hence,
-
-$$
-X[z] = \frac{-z(z+0.4)}{(z-0.8)(z-2)} \qquad 0.8 < |z| < 2
-$$
-
-Therefore,
-
-$$
-Y[z] = X[z]H[z] = \frac{-z^2(z+0.4)}{(z-0.5)(z-0.8)(z-2)}
-$$
-
-Since the system is causal, the ROC of *H*[*z*] is |*z*| > 0.5. The ROC of *X*[*z*] is 0.8 < |*z*| < 2. The common region of convergence for *X*[*z*] and *H*[*z*] is 0.8 < |*z*| < 2. Therefore,
-
-$$
-Y[z] = \frac{-z^2(z+0.4)}{(z-0.5)(z-0.8)(z-2)} \qquad 0.8 < |z| < 2
-$$
-
-Expanding *Y*[*z*] into modified partial fractions yields
-
-$$
-Y[z] = -\frac{z}{z - 0.5} + \frac{8}{3} \left( \frac{z}{z - 0.8} \right) - \frac{8}{3} \left( \frac{z}{z - 2} \right) \qquad 0.8 < |z| < 2
-$$
-
-Since the ROC extends outward from the pole at 0.8, both poles at 0.5 and 0.8 correspond to causal sequence. The ROC extends inward from the pole at 2. Hence, the pole at 2 corresponds to anticausal sequence. Therefore,
-
-$$
-y[n] = \left[ -(0.5)^n + \frac{8}{3}(0.8)^n \right] u[n] + \frac{8}{3}(2)^n u[-(n+1)]
-$$
-
-### **EXAMPLE 5.20 Zero-State Response for an Input with No** *z***-Transform**
-
-For the system in Ex. 5.19, find the zero-state response to input
-
-$$
-x[n] = \underbrace{(0.8)^n u[n]}_{x_1[n]} + \underbrace{(0.6)^n u[-(n+1)]}_{x_2[n]}
-$$
-
-The *z*-transforms of the causal and anticausal components *x*1[*n*] and *x*2[*n*] of the output are
-
-$$
-X_1[z] = \frac{z}{z - 0.8} \qquad |z| > 0.8
-$$
-  
-$$
-X_2[z] = \frac{-z}{z - 0.6} \qquad |z| < 0.6
-$$
-
-Observe that a common ROC for *X*1[*z*] and *X*2[*z*] does not exist. Therefore, *X*[*z*] does not exist. In such a case we take advantage of the superposition principle and find *y*1[*n*] and *y*2[*n*], the system responses to *x*1[*n*] and *x*2[*n*], separately. The desired response *y*[*n*] is the sum of *y*1[*n*] and *y*2[*n*]. Now
-
-$$
-H[z] = \frac{z}{z - 0.5}
-$$
-  $|z| > 0.5$   
-\n
-$$
-Y_1[z] = X_1[z]H[z] = \frac{z^2}{(z - 0.5)(z - 0.8)}
-$$
-  $|z| > 0.8$   
-\n
-$$
-Y_2[z] = X_2[z]H[z] = \frac{-z^2}{(z - 0.5)(z - 0.6)}
-$$
-  $0.5 < |z| < 0.6$ 
-
-Expanding *Y*1[*z*] and *Y*2[*z*] into modified partial fractions yields
-
-$$
-Y_1[z] = -\frac{5}{3} \left( \frac{z}{z - 0.5} \right) + \frac{8}{3} \left( \frac{z}{z - 0.8} \right) \qquad |z| > 0.8
-$$
-  
-$$
-Y_2[z] = 5 \left( \frac{z}{z - 0.5} \right) - 6 \left( \frac{z}{z - 0.6} \right) \qquad 0.5 < |z| < 0.6
-$$
-
-Therefore,
-
-$$
-y_1[n] = \left[ -\frac{5}{3}(0.5)^n + \frac{8}{3}(0.8)^n \right] u[n]
-$$
-  
-$$
-y_2[n] = 5(0.5)^n u[n] + 6(0.6)^n u[-(n+1)]
-$$
-
-and
-
-$$
-y[n] = y_1[n] + y_2[n] = \left[\frac{10}{3}(0.5)^n + \frac{8}{3}(0.8)^n\right]u[n] + 6(0.6)^nu[-(n+1)]
-$$
-
-### <span id="page-582-0"></span>**DR ILL 5.24 Zero-State Response by Bilateral** *z***-Transform**
-
-For the causal system in Ex. 5.19, find the zero-state response to input
-
-$$
-x[n] = \left(\frac{1}{4}\right)^n u[n] + 5(3)^n u[-(n+1)]
-$$
-
-**ANSWER**
-
-<sup>−</sup> <sup>1</sup> 4 *n* +3 1 2 *<sup>n</sup> u*[*n*] +6(3)*nu*[−(*n*+1)]

engineering/linear-systems-and-signals-3rd-edition-bp-lathi/074_5.9 CONNECTING THE LAPLACE AND z-TRANSFORMS.md

@@ -1,75 +0,0 @@
-## **[5.9 CONNECTING THE](#page-11-0) LAPLACE AND** *z***-TRANSFORMS**
-
-We now show that discrete-time systems also can be analyzed by means of the Laplace transform. In fact, we shall see that *the z-transform is the Laplace transform in disguise* and that discrete-time systems can be analyzed as if they were continuous-time systems.
-
-So far we have considered the discrete-time signal as a sequence of numbers and not as an electrical signal (voltage or current). Similarly, we considered a discrete-time system as a mechanism that processes a sequence of numbers (input) to yield another sequence of numbers (output). The system was built by using delays (along with adders and multipliers) that delay sequences of numbers. A digital computer is a perfect example: every signal is a sequence of numbers, and the processing involves delaying sequences of numbers (along with addition and multiplication).
-
-Now suppose we have a discrete-time system with transfer function *H*[*z*] and input *x*[*n*]. Consider a continuous-time signal *x*(*t*) such that its *n*th sample value is *x*[*n*], as shown in Fig. 5.30.† Let the sampled signal be *x*(*t*), consisting of impulses spaced *T* seconds apart with the *n*th impulse of strength *x*[*n*]. Thus,
-
-$$
-\bar{x}(t) = \sum_{n=0}^{\infty} x[n]\delta(t - nT)
-$$
-
-Figure 5.30 shows *x*[*n*] and the corresponding *x*(*t*). The signal *x*[*n*] is applied to the input of a discrete-time system with transfer function *H*[*z*], which is generally made up of delays, adders, and scalar multipliers. Hence, processing *x*[*n*] through *H*[*z*] amounts to operating on the sequence *x*[*n*] by means of delays, adders, and scalar multipliers. Suppose for *x*(*t*) samples, we perform operations identical to those performed on the samples of *x*[*n*] by *H*[*z*]. For this purpose, we need a continuous-time system with transfer function *H*(*s*) that is identical in structure to the discrete-time system *H*[*z*] except that the delays in *H*[*z*] are replaced by elements that delay continuous-time signals (such as voltages or currents). There is no other difference between realizations of *H*[*z*] and *H*(*s*). If a continuous-time impulse δ(*t*) is applied to such a delay of *T* seconds, the output will be δ(*t* −*T*). The continuous-time transfer function of such a delay is *e*−*sT* [see Eq. (4.30)]. Hence, the delay elements with transfer function 1/*z* in the realization of *H*[*z*] will be replaced by the delay elements with transfer function *e*−*sT* in the realization of the corresponding *H*(*s*). This is the same
-
-<sup>†</sup> We can construct such *x*(*t*) from the sample values, as will be explained in Ch. 8.
-
-**Figure 5.30** Connection between the Laplace transform and the *z*-transform.
-
-as *z* being replaced by *esT* . Therefore, *H*(*s*) = *H*[*esT* ]. Let us now apply *x*[*n*] to the input of *H*[*z*] and apply *x*(*t*) at the input of *H*[*esT* ]. Whatever operations are performed by the discrete-time system *H*[*z*] on *x*[*n*] (Fig. 5.30a) are also performed by the corresponding continuous-time system *H*[*esT* ] on the impulse sequence *x*(*t*) (Fig. 5.30b). The delaying of a sequence in *H*[*z*] would amount to delaying of an impulse train in *H*[*esT* ]. Adding and multiplying operations are the same in both cases. In other words, one-to-one correspondence of the two systems is preserved in every aspect. Therefore if *y*[*n*] is the output of the discrete-time system in Fig. 5.30a, then *y*(*t*), the output of the continuous-time system in Fig. 5.30b, would be a sequence of impulse whose *n*th impulse strength is *y*[*n*]. Thus,
-
-$$
-\bar{y}(t) = \sum_{n=0}^{\infty} y[n]\delta(t - nT)
-$$
-
-The system in Fig. 5.30b, being a continuous-time system, can be analyzed via the Laplace transform. If
-
-$$
-\overline{x}(t) \Longleftrightarrow \overline{X}(s) \quad \text{and} \quad \overline{y}(t) \Longleftrightarrow \overline{Y}(s)
-$$
-
-<span id="page-584-0"></span>then
-
-$$
-\overline{Y}(s) = H[e^{sT}]\overline{X}(s)
-$$
-\n(5.53)
-
-Also,
-
-$$
-\overline{X}(s) = \mathcal{L}\left[\sum_{n=0}^{\infty} x[n]\delta(t - nT)\right]
-$$
-
-Now because the Laplace transform of δ(*t* −*nT*) is *e*−*snT* ,
-
-$$
-\overline{X}(s) = \sum_{n=0}^{\infty} x[n]e^{-snT} \quad \text{and} \quad \overline{Y}(s) = \sum_{n=0}^{\infty} y[n]e^{-snT}
-$$
-
-Substitution of these expressions into Eq. (5.53) yields
-
-$$
-\sum_{n=0}^{\infty} y[n]e^{-snT} = H[e^{sT}]\left[\sum_{n=0}^{\infty} x[n]e^{-snT}\right]
-$$
-
-By introducing a new variable *z* = *esT* , this equation can be expressed as
-
-$$
-\sum_{n=0}^{\infty} y[n]z^{-n} = H[z] \sum_{n=0}^{\infty} x[n]z^{-n}
-$$
-
-or
-
-$$
-Y[z] = H[z]X[z]
-$$
-
-where
-
-$$
-X[z] = \sum_{n=0}^{\infty} x[n]z^{-n}
-$$
- and  $Y[z] = \sum_{n=0}^{\infty} y[n]z^{-n}$ 
-
-It is clear from this discussion that the *z*-transform can be considered to be the Laplace transform with a change of variable *z* = *esT* or *s* = (1/*T*)ln*z*. Note that the transformation *z* = *esT* transforms the imaginary axis in the *s* plane (*s* = *j*ω) into a unit circle in the *z* plane (*z* = *esT* = *ej*ω*<sup>T</sup>* , or |*z*| = 1). The LHP and RHP in the *s*-plane map into the inside and the outside, respectively, of the unit circle in the *z* plane.

engineering/linear-systems-and-signals-3rd-edition-bp-lathi/075_5.10 MATLAB - DISCRETE-TIME IIR FILTERS.md

@@ -1,321 +0,0 @@
-## **[5.10 MATLAB: DISCRETE-TIME](#page-11-0) IIR FILTERS**
-
-Recent technological advancements have dramatically increased the popularity of discrete-time filters. Unlike their continuous-time counterparts, the performance of discrete-time filters is not affected by component variations, temperature, humidity, or age. Furthermore, digital hardware is easily reprogrammed, which allows convenient change of device function. For example, certain digital hearing aids are individually programmed to match the required response of a user.
-
-Typically, discrete-time filters are categorized as infinite-impulse response (IIR) or finite-impulse response (FIR). A popular method to obtain a discrete-time IIR filter is by transformation of a corresponding continuous-time filter design. MATLAB greatly assists this process. Although discrete-time IIR filter design is the emphasis of this section, methods for discrete-time FIR filter design are considered in Sec. 9.7.
-
-### <span id="page-585-0"></span>**[5.10-1 Frequency Response and Pole-Zero Plots](#page-11-0)**
-
-Frequency response and pole-zero plots help characterize filter behavior. Similar to continuous-time systems, rational transfer functions for realizable LTID systems are represented in the *z*-domain as
-
-$$
-H[z] = \frac{Y[z]}{X[z]} = \frac{B[z]}{A[z]} = \frac{\sum_{k=0}^{N} b_k z^{-k}}{\sum_{k=0}^{N} a_k z^{-k}} = \frac{\sum_{k=0}^{N} b_k z^{N-k}}{\sum_{k=0}^{N} a_k z^{N-k}}
-$$
-(5.54)
-
-When only the first (*N*<sup>1</sup> + 1) numerator coefficients are nonzero and only the first (*N*<sup>2</sup> + 1) denominator coefficients are nonzero, Eq. (5.54) simplifies to
-
-$$
-H[z] = \frac{Y[z]}{X[z]} = \frac{B[z]}{A[z]} = \frac{\sum_{k=0}^{N_1} b_k z^{-k}}{\sum_{k=0}^{N_2} a_k z^{-k}} = \frac{\sum_{k=0}^{N_1} b_k z^{N_1 - k}}{\sum_{k=0}^{N_2} a_k z^{N_2 - k}} z^{N_2 - N_1}
-$$
-(5.55)
-
-The form of Eq. (5.55) has many advantages. It can be more efficient than Eq. (5.54); it still works when *N*<sup>1</sup> = *N*<sup>2</sup> = *N*; and it more closely conforms to the notation of built-in MATLAB discrete-time signal-processing functions.
-
-The right-hand side of Eq. (5.55) is a form that is convenient for MATLAB computations. The frequency response *H*[*ej*] is obtained by letting *z* = *ej*, where has units of radians. Often, = ω*T*, where ω is the continuous-time frequency in radians per second and *T* is the sampling period in seconds. Defining length-(*N*<sup>2</sup> + 1) coefficient vector **A** = [*a*0,*a*1,...,*aN*<sup>2</sup> ] and length-(*N*1+1) coefficient vector **B** = [*b*0,*b*1,...,*bN*<sup>1</sup> ], program CH5MP1 computes *H*[*ej*] by using Eq. (5.55) for each frequency in the input vector .
-
-```
-function [H] = CH5MP1(B,A,Omega);
-% CH5MP1.m : Chapter 5, MATLAB Program 1
-% Function M-file computes frequency response for LTID systems
-% INPUTS: B = vector of feedforward coefficients
-% A = vector of feedback coefficients
-% Omega = vector of frequencies [rad], typically -pi<=Omega<=pi
-% OUTPUTS: H = frequency response
-N_1 = length(B)-1; N_2 = length(A)-1;
-H = polyval(B,exp(1j*Omega))./polyval(A,exp(1j*Omega)).*...
-   exp(1j*Omega*(N_2-N_1));
-```
-
-Note that owing to MATLAB's indexing scheme, A(k) corresponds to coefficient *ak*−<sup>1</sup> and B(k) corresponds to coefficient *bk*−1. It is also possible to use the signal-processing toolbox function freqz to evaluate the frequency response of a system described by Eq. (5.55). Under special circumstances, the control system toolbox function bode can also be used.
-
-<span id="page-586-0"></span>Program CH5MP2 computes and plots the poles and zeros of an LTID system described by Eq. (5.55), again using vectors **B** and **A**.
-
-```
-function [p,z] = CH5MP2(B,A);
-% CH5MP2.m : Chapter 5, MATLAB Program 2
-% Function M-file computes and plots poles and zeros for LTID systems
-% INPUTS: B = vector of feedforward coefficients
-% A = vector of feedback coefficients
-N_1 = length(B)-1; N_2 = length(A)-1;
-p = roots([A,zeros(1,N_1-N_2)]); z = roots([B,zeros(1,N_2-N_1)]);
-ucirc = exp(1j*linspace(0,2*pi,200)); % Compute unit circle for plot
-plot(real(p),imag(p),'xk',real(z),imag(z),'ok',real(ucirc),imag(ucirc),'k:');
-xlabel('Real'); ylabel('Imag');
-ax = axis; dx = 0.05*(ax(2)-ax(1)); dy = 0.05*(ax(4)-ax(3));
-axis(ax+[-dx,dx,-dy,dy]); axis equal;
-```
-
-The right-hand side of Eq. (5.55) helps explain how the roots are computed. When *N*<sup>1</sup> = *N*2, the term *zN*2−*N*<sup>1</sup> implies additional roots at the origin. If *N*<sup>1</sup> > *N*2, the roots are poles, which are added by concatenating A with zeros(N\_1-N\_2,1); since *N*<sup>2</sup> − *N*<sup>1</sup> ≤ 0, zeros(N\_2-N\_1,1) produces the empty set and B is unchanged. If *N*<sup>2</sup> > *N*1, the roots are zeros, which are added by concatenating B with zeros(N\_2-N\_1,1); since *N*<sup>1</sup> − *N*<sup>2</sup> ≤ 0, zeros(N\_1-N\_2,1) produces the empty set and A is unchanged. Poles and zeros are indicated with black x's and o's, respectively. For visual reference, the unit circle is also plotted. The last two lines in CH5MP2 expand the plot axis box so that root locations are not obscured and also ensure that the real and imaginary axes are drawn to the same scale.
-
-### **[5.10-2 Transformation Basics](#page-11-0)**
-
-Transformation of a continuous-time filter to a discrete-time filter begins with the desired continuous-time transfer function
-
-$$
-H(s) = \frac{Y(s)}{X(s)} = \frac{B(s)}{A(s)} = \frac{\sum_{k=0}^{M} b_{k+N-M} s^{M-k}}{\sum_{k=0}^{N} a_k s^{N-k}}
-$$
-
-As a matter of convenience, *H*(*s*) is represented in factored form as
-
-$$
-H(s) = \frac{b_{N-M}}{a_0} \frac{\prod_{k=1}^{M} (s - z_k)}{\prod_{k=1}^{N} (s - p_k)}
-$$
-(5.56)
-
-where *zk* and *pk* are the system poles and zeros, respectively.
-
-A mapping rule converts the rational function *H*(*s*) to a rational function *H*[*z*]. Requiring that the result be rational ensures that the system realization can proceed with only delay, sum, and multiplier blocks. There are many possible mapping rules. For obvious reasons, good transformations tend to map the ω axis to the unit circle, ω = 0 to *z* = 1, ω = ∞ to *z* = −1, and the left half-plane to the interior of the unit circle. Put another way, sinusoids map to sinusoids, zero frequency maps to zero frequency, high frequency maps to high frequency, and stable systems map to stable systems.
-
-### <span id="page-587-0"></span>568 CHAPTER 5 DISCRETE-TIME SYSTEM ANALYSIS USING THE *Z*-TRANSFORM
-
-Section 5.9 suggests that the *z*-transform can be considered to be a Laplace transform with a change of variable *z* = *esT* or *s* = (1/*T*)ln*z*, where *T* is the sampling interval. It is tempting, therefore, to convert a continuous-time filter to a discrete-time filter by substituting *s* = (1/*T*)ln*z* into *H*(*s*), or *H*[*z*] = *H*(*s*)|*<sup>s</sup>*=(1/*T*)ln*<sup>z</sup>*. This approach is impractical, however, since the resulting *H*[*z*] is not rational and therefore cannot be implemented by using standard blocks. Although not considered here, the so-called matched-*z* transformation relies on the relationship *z* = *esT* to transform system poles and zeros, so the connection is not completely without merit.
-
-### **[5.10-3 Transformation by First-Order Backward Difference](#page-11-0)**
-
-Consider the transfer function *H*(*s*) = *Y*(*s*)/*X*(*s*) = *s*, which corresponds to the first-order continuous-time differentiator
-
-$$
-y(t) = \frac{d}{dt}x(t)
-$$
-
-An approximation that resembles the fundamental theorem of calculus is the first-order backward difference
-
-$$
-y(t) = \frac{x(t) - x(t - T)}{T}
-$$
-
-For sampling interval *T* and *t* = *nT*, the corresponding discrete-time approximation is
-
-$$
-y[n] = \frac{x[n] - x[n-1]}{T}
-$$
-
-which has transfer function
-
-$$
-H[z] = Y[z]/X[z] = \frac{1 - z^{-1}}{T}
-$$
-
-This implies a transformation rule that uses the change of variable *s* = (1−*z*−<sup>1</sup>)/*T* or *z* = 1/(1−*sT*). This transformation rule is appealing since the resulting *H*[*z*] is rational and has the same number of poles and zeros as *H*(*s*). Section 3.4 discusses this transformation strategy in a different way in describing the kinship of difference equations to differential equations.
-
-After some algebra, substituting *s* = (1−*z*−<sup>1</sup>)/*T* into Eq. (5.56) yields
-
-$$
-H[z] = \left(\frac{b_{N-M} \prod_{k=1}^{M} (1/T - z_k)}{a_0 \prod_{k=1}^{N} (1/T - p_k)}\right) \frac{\prod_{k=1}^{M} \left(1 - \frac{1}{1 - Tz_k} z^{-1}\right)}{\prod_{k=1}^{N} \left(1 - \frac{1}{1 - Tp_k} z^{-1}\right)}
-$$
-(5.57)
-
-The discrete-time system has *M* zeros at 1/(1−*Tzk*) and *N* poles at 1/(1−*Tpk*). This transformation rule preserves system stability but does not map the ω axis to the unit circle (see Prob. 5.7-10).
-
-MATLAB program CH5MP3 uses the first-order backward difference method of Eq. (5.57) to convert a continuous-time filter described by coefficient vectors **A** = [*a*0,*a*1,...,*aN*] and **B** = [*bN*−*<sup>M</sup>*,*bN*−*M*+1,...,*bN*] into a discrete-time filter. The form of the discrete-time filter follows Eq. (5.55).
-
-function [Bd,Ad] = CH5MP3(B,A,T); % CH5MP3.m : Chapter 5, MATLAB Program 3 % Function M-file first-order backward difference transformation
-
-```
-% of a continuous-time filter described by B and A into a discrete-time filter.
-% INPUTS: B = vector of continuous-time filter feedforward coefficients
-% A = vector of continuous-time filter feedback coefficients
-% T = sampling interval
-% OUTPUTS: Bd = vector of discrete-time filter feedforward coefficients
-% Ad = vector of discrete-time filter feedback coefficients
-z = roots(B); p = roots(A); % s-domain roots
-gain = B(1)/A(1)*prod(1/T-z)/prod(1/T-p);
-zd = 1./(1-T*z); pd = 1./(1-T*p); % z-domain roots
-Bd = gain*poly(zd); Ad = poly(pd);
-```
-
-### **[5.10-4 Bilinear Transformation](#page-11-0)**
-
-The bilinear transformation is based on a better approximation than first-order backward differences. Again, consider the continuous-time integrator
-
-$$
-y(t) = \frac{d}{dt}x(t)
-$$
-
-Represent signal *x*(*t*) as
-
-$$
-x(t) = \int_{t-T}^{t} \frac{d}{d\tau} x(\tau) d\tau + x(t-T)
-$$
-
-Letting *t* = *nT* and replacing the integral with a trapezoidal approximation yield
-
-$$
-x(nT) = \frac{T}{2} \left[ \frac{d}{dt} x(nT) + \frac{d}{dt} x(nT - T) \right] + x(nT - T)
-$$
-
-Substituting *y*(*t*) for (*d*/*dt*)*x*(*t*), the equivalent discrete-time system is
-
-$$
-x[n] = \frac{T}{2}(y[n] + y[n-1]) + x[n-1]
-$$
-
-From *z*-transforms, the transfer function is
-
-$$
-H[z] = \frac{Y[z]}{X[z]} = \frac{2(1 - z^{-1})}{T(1 + z^{-1})}
-$$
-
-The implied change of variable *s* = 2(1−*z*−1)/*T*(1+*z*−1) or *z* = (1+*sT*/2)/(1−*sT*/2) is called the bilinear transformation. Not only does the bilinear transformation result in a rational function *H*[*z*], the ω axis is correctly mapped to the unit circle (see Prob. 5.6-18a).
-
-After some algebra, substituting *s* = 2(1−*z*−<sup>1</sup>)/*T*(1+*z*−<sup>1</sup>) into Eq. (5.56) yields
-
-$$
-H[z] = \left(\frac{b_{N-M} \prod_{k=1}^{M} (2/T - z_k)}{a_0 \prod_{k=1}^{N} (2/T - p_k)}\right) \frac{\prod_{k=1}^{M} \left(1 - \frac{1 + z_k T/2}{1 - z_k T/2} z^{-1}\right)}{\prod_{k=1}^{N} \left(1 - \frac{1 + p_k T/2}{1 - p_k T/2} z^{-1}\right)} (1 + z^{-1})^{N-M} \tag{5.58}
-$$
-
-In addition to the *M* zeros at (1+*zkT*/2)/(1−*zkT*/2) and *N* poles at (1+*pkT*/2)/(1−*pkT*/2), there are *N*−*M* zeros at minus 1. Since practical continuous-time filters require *M* ≤ *N* for stability, the number of added zeros is thankfully always nonnegative.
-
-<span id="page-589-0"></span>MATLAB program CH5MP4 converts a continuous-time filter described by coefficient vectors **A** = [*a*0,*a*1,...,*aN*] and **B** = [*bN*−*<sup>M</sup>*,*bN*−*M*+1,...,*bN*] into a discrete-time filter by using the bilinear transformation of Eq. (5.58). The form of the discrete-time filter follows Eq. (5.55). If available, it is also possible to use the signal-processing toolbox function bilinear to perform the bilinear transformation.
-
-```
-function [Bd,Ad] = CH5MP4(B,A,T);
-% CH5MP4.m : Chapter 5, MATLAB Program 4
-% Function M-file bilinear transformation of a continuous-time filter
-% described by vectors B and A into a discrete-time filter.
-% Length of B must not exceed A.
-% INPUTS: B = vector of continuous-time filter feedforward coefficients
-% A = vector of continuous-time filter feedback coefficients
-% T = sampling interval
-% OUTPUTS: Bd = vector of discrete-time filter feedforward coefficients
-% Ad = vector of discrete-time filter feedback coefficients
-if (length(B)>length(A)),
-    disp('Numerator order must not exceed denominator order.');
-    return
-end
-z = roots(B); p = roots(A); % s-domain roots
-gain = real(B(1)/A(1)*prod(2/T-z)/prod(2/T-p));
-zd = (1+z*T/2)./(1-z*T/2); pd = (1+p*T/2)./(1-p*T/2); % z-domain roots
-Bd = gain*poly([zd;-ones(length(A)-length(B),1)]); Ad = poly(pd);
-As with most high-level languages, MATLAB supports general if-structures:
-```
-
-```
-if expression,
-       statements;
-elseif expression,
-       statements;
-else,
-       statements;
-end
-```
-
-In the program CH5MP4, the if statement tests *M* > *N*. When true, an error message is displayed and the return command terminates program execution to prevent errors.
-
-### **[5.10-5 Bilinear Transformation with Prewarping](#page-11-0)**
-
-The bilinear transformation maps the entire infinite-length ω axis onto the finite-length unit circle (*z* = *ej*) according to ω = (2/*T*)tan(/2) (see Prob. 5.6-18b). Equivalently, = 2arctan(ω*T*/2). The nonlinearity of the tangent function causes a frequency compression, commonly called frequency warping, that distorts the transformation.
-
-To illustrate the warping effect, consider the bilinear transformation of a continuous-time lowpass filter with cutoff frequency ω*<sup>c</sup>* = 2π3000 rad/s. If the target digital system uses a sampling <span id="page-590-0"></span>rate of 10 kHz, then *T* = 1/(10,000) and ω*<sup>c</sup>* maps to *<sup>c</sup>* = 2arctan(ω*cT*/2) = 1.5116. Thus, the transformed cutoff frequency is short of the desired *<sup>c</sup>* = ω*cT* = 0.6π = 1.8850.
-
-Cutoff frequencies are important and need to be as accurate as possible. By adjusting the parameter *T* used in the bilinear transform, one continuous-time frequency can be exactly mapped to one discrete-time frequency; the process is called prewarping. Continuing the last example, adjusting *T* = (2/ω*c*)tan(*c*/2) ≈ 1/6848 achieves the appropriate prewarping to ensure ω*<sup>c</sup>* = 2π3000 maps to *<sup>c</sup>* = 0.6π.
-
-### **[5.10-6 Example: Butterworth Filter Transformation](#page-11-0)**
-
-To illustrate the transformation techniques, consider a continuous-time 10th-order Butterworth lowpass filter with cutoff frequency ω*<sup>c</sup>* = 2π3000, as designed in Sec. 4.12. First, we determine continuous-time coefficient vectors **A** and **B**.
-
-```
->> omega_c = 2*pi*3000; N=10;
->> poles = roots([(1j*omega_c)^(-2*N),zeros(1,2*N-1),1]);
->> poles = poles(find(poles<0));
->> B = 1; A = poly(poles); A = A/A(end);
-```
-
-Programs CH5MP3 and CH5MP4 are used to perform first-order forward difference and bilinear transformations, respectively.
-
-```
->> Omega = linspace(0,pi,200); T = 1/10000; Omega_c = omega_c*T;
->> [B1,A1] = CH5MP3(B,A,T); % First-order backward difference transformation
->> [B2,A2] = CH5MP4(B,A,T); % Bilinear transformation
->> [B3,A3] = CH5MP4(B,A,2/omega_c*tan(Omega_c/2)); % Bilinear with prewarping
-```
-
-Magnitude responses are computed using CH5MP1 and then plotted.
-
-```
->> H1mag = abs(CH5MP1(B1,A1,Omega));
->> H2mag = abs(CH5MP1(B2,A2,Omega));
->> H3mag = abs(CH5MP1(B3,A3,Omega));
->> plot(Omega,(Omega<=Omega_c),'k',Omega,H1mag,'k-.',...
->> Omega,H2mag,'k--',Omega,H3mag,'k:');
->> axis([0 pi -.05 1.5]);
->> xlabel('\Omega [rad]'); ylabel('Magnitude Response');
->> legend('Ideal','FOBD','BLT','Prewarp BLT','location','best');
-```
-
-The result of each transformation method is shown in Fig. 5.31, where FOBD and BLT stand for first-order backward difference and bilinear transformation, respectively.
-
-Although the first-order backward difference results in a lowpass filter, the method causes significant distortion that makes the resulting filter unacceptable with regard to cutoff frequency. The bilinear transformation is better, but, as predicted, the cutoff frequency falls short of the desired value. Bilinear transformation with prewarping properly locates the cutoff frequency and produces a very acceptable filter response.
-
-<span id="page-591-0"></span>**Figure 5.31** Comparison of various transformation techniques.
-
-### **[5.10-7 Problems Finding Polynomial Roots](#page-12-0)**
-
-Numerically, it is difficult to accurately determine the roots of a polynomial. Consider, for example, a simple polynomial that has a root at minus 1 repeated four times, (*s* + 1)<sup>4</sup> = *s*<sup>4</sup> + 4*s*<sup>3</sup> +6*s*<sup>2</sup> +4*s*+1. The MATLAB roots command returns a surprising result:
-
-```
->> roots([1464 1])'
-   ans = -1.0002 -1.0000-0.0002i -1.0000+0.0002i -0.9998
-```
-
-Even for this low-degree polynomial, MATLAB does not return the true roots.
-
-The problem worsens as polynomial degree increases. The bilinear transformation of the 10th-order Butterworth filter, for example, should have 10 zeros at minus 1. Figure 5.32 shows that the zeros, computed by CH5MP2 with the roots command, are not correctly located.
-
-When possible, programs should avoid root computations that may limit accuracy. For example, results from the transformation programs CH5MP3 and CH5MP4 are more accurate if the true transfer function poles and zeros are passed directly as inputs rather than the polynomial coefficient vectors. When roots must be computed, result accuracy should always be verified.
-
-### **[5.10-8 Using Cascaded Second-Order Sections to Improve Design](#page-12-0)**
-
-The dynamic range of high-degree polynomial coefficients is often large. Adding the difficulties associated with factoring a high-degree polynomial, it is little surprise that high-order designs are difficult.
-
-As with continuous-time filters, performance is improved by using a cascade of second-order sections to design and realize a discrete-time filter. Cascades of second-order sections are also more robust to the coefficient quantization that occurs when discrete-time filters are implemented on fixed-point digital hardware.
-
-To illustrate the performance possible with a cascade of second-order sections, consider a 180th-order transformed Butterworth discrete-time filter with cutoff frequency *<sup>c</sup>* = 0.6π ≈ 1.8850. Program CH5MP5 completes this design, taking care to initially locate poles and zeros without root computations.
-
-**Figure 5.32** Pole-zero plot computed by using roots.
-
-```
-% CH5MP5.m : Chapter 5, MATLAB Program 5
-% Script M-file designs a 180th-order Butterworth lowpass discrete-time filter
-% with cutoff Omega_c = 0.6*pi using 90 cascaded second-order filter sections.
-omega_0 = 1; % Use normalized cutoff frequency for analog prototype
-psi = [0.5:1:90]*pi/180; % Butterworth pole angles
-Omega_c = 0.6*pi; % Discrete-time cutoff frequency
-Omega = linspace(0,pi,1000); % Frequency range for magnitude response
-Hmag = zeros(90,1000); p = zeros(1,180); z = zeros(1,180); % Pre-allocation
-for stage = 1:90,
-    Q = 1/(2*cos(psi(stage))); % Compute Q for stage
-    B = omega_0^2; A = [1 omega_0/Q omega_0^2]; % Compute stage coefficients
-    [B1,A1] = CH5MP4(B,A,2/omega_0*tan(0.6*pi/2)); % Transform stage to DT
-    p(stage*2-1:stage*2) = roots(A1); % Compute z-domain poles for stage
-    z(stage*2-1:stage*2) = roots(B1); % Compute z-domain zeros for stage
-    Hmag(stage,:) = abs(CH5MP1(B1,A1,Omega)); % Compute stage mag response
-end
-ucirc = exp(j*linspace(0,2*pi,200)); % Compute unit circle for pole-zero plot
-figure;
-plot(real(p),imag(p),'kx',real(z),imag(z),'ok',real(ucirc),imag(ucirc),'k:');
-axis equal; xlabel('Real'); ylabel('Imag');
-figure; plot(Omega,prod(Hmag),'k'); axis([0 pi -0.05 1.05]);
-xlabel('\Omega [rad]'); ylabel('Magnitude Response');
-```
-
-The figure command preceding each plot command opens a separate window for each plot.
-
-The filter's pole-zero plot is shown in Fig. 5.33, along with the unit circle, for reference. All 180 zeros of the cascaded design are properly located at minus 1. The wall of poles provides an amazing approximation to the desired brick-wall response, as shown by the magnitude response in Fig. 5.34. It is virtually impossible to realize such high-order designs with continuous-time filters, which adds another reason for the popularity of discrete-time filters. Still, the design is not
-
-<span id="page-593-0"></span>**Figure 5.33** Pole-zero plot for 180th-order discrete-time Butterworth filter.
-
-**Figure 5.34** Magnitude response for a 180th-order discrete-time Butterworth filter.
-
-trivial; even functions from the MATLAB signal-processing toolbox fail to properly design such a high-order discrete-time Butterworth filter.

engineering/linear-systems-and-signals-3rd-edition-bp-lathi/076_5.11 SUMMARY.md

@@ -1,59 +0,0 @@
-## **[5.11 SUMMARY](#page-12-0)**
-
-In this chapter we discussed the analysis of linear, time-invariant, discrete-time (LTID) systems by means of the *z*-transform. The *z*-transform changes the difference equations of LTID systems into algebraic equations. Therefore, solving these difference equations reduces to solving algebraic equations.
-
-The transfer function *H*[*z*] of an LTID system is equal to the ratio of the *z*-transform of the output to the *z*-transform of the input when all initial conditions are zero. Therefore, if *X*[*z*] is the *z*-transform of the input *x*[*n*] and *Y*[*z*] is the *z*-transform of the corresponding output *y*[*n*] <span id="page-594-0"></span>(when all initial conditions are zero), then *Y*[*z*] = *H*[*z*]*X*[*z*]. For an LTID system specified by the difference equation *Q*[*E*]*y*[*n*] = *P*[*E*]*x*[*n*], the transfer function *H*[*z*] = *P*[*z*]/*Q*[*z*]. Moreover, *H*[*z*] is the *z*-transform of the system impulse response *h*[*n*]. We showed in Ch. 3 that the system response to an everlasting exponential *zn* is *H*[*z*]*z<sup>n</sup>*.
-
-We may also view the *z*-transform as a tool that expresses a signal *x*[*n*] as a sum of exponentials of the form *z<sup>n</sup>* over a continuum of the values of *z*. Using the fact that an LTID system response to *z<sup>n</sup>* is *H*[*z*]*z<sup>n</sup>*, we find the system response to *x*[*n*] as a sum of the system's responses to all the components of the form *z<sup>n</sup>* over the continuum of values of *z*.
-
-LTID systems can be realized by scalar multipliers, adders, and time delays. A given transfer function can be synthesized in many different ways. We discussed canonical, transposed canonical, cascade, and parallel forms of realization. The realization procedure is identical to that for continuous-time systems with 1/*s* (integrator) replaced by 1/*z* (unit delay).
-
-The majority of the input signals and practical systems are causal. Consequently, we are required to deal with causal signals most of the time. Restricting all signals to the causal type greatly simplifies *z*-transform analysis; the ROC of a signal becomes irrelevant to the analysis process. This special case of *z*-transform (which is restricted to causal signals) is called the unilateral *z*-transform. Much of the chapter deals with this transform. Section 5.8 discusses the general variety of the *z*-transform (bilateral *z*-transform), which can handle causal and noncausal signals and systems. In the bilateral transform, the inverse transform of *X*[*z*] is not unique, but depends on the ROC of *X*[*z*]. Thus, the ROC plays a crucial role in the bilateral *z*-transform.
-
-In Sec. 5.9, we showed that discrete-time systems can be analyzed by the Laplace transform as if they were continuous-time systems. In fact, we showed that the *z*-transform is the Laplace transform with a change in variable.
-
-### **[REFERENCES](#page-12-0)**
-
-- 1. Lyons, R. G. *Understanding Digital Signal Processing*. Addison-Wesley, Reading, MA, 1997.
-- 2. Oppenheim, A. V., and R. W. Schafer. *Discrete-Time Signal Processing*, 2nd ed. Prentice-Hall, Upper Saddle River, NJ, 1999.
-- 3. Mitra, S. K. *Digital Signal Processing*, 2nd ed. McGraw-Hill, New York, 2001.
-
-## **[PROBLEMS](#page-12-0)**
-
-- **5.1-1** Using the definition, compute the *z*-transform of *<sup>x</sup>*[*n*] = (−1)*n*(*u*[*n*] − *<sup>u</sup>*[*<sup>n</sup>* <sup>−</sup> <sup>8</sup>]). Sketch the poles and zeros of *X*[*z*] in the *z* plane. No calculator is needed to do this problem!
-- **5.1-2** Determine the unilateral *z*-transform *X*[*z*] of the signal *x*[*n*] shown in Fig. P5.1-2. As the picture suggests, *x*[*n*]=−3 for all *n* ≥ 9 and *x*[*n*] = 0 for all *n* < 3.
-- **5.1-3** (a) A causal signal has *z*-transform given by *<sup>X</sup>*[*z*] = *<sup>z</sup>*<sup>2</sup> *<sup>z</sup>*3−<sup>1</sup> . Determine the time-domain signal *x*[*n*] and sketch *x*[*n*] over −4 ≤ *n* ≤ 11. [*Hint:* No complex arithmetic is needed to solve this problem!]
-
-**Figure P5.1-2**
-
-### 576 CHAPTER 5 DISCRETE-TIME SYSTEM ANALYSIS USING THE *Z*-TRANSFORM
-
-- (b) Consider the causal semiperiodic signal *y*[*n*] shown in Fig. P5.1-3. Notice, *y*[*n*] continually repeats the sequence [1, 2, 3] for *n* ≥ 0. Determine the unilateral *z*-transform *Y*[*z*] of this signal. If possible, express your result as a rational function in standard form.
-- **5.1-4** Using the definition of the *z*-transform, find the *z*-transform and the ROC for each of the following signals.
-  - (a) *u*[*n*− *m*]
-  - (b) <sup>γ</sup> *<sup>n</sup>* sinπ*n u*[*n*]
-  - (c) <sup>γ</sup> *<sup>n</sup>* cosπ*n u*[*n*]
-
-**Figure P5.1-3**
-
-(d) 
-$$
-\gamma^n \sin \frac{\pi n}{2} u[n]
-$$
-  
-\n(e)  $\gamma^n \cos \frac{\pi n}{2} u[n]$   
-\n(f)  $\sum_{k=0}^{\infty} 2^{2k} \delta[n-2k]$   
-\n(g)  $\gamma^{n-1} u[n-1]$ 
-
-(h) *<sup>n</sup>*<sup>γ</sup> *<sup>n</sup> <sup>u</sup>*[*n*]
-
-$$
-(i) \; n \, u[n]
-$$
-
-(j) 
-$$
-\frac{\gamma^n}{n!}u[n]
-$$
-
-(k) 

engineering/linear-systems-and-signals-3rd-edition-bp-lathi/077_REFERENCES.md

@@ -1,951 +0,0 @@
-$$
-[2^{n-1} - (-2)^{n-1}]u[n]
-$$
-  
-(l)  $\frac{(\ln \alpha)^n}{n!}u[n]$ 
-
-- **5.1-5** Showing all work, evaluate \$<sup>∞</sup> *<sup>n</sup>*=<sup>0</sup> *<sup>n</sup>*(−3/2)−*n*.
-- **5.1-6** Using only the *z*-transforms of Table 5.1, determine the *z*-transform of each of the following signals.
-
-(a) *u*[*n*] −*u*[*n*−2] (b) <sup>γ</sup> *<sup>n</sup>*−2*u*[*n*<sup>−</sup> <sup>2</sup>] (c) 2*n*+1*u*[*n*<sup>−</sup> <sup>1</sup>] +*en*−1*u*[*n*] (d) <sup>2</sup>−*<sup>n</sup>* cos<sup>π</sup> 3 *n u*[*n*−1] (e) *<sup>n</sup>*<sup>γ</sup> *nu*[*n*<sup>−</sup> <sup>1</sup>] (f) *<sup>n</sup>*(*n*−1)(*n*−2)2*n*−3*u*[*n*−*m*] for *m* = 0, 1, 2, 3 (g) (−1)*nnu*[*n*]
-
-(h) 
-$$
-\sum_{k=0}^{\infty} k\delta(n-2k+1)
-$$
-
-**5.1-7** Find the inverse unilateral *z*-transform of each of the following: (a) *<sup>z</sup>*(*<sup>z</sup>* <sup>−</sup>4)
-
-(a) 
-$$
-z^2 - 5z + 6
-$$
-  
-\n(b)  $\frac{z-4}{z^2 - 5z + 6}$   
-\n(c)  $\frac{(e^{-2} - 2)z}{(z - e^{-2})(z - 2)}$   
-\n(d)  $\frac{(z - 1)^2}{z^3}$   
-\n(e)  $\frac{z(2z + 3)}{(z - 1)(z^2 - 5z + 6)}$   
-\n(f)  $\frac{z(-5z + 22)}{(z + 1)(z - 2)^2}$   
-\n(g)  $\frac{z(1.4z + 0.08)}{(z - 0.2)(z - 0.8)^2}$ 
-
-(h) 
-$$
-\frac{z(z-2)}{z^2 - z + 1}
-$$
-  
-\n(i) 
-$$
-\frac{2z^2 - 0.3z + 0.25}{z^2 + 0.6z + 0.25}
-$$
-  
-\n(j) 
-$$
-\frac{2z(3z-23)}{(z-1)(z^2 - 6z + 25)}
-$$
-  
-\n(k) 
-$$
-\frac{z(3.83z + 11.34)}{(z-2)(z^2 - 5z + 25)}
-$$
-  
-\n(k) 
-$$
-\frac{z^2(-2z^2 + 8z - 7)}{(z^2 - 2z^2 + 8z - 7)}
-$$
-
-$$
-(1) \frac{z^2(-2z^2+8z-7)}{(z-1)(z-2)^3}
-$$
-
-**5.1-8** (a) Expanding *X*[*z*] as a power series in *z*−1, find the first three terms of *x*[*n*] if
-
-$$
-X[z] = \frac{2z^3 + 13z^2 + z}{z^3 + 7z^2 + 2z + 1}
-$$
-
-(b) Extend the procedure used in part (a) to find the first four terms of *x*[*n*] if
-
-$$
-X[z] = \frac{2z^4 + 16z^3 + 17z^2 + 3z}{z^3 + 7z^2 + 2z + 1}
-$$
-
-- **5.1-9** A right-sided signal *x*[*n*] has *z*-transform given by *<sup>X</sup>*[*z*] = *<sup>z</sup>*6+2*z*5+3*z*4+4*z*<sup>3</sup> *<sup>z</sup>*4−<sup>1</sup> . Using a power series expansion of *X*[*z*], determine *x*[*n*] over −5 ≤ *n* ≤ 5.
-- **5.1-10** Find *x*[*n*] by expanding
-
-$$
-X[z] = \frac{\gamma z}{(z - \gamma)^2}
-$$
-
-as a power series in *z*−1.
-
-- **5.1-11** (a) In Table 5.1, if the numerator and the denominator powers of *X*[*z*] are *M* and *N*, respectively, explain why in some cases *N* − *M* = 0, while in others *N* − *M* = 1 or *N* − *M* = *m* (*m* any positive integer).
-  - (b) Without actually finding the *z*-transform, state what is *N* − *M* for *X*[*z*] corresponding to *<sup>x</sup>*[*n*] = <sup>γ</sup> *nu*[*n*−4].
-- **5.2-1** For a discrete-time signal shown in Fig. P5.2-1, show that
-
-$$
-X[z] = \frac{1 - z^{-m}}{1 - z^{-1}}
-$$
-
-Find your answer by using the definition in Eq. (5.1) and by using Table 5.1 and an appropriate property of the *z*-transform.
-
-**Figure P5.2-1**
-
-- **5.2-2** Determine the unilateral *z*-transform of signal *<sup>x</sup>*[*n*] = (1−*n*) cos <sup>π</sup> <sup>2</sup> (*n*−1) *u*[*n*−1].
-- **5.2-3** Suppose a DT signal *x*[*n*] = 2(*u*[*n*−10] −*u*[*n*− <sup>6</sup>]) has a transform *<sup>X</sup>*(*z*). Define *<sup>Y</sup>*(*z*) <sup>=</sup> <sup>1</sup> 2*z*−3 *d dzX*(2*z*). Using graphic plot or vector notation, determine the corresponding signal *y*[*n*].
-- **5.2-4** Suppose a DT signal *x*[*n*] = 3(*u*[*n*] −*u*[*n*−5]) has a transform *X*(*z*). Define *Y*(*z*) = 2*z*−<sup>4</sup> *<sup>d</sup> dzX z* 2 . Using graphic plot or vector notation, determine the corresponding signal *y*[*n*].
-- **5.2-5** Find the *z*-transform of the signal illustrated in Fig. P5.2-5. Solve this problem in two ways, as in Exs. 5.2d and 5.4. Verify that the two answers are equivalent.
-- **5.2-6** Using *z*-transform techniques and properties (no time-domain convolution sum!), determine the convolution *<sup>y</sup>*[*n*] =( <sup>1</sup> <sup>2</sup> )*nu*[*n*−3]∗( <sup>1</sup> <sup>3</sup> )*n*−6*u*[*n*−4]. Express your answer in the form *y*[*n*] = *c*1γ *<sup>n</sup>*−*N*<sup>1</sup> <sup>1</sup> *<sup>u</sup>*[*<sup>n</sup>* <sup>−</sup> *<sup>N</sup>*1] + *<sup>c</sup>*2<sup>γ</sup> *<sup>n</sup>*−*N*<sup>2</sup> <sup>2</sup> *u*[*n* − *N*2], making sure to clearly identify the constants *c*1, *c*2, γ1, γ2, *N*1, and *N*2.
-
-**Figure P5.2-5**
-
-**5.2-7** Determine the inverse unilateral *z*-transform *x*[*n*] of the signal
-
-$$
-X[z] = \frac{d^7}{dz^7} \left[ \frac{z^{-4}}{(z - \frac{1}{2})(z + 3)} \right]
-$$
-
-- **5.2-8** Using only the fact that <sup>γ</sup> *nu*[*n*]⇐⇒*z*/(*z*−γ ) and properties of the *z*-transform, find the *z*-transform of each of the following:
-  - (a) *n*2*u*[*n*]
-  - (b) *<sup>n</sup>*2<sup>γ</sup> *nu*[*n*]
-  - (c) *n*3*u*[*n*]
-  - (d) *an*[*u*[*n*] −*u*[*n*−*m*]]
-  - (e) *ne*−<sup>2</sup>*nu*[*n*−*m*]
-  - (f) (*n*−2)(0.5)*n*−<sup>3</sup> *<sup>u</sup>*[*n*−4]
-- **5.2-9** Using only pair 1 in Table 5.1 and appropriate properties of the *z*-transform, derive iteratively pairs 2 through 9. In other words, first derive pair 2. Then use pair 2 (and pair 1, if needed) to derive pair 3, and so on.
-- **5.2-10** Find the *z*-transform of cos(π*n*/4)*u*[*n*] using only pairs 1 and 11b in Table 5.1 and a suitable property of the *z*-transform.
-- **5.2-11** Apply the time-reversal property to pair 6 of Table 5.1 to show that <sup>γ</sup> *nu*[−(*<sup>n</sup>* <sup>+</sup> <sup>1</sup>)] ⇐⇒ −*z*/(*z*−γ ) and the ROC is given by |*z*| < |γ |.
-- **5.2-12** (a) If *x*[*n*] ⇐⇒ *X*[*z*], then show that (−1)*nx*[*n*] ⇐⇒ *<sup>X</sup>*[−*z*].
-  - (b) Use this result to show that (−γ )*nu*[*n*] ⇐⇒ *z*/(*z* +γ ).
-  - (c) Use these results to find the *z*-transforms of *<sup>x</sup>*i[*n*]=[2*n*−<sup>1</sup> <sup>−</sup>(−2)*n*−1]*u*[*n*] and *<sup>x</sup>*−ii[*n*] = <sup>γ</sup> *<sup>n</sup>* cosπ*n u*[*n*]
-- **5.2-13** (a) If *x*[*n*] ⇐⇒ *X*[*z*], then show that
-
-$$
-\sum_{k=0}^{n} x[k] \Longleftrightarrow \frac{zX[z]}{z-1}
-$$
-
-### 578 CHAPTER 5 DISCRETE-TIME SYSTEM ANALYSIS USING THE *Z*-TRANSFORM
-
-- (b) Use this result to derive pair 2 from pair 1 in Table 5.1.
-- **5.2-14** A number of causal time-domain functions are shown in Fig. P5.2-14. List the function of time that corresponds to each of the following functions of *z*. Few or no calculations are necessary! Be careful, the graphs may be scaled differently.
-
-(a) 
-$$
-\frac{z^2}{(z-0.75)^2}
-$$
-  
-\n(b)  $\frac{z^2 - 0.9z/\sqrt{2}}{z^2 - 0.9\sqrt{2}z + 0.81}$   
-\n(c)  $\sum_{k=1}^{4} z^{-2k}$ 
-
-$$
-k=0
-$$
-\n(d)
-
-\n
-$$
-\frac{z^{-5}}{1-z^{-1}}
-$$
-
-(e) 
-$$
-\frac{z^2}{z^4 - 1}
-$$
-  
-(c) 0.75z
-
-(f) 
-$$
-\frac{1}{(z-0.75)^2}
-$$
-  
-\n(g)  $\frac{z^2 - z/\sqrt{2}}{z-\sqrt{2}}$ 
-
-(g) 
-$$
-\frac{z^2 - \sqrt{2}z + 1}{z^2 - 5z^{-5} + 4z^{-6}}
-$$
-  
-\n(h) 
-$$
-\frac{z^{-1} - 5z^{-5} + 4z^{-6}}{5(1 - z^{-1})^2}
-$$
-  
-\n(i) 
-$$
-\frac{z}{z - 1.1}
-$$
-
-(j) 
-$$
-\frac{0.25z^{-1}}{(1-z^{-1})(1-0.75z^{-1})}
-$$
-
-- **5.2-15** Suppose we upsample a causal signal *x*[*n*] by factor *N* to produce signal *y*[*n*]. Express *Y*(*z*) in terms of *X*(*z*), taking care to mathematically justify your result.
-- **5.3-1** Using *z*-transform techniques, find the output *y*[*n*] of an LTID system specified by the equation *<sup>y</sup>*[*n*]− <sup>1</sup> <sup>3</sup> *y*[*n*−1] = *x*[*n*−1] if the initial condition is *y*[−1] = 2 and the input is *x*[*n*]=−*u*[*n*].
-- **5.3-2** Consider an LTID system *y*[*n*] − *y*[*n* − 2] = *x*[*n*] with *y*[−1] = 0, *y*[−2] = 1, and *x*[*n*] = *u*[*n*].
-  - (a) Determine *Y*[*z*], expressed as a rational function in standard factored form.
-  - (b) Use *Y*[*z*] from part (a) to solve for the system output *y*[*n*].
-- **5.3-3** Solve Prob. 3.8-23 by the *z*-transform method.
-- **5.3-4** Consider a DT system with transfer function *<sup>H</sup>*[*z*] = <sup>2</sup>*z*−<sup>2</sup> *<sup>z</sup>*−0.5 . Assuming the system is both controllable and observable, determine the ZIR *y*zir[*n*] given *y*[−1] = 1.
-- **5.3-5** (a) Solve
-
-$$
-y[n+1] + 2y[n] = x[n+1]
-$$
-
-when 
-$$
-y[0] = 1
-$$
- and  $x[n] = e^{-(n-1)}u[n]$ 
-
-- (b) Find the zero-input and the zero-state components of the response.
-- **5.3-6** Consider a LTID system that is described by the difference equation *<sup>y</sup>*[*n*] − <sup>1</sup> <sup>4</sup> *y*[*n* − 2] = *x*[*n*−1].
-
-- (a) Use transform-domain techniques to determine the zero-state response *y*zsr[*n*] to input *x*[*n*] = 3*u*[*n*−5].
-- (b) Use transform-domain techniques to determine the zero-input response *y*zir[*n*] given *y*zir[−2] = *y*zir[−1] = 1.
-- **5.3-7** (a) Find the output *y*[*n*] of an LTID system specified by the equation
-
-$$
-2y[n+2] - 3y[n+1] + y[n]
-$$
-  
-=  $4x[n+2] - 3x[n+1]$ 
-
-for input *<sup>x</sup>*[*n*] = (4)−*nu*[*n*] and initial conditions *y*[−1] = 0 and *y*[−2] = 1.
-
-- (b) Find the zero-input and the zero-state components of the response.
-- (c) Find the transient and the steady-state components of the response.
-- **5.3-8** Solve Prob. 5.3-7 if initial conditions *y*[−1] and *y*[−2] are instead replaced with auxiliary conditions *y*[0] = 3/2 and *y*[1] = 35/4.
-- **5.3-9** (a) Solve
-
-$$
-4y[n+2] + 4y[n+1] + y[n] = x[n+1]
-$$
-
-with *y*[−1] = 0, *y*[−2] = 1, and *x*[*n*] = *u*[*n*].
-
-- (b) Find the zero-input and the zero-state components of the response.
-- (c) Find the transient and the steady-state components of the response.
-- **5.3-10** Solve
-
-$$
-y[n+2] - 3y[n+1] + 2y[n] = x[n+1]
-$$
-
-if *<sup>y</sup>*[−1] = 2, *<sup>y</sup>*[−2] = 3, and *<sup>x</sup>*[*n*] = (3)*nu*[*n*].
-
-**5.3-11** Solve
-
-$$
-y[n+2] - 2y[n+1] + 2y[n] = x[n]
-$$
-
-with *y*[−1] = 1, *y*[−2] = 0, and *x*[*n*] = *u*[*n*].
-
-- **5.3-12** Consider a causal LTID system described as *<sup>H</sup>*(*z*) <sup>=</sup> <sup>21</sup>(*z*2+1) <sup>16</sup>(*z*2<sup>+</sup> <sup>1</sup> <sup>4</sup> *<sup>z</sup>*<sup>−</sup> <sup>3</sup> 8 ) .
-  - (a) Determine the standard delay-form difference equation description of this system.
-  - (b) Using transform-domain techniques, determine the system impulse response *h*[*n*].
-  - (c) Using transform-domain techniques, determine *y*zir[*n*] given *y*[−1] = 16 and *y*[−2] = 8.
-
-- **5.3-13** Consider a causal LTID system described as *<sup>y</sup>*[*n*] − <sup>5</sup> <sup>6</sup> *<sup>y</sup>*[*<sup>n</sup>* <sup>−</sup> <sup>1</sup>] + <sup>1</sup> <sup>6</sup> *<sup>y</sup>*[*<sup>n</sup>* <sup>−</sup> <sup>2</sup>] = <sup>3</sup> <sup>2</sup> *<sup>x</sup>*[*<sup>n</sup>* <sup>−</sup> <sup>1</sup>] + <sup>3</sup> <sup>2</sup> *x*[*n*−2].
-  - (a) Determine the (standard-form) system transfer function *H*(*z*) and sketch the system pole-zero plot.
-  - (b) Using transform-domain techniques, determine *y*zir[*n*] given *y*[−1] = 2 and *y*[−2]=−2.
-- **5.3-14** Solve
-
-$$
-y[n] + 2y[n-1] + 2y[n-2]
-$$
-  
-=  $x[n-1] + 2x[n-2]$ 
-
-with *<sup>y</sup>*[0] = 0, *<sup>y</sup>*[1] = 1, and *<sup>x</sup>*[*n*] = *enu*[*n*].
-
-- **5.3-15** A system with impulse response *h*[*n*] = <sup>2</sup>(1/3)*nu*[*<sup>n</sup>* <sup>−</sup> <sup>1</sup>] produces an output *<sup>y</sup>*[*n*] = (−2)*nu*[*<sup>n</sup>* <sup>−</sup> <sup>1</sup>]. Determine the corresponding input *x*[*n*].
-- **5.3-16** A professor recently received an unexpected \$10 (a futile bribe attached to a test). Being the savvy investor that she is, the professor decides to invest the \$10 into a savings account that earns 0.5% interest compounded monthly (6.17% APY). Furthermore, she decides to supplement this initial investment with an additional \$5 deposit made every month, beginning the month immediately following her initial investment.
-  - (a) Model the professor's savings account as a constant coefficient linear difference equation. Designate *y*[*n*] as the account balance at month *n*, where *n* = 0 corresponds to the first month that interest is awarded (and that her \$5 deposits begin).
-  - (b) Determine a closed-form solution for *y*[*n*]. That is, you should express *y*[*n*] as a function only of *n*.
-  - (c) If we consider the professor's bank account as a system, what is the system impulse response *h*[*n*]? What is the system transfer function *H*[*z*]?
-  - (d) Explain this fact: if the input to the professor's bank account is the everlasting exponential *<sup>x</sup>*[*n*] = <sup>1</sup>*<sup>n</sup>* <sup>=</sup> 1, then the output is **not** *<sup>y</sup>*[*n*] = <sup>1</sup>*nH*[1] = *<sup>H</sup>*[1].
-- **5.3-17** Sally deposits \$100 into her savings account on the first day of every month except for each December, when she uses her money to buy
-
-holiday gifts. Define *b*[*m*] as the balance in Sally's account on the first day of month *m*. Assume Sally opens her account in January (*m* = 0), continues making monthly payments forever (except each December!), and that her monthly interest rate is 1%. Sally's account balance satisfies a simple difference equation *b*[*m*] = (1.01)*b*[*m* − 1] + *p*[*m*], where *p*[*m*] designates Sally's monthly deposits. Determine a closed-form expression for *b*[*m*] that is only a function of the month *m*.
-
-- **5.3-18** For each impulse response, determine the number of system poles, whether the poles are real or complex, and whether the system is BIBO-stable.
-  - (a) *<sup>h</sup>*1[*n*] = (−1+(0.5)*n*)*u*[*n*]
-  - (b) *<sup>h</sup>*2[*n*] = (*j*)*n*(*u*[*n*] −*u*[*n*−10])
-- **5.3-19** Find the following sums:
-
-(a) 
-$$
-\sum_{k=0}^{n} k
-$$
-  
-(b) 
-$$
-\sum_{k=0}^{n} k^2
-$$
-
-*k*=0 [*Hint:* Consider a system whose output *y*[*n*] is the desired sum. Examine the relationship between *y*[*n*] and *y*[*n* − 1]. Note also that *y*[0] = 0.]
-
-### **5.3-20** Find the following sum:
-
-$$
-\sum_{k=0}^{n} k^3
-$$
-
-[*Hint:* See the hint for Prob. 5.3-19.]
-
-**5.3-21** Find the following sum:
-
-$$
-\sum_{k=0}^{n} ka^k \qquad a \neq 1
-$$
-
-[*Hint:* See the hint for Prob. 5.3-19.]
-
-- **5.3-22** Redo Prob. 5.3-19 using the result in Prob. 5.2-13a.
-- **5.3-23** Redo Prob. 5.3-20 using the result in Prob. 5.2-13a.
-- **5.3-24** Redo Prob. 5.3-21 using the result in Prob. 5.2-13a.
-
-**5.3-25** (a) Find the zero-state response of an LTID system with transfer function
-
-$$
-H[z] = \frac{z}{(z+0.2)(z-0.8)}
-$$
-
-and the input *<sup>x</sup>*[*n*] = *<sup>e</sup>*(*n*+1) *u*[*n*].
-
-(b) Write the difference equation relating the output *y*[*n*] to input *x*[*n*].
-
-**5.3-26** Repeat Prob. 5.3-25 for *x*[*n*] = *u*[*n*] and
-
-$$
-H[z] = \frac{2z+3}{(z-2)(z-3)}
-$$
-
-**5.3-27** Repeat Prob. 5.3-25 for
-
-$$
-H[z] = \frac{6(5z - 1)}{6z^2 - 5z + 1}
-$$
-
-- and the input *x*[*n*] is (a) (4)−*nu*[*n*] (b) (4)−(*n*−2) *u*[*n*−2] (c) (4)−(*n*−2) *u*[*n*] (d) (4)−*nu*[*n*−2]
-- **5.3-28** Repeat Prob. 5.3-25 for *x*[*n*] = *u*[*n*] and
-
-$$
-H[z] = \frac{2z - 1}{z^2 - 1.6z + 0.8}
-$$
-
-- **5.3-29** Find the transfer functions corresponding to each of the systems specified by difference equations in Probs. 5.3-5, 5.3-7, 5.3-9, and 5.3-14.
-- **5.3-30** Find *h*[*n*], the unit impulse response of the systems described by the following equations:
-  - (a) *y*[*n*] +3*y*[*n*−1] +2*y*[*n*−2] = *x*[*n*] +3*x*[*n*− 1] +3*x*[*n*−2]
-  - (b) *y*[*n* + 2] + 2*y*[*n* + 1] + *y*[*n*] = 2*x*[*n* + 2] − *x*[*n*+1]
-  - (c) *y*[*n*]−*y*[*n*−1]+0.5*y*[*n*−2] = *x*[*n*]+2*x*[*n*− 1]
-- **5.3-31** Find *h*[*n*], the unit impulse response of the systems in Probs. 5.3-25, 5.3-26, and 5.3-28.
-- **5.3-32** A system has impulse response *h*[*n*] = *u*[*n* − 3].
-  - (a) Determine the impulse response of the inverse system *h*−1[*n*].
-  - (b) Is the inverse stable? Is the inverse causal?
-  - (c) Your boss asks you to implement *h*−1[*n*] to the best of your ability. Describe your realizable design, taking care to identify any deficiencies.
-
-**5.4-1** A system has impulse response given by
-
-$$
-h[n] = \left[ \left( \frac{1+j}{\sqrt{8}} \right)^n + \left( \frac{1-j}{\sqrt{8}} \right)^n \right] u[n]
-$$
-
-This system can be implemented according to Fig. P5.4-1.
-
-**Figure P5.4-1**
-
-- (a) Determine the coefficients *A*<sup>1</sup> and *A*<sup>2</sup> to implement *h*[*n*] using the structure shown in Fig. P5.4-1.
-- (b) What is the zero-state response *y*0[*n*] of this system, given a shifted unit step input *x*[*n*] = *u*[*n*+3]?
-- **5.4-2** (a) Show the canonic direct form, a cascade, and a parallel realization of
-
-$$
-H[z] = \frac{z(3z - 1.8)}{z^2 - z + 0.16}
-$$
-
-- (b) Find the transpose of the realizations obtained in part (a).
-- **5.4-3** Repeat Prob. 5.4-2 for
-
-$$
-H[z] = \frac{5z + 2.2}{z^2 + z + 0.16}
-$$
-
-**5.4-4** Repeat Prob. 5.4-2 for
-
-$$
-H[z] = \frac{3.8z - 1.1}{(z - 0.2)(z^2 - 0.6z + 0.25)}
-$$
-
-**5.4-5** Repeat Prob. 5.4-2 for
-
-$$
-H[z] = \frac{z(1.6z - 1.8)}{(z - 0.2)(z^2 + z + 0.5)}
-$$
-
-**5.4-6** Repeat Prob. 5.4-2 for
-
-$$
-H[z] = \frac{z(2z^2 + 1.3z + 0.96)}{(z + 0.5)(z - 0.4)^2}
-$$
-
-**5.4-7** Consider the LTID system shown in Fig. P5.4-7.
-
-$$
-x[n]
-$$
-  $\rightarrow$   $z^{-1}$   $\rightarrow$   $z^{-1}$   $\rightarrow$   $z^{-1}$   $\rightarrow$   $y[n]$ 
-
-- **Figure P5.4-7**
-- (a) Determine the standard delay-form difference equation description of this system.
-- (b) Determine the impulse response *h*[*n*] of this system.
-- (c) Is this realization canonical? Explain.
-- (d) Is this system stable? Explain.
-- (e) Is this system causal? Explain.
-- **5.4-8** Realize a system whose transfer function is
-
-$$
-H[z] = \frac{2z^4 + z^3 + 0.8z^2 + 2z + 8}{z^4}
-$$
-
-**5.4-9** Realize a system whose transfer function is given by
-
-$$
-H[z] = \sum_{n=0}^{6} nz^{-n}
-$$
-
-**5.4-10** Consider the LTID system shown in Fig. P5.4-10, where parameter *c* is an arbitrary, real constant.
-
-**Figure P5.4-10**
-
-- (a) Determine the system transfer function *H*[*z*], expressed in standard rational form.
-- (b) Determine all system poles and all system zeros.
-- (c) Is the system of Fig. P5.4-10 canonical? Explain.
-- (d) What constraints, if any, exist on parameter *c* to ensure that the system is stable?
-- **5.4-11** This problem demonstrates the enormous number of ways of implementing even a relatively low-order transfer function. A second-order transfer function has two real zeros and two real poles. Discuss various ways of realizing
-
-such a transfer function. Consider canonic direct, cascade, parallel, and the corresponding transposed forms. Note also that interchange of cascaded sections yields a different realization.
-
-**5.4-12** Consider a digital audio system: an input analog-to-digital converter (ADC) is used to collect input samples at a CD-quality rate *Fs* = 44 kHz. Input samples are processed with a digital filter to generate output samples, which are sent to a digital-to-analog converter (DAC) at the same rate *Fs*. Every sample interval *T* = 1/*Fs*, the digital processor executes the following MATLAB-compatible code:
-
-```
-% Read input sample from the ADC
-x = read_ADC;
-% Process input and...
-mem(1) = x - mem(3)*9/16;
-% ...compute output sample
-y = mem(1)*7/16 - mem(3)*7/16;
-% Send output sample to the DAC
-write_DAC = y;
-% Update memory for next iteration
-mem(3) = mem(2);
-mem(2) = mem(1);
-```
-
-- (a) Does the code implement DFI, DFII, TDFI, or TDFII? Support your answer by drawing the appropriate block diagram labeled in a manner that is consistent with the code.
-- (b) Determine the transfer function *H*[*z*] of this system.
-- (c) What is the basic filtering function of this system: LP, HP, BP, or BS? Justify your answer.
-- (d) Determine the transfer function *H*−1[*z*] of the inverse system to *H*[*z*] and draw its DFI block implementation. How well will the inverse system operate?
-- **5.4-13** Repeat Prob. 5.4-12 but instead use the code:
-
-```
-% Read input sample from the ADC
-x = read_ADC;
-% Compute output sample
-y = x*7/32+mem(1);
-% Send output sample to the DAC
-write_DAC = y;
-% Update memory for next iteration
-mem(1) = mem(2);
-mem(2) = x*7/32 + y*9/16;
-```
-
-- **5.5-1** A CT sinusoid *x*(*t*) = cos(ω*t*) is sampled at a greater-than-Nyquist rate *F*<sup>s</sup> = 1000 Hz to produce a DT sinusoid *x*(*t*) = cos(*n*). Determine the analog frequency ω if (a) <sup>=</sup> <sup>π</sup> 4
-  - (b) <sup>=</sup> <sup>2</sup><sup>π</sup> 3
-  - (c) <sup>=</sup> <sup>7</sup> 8
-- **5.5-2** Find the amplitude and phase response of the digital filters depicted in Fig. P5.5-2.
-- **5.5-3** A causal LTID system *<sup>H</sup>*(*z*) <sup>=</sup> <sup>21</sup>(*z*−*j*)(*z*+*j*) <sup>16</sup>(*z*<sup>−</sup> <sup>1</sup> <sup>2</sup> )(*z*<sup>+</sup> <sup>3</sup> 4 ) has a periodic input *x*[*n*] that toggles between the values 1 and 2. That is, *x*[*n*]=[..., 1, 2, 1, ↓ 2 , 1, 2, 1, ...], where *x*[0] = 2.
-  - (a) Plot the magnitude response <sup>|</sup>*H*(*ej*)<sup>|</sup> over −2π ≤ ≤ 2π.
-  - (b) Plot the phase response *H*(*ej*) over −2π ≤ ≤ 2π.
-  - (c) Determine the system output *y*[*n*] in response to the periodic input *x*[*n*].
-
-**5.5-4** A causal LTID system *<sup>H</sup>*(*z*) <sup>=</sup> <sup>−</sup>7(*z*+1) <sup>32</sup>(*z*−*<sup>j</sup>* <sup>3</sup> <sup>4</sup> )(*z*+*<sup>j</sup>* <sup>3</sup> 4 ) has a periodic input *x*[*n*] that cycles through the 4 values 3, 2, 1, and 2. That is, *x*[*n*] = [..., 3, 2, 1, 2, ↓ 3, 2, 1, 2, ...], where *x*[0] = 3.
-
-- (a) Plot the magnitude response <sup>|</sup>*H*(*ej*)<sup>|</sup> over −2π ≤ ≤ 2π.
-- (b) Plot the phase response *H*(*ej*) over −2π ≤ ≤ 2π.
-- (c) Determine the system output *y*[*n*] in response to the periodic input *x*[*n*].
-- **5.5-5** Find the amplitude and the phase response of the filters shown in Fig. P5.5-5. [*Hint:* Express *<sup>H</sup>*[*ej*] as *<sup>e</sup>*−*j*2.5*Ha*[*ej*].]
-- **5.5-6** Find the frequency response for the moving-average system in Prob. 3.4-3. The input–output equation of this system is given by
-
-$$
-y[n] = \frac{1}{5} \sum_{k=0}^{4} x[n-k]
-$$
-
-**5.5-7** (a) Input–output relationships of two filters are described by
-
-(i) *y*[*n*]=−0.9*y*[*n*−1] +*x*[*n*]
-
-(ii) *y*[*n*] = 0.9*y*[*n*−1] +*x*[*n*]
-
-For each case, find the transfer function, the amplitude response, and the phase response. Sketch the amplitude response, and state the type (highpass, lowpass, etc.) of each filter.
-
-- (b) Find the response of each of these filters to a sinusoid *x*[*n*] = cos*n* for = 0.01π and 0.99π. In general, show that the gain (amplitude response) of filter (i) at frequency <sup>0</sup> is the same as the gain of filter (ii) at frequency π −0.
-- **5.5-8** For an LTID system specified by the equation
-
-$$
-y[n+1] - 0.5y[n] = x[n+1] + 0.8x[n]
-$$
-
-(a) Find the amplitude and the phase response.
-
-- (b) Find the system response *y*[*n*] for the input *x*[*n*] = cos(0.5*k* −(π/3)).
-- **5.5-9** For an asymptotically stable LTID system, show that the steady-state response to input *ejnu*[*n*] is *<sup>H</sup>*[*ej*]*ejnu*[*n*]. The steady-state response is that part of the response which does not decay with time and persists forever.
-- **5.5-10** Express the following signals in terms of apparent frequencies:
-  - (a) cos(0.8π*n*+θ )
-  - (b) sin(1.2π*n*+θ )
-  - (c) cos(6.9*n*+θ )
-
-**Figure P5.5-2**
-
-(b)
-
-**Figure P5.5-5**
-
-- (d) cos(2.8π*n*+θ ) +2 sin(3.7π*n*+θ )
-- (e) sinc(π*n*/2)
-- (f) sinc(3π*n*/2)
-- (g) sinc(2π*n*)
-- **5.5-11** Show that cos √ (0.6π*n* + (π/6)) + 3 cos(1.4π*n* + (π/3)) = 2 cos(0.6π*n* − (π/6)).
-- **5.5-12** (a) A digital filter has the sampling interval *T* = 50µs. Determine the highest frequency that can be processed by this filter without aliasing.
-  - (b) If the highest frequency to be processed is 50 kHz, determine the minimum value of the sampling frequency *F<sup>s</sup>* and the maximum value of the sampling interval *T* that can be used.
-- **5.5-13** Consider the discrete-time system represented by
-
-$$
-y[n] = \sum_{k=0}^{\infty} (0.5)^k x[n-k]
-$$
-
-- (a) Determine and plot the magnitude response <sup>|</sup>*H*[*ej*]| of the system.
-- (b) Determine and plot the phase response *<sup>H</sup>*[*ej*] of the system.
-- (c) Find an efficient block representation that implements this system.
-- **5.6-1** Pole-zero configurations of certain filters are shown in Fig. P5.6-1. Sketch roughly the amplitude response of these filters.
-- **5.6-2** Figure P5.6-2 displays the pole-zero plot of a second-order real, causal LTID system that has *H*[−1]=−1.
-
-**Figure P5.6-2**
-
-- (a) Determine the five constants *b*0, *b*1, *b*2, *a*1, and *a*<sup>2</sup> that specify the transfer function *<sup>H</sup>*[*z*] = *<sup>b</sup>*0*z*2+*b*1*z*+*b*<sup>2</sup> *<sup>z</sup>*2+*a*1*z*+*a*<sup>2</sup> .
-- (b) Using the techniques of Sec. 5.6, accurately hand-sketch the system magnitude response <sup>|</sup>*H*[*ej*]| over the range (−2<sup>π</sup> <sup>≤</sup> <sup>≤</sup> <sup>0</sup>).
-- (c) Determine the output *y*[*n*] of this system if the input is *x*[*n*] = sin <sup>π</sup>*<sup>n</sup>* 2 .
-- **5.6-3** Repeat Prob. 5.6-2 if the zero at *z* = 1 is moved to *z* = −1 and *H*[1]=−1 is specified rather than *H*[−1]=−1.
-- **5.6-4** Figure P5.6-4 displays the pole-zero plot of a second-order real, causal LTID system that has *H*[−1] = 1.
-  - (a) Determine the five constants *k*, *b*1, *b*2, *a*1, and *a*<sup>2</sup> that specify the transfer function *<sup>H</sup>*[*z*] = *<sup>k</sup> <sup>z</sup>*2+*b*1*z*+*b*<sup>2</sup> *<sup>z</sup>*2+*a*1*z*+*a*<sup>2</sup> .
-
-**Figure P5.6-4**
-
-- (b) Using the techniques of Sec. 5.6, accurately hand-sketch the system magnitude response <sup>|</sup>*H*[*ej*]| over the range (−<sup>π</sup> <sup>≤</sup> <sup>≤</sup> π ).
-- (c) A signal *x*(*t*) = cos(100π*t*) + cos(500π*t*) is sampled at a greater than Nyquist rate *F*<sup>s</sup> Hz and then input into the above LTID system to produce DT output *y*[*n*] = β cos(0*n* + θ ). Determine *F*<sup>s</sup> and 0. You do not need to find constants β and θ.
-- (d) Is the inpulse response *h*[*n*] of this system absolutely summable? Justify your answer.
-- **5.6-5** Figure P5.6-5 displays the pole-zero plot of a second-order real, causal LTID system that has *H*[1]=−1.
-
-(a) Determine the five constants *k*, *b*1, *b*2, *a*1, and *a*<sup>2</sup> that specify the transfer function *<sup>H</sup>*[*z*] = *<sup>k</sup> <sup>z</sup>*2+*b*1*z*+*b*<sup>2</sup> *<sup>z</sup>*2+*a*1*z*+*a*<sup>2</sup> .
-
-**Figure P5.6-5**
-
-- (b) Using the techniques of Sec. 5.6, accurately hand-sketch the system magnitude response <sup>|</sup>*H*[*ej*]| over the range (−<sup>π</sup> <sup>≤</sup> <sup>≤</sup> π ).
-- (c) A signal *x*(*t*) = cos(2π*ft*) is sampled at a rate *F*<sup>s</sup> = 1 kHz and then input into the above LTID system to produce DT output *y*[*n*]. Determine, if possible, the frequency or frequencies *f* that will produce zero output, *y*[*n*] = 0.
-- **5.6-6** The system *y*[*n*] − *y*[*n* − 1] = *x*[*n*] − *x*[*n* − 1] is an all-pass system that has zero phase response. Is there any difference between this system and the system *y*[*n*] = *x*[*n*]? Justify your answer.
-- **5.6-7** Figure P5.6-7 displays the pole-zero plot of a second-order real, causal LTID system that has a repeated zero and *H*[1] = 4. The solid circle is the unit circle.
-
-### **Figure P5.6-7**
-
-- (a) Determine the five constants *k*, *b*1, *b*2, *a*1, and *a*<sup>2</sup> that specify the transfer function *<sup>H</sup>*[*z*] = *<sup>k</sup> <sup>z</sup>*2+*b*1*z*+*b*<sup>2</sup> *<sup>z</sup>*2+*a*1*z*+*a*<sup>2</sup> .
-- (b) Using the techniques of Sec. 5.6, accurately hand-sketch the system magnitude response <sup>|</sup>*H*[*ej*]| over the range (−<sup>π</sup> <sup>≤</sup> <sup>≤</sup> π ).
-- (c) Determine the steady-state output *y*ss[*n*] of this system if the input is *x*[*n*] = cos( <sup>3</sup>π*<sup>n</sup>* <sup>4</sup> )*u*[*n*].
-- (d) State whether this system is LP, HP, BP, BS, or other. If the digital system operates at *F*<sup>s</sup> = 8 kHz, what is the approximate hertzian cutoff frequency (or frequencies) of this system?
-- **5.6-8** The magnitude and phase responses of a real, stable, LTI system are shown in Fig. P5.6-8.
-  - (a) What type of system is this: lowpass, highpass, bandpass, or bandstop?
-  - (b) What is the output of this system in response to
-
-$$
-x_1[n] = 2\sin\left(\frac{\pi}{2}n + \frac{\pi}{4}\right)
-$$
-
-(c) What is the output of this system in response to
-
-$$
-x_2[n] = \cos\left(\frac{7\pi}{4}n\right)
-$$
-
-- **5.6-9** Consider an LTID system with system function *<sup>H</sup>*[*z*] = *<sup>b</sup>*<sup>0</sup> *<sup>z</sup>*2+<sup>1</sup> *<sup>z</sup>*2−9/<sup>16</sup> .
-  - (a) Determine the constant *b*<sup>0</sup> so that the system frequency response at = −π is −1.
-  - (b) Accurately sketch the system poles and zeros.
-  - (c) Using the locations of the system poles and zeros, sketch <sup>|</sup>*H*[*ej*]| over 0 <sup>≤</sup> <sup>≤</sup> <sup>2</sup>π.
-
-- (d) Determine the response *y*[*n*] to the input *x*[*n*] = (−1+*j*)+*j <sup>n</sup>* <sup>+</sup>(1−*j*)sin(π*n*+1).
-- (e) Draw an appropriate block diagram representation of this system.
-- **5.6-10** Do Prob. 5.10-3 by graphical procedure. Do the sketches approximately, without using MATLAB.
-- **5.6-11** Do Prob. 5.10-8 by graphical procedure. Do the sketches approximately, without using MATLAB.
-- **5.6-12** (a) Realize a digital filter whose transfer function is given by
-
-$$
-H[z] = K \frac{z+1}{z-a}
-$$
-
-- (b) Sketch the amplitude response of this filter, assuming |*a*| < 1.
-- (c) The amplitude response of this lowpass filter is maximum at = 0. The 3 dB bandwidth is the frequency at which the amplitude response drops to 0.707 (or 1/ <sup>√</sup>2) times its maximum value. Determine the 3 dB bandwidth of this filter when *a* = 0.2.
-- **5.6-13** Design a digital notch filter to reject frequency 5000 Hz completely and to have a sharp recovery on either side of 5000 Hz to a gain of unity. The highest frequency to be processed is 20 kHz (*F<sup>h</sup>* = 20,000). [*Hint:* See Ex. 5.15. The zeros should be at *e*±*j*ω*<sup>T</sup>* for ω corresponding to 5000 Hz, and the poles are at *ae*±*j*ω*<sup>T</sup>* with *a* < 1. Leave your answer in terms of *a*. Realize this filter using the canonical form. Find the amplitude response of the filter.]
-
-**5.6-14** Consider the desired DT system magnitude response <sup>|</sup>*H*[*ej*]| in Fig. P5.6-14.
-
-**Figure P5.6-14**
-
-- (a) Is the filter LP, HP, BP, BS, or other? Explain.
-- (b) Sketch the pole-zero plot of a 2nd-order system that behaves as a reasonable approximation of Fig. P5.6-14. What is the coefficient *b*0?
-- **5.6-15** Show that a first-order LTID system with a pole at *z* = *r* and a zero at *z* = 1/*r* (*r* ≤ 1) is an allpass filter. In other words, show that the amplitude response <sup>|</sup>*H*[*ej*]| of a system with the transfer function
-
-$$
-H[z] = \frac{z - \frac{1}{r}}{z - r} \qquad r \le 1
-$$
-
-is constant with frequency. This is a first-order allpass filter. [*Hint:* Show that the ratio of the distances of any point on the unit circle from the zero (at *z* = 1/*r*) and the pole (at *z* = *r*) is a constant 1/*r*.]
-
-Generalize this result to show that an LTID system with two poles at *<sup>z</sup>* <sup>=</sup> *re*±*j*<sup>θ</sup> and two zeros at *<sup>z</sup>* <sup>=</sup> (1/*r*)*e*±*j*<sup>θ</sup> (*<sup>r</sup>* <sup>≤</sup> <sup>1</sup>) is an allpass filter. In other words, show that the amplitude response of a system with the transfer function
-
-$$
-H[z] = \frac{\left(z - \frac{1}{r}e^{j\theta}\right)\left(z - \frac{1}{r}e^{-j\theta}\right)}{(z - re^{j\theta})(z - re^{-j\theta})}
-$$
-$$
-= \frac{z^2 - \left(\frac{2}{r}\cos\theta\right)z + \frac{1}{r^2}}{z^2 - (2r\cos\theta)z + r^2} \qquad r \le 1
-$$
-
-is constant with frequency.
-
-**5.6-16** (a) If *h*1[*n*] and *h*2[*n*], the impulse responses of two LTID systems are related by *h*2[*n*] = (−1)*nh*1[*n*], then show that
-
-$$
-H_2[e^{j\Omega}] = H_1[e^{j(\Omega \pm \pi)}]
-$$
-
-How is the frequency response spectrum *<sup>H</sup>*2[*ej*] related to the *<sup>H</sup>*1[*ej*]?
-
-- (b) If *H*1[*z*] represents an ideal lowpass filter with cutoff frequency *c*, sketch *<sup>H</sup>*2[*ej*]. What type of filter is *<sup>H</sup>*2[*ej*]?
-- **5.6-17** Mappings such as the bilinear transformation are useful in the conversion of continuous-time filters to discrete-time filters. Another useful type of transformation is one that converts a discrete-time filter into a different type of discrete-time filter. Consider a transformation that replaces *z* with −*z*.
-  - (a) Show that this transformation converts lowpass filters into highpass filters and highpass filters into lowpass filters.
-  - (b) If the original filter is an FIR filter with impulse response *h*[*n*], what is the impulse response of the transformed filter?
-- **5.6-18** The bilinear transformation is defined by the rule *s* = 2(1−*z*−1)/*T*(1+*z*−1).
-  - (a) Show that this transformation maps the ω axis in the *<sup>s</sup>* plane to the unit circle *<sup>z</sup>* <sup>=</sup> *ej* in the *z* plane.
-  - (b) Show that this transformation maps to 2 arctan(ω*T*/2).
-- **5.7-1** In Ch. 3, we used another approximation to find a digital system to realize an analog system. We showed that an analog system specified by Eq. (3.12) can be realized by using the digital system specified by Eq. (3.13). Compare that solution with the one resulting from the
-
-impulse-invariance method. Show that one result is a close approximation of the other and that the approximation improves as *T* → 0.
-
-- **5.7-2** A CT system has impulse response *h*ct(*t*) = *e*−*<sup>t</sup> u*(*t*). Draw the DFI realization of the corresponding DT system designed by the impulse-invariance method with *T* = 0.1.
-- **5.7-3** (a) Using the impulse-invariance criterion, design a digital filter to realize an analog filter with transfer function
-
-$$
-H_a(s) = \frac{7s + 20}{2(s^2 + 7s + 10)}
-$$
-
-- (b) Show a canonical and a parallel realization of the filter. Use a 1% criterion for the choice of *T*.
-- **5.7-4** Use the impulse-invariance criterion to design a digital filter to realize the second-order analog Butterworth filter with transfer function
-
-$$
-H_a(s) = \frac{1}{s^2 + \sqrt{2}s + 1}
-$$
-
-Use a 1% criterion for the choice of *T*.
-
-- **5.7-5** Design a digital integrator using the impulse-invariance method. Find and give a rough sketch of the amplitude response, and compare it with that of the ideal integrator. If this integrator is used primarily for integrating audio signals (whose bandwidth is 20 kHz), determine a suitable value for *T*.
-- **5.7-6** An oscillator by definition is a source (no input) that generates a sinusoid of a certain frequency ω0. Therefore, an oscillator is a system whose zero-input response is a sinusoid of the desired frequency. Find the transfer function of a digital oscillator to oscillate at 10 kHz by the methods described in parts (a) and (b). In both methods, select *T* so that there are 10 samples in each cycle of the sinusoid.
-  - (a) Choose *H*[*z*] directly so that its zero-input response is a discrete-time sinusoid of frequency =ω*T* corresponding to 10 kHz.
-  - (b) Choose *Ha*(*s*) whose zero-input response is an analog sinusoid of 10 kHz. Now use the impulse invariance method to determine *H*[*z*].
-  - (c) Show a canonical realization of the oscillator.
-
-- **5.7-7** A variant of the impulse invariance method is the *step-invariance* method of digital filter synthesis. In this method, for a given *Ha*(*s*), we design *H*[*z*] in Fig. 5.24a such that *y*(*nT*) in Fig. 5.24b is identical to *y*[*n*] in Fig. 5.24a when *x*(*t*) = *u*(*t*).
-  - (a) Show that, in general,
-
-$$
-H[z] = \frac{z-1}{z} \mathcal{Z} \left[ \left( \mathcal{L}^{-1} \frac{H_a(s)}{s} \right)_{t=kT} \right]
-$$
-
-(b) Use this method to design *H*[*z*] for
-
-$$
-H_a(s) = \frac{\omega_c}{s + \omega_c}
-$$
-
-- (c) Use the step-invariance method to synthesize a discrete-time integrator and compare its amplitude response with that of the ideal integrator.
-- **5.7-8** Use the *ramp-invariance* method to synthesize a discrete-time differentiator and integrator. In this method, for a given *Ha*(*s*), we design *H*[*z*] such that *y*(*nT*) in Fig. 5.24b is identical to *y*[*n*] in Fig. 5.24a when *x*(*t*) = *tu*(*t*).
-- **5.7-9** In an impulse-invariance design, show that if *Ha*(*s*) is a transfer function of a stable system, the corresponding *H*[*z*] is also a transfer function of a stable system.
-- **5.7-10** First-order backward differences provide the transformation rule *s* = (1−*z*−1)/*T*.
-  - (a) Show that this transformation maps the ω axis in the *s* plane to a circle of radius 1/2 centered at (1/2, 0) in the *z* plane.
-  - (b) Show that this transformation maps the left-half *s* plane to the interior of the unit circle in the *z* plane, which ensures that stability is preserved.
-- **5.8-1** Find the *z*-transform (if it exists) and the corresponding ROC for each of the following signals:
-  - (a) (0.8)*nu*[*n*] +2*nu*[−(*n*+1)]
-
-(b) 
-$$
-2^n u[n] - 3^n u[-(n+1)]
-$$
-
-(c) 
-$$
-(-2)^{n+3}u[-n] + \sum_{k=0}^{\infty} (0.5)^{k-1} \delta(n-2k)
-$$
-
-(d) 
-$$
-(0.8)^n u[n] + (0.9)^n u[-(n+1)]
-$$
-
-(e) 
-$$
-[(0.8)^n + 3(0.4)^n]u[-(n+1)]
-$$
-
-(f) [(0.8)*<sup>n</sup>* <sup>+</sup>3(0.4)*n*]*u*[*n*]
-
-(g) 
-$$
-(0.8)^n u[n] + 3(0.4)^n u[-(n+1)]
-$$
-
-$$
-(0.6) \quad u[n] + 5(0.4) \quad u[
-$$
-
-(h) 
-$$
-(0.5)^{|n|}
-$$
-
-$$
-(i) \t n u[-(n+1)]
-$$
-
-- **5.8-2** Using the definition, compute the bilateral *z*-transform *X*(*z*) of (a) *<sup>x</sup>*[*n*] = <sup>3</sup>*nu*[−*n*] (b) *<sup>x</sup>*[*n*] = ( <sup>1</sup> <sup>3</sup> )*nu*[*n*] Express your answers in standard rational form.
-- **5.8-3** Determine the inverse *z*-transform *x*[*n*] of *<sup>X</sup>*[*z*] = *<sup>z</sup>*2<sup>−</sup> <sup>1</sup> 3 *z* (*z*−1)(*z*+2) with ROC 1 <sup>&</sup>lt; <sup>|</sup>*z*<sup>|</sup> <sup>&</sup>lt; 2.
-
-**5.8-4** Find the inverse 
-$$
-z
-$$
--transform of
-
-$$
-X[z] = \frac{(e^{-2} - 2)z}{(z - e^{-2})(z - 2)}
-$$
-
-when the ROC is,  
-\n(a) 
-$$
-|z| > 2
-$$
-  
-\n(b)  $e^{-2} < |z| < 2$   
-\n(c)  $|z| < e^{-2}$ 
-
-**5.8-5** Use partial fraction expansions, *z*-transform tables, and a region of convergence (|*z*| < 1/2) to determine the inverse *z*-transform of
-
-$$
-X(z) = \frac{1}{(2z+1)(z+1)(z+\frac{1}{2})}
-$$
-
-- **5.8-6** Using *z*-transform techniques and properties (no time-domain convolution sum!), determine the convolution *<sup>y</sup>*[*n*] = ( <sup>1</sup> <sup>3</sup> )*n*−3*u*[*<sup>n</sup>* <sup>−</sup> <sup>2</sup>] ∗ (2)*nu*[−*n*]. Express your answer in the form *y*[*n*] = *c*1γ *<sup>n</sup>* <sup>1</sup> *<sup>u</sup>*[*<sup>n</sup>* <sup>+</sup> *<sup>N</sup>*1] + *<sup>c</sup>*2<sup>γ</sup> *<sup>n</sup>* <sup>2</sup> *u*[−*n* + *N*2], making sure to clearly identify the constants *c*1, *c*2, γ1, γ2, *N*1, and *N*2.
-- **5.8-7** Using partial fraction expansions, *z*-transform tables, and the fact that *h*[*n*] is stable, determine the inverse *z*-transform of
-
-$$
-H[z] = \frac{z^4 + z^3}{(z - 2)(z + \frac{1}{2})}.
-$$
-
-**5.8-8** Consider the system
-
-$$
-H[z] = \frac{z(z - \frac{1}{2})}{(z^3 - \frac{27}{8})}
-$$
-
-- (a) Draw the pole-zero diagram for *H*[*z*] and identify all possible regions of convergence.
-- (b) Draw the pole-zero diagram for *H*−1[*z*] and identify all possible regions of convergence.
-- **5.8-9** A discrete-time signal *x*[*n*] has a rational *z*-transform that contains a pole at *z* = 0.5. Given *<sup>x</sup>*1[*n*] = (1/3)*nx*[*n*] is absolutely summable and
-
-Problems 589
-
-*<sup>x</sup>*2[*n*] = (1/4)*nx*[*n*] is **not** absolutely summable, determine whether *x*[*n*] is left-sided, right-sided, or two-sided. Justify your answer!
-
-- **5.8-10** Let *x*[*n*] be an absolutely summable signal with rational *z*-transform *X*[*z*]. *X*[*z*] is known to have a pole at *z* = (0.75+0.75*j*), and other poles may be present. Recall that an absolutely summable signal satisfies %<sup>∞</sup> −∞ |*x*[*n*]| < ∞.
-  - (a) Can *x*[*n*] be left-sided? Explain.
-  - (b) Can *x*[*n*] be right-sided? Explain.
-  - (c) Can *x*[*n*] be two-sided? Explain.
-  - (d) Can *x*[*n*] be of finite duration? Explain.
-- **5.8-11** Consider a causal system that has transfer function
-
-$$
-H[z] = \frac{z - 0.5}{z + 0.5}
-$$
-
-When appropriate, assume initial conditions of zero.
-
-- (a) Determine the output *y*1[*n*] of this system in response to *<sup>x</sup>*1[*n*] = (3/4)*nu*[*n*].
-- (b) Determine the output *y*2[*n*] of this system in response to *<sup>x</sup>*2[*n*] = (3/4)*n*.
-- (c) Determine the output *y*3[*n*] of this system in response to *<sup>x</sup>*3[*n*] = (3/4)*nu*[−*n*−1].
-- **5.8-12** Let *<sup>x</sup>*[*n*] =(−1)*nu*[*n*−*n*0]+α*nu*[−*n*]. Determine the constraints on the complex number α and the integer *n*<sup>0</sup> so that the *z*-transform *X*[*z*] exists with region of convergence 1 < |*z*| < 2.
-- **5.8-13** Using the definition, compute the bilateral *z*-transform, including the region of convergence (ROC), of the following complex-valued functions:
-  - (a) *<sup>x</sup>*1[*n*] = (−*j*)−*nu*[−*n*] +δ[−*n*]
-  - (b) *<sup>x</sup>*2[*n*] = (*j*)*<sup>n</sup>* cos(*n*+1)*u*[*n*]
-  - (c) *x*3[*n*] = *j*sinh(*n*)*u*[−*n*+1]
-  - (d) *<sup>x</sup>*4[*n*] = %<sup>0</sup> *<sup>k</sup>*=−∞(2*j*)*n*δ[*n*−2*k*]
-- **5.8-14** Use partial fraction expansions, *z*-transform tables, and a region of convergence (0.5 < |*z*| < 2) to determine the inverse *z*-transform of
-  - (a) *X*1[*z*] = 1 <sup>1</sup><sup>+</sup> <sup>13</sup> <sup>6</sup> *<sup>z</sup>*−<sup>1</sup> <sup>+</sup> <sup>1</sup> <sup>6</sup> *<sup>z</sup>*−<sup>2</sup> <sup>−</sup> <sup>1</sup> <sup>3</sup> *z*−<sup>3</sup> 1
-
-(b) 
-$$
-X_2[z] = \frac{1}{z^{-3}(2 - z^{-1})(1 + 2z^{-1})}
-$$
-
-**5.8-15** Use partial fraction expansions, *z*-transform tables, and the fact that the systems are stable to determine the inverse *z*-transform of
-
-(a) 
-$$
-H_1[z] = \frac{z^{-1}}{(z - \frac{1}{2})(1 + \frac{1}{2}z^{-1})}
-$$
-
-(b) 
-$$
-H_2[z] = \frac{z+1}{z^3(z-2)(z+\frac{1}{2})}
-$$
-
-**5.8-16** By inserting *N* − 1 zeros between every sample of a unit step, we obtain a signal
-
-$$
-h[n] = \sum_{k=0}^{\infty} \delta[n - Nk]
-$$
-
-Determine *H*[*z*], the bilateral *z*-transform of *h*[*n*]. Identify the number and location(s) of the poles of *H*[*z*].
-
-- **5.8-17** Using transform-domain techniques, determine the zero-state response *y*zsr[*n*] of LTID system *<sup>y</sup>*[*n*] − <sup>1</sup> <sup>4</sup> *y*[*n* − 2] = *x*[*n*] to the noncausal input *<sup>x</sup>*[*n*] = <sup>2</sup>*nu*[2−*n*].
-- **5.8-18** Determine the zero-state response of a system having a transfer function
-
-$$
-H[z] = \frac{z}{(z+0.2)(z-0.8)} \qquad |z| > 0.8
-$$
-
-and an input *x*[*n*] given by
-
-- (a) *<sup>x</sup>*[*n*] = *enu*[*n*]
-- (b) *<sup>x</sup>*[*n*] = <sup>2</sup>*nu*[−(*n*+1)]
-- (c) *<sup>x</sup>*[*n*] = *enu*[*n*] +2*nu*[−(*n*+1)]
-- (d) *<sup>x</sup>*[*n*] = <sup>2</sup>*nu*[*n*] +*u*[−(*n*+1)]
-- (e) *<sup>x</sup>*[*n*] = *<sup>e</sup>*−<sup>2</sup>*nu*[−(*n*+1)]
-- **5.8-19** The discrete cross-correlation between real signal *x*[*n*] and real signal *y*[*n*] is
-
-$$
-c_{xy}[n] = \sum_{k=-\infty}^{\infty} x[k]y[k-n]
-$$
-
-Let signal *x*[*n*] have *z*-transform *X*[*z*] with ROC *Rx*, and let signal *y*[*n*] have *z*-transform *Y*[*z*] with ROC *Ry*. Determine *CXY* [*z*] (the bilateral *z*-transform of *cxy*[*n*]) in terms of the *z*-transforms of *x*[*n*] and *y*[*n*].
-
-**5.8-20** Transform properties can be very useful. The accumulation property states
-
-$$
-\sum_{k=-\infty}^{n} x[k] \Longleftrightarrow \frac{z}{z-1} X[z]
-$$
-
-This property is the DT version of the Laplace transform "integration in time" property. Given *x*[*n*] ⇐⇒ *X*[*z*], prove the accumulation property. [*Hint:* Polynomial long division can be helpful.] **5.10-1** Use MATLAB to generate pole-zero plots for the causal systems with the following transfer functions:
-
-(a) 
-$$
-H_a[z] = \frac{z^4 - \sqrt{2}z^2 + 1}{z^4 + 0.4096}
-$$
-  
-(b)  $H_b[z] = \frac{-3z^{-1} + \frac{3}{4}z^{-3}}{3 + \frac{3}{8}z^{-2} + 2z^{-4}}$ 
-
-<sup>3</sup><sup>+</sup> <sup>3</sup> <sup>2</sup> *<sup>z</sup>*−2+2*z*−<sup>4</sup> In each case, determine whether the system is
-
-- stable.
-- **5.10-2** For each of the following stable LTID systems, use MATLAB to generate magnitude and phase response plots over −π ≤ <π. (a) *<sup>H</sup>*a[*z*] = cos(*z*) *z*−0.5
-  - (b) *H*b[*z*] = *z*<sup>3</sup> sin(*z*−1)
-- **5.10-3** Consider an LTID system described by the difference equation 4*y*[*n* + 2] − *y*[*n*] = *x*[*n* + 2] +*x*[*n*].
-  - (a) Plot the pole-zero diagram for this system.
-  - (b) Plot the system's magnitude response <sup>|</sup>*H*[*ej*]| over <sup>−</sup><sup>π</sup> <sup>≤</sup> <sup>≤</sup> <sup>π</sup>.
-  - (c) What type of system is this: lowpass, highpass, bandpass, or bandstop?
-  - (d) Is this system stable? Justify your answer.
-  - (e) Is this system real? Justify your answer.
-  - (f) If the system input is of the form *x*[*n*] = cos(*n*), what is the greatest possible amplitude of the output? Justify your answer.
-  - (g) Draw an efficient, causal implementation of this system using only add, scale, and delay blocks.
-- **5.10-4** Consider an LTID system with system function *<sup>H</sup>*(*z*) <sup>=</sup> *<sup>b</sup>*<sup>0</sup> *<sup>z</sup>*2+<sup>1</sup> *<sup>z</sup>*2−4/<sup>9</sup> .
-  - (a) Determine the constant *b*<sup>0</sup> so that the system frequency response at = −π is −1.
-  - (b) Plot the pole-zero diagram for this system.
-  - (c) Plot the system's magnitude response <sup>|</sup>*H*[*ej*]| over <sup>−</sup><sup>π</sup> <sup>≤</sup> <sup>≤</sup> <sup>π</sup>.
-  - (d) Plot the system's phase response *<sup>H</sup>*[*ej*] over −π ≤ ≤ π.
-  - (e) What type of system is this: lowpass, highpass, bandpass, or bandstop?
-  - (f) Determine the response *y*[*n*] to the input *<sup>x</sup>*[*n*] = (−<sup>1</sup> <sup>−</sup> *<sup>j</sup>*) <sup>+</sup> (−*j*)*<sup>n</sup>* <sup>+</sup> (<sup>1</sup> <sup>−</sup> *<sup>j</sup>*) cos(π*<sup>n</sup>* <sup>+</sup> <sup>1</sup> 3 ).
-  - (g) Draw a TDFII block diagram representation of this system.
-
-**5.10-5** Consider the LTID system shown in Fig. P5.10-5, where parameters *c*<sup>1</sup> and *c*<sup>2</sup> are arbitrary constants.
-
-### **Figure P5.10-5**
-
-- (a) Determine the system function *H*[*z*], expressed in standard rational form.
-- (b) What is the order *N* of this system?
-- (c) Determine the *N* poles and *N* zeros of this system. Use MATLAB to create the corresponding pole-zero plot.
-- (d) What constraints, if any, exist on parameters *c*<sup>1</sup> and *c*<sup>2</sup> to ensure that the system is stable?
-- (e) Determine *c*<sup>1</sup> and *c*<sup>2</sup> so that this system functions as an LPF with narrow passband. Use MATLAB to generate the corresponding magnitude response <sup>|</sup>*H*[*ej*]| over −π ≤ ≤ π.
-- (f) Ms. Zeroine, the heroine of DT systems, believes that if *x*[*n*] = 0, then *y*[*n*] = 0 also. Is Ms. Zeroine correct? Fully justify your answer.
-- (g) Dr. Strange suggests that by setting *c*<sup>1</sup> = −1 and *c*<sup>2</sup> = −2, the system will act as a highpass filter. Is Dr. Strange right? Fully justify your answer.
-- **5.10-6** One interesting and useful application of discrete systems is the implementation of complex (rather than real) systems. A complex system is one in which a real-valued input can produce a complex-valued output. Complex systems that are described by constant coefficient difference equations require at least one complex-valued coefficient, and they are capable of operating on complex-valued inputs. Consider the complex discrete-time system
-
-$$
-H[z] = \frac{z^2 - j}{z - 0.9e^{j3\pi/4}}
-$$
-
-(a) Determine and plot the system zeros and poles.
-
-- (b) Sketch the magnitude response <sup>|</sup>*H*[*ej*ω]| of this system over −2π ≤ ω ≤ 2π. Comment on the system's behavior.
-- **5.10-7** Consider the complex system
-
-$$
-H[z] = \frac{z^4 - 1}{2(z^2 + 0.81j)}
-$$
-
-Refer to Prob. 5.10-6 for an introduction to complex systems.
-
-- (a) Plot the pole-zero diagram for *H*[*z*].
-- (b) Plot the system's magnitude response <sup>|</sup>*H*[*ej*]| over <sup>−</sup><sup>π</sup> <sup>≤</sup> <sup>≤</sup> <sup>π</sup>.
-- (c) Explain why *H*[*z*] is a noncausal system. Do not give a general definition of causality; specifically identify what makes this system noncausal.
-- (d) One way to make this system causal is to add two poles to *H*[*z*]. That is,
-
-$$
-H_{\text{causal}}[z] = H[z] \frac{1}{(z-a)(z-b)}
-$$
-
-Find poles *<sup>a</sup>* and *<sup>b</sup>* such that <sup>|</sup>*H*causal[*ej*]| = <sup>|</sup>*H*[*ej*]|.
-
-- (e) Draw an efficient block implementation of *H*causal[*z*].
-- **5.10-8** A discrete-time LTI system is shown in Fig. P5.10-8.
-  - (a) Determine the difference equation that describes this system.
-  - (b) Determine the magnitude response <sup>|</sup>*H*[*ej*]| for this system and simplify your answer. Plot the magnitude response over −π ≤ ≤ π. What type of standard filter (lowpass, highpass, bandpass, or bandstop) best describes this system?
-  - (c) Determine the impulse response *h*[*n*] of this system.
-
-**Figure P5.10-8**
-
-- **5.10-9** Determine the impulse response *h*[*n*] for the system shown in Fig. P5.10-9. Is the system stable? Is the system causal?
-- **5.10-10** An LTID filter has an impulse response function given by *h*[*n*] = δ[*n* − 1] + δ[*n* + 1]. Determine and carefully sketch the magnitude response <sup>|</sup>*H*[*ej*]| over the range <sup>−</sup><sup>π</sup> <sup>≤</sup> <sup>≤</sup> <sup>π</sup>. For this range of frequencies, is this filter lowpass, highpass, bandpass, or bandstop?
-- **5.10-11** A causal, stable discrete system has the rather strange transfer function *H*[*z*] = cos(*z*−1).
-  - (a) Write MATLAB code that will compute and plot the magnitude response of this system over an appropriate range of digital frequencies . Comment on the system.
-  - (b) Determine the impulse response *h*[*n*]. Plot *h*[*n*] over (0 ≤ *n* ≤ 10).
-  - (c) Determine a difference equation description for an FIR filter that closely approximates the system *H*[*z*] = cos(*z*−1). To verify proper behavior, plot the FIR filter's magnitude response and compare it with the magnitude response computed in part (a).
-- **5.10-12** The MATLAB signal-processing toolbox function butter helps design digital Butterworth filters. Use MATLAB help to learn how butter works. For each of the following cases, design the filter, plot the filter's poles and zeros in the complex *z* plane, and plot the decibel magnitude response 20log10 <sup>|</sup>*H*[*ej*]|.
-  - (a) Design an eighth-order digital lowpass filter with *<sup>c</sup>* = π/3.
-  - (b) Design an eighth-order digital highpass filter with *<sup>c</sup>* = π/3.
-  - (c) Design an eighth-order digital bandpass filter with passband between 5π/24 and 11π/24.
-  - (d) Design an eighth-order digital bandstop filter with stopband between 5π/24 and 11π/24.
-- **5.10-13** The MATLAB signal-processing toolbox function cheby1 helps design digital Chebyshev
-
-type I filters. A Chebyshev type I filter has passband ripple and smooth stopband. Setting the passband ripple to *Rp* = 3 dB, repeat Prob. 5.10-12 using the cheby1 command. With all other parameters held constant, what is the general effect of reducing *Rp*, the allowable passband ripple?
-
-**5.10-14** The MATLAB signal-processing toolbox function cheby2 helps design digital Chebyshev type II filters. A Chebyshev type II filter has smooth passband and ripple in the stopband. Setting the stopband ripple *Rs* = 20 dB down, repeat Prob. 5.10-12 using the cheby2 command. With all other parameters held constant, what is the general effect of increasing *Rs*, the minimum stopband attenuation?
-
-**5.10-15** The MATLAB signal-processing toolbox function ellip helps design digital elliptic filters. An elliptic filter has ripple in both the passband and the stopband. Setting the passband ripple to *Rp* = 3 dB and the stopband ripple *Rs* = 20 dB down, repeat Prob. 5.10-12 using the ellip command.
-
-<span id="page-612-0"></span>

engineering/linear-systems-and-signals-3rd-edition-bp-lathi/078_6 CONTINUOUS-TIME SIGNAL ANALYSIS - THE FOURIER SERIES.md

@@ -1,5 +0,0 @@
-# **[CONTINUOUS-TIME](#page-12-0) SIGNAL ANALYSIS: THE FOURIER SERIES**
-
-Electrical engineers instinctively think of signals in terms of their frequency spectra and think of systems in terms of their frequency response. Most teenagers know about the audible portion of audio signals having a bandwidth of about 20 kHz and the need for good-quality speakers to respond up to 20 kHz. This is basically thinking in the frequency domain. In Chs. 4 and 5 we discussed extensively the frequency-domain representation of systems and their spectral response (system response to signals of various frequencies). In Chs. 6 through 9, we discuss spectral representation of signals, where signals are expressed as a sum of sinusoids or exponentials. Actually, we touched on this topic in Chs. 4 and 5. Recall that the Laplace transform of a continuous-time signal is its spectral representation in terms of exponentials (or sinusoids) of complex frequencies. Similarly the *z*-transform of a discrete-time signal is its spectral representation in terms of discrete-time exponentials. However, in the earlier chapters we were concerned mainly with system representation; the spectral representation of signals was incidental to the system analysis. Spectral analysis of signals is an important topic in its own right, and now we turn to this subject.
-
-In this chapter we show that a periodic signal can be represented as a sum of sinusoids (or exponentials) of various frequencies. These results are extended to aperiodic signals in Ch. 7 and to discrete-time signals in Ch. 9. The fascinating subject of sampling of continuous-time signals is discussed in Ch. 8, leading to A/D (analog-to-digital) and D/A conversion. Chapter 8 forms the bridge between the continuous-time and the discrete-time worlds.

engineering/linear-systems-and-signals-3rd-edition-bp-lathi/079_6.1 PERIODIC SIGNAL REPRESENTATION BY TRIGONOMETRIC FOURIER SERIES.md

@@ -1,599 +0,0 @@
-## **6.1 PERIODIC SIGNAL [REPRESENTATION](#page-12-0) BY TRIGONOMETRIC FOURIER SERIES**
-
-As seen in Sec. 1.3-3 [Eq. (1.7)], a periodic signal *x*(*t*) with period *T*<sup>0</sup> (Fig. 6.1) has the property
-
-$$
-x(t) = x(t + T_0) \qquad \text{for all } t
-$$
-
-The *smallest* value of *T*<sup>0</sup> that satisfies this periodicity condition is the *fundamental period* of *x*(*t*). As argued in Sec. 1.3-3, this equation implies that *x*(*t*) starts at −∞ and continues to ∞. Moreover, the area under a periodic signal *x*(*t*) over any interval of duration *T*<sup>0</sup> is the same; that is, for any
-
-**Figure 6.1** A periodic signal of period *T*0.
-
-real numbers *a* and *b*
-
-$$
-\int_{a}^{a+T_0} x(t) dt = \int_{b}^{b+T_0} x(t) dt
-$$
-
-This result follows from the fact that a periodic signal takes the same values at intervals of *T*0. Hence, the values over any segment of duration *T*<sup>0</sup> are repeated in any other interval of the same duration. For convenience, the area under *x*(*t*) over any interval of duration *T*<sup>0</sup> will be denoted by
-
-$$
-\int_{T_0} x(t) \, dt
-$$
-
-The frequency of a sinusoid cos 2π*f*0*t* or sin 2π*f*0*t* is *f*0, and the period is *T*<sup>0</sup> = 1/*f*0. These sinusoids can also be expressed as cosω0*t* or sinω0*t*, where ω<sup>0</sup> = 2π*f*<sup>0</sup> is the *radian frequency*, although for brevity, it is often referred to as frequency (see Sec. B.2). A sinusoid of frequency *nf*<sup>0</sup> is said to be the *nth harmonic* of the sinusoid of frequency *f*0.
-
-Let us consider a signal *x*(*t*) made up of a sines and cosines of frequency ω<sup>0</sup> and all of its harmonics (including the zeroth harmonic; i.e., dc) with arbitrary amplitudes† :
-
-$$
-x(t) = a_0 + \sum_{n=1}^{\infty} a_n \cos n\omega_0 t + b_n \sin n\omega_0 t \tag{6.1}
-$$
-
-The frequency ω<sup>0</sup> is called the *fundamental frequency.*
-
-We now prove an extremely important property: *x*(*t*) in Eq. (6.1) is a periodic signal with the same period as that of the fundamental, regardless of the values of the amplitudes *an* and *bn*. Note that the period *T*<sup>0</sup> of the fundamental satisfies
-
-$$
-T_0 = \frac{1}{f_0} = \frac{2\pi}{\omega_0} \quad \text{and} \quad \omega_0 T_0 = 2\pi \tag{6.2}
-$$
-
-<sup>†</sup> In Eq. (6.1), the constant term *<sup>a</sup>*<sup>0</sup> corresponds to the cosine term for *<sup>n</sup>* <sup>=</sup> 0 because cos(<sup>0</sup> <sup>×</sup> <sup>ω</sup>0)*<sup>t</sup>* <sup>=</sup> 1. However, sin(0×ω0)*t* = 0. Hence, the sine term for *n* = 0 is nonexistent.
-
-To prove the periodicity of *x*(*t*), all we need is to show that *x*(*t*) = *x*(*t* +*T*0). From Eq. (6.1),
-
-$$
-x(t+T_0) = a_0 + \sum_{n=1}^{\infty} a_n \cos n\omega_0 (t+T_0) + b_n \sin n\omega_0 (t+T_0)
-$$
-  
-=  $a_0 + \sum_{n=1}^{\infty} a_n \cos(n\omega_0 t + n\omega_0 T_0) + b_n \sin(n\omega_0 t + n\omega_0 T_0)$ 
-
-From Eq. (6.2), we have *n*ω0*T*<sup>0</sup> = 2π*n*, and
-
-$$
-x(t + T_0) = a_0 + \sum_{n=1}^{\infty} a_n \cos(n\omega_0 t + 2\pi n) + b_n \sin(n\omega_0 t + 2\pi n)
-$$
-  
-=  $a_0 + \sum_{n=1}^{\infty} a_n \cos n\omega_0 t + b_n \sin n\omega_0 t = x(t)$ 
-
-We could also infer this result intuitively. In one fundamental period *T*0, the *n*th harmonic executes *n* complete cycles. Hence, every sinusoid on the right-hand side of Eq. (6.1) executes a complete number of cycles in one fundamental period *T*0. Therefore, at *t* = *T*0, every sinusoid starts as if it were the origin and repeats the same drama over the next *T*<sup>0</sup> seconds, and so on, ad infinitum. Hence, the sum of such harmonics results in a periodic signal of period *T*0.
-
-This result shows that any combination of sinusoids of frequencies 0, *f*0, 2*f*0, ..., *kf*<sup>0</sup> is a periodic signal of period *T*<sup>0</sup> = 1/*f*<sup>0</sup> regardless of the values of amplitudes *ak* and *bk* of these sinusoids. By changing the values of *ak* and *bk* in Eq. (6.1), we can construct a variety of periodic signals, all of the same period *T*<sup>0</sup> (*T*<sup>0</sup> = 1/*f*<sup>0</sup> = 2π/ω0).
-
-The converse of this result is also true. We shall show in Sec. 6.5-4 that *a periodic signal x*(*t*) *with a period T*<sup>0</sup> *can be expressed as a sum of a sinusoid of frequency f*<sup>0</sup> *(f*<sup>0</sup> = 1/*T*0*) and all its harmonics, as shown in Eq. (6.1)*. † The infinite series on the right-hand side of Eq. (6.1) is known as the *trigonometric Fourier series* of a periodic signal *x*(*t*).
-
-### COMPUTING THE COEFFICIENTS OF A FOURIER SERIES
-
-To determine the coefficients of a Fourier series, consider an integral *I* defined by
-
-$$
-I = \int_{T_0} \cos n\omega_0 t \cos m\omega_0 t \, dt
-$$
-
-where \$ *<sup>T</sup>*<sup>0</sup> stands for integration over any contiguous interval of *T*<sup>0</sup> seconds. By using a trigonometric identity (see Sec. B.8-6), this integral can be expressed as
-
-$$
-I = \frac{1}{2} \left[ \int_{T_0} \cos(n+m)\omega_0 t \, dt + \int_{T_0} \cos(n-m)\omega_0 t \, dt \right] \tag{6.3}
-$$
-
-<sup>†</sup> Strictly speaking, this statement applies only if a periodic signal *x*(*t*) is a continuous function of *t*. However, Sec. 6.5-4 shows that it can be applied even for discontinuous signals, if we interpret the equality in Eq. (6.1) in the mean-square sense instead of in the ordinary sense. This means that the power of the difference between the periodic signal *x*(*t*) and its Fourier series on the right-hand side of Eq. (6.1) approaches zero as the number of terms in the series approaches infinity.
-
-#### 596 CHAPTER 6 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES
-
-Because cos ω0*t* executes one complete cycle during any interval of duration *T*0, cos(*n* + *m*)ω0*t* executes (*n* + *m*) complete cycles during any interval of duration *T*0. Therefore, the first integral in Eq. (6.3), which represents the area under *n*+*m* complete cycles of a sinusoid, equals zero. The same argument shows that the second integral in Eq. (6.3) is also zero, except when *n* = *m*. Hence, *I* in Eq. (6.3) is zero for all *n* = *m*. When *n* = *m*, the first integral in Eq. (6.3) is still zero, but the second integral yields
-
-$$
-I = \frac{1}{2} \int_{T_0} dt = \frac{T_0}{2}
-$$
-
-Thus,
-
-$$
-\int_{T_0} \cos n\omega_0 t \cos m\omega_0 t \, dt = \begin{cases} 0 & n \neq m \\ \frac{T_0}{2} & m = n \neq 0 \end{cases} \tag{6.4}
-$$
-
-Using similar arguments, we can show that
-
-$$
-\int_{T_0} \sin n\omega_0 t \sin m\omega_0 t \, dt = \begin{cases} 0 & n \neq m \\ \frac{T_0}{2} & n = m \neq 0 \end{cases} \tag{6.5}
-$$
-
-and
-
-$$
-\int_{T_0} \sin n\omega_0 t \cos m\omega_0 t \, dt = 0 \qquad \text{for all } n \text{ and } m \tag{6.6}
-$$
-
-To determine *a*<sup>0</sup> in Eq. (6.1), we integrate both sides of Eq. (6.1) over one period *T*<sup>0</sup> to yield
-
-$$
-\int_{T_0} x(t) dt = a_0 \int_{T_0} dt + \sum_{n=1}^{\infty} \left[ a_n \int_{T_0} \cos n\omega_0 t dt + b_n \int_{T_0} \sin n\omega_0 t dt \right]
-$$
-
-Recall that *T*<sup>0</sup> is the period of a sinusoid of frequency ω0. Therefore, functions cos *n*ω0*t* and sin *n*ω0*t* execute *n* complete cycles over any interval of *T*<sup>0</sup> seconds so that the area under these functions over an interval *T*<sup>0</sup> is zero, and the last two integrals on the right-hand side of the foregoing equation are zero. This yields
-
-$$
-\int_{T_0} x(t) dt = a_0 \int_{T_0} dt = a_0 T_0 \quad \text{and} \quad a_0 = \frac{1}{T_0} \int_{T_0} x(t) dt
-$$
-
-Next we multiply both sides of Eq. (6.1) by cos*m*ω0*t* and integrate the resulting equation over an interval *T*0:
-
-$$
-\int_{T_0} x(t) \cos m\omega_0 t \, dt = a_0 \int_{T_0} \cos m\omega_0 t \, dt + \sum_{n=1}^{\infty} \left[ a_n \int_{T_0} \cos n\omega_0 t \cos m\omega_0 t \, dt + b_n \int_{T_0} \sin n\omega_0 t \cos m\omega_0 t \, dt \right]
-$$
-
-The first integral on the right-hand side is zero because it is an area under *m* integral number of cycles of a sinusoid. Also, the last integral on the right-hand side vanishes because of Eq. (6.6). This leaves only the middle integral, which is also zero for all *n* = *m* because of Eq. (6.4). But *n* takes on all values from 1 to ∞, including *m*. When *n* = *m*, this integral is *T*0/2, according to Eq. (6.4). Therefore, from the infinite number of terms on the right-hand side, only one term survives to yield *anT*0/2 = *amT*0/2 (recall that *n* = *m*). Therefore,
-
-$$
-\int_{T_0} x(t) \cos m\omega_0 t \, dt = \frac{a_m T_0}{2} \qquad \text{and} \qquad a_m = \frac{2}{T_0} \int_{T_0} x(t) \cos m\omega_0 t \, dt
-$$
-
-Similarly, by multiplying both sides of Eq. (6.1) by sin *n*ω0*t* and then integrating over an interval *T*0, we obtain
-
-$$
-b_m = \frac{2}{T_0} \int_{T_0} x(t) \sin m\omega_0 t \, dt
-$$
-
-To sum up our discussion, which applies to real or complex *x*(*t*), we have shown that a periodic signal *x*(*t*) with period *T*<sup>0</sup> can be expressed as a sum of a sinusoid of period *T*<sup>0</sup> and its harmonics:
-
-$$
-x(t) = a_0 + \sum_{n=1}^{\infty} a_n \cos n\omega_0 t + b_n \sin n\omega_0 t \tag{6.7}
-$$
-
-where <sup>ω</sup><sup>0</sup> <sup>=</sup> <sup>2</sup>π*f*<sup>0</sup> <sup>=</sup> <sup>2</sup><sup>π</sup> *<sup>T</sup>*<sup>0</sup> and
-
-$$
-a_0 = \frac{1}{T_0} \int_{T_0} x(t) dt, \quad a_n = \frac{2}{T_0} \int_{T_0} x(t) \cos n\omega_0 t dt, \quad \text{and} \quad b_n = \frac{2}{T_0} \int_{T_0} x(t) \sin n\omega_0 t dt \tag{6.8}
-$$
-
-### COMPACT FORM OF FOURIER SERIES
-
-The results derived so far are general and apply whether *x*(*t*) is a real or a complex function of *t*. However, when *x*(*t*) is real, coefficients *an* and *bn* are real for all *n*, and the trigonometric Fourier series can be expressed in a *compact form,* using the results in Eq. (B.16):
-
-$$
-x(t) = C_0 + \sum_{n=1}^{\infty} C_n \cos(n\omega_0 t + \theta_n)
-$$
-\n(6.9)
-
-where *Cn* and θ*<sup>n</sup>* are related to *an* and *bn*, as [see Eq. (B.17)]
-
-$$
-C_0 = a_0
-$$
-,  $C_n = \sqrt{a_n^2 + b_n^2}$ , and  $\theta_n = \tan^{-1} \left( \frac{-b_n}{a_n} \right)$  (6.10)
-
-These results are summarized in Table 6.1.
-
-The compact form in Eq. (6.9) uses the cosine form. We could just as well have used the sine form, with terms sin(*n*ω0*t* + θ*n*) instead of cos(*n*ω0*t* + θ*n*). The literature overwhelmingly favors the cosine form, for no apparent reason except possibly that the cosine phasor is represented by the horizontal axis, which happens to be the reference axis in phasor representation.
-
-Equation (6.8) shows that *a*<sup>0</sup> (or *C*0) is the average value of *x*(*t*) (averaged over one period). This value can often be determined by inspection of *x*(*t*).
-
-Because *an* and *bn* are real, *Cn* and θ*<sup>n</sup>* are also real. In the following discussion of trigonometric Fourier series, we shall assume real *x*(*t*), unless mentioned otherwise.
-
-| Series Form                                            | Coefficient Computation                            | Conversion Formulas           |
-|--------------------------------------------------------|----------------------------------------------------|-------------------------------|
-| Trigonometric                                          | #<br>1<br>a0<br>f(t)dt<br>=<br>T0<br>T0            | a0<br>= C0<br>= D0            |
-| "∞<br>f(t) = a0+<br>an<br>cosnω0t+bn<br>sinnω0t<br>n=1 | #<br>2<br>an<br>f(t) cosnω0t dt<br>=<br>T0<br>T0   | = Cnejθn =<br>an−jbn<br>2Dn   |
-|                                                        | #<br>2<br>bn<br>f(t)sinnω0t dt<br>=<br>T0          | = Cne−jθn =<br>an+jbn<br>2D−n |
-| Compact trigonometric                                  | T0<br>C0<br>= a0                                   | C0<br>= D0                    |
-| "∞<br>f(t) = C0<br>Cn<br>cos(nω0t +θn)<br>+            | =<br>2<br>2 +bn<br>Cn<br>an                        | Cn<br>= 2 Dn <br>n ≥ 1        |
-| n=1                                                    | −bn<br>= tan−1<br>θn<br>an                         | θn<br>=  Dn                   |
-| Exponential                                            |                                                    |                               |
-| f(t) = "∞<br>Dnejnω0t<br>n=−∞                          | #<br>1<br>f(t)e−jnω0t<br>Dn<br>dt<br>=<br>T0<br>T0 |                               |
-
-<span id="page-617-0"></span>**TABLE 6.1** Fourier Series Representation of a Periodic Signal of Period *T*<sup>0</sup> (ω<sup>0</sup> = 2π/*T*0)
-
-### **[6.1-1 The Fourier Spectrum](#page-12-0)**
-
-The compact trigonometric Fourier series in Eq. (6.9) indicates that a periodic signal *x*(*t*) can be expressed as a sum of sinusoids of frequencies 0 (dc), ω0, 2ω0, ..., *n*ω0, ..., whose amplitudes are *C*0, *C*1, *C*2, ..., *Cn*, ..., and whose phases are 0, θ1, θ2, ..., θ*n*, ..., respectively. We can readily plot amplitude *Cn* versus *n* (*the amplitude spectrum*) and θ*<sup>n</sup>* versus *n* (the *phase spectrum*).† Because *n* is proportional to the frequency *n*ω0, these plots are scaled plots of *Cn* versus ω and θ*<sup>n</sup>* versus ω. The two plots together are the *frequency spectra* of *x*(*t*). These spectra show at a glance the frequency contents of the signal *x*(*t*) with their amplitudes and phases. Knowing these spectra, we can reconstruct or synthesize the signal *x*(*t*) according to Eq. (6.9). Therefore, frequency spectra, which are an alternative way of describing a periodic signal *x*(*t*), are in every way equivalent to the plot of *x*(*t*) as a function of *t*. The frequency spectra of a signal constitute the *frequency-domain description* of *x*(*t*), in contrast to the *time-domain description,* where *x*(*t*) is specified as a function of time.
-
-In computing θ*n*, the phase of the *n*th harmonic from Eq. (6.10), the quadrant in which θ*<sup>n</sup>* lies should be determined from the signs of *an* and *bn*. For example, if *an* = −1 and *bn* = 1, θ*<sup>n</sup>* lies in the third quadrant, and
-
-$$
-\theta_n = \tan^{-1}\left(\frac{-1}{-1}\right) = -135^\circ
-$$
-
-Observe that
-
-$$
-\tan^{-1}\left(\frac{-1}{-1}\right) \neq \tan^{-1}(1) = 45^{\circ}
-$$
-
-<sup>†</sup> The amplitude *Cn*, by definition here, is nonnegative. Some authors define amplitude *An* that can take positive or negative values and magnitude *Cn* = |*An*| that can only be nonnegative. Thus, what we call amplitude spectrum becomes magnitude spectrum. The distinction between amplitude and magnitude, although useful, is avoided in this book in the interest of keeping definitions of essentially similar entities to a minimum.
-
-Although *Cn*, the amplitude of the *n*th harmonic as defined in Eq. (6.10), is positive, we shall find it convenient to allow *Cn* to take on negative values when *bn* = 0. This will become clear in later examples.
-
-## **EXAMPLE 6.1 Compact Trigonometric Fourier Series of Periodic Exponential Wave**
-
-Find the compact trigonometric Fourier series for the periodic signal *x*(*t*) shown in Fig. 6.2a. Sketch the amplitude and phase spectra for *x*(*t*).
-
-**Figure 6.2 (a)** A periodic signal and **(b, c)** its Fourier spectra.
-
-In this case the period *T*<sup>0</sup> = π and the fundamental frequency *f*<sup>0</sup> = 1/*T*<sup>0</sup> = 1/π Hz, and
-
-$$
-\omega_0 = \frac{2\pi}{T_0} = 2 \,\text{rad/s}
-$$
-
-Therefore,
-
-$$
-x(t) = a_0 + \sum_{n=1}^{\infty} a_n \cos 2nt + b_n \sin 2nt
-$$
-
-where
-
-$$
-a_0 = \frac{1}{\pi} \int_{T_0} x(t) dt
-$$
-
-In this example the obvious choice for the interval of integration is from 0 to π. Hence,
-
-$$
-a_0 = \frac{1}{\pi} \int_0^{\pi} e^{-t/2} dt = 0.504
-$$
-  
-$$
-a_n = \frac{2}{\pi} \int_0^{\pi} e^{-t/2} \cos 2nt dt = 0.504 \left(\frac{2}{1 + 16n^2}\right)
-$$
-
-and
-
-$$
-b_n = \frac{2}{\pi} \int_0^{\pi} e^{-t/2} \sin 2nt \, dt = 0.504 \left( \frac{8n}{1 + 16n^2} \right)
-$$
-
-Therefore,
-
-$$
-x(t) = 0.504 \left[ 1 + \sum_{n=1}^{\infty} \frac{2}{1 + 16n^2} (\cos 2nt + 4n \sin 2nt) \right]
-$$
-
-Also from Eq. (6.10),
-
-$$
-C_0 = a_0 = 0.504
-$$
-  
-\n
-$$
-C_n = \sqrt{a_n^2 + b_n^2} = 0.504 \sqrt{\frac{4}{(1 + 16n^2)^2} + \frac{64n^2}{(1 + 16n^2)^2}} = 0.504 \left(\frac{2}{\sqrt{1 + 16n^2}}\right)
-$$
-  
-\n
-$$
-\theta_n = \tan^{-1}\left(\frac{-b_n}{a_n}\right) = \tan^{-1}(-4n) = -\tan^{-1} 4n
-$$
-
-Amplitude and phases of the dc and the first seven harmonics are computed from the above equations as
-
-| n  | 0     | 1       | 2       | 3       | 4       | 5       | 6       | 7       |
-|----|-------|---------|---------|---------|---------|---------|---------|---------|
-| Cn | 0.504 | 0.244   | 0.125   | 0.084   | 0.063   | 0.0504  | 0.042   | 0.036   |
-| θn | 0◦    | −75.96◦ | −82.87◦ | −85.24◦ | −86.42◦ | −87.14◦ | −87.61◦ | −87.95◦ |
-
-We can use these numerical values to express *x*(*t*) as
-
-$$
-x(t) = 0.504 + 0.504 \sum_{n=1}^{\infty} \frac{2}{\sqrt{1 + 16n^2}} \cos(2nt - \tan^{-1} 4n)
-$$
-  
-= 0.504 + 0.244 cos (2t - 75.96°) + 0.125 cos (4t - 82.87°)  
-+ 0.084 cos (6t - 85.24°) + 0.063 cos (8t - 86.42°) + ... (6.11)
-
-### PLOTTING FOURIER SERIES SPECTRA USING MATLAB
-
-MATLAB is well suited to compute and plot Fourier series spectra. The results in Fig. 6.3, which plot *Cn* and θ*<sup>n</sup>* as functions of *n*, match Figs. 6.2b and 6.2c, which plot *Cn* and θ*<sup>n</sup>* as functions of ω = *n*ω<sup>0</sup> = 2*n*. Plots of *an* and *bn* are similarly simple to generate.
-
-The amplitude and phase spectra for *x*(*t*), in Figs. 6.2b and 6.2c, tell us at a glance the frequency composition of *x*(*t*), that is, the amplitudes and phases of various sinusoidal components of *x*(*t*). Knowing the frequency spectra, we can reconstruct *x*(*t*), as shown on the right-hand side of Eq. (6.11). Therefore the frequency spectra (Figs. 6.2b, 6.2c) provide an alternative description—the frequency-domain description of *x*(*t*). The time-domain description of *x*(*t*) is shown in Fig. 6.2a. *A signal, therefore, has a dual identity: the time-domain identity x*(*t*) *and the frequency-domain identity (Fourier spectra). The two identities complement each other; taken together, they provide a better understanding of a signal.*
-
-An interesting aspect of Fourier series is that whenever there is a jump discontinuity in *x*(*t*), the series at the point of discontinuity converges to an average of the left-hand and right-hand limits of *x*(*t*) at the instant of discontinuity.† In the present example, for instance, *x*(*t*) is discontinuous at *t* = 0 with *x*(0+) = 1 and *x*(0−) = *x*(π ) = *e*−π/<sup>2</sup> = 0.208. The corresponding Fourier series converges to a value (1 + 0.208)/2 = 0.604 at *t* = 0. This is easily verified from Eq. (6.11) by setting *t* = 0.
-
-<sup>†</sup> This behavior of the Fourier series is dictated by its convergence in the mean, discussed later in Secs. 6.2 and 6.5.
-
-### 602 CHAPTER 6 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES
-
-### **EXAMPLE 6.2 Compact Trigonometric Fourier Series of a Periodic Triangle Wave**
-
-Find the compact trigonometric Fourier series for the triangular periodic signal *x*(*t*) shown in Fig. 6.4a, and sketch the amplitude and phase spectra for *x*(*t*).
-
-**Figure 6.4 (a)** A triangular periodic signal and **(b, c)** its Fourier spectra.
-
-In this case the period *T*<sup>0</sup> = 2. Hence,
-
-$$
-\omega_0 = \frac{2\pi}{2} = \pi
-$$
-
-and
-
-$$
-x(t) = a_0 + \sum_{n=1}^{\infty} a_n \cos n\pi t + b_n \sin n\pi t
-$$
-
-where
-
-$$
-x(t) = \begin{cases} 2At & |t| < \frac{1}{2} \\ 2A(1-t) & \frac{1}{2} < t < \frac{3}{2} \end{cases}
-$$
-
-Here it will be advantageous to choose the interval of integration from −1/2 to 3/2 rather than 0 to 2.
-
-A glance at Fig. 6.4a shows that the average value (dc) of *x*(*t*) is zero so that *a*<sup>0</sup> = 0. Also,
-
-$$
-a_n = \frac{2}{2} \int_{-1/2}^{3/2} x(t) \cos n\pi t dt
-$$
-  
-= 
-$$
-\int_{-1/2}^{1/2} 2A t \cos n\pi t dt + \int_{1/2}^{3/2} 2A(1-t) \cos n\pi t dt
-$$
-
-Detailed evaluation of these integrals shows that both have a value of zero. Therefore *an* = 0. Next,
-
-$$
-b_n = \int_{-1/2}^{1/2} 2At \sin n\pi t \, dt + \int_{1/2}^{3/2} 2A(1-t) \sin n\pi t \, dt
-$$
-
-Detailed evaluation of these integrals yields, in turn,
-
-$$
-b_n = \frac{8A}{n^2 \pi^2} \sin\left(\frac{n\pi}{2}\right) = \begin{cases} 0 & n \text{ even} \\ \frac{8A}{n^2 \pi^2} & n = 1, 5, 9, 13, \dots \\ -\frac{8A}{n^2 \pi^2} & n = 3, 7, 11, 15, \dots \end{cases}
-$$
-
-Therefore,
-
-$$
-x(t) = \frac{8A}{\pi^2} \left[ \sin \pi t - \frac{1}{9} \sin 3\pi t + \frac{1}{25} \sin 5\pi t - \frac{1}{49} \sin 7\pi t + \dots \right]
-$$
- (6.12)
-
-To plot Fourier spectra, the series must be converted into compact trigonometric form as in Eq. (6.9). In this case this is readily done by converting sine terms into cosine terms with a suitable phase shift. For example,
-
-$$
-\sin kt = \cos (kt - 90^\circ) \qquad \text{and} \qquad -\sin kt = \cos (kt + 90^\circ)
-$$
-
-By using these identities, Eq. (6.12) can be expressed as
-
-$$
-x(t) = \frac{8A}{\pi^2} \bigg[ \cos(\pi t - 90^\circ) + \frac{1}{9} \cos(3\pi t + 90^\circ) + \frac{1}{25} \cos(5\pi t - 90^\circ) + \frac{1}{49} \cos(7\pi t + 90^\circ) + \cdots \bigg]
-$$
-
-In this series all the even harmonics are missing. The phases of the odd harmonics alternate from −90◦ to 90◦. Figure 6.4 shows amplitude and phase spectra for *x*(*t*).
-
-## **EXAMPLE 6.3 Converting a Trigonometric FS to a Compact Trigonometric FS**
-
-A periodic signal *x*(*t*) is represented by a trigonometric Fourier series
-
-$$
-x(t) = 2 + 3\cos 2t + 4\sin 2t + 2\sin (3t + 30^\circ) - \cos (7t + 150^\circ)
-$$
-
-Express this series as a compact trigonometric Fourier series, and sketch amplitude and phase spectra for *x*(*t*).
-
-In compact trigonometric Fourier series, the sine and cosine terms of the same frequency are combined into a single term and all terms are expressed as cosine terms with positive amplitudes. Using Eqs. (6.9) and (6.10), we have
-
-$$
-3\cos 2t + 4\sin 2t = 5\cos (2t - 53.13^{\circ})
-$$
-
-Also,
-
-$$
-\sin(3t + 30^{\circ}) = \cos(3t + 30^{\circ} - 90^{\circ}) = \cos(3t - 60^{\circ})
-$$
-
-and
-
-$$
--\cos(7t + 150^{\circ}) = \cos(7t + 150^{\circ} - 180^{\circ}) = \cos(7t - 30^{\circ})
-$$
-
-Therefore,
-
-$$
-x(t) = 2 + 5\cos(2t - 53.13^{\circ}) + 2\cos(3t - 60^{\circ}) + \cos(7t - 30^{\circ})
-$$
-
-**Figure 6.5** Fourier spectra of the signal.
-
-In this case only four components (including dc) are present. The amplitude of dc is 2. The remaining three components are of frequencies ω = 2, 3, and 7 with amplitudes 5, 2, and 1 and phases −53.13◦, −60◦, and −30◦, respectively. The amplitude and phase spectra for this signal are shown in Figs. 6.5a and 6.5b, respectively.
-
-### **EXAMPLE 6.4 Compact Trigonometric Fourier Series of a Periodic Square Wave**
-
-Find the compact trigonometric Fourier series for the square-pulse periodic signal shown in Fig. 6.6a and sketch its Fourier spectrum.
-
-**Figure 6.6 (a)** A square pulse periodic signal and **(b)** its Fourier spectrum.
-
-Here the period is *T*<sup>0</sup> = 2π and ω<sup>0</sup> = 2π/*T*<sup>0</sup> = 1. Therefore,
-
-$$
-x(t) = a_0 + \sum_{n=1}^{\infty} a_n \cos nt + b_n \sin nt
-$$
-
-where
-
-$$
-a_0 = \frac{1}{T_0} \int_{T_0} x(t) dt
-$$
-
-From Fig. 6.6a, it is clear that a proper choice of region of integration is from −π to π. But since *x*(*t*) = 1 only over (−π/2, π/2), and *x*(*t*) = 0 over the remaining segment,
-
-$$
-a_0 = \frac{1}{2\pi} \int_{-\pi/2}^{\pi/2} dt = \frac{1}{2}
-$$
-
-We could have found *a*0, the average value of *x*(*t*), to be 1/2 merely by inspection of *x*(*t*) in Fig. 6.6a. Also,
-
-$$
-a_n = \frac{1}{\pi} \int_{-\pi/2}^{\pi/2} \cos nt \, dt = \frac{2}{n\pi} \sin\left(\frac{n\pi}{2}\right)
-$$
-  
-= 
-$$
-\begin{cases} 0 & n \text{ even} \\ \frac{2}{\pi n} & n = 1, 5, 9, 13, \dots \\ -\frac{2}{\pi n} & n = 3, 7, 11, 15, \dots \end{cases}
-$$
-  
-$$
-b_n = \frac{1}{\pi} \int_{-\pi/2}^{\pi/2} \sin nt \, dt = 0
-$$
-
-Therefore
-
-$$
-x(t) = \frac{1}{2} + \frac{2}{\pi} \left( \cos t - \frac{1}{3} \cos 3t + \frac{1}{5} \cos 5t - \frac{1}{7} \cos 7t + \cdots \right)
-$$
-(6.13)
-
-Observe that *bn* = 0 and all the sine terms are zero. Only the cosine terms appear in the trigonometric series. The series is therefore already in the compact form except that the amplitudes of alternating harmonics are negative. Now by definition, amplitudes *Cn* are positive [see Eq. (6.10)]. The negative sign can be accommodated by associating a proper phase, as seen from the trigonometric identity†
-
-−cos *x* = cos(*x* −π )
-
-Using this fact, we can express the series in Eq. (6.13) as
-
-$$
-x(t) = \frac{1}{2} + \frac{2}{\pi} \left[ \cos \omega_0 t + \frac{1}{3} \cos (3\omega_0 t - \pi) + \frac{1}{5} \cos 5\omega_0 t + \frac{1}{7} \cos (7\omega_0 t - \pi) + \frac{1}{9} \cos 9\omega_0 t + \cdots \right]
-$$
-
-<sup>†</sup> Because cos(*x*±π ) = −cos *<sup>x</sup>*, we could have chosen the phase <sup>π</sup> or <sup>−</sup>π. In fact, cos(*x*±*N*π ) = −cos *<sup>x</sup>* for any odd integral value of *N*. Therefore the phase can be chosen as ±*N*π, where *N* is any convenient odd integer.
-
-<span id="page-626-0"></span>This is the desired form of the compact trigonometric Fourier series. The amplitudes are
-
-$$
-C_0 = \frac{1}{2}
-$$
- and  $C_n = \begin{cases} 0 & n \text{ even} \\ \frac{2}{\pi n} & n \text{ odd} \end{cases}$ 
-
-The phases are
-
-$$
-\theta_n = \begin{cases} 0 & \text{for all } n \neq 3, 7, 11, 15, \dots \\ -\pi & n = 3, 7, 11, 15, \dots \end{cases}
-$$
-
-We might use these values to plot amplitude and phase spectra. However, we can simplify our task in this special case if we allow amplitude *Cn* to take on negative values. If this is allowed, we do not need a phase of −π to account for the sign as seen from Eq. (6.13). This means that phases of all components are zero, and we can discard the phase spectrum and manage with only the amplitude spectrum, as shown in Fig. 6.6b. Observe that there is no loss of information in doing so and that the amplitude spectrum in Fig. 6.6b has the complete information about the Fourier series in Eq. (6.13). *Therefore, whenever all sine terms vanish (bn* = 0*), it is convenient to allow Cn to take on negative values*. This permits the spectral information to be conveyed by a single spectrum.†
-
-Let us investigate the behavior of the series at the points of discontinuities. For the discontinuity at *t* =π/2, the values of *x*(*t*) on either sides of the discontinuity are *x*((π/2)−)=1 and *x*((π/2)+) = 0. We can verify by setting *t* = π/2 in Eq. (6.13) that *x*(π/2) = 0.5, which is a value midway between the values of *x*(*t*) on either side of the discontinuity at *t* = π/2.
-
-### **[6.1-2 The Effect of Symmetry](#page-12-0)**
-
-The Fourier series for the signal *x*(*t*) in Fig. 6.2a (Ex. 6.1) consists of sine and cosine terms, but the series for the signal *x*(*t*) in Fig. 6.4a (Ex. 6.2) consists of sine terms only, and the series for the signal *x*(*t*) in Fig. 6.6a (Ex. 6.4) consists of cosine terms only. This is no accident. We can show that the Fourier series of any even periodic function *x*(*t*) consists of cosine terms only and the series for any odd periodic function *x*(*t*) consists of sine terms only. Moreover, because of symmetry (even or odd), the information of one period of *x*(*t*) is implicit in only half the period, as seen in Figs. 6.4a and 6.6a. In these cases, knowing the signal over a half-period and knowing the kind of symmetry (even or odd), we can determine the signal waveform over a complete period. For this reason, the Fourier coefficients in these cases can be computed by integrating over only half the period rather than a complete period. To prove this result, recall that
-
-$$
-a_0 = \frac{1}{T_0} \int_{-T_0/2}^{T_0/2} x(t) dt, \quad a_n = \frac{2}{T_0} \int_{-T_0/2}^{T_0/2} x(t) \cos n\omega_0 t dt, \quad \text{and} \quad b_n = \frac{2}{T_0} \int_{-T_0/2}^{T_0/2} x(t) \sin n\omega_0 t dt
-$$
-
-Recall also that cos*n*ω0*t* is an even function and sin*n*ω0*t* is an odd function of *t*. If *x*(*t*) is an even function of *t*, then *x*(*t*) cos*n*ω0*t* is also an even function and *x*(*t*)sin *n*ω0*t* is an odd function of *t*
-
-<sup>†</sup> Here, the distinction between amplitude *An* and magnitude *Cn* = |*An*<sup>|</sup> would have been useful. But, for the reasons mentioned in the footnote on page 598, we refrain from this distinction formally.
-
-#### 608 CHAPTER 6 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES
-
-(see Sec. 1.5-1). Therefore, following from Eq. (1.16),
-
-$$
-a_0 = \frac{2}{T_0} \int_0^{T_0/2} x(t) dt, \quad a_n = \frac{4}{T_0} \int_0^{T_0/2} x(t) \cos n\omega_0 t dt, \quad \text{and} \quad b_n = 0 \tag{6.14}
-$$
-
-Similarly, if *x*(*t*) is an odd function of *t*, then *x*(*t*) cos *n*ω0*t* is an odd function of *t* and *x*(*t*)sin *n*ω0*t* is an even function of *t*. Therefore,
-
-$$
-a_n = 0
-$$
- and  $b_n = \frac{4}{T_0} \int_0^{T_0/2} x(t) \sin n\omega_0 t dt$  (6.15)
-
-Observe that because of symmetry, the integration required to compute the coefficients need be performed over only half the period.
-
-If a periodic signal *x*(*t*) shifted by half the period remains unchanged except for a sign—that is, if
-
-$$
-x\left(t - \frac{T_0}{2}\right) = -x(t)
-$$
-
-then the signal is said to have a *half-wave* symmetry. It can be shown that for a signal with a half-wave symmetry, all the even-numbered harmonics vanish (see Prob. 6.1-6). The signal in Fig. 6.4a is an example of such a symmetry. The signal in Fig. 6.6a also has this symmetry, although it is not obvious owing to a dc component. If we subtract the dc component of 0.5 from this signal, the remaining signal has half-wave symmetry. For this reason, this signal has only odd harmonics and a dc component of 0.5.
-
-### **DR ILL 6.1 Compact Trigonometric Fourier Series**
-
-Find the compact trigonometric Fourier series for periodic signals shown in Fig. 6.7. Sketch their amplitude and phase spectra. Allow *Cn* to take on negative values if *bn* = 0 so that the phase spectrum can be eliminated. [*Hint:* Use Eqs. (6.14) and (6.15) for appropriate symmetry conditions.]
-
-### **ANSWERS**
-
-(a) 
-$$
-x(t) = \frac{1}{3} - \frac{4}{\pi^2} \left( \cos \pi t - \frac{1}{4} \cos 2\pi t + \frac{1}{9} \cos 3\pi t - \frac{1}{16} \cos 4\pi t + \cdots \right)
-$$
-  
-\n
-$$
-= \frac{1}{3} + \frac{4}{\pi^2} \sum_{n=1}^{\infty} \frac{(-1)^n}{n^2} \cos n\pi t
-$$
-\n(b)  $x(t) = \frac{2A}{\pi} \left[ \sin \pi t - \frac{1}{2} \sin 2\pi t + \frac{1}{3} \sin 3\pi t - \frac{1}{4} \sin 4\pi t + \cdots \right]$   
-\n
-$$
-= \frac{2A}{\pi} \left[ \cos (\pi t - 90^\circ) + \frac{1}{2} \cos (2\pi t + 90^\circ) + \frac{1}{3} \cos (3\pi t - 90^\circ) + \frac{1}{4} \cos (4\pi t + 90^\circ) + \cdots \right]
-$$
-
-<span id="page-628-0"></span>
-
-### **[6.1-3 Determining the Fundamental Frequency and Period](#page-12-0)**
-
-We have seen that every periodic signal can be expressed as a sum of sinusoids of a fundamental frequency ω<sup>0</sup> and its harmonics. One may ask whether a sum of sinusoids of *any* frequencies represents a periodic signal. If so, how does one determine the period? Consider the following three functions:
-
-$$
-x_1(t) = 2 + 7\cos(\frac{1}{2}t + \theta_1) + 3\cos(\frac{2}{3}t + \theta_2) + 5\cos(\frac{7}{6}t + \theta_3)
-$$
-  
-\n
-$$
-x_2(t) = 2\cos(2t + \theta_1) + 5\sin(\pi t + \theta_2)
-$$
-  
-\n
-$$
-x_3(t) = 3\sin(3\sqrt{2}t + \theta) + 7\cos(6\sqrt{2}t + \phi)
-$$
-
-Recall that every frequency in a periodic signal is an integer multiple of the fundamental frequency ω0. Therefore the ratio of any two frequencies is of the form *m*/*n*, where *m* and *n* are integers. This means that the ratio of any two frequencies is a rational number. When the ratio of two frequencies is a rational number, the frequencies are said to be *harmonically* related.
-
-The largest number of which all the frequencies are integer multiples is the fundamental frequency. In other words, the fundamental frequency is the *greatest common factor* (GCF) of all the frequencies in the series. The frequencies in the spectrum of *x*1(*t*) are 1/2, 2/3, and 7/6 (we do not consider dc). The ratios of the successive frequencies are 3:4 and 4:7, respectively. Because both these numbers are rational, all the three frequencies in the spectrum are harmonically related, and the signal *x*1(*t*) is periodic. The GCF, that is, the greatest number of which 1/2, 2/3, and 7/6 are integer multiples, is 1/6.† Moreover, 3(1/6) = 1/2, 4(1/6) = 2/3, and 7(1/6) = 7/6. Therefore the fundamental frequency is 1/6, and the three frequencies in the spectrum are the third, fourth, and seventh harmonics. Observe that the fundamental frequency component is absent in this Fourier series.
-
-<sup>†</sup> The greatest common factor of *a*1/*b*1, *a*2/*b*2, ..., *am*/*bm* is the ratio of the GCF of the numerators set (*a*1,*a*2,...,*am*) to the LCM (least common multiple) of the denominator set (*b*1,*b*2,...,*bm*). For instance, for the set (2/3, 6/7, 2), the GCF of the numerator set (2, 6, 2) is 2; the LCM of the denominator set (3, 7, 1) is 21. Therefore, 2/21 is the largest number of which 2/3, 6/7, and 2 are integer multiples.
-
-#### 610 CHAPTER 6 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES
-
-The signal *x*2(*t*) is not periodic because the ratio of two frequencies in the spectrum is 2/π, which is not a rational number. The signal *x*3(*t*) is periodic because the ratio of frequencies 3 √ 2 and 6<sup>√</sup> 2 is 1/2, a rational number. The greatest common factor of 3√2 and 6√2 is 3√2. Therefore, the fundamental frequency ω<sup>0</sup> = 3 <sup>√</sup>2, and the period
-
-$$
-T_0 = \frac{2\pi}{(3\sqrt{2})} = \frac{\sqrt{2}}{3}\pi
-$$
-
-### **DR ILL 6.2 Determining Periodicity, Fundamental Frequency, and Harmonic Content**
-
-Determine whether the signal
-
-$$
-x(t) = \cos\left(\frac{2}{3}t + 30^{\circ}\right) + \sin\left(\frac{4}{5}t + 45^{\circ}\right)
-$$
-
-is periodic. If it is periodic, find the fundamental frequency and the period. What harmonics are present in *x*(*t*)?
-
-### **ANSWERS**
-
-Periodic with ω<sup>0</sup> =2/15 and period *T*<sup>0</sup> =15π. Signal *x*(*t*) contains the fifth and sixth harmonics.
-
-## A HISTORICAL NOTE: BARON JEAN-BAPTISTE-JOSEPH FOURIER (1768–1830)
-
-The Fourier series and integral comprise a most beautiful and fruitful development, which serves as an indispensable instrument in the treatment of many problems in mathematics, science, and engineering. Maxwell was so taken by the beauty of the Fourier series that he called it a great mathematical poem. In electrical engineering, it is central to the areas of communication, signal processing, and several other fields, including antennas, but its initial reception by the scientific world was not enthusiastic. In fact, Fourier could not get his results published as a paper.
-
-Fourier, a tailor's son, was orphaned at age 8 and educated at a local military college (run by Benedictine monks), where he excelled in mathematics. The Benedictines prevailed upon the young genius to choose the priesthood as his vocation, but revolution broke out before he could take his vows. Fourier joined the people's party. But in its early days, the French Revolution, like most such upheavals, liquidated a large segment of the intelligentsia, including prominent scientists such as Lavoisier. Observing this trend, many intellectuals decided to leave France to save themselves from a rapidly rising tide of barbarism. Fourier, despite his early enthusiasm for the Revolution, narrowly escaped the guillotine twice. It was to the everlasting credit of Napoleon that he stopped the persecution of the intelligentsia and founded new schools to replenish their ranks. The 26-year-old Fourier was appointed chair of mathematics at the newly created École Normale in 1794 [1].
-
-Jean-Baptiste-Joseph Fourier and Napoleon
-
-Napoleon was the first modern ruler with a scientific education, and he was one of the rare persons who are equally comfortable with soldiers and scientists. The age of Napoleon was one of the most fruitful in the history of science. Napoleon liked to sign himself as "member of *Institut de France*" (a fraternity of scientists), and he once expressed to Laplace his regret that "force of circumstances has led me so far from the career of a scientist" [2]. Many great figures in science and mathematics, including Fourier and Laplace, were honored and promoted by Napoleon. In 1798 he took a group of scientists, artists, and scholars—Fourier among them—on his Egyptian expedition, with the promise of an exciting and historic union of adventure and research. Fourier proved to be a capable administrator of the newly formed Institut d'Égypte, which, incidentally, was responsible for the discovery of the Rosetta Stone. The inscription on this stone in two languages and three scripts (hieroglyphic, demotic, and Greek) enabled Thomas Young and Jean-François Champollion, a protégé of Fourier, to invent a method of translating hieroglyphic writings of ancient Egypt—the only significant result of Napoleon's Egyptian expedition.
-
-Back in France in 1801, Fourier briefly served in his former position as professor of mathematics at the École Polytechnique in Paris. In 1802 Napoleon appointed him the prefect of Isère (with its headquarters in Grenoble), a position in which Fourier served with distinction. Fourier was named Baron of the Empire by Napoleon in 1809. Later, when Napoleon was exiled to Elba, his route was to take him through Grenoble. Fourier had the route changed to avoid meeting Napoleon, which would have displeased Fourier's new master, King Louis XVIII. Within a year, Napoleon escaped from Elba and returned to France. At Grenoble, Fourier was brought before him in chains. Napoleon scolded Fourier for his ungrateful behavior but reappointed him the prefect of Rhône at Lyons. Within four months Napoleon was defeated at Waterloo and was exiled to St. Helena, where he died in 1821. Fourier once again was in disgrace as a Bonapartist and had <span id="page-631-0"></span>to pawn his possessions to keep himself alive. But through the intercession of a former student, who was now a prefect of Paris, he was appointed director of the statistical bureau of the Seine, a position that allowed him ample time for scholarly pursuits. Later, in 1827, he was elected to the powerful position of perpetual secretary of the Paris Academy of Science, a section of the Institut de France [3].
-
-While serving as the prefect of Grenoble, Fourier carried on his elaborate investigation of the propagation of heat in solid bodies, which led him to the Fourier series and the Fourier integral. On December 21, 1807, he announced these results in a prize paper on the theory of heat. Fourier claimed that an arbitrary function (continuous or with discontinuities) defined in a finite interval by an arbitrarily capricious graph can always be expressed as a sum of sinusoids (Fourier series). The judges, who included the great French mathematicians Laplace, Lagrange, Legendre, Monge, and LaCroix, admitted the novelty and importance of Fourier's work but criticized it for lack of mathematical rigor and generality. Lagrange thought it incredible that a sum of sines and cosines could add up to anything but an infinitely differentiable function. Moreover, one of the properties of an infinitely differentiable function is that if we know its behavior over an arbitrarily small interval, we can determine its behavior over the entire range (the Taylor–Maclaurin series). Such a function is far from an arbitrary or a capriciously drawn graph [4]. Laplace had additional reason to criticize Fourier's work. Laplace and his students had already approached the problem of heat conduction from a different angle, and Laplace was reluctant to accept the superiority of Fourier's method [5]. Fourier thought the criticism unjustified but was unable to prove his claim because the tools required for operations with infinite series were not available at the time. However, posterity has proved Fourier to be closer to the truth than his critics. This is the classic conflict between pure mathematicians and physicists or engineers, as we saw earlier (Ch. 4) in the life of Oliver Heaviside. In 1829 Dirichlet proved Fourier's claim concerning capriciously drawn functions with a few restrictions (Dirichlet conditions).
-
-Although three of the four judges were in favor of publication, Fourier's paper was rejected because of vehement opposition by Lagrange. Fifteen years later, after several attempts and disappointments, Fourier published the results in expanded form as a text, *Théorie analytique de la chaleur,* which is now a classic.

engineering/linear-systems-and-signals-3rd-edition-bp-lathi/080_6.2 EXISTENCE AND CONVERGENCE OF THE FOURIER SERIES.md

@@ -1,261 +0,0 @@
-## **[6.2 EXISTENCE AND](#page-12-0) CONVERGENCE OF THE FOURIER SERIES**
-
-For the existence of the Fourier series, coefficients *a*0,*an*, and *bn* in Eq. (6.8) must be finite. It follows from Eq. (6.8) that the existence of these coefficients is guaranteed if *x*(*t*) is absolutely integrable over one period; that is,
-
-$$
-\int_{T_0} |x(t)| \, dt < \infty \tag{6.16}
-$$
-
-However, existence, by itself, does not inform us about the nature and the manner in which the series converges. We shall first discuss the notion of convergence.
-
-### <span id="page-632-0"></span>**[6.2-1 Convergence of a Series](#page-12-0)**
-
-The key to many puzzles lies in the nature of the convergence of the Fourier series. Convergence of infinite series is a complex problem. It took mathematicians several decades to understand the convergence aspect of the Fourier series. We shall barely scratch the surface here.
-
-Nothing annoys a student more than the discussion of convergence. "Have we not proved," they ask, "that a periodic signal *x*(*t*) can be expressed as a Fourier series"? Then why spoil the fun by this annoying discussion? All we have shown so far is that a signal represented by a Fourier series in Eq. (6.1) is periodic. We have not proved the converse, that every periodic signal can be expressed as a Fourier series. This issue will be tackled later, in Sec. 6.5-4, where it will be shown that a periodic signal can be represented by a Fourier series, as in Eq. (6.1), where the equality of the two sides of the equation is not in the ordinary sense, but in the mean-square sense (explained later in this discussion). But the astute reader should have been skeptical of the claims of the Fourier series to represent discontinuous functions in Figs. 6.2a and 6.6a. If *x*(*t*) has a jump discontinuity, say, at *t* = 0, then *x*(0+), *x*(0), and *x*(0−) are generally different. How could a series consisting of the sum of continuous functions of the smoothest type (sinusoids) add to one value at *t* = 0<sup>−</sup> and a different value at *t* = 0 and yet another value at *t* = 0+? The demand is impossible to satisfy unless the math involved executes some spectacular acrobatics. How does a Fourier series act under such conditions? Precisely for this reason, the great mathematicians Lagrange and Laplace, two of the judges examining Fourier's paper, were skeptical of Fourier's claims and voted against publication of the paper that later became a classic.
-
-There are also other issues. In any practical application, we can use only a finite number of terms in a series. If, with a fixed number of terms, the series guarantees convergence within an arbitrarily small error at every value of *t*, such a series is highly desirable and is called a *uniformly convergent* series. If a series converges at every value of *t*, but to guarantee convergence within a given error requires a different number of terms at different *t*, then the series is still convergent, but less desirable. It goes under the name *pointwise convergent* series.
-
-Finally, we have the case of a series that refuses to converge at some *t*, no matter how many terms are added. But the series may *converge in the mean*; that is, the energy of the difference between *x*(*t*) and the corresponding finite term series approaches zero as the number of terms approaches infinity.† To explain this concept, let us consider representation of a function *x*(*t*) by an infinite series
-
-$$
-x(t) = \sum_{n=1}^{\infty} z_n(t)
-$$
-
-Let the partial sum of the first *N* terms of the series on the right-hand side be denoted by *xN*(*t*), that is,
-
-$$
-x_N(t) = \sum_{n=1}^N z_n(t)
-$$
-
-<sup>†</sup> The behavior is called "convergence in the mean" because minimizing the error energy over a certain interval is equivalent to minimizing the mean-square value of the error over the same interval.
-
-#### 614 CHAPTER 6 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES
-
-If we approximate *x*(*t*) by *xN*(*t*) (the partial sum of the first *N* terms of the series), the *error* in the approximation is the difference *x*(*t*) − *xN*(*t*). The series converges *in the mean* to *x*(*t*) in the interval (0, *T*0) if
-
-$$
-\int_0^{T_0} |x(t) - x_N(t)|^2 dt \to 0 \quad \text{as} \quad N \to \infty
-$$
-
-Hence, the energy of the error *x*(*t*)−*xN*(*t*) approaches zero as *N* → ∞. This form of convergence does not require the series to be equal to *x*(*t*) for all *t*. It just requires the energy of the difference (area under |*x*(*t*)−*xN*(*t*)| 2) to vanish as *N* → ∞. Superficially it may appear that if the energy of a signal over an interval is zero, the signal (the error) must be zero everywhere. This is not true. The signal energy can be zero even if there are nonzero values at a finite number of isolated points. This is because although the signal is nonzero at a point (and zero everywhere else), the area under its square is still zero. Thus, a series that converges in the mean to *x*(*t*) need not converge to *x*(*t*) at a finite number of points. This is precisely what happens to the Fourier series when *x*(*t*) has jump discontinuities. This is also what makes Fourier series convergence compatible with the Gibbs phenomenon, to be discussed later in this section.
-
-There is a simple criterion for ensuring that a periodic signal *x*(*t*) has a Fourier series that converges in the mean. The Fourier series for *x*(*t*) converges to *x*(*t*) in the mean if *x*(*t*) has a finite energy over one period, that is,
-
-$$
-\int_{T_0} |x(t)|^2 dt < \infty \tag{6.17}
-$$
-
-Thus, the periodic signal *x*(*t*), having a finite energy over one period, guarantees the convergence in the mean of its Fourier series. In all the examples discussed so far, Eq. (6.17) is satisfied; hence the corresponding Fourier series converges in the mean. Equation (6.17), like Eq. (6.16), guarantees that the Fourier coefficients are finite.
-
-We shall now discuss an alternate set of criteria, due to Dirichlet, for convergence of the Fourier series.
-
-### DIRICHLET CONDITIONS
-
-Dirichlet showed that if *x*(*t*) satisfies certain conditions (*Dirichlet conditions*), its Fourier series is guaranteed to converge pointwise at all points where *x*(*t*) is continuous. Moreover, at the points of discontinuities, *x*(*t*) converges to the value midway between the two values of *x*(*t*) on either side of the discontinuity. These conditions are:
-
-- 1. The function *x*(*t*) must be absolutely integrable; that is, it must satisfy Eq. (6.16).
-- 2. The function *x*(*t*) must have only a finite number of finite discontinuities in one period.
-- 3. The function *x*(*t*) must contain only a finite number of maxima and minima in one period.
-
-All practical signals, including those in Exs. 6.1, 6.2, 6.3, and 6.4, satisfy these conditions.
-
-### <span id="page-634-0"></span>**[6.2-2 The Role of Amplitude and Phase Spectra in Waveshaping](#page-12-0)**
-
-The trigonometric Fourier series of a signal *x*(*t*) shows explicitly the sinusoidal components of *x*(*t*). We can synthesize *x*(*t*) by adding the sinusoids in the spectrum of *x*(*t*). Let us synthesize the square-pulse periodic signal *x*(*t*) of Fig. 6.6a by adding successive harmonics in its spectrum step by step and observing the similarity of the resulting signal to *x*(*t*). The Fourier series for this function as found in Eq. (6.13) is
-
-$$
-x(t) = \frac{1}{2} + \frac{2}{\pi} \left( \cos t - \frac{1}{3} \cos 3t + \frac{1}{5} \cos 5t - \frac{1}{7} \cos 7t + \cdots \right)
-$$
-
-We start the synthesis with only the first term in the series (*n* = 0), a constant 1/2 (dc); this is a gross approximation of the square wave, as shown in Fig. 6.8a. In the next step we add the dc (*n* = 0) and the first harmonic (fundamental), which results in a signal shown in Fig. 6.8b. Observe that the synthesized signal somewhat resembles *x*(*t*). It is a smoothed-out version of *x*(*t*). The sharp corners in *x*(*t*) are not reproduced in this signal because sharp corners mean rapid changes, and their reproduction requires rapidly varying (i.e., higher-frequency) components, which are excluded. Figure 6.8c shows the sum of dc, first, and third harmonics (even harmonics are absent). As we increase the number of harmonics progressively, as shown in Figs. 6.8d (sum up to the fifth harmonic) and 6.8e (sum up to the nineteenth harmonic), the edges of the pulses become sharper and the signal resembles *x*(*t*) more closely.
-
-### ASYMPTOTIC RATE OF AMPLITUDE SPECTRUM DECAY
-
-Figure 6.8 brings out one interesting aspect of the Fourier series. Lower frequencies in the Fourier series affect the large-scale behavior of *x*(*t*), whereas the higher frequencies determine the fine structure such as rapid wiggling. Hence, sharp changes in *x*(*t*), being a part of fine structure, necessitate higher frequencies in the Fourier series. The sharper the change [the higher the time derivative *x*˙(*t*)], the higher are the frequencies needed in the series.
-
-The amplitude spectrum indicates the amounts (amplitudes) of various frequency components of *x*(*t*). If *x*(*t*) is a smooth function, its variations are less rapid. Synthesis of such a function requires predominantly lower-frequency sinusoids and relatively small amounts of rapidly varying (higher-frequency) sinusoids. The amplitude spectrum of such a function would decay swiftly with frequency. To synthesize such a function, we require fewer terms in the Fourier series for a good approximation. On the other hand, a signal with sharp changes, such as jump discontinuities, contains rapid variations, and its synthesis requires a relatively large amount of high-frequency components. The amplitude spectrum of such a signal would decay slowly with frequency, and to synthesize such a function, we require many terms in its Fourier series for a good approximation. The square wave *x*(*t*) is a discontinuous function with jump discontinuities, and therefore its amplitude spectrum decays rather slowly, as 1/*n* [see Eq. (6.13)]. On the other hand, the triangular-pulse periodic signal in Fig. 6.4a is smoother because it is a continuous function (no jump discontinuities). Its spectrum decays rapidly with frequency as 1/*n*<sup>2</sup> [see Eq. (6.12)].
-
-We can show that if the first *k* − 1 derivatives of a periodic signal *x*(*t*) are continuous and the *k*th derivative is discontinuous, then its amplitude spectrum *Cn* decays with frequency at least as rapidly as 1/*nk*+<sup>1</sup> [6]. This result provides a simple and useful means for predicting the asymptotic
-
-**Figure 6.8** Synthesis of a square-pulse periodic signal by successive addition of its harmonics.
-
-rate of convergence of the Fourier series. In the case of the square-wave signal (Fig. 6.6a), the zeroth derivative of the signal (the signal itself) is discontinuous so that *k* = 0. For the triangular periodic signal in Fig. 6.4a, the first derivative is discontinuous; that is, *k* = 1. For this reason, the spectra of these signals decay as 1/*n* and 1/*n*2, respectively.
-
-## **EXAMPLE 6.5 Square-Wave Synthesis by Truncated Fourier Series Using MATLAB**
-
-Use MATLAB to synthesize and plot the square wave of Fig. 6.8a using a Fourier series that is truncated to the 19th harmonic. The result should match Fig. 6.8e.
-
-To synthesize the waveform, we use the Fourier series of Eq. (6.13).
-
-```
->> x = @(t) 1.0*(mod(t+pi/2,2*pi)<=pi);
->> t = linspace(-2*pi,2*pi,10001);
->> x19 = 0.5*ones(size(t));
->> for n=1:19, x19 = x19+2/(pi*n)*sin(pi*n/2)*cos(n*t); end
->> plot(t,x19,'k-'); axis([-2*pi 2*pi -0.2 1.2]);
->> xlabel('t'); ylabel('x_{19}(t)');
-```
-
-As expected, the result of Fig. 6.9 matches Fig. 6.8e.
-
-### PHASE SPECTRUM: THE WOMAN BEHIND A SUCCESSFUL MAN
-
-The role of the amplitude spectrum in shaping the waveform *x*(*t*) is quite clear. However, the role of the phase spectrum in shaping this waveform is less obvious. Yet, the phase spectrum, like the woman behind a successful man,† plays an equally important role in waveshaping. We can explain this role by considering a signal *x*(*t*) that has rapid changes, such as jump discontinuities. To synthesize an instantaneous change at a jump discontinuity, the phases of the various sinusoidal components in its spectrum must be such that all (or most) of the harmonic components will have
-
-<sup>†</sup> Or, to keep up with the times, the man behind a successful woman.
-
-#### 618 CHAPTER 6 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES
-
-one sign before the discontinuity and the opposite sign after the discontinuity. This will result in a sharp change in *x*(*t*) at the point of discontinuity. We can verify this fact in any waveform with jump discontinuity. Consider, for example, the sawtooth waveform in Fig. 6.7b. This waveform has a discontinuity at *t* = 1. The Fourier series for this waveform, as given in Drill 6.1b, is
-
-$$
-x(t) = \frac{2A}{\pi} \left[ \cos(\pi t - 90^\circ) + \frac{1}{2} \cos(2\pi t + 90^\circ) + \frac{1}{3} \cos(3\pi t - 90^\circ) + \frac{1}{4} \cos(4\pi t + 90^\circ) + \cdots \right]
-$$
-
-Figure 6.10 shows the first three components of this series. The phases of all the (infinite) components are such that all the components are positive just before *t* = 1 and turn negative just after *t* = 1, the point of discontinuity. The same behavior is also observed at *t* = −1, where a similar discontinuity occurs. This sign change in all the harmonics adds up to produce very nearly a jump discontinuity. The role of the phase spectrum is crucial in achieving a sharp change in the waveform. If we ignore the phase spectrum when trying to reconstruct this signal, the result will be a smeared and spread-out waveform. In general, the phase spectrum is just as crucial as the amplitude spectrum in determining the waveform. *The synthesis of any signal x*(*t*) *is achieved by using a proper combination of amplitudes and phases of various sinusoids. This unique combination is the Fourier spectrum of x*(*t*)*.*
-
-**Figure 6.10** Role of the phase spectrum in shaping a periodic signal.
-
-### FOURIER SYNTHESIS OF DISCONTINUOUS FUNCTIONS: THE GIBBS PHENOMENON
-
-Figure 6.8 showed the square function *x*(*t*) and its approximation by a truncated trigonometric Fourier series that includes only the first *N* harmonics for *N* = 1, 3, 5, and 19. The plot of the truncated series approximates closely the function *x*(*t*) as *N* increases, and we expect that the series will converge exactly to *x*(*t*) as *N* → ∞. Yet the curious fact, as seen from Fig. 6.8, is that even for large *N*, the truncated series exhibits an oscillatory behavior and an overshoot approaching a value of about 9% in the vicinity of the discontinuity at the nearest peak of oscillation.† Regardless of the value of *N*, the overshoot remains at about 9%. Such strange behavior certainly would undermine anyone's faith in the Fourier series. In fact, this behavior puzzled many scholars at the turn of the century. Josiah Willard Gibbs, an eminent mathematical physicist who was the inventor of vector analysis, gave a mathematical explanation of this behavior (now called the *Gibbs phenomenon*).
-
-We can reconcile the apparent aberration in the behavior of the Fourier series by observing from Fig. 6.8 that the frequency of oscillation of the synthesized signal is *Nf*0, so the width of the spike with 9% overshoot is approximately 1/2*Nf*0. As we increase *N*, the frequency of oscillation increases and the spike width 1/2*Nf*<sup>0</sup> diminishes. As *N* → ∞, the error power → 0 because the error consists mostly of the spikes, whose widths → 0. Therefore, as *N* → ∞, the corresponding Fourier series differs from *x*(*t*) by about 9% at the immediate left and right of the points of discontinuity, and yet the error power →0. The reason for all this confusion is that in this case, the Fourier series converges in the mean. When this happens, all we promise is that the error energy (over one period) → 0 as *N* → ∞. Thus, the series may differ from *x*(*t*) at some points and yet have the error signal power zero, as verified earlier. Note that the series, in this case, also converges pointwise at all points except the points of discontinuity. It is precisely at the discontinuities that the series differs from *x*(*t*) by 9%.‡
-
-When we use only the first *N* terms in the Fourier series to synthesize a signal, we are abruptly terminating the series, giving a unit weight to the first *N* harmonics and zero weight to all the remaining harmonics beyond *N*. This abrupt termination of the series causes the Gibbs phenomenon in synthesis of discontinuous functions. Section 7.8 offers more discussion on the Gibbs phenomenon, its ramifications, and cure.
-
-The Gibbs phenomenon is present only when there is a jump discontinuity in *x*(*t*). When a continuous function *x*(*t*) is synthesized by using the first *N* terms of the Fourier series, the synthesized function approaches *x*(*t*) for all *t* as *N* → ∞. No Gibbs phenomenon appears. This can be seen in Fig. 6.11, which shows one cycle of a continuous periodic signal being synthesized from the first 19 harmonics. Compare the similar situation for a discontinuous signal in Fig. 6.8.
-
-### **DR ILL 6.3 Rate of Spectral Decay**
-
-By inspection of signals in Figs. 6.2a, 6.7a, and 6.7b, determine the asymptotic rate of decay of their amplitude spectra.
-
-<sup>†</sup> There is also an undershoot of 9% at the other side [at *<sup>t</sup>* <sup>=</sup> (π/2)+] of the discontinuity.
-
-<sup>‡</sup> Actually, at discontinuities, the series converges to a value midway between the values on either side of the discontinuity. The 9% overshoot occurs at *t* = (π/2)<sup>−</sup> and 9% undershoot occurs at *t* = (π/2)+.
-
-**Figure 6.11** Fourier synthesis of a continuous signal using first 19 harmonics.
-
-### **ANSWERS**
-
-1/*n*, 1/*n*2, and 1/*n*, respectively.
-
-### A HISTORICAL NOTE ON THE GIBBS PHENOMENON
-
-Normally speaking, troublesome functions with strange behavior are invented by mathematicians; we rarely see such oddities in practice. In the case of the Gibbs phenomenon, however, the tables were turned. A rather puzzling behavior was observed in a mundane object, a mechanical wave synthesizer, and then well-known mathematicians of the day were dispatched on the scent of it to discover its hideout.
-
-Albert Michelson (of Michelson–Morley fame) was an intense, practical man who developed ingenious physical instruments of extraordinary precision, mostly in the field of optics. His harmonic analyzer, developed in 1898, could compute the first 80 coefficients of the Fourier series of a signal *x*(*t*) specified by any graphical description. The instrument could also be used as a harmonic synthesizer, which could plot a function *x*(*t*) generated by summing the first 80 harmonics (Fourier components) of arbitrary amplitudes and phases. This analyzer, therefore, had the ability of self-checking its operation by analyzing a signal *x*(*t*) and then adding the resulting 80 components to see whether the sum yielded a close approximation of *x*(*t*).
-
-Michelson found that the instrument checked very well with most of signals analyzed. However, when he tried a discontinuous function, such as a square wave,† a curious behavior was observed. The sum of 80 components showed oscillatory behavior (ringing), with an overshoot of 9% in the vicinity of the points of discontinuity. Moreover, this behavior was a constant feature regardless of the number of terms added. A larger number of terms made the oscillations proportionately faster, but regardless of the number of terms added, the overshoot remained 9%. This puzzling behavior caused Michelson to suspect some mechanical defect in his synthesizer. He wrote about his observation in a letter to *Nature* (December 1898). Josiah Willard Gibbs, who was a professor at Yale, investigated and clarified this behavior for a sawtooth periodic signal in a letter to *Nature* [7]. Later, in 1906, Bôcher generalized the result for any function with discontinuity [8].
-
-<sup>†</sup> Actually, it was a periodic sawtooth signal.
-
-<span id="page-640-0"></span>Albert Michelson and Josiah Willard Gibbs
-
-It was Bôcher who gave the name *Gibbs phenomenon* to this behavior. Gibbs showed that the peculiar behavior in the synthesis of a square wave was inherent in the behavior of the Fourier series because of nonuniform convergence at the points of discontinuity.
-
-This, however, is not the end of the story. Both Bôcher and Gibbs were under the impression that this property had remained undiscovered until Gibbs's work published in 1899. It is now known that what is called the Gibbs phenomenon had been observed in 1848 by Wilbraham of Trinity College, Cambridge, who clearly saw the behavior of the sum of the Fourier series components in the periodic sawtooth signal later investigated by Gibbs [9]. Apparently, this work was not known to most people, including Gibbs and Bôcher.
-
-## **[6.3 EXPONENTIAL](#page-12-0) FOURIER SERIES**
-
-By using Euler's equality, we can express cos *n*ω0*t* and sin *n*ω0*t* in terms of exponentials *ejn*<sup>ω</sup>0*<sup>t</sup>* and *e*−*jn*ω0*<sup>t</sup>* . Clearly, we should be able to express the trigonometric Fourier series in Eq. (6.7) in terms of exponentials of the form *ejn*<sup>ω</sup>0*<sup>t</sup>* with the index *n* taking on all integer values from −∞ to ∞, including zero. Derivation of the exponential Fourier series from the results already derived for the trigonometric Fourier series is straightforward, involving conversion of sinusoids to exponentials. We shall, however, derive them here independently, without using the prior results of the trigonometric series.
-
-This discussion shows that the *exponential Fourier series* for a periodic signal *x*(*t*) can be expressed as
-
-$$
-x(t) = \sum_{n = -\infty}^{\infty} D_n e^{jn\omega_0 t}
-$$
-
-#### 622 CHAPTER 6 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES
-
-To derive the coefficients *Dn*, we multiply both sides of this equation by *e*−*jm*ω0*<sup>t</sup>* (*m* integer) and integrate over one period. This yields
-
-$$
-\int_{T_0} x(t) e^{-jm\omega_0 t} dt = \sum_{n=-\infty}^{\infty} D_n \int_{T_0} e^{j(n-m)\omega_0 t} dt
-$$
-
-To simplify this expression, we use the *orthogonality* property of exponentials, which states that†
-
-$$
-\int_{T_0} e^{jn\omega_0 t} e^{-jm\omega_0 t} dt = \begin{cases} 0 & m \neq n \\ T_0 & m = n \end{cases}
-$$
-\n(6.18)
-
-Thus,
-
-$$
-\int_{T_0} x(t) e^{-jm\omega_0 t} dt = D_m T_0
-$$
-
-from which we obtain
-
-$$
-D_m = \frac{1}{T_0} \int_{T_0} x(t) e^{-jm\omega_0 t} dt.
-$$
-
-To summarize, the exponential Fourier series can be expressed as
-
-$$
-x(t) = \sum_{n = -\infty}^{\infty} D_n e^{jn\omega_0 t} \qquad \text{where} \qquad D_n = \frac{1}{T_0} \int_{T_0} x(t) e^{-jn\omega_0 t} dt \tag{6.19}
-$$
-
-Observe the compactness of Eq. (6.19) and compare it with the trigonometric Fourier series expression. Such a comparison demonstrates very clearly the principal virtue of the exponential Fourier series. First, the form of the series is most compact. Second, the mathematical expression for deriving the coefficients of the series is also compact. It is much more convenient to handle the exponential series than the trigonometric one. For these reasons we shall use the exponential (rather than trigonometric) representation of signals in the rest of the book.
-
-We can now relate *Dn* to trigonometric series coefficients *an* and *bn*. Setting *n*=0 in Eq. (6.19), we obtain
-
-$$
-D_0=a_0
-$$
-
-Moreover, for *n* = 0,
-
-$$
-D_n = \frac{1}{T_0} \int_{T_0} x(t) \cos n\omega_0 t \, dt - \frac{j}{T_0} \int_{T_0} x(t) \sin n\omega_0 t \, dt = \frac{1}{2} (a_n - jb_n) \tag{6.20}
-$$
-
-$$
-\int_{T_0} e^{j(n-m)\omega_0 t} dt = \int_{T_0} \cos{(n-m)\omega_0 t} dt + j \int_{T_0} \sin{(n-m)\omega_0 t} dt
-$$
-
-Both the integrals on the right-hand side represent area under *n* − *m* number of cycles. Because *n* − *m* is an integer, both the areas are zero. Hence, Eq. (6.18) follows.
-
-<sup>†</sup> We can readily prove this property as follows. For the case of *<sup>m</sup>* <sup>=</sup> *<sup>n</sup>*, the integrand in Eq. (6.18) is unity and the integral is *T*0. When *m* = *n*, the integral on the left-hand side of Eq. (6.18) can be expressed as
-
-and
-
-$$
-D_{-n} = \frac{1}{T_0} \int_{T_0} x(t) \cos n\omega_0 t \, dt + \frac{j}{T_0} \int_{T_0} x(t) \sin n\omega_0 t \, dt = \frac{1}{2} (a_n + jb_n) \tag{6.21}
-$$
-
-These results are valid for general *x*(*t*), real or complex. When *x*(*t*) is real, *an* and *bn* are real, and Eqs. (6.20) and (6.21) show that *Dn* and *D*−*<sup>n</sup>* are conjugates.
-
-$$
-D_{-n}=D_n^*
-$$
-
-Moreover, from Eq. (6.10), we observe that
-
-$$
-a_n - jb_n = \sqrt{a_n^2 + b_n^2} \, e^{j \tan^{-1} \left( \frac{-b_n}{a_n} \right)} = C_n e^{j \theta_n}
-$$
-
-Hence,
-
-$$
-D_0=a_0=C_0
-$$
-
-and
-
-$$
-D_n = \frac{1}{2} C_n e^{j\theta_n} \qquad D_{-n} = \frac{1}{2} C_n e^{-j\theta_n}
-$$
-
-Therefore, for *n* = 0,
-
-$$
-|D_n| = |D_{-n}| = \frac{1}{2}C_n, \quad \angle D_n = \theta_n, \quad \text{and} \angle D_{-n} = -\theta_n \tag{6.22}
-$$
-
-Note that |*Dn*| are the amplitudes and *Dn* are the angles of various exponential components. From Eq. (6.22) it follows that when *x*(*t*) is real, the amplitude spectrum (|*Dn*| versus ω) is an even function of ω and the angle spectrum ( *Dn* versus ω) is an odd function of ω. For complex *x*(*t*), *Dn* and *D*−*<sup>n</sup>* are generally not conjugates.

engineering/linear-systems-and-signals-3rd-edition-bp-lathi/081_6.3 EXPONENTIAL FOURIER SERIES.md

@@ -1,417 +0,0 @@
-### **EXAMPLE 6.6 Exponential Fourier Series of Periodic Exponential Wave**
-
-Find the exponential Fourier series for the signal of Fig. 6.2a from Ex. 6.1.
-
-In this case *T*<sup>0</sup> = π, ω<sup>0</sup> = 2π/*T*<sup>0</sup> = 2, and
-
-$$
-x(t) = \sum_{n = -\infty}^{\infty} D_n e^{j2nt}
-$$
-
-where
-
-$$
-D_n = \frac{1}{T_0} \int_{T_0} x(t) e^{-j2nt} dt = \frac{1}{\pi} \int_0^{\pi} e^{-t/2} e^{-j2nt} dt = \frac{1}{\pi} \int_0^{\pi} e^{-(1/2 + j2n)t} dt
-$$
-  
-=  $\frac{-1}{\pi (\frac{1}{2} + j2n)} e^{-(1/2 + j2n)t} \Big|_0^{\pi} = \frac{0.504}{1 + j4n}$ 
-
-$$
-x(t) = 0.504 \sum_{n=-\infty}^{\infty} \frac{1}{1+j4n} e^{j2nt}
-$$
-  
-= 0.504  $\left[ 1 + \frac{1}{1+j4} e^{j2t} + \frac{1}{1+j8} e^{j4t} + \frac{1}{1+j12} e^{j6t} + \cdots + \frac{1}{1-j4} e^{-j2t} + \frac{1}{1-j8} e^{-j4t} + \frac{1}{1-j12} e^{-j6t} + \cdots \right]$ 
-
-Observe that the coefficients *Dn* are complex. Moreover, *Dn* and *D*−*<sup>n</sup>* are conjugates, as expected.
-
-### **[6.3-1 Exponential Fourier Spectra](#page-12-0)**
-
-In exponential spectra, we plot coefficients *Dn* as a function of ω. But since *Dn* is complex in general, we need both parts of one of two sets of plots: the real and the imaginary parts of *Dn*, or the magnitude and the angle of *Dn*. We prefer the latter because of its close connection to the amplitudes and phases of corresponding components of the trigonometric Fourier series. We therefore plot |*Dn*| versus ω and *Dn* versus ω. This requires that the coefficients *Dn* be expressed in polar form as |*Dn*|*ej Dn* , where |*Dn*| are the amplitudes and *Dn* are the angles of various exponential components. Equation (6.22) shows that for real *x*(*t*), the amplitude spectrum (|*Dn*| versus ω) is an even function of ω and the angle spectrum ( *Dn* versus ω) is an odd function of ω.
-
-For the series in Ex. 6.6, for instance,
-
-$$
-D_0 = 0.504
-$$
-  
-\n
-$$
-D_1 = \frac{0.504}{1+j4} = 0.122e^{-j75.96^\circ} \implies |D_1| = 0.122, \angle D_1 = -75.96^\circ
-$$
-  
-\n
-$$
-D_{-1} = \frac{0.504}{1-j4} = 0.122e^{j75.96^\circ} \implies |D_{-1}| = 0.122, \angle D_{-1} = 75.96^\circ
-$$
-
-and
-
-$$
-D_2 = \frac{0.504}{1+j8} = 0.0625e^{-j82.87^\circ} \implies |D_2| = 0.0625, \ \angle D_2 = -82.87^\circ
-$$
-  
-$$
-D_{-2} = \frac{0.504}{1-j8} = 0.0625e^{j82.87^\circ} \implies |D_{-2}| = 0.0625, \ \angle D_{-2} = 82.87^\circ
-$$
-
-and so on. Note that *Dn* and *D*−*<sup>n</sup>* are conjugates, as expected [see Eq. (6.22)].
-
-Figure 6.12 shows the frequency spectra (amplitude and angle) of the exponential Fourier series for the periodic signal *x*(*t*) in Fig. 6.2a.
-
-We notice some interesting features of these spectra. First, the spectra exist for positive as well as negative values of ω (the frequency). Second, the amplitude spectrum is an even function of ω and the angle spectrum is an odd function of ω.
-
-<span id="page-643-0"></span>and
-
-**Figure 6.12** Exponential Fourier spectra for the signal in Fig. 6.2a.
-
-At times it may appear that the phase spectrum of a real periodic signal fails to satisfy the odd symmetry: for example, when *Dk* = *D*−*<sup>k</sup>* = −10. In this case, *Dk* = 10*ej*<sup>π</sup> , and therefore, *D*−*<sup>k</sup>* = 10*e*−*j*<sup>π</sup> . Recall that *e*±*j*<sup>π</sup> = −1. Here, although *Dk* = *D*−*<sup>k</sup>*, their phases should be taken as π and −π.
-
-### **EXAMPLE 6.7 Plotting Fourier Series Spectra with MATLAB**
-
-Using MATLAB and the results of Ex. 6.6, compute and plot the exponential Fourier spectra for the periodic signal *x*(*t*) shown in Fig. 6.2a. The result should match Fig. 6.12.
-
-The expression for *Dn* is derived in Ex. 6.6.
-
-```
->> clf; n = (-5:5); D_n = 0.504./(1+4j*n);
-```
-
-- >> subplot(1,2,1); stem(n,abs(D\_n),'.k');
-- >> xlabel('n'); ylabel('|D\_n|');
-- >> subplot(1,2,2); stem(n,angle(D\_n),'.k');
-- >> xlabel('n'); ylabel('\angle D\_n [rad]');
-
-Except that it is plotted as a function of *n* rather than ω, the result in Fig. 6.13 matches Fig. 6.12.
-
-### WHAT IS A NEGATIVE FREQUENCY?
-
-The existence of the spectrum at negative frequencies is somewhat disturbing because, by definition, the frequency (number of repetitions per second) is a positive quantity. How do we interpret a negative frequency? We can use a trigonometric identity to express a sinusoid of a negative frequency −ω<sup>0</sup> as
-
-$$
-\cos(-\omega_0 t + \theta) = \cos(\omega_0 t - \theta)
-$$
-
-This equation clearly shows that the frequency of a sinusoid cos(ω0*t* + θ ) is |ω0|, which is a positive quantity. The same conclusion is reached by observing that
-
-$$
-e^{\pm j\omega_0 t} = \cos \omega_0 t \pm j \sin \omega_0 t
-$$
-
-Thus, the frequency of exponentials *e*±*j*ω0*<sup>t</sup>* is indeed |ω0|. How do we then interpret the spectral plots for negative values of ω? A more satisfying way of looking at the situation is to say that *exponential spectra are a graphical representation of coefficients Dn as a function of* ω*. Existence of the spectrum at* ω = −*n*ω<sup>0</sup> *is merely an indication that an exponential component e*−*jn*ω0*<sup>t</sup> exists in the series*. We know that a sinusoid of frequency *n*ω<sup>0</sup> can be expressed in terms of a pair of exponentials *ejn*<sup>ω</sup>0*<sup>t</sup>* and *e*−*jn*ω0*<sup>t</sup>* .
-
-We see a close connection between the exponential spectra in Fig. 6.12 and the spectra of the corresponding trigonometric Fourier series for *x*(*t*) (Figs. 6.2b, 6.2c). Equation (6.22) explains the reason for the close connection, for real *x*(*t*), between the trigonometric spectra (*Cn* and θ*n*) with exponential spectra (|*Dn*| and *Dn*). The dc components *D*<sup>0</sup> and *C*<sup>0</sup> are identical in both spectra. Moreover, the exponential amplitude spectrum |*Dn*| is half the trigonometric amplitude spectrum *Cn* for *n* ≥ 1. The exponential angle spectrum *Dn* is identical to the trigonometric phase spectrum θ*<sup>n</sup>* for *n* ≥ 0. We can therefore produce the exponential spectra merely by inspection of trigonometric spectra, and vice versa. The following example demonstrates this feature.
-
-### **EXAMPLE 6.8 Relating Exponential to Trigonometric Fourier Series Spectra**
-
-The trigonometric Fourier spectra of a certain periodic signal *x*(*t*) are shown in Fig. 6.14a. After inspecting these spectra, sketch the corresponding exponential Fourier spectra and verify your results analytically.
-
-**Figure 6.14** Fourier series spectra for Ex. 6.8.
-
-The trigonometric spectral components exist at frequencies 0, 3, 6, and 9. The exponential spectral components exist at 0, 3, 6, 9, and −3, −6, −9. Consider first the amplitude spectrum. The dc component remains unchanged: that is, *D*<sup>0</sup> = *C*<sup>0</sup> = 16. Now |*Dn*| is an even function of ω and |*Dn*|=|*D*−*<sup>n</sup>*| = *Cn*/2. Thus, all the remaining spectrum |*Dn*| for positive *n* is half the trigonometric amplitude spectrum *Cn*, and the spectrum |*Dn*| for negative *n* is a reflection about the vertical axis of the spectrum for positive *n*, as shown in Fig. 6.14b.
-
-The angle spectrum is *Dn* = θ*<sup>n</sup>* for positive *n* and is −θ*<sup>n</sup>* for negative *n*, as depicted in Fig. 6.14b. We shall now verify that both sets of spectra represent the same signal.
-
-Signal *x*(*t*), whose trigonometric spectra are shown in Fig. 6.14a, has four spectral components of frequencies 0, 3, 6, and 9. The dc component is 16. The amplitude and the phase of the component of frequency 3 are 12 and −π/4, respectively. Therefore, this component can be expressed as 12cos(3*t* − π/4). Proceeding in this manner, we can write the Fourier series for *x*(*t*) as
-
-$$
-x(t) = 16 + 12\cos\left(3t - \frac{\pi}{4}\right) + 8\cos\left(6t - \frac{\pi}{2}\right) + 4\cos\left(9t - \frac{\pi}{4}\right)
-$$
-
-Consider now the exponential spectra in Fig. 6.14b. They contain components of frequencies 0 (dc), ±3, ±6, and ±9. The dc component is *D*<sup>0</sup> = 16. The component *ej*3*<sup>t</sup>* (frequency 3) has magnitude 6 and angle −π/4. Therefore, this component strength is 6*e*−*j*π/4, and it can be expressed as (6*e*−*j*π/4)*ej*3*<sup>t</sup>* . Similarly, the component of frequency −3 is (6*ej*π/4)*e*−*j*3*<sup>t</sup>* . Proceeding in this manner, *x*ˆ(*t*), the signal corresponding to the spectra in Fig. 6.14b, is
-
-$$
-\hat{x}(t) = 16 + \left[6e^{-j\pi/4}e^{j3t} + 6e^{j\pi/4}e^{-j3t}\right] + \left[4e^{-j\pi/2}e^{j6t} + 4e^{j\pi/2}e^{-j6t}\right]
-$$
-\n
-$$
-+ \left[2e^{-j\pi/4}e^{j9t} + 2e^{j\pi/4}e^{-j9t}\right]
-$$
-\n
-$$
-= 16 + 6\left[e^{j(3t - \pi/4)} + e^{-j(3t - \pi/4)}\right] + 4\left[e^{j(6t - \pi/2)} + e^{-j(6t - \pi/2)}\right]
-$$
-\n
-$$
-+ 2\left[e^{j(9t - \pi/4)} + e^{-j(9t - \pi/4)}\right]
-$$
-\n
-$$
-= 16 + 12\cos\left(3t - \frac{\pi}{4}\right) + 8\cos\left(6t - \frac{\pi}{2}\right) + 4\cos\left(9t - \frac{\pi}{4}\right)
-$$
-
-Clearly both sets of spectra represent the same periodic signal.
-
-### BANDWIDTH OF A SIGNAL
-
-The difference between the highest and the lowest frequencies of the spectral components of a signal is the *bandwidth* of the signal. The bandwidth of the signal whose exponential spectra are shown in Fig. 6.14b is 9 (in radians). The highest and lowest frequencies are 9 and 0, respectively. Note that the component of frequency 12 has zero amplitude and is nonexistent. Moreover, the lowest frequency is 0, not −9. Recall that the frequencies (in the conventional sense) of the spectral components at ω = −3, −6, and −9 in reality are 3, 6, and 9.† The bandwidth can be more readily seen from the trigonometric spectra in Fig. 6.14a.
-
-### **EXAMPLE 6.9 Fourier Series Spectra of an Impulse Train**
-
-Find the exponential Fourier series and sketch the corresponding spectra for the impulse train δ*<sup>T</sup>*<sup>0</sup> (*t*) depicted in Fig. 6.15a. From this result, sketch the trigonometric spectrum and write the trigonometric Fourier series for δ*<sup>T</sup>*<sup>0</sup> (*t*).
-
-<sup>†</sup> Some authors *do* define bandwidth as the difference between the highest and the lowest (negative) frequency in the exponential spectrum. The bandwidth according to this definition is twice that defined here. In reality, this phrasing defines not the signal bandwidth but the *spectral width* (width of the exponential spectrum of the signal).
-
-The unit impulse train shown in Fig. 6.15a can be expressed as
-
-$$
-\sum_{n=-\infty}^{\infty} \delta(t - nT_0)
-$$
-
-Following Papoulis, we shall denote this function as δ*<sup>T</sup>*<sup>0</sup> (*t*) for the sake of notational brevity. The exponential Fourier series is given by
-
-$$
-\delta_{T_0}(t) = \sum_{n = -\infty}^{\infty} D_n e^{jn\omega_0 t} \qquad \omega_0 = \frac{2\pi}{T_0}
-$$
-\n(6.23)
-
-where
-
-$$
-D_n = \frac{1}{T_0} \int_{T_0} \delta_{T_0}(t) e^{-jn\omega_0 t} dt
-$$
-
-Choosing the interval of integration (−*T*0/2,*T*0/2) and recognizing that over this interval δ*<sup>T</sup>*<sup>0</sup> (*t*) = δ(*t*), we get
-
-$$
-D_n = \frac{1}{T_0} \int_{-T_0/2}^{T_0/2} \delta(t) e^{-jn\omega_0 t} dt
-$$
-
-### 630 CHAPTER 6 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES
-
-In this integral, the impulse is located at *t* = 0. From the sampling property of Eq. (1.11), the integral on the right-hand side is the value of *e*−*jn*ω0*<sup>t</sup>* at *t* = 0 (where the impulse is located). Therefore,
-
-$$
-D_n = \frac{1}{T_0} \tag{6.24}
-$$
-
-From this result, we see that the exponential spectrum is constant for all frequencies, as shown in Fig. 6.15b. The spectrum, being real, requires only the amplitude plot. All phases are zero.
-
-Substituting *Dn* <sup>=</sup> <sup>1</sup> *<sup>T</sup>*<sup>0</sup> into Eq. (6.23) yields the desired exponential Fourier series
-
-$$
-\delta_{T_0}(t) = \frac{1}{T_0} \sum_{n = -\infty}^{\infty} e^{jn\omega_0 t} \qquad \omega_0 = \frac{2\pi}{T_0}
-$$
-
-To sketch the trigonometric spectrum, we use Eq. (6.22) to obtain
-
-$$
-C_0 = D_0 = \frac{1}{T_0}
-$$
-  
-\n
-$$
-C_n = 2|D_n| = \frac{2}{T_0} \qquad n = 1, 2, 3, ...
-$$
-  
-\n
-$$
-\theta_n = 0
-$$
-
-Figure 6.15c shows the trigonometric Fourier spectrum. From this spectrum we can express δ*<sup>T</sup>*<sup>0</sup> (*t*) as
-
-$$
-\delta_{T_0}(t) = \frac{1}{T_0} \left[ 1 + 2(\cos \omega_0 t + \cos 2\omega_0 t + \cos 3\omega_0 t + \cdots) \right] \qquad \omega_0 = \frac{2\pi}{T_0} \tag{6.25}
-$$
-
-### EFFECT OF SYMMETRY IN EXPONENTIAL FOURIER SERIES
-
-When *x*(*t*) has an even symmetry, *bn* = 0, and from Eq. (6.20), *Dn* = *an*/2, which is real (positive or negative). Hence, *Dn* can only be 0 or ±π. Moreover, we may compute *Dn* = *an*/2 by using Eq. (6.14), which requires integration over a half-period only. Similarly, when *x*(*t*) has an odd symmetry, *an* = 0, and *Dn* = −*jbn*/2 is imaginary (positive or negative). Hence, *Dn* can only be 0 or ±π/2. Moreover, we may compute *Dn* = −*jbn*/2 by using Eq. (6.15), which requires integration over a half-period only. Note, however, that in the exponential case, we are using the symmetry property indirectly by finding the trigonometric coefficients. We cannot apply it directly in finding *Dn* from Eq. (6.19) since the function *ejn*<sup>ω</sup>0*<sup>t</sup>* is neither even nor odd.
-
-## **DR ILL 6.4 Relating Trigonometric to Exponential Fourier Series Spectra**
-
-The exponential Fourier spectra of a certain periodic signal *x*(*t*) are shown in Fig. 6.16. Determine and sketch the trigonometric Fourier spectra of *x*(*t*) by inspection of Fig. 6.16. Now write the (compact) trigonometric Fourier series for *x*(*t*).
-
-## **DR ILL 6.5 Fourier Series Spectrum of a Full-Wave Rectified Sine Wave**
-
-Find the exponential Fourier series and sketch the corresponding Fourier spectrum *Dn* versus ω for the full-wave rectified sine wave depicted in Fig. 6.17.
-
-### **ANSWER**
-
-### **DR ILL 6.6 Exponential Fourier Series and Spectra**
-
-Find the exponential Fourier series and sketch the corresponding Fourier spectra for the periodic signals shown in Fig. 6.7.
-
-#### <span id="page-651-0"></span>**ANSWERS (a)** *<sup>x</sup>*(*t*) <sup>=</sup> <sup>1</sup> <sup>3</sup> <sup>+</sup> 2 π2 "∞ *n*=−∞(*n*=0) (−1)*<sup>n</sup> <sup>n</sup>*<sup>2</sup> *ejn*π*<sup>t</sup>* **(b)** *<sup>x</sup>*(*t*) <sup>=</sup> *jA* π "∞ *n*=−∞(*n*=0) (−1)*<sup>n</sup> <sup>n</sup> ejn*π*<sup>t</sup>*
-
-### **[6.3-2 Parseval's Theorem](#page-12-0)**
-
-The trigonometric Fourier series of a periodic signal *x*(*t*) is given by
-
-$$
-x(t) = C_0 + \sum_{n=1}^{\infty} C_n \cos(n\omega_0 t + \theta_n)
-$$
-
-Every term on the right-hand side of this equation is a power signal. As shown in Ex. 1.2, Eq. (1.3), the power of *x*(*t*) is equal to the sum of the powers of all the sinusoidal components on the right-hand side.
-
-$$
-P_x = C_0^2 + \frac{1}{2} \sum_{n=1}^{\infty} C_n^2
-$$
-\n(6.26)
-
-This result is one form of *Parseval's theorem,* as applied to power signals. It states that the power of a periodic signal is equal to the sum of the powers of its Fourier components.
-
-We can apply the same argument to the exponential Fourier series (see Prob. 1.1-11). The power of a periodic signal *x*(*t*) can be expressed as a sum of the powers of its exponential components. In Eq. (1.4), we showed that the power of an exponential *Dej*<sup>ω</sup>0*<sup>t</sup>* is |*D*2|. We can use this result to express the power of a periodic signal *x*(*t*) in terms of its exponential Fourier series coefficients as
-
-$$
-P_x = \sum_{n=-\infty}^{\infty} |D_n|^2 \tag{6.27}
-$$
-
-For a real *x*(*t*), |*D*−*<sup>n</sup>*|=|*Dn*|. Therefore,
-
-$$
-P_x = D_0^2 + 2\sum_{n=1}^{\infty} |D_n|^2
-$$
-\n(6.28)
-
-### **EXAMPLE 6.10 Harmonic Distortion of Clipped Sinusoid**
-
-The input signal to an audio amplifier of gain 100 is given by *x*(*t*) = 0.1 cosω0*t*. Hence, the output is a sinusoid 10 cos ω0*t*. However, the amplifier, being nonlinear at higher amplitude levels, clips all amplitudes beyond ±8 volts, as shown in Fig. 6.18a. We shall determine the harmonic distortion incurred in this operation.
-
-**Figure 6.18 (a)** A clipped sinusoid cos ω0*t*. **(b)** The distortion component *xd*(*t*) of the signal in (a).
-
-The output *y*(*t*) is the clipped signal in Fig. 6.18a. The distortion signal *yd*(*t*), shown in Fig. 6.18b, is the difference between the undistorted sinusoid 10cosω0*t* and the output signal *y*(*t*). The signal *yd*(*t*), whose period is *T*<sup>0</sup> [the same as that of *y*(*t*)], can be described over the first cycle as
-
-$$
-y_d(t) = \begin{cases} 10 \cos \omega_0 t - 8 & |t| \le 0.1024T_0 \\ 10 \cos \omega_0 t + 8 & \frac{T_0}{2} - 0.1024T_0 \le |t| \le \frac{T_0}{2} + 0.1024T_0 \\ 0 & \text{everywhere else} \end{cases}
-$$
-
-Observe that *yd*(*t*) is an even function of *t* and its mean value is zero. Hence, *a*<sup>0</sup> = *C*<sup>0</sup> = 0, and *bn* = 0. Thus, *Cn* = *an* and the Fourier series for *yd*(*t*) can be expressed as
-
-$$
-y_d(t) = \sum_{n=1}^{\infty} C_n \cos n\omega_0 t
-$$
-
-As usual, we can compute the coefficients *Cn* (which is equal to *an*) by integrating *yd*(*t*) cos *n*ω0*t* over one cycle (and then dividing by 2/*T*0). Because *yd*(*t*) has even symmetry, we can find *an* by integrating the expression over a half-cycle only using Eq. (6.14). The
-
-#### 634 CHAPTER 6 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES
-
-straightforward evaluation of the appropriate integral yields†
-
-$$
-C_n = \begin{cases} \frac{20}{\pi} \left[ \frac{\sin\left[0.6435(n+1)\right]}{n+1} + \frac{\sin\left[0.6435(n-1)\right]}{n-1} \right] - \frac{32}{\pi} \left[ \frac{\sin\left(0.6435n\right)}{n} \right] & n \text{ odd} \\ 0 & n \text{ even} \end{cases}
-$$
-
-Computing the coefficients *C*1, *C*2, *C*3, ... from this expression, we can write
-
-*yd*(*t*) = 1.04 cos ω0*t* +0.733 cos 3ω0*t* +0.311 cos 5ω0*t* +···
-
-### COMPUTING HARMONIC DISTORTION
-
-We can compute the amount of harmonic distortion in the output signal by computing the power of the distortion component *yd*(*t*). Because *yd*(*t*) is an even function of *t* and because the energy in the first half-cycle is identical to the energy in the second half-cycle, we can compute the power by averaging the energy over a quarter-cycle. Thus,
-
-$$
-P_{y_d} = \frac{1}{T_0} \int_{-T_0/2}^{T_0/2} y_d^2(t) dt = \frac{1}{T_0/4} \int_0^{T_0/4} y_d^2(t) dt
-$$
-  
-=  $\frac{4}{T_0} \int_0^{0.1024T_0} (10 \cos \omega_0 t - 8)^2 dt = 0.865$ 
-
-The power of the desired signal 10cos ω0*t* is (10)<sup>2</sup>/2 = 50. Hence, the total harmonic distortion is‡
-
-$$
-D_{\text{tot}} = \frac{0.865}{50} \times 100 = 1.73\%
-$$
-
-The power of the third harmonic components of *yd*(*t*) is (0.733)<sup>2</sup>/2 = 0.2686. The third harmonic distortion is
-
-$$
-D_3 = \frac{0.2686}{50} \times 100 = 0.5372\%
-$$
-
-$$
-C_n = a_n = \frac{8}{T_0} \int_0^{0.1024T_0} [10 \cos \omega_0 t - 8] \cos n\omega_0 t dt
-$$
-
-<sup>†</sup> In addition, *yd*(*t*) exhibits half-wave symmetry (see Prob. 6.1-6), where the second half-cycle is the negative of the first. Because of this property, all the even harmonics vanish, and the odd harmonics can be computed by integrating the appropriate expressions over the first half-cycle only (from −*T*0/4 to *T*0/4) and doubling the resulting values. Moreover, because of even symmetry, we can integrate the appropriate expressions over 0 to *T*0/4 (instead of from −*T*0/4 to *T*0/4) and double the resulting values. In essence, this allows us to compute *Cn* by integrating the expression over the quarter-cycle only and then quadrupling the resulting values. Thus,
-
-<sup>‡</sup> In the literature, the harmonic distortion often refers to the rms distortion rather than the power distortion. The rms values are the square-root values of the corresponding powers. Thus, the third harmonic distortion in this sense is <sup>√</sup>(0.2686/50) <sup>×</sup> <sup>100</sup> <sup>=</sup> 7.33%. Alternately, we may also compute this value directly from the amplitudes of the third harmonic 0.733 and that of the fundamental as 10. The ratio of the rms values is (0.733/ <sup>√</sup>2) : (10/ √ 2) = 0.0733 and the percentage distortion is 7.33%.
-
-### <span id="page-654-0"></span>**[6.3-3 Properties of the Fourier Series](#page-12-0)**
-
-As with the Laplace and *z*-transforms, the Fourier series has a variety of properties that can simplify work and help provide a more intuitive understanding of signals. Table 6.2 provides the most important properties of the Fourier series for a periodic signal *x*(*t*) and its spectrum *Dn*. Properties that involve two signals require that the two signals have a common fundamental frequency ω0. While not given here, the proofs of these properties are straightforward and parallel the proofs of the Fourier transform properties given in Ch. 7.
-
-To demonstrate the utility of Fourier series properties, let us consider an example where we use a selection of properties to simplify the work of finding a piecewise polynomial signal's spectrum.
-
-| Operation             | x(t)                         | Dn             |
-|-----------------------|------------------------------|----------------|
-| Scalar multiplication | kx(t)                        | kDn            |
-| Addition              | x1(t)+x2(t)                  | D1,n<br>+D2,n  |
-|                       | x1(t), x2(t) require same ω0 |                |
-| Conjugation           | x∗(t)                        | D∗<br>−n       |
-| Reversal              | x(−t)                        | D−n            |
-| Time shifting         | x(t −t0)                     | Dne−jnω0t0     |
-| Frequency shifting    | x(t)ejn0ω0t                  | Dn−n0          |
-| Frequency convolution | x1(t)x2(t)                   | D1,n<br>∗ D2,n |
-|                       | x1(t), x2(t) require same ω0 |                |
-| Time differentiation  | dkx(t)<br>dtk                | (jnω0)kDn      |
-
-**TABLE 6.2** Selected Fourier Series Properties
-
-### **EXAMPLE 6.11 Using Fourier Series Properties**
-
-Use properties rather than integration to compute the exponential Fourier series coefficients *Dn* of the triangular signal *x*(*t*) shown in Fig. 6.4. Verify the correctness of *Dn* for *A* = 1 by synthesizing *x*(*t*) with a suitable truncation of Eq. (6.19).
-
-From Fig. 6.4, we see that *x*(*t*) is a piecewise linear function that is *T*<sup>0</sup> = 2 periodic. To compute *Dn* directly using Eq. (6.19) would therefore require tedious integration by parts. Fortunately, we can compute *Dn* without integration by instead using Fourier series properties. First, however, we must compute the dc component *D*<sup>0</sup> separately from other *Dn*. By simple inspection of Fig. 6.4, we see that *x*(*t*) has no dc component, so *D*<sup>0</sup> = 0.
-
-To determine the remaining *Dn*, we begin by noting that *x*(*t*) has a constant slope of either 2*A* or −2*A*. Thus, differentiating *x*(*t*) once yields a square wave with amplitudes ±2*A*. Here, differentiation reduces *x*(*t*) from a piecewise linear to a piecewise constant function,
-
-thereby eliminating the need for integration by parts.† Differentiating *x*(*t*) twice yields a pair of shifted impulse trains, weighted by ±4*A*. This second differentiation eliminates the need for any integration whatsoever, since the Fourier series coefficients of an impulse train are known to be <sup>1</sup> *<sup>T</sup>*<sup>0</sup> (see Ex. 6.9).
-
-Stated mathematically, we see that
-
-$$
-\frac{d^2}{dt^2}x(t) = 4A\delta_2(t + \frac{1}{2}) - 4A\delta_2(t - \frac{1}{2})
-$$
-
-Transforming this expression and using the Fourier series properties of scalar multiplication, addition, frequency shifting, and time differentiation yield (for *n* = 0)
-
-$$
-(in\pi)^2 D_n = 4A \frac{e^{jn\omega_0/2}}{2} - 4A \frac{e^{-jn\omega_0/2}}{2}
-$$
-
-Substituting ω<sup>0</sup> = 2π/*T*<sup>0</sup> = π and solving for *Dn* yield
-
-$$
-D_n = 4A \frac{e^{in\pi/2} - e^{-in\pi/2}}{-2n^2\pi^2}
-$$
-
-Combining with the dc component and simplifying expressions, the final result is
-
-$$
-D_n = \begin{cases} 0 & n = 0\\ \frac{-A4j\sin(n\pi/2)}{n^2\pi^2} & n \neq 0 \end{cases}
-$$
-
-Overall, this is a neat way to determine the signal's spectrum. Through the selective use of properties, we have determined *Dn* without any integration. With a little care, this basic approach can yield the spectrum of *any piecewise polynomial periodic function*. Since all periodic functions of practical interest to engineers can, to an arbitrary level of accuracy, be represented as piecewise polynomial functions, we can always find their spectra without integration except for the dc term *D*0.
-
-### USING A TRUNCATED FOURIER SERIES TO VERIFY SPECTRUM CORRECTNESS
-
-While a signal's spectrum *Dn* provides useful insight into signal character, it can be difficult to look at *Dn* and know that it is correct for a particular signal *x*(*t*). For example, is it at all obvious that *Dn* <sup>=</sup> <sup>4</sup>*Aj*sin(*n*π/2) <sup>−</sup>*n*2π<sup>2</sup> is the spectrum for the triangle wave of Fig. 6.4? Probably not.
-
-There is an easy way, however, to verify a signal's spectrum: synthesize *x*(*t*) using a suitable truncation of Eq. (6.19). If the synthesized signal closely matches the original, we can be relatively certain that the spectrum *Dn* is correct. What is a suitable truncation? Well, it depends. All significant *Dn* terms need to be included, but not so many as to make the reconstruction impractical to compute. Since in the present case *Dn* decays at a rate 1/*n*2, a good approximation is possible with a relatively few number of terms; a 10-harmonic truncation should be just fine. Let us use MATLAB to synthesize *x*(*t*) using a 10-harmonic truncated Fourier series.
-
-<sup>†</sup> Differentiation also destroys the dc component of the signal, providing further justification as to why the dc component needs to be separately computed.
-
-<span id="page-656-0"></span>To begin, we define *A*, *Dn*, *T*0, ω0, and a time vector that spans two periods of the waveform.
-
-```
->> A = 1; D = @(n) -A*4j*sin(n*pi/2)./(n.^2*pi^2);
->> T0 = 2; omega0 = 2*pi/T0; t = (-T0:.001:T0);
-```
-
-Next, we set the dc portion of the signal.
-
->> D0 = 0; x10 = D0\*ones(size(t));
-
-To add the desired 10 harmonics, we enter a loop for 1 ≤ *n* ≤ 10 and add in the *Dn* and *D*−*<sup>n</sup>* terms. Although *x*(*t*) should be real, small round-off errors cause the reconstruction to be complex. These small imaginary parts are removed using the real command.
-
-```
->> for n = 1:10,
->> x10 = x10+real(D(n)*exp(1j*omega0*n*t)+D(-n)*exp(-1j*omega0*n*t));
->> end
-```
-
-Lastly, we plot the resulting truncated Fourier series synthesis of *x*(*t*).
-
->> plot(t,x10,'k'); xlabel('t'); ylabel('x\_{10}(t)');
-
-Since the synthesized waveform shown in Fig. 6.19 closely matches the original waveform in Fig. 6.4, we have high confidence that the computed *Dn* are correct.

engineering/linear-systems-and-signals-3rd-edition-bp-lathi/082_6.4 LTIC SYSTEM RESPONSE TO PERIODIC INPUTS.md

@@ -1,120 +0,0 @@
-## **[6.4 LTIC SYSTEM](#page-12-0) RESPONSE TO PERIODIC INPUTS**
-
-A periodic signal can be expressed as a sum of everlasting exponentials (or sinusoids). We also know how to find the response of an LTIC system to an everlasting exponential. From this information, we can readily determine the response of an LTIC system to periodic inputs. A periodic signal *x*(*t*) with period *T*<sup>0</sup> can be expressed as an exponential Fourier series
-
-$$
-x(t) = \sum_{n = -\infty}^{\infty} D_n e^{jn\omega_0 t} \qquad \omega_0 = \frac{2\pi}{T_0}
-$$
-
-### 638 CHAPTER 6 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES
-
-In Sec. 4.8, we showed that the response of an LTIC system with transfer function *H*(*s*) to an everlasting exponential input *ej*ω*<sup>t</sup>* is an everlasting exponential *H*(*j*ω)*ej*ω*<sup>t</sup>* . This input–output pair can be displayed as†
-
-$$
-\underbrace{e^{j\omega t}}_{\text{input}} \Longrightarrow \underbrace{H(j\omega)e^{j\omega t}}_{\text{output}}
-$$
-
-Therefore, from the linearity property,
-
-$$
-\underbrace{\sum_{n=-\infty}^{\infty} D_n e^{jn\omega_0 t}}_{\text{input } x(t)} \Longrightarrow \underbrace{\sum_{n=-\infty}^{\infty} D_n H(jn\omega_0) e^{jn\omega_0 t}}_{\text{response } y(t)}
-$$
-(6.29)
-
-The response *y*(*t*) is obtained in the form of an exponential Fourier series and is therefore a periodic signal of the same period as that of the input.
-
-We shall demonstrate the utility of these results by the following example.
-
-### **EXAMPLE 6.12 Full-Wave Rectifier**
-
-A full-wave rectifier (Fig. 6.20a) is used to obtain a dc signal from a sinusoid sin *t*. The rectified signal *x*(*t*), depicted in Fig. 6.17, is applied to the input of a lowpass *RC* filter, which suppresses the time-varying component and yields a dc component with some residual ripple. Find the filter output *y*(*t*). Find also the dc output and the rms value of the ripple voltage.
-
-First, we shall find the Fourier series for the rectified signal *x*(*t*), whose period is *T*<sup>0</sup> = π. Consequently, ω<sup>0</sup> = 2, and
-
-$$
-x(t) = \sum_{n = -\infty}^{\infty} D_n e^{j2nt}
-$$
-
-where
-
-$$
-D_n = \frac{1}{\pi} \int_0^{\pi} \sin t e^{-j2nt} dt = \frac{2}{\pi (1 - 4n^2)}
-$$
-(6.30)
-
-Therefore,
-
-$$
-x(t) = \sum_{n = -\infty}^{\infty} \frac{2}{\pi (1 - 4n^2)} e^{j2nt}
-$$
-
-<sup>†</sup> This result applies only to asymptotically stable systems. This is because when *<sup>s</sup>* <sup>=</sup> *<sup>j</sup>*ω, the integral on the right-hand side of Eq. (2.39) does not converge for unstable systems. Moreover, for marginally stable systems also, that integral does not converge in the ordinary sense, and *H*(*j*ω) cannot be obtained from *H*(*s*) by replacing *s* with *j*ω.
-
-**Figure 6.20 (a)** Full-wave rectifier with a lowpass filter and **(b)** its output.
-
-Next, we find the transfer function of the *RC* filter in Fig. 6.20a. This filter is identical to the *RC* circuit in Ex. 1.17 (Fig. 1.35) for which the differential equation relating the output (capacitor voltage) to the input *x*(*t*) was found to be [Eq. (1.31)]:
-
-$$
-(3D+1)y(t) = x(t)
-$$
-
-The transfer function *H*(*s*) for this system is found from Eq. (2.41) as
-
-$$
-H(s) = \frac{1}{3s+1}
-$$
-
-and
-
-$$
-H(j\omega) = \frac{1}{3j\omega + 1} \tag{6.31}
-$$
-
-From Eq. (6.29), the filter output *y*(*t*) can be expressed as (with ω<sup>0</sup> = 2)
-
-$$
-y(t) = \sum_{n = -\infty}^{\infty} D_n H(jn\omega_0) e^{jn\omega_0 t} = \sum_{n = -\infty}^{\infty} D_n H(j2n) e^{j2nt}
-$$
-
-Substituting *Dn* and *H*(*j*2*n*) from Eqs. (6.30) and (6.31) in the foregoing equation, we obtain
-
-$$
-y(t) = \sum_{n = -\infty}^{\infty} \frac{2}{\pi (1 - 4n^2)(j6n + 1)} e^{j2nt}
-$$
-
-#### 640 CHAPTER 6 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES
-
-Note that the output *y*(*t*) is also a periodic signal given by the exponential Fourier series on the right-hand side. The output is shown in Fig. 6.20b.
-
-The output Fourier series coefficient corresponding to *n* = 0 is the dc component of the output, given by 2/π. The remaining terms in the Fourier series constitute the unwanted component called the ripple. We can determine the rms value of the ripple voltage by using Eq. (6.27) to find the power of the ripple component. The power of the ripple is the power of all the components except the dc (*n* = 0). Note that *D*ˆ *<sup>n</sup>*, the exponential Fourier coefficient for the output *y*(*t*), is
-
-$$
-\hat{D}_n = \frac{2}{\pi (1 - 4n^2)(j6n + 1)}
-$$
-
-Therefore, from Eq. (6.28), we have
-
-$$
-P_{\text{right}} = 2 \sum_{n=1}^{\infty} |D_n|^2 = 2 \sum_{n=1}^{\infty} \left| \frac{2}{\pi (1 - 4n^2)(j6n + 1)} \right|^2 = \frac{8}{\pi^2} \sum_{n=1}^{\infty} \frac{1}{(1 - 4n^2)^2 (36n^2 + 1)}
-$$
-
-Numerical computation of the right-hand side yields *P*ripple = 0.0025, and the ripple rms value <sup>=</sup> *P*ripple <sup>=</sup> 0.05. This shows that the rms ripple voltage is 5% of the amplitude of the input sinusoid.
-
-### WHY USE EXPONENTIALS?
-
-The exponential Fourier series is just another way of representing trigonometric Fourier series (or vice versa). The two forms carry identical information—no more, no less. The reasons for preferring the exponential form have already been mentioned: this form is more compact, and the expression for deriving the exponential coefficients is also more compact than those in the trigonometric series. Furthermore, the LTIC system response to exponential signals is also simpler (more compact) than the system response to sinusoids. In addition, the exponential form proves to be much easier than the trigonometric form to manipulate mathematically and otherwise handle in the area of signals as well as systems. Moreover, exponential representation proves much more convenient for analysis of complex *x*(*t*). For these reasons, in our future discussion we shall use the exponential form exclusively.
-
-A minor disadvantage of the exponential form is that it cannot be visualized as easily as sinusoids. For intuitive and qualitative understanding, the sinusoids have the edge over exponentials. Fortunately, this difficulty can be overcome readily because of the close connection between exponential and Fourier spectra. For the purpose of mathematical analysis, we shall continue to use exponential signals and spectra; but to understand the physical situation intuitively or qualitatively, we shall speak in terms of sinusoids and trigonometric spectra. Thus, although all mathematical manipulation will be in terms of exponential spectra, we shall now speak of exponential and sinusoids interchangeably when we discuss intuitive and qualitative insights in attempting to arrive at an understanding of physical situations. This is an important point; readers should make an extra effort to familiarize themselves with the two forms of spectra, their relationships, and their convertibility.
-
-### <span id="page-660-0"></span>DUAL PERSONALITY OF A SIGNAL
-
-The discussion so far shows that a periodic signal has a dual personality—the time domain and the frequency domain. It can be described by its waveform or by its Fourier spectra. The timeand frequency-domain descriptions provide complementary insights into a signal. For in-depth perspective, we need to understand both these identities. It is important to learn to think of a signal from both perspectives. In the next chapter, we shall see that aperiodic signals also have this dual personality. Moreover, we shall show that even LTI systems have this dual personality, which offers complementary insights into the system behavior.
-
-### LIMITATIONS OF THE FOURIER SERIES METHOD OF ANALYSIS
-
-We have developed here a method of representing a periodic signal as a weighted sum of everlasting exponentials whose frequencies lie along the ω axis in the *s* plane. This representation (Fourier series) is valuable in many applications. However, as a tool for analyzing linear systems, it has serious limitations and consequently has limited utility for the following reasons:
-
-- 1. The Fourier series can be used only for periodic inputs. All practical inputs are aperiodic (remember that a periodic signal starts at *t* = −∞).
-- 2. The Fourier methods can be applied readily to BIBO-stable (or asymptotically stable) systems. It cannot handle unstable or even marginally stable systems.
-
-The first limitation can be overcome by representing aperiodic signals in terms of everlasting exponentials. This representation can be achieved through the Fourier integral, which may be considered to be an extension of the Fourier series. We shall therefore use the Fourier series as a stepping-stone to the Fourier integral developed in the next chapter. The second limitation can be overcome by using exponentials *est*, where *s* is not restricted to the imaginary axis but is free to take on complex values. This generalization leads to the Laplace integral, discussed in Ch. 4 (the Laplace transform).

engineering/linear-systems-and-signals-3rd-edition-bp-lathi/083_6.5 GENERALIZED FOURIER SERIES - SIGNALS AS VECTORS.md

@@ -1,707 +0,0 @@
-## **[6.5 GENERALIZED](#page-12-0) FOURIER SERIES: SIGNALS AS VECTORS**
-
-We now consider a very general approach to signal representation with far-reaching consequences.† There is a perfect analogy between signals and vectors; the analogy is so strong that the term *analogy* understates the reality. Signals are not just *like* vectors. Signals *are* vectors! A vector can be represented as a sum of its components in a variety of ways, depending on the choice of coordinate system. A signal can also be represented as a sum of its components in a variety of ways. Let us begin with some basic vector concepts and then apply these concepts to signals.
-
-<sup>†</sup> This section closely follows the material from the author's earlier book [10]. Omission of this section will not cause any discontinuity in understanding the rest of the book. Derivation of Fourier series through the signal-vector analogy provides an interesting insight into signal representation and other topics such as signal correlation, data truncation, and signal detection.
-
-### <span id="page-661-0"></span>**[6.5-1 Component of a Vector](#page-12-0)**
-
-A vector is specified by its magnitude and its direction. We shall denote all vectors by boldface. For example, **x** is a certain vector with magnitude or length |**x**|. For the two vectors **x** and **y** shown in Fig. 6.21, we define their dot (inner or scalar) product as
-
-$$
-\mathbf{x} \cdot \mathbf{y} = |\mathbf{x}| |\mathbf{y}| \cos \theta
-$$
-
-where θ is the angle between these vectors. Using this definition, we can express |**x**|, the length of a vector **x**, as
-
-$$
-|\mathbf{x}|^2 = \mathbf{x} \cdot \mathbf{x}
-$$
-
-Let the component of **x** along **y** be *c***y** as depicted in Fig. 6.21. Geometrically, the component of **x** along **y** is the projection of **x** on **y** and is obtained by drawing a perpendicular from the tip of **x** on the vector **y**, as illustrated in Fig. 6.21. What is the mathematical significance of a component of a vector along another vector? As seen from Fig. 6.21, the vector **x** can be expressed in terms of vector **y** as
-
-$$
-\mathbf{x} = c\mathbf{y} + \mathbf{e}
-$$
-
-However, this is not the only way to express **x** in terms of **y**. From Fig. 6.22, which shows two of the infinite other possibilities, we have
-
-$$
-\mathbf{x} = c_1 \mathbf{y} + \mathbf{e}_1 = c_2 \mathbf{y} + \mathbf{e}_2
-$$
-
-In each of these three representations, **x** is represented in terms of **y** plus another vector called the *error vector*. If we approximate **x** by *c***y**,
-
-$$
-\mathbf{x} \simeq c\mathbf{y}
-$$
-
-the error in the approximation is the vector **e** = **x** − *c***y**. Similarly, the errors in approximations in these drawings are **e**<sup>1</sup> (Fig. 6.22a) and **e**<sup>2</sup> (Fig. 6.22b). What is unique about the approximation in Fig. 6.21 is that the error vector is the smallest. We can now define mathematically the component of a vector **x** along vector **y** to be *c***y** where *c* is chosen to minimize the length of the error vector **e** = **x** − *c***y**. Now, the length of the component of **x** along **y** is |**x**| cos θ. But it is also *c*|**y**|, as seen from Fig. 6.21. Therefore,
-
-$$
-c|\mathbf{y}| = |\mathbf{x}| \cos \theta
-$$
-
-Multiplying both sides by |**y**| yields
-
-$$
-c|\mathbf{y}|^2 = |\mathbf{x}||\mathbf{y}|\cos\theta = \mathbf{x}\cdot\mathbf{y}
-$$
-
-**Figure 6.21** Component (projection) of a vector along another vector.
-
-<span id="page-662-0"></span>**Figure 6.22** Approximation of a vector in terms of another vector.
-
-Therefore,
-
-$$
-c = \frac{\mathbf{x} \cdot \mathbf{y}}{\mathbf{y} \cdot \mathbf{y}} = \frac{1}{|\mathbf{y}|^2} \mathbf{x} \cdot \mathbf{y}
-$$
- (6.32)
-
-From Fig. 6.21, it is apparent that when **x** and **y** are perpendicular, or orthogonal, then **x** has a zero component along **y**; consequently, *c* = 0. Keeping an eye on Eq. (6.32), we therefore define **x** and **y** to be *orthogonal* if the inner (scalar or dot) product of the two vectors is zero, that is, if
-
-**x** · **y** = 0
-
-### **[6.5-2 Signal Comparison and Component of a Signal](#page-12-0)**
-
-The concept of a vector component and orthogonality can be extended to signals. Consider the problem of approximating a real signal *x*(*t*) in terms of another real signal *y*(*t*) over an interval (*t*1, *t*2):
-
-$$
-x(t) \simeq cy(t) \qquad t_1 < t < t_2
-$$
-
-The error *e*(*t*) in this approximation is
-
-$$
-e(t) = \begin{cases} x(t) - cy(t) & t_1 < t < t_2 \\ 0 & \text{otherwise} \end{cases}
-$$
-
-We now select a criterion for the "best approximation." We know that the signal energy is one possible measure of a signal size. For best approximation, we shall use the criterion that minimizes the size or energy of the error signal *e*(*t*) over the interval (*t*1,*t*2). This energy *Ee* is given by
-
-$$
-E_e = \int_{t_1}^{t_2} e^2(t) dt = \int_{t_1}^{t_2} [x(t) - cy(t)]^2 dt
-$$
-
-Note that the right-hand side is a definite integral with *t* as the dummy variable. Hence, *Ee* is a function of the parameter *c* (not *t*) and *Ee* is minimum for some choice of *c*. To minimize *Ee*, a necessary condition is
-
-$$
-\frac{dE_e}{dc} = 0
-$$
-$$
-\frac{d}{dc} \left[ \int_{t_1}^{t_2} [x(t) - cy(t)]^2 dt \right] = 0
-$$
-
-*d*
-
-or
-
-#### 644 CHAPTER 6 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES
-
-Expanding the squared term inside the integral, we obtain
-
-$$
-\frac{d}{dc} \left[ \int_{t_1}^{t_2} x^2(t) dt \right] - \frac{d}{dc} \left[ 2c \int_{t_1}^{t_2} x(t) y(t) dt \right] + \frac{d}{dc} \left[ c^2 \int_{t_1}^{t_2} y^2(t) dt \right] = 0
-$$
-
-from which we get
-
-$$
--2\int_{t_1}^{t_2} x(t)y(t) dt + 2c \int_{t_1}^{t_2} y^2(t) dt = 0
-$$
-  
-$$
-c = \frac{\int_{t_1}^{t_2} x(t)y(t) dt}{\int_{t_1}^{t_2} y^2(t) dt} = \frac{1}{E_y} \int_{t_1}^{t_2} x(t)y(t) dt
-$$
- (6.33)
-
-We observe a remarkable similarity between the behavior of vectors and signals, as indicated by Eqs. (6.32) and (6.33). It is evident from these two parallel expressions that *the area under the product of two signals corresponds to the inner (scalar or dot) product of two vectors*. In fact, the area under the product of *x*(*t*) and *y*(*t*) is called the *inner product* of *x*(*t*) and *y*(*t*), and is denoted by (*x*, *y*). The energy of a signal is the inner product of a signal with itself, and corresponds to the vector length square (which is the inner product of the vector with itself).
-
-To summarize our discussion, if a signal *x*(*t*) is approximated by another signal *y*(*t*) as
-
-*x*(*t*) *cy*(*t*)
-
-then the optimum value of *c* that minimizes the energy of the error signal in this approximation is given by Eq. (6.33).
-
-Taking our clue from vectors, we say that a signal *x*(*t*) contains a component *cy*(*t*), where *c* is given by Eq. (6.33). Note that in vector terminology, *cy*(*t*) is the projection of *x*(*t*) on *y*(*t*). Continuing with the analogy, we say that if the component of a signal *x*(*t*) of the form *y*(*t*) is zero (i.e., *c* = 0), the signals *x*(*t*) and *y*(*t*) are orthogonal over the interval (*t*1, *t*2). Therefore, we define the real signals *x*(*t*) and *y*(*t*) to be orthogonal over the interval (*t*1, *t*2) if†
-
-$$
-\int_{t_1}^{t_2} x(t)y(t) dt = 0
-$$
-\n(6.34)
-
-### **EXAMPLE 6.13 Sine-Wave Approximation of a Square Wave**
-
-For the square signal *x*(*t*) shown in Fig. 6.23, find the component in *x*(*t*) of the form sin*t*. In other words, approximate *x*(*t*) in terms of sin *t*
-
-*x*(*t*) *c*sin*t* 0 < *t* < 2π
-
-so that the energy of the error signal is minimum.
-
-<sup>†</sup> For complex signals, the definition is modified as in Eq. (6.37), in Sec. 6.5-3.
-
-<span id="page-664-0"></span>**Figure 6.23** Approximation of a square wave in terms of a single sinusoid.
-
-In this case,
-
-$$
-y(t) = \sin t
-$$
- and  $E_y = \int_0^{2\pi} \sin^2(t) dt = \pi$ 
-
-From Eq. (6.33), we find
-
-$$
-c = \frac{1}{\pi} \int_0^{2\pi} x(t) \sin t \, dt = \frac{1}{\pi} \left[ \int_0^{\pi} \sin t \, dt + \int_{\pi}^{2\pi} -\sin t \, dt \right] = \frac{4}{\pi}
-$$
-
-Thus,
-
-$$
-x(t) \simeq \frac{4}{\pi} \sin t
-$$
-
-represents the best approximation of *x*(*t*) by the function sin*t*, which will minimize the error energy. This sinusoidal component of *x*(*t*) is shaded in Fig. 6.23. By analogy with vectors, we say that the square function *x*(*t*) depicted in Fig. 6.23 has a component of signal sin*t* and that the magnitude of this component is 4/π.
-
-### **DR ILL 6.7 Sine Wave Approximation of a Ramp Function**
-
-Show that over an interval (−π < *t* < π), the "best" approximation of the signal *x*(*t*) = *t* in terms of the function sin*t* is 2 sin*t*. Verify that the error signal *e*(*t*) = *t* − 2 sin*t* is orthogonal to the signal sin*t* over the interval −π < *t* < π. Sketch the signals *t* and 2 sin*t* over the interval −π < *t* < π.
-
-### **[6.5-3 Extension to Complex Signals](#page-12-0)**
-
-So far we have restricted ourselves to real functions of *t*. To generalize the results to complex functions of *t*, consider again the problem of approximating a signal *x*(*t*) by a signal *y*(*t*) over an
-
-#### 646 CHAPTER 6 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES
-
-interval (*t*<sup>1</sup> < *t* < *t*2):
-
-$$
-x(t) \simeq cy(t)
-$$
-
-where *x*(*t*) and *y*(*t*) now can be complex functions of *t*. Recall that the energy *Ey* of the complex signal *y*(*t*) over an interval (*t*1, *t*2) is
-
-$$
-E_{y} = \int_{t_1}^{t_2} |y(t)|^2 dt
-$$
-
-In this case, both the coefficient *c* and the error
-
-$$
-e(t) = x(t) - cy(t)
-$$
-
-are complex (in general). For the "best" approximation, we choose *c* to minimize the energy *Ee* of the error signal *e*(*t*). Now,
-
-$$
-E_e = \int_{t_1}^{t_2} |x(t) - cy(t)|^2 dt
-$$
-\n(6.35)
-
-Recall also that
-
-$$
-|u + v|^2 = (u + v)(u^* + v^*) = |u|^2 + |v|^2 + u^*v + uv^* \tag{6.36}
-$$
-
-After some manipulation, we can use this result to rearrange Eq. (6.35) as
-
-$$
-E_e = \int_{t_1}^{t_2} |x(t)|^2 dt - \left| \frac{1}{\sqrt{E_y}} \int_a^{t_2} x(t) y^*(t) dt \right|^2 + \left| c \sqrt{E_y} - \frac{1}{\sqrt{E_y}} \int_{t_1}^{t_2} x(t) y^*(t) dt \right|^2
-$$
-
-Since the first two terms on the right-hand side are independent of *c*, it is clear that *Ee* is minimized by choosing *c* so that the third term on the right-hand side is zero. This yields
-
-$$
-c = \frac{1}{E_y} \int_{t_1}^{t_2} x(t) y^*(t) dt
-$$
-
-In light of this result, we need to redefine orthogonality for the complex case as follows: two complex functions *x*1(*t*) and *x*2(*t*) are orthogonal over an interval (*t*<sup>1</sup> < *t* < *t*2) if
-
-$$
-\int_{t_1}^{t_2} x_1(t) x_2^*(t) dt = 0 \qquad \text{or} \qquad \int_{t_1}^{t_2} x_1^*(t) x_2(t) dt = 0 \tag{6.37}
-$$
-
-Either equality suffices. This is a general definition of orthogonality, which reduces to Eq. (6.34) when the functions are real.
-
-### **DR ILL 6.8 Complex Exponential Approximation of a Square Wave**
-
-Show that over an interval (0 < *t* < 2π ), the "best" approximation of the square signal *x*(*t*) in Fig. 6.23 in terms of the signal *ejt* is given by (2/*j*π ) *ejt*. Verify that the error signal *e*(*t*) = *x*(*t*)−(2/*j*π )*ejt* is orthogonal to the signal *ejt*.
-
-### <span id="page-666-0"></span>ENERGY OF THE SUM OF ORTHOGONAL SIGNALS
-
-We know that the square of the length of a sum of two orthogonal vectors is equal to the sum of the squares of the lengths of the two vectors. Thus, if vectors **x** and **y** are orthogonal, and if **z** = **x**+**y**, then
-
-$$
-|\mathbf{z}|^2 = |\mathbf{x}|^2 + |\mathbf{y}|^2
-$$
-
-We have a similar result for signals. The energy of the sum of two orthogonal signals is equal to the sum of the energies of the two signals. Thus, if signals *x*(*t*) and *y*(*t*) are orthogonal over an interval (*t*1, *t*2), and if *z*(*t*) = *x*(*t*)+*y*(*t*), then
-
-$$
-E_z = E_x + E_y
-$$
-
-We now prove this result for complex signals, of which real signals are a special case. From Eq. (6.36), it follows that
-
-$$
-\int_{t_1}^{t_2} |x(t) + y(t)|^2 dt = \int_{t_1}^{t_2} |x(t)|^2 dt + \int_{t_1}^{t_2} |y(t)|^2 dt + \int_{t_1}^{t_2} x(t)y^*(t) dt + \int_{t_1}^{t_2} x^*(t)y(t) dt
-$$
-  
-= 
-$$
-\int_{t_1}^{t_2} |x(t)|^2 dt + \int_{t_1}^{t_2} |y(t)|^2 dt
-$$
-
-The last result follows from the fact that because of orthogonality, the two integrals of the products *x*(*t*)*y*∗(*t*) and *x*∗(*t*)*y*(*t*) are zero [see Eq. (6.37)]. This result can be extended to the sum of any number of mutually orthogonal signals.
-
-### **[6.5-4 Signal Representation by an Orthogonal Signal Set](#page-12-0)**
-
-In this section we show a way of representing a signal as a sum of orthogonal signals. Here again we can benefit from the insight gained from a similar problem in vectors. We know that a vector can be represented as a sum of orthogonal vectors, which form the coordinate system of a vector space. The problem in signals is analogous, and the results for signals are parallel to those for vectors. So, let us review the case of vector representation.
-
-### ORTHOGONAL VECTOR SPACE
-
-Let us investigate a three-dimensional Cartesian vector space described by three mutually orthogonal vectors **x**1, **x**2, and **x**3, as illustrated in Fig. 6.24. First, we shall seek to approximate a three-dimensional vector **x** in terms of two mutually orthogonal vectors **x**<sup>1</sup> and **x**2:
-
-$$
-\mathbf{x} \simeq c_1 \mathbf{x}_1 + c_2 \mathbf{x}_2
-$$
-
-The error **e** in this approximation is
-
-$$
-\mathbf{e} = \mathbf{x} - (c_1 \mathbf{x}_1 + c_2 \mathbf{x}_2)
-$$
-
-or
-
-$$
-\mathbf{x} = c_1 \mathbf{x}_1 + c_2 \mathbf{x}_2 + \mathbf{e}
-$$
-
-**Figure 6.24** Representation of a vector in three-dimensional space.
-
-As in the earlier geometrical argument, we see from Fig. 6.24 that the length of **e** is minimum when **e** is perpendicular to the **x**1–**x**<sup>2</sup> plane, and *c*1**x**<sup>1</sup> and *c*2**x**<sup>2</sup> are the projections (components) of **x** on **x**<sup>1</sup> and **x**2, respectively. Therefore, the constants *c*<sup>1</sup> and *c*<sup>2</sup> are given by Eq. (6.32). Observe that the error vector is orthogonal to both the vectors **x**<sup>1</sup> and **x**2.
-
-Now, let us determine the "best" approximation to **x** in terms of all three mutually orthogonal vectors **x**1, **x**2, and **x**3:
-
-$$
-\mathbf{x} \simeq c_1 \mathbf{x}_1 + c_2 \mathbf{x}_2 + c_3 \mathbf{x}_3 \tag{6.38}
-$$
-
-Figure 6.24 shows that a unique choice of *c*1, *c*2, and *c*<sup>3</sup> exists, for which Eq. (6.38) is no longer an approximation but an equality
-
-$$
-\mathbf{x} = c_1 \mathbf{x}_1 + c_2 \mathbf{x}_2 + c_3 \mathbf{x}_3
-$$
-
-In this case, *c*1**x**1, *c*2**x**2, and *c*3**x**<sup>3</sup> are the projections (components) of **x** on **x**1,**x**2, and **x**3, respectively; that is,
-
-$$
-c_i = \frac{\mathbf{x} \cdot \mathbf{x}_i}{\mathbf{x}_i \cdot \mathbf{x}_i} = \frac{1}{|\mathbf{x}_i|^2} \mathbf{x} \cdot \mathbf{x}_i \qquad i = 1, 2, 3
-$$
- (6.39)
-
-Note that the error in the approximation is zero when **x** is approximated in terms of three mutually orthogonal vectors: **x**1, **x**2, and **x**3. The reason is that **x** is a three-dimensional vector, and the vectors **x**1, **x**2, and **x**<sup>3</sup> represent a *complete set* of orthogonal vectors in three-dimensional space. Completeness here means that it is impossible to find another vector **x**<sup>4</sup> in this space, which is orthogonal to all three vectors, **x**1,**x**2, and **x**3. Any vector in this space can then be represented (with zero error) in terms of these three vectors. Such vectors are known as *basis* vectors. If a set of vectors {**x***i*} is not complete, the error in the approximation will generally not be zero. Thus, in the three-dimensional case discussed earlier, it is generally not possible to represent a vector **x** in terms of only two basis vectors without an error.
-
-The choice of basis vectors is not unique. In fact, a set of basis vectors corresponds to a particular choice of coordinate system. Thus, a three-dimensional vector **x** may be represented in many different ways, depending on the coordinate system used.
-
-### ORTHOGONAL SIGNAL SPACE
-
-We start with real signals and then extend the discussion to complex signals. We proceed with our signal approximation problem, using clues and insights developed for vector approximation. As before, we define orthogonality of a real signal set *x*1(*t*), *x*2(*t*), ..., *xN*(*t*) over interval (*t*1,*t*2) as
-
-$$
-\int_{t_1}^{t_2} x_m(t) x_n(t) dt = \begin{cases} 0 & m \neq n \\ E_n & m = n \end{cases}
-$$
- (6.40)
-
-If the energies *En* = 1 for all *n*, then the set is *normalized* and is called an *orthonormal set*. An orthogonal set can always be normalized by dividing *xn*(*t*) by <sup>√</sup>*En* for all *<sup>n</sup>*.
-
-Now, consider approximating a signal *x*(*t*) over the interval (*t*1, *t*2) by a set of *N* real, mutually orthogonal signals *x*1(*t*), *x*2(*t*),..., *xN*(*t*) as
-
-$$
-x(t) \simeq c_1 x_1(t) + c_2 x_2(t) + \dots + c_N x_N(t) \simeq \sum_{n=1}^N c_n x_n(t)
-$$
- (6.41)
-
-In the approximation of Eq. (6.41), the error *e*(*t*) is
-
-$$
-e(t) = x(t) - \sum_{n=1}^{N} c_n x_n(t)
-$$
-
-and *Ee*, the error signal energy, is
-
-$$
-E_e = \int_{t_1}^{t_2} e^2(t) dt = \int_{t_1}^{t_2} \left[ x(t) - \sum_{n=1}^{N} c_n x_n(t) \right]^2 dt
-$$
- (6.42)
-
-According to our criterion for best approximation, we select the values of *ci* that minimize *Ee*. Hence, the necessary condition is ∂*Ee*/*dci* = 0 for *i* = 1, 2,...,*N*, that is,
-
-$$
-\frac{\partial}{\partial c_i} \int_{t_1}^{t_2} \left[ x(t) - \sum_{n=1}^N c_n x_n(t) \right]^2 dt = 0
-$$
-
-When we expand the integrand, we find that all the cross-multiplication terms arising from the orthogonal signals are zero by virtue of orthogonality: that is, all terms of the form \$ *xm*(*t*)*xn*(*t*)*dt* with *m* = *n* vanish. Similarly, the derivative with respect to *ci* of all terms that do not contain *ci* is zero. For each *i*, this leaves only two nonzero terms:
-
-$$
-\frac{\partial}{\partial c_i} \int_{t_1}^{t_2} \left[ -2c_i x(t) x_i(t) + c_i^2 x_i^2(t) \right] dt = 0
-$$
-
-or
-
-$$
--2\int_{t_1}^{t_2} x(t)x_i(t) dt + 2c_i \int_{t_1}^{t_2} x_i^2(t) dt = 0 \qquad i = 1, 2, \dots, N
-$$
-
-Therefore,
-
-$$
-c_i = \frac{\int_{t_1}^{t_2} x(t) x_i(t) dt}{\int_{t_1}^{t_2} x_i^2(t) dt} = \frac{1}{E_i} \int_{t_1}^{t_2} x(t) x_i(t) dt \qquad i = 1, 2, ..., N
-$$
- (6.43)
-
-A comparison of Eq. (6.43) with Eq. (6.39) forcefully brings out the analogy of signals with vectors.
-
-**Finality Property.** Equation (6.43) shows one interesting property of the coefficients of *c*1, *c*2, ..., *cN*: the optimum value of any coefficient in Eq. (6.41) is independent of the number of terms used in the approximation. For example, if we used only one term (*N* = 1) or two terms (*N* = 2) or any number of terms, the optimum value of the coefficient *c*<sup>1</sup> would be the same [as given by Eq. (6.43)]. The advantage of this approximation of a signal *x*(*t*) by a set of mutually orthogonal signals is that we can continue to add terms to the approximation without disturbing the previous terms. This property of *finality* of the values of the coefficients is very important from a practical point of view.†
-
-### ENERGY OF THE ERROR SIGNAL
-
-When the coefficients *ci* in Eq. (6.41) are chosen according to Eq. (6.43), the error signal energy is minimized. This minimum value of *Ee* is given by Eq. (6.42):
-
-$$
-E_e = \int_{t_1}^{t_2} \left[ x(t) - \sum_{n=1}^{N} c_n x_n(t) \right]^2 dt
-$$
-  
-= 
-$$
-\int_{t_1}^{t_2} x^2(t) dt + \sum_{n=1}^{N} c_n^2 \int_{t_1}^{t_2} x_n^2(t) dt - 2 \sum_{n=1}^{N} c_n \int_{t_1}^{t_2} x(t) x_n(t) dt
-$$
-
-$$
-x(t_1) = a_0 + a_1t_1
-$$
- and  $x(t_2) = a_0 + a_1t_2$ 
-
-Solution of these equations yields the desired values of *a*<sup>0</sup> and *a*1. For a three-point approximation, we must choose the polynomial *a*<sup>0</sup> +*a*1*t* +*a*2*t* <sup>2</sup> with
-
-$$
-x(t_i) = a_0 + a_1t_i + a_2t_i^2
-$$
-  $i = 1, 2, and 3$ 
-
-The approximation improves with a larger number of points (higher-order polynomial), but the coefficients *a*0, *a*1, *a*2, ... do not have the finality property. Every time we increase the number of terms in the polynomial, we need to recalculate the coefficients.
-
-<sup>†</sup> Contrast this situation with a polynomial approximation of *x*(*t*). Suppose we wish to find a two-point approximation of *x*(*t*) by a polynomial in *t*; that is, the polynomial is to be equal to *x*(*t*) at two points *t*<sup>1</sup> and *t*2. This can be done by choosing a first-order polynomial *a*<sup>0</sup> +*a*1*t* with
-
-Substitution of Eqs. (6.40) and (6.43) in this equation yields
-
-$$
-E_e = \int_{t_1}^{t_2} x^2(t) dt + \sum_{n=1}^{N} c_n^2 E_n - 2 \sum_{n=1}^{N} c_n^2 E_n = \int_{t_1}^{t_2} x^2(t) dt - \sum_{n=1}^{N} c_n^2 E_n
-$$
- (6.44)
-
-Observe that because the term *c*<sup>2</sup> *<sup>k</sup>Ek* is nonnegative, the error energy *Ee* generally decreases as *N*, the number of terms, is increased. Hence, it is possible that the error energy →0 as *N* → ∞. When this happens, the orthogonal signal set is said to be *complete*. In this case, Eq. (6.41) is no more an approximation but an equality
-
-$$
-x(t) = c_1 x_1(t) + c_2 x_2(t) + \dots + c_n x_n(t) + \dots = \sum_{n=1}^{\infty} c_n x_n(t) \qquad t_1 < t < t_2 \tag{6.45}
-$$
-
-where the coefficients *cn* are given by Eq. (6.43). Because the error signal energy approaches zero, it follows that the energy of *x*(*t*) is now equal to the sum of the energies of its orthogonal components *c*1*x*1(*t*), *c*2*x*2(*t*), *c*3*x*3(*t*), ....
-
-The series on the right-hand side of Eq. (6.45) is called the *generalized Fourier series* of *x*(*t*) with respect to the set {*xn*(*t*)}. When the set {*xn*(*t*)} is such that the error energy *Ee* → 0 as *N* → ∞ for every member of some particular class, we say that the set {*xn*(*t*)} is complete on (*t*1, *t*2) for that class of *x*(*t*), and the set {*xn*(*t*)} is called a set of *basis functions* or *basis signals*. Unless otherwise mentioned, in the future we shall consider only the class of energy signals.
-
-Thus, when the set {*xn*(*t*)} is complete, we have the equality of Eq. (6.45). One subtle point that must be understood clearly is the meaning of equality in Eq. (6.45). *The equality here is not an equality in the ordinary sense, but in the sense that the error energy, that is, the energy of the difference between the two sides of Eq. (6.45), approaches zero*. If the equality exists in the ordinary sense, the error energy is always zero, but the converse is not necessarily true. The error energy can approach zero even though *e*(*t*), the difference between the two sides, is nonzero at some isolated instants. The reason is that even if *e*(*t*) is nonzero at such instants, the area under *e*<sup>2</sup>(*t*) is still zero; thus the Fourier series on the right-hand side of Eq. (6.45) may differ from *x*(*t*) at a finite number of points.
-
-In Eq. (6.45), the energy of the left-hand side is *Ex*, and the energy of the right-hand side is the sum of the energies of all the orthogonal components.† Thus,
-
-$$
-\int_{t_1}^{t_2} x^2(t) dt = c_1^2 E_1 + c_2^2 E_2 + \dots = \sum_{n=1}^{\infty} c_n^2 E_n
-$$
- (6.46)
-
-This is *Parseval's theorem* expressed for energy signals. In Eqs. (6.26) and (6.27), we have already encountered Parseval's theorem for power signals. Recall that the signal energy (area under the squared value of a signal) is analogous to the square of the length of a vector in the vector-signal analogy. In vector space, we know that the square of the length of a vector is equal to the sum of
-
-<sup>†</sup> Note that the energy of a signal *cx*(*t*) is *c*<sup>2</sup>*Ex*.
-
-### 652 CHAPTER 6 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES
-
-the squares of the lengths of its orthogonal components. Parseval's theorem of Eq. (6.46) is the statement of this fact as it applies to signals.
-
-### GENERALIZATION TO COMPLEX SIGNALS
-
-The foregoing results can be generalized to complex signals as follows: a set of functions *x*1(*t*), *x*2(*t*), ..., *xN*(*t*) is mutually orthogonal over the interval (*t*1, *t*2) if
-
-$$
-\int_{t_1}^{t_2} x_m(t) x_n^*(t) dt = \begin{cases} 0 & m \neq n \\ E_n & m = n \end{cases}
-$$
-
-If this set is complete for a certain class of functions, then a function *x*(*t*) in this class can be expressed as
-
-$$
-x(t) = c_1 x_1(t) + c_2 x_2(t) + \cdots + c_i x_i(t) + \cdots
-$$
-
-where
-
-$$
-c_n = \frac{1}{E_n} \int_{t_1}^{t_2} x(t) x_n^*(t) dt
-$$
-\n(6.47)
-
-### **EXAMPLE 6.14 Approximating a Square Wave with a Set of Harmonic Sine Waves**
-
-In Ex. 6.13, the square signal *x*(*t*) in Fig. 6.23 is approximated by a single sinusoid sin *t*. In this example, we approximate *x*(*t*) using the set of harmonic sine waves sin *t*, sin 2*t*, ..., sin *nt*, ..., and see how the approximation improves with the number of terms.
-
-To begin, we note that the set of harmonic sine waves sin *t*, sin 2*t*,..., sin *nt*,... is orthogonal over any interval of duration 2π. † The reader can verify this fact by showing that for any real number *a*,
-
-$$
-\int_{a}^{a+2\pi} \sin mt \sin nt dt = \begin{cases} 0 & m \neq n \\ \pi & m = n \end{cases}
-$$
- (6.48)
-
-Using this set, we approximate *x*(*t*) as
-
-*x*(*t*) *c*<sup>1</sup> sin *t* +*c*<sup>2</sup> sin 2*t* +···+*cn* sin *Nt*
-
-<sup>†</sup> This sine set, along with the cosine set cos 0*t*, cos *t*, cos 2*t*,..., cos*nt*,..., forms a complete set. In this case, however, the coefficients *ci* corresponding to the cosine terms are zero. For this reason, we have omitted cosine terms in this example. This composite sine and cosine set is the basis set for the trigonometric Fourier series.
-
-**Figure 6.25** Approximation of a square wave by a sum of harmonic sinusoids.
-
-Therefore,
-
-$$
-x(t) \simeq \frac{4}{\pi} \left( \sin t + \frac{1}{3} \sin 3t + \frac{1}{5} \sin 5t + \dots + \frac{1}{N} \sin Nt \right)
-$$
- (6.49)
-
-Note that coefficients of terms sin*kt* are zero for even values of *k*. Figure 6.25 shows how the approximation improves as we increase the number of terms in the series.
-
-Let us investigate the error signal energy as *N* → ∞. From Eq. (6.44),
-
-$$
-E_e = \int_0^{2\pi} x^2(t) dt - \sum_{n=1}^{\infty} c_n^2 E_n
-$$
-
-Note that
-
-$$
-\int_0^{2\pi} x^2(t) dt = \int_0^{\pi} 1^2 dt + \int_{\pi}^{2\pi} -1^2 dt = 2\pi
-$$
-$$
-c_n^2 = \begin{cases} \frac{16}{n^2 \pi^2} & n \text{ odd} \\ 0 & n \text{ even} \end{cases}
-$$
-
-and from Eq. (6.48),
-
-$$
-E_n=\pi
-$$
-
-Therefore,
-
-$$
-E_e = 2\pi - \sum_{n=1,3,5,\dots}^{N} \frac{16}{n^2 \pi^2} \pi = 2\pi - \frac{16}{\pi} \sum_{n=1,3,5,\dots}^{N} \frac{1}{n^2}
-$$
-
-For a single-term approximation (*N* = 1),
-
-$$
-E_e = 2\pi - \frac{16}{\pi} = 1.1938
-$$
-
-For a two-term approximation (*N* = 3),
-
-$$
-E_e = 2\pi - \frac{16}{\pi} \left( 1 + \frac{1}{9} \right) = 0.6243
-$$
-
-Continuing this process, we compute the error energy *Ee* for various values of *N* as
-
-| N     | 1      | 3      | 5      | 7      | 99      | $\infty$ |
-|-------|--------|--------|--------|--------|---------|----------|
-| $E_e$ | 1.1938 | 0.6243 | 0.4206 | 0.3166 | 0.02545 | 0        |
-
-Clearly, *x*(*t*) can be represented by the infinite series
-
-$$
-x(t) = \frac{4}{\pi} \left( \sin t + \frac{1}{3} \sin 3t + \frac{1}{5} \sin 5t + \cdots \right) = \frac{4}{\pi} \sum_{n=1,3,5,\dots}^{\infty} \frac{1}{n} \sin nt
-$$
-
-The equality exists in the sense that the error signal energy → 0 as *N* → ∞. In this case, the error energy decreases rather slowly with *N*, indicating that the series converges slowly. This is to be expected because *x*(*t*) has jump discontinuities and consequently, according to discussion in Sec. 6.2-2, the series converges asymptotically as 1/*n*.
-
-### **DR ILL 6.9 Approximating a Ramp Signal with a Set of Harmonic Sine Waves**
-
-Approximate the signal *x*(*t*) = *t* − π (Fig. 6.26) over the interval (0, 2π ) in terms of the set of sinusoids {sin *nt*}, *n* = 0, 1, 2,..., used in Ex. 6.14. Find *Ee*, the error energy. Show that *Ee* → 0 as *N* → ∞.
-
-4π
-
-### **ANSWERS**
-
-%*<sup>N</sup>*
-
-1
-
-*x*(*t*) −2 *n*=1 *<sup>N</sup>* sin *nt* and *Ee* <sup>=</sup> <sup>2</sup> 3π<sup>3</sup> <sup>−</sup>%*<sup>N</sup> n*=1 *n*2 *x*(*t*) *p p p* 2*p t* **Figure 6.26** Ramp signal for Drill 6.9.
-
-### SOME EXAMPLES OF GENERALIZED FOURIER SERIES
-
-Signals are vectors in every sense. Like a vector, a signal can be represented as a sum of its components in a variety of ways. Just as vector coordinate systems are formed by mutually orthogonal vectors (rectangular, cylindrical, spherical), we also have signal coordinate systems (basis signals) formed by a variety of sets of mutually orthogonal signals. There exist a large number of orthogonal signal sets that can be used as basis signals for generalized Fourier series. Some well-known signal sets are trigonometric (sinusoid) functions, exponential functions, Walsh functions, Bessel functions, Legendre polynomials, Laguerre functions, Jacobi polynomials, Hermite polynomials, and Chebyshev polynomials. The functions that concern us most in this book are the trigonometric and the exponential sets discussed earlier in this chapter.
-
-### LEGENDRE FOURIER SERIES
-
-A set of Legendre polynomials *Pn*(*t*) (*n* = 0, 1, 2, 3,...) forms a complete set of mutually orthogonal functions over an interval (−1 < *t* < 1). These polynomials can be defined by the Rodrigues formula:
-
-$$
-P_n(t) = \frac{1}{2^n n!} \frac{d^n}{dt^n} (t^2 - 1)^n \qquad n = 0, 1, 2, \dots
-$$
-
-It follows from this equation that
-
-$$
-P_0(t) = 1, P_1(t) = t, P_2(t) = \left(\frac{3}{2}t^2 - \frac{1}{2}\right), P_3(t) = \left(\frac{5}{2}t^3 - \frac{3}{2}t\right)
-$$
-, and so on
-
-We may verify the orthogonality of these polynomials by showing that
-
-$$
-\int_{-1}^{1} P_m(t) P_n(t) dt = \begin{cases} 0 & m \neq n \\ \frac{2}{2m+1} & m = n \end{cases}
-$$
-
-We can express a function *x*(*t*) in terms of Legendre polynomials over an interval (−1 < *t* < 1) as
-
-$$
-x(t) = c_0 P_0(t) + c_1 P_1(t) + \dots + c_r P_r(t) + \dots
-$$
-\n(6.50)
-
-where
-
-$$
-c_r = \frac{\int_{-1}^{1} x(t)P_r(t) dt}{\int_{-1}^{1} P_r^2(t) dt} = \frac{2r+1}{2} \int_{-1}^{1} x(t)P_r(t) dt
-$$
-\n(6.51)
-
-Note that although the series representation is valid over the interval (−1, 1), it can be extended to any interval by the appropriate time scaling (see Prob. 6.5-8).
-
-### **EXAMPLE 6.15 Legendre Fourier Series**
-
-Determine the Legendre Fourier series of the square signal shown in Fig. 6.27.
-
-**Figure 6.27** Square signal for Ex. 6.15.
-
-From Eq. (6.50), we know that the Legendre Fourier series takes the form
-
-$$
-x(t) = c_0 P_0(t) + c_1 P_1(t) + \dots + c_r P_r(t) + \dots
-$$
-
-The coefficients *c*0, *c*1, *c*2,..., *cr* may be found from Eq. (6.51). We have
-
-$$
-x(t) = \begin{cases} 1 & \cdots -1 < t < 0 \\ -1 & \cdots 0 < t < 1 \end{cases}
-$$
-
-and
-
-$$
-c_0 = \frac{1}{2} \int_{-1}^{1} x(t) dt = 0
-$$
-  
-\n
-$$
-c_1 = \frac{3}{2} \int_{-1}^{1} tx(t) dt = \frac{3}{2} \left( \int_{-1}^{0} t dt - \int_{0}^{1} t dt \right) = -\frac{3}{2}
-$$
-  
-\n
-$$
-c_2 = \frac{5}{2} \int_{-1}^{1} x(t) \left( \frac{3}{2} t^2 - \frac{1}{2} \right) dt = 0
-$$
-
-This result follows immediately from the fact that the integrand is an odd function of *t*. In fact, this is true of all *cr* for even values of *r*, that is,
-
-$$
-c_0 = c_2 = c_4 = c_6 = \cdots = 0
-$$
-
-Also,
-
-$$
-c_3 = \frac{7}{2} \int_{-1}^{1} x(t) \left( \frac{5}{2} t^3 - \frac{3}{2} t \right) dt = \frac{7}{2} \left[ \int_{-1}^{0} \left( \frac{5}{2} t^3 - \frac{3}{2} t \right) dt - \int_{0}^{1} \left( \frac{5}{2} t^3 - \frac{3}{2} t \right) dt \right] = \frac{7}{8}
-$$
-
-In a similar way, coefficients *c*5, *c*7,... can be evaluated. We now have
-
-$$
-x(t) = -\frac{3}{2}t + \frac{7}{8}\left(\frac{5}{2}t^3 - \frac{3}{2}t\right) + \cdots
-$$
-
-### TRIGONOMETRIC FOURIER SERIES
-
-We have already proved [see Eqs. (6.4), (6.5), and (6.6)] that the trigonometric signal set
-
-{1, cosω0*t*, cos 2ω0*t*, ..., cos*n*ω0*t*, ...; sinω0*t*, sin 2ω0*t*, ..., sin*n*ω0*t*, ...}
-
-is orthogonal over any interval of duration *T*0, where *T*<sup>0</sup> = 1/*f*<sup>0</sup> is the period of the sinusoid of frequency *f*0. This is a complete set for a class of signals with finite energies [11, 12]. Therefore, we can express a signal *x*(*t*) by a trigonometric Fourier series over any interval of duration *T*<sup>0</sup> seconds as
-
-$$
-x(t) = a_0 + a_1 \cos \omega_0 t + a_2 \cos 2\omega_0 t + \cdots
-$$
-$$
-+ b_1 \sin \omega_0 t + b_2 \sin 2\omega_0 t + \cdots
-$$
-
-or
-
-$$
-x(t) = a_0 + \sum_{n=1}^{\infty} a_n \cos n\omega_0 t + b_n \sin n\omega_0 t \qquad t_1 < t < t_1 + T_0
-$$
-
-where
-
-$$
-\omega_0 = 2\pi f_0 = \frac{2\pi}{T_0}
-$$
-
-We can use Eq. (6.43) to determine the Fourier coefficients *a*0, *an*, and *bn*. Thus,
-
-$$
-a_n = \frac{\int_{t_1}^{t_1 + T_0} x(t) \cos n\omega_0 t \, dt}{\int_{t_1}^{t_1 + T_0} \cos^2 n\omega_0 t \, dt}
-$$
-\n(6.52)
-
-The integral in the denominator of Eq. (6.52) has already been found to be *T*0/2 when *n* = 0 [Eq. (6.4) with *m* = *n*]. For *n* = 0, the denominator is *T*0. Hence,
-
-$$
-a_0 = \frac{1}{T_0} \int_{t_1}^{t_1+T_0} x(t) dt \quad \text{and} \quad a_n = \frac{2}{T_0} \int_{t_1}^{t_1+T_0} x(t) \cos n\omega_0 t dt \qquad n = 1, 2, 3, \dots \tag{6.53}
-$$
-
-Similarly, we find that
-
-$$
-b_n = \frac{2}{T_0} \int_{t_1}^{t_1 + T_0} x(t) \sin n\omega_0 t \, dt \qquad n = 1, 2, 3, \dots \tag{6.54}
-$$
-
-Note that the Fourier series in Eq. (6.49) of Ex. 6.14 is indeed the trigonometric Fourier series with *T*<sup>0</sup> = 2π and ω<sup>0</sup> = 2π/*T*0. In this particular example, it is easy to verify from Eq. (6.53) that *an* = 0 for all *n*, including *n* = 0. Hence, the Fourier series in that example consisted only of sine terms.
-
-### EXPONENTIAL FOURIER SERIES
-
-As shown in the footnote on page 622, the set of exponentials *ejn*<sup>ω</sup>0*<sup>t</sup>* (*n* = 0,±1,±2,...) is a set of functions orthogonal over any interval of duration *T*<sup>0</sup> = 2π/ω0. An arbitrary signal *x*(*t*) can now be expressed over an interval (*t*1,*t*<sup>1</sup> +*T*0) as
-
-$$
-x(t) = \sum_{n = -\infty}^{\infty} D_n e^{jn\omega_0 t} \qquad t_1 < t < t_1 + T_0
-$$
-
-where [see Eq. (6.47)]
-
-$$
-D_n = \frac{1}{T_0} \int_{t_1}^{t_1+T_0} x(t) e^{-jn\omega_0 t} dt
-$$
-
-### WHY USE THE EXPONENTIAL SET?
-
-If *x*(*t*) can be represented in terms of hundreds of different orthogonal sets, why do we exclusively use the exponential (or trigonometric) set for the representation of signals or LTI systems? It so
-
-<span id="page-678-0"></span>happens that the exponential signal is an eigenfunction of LTI systems. In other words, for an LTI system, only an exponential input *est* yields the response that is also an exponential of the same form, given by *H*(*s*)*est*. The same is true of the trigonometric set. This fact makes the use of exponential signals natural for LTI systems in the sense that the system analysis using exponentials as the basis signals is greatly simplified.

engineering/linear-systems-and-signals-3rd-edition-bp-lathi/084_6.6 NUMERICAL COMPUTATION OF D_n.md

@@ -1,67 +0,0 @@
-## **6.6 NUMERICAL [COMPUTATION OF](#page-12-0)** *Dn*
-
-We can compute *Dn* numerically by using the DFT (the discrete Fourier transform discussed in Sec. 8.5), which uses the samples of a periodic signal *x*(*t*) over one period. The sampling interval is *T* seconds. Hence, there are *N*<sup>0</sup> =*T*0/*T* number of samples in one period *T*0. To find the relationship between *Dn* and the samples of *x*(*t*), consider Eq. (6.19) and write
-
-$$
-D_n = \frac{1}{T_0} \int_{T_0} x(t)e^{-jn\omega_0 t} dt
-$$
-  
-= 
-$$
-\lim_{T \to 0} \frac{1}{N_0 T} \sum_{k=0}^{N_0 - 1} x(kT)e^{-jn\omega_0 kT} T
-$$
-  
-= 
-$$
-\lim_{T \to 0} \frac{1}{N_0} \sum_{k=0}^{N_0 - 1} x(kT)e^{-jn\Omega_0 k}
-$$
-(6.55)
-
-where *x*(*kT*) is the *k*th sample of *x*(*t*) and
-
-$$
-N_0 = \frac{T_0}{T} \quad \text{and} \quad \Omega_0 = \omega_0 T = \frac{2\pi}{N_0}
-$$
-
-In practice, it is impossible to make *T* → 0 in computing the right-hand side of Eq. (6.55). We can make *T* small, but not zero, which will cause the data to increase without limit. Thus, we shall ignore the limit on *T* in Eq. (6.55) with the implicit understanding that *T* is reasonably small. Nonzero *T* will result in some computational error, which is inevitable in any numerical evaluation of an integral. The error resulting from nonzero *T* is called the *aliasing error,* which is discussed in more detail in Ch. 8. Thus, we can express Eq. (6.55) as
-
-$$
-D_n \approx \frac{1}{N_0} \sum_{k=0}^{N_0 - 1} x(kT) e^{-jn\Omega_0 k}
-$$
- (6.56)
-
-Since 0*N*<sup>0</sup> = 2π, we know that *ejn*0(*k*+*N*0) = *ejn*0*<sup>k</sup>*, and it follows that
-
-$$
-D_{n+N_0}=D_n
-$$
-
-The periodicity property *Dn*+*N*<sup>0</sup> = *Dn* means that beyond *n* = *N*0/2, the coefficients represent the values for negative *n*. For instance, when *N*<sup>0</sup> = 32, *D*<sup>17</sup> = *D*−15, *D*<sup>18</sup> = *D*−14,...,*D*<sup>31</sup> = *D*−1. The cycle repeats again from *n* = 32 on.
-
-We can use the efficient FFT (the *fast Fourier transform* discussed in Sec. 8.6) to compute the right-hand side of Eq. (6.56). We shall use MATLAB to implement the FFT algorithm. For
-
-### 660 CHAPTER 6 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES
-
-this purpose, we need samples of *x*(*t*) over one period starting at *t* = 0. In this algorithm, it is also preferable (although not necessary) that *N*<sup>0</sup> be a power of 2, (i.e., *N*<sup>0</sup> = 2*<sup>m</sup>* where *m* is an integer).
-
-### **EXAMPLE 6.16 Numerical Computation of Fourier Spectra**
-
-Numerically compute and then plot the exponential Fourier spectra for the periodic signal in Fig. 6.2a (Ex. 6.1).
-
-The samples of *x*(*t*) start at *t* = 0 and the last (*N*0th) sample is at *t* = *T*<sup>0</sup> − *T*. At the points of discontinuity, the sample value is taken as the average of the values of the function on two sides of the discontinuity. Thus, the sample at *t* = 0 is not 1 but (*e*−π/<sup>2</sup> +1)/2 = 0.604. To determine *N*0, we require that *Dn* for *n* ≥ *N*0/2 be negligible. Because *x*(*t*) has a jump discontinuity, *Dn* decays rather slowly as 1/*n*. Hence, a choice of *N*<sup>0</sup> = 200 is acceptable because the (*N*0/2)nd (100th) harmonic is about 1% of the fundamental. However, we also require *N*<sup>0</sup> to be a power of 2. Hence, we shall take *N*<sup>0</sup> = 256 = 28.
-
-First, the basic parameters are established.
-
->> T\_0 = pi; N\_0 = 256; T = T\_0/N\_0; t = (0:T:T\*(N\_0-1))'; >> x = exp(-t/2); x(1) = (exp(-pi/2)+1)/2;
-
-Next, the DFT, computed by means of the fft function, is used to approximate the exponential Fourier spectra up to *n* = *N*0/2. To facilitate comparison with previous plots of *Dn*, we only plot the results over −5 ≤ *n* ≤ 5.
-
-```
->> D_n = fft(x)/N_0; n = [-N_0/2:N_0/2-1]';
->> clf; subplot(1,2,1); stem(n,abs(fftshift(D_n)),'.k');
->> axis([-5 5 0 .6]); xlabel('n'); ylabel('|D_n|');
->> subplot(1,2,2); stem(n,angle(fftshift(D_n)),'.k');
->> axis([-5 5 -2 2]); xlabel('n'); ylabel('\angle D_n [rad]');
-```
-
-As shown in Fig. 6.28, the resulting approximation is visually indistinguishable from the true Fourier series spectra shown in Fig. 6.12 or Fig. 6.13.

engineering/linear-systems-and-signals-3rd-edition-bp-lathi/085_6.7 MATLAB - FOURIER SERIES APPLICATIONS.md

@@ -1,190 +0,0 @@
-## <span id="page-680-0"></span>**[6.7 MATLAB: FOURIER](#page-12-0) SERIES APPLICATIONS**
-
-Computational packages such as MATLAB simplify the Fourier-based analysis, design, and synthesis of periodic signals. MATLAB permits rapid and sophisticated calculations, which promote practical application and intuitive understanding of the Fourier series.
-
-### **[6.7-1 Periodic Functions and the Gibbs Phenomenon](#page-12-0)**
-
-It is sufficient to define any *T*0-periodic function over the interval (0 ≤ *t* < *T*0). For example, consider the 2π-periodic function given by
-
-$$
-x(t) = \begin{cases} t/A & 0 \le t < A \\ 1 & A \le t < \pi \\ 0 & \pi \le t < 2\pi \\ x(t + 2\pi) & \text{otherwise} \end{cases}
-$$
-
-Although similar to a square wave, *x*(*t*) has a linearly rising edge of width *A*, where (0 < *A* < π). As *A* → 0, *x*(*t*) approaches a square wave; as *A* → π, *x*(*t*) approaches a type of sawtooth wave.
-
-In MATLAB, the mod command helps represent periodic functions such as *x*(*t*).
-
->> x = @(t,A) mod(t,2\*pi)/A.\*(mod(t,2\*pi)<A)+((mod(t,2\*pi)>=A)&(mod(t,2\*pi)<pi));
-
-Sometimes referred to as the signed remainder after division, mod(t,2\*pi) returns the value *t* modulo 2π. Thought of another way, the mod operator appropriately shifts *t* into [0,*T*0), where *x*(*t*) is conveniently defined. Notice also that an anonymous function can be defined with multiple input variables.
-
-The exponential Fourier series coefficients for *x*(*t*) (see Prob. 6.3-2) are given by
-
-$$
-D_n = \begin{cases} \frac{2\pi - A}{4\pi} & n = 0\\ \frac{1}{2\pi n} \left( \frac{e^{-jnA} - 1}{nA} + je^{-jn\pi} \right) & \text{otherwise} \end{cases}
-$$
-
-Since *x*(*t*) is real, *D*−*<sup>n</sup>* = *D*<sup>∗</sup> *<sup>n</sup>*. Truncating the Fourier series at |*n*| = *N* yields the approximation
-
-$$
-x(t) \approx x_N(t) = D_0 + \sum_{n=1}^{N} \left( D_n e^{jnt} + D_n^* e^{-jnt} \right)
-$$
-\n(6.57)
-
-For a user-specified *N*, program CH6MP1 uses Eq. (6.57) to compute *xN*(*t*) over (−π/4 ≤ *t* < 2π +π/4).
-
-```
-function [x_N,t] = CH6MP1(A,N);
-% CH6MP1.m : Chapter 6, MATLAB Program 1
-% Function M-file approximates x(t) using Fourier series truncated at |n|=N
-% INPUTS: A = width of rising edge
-% N = largest harmonic of truncated Fourier series
-% OUTPUTS: x_N = Nth harmonic truncated Fourier series
-```
-
-```
-% t = time vector for x_N
-```
-
-```
-% Define FS coefficients for signal x(t)
-D = @(n) 1/(2*pi*n)*((exp(-1j*n*A)-1)/(n*A) + 1j*exp(-1j*n*pi));
-% Construct truncated FS approximation of x(t) using N harmonics
-t = linspace(-pi/4,2*pi+pi/4,10000); % Time vector exceeds one period.
-x_N = (2*pi-A)/(4*pi)*ones(size(t)); % Compute dc term
-for n = 1:N, % Compute N remaining terms
-   x_N = x_N+real(D(n)*exp(1j*n*t) + conj(D(n))*exp(-1j*n*t));
-end
-```
-
-Although theoretically not required, the real command ensures that small computer round-off errors do not cause a complex-valued result.
-
-Using program CH6MP1 with *A* = π/2 and *N* = 20, Fig. 6.29 compares *x*(*t*) and *x*20(*t*).
-
->> A = pi/2; [x\_20,t] = CH6MP1(A,20); >> plot(t,x\_20,'k',t,x(t,A),'k:'); axis([-pi/4,2\*pi+pi/4,-0.1,1.1]); >> xlabel('t'); ylabel('x\_{20}(t)');
-
-As expected, the falling edge is accompanied by the overshoot that is characteristic of the Gibbs phenomenon.
-
-Increasing *N* to 100, as shown in Fig. 6.30, improves the approximation but does not reduce the overshoot.
-
-```
->> [x_100,t] = CH6MP1(A,100);
-```
-
->> plot(t,x\_100,'k',t,x(t,A),'k:'); axis([-pi/4,2\*pi+pi/4,-0.1,1.1]);
-
-```
->> xlabel('t'); ylabel('x_{100}(t)');
-```
-
-Reducing *A* to π/64 produces a curious result. For *N* = 20, both the rising and falling edges are accompanied by roughly 9% of overshoot, as shown in Fig. 6.31. As the number of terms is increased, overshoot persists only in the vicinity of jump discontinuities. For *xN*(*t*), increasing *N* decreases the overshoot near the rising edge but not near the falling edge. Remember that it is a
-
-**Figure 6.29** Comparison of *x*20(*t*) and *x*(*t*) when *A* = π/2.
-
-**Figure 6.30** Comparison of *x*100(*t*) and *x*(*t*) when *A* = π/2.
-
-**Figure 6.31** Comparison of *x*20(*t*) and *x*(*t*) when *A* = π/64.
-
-**Figure 6.32** Comparison of *x*100(*t*) and *x*(*t*) when *A* = π/64.
-
-true jump discontinuity that causes the Gibbs phenomenon. A continuous signal, no matter how sharply it rises, can always be represented by a Fourier series at every point within any small error by increasing *N*. This is not the case when a true jump discontinuity is present. Figure 6.32 illustrates this behavior using *N* = 100.
-
-### <span id="page-683-0"></span>**[6.7-2 Optimization and Phase Spectra](#page-12-0)**
-
-Although magnitude spectra typically receive the most attention, phase spectra are critically important in some applications. Consider the problem of characterizing the frequency response of an unknown system. By applying sinusoids one at a time, the frequency response is empirically measured one point at a time. This process is tedious at best. Applying a superposition of many sinusoids, however, allows simultaneous measurement of many points of the frequency response. Such measurements can be taken by a spectrum analyzer equipped with a transfer function mode or by applying Fourier analysis techniques, which are discussed in later chapters.
-
-A multitone test signal *m*(*t*) is constructed as a superposition of *N* real sinusoids
-
-$$
-m(t) = \sum_{n=1}^{N} M_n \cos{(\omega_n t + \theta_n)}
-$$
-
-where *Mn* and θ*<sup>n</sup>* establish the relative magnitude and phase of each sinusoidal component. It is sensible to constrain all gains to be equal, *Mn* = *M* for all *n*. This ensures equal treatment at each point of the measured frequency response. Although the value *M* is normally chosen to set the desired signal power, we set *M* = 1 for convenience.
-
-While not required, it is also sensible to space the sinusoidal components uniformly in frequency.
-
-$$
-m(t) = \sum_{n=1}^{N} \cos(n\omega_0 t + \theta_n)
-$$
-\n(6.58)
-
-Another sensible alternative, which spaces components logarithmically in frequency, is treated in Prob. 6.7-4.
-
-Equation (6.58) is now a truncated compact-form Fourier series with a flat magnitude spectrum. Frequency resolution and range are set by ω<sup>0</sup> and *N*, respectively. For example, a 2 kHz range with a resolution of 100 Hz requires ω<sup>0</sup> = 2π100 and *N* = 20. The only remaining unknowns are the θ*n*.
-
-While it is tempting to set θ*<sup>n</sup>* = 0 for all *n*, the results are quite unsatisfactory. MATLAB helps demonstrate the problem by using ω<sup>0</sup> = 2π100 and *N* = 20 sinusoids, each with a peak-to-peak voltage of 1 volt.
-
-```
->> m = @(theta,t,omega) sum(cos(omega*t+theta*ones(size(t))));
->> N = 20; omega = 2*pi*100*[1:N]'; theta = zeros(size(omega));
->> t = linspace(-0.01,0.01,10000);
->> plot(t,m(theta,t,omega),'k'); xlabel('t [sec]'); ylabel('m(t) [volts]');
-```
-
-As shown in Fig. 6.33, θ*<sup>n</sup>* = 0 causes each sinusoid to constructively add. The resulting 20 volt peak can saturate system components, such as operational amplifiers operating with ±12 volt rails. To improve signal performance, the maximum amplitude of *m*(*t*) over *t* needs to be reduced.
-
-One way to reduce max*t*(|*m*(*t*)|) is to reduce *M*, the strength of each component. Unfortunately, this approach reduces the system's signal-to-noise ratio and ultimately degrades measurement quality. Therefore, reducing *M* is not a smart decision. The phases θ*n*, however, can be adjusted to reduce max*t*(|*m*(*t*)|) while preserving signal power. In fact, since θ*<sup>n</sup>* = 0 maximizes max*t*(|*m*(*t*)|), just about any other choice of θ*<sup>n</sup>* will improve the situation. Even a random choice should improve performance.
-
-**Figure 6.33** Test signal *m*(*t*) with θ*<sup>n</sup>* = 0.
-
-As with any computer, MATLAB cannot generate truly random numbers. Rather, it generates pseudo-random numbers. Pseudo-random numbers are deterministic sequences that appear to be random. The particular sequence of numbers that is realized depends entirely on the initial state of the pseudo-random number generator. Setting the generator's initial state to a known value allows a "random" experiment with reproducible results. The command rng(0) initializes the state of the pseudo-random number generator to a known condition of zero, and the MATLAB command rand(a,b) generates an a-by-b matrix of pseudo-random numbers that are uniformly distributed over the interval (0, 1). Radian phases occupy the wider interval (0, 2π ), so the results from rand need to be appropriately scaled.
-
->> rng(0); theta\_rand0 = 2\*pi\*rand(N,1);
-
-Next, we recompute and plot *m*(*t*) using the randomly chosen θ*n*.
-
-```
->> m_rand0 = m(theta_rand0,t,omega);
-```
-
-```
->> plot(t,m_rand0,'k'); axis([-0.01,0.01,-10,10]);
-```
-
-```
->> xlabel('t [sec]'); ylabel('m(t) [volts]');
-```
-
->> set(gca,'ytick',[min(m\_rand0),max(m\_rand0)]); grid on;
-
-For a vector input, the min and max commands return the minimum and maximum values of the vector. Using these values to set *y* axis tick marks makes it easy to identify the extreme values of the *m*(*t*). As seen from Fig. 6.34, the maximum amplitude is now 7.6307, which is significantly smaller than the maximum of 20 when θ*<sup>n</sup>* = 0.
-
-Randomly chosen phases suffer a fatal fault: there is little guarantee of optimal performance. For example, repeating the experiment with rng(5) produces a maximum magnitude of 8.2399 volts, as shown in Fig. 6.35. This value is significantly higher than the previous maximum of 7.6307 volts. Clearly, it is better to replace a random solution with an optimal solution.
-
-What constitutes "optimal"? Many choices exist, but desired signal criteria naturally suggest that optimal phases minimize the maximum magnitude of *m*(*t*) over all *t*. To find these optimal phases, MATLAB's fminsearch command is useful. First, the function to be minimized, called the objective function, is defined.
-
->> maxmagm = @(theta,t,omega) max(abs(sum(cos(omega\*t+theta\*ones(size(t))))));
-
-666 CHAPTER 6 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES
-
-**Figure 6.34** Test signal *m*(*t*) with random θ*<sup>n</sup>* found by using rng(0).
-
-**Figure 6.35** Test signal *m*(*t*) with random θ*<sup>n</sup>* found by using rand('state',1).
-
-The anonymous function argument order is important; fminsearch uses the first input argument as the variable of minimization. To minimize over θ, as desired, θ must be the first argument of the objective function maxmagm.
-
-Next, the time vector is shortened to include only one period of *m*(*t*).
-
->> t = linspace(0,0.01,401);
-
-A full period ensures that all values of *m*(*t*) are considered; the short length of t helps ensure that functions execute quickly. An initial value of θ is randomly chosen to begin the search.
-
-```
->> rng(0); theta_init = 2*pi*rand(N,1);
->> theta_opt = fminsearch(maxmagm,theta_init,[],t,omega);
-```
-
-Notice that fminsearch finds the minimizer to maxmagm over θ by using an initial value theta\_init. Most numerical minimization techniques are capable of finding only local minima, and fminsearch is no exception. As a result, fminsearch does not always produce a unique solution. The empty square brackets indicate no special options are requested, and the remaining ordered arguments are secondary inputs for the objective function. Full format details for fminsearch are available from MATLAB's help facilities.
-
-<span id="page-686-0"></span>**Figure 6.36** Test signal *m*(*t*) with optimized phases.
-
-Figure 6.36 shows the phase-optimized test signal. The maximum magnitude is reduced to a value of 5.3632 volts, which is a significant improvement over the original peak of 20 volts.
-
-Although the signals shown in Figs. 6.33 through 6.36 look different, they all possess the same magnitude spectra. The signals differ only in phase spectra. It is interesting to investigate the similarities and differences of these signals in ways other than graphs and mathematics. For example, is there an audible difference between the signals? For computers equipped with sound capability, the MATLAB sound command can be used to find out.
-
-```
->> Fs = 8000; t = [0:1/Fs:2]; % Two second records at a sampling rate of 8kHz
->> sound(m(theta,t,omega)/20,Fs); % Play (scaled) m(t) constructed using zero phases
-```
-
-Since the sound command clips magnitudes that exceed 1, the input vector is scaled by 1/20 to avoid clipping and the resulting sound distortion. The signals using other phase assignments are created and played in a similar fashion. How well does the human ear discern the differences in phase spectra? If you are like most people, you will not be able to discern any differences in how these waveforms sound.

engineering/linear-systems-and-signals-3rd-edition-bp-lathi/086_6.8 SUMMARY.md

@@ -1,17 +0,0 @@
-## **[6.8 SUMMARY](#page-12-0)**
-
-In this chapter we showed how a periodic signal can be represented as a sum of sinusoids or exponentials. If the frequency of a periodic signal is *f*0, then it can be expressed as a weighted sum of a sinusoid of frequency *f*<sup>0</sup> and its harmonics (the trigonometric Fourier series). We can reconstruct the periodic signal from a knowledge of the amplitudes and phases of these sinusoidal components (amplitude and phase spectra).
-
-If a periodic signal *x*(*t*) has an even symmetry, its Fourier series contains only cosine terms (including dc). In contrast, if *x*(*t*) has an odd symmetry, its Fourier series contains only sine terms. If *x*(*t*) has neither type of symmetry, its Fourier series contains both sine and cosine terms.
-
-At points of discontinuity, the Fourier series for *x*(*t*) converges to the mean of the values of *x*(*t*) on either side of the discontinuity. For signals with discontinuities, the Fourier series converges in the mean and exhibits Gibbs phenomenon at the points of discontinuity. The amplitude spectrum of the Fourier series for a periodic signal *x*(*t*) with jump discontinuities decays slowly (as 1/*n*) with frequency. We need a large number of terms in the Fourier series to approximate *x*(*t*) within
-
-### <span id="page-687-0"></span>668 CHAPTER 6 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES
-
-a given error. In contrast, the amplitude spectrum of a smoother periodic signal decays faster with frequency and we require a smaller number of terms in the series to approximate *x*(*t*) within a given error.
-
-A sinusoid can be expressed in terms of exponentials. Therefore, the Fourier series of a periodic signal can also be expressed as a sum of exponentials (the exponential Fourier series). The exponential form of the Fourier series and the expressions for the series coefficients are more compact than those of the trigonometric Fourier series. Also, the response of LTIC systems to an exponential input is much simpler than that for a sinusoidal input. Moreover, the exponential form of representation lends itself better to mathematical manipulations than does the trigonometric form. This includes the establishment of useful Fourier series properties that simplify work and help provide a more intuitive understanding of signals. For these reasons, the exponential form of the series is preferred in modern practice in the areas of signals and systems.
-
-The plots of amplitudes and angles of various exponential components of the Fourier series as functions of the frequency are the exponential Fourier spectra (amplitude and angle spectra) of the signal. Because a sinusoid cosω0*t* can be represented as a sum of two exponentials, *ej*<sup>ω</sup>0*<sup>t</sup>* and *e*−*j*ω0*<sup>t</sup>* , the frequencies in the exponential spectra range from ω = −∞ to ∞. By definition, frequency of a signal is always a positive quantity. Presence of a spectral component of a negative frequency −*n*ω<sup>0</sup> merely indicates that the Fourier series contains terms of the form *e*−*jn*ω0*<sup>t</sup>* . The spectra of the trigonometric and exponential Fourier series are closely related, and one can be found by the inspection of the other.
-
-In Sec. 6.5 we discuss a method of representing signals by the generalized Fourier series, of which the trigonometric and exponential Fourier series are special cases. Signals are vectors in every sense. Just as a vector can be represented as a sum of its components in a variety of ways, depending on the choice of the coordinate system, a signal can be represented as a sum of its components in a variety of ways, of which the trigonometric and exponential Fourier series are only two examples. Just as we have vector coordinate systems formed by mutually orthogonal vectors, we also have signal coordinate systems (basis signals) formed by mutually orthogonal signals. Any signal in this signal space can be represented as a sum of the basis signals. Each set of basis signals yields a particular Fourier series representation of the signal. The signal is equal to its Fourier series, not in the ordinary sense, but in the special sense that the energy of the difference between the signal and its Fourier series approaches zero. This allows for the signal to differ from its Fourier series at some isolated points.

engineering/linear-systems-and-signals-3rd-edition-bp-lathi/087_REFERENCES.md

@@ -1,15 +0,0 @@
-### **[REFERENCES](#page-12-0)**
-
-- 1. Bell, E. T. *Men of Mathematics.* Simon & Schuster, New York, 1937.
-- 2. Durant, W., and Durant, A. *The Age of Napoleon,* Part XI in *The Story of Civilization Series.* Simon & Schuster, New York, 1975.
-- 3. Calinger, R. *Classics of Mathematics,* 4th ed. Moore Publishing, Oak Park, IL, 1982.
-- 4. Lanczos, C. *Discourse on Fourier Series*. Oliver Boyd, London, 1966.
-- 5. Körner, T. W. *Fourier Analysis.* Cambridge University Press, Cambridge, UK, 1989.
-- 6. Guillemin, E. A. *Theory of Linear Physical Systems.* Wiley, New York, 1963.
-- 7. Gibbs, W. J. *Nature,* vol. 59, p. 606, April 1899.
-- 8. Bôcher, M. *Annals of Mathematics,* vol. 7, no. 2, 1906.
-
-- <span id="page-688-0"></span>9. Carslaw, H. S. *Bulletin of the American Mathematical Society,* vol. 31, pp. 420–424, October 1925.
-- 10. Lathi, B. P. *Signals, Systems, and Communication.* Wiley, New York, 1965.
-- 11. Walker P. L. *The Theory of Fourier Series and Integrals.* Wiley, New York, 1986.
-- 12. Churchill, R. V., and Brown, J. W. *Fourier Series and Boundary Value Problems,* 3rd ed. McGraw-Hill, New York, 1978.

engineering/linear-systems-and-signals-3rd-edition-bp-lathi/088_PROBLEMS.md

@@ -1,851 +0,0 @@
-## **[PROBLEMS](#page-12-0)**
-
-- **6.1-1** For each of the periodic signals shown in Fig. P6.1-1, find the compact trigonometric Fourier series and sketch the amplitude and phase spectra. If either the sine or cosine terms are absent in the Fourier series, explain why.
-- **6.1-2** (a) Find the trigonometric Fourier series for *y*(*t*) shown in Fig. P6.1-2.
-  - (b) The signal *y*(*t*) can be obtained by time reversal of *x*(*t*) shown in Fig. 6.2a. Use this fact to obtain the Fourier series for *y*(*t*) from the results in Ex. 6.1. Verify that the Fourier series thus obtained is identical to that found in part (a).
-  - (c) Show that, in general, time reversal of a periodic signal does not affect the amplitude spectrum, and the phase spectrum is also unchanged except for the change of sign.
-- **6.1-3** (a) Find the trigonometric Fourier series for the periodic signal *y*(*t*) depicted in Fig. P6.1-3.
-  - (b) The signal *y*(*t*) can be obtained by time compression of *x*(*t*) shown in Fig. 6.2a by a factor 2. Use this fact to obtain the Fourier series for *y*(*t*) from the results in Ex. 6.1. Verify that the Fourier series thus obtained is identical to that found in part (a).
-  - (c) Show that, in general, time compression of a periodic signal by a factor *a* expands the Fourier spectra along the ω axis by the same factor *a*. In other words *C*0,*Cn*, and θ*<sup>n</sup>* remain unchanged, but the fundamental frequency is increased by the factor *a*, thus expanding the spectrum. Similarly, time expansion of a periodic signal by a factor *a* compresses its Fourier spectra along the ω axis by the factor *a*.
-- **6.1-4** (a) Find the trigonometric Fourier series for the periodic signal *g*(*t*) in Fig. P6.1-4. Take advantage of the symmetry.
-
-- (b) Observe that *g*(*t*) is identical to *x*(*t*) in Fig. 6.4a left-shifted by 0.5 second. Use this fact to obtain the Fourier series for *g*(*t*) from the results in Ex. 6.2. Verify that the Fourier series thus obtained is identical to that found in part (a).
-- (c) Show that, in general, a time shift of *T* seconds of a periodic signal does not affect the amplitude spectrum. However, the phase of the *n*th harmonic is increased or decreased *n*ω0*T* depending on whether the signal is advanced or delayed by *T* seconds.
-- **6.1-5** Determine the trigonometric Fourier series coefficients *an* and *bn* for the following signals. In each case, also determine the signals' fundamental radian frequency ω0. No integration is required to solve this problem.
-  - (a) *x*a(*t*) = cos(3π*t*)
-  - (b) *x*b(*t*) = sin(7π*t*)
-  - (c) *x*c(*t*) = 2+4 cos(3π*t*)−2*j*sin(7π*t*)
-  - (d) *x*d(*t*) = (1+*j*)sin(3π*t*)+(2−*j*) cos(7π*t*)
-  - (e) *x*e(*t*) = sin(3π*t* +1)+2 cos(7π*t* −2)
-  - (f) *x*f(*t*) = sin(6π*t*)+2 cos(14π*t*)
-- **6.1-6** If the two halves of one period of a periodic signal are identical in shape except that one is the negative of the other, the periodic signal is said to have a *half-wave symmetry*. If a periodic signal *x*(*t*) with a period *T*<sup>0</sup> satisfies the half-wave symmetry condition, then
-
-$$
-x\left(t - \frac{T_0}{2}\right) = -x(t)
-$$
-
-In this case, show that all the even-numbered harmonics vanish and that the odd-numbered harmonic coefficients are given by
-
-$$
-a_n = \frac{4}{T_0} \int_0^{T_0/2} x(t) \cos n\omega_0 t \, dt
-$$
-
-**Figure P6.1-2**
-
-(b)
-
-**Figure P6.1-6**
-
-and
-
-$$
-b_n = \frac{4}{T_0} \int_0^{T_0/2} x(t) \sin n\omega_0 t \, dt
-$$
-
-Using these results, find the Fourier series for the periodic signals in Fig. P6.1-6.
-
-**6.1-7** Over a finite interval, a signal can be represented by more than one trigonometric (or exponential) Fourier series. For instance, if we wish to represent *x*(*t*) = *t* over an interval 0 < *t* < 1 by a Fourier series with fundamental frequency ω<sup>0</sup> = 2, we can draw a pulse *x*(*t*) = *t* over the interval 0 < *t* < 1 and repeat the pulse every π seconds so that *T*<sup>0</sup> = π and ω<sup>0</sup> = 2 (Fig. P6.1-7a). If we want the fundamental frequency ω<sup>0</sup> to be 4, we repeat the pulse every π/2 seconds. If we want the series to contain only cosine terms with ω<sup>0</sup> = 2, we construct a pulse *x*(*t*) = |*t*| over −1 < *t* < 1, and repeat it every π seconds (Fig. P6.1-7b). The resulting signal is an even function with period π. Hence, its Fourier series will have only cosine terms with ω<sup>0</sup> = 2. The resulting Fourier series represents *x*(*t*) = *t* over 0 < *t* < 1, as desired. We do not care what it represents outside this interval.
-
-Sketch the periodic signal *x*(*t*) such that *x*(*t*) = *t* for 0 < *t* < 1 and the Fourier series for *x*(*t*) satisfies the following conditions.
-
-- (a) ω<sup>0</sup> = π/2 and contains all harmonics, but cosine terms only
-- (b) ω<sup>0</sup> = 2 and contains all harmonics, but sine terms only
-- (c) ω<sup>0</sup> = π/2 and contains all harmonics, which are exclusively neither sine nor cosine
-- (d) ω<sup>0</sup> = 1 and contains only odd harmonics and cosine terms
-- (e) ω<sup>0</sup> = π/2 and contains only odd harmonics and sine terms
-
-**Figure P6.1-7**
-
-(f) ω<sup>0</sup> = 1 and contains only odd harmonics, which are exclusively neither sine nor cosine.
-
-[*Hint:* For parts (d), (e), and (f), you need to use half-wave symmetry discussed in Prob. 6.1-6. Cosine terms imply a possible dc component.] You are asked only to sketch the periodic signal *x*(*t*) satisfying the given conditions. Do not find the values of the Fourier coefficients.
-
-- **6.1-8** State with reasons whether the following signals are periodic or aperiodic. For periodic signals, find the period and state which harmonics are present in the series.
-  - (a) 3 sin *t* +2 sin 3*t*
-  - (b) 2+5 sin 4*t* +4 cos 7*t*
-  - (c) 2 sin 3*t* +7 cos π*t*
-  - (d) 7 cos π*t* +5 sin 2π*t*
-  - (e) 3 cos <sup>√</sup>2*<sup>t</sup>* <sup>+</sup>5 cos 2*<sup>t</sup>*
-  - (f) sin 5*t* <sup>2</sup> <sup>+</sup>3 cos 6*t* <sup>5</sup> <sup>+</sup>3 sin *<sup>t</sup>* <sup>7</sup> <sup>+</sup>30◦ *t*
-
-(g) 
-$$
-\sin 3t + \cos \frac{15}{4}
-$$
-
-- (h) (3 sin 2*t* +sin 5*t*)<sup>2</sup>
-- (i) (5 sin 2*t*)<sup>3</sup>
-- **6.3-1** For each of the periodic signals in Fig. P6.1-1, find exponential Fourier series and sketch the corresponding spectra.
-- **6.3-2** A 2π-periodic signal *x*(*t*) is specified over one period as
-
-$$
-x(t) = \begin{cases} \frac{1}{A}t & 0 \le t < A \\ 1 & A \le t < \pi \\ 0 & \pi \le t < 2\pi \end{cases}
-$$
-
-Sketch *x*(*t*) over two periods from *t* = 0 to 4π. Show that the exponential Fourier series coefficients *Dn* for this series are given by
-
-$$
-D_n = \begin{cases} \frac{2\pi - A}{4\pi} & n = 0\\ \frac{1}{2\pi n} \left( \frac{e^{-jAn} - 1}{An} + je^{-jn\pi} \right) & n \neq 0 \end{cases}
-$$
-
-**6.3-3** A periodic signal *x*(*t*) is expressed by the following Fourier series:
-
-$$
-x(t) = 3\cos t + \sin\left(t - \frac{\pi}{6}\right) - 2\cos\left(t - \frac{\pi}{3}\right)
-$$
-
-- (a) Sketch the amplitude and phase spectra for the trigonometric series.
-- (b) By inspection of spectra in part (a), sketch the exponential Fourier series spectra.
-- (c) By inspection of spectra in part (b), write the exponential Fourier series for *x*(*t*).
-- (d) Show that the series found in part (c) is equivalent to the trigonometric series for *x*(*t*).
-- **6.3-4** The trigonometric Fourier series of a certain periodic signal is given by
-
-$$
-x(t) = 3 + \sqrt{3}\cos 2t + \sin 2t
-$$
-$$
-+ \sin 3t - \frac{1}{2}\cos\left(5t + \frac{\pi}{3}\right)
-$$
-
-- (a) Sketch the trigonometric Fourier spectra.
-- (b) By inspection of spectra in part (a), sketch the exponential Fourier series spectra.
-- (c) By inspection of spectra in part (b), write the exponential Fourier series for *x*(*t*).
-- (d) Show that the series found in part (c) is equivalent to the trigonometric series for *x*(*t*).
-- **6.3-5** The exponential Fourier series of a certain function is given as
-
-$$
-x(t) = (2+j2)e^{-j3t} + j2e^{-jt} + 3 - j2e^{jt} + (2-j2)e^{j3t}
-$$
-
-- (a) Sketch the exponential Fourier spectra.
-- (b) By inspection of the spectra in part (a), sketch the trigonometric Fourier spectra for *x*(*t*). Find the compact trigonometric Fourier series from these spectra.
-- (c) Show that the trigonometric series found in part (b) is equivalent to the exponential series for *x*(*t*).
-- (d) Find the signal bandwidth.
-- **6.3-6** Figure P6.3-6 shows the trigonometric Fourier spectra of a periodic signal *x*(*t*).
-  - (a) By inspection of Fig. P6.3-6, find the trigonometric Fourier series representing *x*(*t*).
-  - (b) By inspection of Fig. P6.3-6, sketch the exponential Fourier spectra of *x*(*t*).
-  - (c) By inspection of the exponential Fourier spectra obtained in part (b), find the exponential Fourier series for *x*(*t*).
-  - (d) Show that the series found in parts (a) and (c) are equivalent.
-- **6.3-7** Figure P6.3-7 shows the exponential Fourier spectra of a periodic signal *x*(*t*).
-
-- (a) By inspection of Fig. P6.3-7, find the exponential Fourier series representing *x*(*t*).
-- (b) By inspection of Fig. P6.3-7, sketch the trigonometric Fourier spectra for *x*(*t*).
-- (c) By inspection of the trigonometric Fourier spectra found in part (b), find the trigonometric Fourier series for *x*(*t*).
-- (d) Show that the series found in parts (a) and (c) are equivalent.
-- **6.3-8** Let periodic signal *x*(*t*) have exponential Fourier series spectrum *Dn*. Prove the following properties.
-  - (a) If *x*(*t*) has even symmetry, then *Dn* also has even symmetry.
-  - (b) If *x*(*t*) has odd symmetry, then *Dn* also has odd symmetry.
-  - (c) If *x*(*t*) is real, then *Dn* is conjugate symmetric (*Dn* = *D*<sup>∗</sup> <sup>−</sup>*n*).
-  - (d) If *x*(*t*) is imaginary, then *Dn* is conjugate antisymmetric (*Dn* = −*D*<sup>∗</sup> <sup>−</sup>*n*).
-- **6.3-9** (a) Find the exponential Fourier series for the signal in Fig. P6.3-9a.
-  - (b) Using the results in part (a), find the Fourier series for the signal *x*ˆ(*t*) in Fig. P6.3-9b,
-
-**Figure P6.3-7**
-
-#### **Figure P6.3-9**
-
-which is a time-shifted version of the signal *x*(*t*).
-
-- (c) Using the results in part (a), find the Fourier series for the signal *x*˜(*t*) in Fig. P6.3-9c, which is a time-scaled version of the signal *x*(*t*).
-- **6.3-10** A periodic signal *x*(*t*) is expressed as an exponential Fourier series
-
-$$
-x(t) = \sum_{n = -\infty}^{\infty} D_n e^{jn\omega_0 t}
-$$
-
-(a) Show that the exponential Fourier series for *x*ˆ(*t*) = *x*(*t* −*T*) is given by
-
-$$
-\hat{x}(t) = \sum_{n = -\infty}^{\infty} \hat{D}_n e^{jn\omega_0 t}
-$$
-
-in which
-
-$$
-|\tilde{D}_n| = |D_n| \quad \text{and} \quad \angle \tilde{D}_n = \angle D_n - n\omega_0 T
-$$
-
-This result shows that time shifting of a periodic signal by *T* seconds merely changes the phase spectrum by *n*ω0*T*. The amplitude spectrum is unchanged.
-
-(b) Show that the exponential Fourier series for *x*˜(*t*) = *x*(*at*) is given by
-
-$$
-\tilde{x}(t) = \sum_{n = -\infty}^{\infty} D_n e^{jn(a\omega_0)t}
-$$
-
-This result shows that time compression of a periodic signal by a factor *a* expands its Fourier spectra along the ω axis by the same factor *a*. Similarly, time expansion of a periodic signal by a factor *a* compresses its Fourier spectra along the ω axis by the factor *a*. Intuitively explain this result.
-
-**6.3-11** (a) The Fourier series for the periodic signal in Fig. 6.7a is given in Drill 6.1. Verify Parseval's theorem for this series, given that
-
-$$
-\sum_{n=1}^{\infty} \frac{1}{n^4} = \frac{\pi^4}{90}
-$$
-
-- (b) If *x*(*t*) is approximated by the first *N* terms in this series, find *N* so that the power of the error signal is less than 1% of *Px*.
-- **6.3-12** (a) The Fourier series for the periodic signal in Fig. 6.7b is given in Drill 6.1. Verify
-
-Problems 675
-
-Parseval's theorem for this series, given that
-
-$$
-\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}
-$$
-
-- (b) If *x*(*t*) is approximated by the first *N* terms in this series, find *N* so that the power of the error signal is less than 10% of *Px*.
-- **6.3-13** The signal *x*(*t*) in Fig. 6.17 is approximated by the first 2*N* + 1 terms (from *n* = −*N* to *N*) in its exponential Fourier series given in Drill 6.5. Determine the value of *N* if this (2*N* + 1)-term Fourier series power is to be no less than 99.75% of the power of *x*(*t*).
-- **6.3-14** (a) A 2 rad/s periodic signal *x*1(*t*) has Fourier series spectrum *D*1[*n*]. Determine the Fourier series spectrum *<sup>X</sup>*2[*n*] of *<sup>x</sup>*2(*t*) <sup>=</sup> <sup>1</sup> <sup>3</sup> *x*1(−*t* −5) in terms of *X*1[*n*].
-  - (b) A 2 rad/s periodic signal *x*1(*t*) has Fourier series spectrum *D*1[*n*]. Determine the Fourier series spectrum *X*2[*n*] of *x*2(*t*) = cos(10*t*)*x*1(*t*) in terms of *X*1[*n*].
-  - (c) A 3 rad/s periodic signal *x*1(*t*) has Fourier series spectrum *D*1[*n*]. Determine the Fourier series spectrum *X*2[*n*] of *x*2(*t*) = *x*1(−*t*)−3*x*1(*t* +2) in terms of *X*1[*n*].
-- **6.3-15** A 2-periodic signal *x*(*t*) is defined as
-
-$$
-x(t) = \begin{cases} -t^2 - t + 0.25 & -1 \le t < 0\\ t^2 - t + 0.25 & 0 \le t < 1\\ x(t+2) & \forall t \end{cases}
-$$
-
-- (a) Plot *x*(*t*) over −2 ≤ *t* ≤ 2.
-- (b) Determine *D*0, the dc content of *x*(*t*).
-- (c) Similar to Ex. 6.11, use properties and not integration to determine *Dn* for *n* = 0.
-- (d) Plot the magnitude spectrum |*Dn*| over a suitable range of *n*.
-- (e) How does the magnitude spectrum |*Dn*| compare to that of signal *y*(*t*) = cos(π*t*)? Note similarities as well as major differences.
-- **6.3-16** A 3-periodic signal *x*(*t*) is defined as
-
-$$
-x(t) = \begin{cases} \n|t| & -1 \le t \le 1\\ \n0 & 1 < |t| \le 1.5\\ \nx(t+3) & \forall t\n\end{cases}
-$$
-
-(a) Plot 
-$$
-x(t)
-$$
- over  $-3 \le t \le 3$ .
-
-- (b) Determine *D*0, the dc content of *x*(*t*).
-- (c) Similar to Ex. 6.11, use properties and not integration to determine *Dn* for *n* = 0.
-- (d) Plot the magnitude spectrum |*Dn*| over a suitable range of *n*. What is the most dominant frequency component of this signal?
-- **6.4-1** Find the response of an LTIC system with transfer function
-
-$$
-H(s) = \frac{s}{s^2 + 2s + 3}
-$$
-
-to the periodic input shown in Fig. 6.2a.
-
-- **6.4-2** A periodic signal *x*(*t*) = 1 + 2 cos(5π*t*) + 3 sin(14π*t*) is applied to an LTIC system to produce output *y*(*t*).
-  - (a) Determine ω0, the fundamental radian frequency of *x*(*t*).
-  - (b) Determine *Dn*, the exponential Fourier series spectrum of *x*(*t*).
-  - (c) If the system is an ideal lowpass filter with cutoff frequency *f*<sup>c</sup> = 2 Hz, what is the output *y*(*t*)?
-  - (d) If the system is an ideal highpass filter with cutoff frequency *f*<sup>c</sup> = 2 Hz, what is the output *y*(*t*)?
-  - (e) If the system is an ideal bandpass filter with a 4 Hz passband centered at 4 Hz, what is the output *y*(*t*)?
-  - (f) If the system is an ideal bandstop filter with a 5 Hz stopband centered at 10 Hz, what is the output *y*(*t*)?
-  - (g) Describe the frequency response of a filter that, in response to *x*(*t*), would produce the output *y*(*t*) = 4 cos(5π*t*) −9 sin(14π*t*).
-- **6.4-3** Consider a *T*<sup>0</sup> = 1 periodic signal *x*(*t*) defined as
-
-$$
-x(t) = \begin{cases} 1 - t^2 & 0 < t \le 1 \\ x(t+1) & \forall t \end{cases}
-$$
-
-- (a) Sketch *x*(*t*) for −2 ≤ *t* ≤ 2.
-- (b) Determine *Dn*, the exponential Fourier series spectrum of *x*(*t*).
-- (c) If *x*(*t*) is applied to an ideal bandpass filter with a 1 Hz passband centered at 3 Hz, determine the output *y*(*t*).
-- **6.4-4** (a) Find the exponential Fourier series for a signal *x*(*t*) = cos 5*t* sin 3*t*. You can do this without evaluating any integrals.
-  - (b) Sketch the Fourier spectra.
-
-(c) The signal *x*(*t*) is applied at the input of an LTIC system with frequency response, as shown in Fig. P6.4-4. Find the output *y*(*t*).
-
-**Figure P6.4-4**
-
-- **6.4-5** (a) Find the exponential Fourier series for a periodic signal *x*(*t*) shown in Fig. P6.4-5a.
-  - (b) The signal *x*(*t*) is applied at the input of an LTIC system shown in Fig. P6.4-5b. Find the expression for the output *y*(*t*).
-- **6.4-6** A *T*-periodic τ/*T* duty-cycle square wave *p*(*t*) is defined as
-
-$$
-p(t) = \begin{cases} 1 & |t| < \frac{\tau}{2} \\ 0 & \frac{\tau}{2} < |t| < \frac{T}{2} \\ p(t+T) & \forall t \end{cases}
-$$
-
-where 0 <τ< *T*. Also consider the frequency response *H*(ω) of a lowpass communications channel with 10 rad/s bandwidth (e.g., |*H*(ω)| ≈ 0 for ω > 10). If *T* and τ are properly chosen, we can estimate *H*(ω) at points ω = *n*ω<sup>0</sup> as *<sup>H</sup>*<sup>ˆ</sup> (*n*ω0) <sup>=</sup> <sup>1</sup> *P*0 *Yn*, where *Yn* is the exponential FS spectrum of the channel output *y*(*t*) in response to input *p*(*t*) and *P*<sup>0</sup> is the dc component of *p*(*t*).
-
-- (a) Using direct integration, determine the exponential Fourier series coefficients *Pn* of signal *p*(*t*).
-- (b) Determine a suitable value *T* so that *p*(*t*) applied to system *H*(ω) has 21 component frequencies over the system bandwidth 0 ≤ ω ≤ 10.
-- (c) Assuming *T* is properly chosen, determine a suitable duty cycle τ/*T* so that *H*(*n*ω0) ≈ *Yn*. Carefully justify your result.
-- (d) From the perspective of using *p*(*t*) to help measure the system frequency response *H*(ω), what happens if *T* is properly chosen but τ/*T* is chosen too small?
-- (e) From the perspective of using *p*(*t*) to help measure the system frequency response *H*(ω), what happens if *T* is properly chosen but τ/*T* is chosen too large?
-- **6.5-1** Derive Eq. (6.32) in an alternate way by observing that **e** = (**x**−*c***y**) and |**e**| <sup>2</sup> = (**x**−*c***y**)·(**x** − *c***y**) = |**x**| <sup>2</sup> +*c*2|**y**| <sup>2</sup> −2*c***x** · **y**.
-- **6.5-2** A signal *x*(*t*) is approximated in terms of a signal *y*(*t*) over an interval (*t*1, *t*2):
-
-$$
-x(t) \simeq cy(t) \qquad t_1 < t < t_2
-$$
-
-where *c* is chosen to minimize the error energy.
-
-**Figure P6.4-5**
-
-- (a) Show that *y*(*t*) and the error *e*(*t*) = *x*(*t*) − *cy*(*t*) are orthogonal over the interval (*t*1, *t*2).
-- (b) If possible, explain the result in terms of a signal-vector analogy.
-- (c) Verify this result for the square signal *x*(*t*) in Fig. 6.23 and its approximation in terms of signal sin *t*.
-- **6.5-3** If *x*(*t*) and *y*(*t*) are orthogonal, then show that the energy of the signal *x*(*t*) + *y*(*t*) is identical to the energy of the signal *x*(*t*) − *y*(*t*) and is given by *Ex* + *Ey*. Explain this result by using the vector analogy. In general, show that for orthogonal signals *x*(*t*) and *y*(*t*) and for any pair of arbitrary real constants *c*<sup>1</sup> and *c*2, the energies of *c*1*x*(*t*) + *c*2*y*(*t*) and *c*1*x*(*t*) − *c*2*y*(*t*) are both given by *c*<sup>2</sup> <sup>1</sup>*Ex* <sup>+</sup>*c*<sup>2</sup> <sup>2</sup>*Ey*.
-- **6.5-4** (a) For the signals *x*(*t*) and *y*(*t*) depicted in Fig. P6.5-4, find the component of the form *y*(*t*) contained in *x*(*t*). In other words, find the optimum value of *c* in the approximation *x*(*t*) ≈ *cy*(*t*) so that the error signal energy is minimum.
-  - (b) Find the error signal *e*(*t*) and its energy *Ee*. Show that the error signal is orthogonal to *y*(*t*), and that *Ex* = *c*<sup>2</sup>*Ey* + *Ee*. Explain this result in terms of vectors.
-
-**Figure P6.5-4**
-
-- **6.5-5** For the signals *x*(*t*) and *y*(*t*) shown in Fig. P6.5-4, find the component of the form *x*(*t*) contained in *y*(*t*). In other words, find the optimum value of *c* in the approximation *y*(*t*) ≈ *cx*(*t*) so that the error signal energy is minimum. What is the error signal energy?
-- **6.5-6** Represent the signal *x*(*t*) shown in Fig. P6.5-4a over the interval from 0 to 1 by a trigonometric Fourier series of fundamental frequency ω<sup>0</sup> = 2π. Compute the error energy in the representation of *x*(*t*) by only the first *N* terms of this series for *N* = 1, 2, 3, and 4.
-
-- **6.5-7** Represent *x*(*t*) = *t* over the interval (0, 1) by a trigonometric Fourier series that has
-  - (a) ω<sup>0</sup> = 2π and only sine terms
-  - (b) ω<sup>0</sup> = π and only sine terms
-  - (c) ω<sup>0</sup> = π and only cosine terms
-
-You may use a dc term in these series if necessary.
-
-- **6.5-8** In Ex. 6.15, we represented the function in Fig. 6.27 by Legendre polynomials.
-  - (a) Use the results in Ex. 6.15 to represent the signal *g*(*t*) in Fig. P6.5-8 by Legendre polynomials.
-  - (b) Compute the error energy for the approximations having one and two (nonzero) terms.
-
-**Figure P6.5-8**
-
-- **6.5-9** Walsh functions, which can take on only two amplitude values, form a complete set of orthonormal functions and are of great practical importance in digital applications because they can be easily generated by logic circuitry and because multiplication with these functions can be implemented by simply using a polarity-reversing switch. Figure P6.5-9 shows the first eight functions in this set. Represent *x*(*t*) in Fig. P6.5-4a over the interval (0, 1) by using a Walsh Fourier series with these eight basis functions. Compute the energy of *e*(*t*), the error in the approximation, using the first *N* nonzero terms in the series for *N* = 1, 2, 3, and 4. In Prob. 6.5-6 we found the trigonometric Fourier series for *x*(*t*). How does the Walsh series compare with the trigonometric series in Prob. 6.5-6 from the viewpoint of the error energy for a given *N*?
-- **6.5-10** For the four-dimensional real space *<sup>R</sup>*4, the so-called Walsh basis is given by: φ<sup>1</sup> = [1, 1, 1, 1], φ<sup>2</sup> = [1, 1,−1,−1], φ<sup>3</sup> = [1,−1,−1, 1], and φ<sup>4</sup> = [1,−1, 1,−1]. Denoting elements *x* = [*x*1, *x*2, *x*3, *x*4] and *y* = [*y*1, *y*2, *y*3, *y*4] (*x*, *<sup>y</sup>* <sup>∈</sup> *<sup>R</sup>*4), we can define orthogonality as
-
-### **Figure P6.5-9**
-
-%<sup>4</sup> *<sup>k</sup>*=<sup>1</sup> *xky*<sup>∗</sup> *<sup>k</sup>* = 0. In linear algebra terminology, orthogonality here means that the inner product of vectors *x* and *y* is zero. Lastly, define a vector *z* = [−4, 0, 1,−7].
-
-- (a) Show that the Walsh basis functions are mutually orthogonal. This requires a total of six calculations.
-- (b) Are the Walsh basis functions normal? That is, does the inner product of each Walsh basis function with itself evaluate to 1?
-- (c) Determine the coefficients [*c*1, *c*2, *c*3, *c*4] to represent *z* using Walsh basis functions as *<sup>z</sup>*<sup>ˆ</sup> <sup>=</sup> %<sup>4</sup> *<sup>k</sup>*=<sup>1</sup> *ck*φ*k*.
-- (d) Determine the best three-dimensional approximation *z*ˆ3*<sup>D</sup>* to *z* in terms of the Walsh basis functions. That is, your estimate can only be a linear combination of three functions from [φ1,φ2,φ3,φ4]. Evaluate the three-term sum to determine the four elements of vector *z*ˆ3*D*.
-- **6.5-11** A function can be expanded in terms of many different types of basis functions, not just the complex exponentials of Fourier analysis. For example, Walsh functions are explored in Prob. 6.5-9. *Laguerre* polynomials *Lk*(*t*), which have support on the interval [0,∞), are another possible set of basis functions. The Laguerre expansion using any number of terms we choose. For the Laguerre expansion, we define orthogonality a little differently, as
-
-$$
-\int_0^\infty e^{-t}x(t)y(t)dt = 0.
-$$
-
-Notice the presence of the *e*−*<sup>t</sup>* term in the integral. Using this definition, Laguerre polynomials are orthonormal.
-
-(a) Show that *L*0(*t*) is normal. That is, show that
-
-$$
-\int_0^\infty e^{-t} L_0(t) L_0^*(t) dt = 1
-$$
-
-(b) Show that *L*1(*t*) is normal. That is, show that
-
-$$
-\int_0^{\infty} e^{-t} L_1(t) L_1^*(t) dt = 1
-$$
-
-- (c) Show that *L*0(*t*) is orthogonal to *L*1(*t*).
-- (d) Compute the coefficient *c*<sup>0</sup> that produces the best approximation *x*ˆ0(*t*) = *c*0*L*0(*t*) of the function *<sup>x</sup>*(*t*) <sup>=</sup> *<sup>e</sup>*−*<sup>t</sup> u*(*t*).
-- (e) For the best estimate *x*ˆ1(*t*) = *c*0*L*0(*t*) + *<sup>c</sup>*1*L*1(*t*) of the function *<sup>x</sup>*(*t*) <sup>=</sup> *<sup>e</sup>*−*<sup>t</sup> u*(*t*), the coefficient *c*<sup>0</sup> remains unchanged from part **()**(d) and *<sup>c</sup>*<sup>1</sup> <sup>=</sup> <sup>1</sup> <sup>4</sup> . Confirm that *<sup>c</sup>*<sup>1</sup> <sup>=</sup> <sup>1</sup> 4 and explain why the coefficient *c*<sup>0</sup> does not change.
-- **6.7-1** A periodic signal has <sup>ω</sup><sup>0</sup> <sup>=</sup> <sup>2</sup> <sup>3</sup>π and exponential Fourier series spectrum *Dn* = *j* cos (π*n*/10)(*u*[*n*+10] −*u*[*n*−11]). [*Hint:* Refer to Prob. 6.3-8 for some useful properties.]
-  - (a) Determine the period *T*<sup>0</sup> of the corresponding signal *x*(*t*).
-  - (b) Is the time-domain signal real, imaginary, or neither? Justify your answer.
-  - (c) Is the time-domain signal even, odd, or neither? Justify your answer.
-  - (d) Use MATLAB to synthesize the timedomain signal *x*(*t*) and plot it over the interval [−*T*0,*T*0].
-- **6.7-2** Repeat Prob. 6.7-1 for a periodic signal with <sup>ω</sup><sup>0</sup> <sup>=</sup> <sup>3</sup> <sup>2</sup>π and *Dn* = 2 sin(π*n*/10)(*u*[*n* + 10] − *u*[*n*−11]).
-- **6.7-3** Consider the (*T*<sup>0</sup> = 1)-periodic signal *x*(*t*):
-
-$$
-x(t) = \begin{cases} 2t - t^2 & 0 < t \le 1 \\ x(t+1) & \forall t \end{cases}.
-$$
-
-- (a) Sketch *x*(*t*) for −2 ≤ *t* ≤ 2.
-- (b) Using properties and minimal integration, determine *Dn*, the exponential Fourier spectrum of *x*(*t*).
-
-- (c) Verify the correctness of *Dn* by using MAT-LAB to synthesize *x*(*t*) with a suitable truncation of Eq. (6.19).
-- (d) Suppose *x*(*t*) is applied to an ideal bandpass filter with passband between 2.5 and 3.5 Hz. Determine the filter output *y*(*t*). Simplify your answer.
-
-[*Hint:* Refer to Ex. 6.11.]
-
-**6.7-4** Section 6.7 discusses the construction of a phase-optimized multitone test signal with linearly spaced frequency components. This problem investigates a similar signal with logarithmically spaced frequency components.
-
-> A multitone test signal *m*(*t*) is constructed by using a superposition of *N* real sinusoids
-
-$$
-m(t) = \sum_{n=1}^{N} \cos(\omega_n t + \theta_n)
-$$
-
-where θ*<sup>n</sup>* establishes the relative phase of each sinusoidal component.
-
-- (a) Determine a suitable set of *N* = 10 frequencies ω*<sup>n</sup>* that logarithmically spans [(2π ) ≤ ω ≤ 100(2π )] yet still results in a periodic test signal *m*(*t*). Determine the period *T*<sup>0</sup> of your signal. Using θ*<sup>n</sup>* = 0, plot the resulting (*T*0)-periodic signal over −*T*0/2 ≤ *t* ≤ *T*0/2.
-- (b) Determine a suitable set of phases θ*<sup>n</sup>* that minimize the maximum magnitude of *m*(*t*). Plot the resulting signal and identify the maximum magnitude that results.
-- (c) Many systems suffer from what is called one-over-*f* noise. The power of this undesirable noise is proportional to 1/*f* . Thus, low-frequency noise is stronger than high-frequency noise. What modifications to *m*(*t*) are appropriate for use in environments with 1/*f* noise? Justify your answer.
-
-# **[CONTINUOUS-TIME](#page-12-0) SIGNAL ANALYSIS: THE FOURIER TRANSFORM**
-
-We can analyze linear systems in many different ways by taking advantage of the property of linearity, whereby the input is expressed as a sum of simpler components. The system response to any complex input can be found by summing the system's response to these simpler components of the input. In time-domain analysis, we separated the input into impulse components. In the frequency-domain analysis in Ch. 4, we separated the input into exponentials of the form *est* (the Laplace transform), where the complex frequency *s* = σ + *j*ω. The Laplace transform, although very valuable for system analysis, proves somewhat awkward for signal analysis, where we prefer to represent signals in terms of exponentials *ej*ω*<sup>t</sup>* instead of *est*. This is accomplished by the Fourier transform. In a sense, the Fourier transform may be considered to be a special case of the Laplace transform with *s* = *j*ω. Although this view is true most of the time, it does not always hold because of the nature of convergence of the Laplace and Fourier integrals.
-
-In Ch. 6, we succeeded in representing periodic signals as a sum of (everlasting) sinusoids or exponentials of the form *ej*ω*<sup>t</sup>* . The Fourier integral developed in this chapter extends this spectral representation to aperiodic signals.
-
-## **7.1 APERIODIC SIGNAL [REPRESENTATION](#page-12-0) BY THE FOURIER INTEGRAL**
-
-Applying a limiting process, we now show that an aperiodic signal can be expressed as a continuous sum (integral) of everlasting exponentials. To represent an aperiodic signal *x*(*t*) such as the one depicted in Fig. 7.1a by everlasting exponentials, let us construct a new periodic signal *xT*<sup>0</sup> (*t*) formed by repeating the signal *x*(*t*) at intervals of *T*<sup>0</sup> seconds, as illustrated in Fig. 7.1b. The period *T*<sup>0</sup> is made long enough to avoid overlap between the repeating pulses. The periodic signal *xT*<sup>0</sup> (*t*) can be represented by an exponential Fourier series. If we let *T*<sup>0</sup> → ∞, the pulses in the periodic signal repeat after an infinite interval and, therefore,
-
-$$
-\lim_{T_0 \to \infty} x_{T_0}(t) = x(t)
-$$
-
-<span id="page-699-0"></span>CHAPTER
-
-**7**
-
-**Figure 7.1** Construction of a periodic signal: (a) signal *x*(*t*) and (b) periodic extension of *x*(*t*).
-
-Thus, the Fourier series representing *xT*<sup>0</sup> (*t*) will also represent *x*(*t*) in the limit *T*<sup>0</sup> → ∞. The exponential Fourier series for *xT*<sup>0</sup> (*t*) is given by
-
-$$
-x_{T_0}(t) = \sum_{n = -\infty}^{\infty} D_n e^{jn\omega_0 t}
-$$
-\n(7.1)
-
-where <sup>ω</sup><sup>0</sup> <sup>=</sup> <sup>2</sup><sup>π</sup> *<sup>T</sup>*<sup>0</sup> and
-
-$$
-D_n = \frac{1}{T_0} \int_{-T_0/2}^{T_0/2} x_{T_0}(t) e^{-jn\omega_0 t} dt
-$$
-\n(7.2)
-
-Observe that integrating *xT*<sup>0</sup> (*t*) over (−*T*0/2,*T*0/2) is the same as integrating *x*(*t*) over (−∞,∞). Therefore, Eq. (7.2) can be expressed as
-
-$$
-D_n = \frac{1}{T_0} \int_{-\infty}^{\infty} x(t) e^{-jn\omega_0 t} dt
-$$
-\n(7.3)
-
-It is interesting to see how the nature of the spectrum changes as *T*<sup>0</sup> increases. To understand this fascinating behavior, let us define *X*(ω), a continuous function of ω, as
-
-$$
-X(\omega) = \int_{-\infty}^{\infty} x(t)e^{-j\omega t}dt
-$$
-\n(7.4)
-
-A glance at Eqs. (7.3) and (7.4) shows that
-
-$$
-D_n = \frac{1}{T_0} X(n\omega_0) \tag{7.5}
-$$
-
-**Figure 7.2** Change in the Fourier spectrum when the period *T*<sup>0</sup> in Fig. 7.1 is doubled.
-
-This means that the Fourier coefficients *Dn* are 1/*T*<sup>0</sup> times the samples of *X*(ω) uniformly spaced at intervals of ω0, as depicted in Fig. 7.2a.† Therefore, (1/*T*0)*X*(ω) is the envelope for the coefficients *Dn*. We now let *T*<sup>0</sup> → ∞ by doubling *T*<sup>0</sup> repeatedly. Doubling *T*<sup>0</sup> halves the fundamental frequency ω<sup>0</sup> so that there are now twice as many components (samples) in the spectrum. However, by doubling *T*0, the envelope (1/*T*0)*X*(ω) is halved, as shown in Fig. 7.2b. If we continue this process of doubling *T*<sup>0</sup> repeatedly, the spectrum progressively becomes denser while its magnitude becomes smaller. Note, however, that the relative shape of the envelope remains the same [proportional to *X*(ω) in Eq. (7.4)]. In the limit as *T*<sup>0</sup> → ∞, ω<sup>0</sup> → 0 and *Dn* → 0. This result makes for a spectrum so dense that the spectral components are spaced at zero (infinitesimal) intervals. At the same time, the amplitude of each component is zero (infinitesimal). We have *nothing of everything, yet we have something!* This paradox sounds like *Alice in Wonderland,* but as we shall see, these are the classic characteristics of a very familiar phenomenon.‡
-
-Substitution of Eq. (7.5) in Eq. (7.1) yields
-
-$$
-x_{T_0}(t) = \sum_{n = -\infty}^{\infty} \frac{X(n\omega_0)}{T_0} e^{jn\omega_0 t}
-$$
-\n(7.6)
-
-As *T*<sup>0</sup> →∞, ω<sup>0</sup> becomes infinitesimal (ω<sup>0</sup> →0). Hence, we shall replace ω<sup>0</sup> by a more appropriate notation, <sup>ω</sup>. In terms of this new notation, <sup>ω</sup><sup>0</sup> <sup>=</sup> <sup>2</sup><sup>π</sup> *<sup>T</sup>*<sup>0</sup> becomes
-
-$$
-\Delta \omega = \frac{2\pi}{T_0}
-$$
-
-<sup>†</sup> For the sake of simplicity, we assume *Dn*, and therefore *X*(ω), in Fig. 7.2, to be real. The argument, however, is also valid for complex *Dn* [or *X*(ω)].
-
-<sup>‡</sup> If nothing else, the reader now has irrefutable proof of the proposition that 0% ownership of everything is better than 100% ownership of nothing.
-
-**Figure 7.3** The Fourier series becomes the Fourier integral in the limit as *T*<sup>0</sup> → ∞.
-
-and Eq. (7.6) becomes
-
-$$
-x_{T_0}(t) = \sum_{n=-\infty}^{\infty} \left[ \frac{X(n\Delta\omega)\Delta\omega}{2\pi} \right] e^{(jn\Delta\omega)t}
-$$
-
-This equation shows that *xT*<sup>0</sup> (*t*) can be expressed as a sum of everlasting exponentials of frequencies 0,±ω,±2ω,±3ω,... (the Fourier series). The amount of the component of frequency *n*ω is [*X*(*n*ω)ω]/2π. In the limit as *T*<sup>0</sup> → ∞, ω → 0 and *xT*<sup>0</sup> (*t*) → *x*(*t*). Therefore,
-
-$$
-x(t) = \lim_{T_0 \to \infty} x_{T_0}(t) = \lim_{\Delta \omega \to 0} \frac{1}{2\pi} \sum_{n = -\infty}^{\infty} X(n\Delta \omega) e^{(jn\Delta \omega)t} \Delta \omega
-$$
-\n(7.7)
-
-The sum on the right-hand side of Eq. (7.7) can be viewed as the area under the function *X*(ω)*ej*ω*<sup>t</sup>* , as illustrated in Fig. 7.3. Therefore,
-
-$$
-x(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} X(\omega) e^{j\omega t} d\omega
-$$
- (7.8)
-
-The integral on the right-hand side is called the *Fourier integral*. We have now succeeded in representing an aperiodic signal *x*(*t*) by a Fourier integral (rather than a Fourier series).† This integral is basically a Fourier series (in the limit) with fundamental frequency ω → 0, as seen from Eq. (7.7). The amount of the exponential *ejn*ω*<sup>t</sup>* is *X*(*n*ω)ω/2π. Thus, the function *X*(ω) given by Eq. (7.4) acts as a spectral function.
-
-We call *X*(ω) the *direct* Fourier transform of *x*(*t*), and *x*(*t*) the *inverse* Fourier transform of *X*(ω). The same information is conveyed by the statement that *x*(*t*) and *X*(ω) are a Fourier transform pair. Symbolically, this statement is expressed as
-
-$$
-X(\omega) = \mathcal{F}[x(t)]
-$$
- and  $x(t) = \mathcal{F}^{-1}[X(\omega)]$ 
-
-<sup>†</sup> This derivation should not be considered to be a rigorous proof of Eq. (7.8). The situation is not as simple as we have made it appear [1].
-
-or
-
-$$
-x(t) \Longleftrightarrow X(\omega)
-$$
-
-To recapitulate,
-
-$$
-X(\omega) = \int_{-\infty}^{\infty} x(t)e^{-j\omega t}dt
-$$
-\n(7.9)
-
-and
-
-$$
-x(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} X(\omega) e^{j\omega t} d\omega
-$$
-\n(7.10)
-
-It is helpful to keep in mind that the Fourier integral in Eq. (7.10) is of the nature of a Fourier series with fundamental frequency ω approaching zero [Eq. (7.7)]. Therefore, most of the discussion and properties of Fourier series apply to the Fourier transform as well. *The transform X*(ω) *is the frequency-domain specification of x*(*t*)*.*
-
-We can plot the spectrum *X*(ω) as a function of ω. Since *X*(ω) is complex, we have both amplitude and angle (or phase) spectra
-
-$$
-X(\omega) = |X(\omega)|e^{j\angle X(\omega)}
-$$
-
-in which |*X*(ω)| is the amplitude and *X*(ω) is the angle (or phase) of *X*(ω). According to Eq. (7.9),
-
-$$
-X(-\omega) = \int_{-\infty}^{\infty} x(t)e^{j\omega t}dt
-$$
-
-Taking the conjugates of both sides yields
-
-$$
-x^*(t) \Longleftrightarrow X^*(-\omega) \tag{7.11}
-$$
-
-This property is known as the *conjugation property.* Now, if *x*(*t*) is a real function of *t*, then *x*(*t*) = *x*∗(*t*), and from the conjugation property, we find that
-
-$$
-X(-\omega) = X^*(\omega)
-$$
-
-This is the *conjugate symmetry* property of the Fourier transform, applicable to real *x*(*t*). Therefore, for real *x*(*t*),
-
-$$
-|X(-\omega)| = |X(\omega)| \quad \text{and} \quad \angle X(-\omega) = -\angle X(\omega) \tag{7.12}
-$$
-
-Thus, for real *x*(*t*), the amplitude spectrum |*X*(ω)| is an even function, and the phase spectrum *X*(ω) is an odd function of ω. These results were derived earlier for the Fourier spectrum of a periodic signal [Eq. (6.22)] and should come as no surprise.
-
-### **EXAMPLE 7.1 Fourier Transform of a Causal Exponential**
-
-Find the Fourier transform of *e*−*atu*(*t*).
-
-By definition [Eq. (7.9)],
-
-$$
-X(\omega) = \int_{-\infty}^{\infty} e^{-at} u(t) e^{-j\omega t} dt = \int_{0}^{\infty} e^{-(a+j\omega)t} dt = \frac{-1}{a+j\omega} e^{-(a+j\omega)t} \Big|_{0}^{\infty}
-$$
-
-But |*e*−*j*ω*<sup>t</sup>* | = 1. Therefore, as *t* → ∞, *e*−(*a*+*j*ω)*<sup>t</sup>* = *e*−*ate*−*j*ω*<sup>t</sup>* = ∞ if *a* < 0, but it is equal to 0 if *a* > 0. Therefore,
-
-$$
-X(\omega) = \frac{1}{a + j\omega} \qquad a > 0
-$$
-
-Expressing *<sup>a</sup>*+*j*<sup>ω</sup> in the polar form as <sup>√</sup> *<sup>a</sup>*<sup>2</sup> <sup>+</sup>ω<sup>2</sup> *ej*tan−1(ω/*a*) , we obtain
-
-$$
-X(\omega) = \frac{1}{\sqrt{a^2 + \omega^2}} e^{-j \tan^{-1}(\omega/a)}
-$$
-
-Therefore,
-
-$$
-|X(\omega)| = \frac{1}{\sqrt{a^2 + \omega^2}} \quad \text{and} \quad \angle X(\omega) = -\tan^{-1}\left(\frac{\omega}{a}\right)
-$$
-
-The amplitude spectrum |*X*(ω)| and the phase spectrum *X*(ω) are depicted in Fig. 7.4b. Observe that |*X*(ω)| is an even function of ω, and *X*(ω) is an odd function of ω, as expected.
-
-### EXISTENCE OF THE FOURIER TRANSFORM
-
-In Ex. 7.1 we observed that when *a* < 0, the Fourier integral for *e*−*atu*(*t*) does not converge. Hence, the Fourier transform for *e*−*atu*(*t*) does not exist if *a* < 0 (growing exponential). Clearly, not all signals are Fourier transformable.
-
-Because the Fourier transform is derived here as a limiting case of the Fourier series, it follows that the basic qualifications of the Fourier series, such as *equality in the mean* and convergence conditions in suitably modified form, apply to the Fourier transform as well. It can be shown that if *x*(*t*) has a finite energy, that is, if
-
-$$
-\int_{-\infty}^{\infty} |x(t)|^2 dt < \infty
-$$
-
-then the Fourier transform *X*(ω) is finite and converges to *x*(*t*) in the mean. This means, if we let
-
-$$
-\hat{x}(t) = \lim_{W \to \infty} \frac{1}{2\pi} \int_{-W}^{W} X(\omega) e^{j\omega t} d\omega
-$$
-
-then Eq. (7.10) implies
-
-$$
-\int_{-\infty}^{\infty} \left| x(t) - \hat{x}(t) \right|^2 dt = 0 \tag{7.13}
-$$
-
-In other words, *x*(*t*) and its Fourier integral [the right-hand side of Eq. (7.10)] can differ at some values of *t* without contradicting Eq. (7.13). We shall now discuss an alternate set of criteria due to Dirichlet for convergence of the Fourier transform.
-
-As with the Fourier series, if *x*(*t*) satisfies certain conditions (*Dirichlet conditions*), its Fourier transform is guaranteed to converge pointwise at all points where *x*(*t*) is continuous. Moreover, at the points of discontinuity, *x*(*t*) converges to the value midway between the two values of *x*(*t*) on either side of the discontinuity. The Dirichlet conditions are as follows:
-
-1. *x*(*t*) should be absolutely integrable, that is,
-
-$$
-\int_{-\infty}^{\infty} |x(t)| \, dt < \infty \tag{7.14}
-$$
-
-If this condition is satisfied, we see that the integral on the right-hand side of Eq. (7.9) is guaranteed to have a finite value.
-
-- 2. *x*(*t*) must have only a finite number of finite discontinuities within any finite interval.
-- 3. *x*(*t*) must contain only a finite number of maxima and minima within any finite interval.
-
-We stress here that although the Dirichlet conditions are sufficient for the existence and pointwise convergence of the Fourier transform, they are not necessary. For example, we saw in Ex. 7.1 that a growing exponential, which violates Dirichlet's first condition in Eq. (7.14), does not have a Fourier transform. But the signal of the form (sin*at*)/*t*, which *does* violate this condition, does have a Fourier transform.
-
-Any signal that can be generated in practice satisfies the Dirichlet conditions and therefore has a Fourier transform. Thus, the physical existence of a signal is a sufficient condition for the existence of its transform.
-
-### LINEARITY OF THE FOURIER TRANSFORM
-
-The Fourier transform is linear; that is, if
-
-$$
-x_1(t) \Longleftrightarrow X_1(\omega) \quad \text{and} \quad x_2(t) \Longleftrightarrow X_2(\omega)
-$$
-
-then
-
-$$
-a_1x_1(t) + a_2x_2(t) \Longleftrightarrow a_1X_1(\omega) + a_2X_2(\omega) \tag{7.15}
-$$
-
-<span id="page-706-0"></span>The proof is trivial and follows directly from Eq. (7.9). This result can be extended to any finite number of terms. It can be extended to an infinite number of terms only if the conditions required for interchangeability of the operations of summation and integration are satisfied.
-
-### **[7.1-1 Physical Appreciation of the Fourier Transform](#page-12-0)**
-
-In understanding any aspect of the Fourier transform, we should remember that Fourier representation is a way of expressing a signal in terms of everlasting sinusoids (or exponentials). The Fourier spectrum of a signal indicates the relative amplitudes and phases of sinusoids that are required to synthesize that signal. A periodic signal Fourier spectrum has finite amplitudes and exists at discrete frequencies (ω<sup>0</sup> and its multiples). Such a spectrum is easy to visualize, but the spectrum of an aperiodic signal is not easy to visualize because it has a continuous spectrum. The continuous spectrum concept can be appreciated by considering an analogous, more tangible phenomenon. One familiar example of a continuous distribution is the loading of a beam. Consider a beam loaded with weights *D*1,*D*2,*D*3,...,*Dn* units at the uniformly spaced points *y*1, *y*2,..., *yn*, as shown in Fig. 7.5a.
-
-The total load *WT* on the beam is given by the sum of these loads at each of the *n* points:
-
-$$
-W_T = \sum_{i=1}^n D_i
-$$
-
-Consider now the case of a continuously loaded beam, as depicted in Fig. 7.5b. In this case, although there appears to be a load at every point, the load at any one point is zero. This does not mean that there is no load on the beam. A meaningful measure of load in this situation is not the load at a point, but rather the loading density per unit length at that point. Let *X*(*y*) be the loading density per unit length of beam. It then follows that the load over a beam length *y*(*y* → 0), at some point *y*, is *X*(*y*)*y*. To find the total load on the beam, we divide the beam into segments of interval *y*(*y* → 0). The load over the *n*th such segment of length *y* is *X*(*ny*)*y*. The total load *WT* is given by
-
-$$
-W_T = \lim_{\Delta y \to 0} \sum_{y_1}^{y_n} X(n\Delta y) \, \Delta y = \int_{y_1}^{y_n} X(y) \, dy
-$$
-
-The load now exists at every point, and *y* is now a continuous variable. In the case of discrete loading (Fig. 7.5a), the load exists only at *n* discrete points. At other points, there is no load. On the other hand, in the continuously loaded case, the load exists at every point, but at any specific
-
-**Figure 7.5** Weight-loading analogy for the Fourier transform.
-
-point *y*, the load is zero. The load over a small interval *y*, however, is [*X*(*ny*)]*y* (Fig. 7.5b). Thus, even though the load at a point *y* is zero, the relative load at that point is *X*(*y*).
-
-An exactly analogous situation exists in the case of a signal spectrum. When *x*(*t*) is periodic, the spectrum is discrete, and *x*(*t*) can be expressed as a sum of discrete exponentials with finite amplitudes:
-
-$$
-x(t) = \sum_{n} D_n e^{jn\omega_0 t}
-$$
-
-For an aperiodic signal, the spectrum becomes continuous; that is, the spectrum exists for every value of ω, but the amplitude of each component in the spectrum is zero. The meaningful measure here is not the amplitude of a component of some frequency but the spectral density per unit bandwidth. From Eq. (7.7), it is clear that *x*(*t*) is synthesized by adding exponentials of the form *ejn*ω*<sup>t</sup>* , in which the contribution by any one exponential component is zero. But the contribution by exponentials in an infinitesimal band ω located at ω = *n*ω is (1/2π )*X*(*n*ω)ω, and the addition of all these components yields *x*(*t*) in the integral form:
-
-$$
-x(t) = \lim_{\Delta\omega \to 0} \frac{1}{2\pi} \sum_{n=-\infty}^{\infty} X(n\Delta\omega)e^{(jn\Delta\omega)t} \Delta\omega = \frac{1}{2\pi} \int_{-\infty}^{\infty} X(\omega)e^{i\omega t} d\omega
-$$
-
-Thus, *n*ω approaches a continuous variable ω. The spectrum now exists at every ω. The contribution by components within a band *d*ω is (1/2π )*X*(ω)*d*ω = *X*(ω)*df* , where *df* is the bandwidth in hertz. Clearly, *X*(ω) is the *spectral density* per unit bandwidth (in hertz).† It also follows that even if the amplitude of any one component is infinitesimal, the relative amount of a component of frequency ω is *X*(ω). Although *X*(ω) is a spectral density, in practice, it is customarily called the *spectrum* of *x*(*t*) rather than the spectral density of *x*(*t*). Deferring to this convention, we shall call *X*(ω) the Fourier spectrum (or Fourier transform) of *x*(*t*).
-
-### A MARVELOUS BALANCING ACT
-
-An important point to remember here is that *x*(*t*) is represented (or synthesized) by exponentials or sinusoids that are everlasting (not causal). Such conceptualization leads to a rather fascinating picture when we try to visualize the synthesis of a timelimited pulse signal *x*(*t*) [Fig. 7.6] by the sinusoidal components in its Fourier spectrum. The signal *x*(*t*) exists only over an interval (*a*,*b*) and is zero outside this interval. The spectrum of *x*(*t*) contains an infinite number of exponentials (or sinusoids), which start at *t* = −∞ and continue forever. The amplitudes and phases of these components add up exactly to *x*(*t*) over the finite interval (*a*,*b*) and to zero everywhere outside this interval. Juggling the amplitudes and phases of an infinite number of components to achieve
-
-<sup>†</sup> To stress that the signal spectrum is a *density* function, we shall shade the plot of <sup>|</sup>*X*(ω)<sup>|</sup> (as in Fig. 7.4b). The representation of *X*(ω), however, will be a line plot, primarily to avoid visual confusion.
-
-<span id="page-708-0"></span>such a perfect and delicate balance boggles the human imagination. Yet the Fourier transform accomplishes it routinely, without much thinking on our part. Indeed, we become so involved in mathematical manipulations that we fail to notice this marvel.
-
-## **[7.2 TRANSFORMS OF](#page-12-0) SOME USEFUL FUNCTIONS**
-
-For convenience, we now introduce a compact notation for the useful gate, triangle, and interpolation functions.
-
-### UNIT GATE FUNCTION
-
-We define a unit gate function rect(*x*) as a gate pulse of unit height and unit width, centered at the origin, as illustrated in Fig. 7.7a† :
-
-rect 
-$$
-(x)
-$$
- = 
-$$
-\begin{cases} 0 & |x| > \frac{1}{2} \\ \frac{1}{2} & |x| = \frac{1}{2} \\ 1 & |x| < \frac{1}{2} \end{cases}
-$$
- (7.16)
-
-The gate pulse in Fig. 7.7b is the unit gate pulse rect(*x*) expanded by a factor τ along the horizontal axis and therefore can be expressed as rect(*x*/τ ) (see Sec. 1.2-2). Observe that τ , the denominator of the argument of rect (*x*/τ ), indicates the width of the pulse.
-
-### UNIT TRIANGLE FUNCTION
-
-We define a unit triangle function (*x*) as a triangular pulse of unit height and unit width, centered at the origin, as shown in Fig. 7.8a
-
-$$
-\Delta(x) = \begin{cases} 0 & |x| \ge \frac{1}{2} \\ 1 - 2|x| & |x| < \frac{1}{2} \end{cases}
-$$
-(7.17)
-
-**Figure 7.7** A gate pulse.
-
-<sup>†</sup> At <sup>|</sup>*x*| = 0.5, we require rect(*x*) <sup>=</sup> 0.5 because the inverse Fourier transform of a discontinuous signal converges to the mean of its two values at the discontinuity.
-
-**Figure 7.8** A triangle pulse.
-
-The pulse in Fig. 7.8b is (*x*/τ ). Observe that here, as for the gate pulse, the denominator τ of the argument of (*x*/τ ) indicates the pulse width.
-
-### INTERPOLATION FUNCTION SINC (*x*)
-
-The function sin*x*/*x* is the "sine over argument" function denoted by sinc (*x*). † This function plays an important role in signal processing. It is also known as the *filtering or interpolating function*. We define
-
-$$
-\operatorname{sinc}(x) = \frac{\sin x}{x} \tag{7.18}
-$$
-
-Inspection of Eq. (7.18) shows the following:
-
-- 1. sinc (*x*) is an even function of *x*.
-- 2. sinc (*x*) = 0 when sin *x* = 0 except at *x* = 0, where it appears to be indeterminate. This means that sinc*x* = 0 for *x* = ±π,±2π,±3π,....
-- 3. Using L'Hôpital's rule, we find sinc (0) = 1.
-- 4. sinc (*x*) is the product of an oscillating signal sin*x* (of period 2π) and a monotonically decreasing function 1/*x*. Therefore, sinc (*x*) exhibits damped oscillations of period 2π, with amplitude decreasing continuously as 1/*x*.
-
-Figure 7.9a shows sinc(*x*). Observe that sinc (*x*) = 0 for values of *x* that are positive and negative integer multiples of π. Figure 7.9b shows sinc (3ω/7). The argument 3ω/7 = π when ω = 7π/3. Therefore, the first zero of this function occurs at ω = 7π/3.
-
-### **DR ILL 7.1 Sketching Basic Functions**
-
-Sketch: **(a)** rect(*x*/8), **(b)** (ω/10), **(c)** sinc (3πω/2), and **(d)** sinc (*t*)rect(*t*/4π ).
-
-$$
-\operatorname{sinc}(x) = \frac{\sin \pi x}{\pi x}
-$$
-
-<sup>†</sup> sinc(*x*) is also denoted by Sa (*x*) in the literature. Some authors define sinc (*x*) as
-
-**Figure 7.9** A sinc pulse.
-
-### **EXAMPLE 7.2 Fourier Transform of a Rectangular Pulse**
-
-Find the Fourier transform of *x*(*t*) = rect(*t*/τ ) (Fig. 7.10a).
-
-$$
-X(\omega) = \int_{-\infty}^{\infty} \text{rect}\left(\frac{t}{\tau}\right) e^{-j\omega t} dt
-$$
-
-Since rect(*t*/τ ) = 1 for |*t*| < τ/2, and since it is zero for |*t*| > τ/2,
-
-$$
-X(\omega) = \int_{-\tau/2}^{\tau/2} e^{-j\omega t} dt
-$$
-  
-=  $-\frac{1}{j\omega} (e^{-j\omega \tau/2} - e^{j\omega \tau/2}) = \frac{2 \sin(\frac{\omega \tau}{2})}{\omega}$   
-=  $\tau \frac{\sin(\frac{\omega \tau}{2})}{(\frac{\omega \tau}{2})} = \tau \operatorname{sinc}(\frac{\omega \tau}{2})$ 
-
-**Figure 7.10 (a)** A gate pulse *x*(*t*), **(b)** its Fourier spectrum *X*(ω), **(c)** its amplitude spectrum |*X*(ω)|, and **(d)** its phase spectrum *X*(ω).
-
-Therefore,
-
-$$
-\text{rect}\left(\frac{t}{\tau}\right) \Longleftrightarrow \tau \text{ sinc}\left(\frac{\omega\tau}{2}\right) \tag{7.19}
-$$
-
-Recall that sinc(*x*) = 0 when *x* = ±*n*π. Hence, sinc(ωτ /2) = 0 when ωτ/2 = ±*n*π; that is, when ω = ±2*n*π/τ ,(*n* = 1, 2, 3,. . .), as depicted in Fig. 7.10b. The Fourier transform *X*(ω) shown in Fig. 7.10b exhibits positive and negative values. A negative amplitude can be considered to be a positive amplitude with a phase of −π or π. We use this observation to plot the amplitude spectrum |*X*(ω)|=|sinc (ωτ /2)| (Fig. 7.10c) and the phase spectrum *X*(ω) (Fig. 7.10d). The phase spectrum, which is required to be an odd function of ω, may be drawn in several other ways because a negative sign can be accounted for by a phase of ±*n*π, where *n* is any odd integer. All such representations are equivalent.
-
-#### BANDWIDTH OF RECT *t τ*
-
-The spectrum *X*(ω) in Fig. 7.10 peaks at ω = 0 and decays at higher frequencies. Therefore, rect(*t*/τ ) is a lowpass signal with most of the signal energy in lower-frequency components. Strictly speaking, because the spectrum extends from 0 to ∞, the bandwidth is ∞. However, much of the spectrum is concentrated within the first lobe (from ω = 0 to ω = 2π/τ ). Therefore, a rough estimate of the bandwidth of a rectangular pulse of width τ seconds is 2π/τ rad/s, or 1/τ Hz.† Note the reciprocal relationship of the pulse width with its bandwidth. We shall observe later that this result is true, in general.
-
-<sup>†</sup> To compute bandwidth, we must consider the spectrum for positive values of ω only. See the discussion in Sec. 6.3.
-
-Find the Fourier transform of the unit impulse δ(*t*).
-
-Using the sampling property of the impulse [Eq. (1.11)], we obtain
-
-$$
-\mathcal{F}[\delta(t)] = \int_{-\infty}^{\infty} \delta(t)e^{-j\omega t}dt = 1 \quad \text{and} \quad \delta(t) \Longleftrightarrow 1
-$$
-
-Figure 7.11 shows δ(*t*) and its spectrum.
-
-### **EXAMPLE 7.4 Inverse Fourier Transform of the Dirac Delta Function**
-
-Find the inverse Fourier transform of δ(ω).
-
-On the basis of Eq. (7.10) and the sampling property of the impulse function,
-
-$$
-\mathcal{F}^{-1}[\delta(\omega)] = \frac{1}{2\pi} \int_{-\infty}^{\infty} \delta(\omega) e^{j\omega t} d\omega = \frac{1}{2\pi}
-$$
-
-Therefore,
-
-$$
-\frac{1}{2\pi} \Longleftrightarrow \delta(\omega) \quad \text{and} \quad 1 \Longleftrightarrow 2\pi \delta(\omega) \tag{7.20}
-$$
-
-This result shows that the spectrum of a constant signal *x*(*t*) = 1 is an impulse 2πδ(ω), as illustrated in Fig. 7.12.
-
-The result [Eq. (7.20)] could have been anticipated on qualitative grounds. Recall that the Fourier transform of *x*(*t*) is a spectral representation of *x*(*t*) in terms of everlasting exponential components of the form *ej*ω*<sup>t</sup>* . Now, to represent a constant signal *x*(*t*) = 1, we need a single
-
-**Figure 7.12 (a)** A constant (dc) signal and **(b)** its Fourier spectrum.
-
-everlasting exponential *ej*ω*<sup>t</sup>* with ω = 0.† This results in a spectrum at a single frequency ω = 0. Another way of looking at the situation is that *x*(*t*) = 1 is a dc signal that has a single frequency ω = 0 (dc).
-
-If an impulse at ω = 0 is a spectrum of a dc signal, what does an impulse at ω = ω<sup>0</sup> represent? We shall answer this question in the next example.
-
-### **EXAMPLE 7.5 Inverse Fourier Transform of a Shifted Dirac Delta Function**
-
-Find the inverse Fourier transform of δ(ω −ω0).
-
-Using the sampling property of the impulse function, we obtain
-
-$$
-\mathcal{F}^{-1}[\delta(\omega - \omega_0)] = \frac{1}{2\pi} \int_{-\infty}^{\infty} \delta(\omega - \omega_0) e^{j\omega t} d\omega = \frac{1}{2\pi} e^{j\omega_0 t}
-$$
-
-Therefore,
-
-$$
-\frac{1}{2\pi}e^{j\omega_0 t} \Longleftrightarrow \delta(\omega - \omega_0) \quad \text{and} \quad e^{j\omega_0 t} \Longleftrightarrow 2\pi\delta(\omega - \omega_0) \tag{7.21}
-$$
-
-This result shows that the spectrum of an everlasting exponential *ej*<sup>ω</sup>0*<sup>t</sup>* is a single impulse at ω = ω0. We reach the same conclusion by qualitative reasoning. To represent the everlasting
-
-<sup>†</sup> The constant multiplier 2<sup>π</sup> in the spectrum [*X*(ω) <sup>=</sup> <sup>2</sup>πδ(ω)] may be a bit puzzling. Since 1 <sup>=</sup> *<sup>e</sup>j*ω*<sup>t</sup>* with ω = 0, it appears that the Fourier transform of *x*(*t*) = 1 should be an impulse of strength unity rather than 2π. Recall, however, that in the Fourier transform *x*(*t*) is synthesized by exponentials not of amplitude *X*(*n*ω)ω but of amplitude 1/2π times *X*(*n*ω)ω, as seen from Eq. (7.7). Had we used variable *f* (hertz) instead of ω, the spectrum would have been the unit impulse.
-
-exponential *ej*<sup>ω</sup>0*<sup>t</sup>* , we need a single everlasting exponential *ej*ω*<sup>t</sup>* with ω = ω0. Therefore, the spectrum consists of a single component at frequency ω = ω0. From Eq. (7.21) it follows that
-
-$$
-e^{-j\omega_0 t} \Longleftrightarrow 2\pi \delta(\omega + \omega_0)
-$$
-
-# **EXAMPLE 7.6 Fourier Transform of a Sinusoid** Find the Fourier transform of the everlasting sinusoid cos ω0*t* (Fig. 7.13a). *v*<sup>0</sup> 0 *v*<sup>0</sup> *x*(*t*) cos *v*0*t X*(*v*) *p <sup>t</sup> <sup>v</sup>* <sup>0</sup> (a) (b) **Figure 7.13 (a)** A cosine signal and **(b)** its Fourier spectrum.
-
-Recall Euler's formula
-
-$$
-\cos \omega_0 t = \frac{1}{2} (e^{j\omega_0 t} + e^{-j\omega_0 t})
-$$
-
-Applying Eq. (7.21), we obtain
-
-$$
-\cos \omega_0 t \Longleftrightarrow \pi [\delta(\omega + \omega_0) + \delta(\omega - \omega_0)]
-$$
-
-The spectrum of cos ω0*t* consists of two impulses at ω<sup>0</sup> and −ω0, as shown in Fig. 7.13b. The result also follows from qualitative reasoning. An everlasting sinusoid cos ω0*t* can be synthesized by two everlasting exponentials, *ej*<sup>ω</sup>0*<sup>t</sup>* and *e*−*j*ω0*<sup>t</sup>* . Therefore, the Fourier spectrum consists of only two components of frequencies ω<sup>0</sup> and −ω0.
-
-### **EXAMPLE 7.7 Fourier Transform of a Periodic Signal**
-
-Determine the Fourier transform of a periodic signal *x*(*t*) using its Fourier series representation.
-
-We can use a Fourier series to express a periodic signal as a sum of exponentials of the form *ejn*<sup>ω</sup>0*<sup>t</sup>* , whose Fourier transform is found in Eq. (7.21). Hence, we can readily find the Fourier transform of a periodic signal by using the linearity property in Eq. (7.15).

engineering/linear-systems-and-signals-3rd-edition-bp-lathi/089_7 CONTINUOUS-TIME SIGNAL ANALYSIS - THE FOURIER TRANSFORM.md

@@ -1,115 +0,0 @@
-### 696 CHAPTER 7 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM
-
-The Fourier series of a periodic signal *x*(*t*) with period *T*<sup>0</sup> is given by
-
-$$
-x(t) = \sum_{n = -\infty}^{\infty} D_n e^{jn\omega_0 t} \qquad \omega_0 = \frac{2\pi}{T_0}
-$$
-
-Taking the Fourier transform of both sides, we obtain†
-
-$$
-X(\omega) = 2\pi \sum_{n=-\infty}^{\infty} D_n \delta(\omega - n\omega_0)
-$$
- (7.22)
-
-**Figure 7.14 (a)** The uniform impulse train and **(b)** its Fourier transform.
-
-As shown in Eq. (6.24) from Ex. 6.9, the Fourier coefficients *Dn* for δ*<sup>T</sup>*<sup>0</sup> (*t*) are constant *Dn* = 1/*T*0. From Eq. (7.22), the Fourier transform of δ*<sup>T</sup>*<sup>0</sup> (*t*) is therefore
-
-$$
-X(\omega) = \frac{2\pi}{T_0} \sum_{n=-\infty}^{\infty} \delta(\omega - n\omega_0) = \omega_0 \delta_{\omega_0}(\omega), \quad \text{where } \omega_0 = \frac{2\pi}{T_0}
-$$
-
-The corresponding spectrum is shown in Fig. 7.14b.
-
-<sup>†</sup> We assume here that the linearity property can be extended to an infinite sum.
-
-### **EXAMPLE 7.9 Fourier Transform of the Unit Step Function**
-
-Find the Fourier transform of the unit step function *u*(*t*).
-
-Trying to find the Fourier transform of *u*(*t*) by direct integration leads to an indeterminate result because
-
-$$
-U(\omega) = \int_{-\infty}^{\infty} u(t)e^{-j\omega t}dt = \int_{0}^{\infty} e^{-j\omega t}dt = \left. \frac{-1}{j\omega}e^{-j\omega t} \right|_{0}^{\infty}
-$$
-
-The upper limit of *e*−*j*ω*<sup>t</sup>* as *t*→∞ yields an indeterminate answer. So we approach this problem by considering *u*(*t*) to be a decaying exponential *e*−*atu*(*t*) in the limit as *a* → 0 (Fig. 7.15a). Thus,
-
-$$
-u(t) = \lim_{a \to 0} e^{-at} u(t)
-$$
-
-and
-
-$$
-U(\omega) = \lim_{a \to 0} \mathcal{F}\lbrace e^{-at} u(t) \rbrace = \lim_{a \to 0} \frac{1}{a + j\omega}
-$$
-
-Expressing the right-hand side in terms of its real and imaginary parts yields
-
-$$
-U(\omega) = \lim_{a \to 0} \left[ \frac{a}{a^2 + \omega^2} - j \frac{\omega}{a^2 + \omega^2} \right] = \lim_{a \to 0} \left[ \frac{a}{a^2 + \omega^2} \right] + \frac{1}{j\omega}
-$$
-
-**Figure 7.15** Derivation of the Fourier transform of the step function.
-
-The function *a*/(*a*<sup>2</sup> + ω<sup>2</sup>) has interesting properties. First, the area under this function (Fig. 7.15b) is π regardless of the value of *a*:
-
-$$
-\int_{-\infty}^{\infty} \frac{a}{a^2 + \omega^2} \, d\omega = \tan^{-1} \frac{\omega}{a} \bigg|_{-\infty}^{\infty} = \pi
-$$
-
-### 698 CHAPTER 7 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM
-
-Second, when *a* → 0, this function approaches zero for all ω = 0, and all its area (π) is concentrated at a single point ω = 0. Clearly, as *a* → 0, this function approaches an impulse of strength π. Thus,
-
-$$
-U(\omega) = \pi \delta(\omega) + \frac{1}{j\omega} \tag{7.23}
-$$
-
-Note that *u*(*t*) is not a "true" dc signal because it is not constant over the interval −∞ to ∞. To synthesize "true" dc, we require only one everlasting exponential with ω = 0 (impulse at ω = 0). The signal *u*(*t*) has a jump discontinuity at *t* = 0. It is impossible to synthesize such a signal with a single everlasting exponential *ej*ω*<sup>t</sup>* . To synthesize this signal from everlasting exponentials, we need, in addition to an impulse at ω = 0, all the frequency components, as indicated by the term 1/*j*ω in Eq. (7.23).
-
-### **EXAMPLE 7.10 Fourier Transform of the Sign Function**
-
-Find the Fourier transform of the sign function sgn(*t*) [pronounced *signum* (*t*)], depicted in Fig. 7.16.
-
-*t* sgn (*t*) 1 0 -<sup>1</sup> **Figure 7.16** The signum function sgn(*t*). Observe that sgn(*t*)+1 = 2*u*(*t*) ⇒ sgn(*t*) = 2*u*(*t*)−1 Using Eqs. (7.20) and (7.23) and the linearity property, we obtain sgn(*t*) ⇐⇒ 2 *j*ω
-
-Table 7.1 provides many common Fourier transform pairs.
-
-| No. | x(t)                     | X(ω)                                                  |                    |
-|-----|--------------------------|-------------------------------------------------------|--------------------|
-| 1   | e−atu(t)                 | 1<br>a+jω                                             | a > 0              |
-| 2   | eatu(−t)                 | 1<br>a−jω                                             | a > 0              |
-| 3   | e−a t                    | 2a<br>a2 +ω2                                          | a > 0              |
-| 4   | te−atu(t)                | 1<br>(a+jω)2                                          | a > 0              |
-| 5   | ne−atu(t)<br>t           | n!<br>(a+jω)n+1                                       | a > 0              |
-| 6   | δ(t)                     | 1                                                     |                    |
-| 7   | 1                        | 2πδ(ω)                                                |                    |
-| 8   | ejω0t                    | 2πδ(ω −ω0)                                            |                    |
-| 9   | cos ω0t                  | π[δ(ω −ω0) +δ(ω +ω0)]                                 |                    |
-| 10  | sin ω0t                  | jπ[δ(ω +ω0)−δ(ω −ω0)]                                 |                    |
-| 11  | u(t)                     | 1<br>πδ(ω)+<br>jω                                     |                    |
-| 12  | sgnt                     | 2<br>jω                                               |                    |
-| 13  | cos ω0t u(t)             | π<br>jω<br>2 [δ(ω −ω0) +δ(ω +ω0)] +<br>ω2<br>0 −ω2    |                    |
-| 14  | sin ω0t u(t)             | π<br>ω0<br>[δ(ω −ω0)−δ(ω +ω0)] +<br>ω2<br>2j<br>0 −ω2 |                    |
-| 15  | e−atsin<br>ω0t u(t)      | ω0<br>(a+jω)2 +ω2<br>0                                | a > 0              |
-| 16  | e−at cos<br>ω0t u(t)     | a+jω<br>(a+jω)2 +ω2<br>0                              | a > 0              |
-| 17  | rect t<br><br>τ          | τ sincωτ<br><br>2                                     |                    |
-| 18  | W<br>π sinc(Wt)          | rect ω<br><br>2W                                      |                    |
-| 19  | t<br><br>τ               | τ<br>ωτ<br><br>sinc2<br>2<br>4                        |                    |
-| 20  | Wt<br>W<br>2π sinc2<br>2 | ω<br><br>2W                                           |                    |
-| 21  | "∞<br>δ(t −nT)           | "∞<br>ω0<br>δ(ω −nω0)                                 | 2π<br>ω0<br>=<br>T |
-| 22  | n=−∞<br>2/2σ2<br>e−t     | n=−∞<br>√2πe−σ2ω2/2<br>σ                              |                    |
-
-**TABLE 7.1** Select Fourier Transform Pairs
-
-### <span id="page-719-0"></span>**DR ILL 7.2 Inverse Fourier Transform of a Rectangular Pulse**
-
-Show that the inverse Fourier transform of *X*(ω) illustrated in Fig. 7.17 is *x*(*t*) = (ω0/π )sinc (ω0*t*). Sketch *x*(*t*).
-
-### **DR ILL 7.3 Fourier Transform of a General Sinusoid**
-
-Show that cos(ω0*t* +θ ) ⇐⇒ π[δ(ω +ω0)*e*−*j*<sup>θ</sup> +δ(ω −ω0)*ej*<sup>θ</sup> ].

engineering/linear-systems-and-signals-3rd-edition-bp-lathi/090_7.2 TRANSFORMS OF SOME USEFUL FUNCTIONS.md

@@ -1,20 +0,0 @@
-### **[7.2-1 Connection Between the Fourier and Laplace Transforms](#page-12-0)**
-
-The general (bilateral) Laplace transform of a signal *x*(*t*), according to Eq. (4.1), is
-
-$$
-X(s) = \int_{-\infty}^{\infty} x(t)e^{-st} dt
-$$
- (7.24)
-
-Setting *s* = *j*ω in this equation yields
-
-$$
-X(j\omega) = \int_{-\infty}^{\infty} x(t)e^{-j\omega t} dt
-$$
-
-where *X*(*j*ω) = *X*(*s*)|*<sup>s</sup>*=*j*ω. But, the right-hand-side integral defines *X*(ω), the Fourier transform of *x*(*t*). Does this mean that the Fourier transform can be obtained from the corresponding Laplace transform by setting *s* = *j*ω? In other words, is it true that *X*(*j*ω) = *X*(ω)? Yes and no. Yes, it is true in most cases. For example, when *x*(*t*) = *e*−*atu*(*t*), its Laplace transform is 1/(*s* + *a*), and *X*(*j*ω) = 1/(*j*ω +*a*), which is equal to *X*(ω) (assuming *a* < 0). However, for the unit step function *u*(*t*), the Laplace transform is
-
-$$
-u(t) \Longleftrightarrow \frac{1}{s} \qquad \text{Re}\, s > 0
-$$

engineering/linear-systems-and-signals-3rd-edition-bp-lathi/091_7.3 SOME PROPERTIES OF THE FOURIER TRANSFORM.md

@@ -1,608 +0,0 @@
-### 7.3 Some Properties of the Fourier Transform 701
-
-<span id="page-720-0"></span>The Fourier transform is given by
-
-$$
-u(t) \Longleftrightarrow \frac{1}{j\omega} + \pi \delta(\omega)
-$$
-
-Clearly, *X*(*j*ω) = *X*(ω) in this case.
-
-To understand this puzzle, consider the fact that we obtain *X*(*j*ω) by setting *s* = *j*ω in Eq. (7.24). This implies that the integral on the right-hand side of Eq. (7.24) converges for *s* = *j*ω, meaning that *s* = *j*ω (the imaginary axis) lies in the ROC for *X*(*s*). The general rule is that only when the ROC for *X*(*s*) includes the ω axis, does setting *s* = *j*ω in *X*(*s*) yield the Fourier transform *X*(ω), that is, *X*(*j*ω) = *X*(ω). This is the case of absolutely integrable *x*(*t*). If the ROC of *X*(*s*) excludes the ω axis, *X*(*j*ω) = *X*(ω). This is the case for exponentially growing *x*(*t*) and also *x*(*t*) that is constant or is oscillating with constant amplitude.
-
-The reason for this peculiar behavior has something to do with the nature of convergence of the Laplace and the Fourier integrals when *x*(*t*) is not absolutely integrable.†
-
-This discussion shows that although the Fourier transform may be considered as a special case of the Laplace transform, we need to circumscribe such a view. This fact can also be confirmed by noting that a periodic signal has the Fourier transform, but the Laplace transform does not exist.
-
-## **7.3 SOME [PROPERTIES OF THE](#page-12-0) FOURIER TRANSFORM**
-
-We now study some of the important properties of the Fourier transform and their implications as well as applications. We have already encountered two important properties, linearity [Eq. (7.15)] and the conjugation property [Eq. (7.11)].
-
-Before embarking on this study, we shall explain an important and pervasive aspect of the Fourier transform: the time-frequency duality.
-
-<sup>†</sup> To explain this point, consider the unit step function and its transforms. Both the Laplace and the Fourier transform synthesize *x*(*t*), using everlasting exponentials of the form *est*. The frequency *s* can be anywhere in the complex plane for the Laplace transform, but it must be restricted to the ω axis in the case of the Fourier transform. The unit step function is readily synthesized in the Laplace transform by a relatively simple spectrum *X*(*s*) = 1/*s*, in which the frequencies *s* are chosen in the RHP [the region of convergence for *u*(*t*) is Re *s* > 0]. In the Fourier transform, however, we are restricted to values of *s* on the ω axis only. The function *u*(*t*) can still be synthesized by frequencies along the ω axis, but the spectrum is more complicated than it is when we are free to choose the frequencies in the RHP. In contrast, when *x*(*t*) is absolutely integrable, the region of convergence for the Laplace transform includes the ω axis, and we can synthesize *x*(*t*) by using frequencies along the ω axis in both transforms. This leads to *X*(*j*ω) = *X*(ω).
-
-We may explain this concept by an example of two countries, X and Y. Suppose these countries want to construct similar dams in their respective territories. Country X has financial resources but not much manpower. In contrast, Y has considerable manpower but few financial resources. The dams will still be constructed in both countries, although the methods used will be different. Country X will use expensive but efficient equipment to compensate for its lack of manpower, whereas Y will use the cheapest possible equipment in a labor-intensive approach to the project. Similarly, both Fourier and Laplace integrals converge for *u*(*t*), but the makeup of the components used to synthesize *u*(*t*) will be very different for two cases because of the constraints of the Fourier transform, which are not present for the Laplace transform.
-
-### TIME-FREQUENCY DUALITY IN THE TRANSFORM OPERATIONS
-
-Equations (7.9) and (7.10) show an interesting fact: the direct and the inverse transform operations are remarkably similar. These operations, required to go from *x*(*t*) to *X*(ω) and then from *X*(ω) to *x*(*t*), are depicted graphically in Fig. 7.18. The inverse transform equation can be obtained from the direct transform equation by replacing *x*(*t*) with *X*(ω), *t* with ω, and ω with *t*. In a similar way, we can obtain the direct from the inverse. There are only two minor differences in these operations: the factor 2π appears only in the inverse operator, and the exponential indices in the two operations have opposite signs. Otherwise the two equations are duals of each other.†
-
-This observation has far-reaching consequences in the study of the Fourier transform. It is the basis of the so-called duality of time and frequency. *The duality principle may be compared with a photograph and its negative. A photograph can be obtained from its negative, and by using an identical procedure, a negative can be obtained from the photograph*. For any result or relationship between *x*(*t*) and *X*(ω), there exists a dual result or relationship, obtained by interchanging the roles of *x*(*t*) and *X*(ω) in the original result (along with some minor modifications arising because of the factor 2π and a sign change). For example, the time-shifting property, to be proved later, states that if *x*(*t*) ⇐⇒ *X*(ω), then
-
-$$
-x(t-t_0) \Longleftrightarrow X(\omega)e^{-j\omega t_0}
-$$
-
-**Figure 7.18** A near symmetry between the direct and the inverse Fourier transforms.
-
-$$
-X(2\pi f) = \int_{-\infty}^{\infty} x(t)e^{-j2\pi ft} dt \quad \text{and} \quad x(t) = \int_{-\infty}^{\infty} X(2\pi f)e^{j2\pi ft} df
-$$
-
-This leaves only one significant difference, that of sign change in the exponential index.
-
-<sup>†</sup> Of the two differences, the former can be eliminated by change of variable from ω to *f* (in hertz). In this case ω = 2π*f* and *d*ω = 2π *df* .
-
-Therefore, the direct and the inverse transforms are given by
-
-The dual of this property (the frequency-shifting property) states that
-
-$$
-x(t)e^{j\omega_0 t} \Longleftrightarrow X(\omega - \omega_0)
-$$
-
-Observe the role reversal of time and frequency in these two equations (with the minor difference of the sign change in the exponential index). The value of this principle lies in the fact that *whenever we derive any result, we can be sure that it has a dual.* This possibility can give valuable insights about many unsuspected properties or results in signal processing.
-
-The properties of the Fourier transform are useful not only in deriving the direct and inverse transforms of many functions, but also in obtaining several valuable results in signal processing. The reader should not fail to observe the ever-present duality in this discussion.
-
-### LINEARITY
-
-The linearity property, already introduced as Eq. (7.15), states that if *x*1(*t*) ⇐⇒ *X*1(ω) and *x*2(*t*) ⇐⇒ *X*2(ω), then *a*1*x*1(*t*)+*a*2*x*2(*t*) ⇐⇒ *a*1*X*1(ω)+*a*2*X*2(ω).
-
-### CONJUGATION AND CONJUGATE SYMMETRY
-
-The conjugation property, which has already been introduced, states that if *x*(*t*) ⇐⇒ *X*(ω), then
-
-$$
-x^*(t) \Longleftrightarrow X^*(-\omega)
-$$
-
-From this property follows the conjugate symmetry property, also introduced earlier, which states that if *x*(*t*) is real, then
-
-$$
-X(-\omega) = X^*(\omega)
-$$
-
-### DUALITY
-
-The duality property states that if
-
-$$
-x(t) \Longleftrightarrow X(\omega)
-$$
-
-then
-
-$$
-X(t) \Longleftrightarrow 2\pi x(-\omega) \tag{7.25}
-$$
-
-**Proof.** From Eq. (7.10) we can write
-
-$$
-x(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} X(u)e^{iut} du
-$$
-
-Hence,
-
-$$
-2\pi x(-t) = \int_{-\infty}^{\infty} X(u)e^{-jut} du
-$$
-
-Changing *t* to ω yields Eq. (7.25).
-
-*t*
-
-From Eq. (7.19) we have
-
-1
-
-$$
-\underbrace{\text{rect}\left(\frac{t}{\tau}\right)}_{x(t)} \Longleftrightarrow \underbrace{\tau \text{ sinc}\left(\frac{\omega \tau}{2}\right)}_{X(\omega)}
-$$
-
-Also, *X*(*t*) is the same as *X*(ω) with ω replaced by *t*, and *x*(−ω) is the same as *x*(*t*) with *t* replaced by −ω. Therefore, the duality property of Eq. (7.25) yields
-
-$$
-\underbrace{\tau \text{ sinc}\left(\frac{\tau t}{2}\right)}_{X(t)} \Longleftrightarrow \underbrace{2\pi \text{ rect}\left(\frac{-\omega}{\tau}\right)}_{2\pi x(-\omega)} = 2\pi \text{ rect}\left(\frac{\omega}{\tau}\right)
-$$
-
-In this result, we used the fact that rect(−*x*) = rect(*x*) because rect is an even function. Figure 7.19b shows this pair graphically. Observe the interchange of the roles of *t* and ω (with the minor adjustment of the factor 2π). This result appears as pair 18 in Table 7.1 (with τ/2 = *W*).
-
-As an interesting exercise, the reader should generate the dual of every pair in Table 7.1 by applying the duality property.
-
-## **DR ILL 7.4 Applying the Duality Property of the Fourier Transform**
-
-Apply the duality property to pairs 1, 3, and 9 (Table 7.1) to show that
-
-- **(a)** 1/(*jt* +*a*) ⇐⇒ 2π*ea*<sup>ω</sup>*u*(−ω)
-- **(b)** 2*a*/(*t* <sup>2</sup> +*a*2) ⇐⇒ 2π*e*−*a*|ω<sup>|</sup>
-- **(c)** δ(*t* +*t*0)+δ(*t* −*t*0) ⇐⇒ 2 cos *t*0ω
-
-### THE SCALING PROPERTY If
-
-*x*(*t*) ⇐⇒ *X*(ω)
-
-then, for any real constant *a*,
-
-$$
-x(at) \Longleftrightarrow \frac{1}{|a|}X\left(\frac{\omega}{a}\right) \tag{7.26}
-$$
-
-**Proof.** For a positive real constant *a*,
-
-$$
-\mathcal{F}[x(at)] = \int_{-\infty}^{\infty} x(at)e^{-j\omega t}dt = \frac{1}{a}\int_{-\infty}^{\infty} x(u)e^{(-j\omega/a)u}du = \frac{1}{a}X\left(\frac{\omega}{a}\right)
-$$
-
-Similarly, we can demonstrate that if *a* < 0,
-
-$$
-x(at) \Longleftrightarrow \frac{-1}{a}X\left(\frac{\omega}{a}\right)
-$$
-
-Hence follows Eq. (7.26).
-
-### SIGNIFICANCE OF THE SCALING PROPERTY
-
-The function *x*(*at*) represents the function *x*(*t*) compressed in time by a factor *a* (see Sec. 1.2-2). Similarly, a function *X*(ω/*a*) represents the function *X*(ω) expanded in frequency by the same factor *a*. *The scaling property states that time compression of a signal results in its spectral expansion, and time expansion of the signal results in its spectral compression*. Intuitively, compression in time by factor *a* means that the signal is varying faster by factor *a*. † To synthesize such a signal, the frequencies of its sinusoidal components must be increased by the factor *a*, implying that its frequency spectrum is expanded by the factor *a*. Similarly, a signal expanded in time varies more slowly; hence the frequencies of its components are lowered, implying that its frequency spectrum is compressed. For instance, the signal cos 2ω0*t* is the same as the signal cosω0*t* time-compressed by a factor of 2. Clearly, the spectrum of the former (impulse at ±2ω0) is an expanded version of the spectrum of the latter (impulse at ±ω0). The effect of this scaling is demonstrated in Fig. 7.20.
-
-<sup>†</sup> We are assuming *a* > 1, although the argument still holds if *a* < 1. In the latter case, compression becomes expansion by factor 1/*a*, and vice versa.
-
-**Figure 7.20** The scaling property of the Fourier transform.
-
-### RECIPROCITY OF SIGNAL DURATION AND ITS BANDWIDTH
-
-The scaling property implies that if *x*(*t*) is wider, its spectrum is narrower, and vice versa. Doubling the signal duration halves its bandwidth, and vice versa. This suggests that the bandwidth of a signal is inversely proportional to the signal duration or width (in seconds).† We have already verified this fact for the gate pulse, where we found that the bandwidth of a gate pulse of width τ seconds is 1/τ Hz. More discussion of this interesting topic can be found in the literature [2].
-
-By letting *a* = −1 in Eq. (7.26), we obtain the *inversion (or reflection) property of time and frequency:*
-
-$$
-x(-t) \Longleftrightarrow X(-\omega) \tag{7.27}
-$$
-
-### **EXAMPLE 7.12 Fourier Transform Reflection Property**
-
-Using the reflection property of the Fourier transform and Table 7.1, find the Fourier transforms of *eatu*(−*t*) and *e*−*a*|*t*<sup>|</sup> .
-
-Application of Eq. (7.27) to pair 1 of Table 7.1 yields
-
-$$
-e^{at}u(-t) \Longleftrightarrow \frac{1}{a-j\omega} \qquad a > 0
-$$
-
-Also,
-
-$$
-e^{-a|t|} = e^{-at}u(t) + e^{at}u(-t)
-$$
-
-<sup>†</sup> When a signal has infinite duration, we must consider its effective or equivalent duration. There is no unique definition of effective signal duration. One possible definition is given in Eq. (2.47).
-
-Therefore,
-
-$$
-e^{-a|t|} \Longleftrightarrow \frac{1}{a+j\omega} + \frac{1}{a-j\omega} = \frac{2a}{a^2 + \omega^2} \qquad a > 0 \tag{7.28}
-$$
-
-The signal *e*−*a*|*t*<sup>|</sup> and its spectrum are illustrated in Fig. 7.21.
-
-## THE TIME-SHIFTING PROPERTY If
-
-*x*(*t*) ⇐⇒ *X*(ω)
-
-then
-
-$$
-x(t - t_0) \Longleftrightarrow X(\omega)e^{-j\omega t_0} \tag{7.29}
-$$
-
-**Proof.** By definition,
-
-$$
-\mathcal{F}[x(t-t_0)] = \int_{-\infty}^{\infty} x(t-t_0)e^{-j\omega t} dt
-$$
-
-Letting *t* −*t*<sup>0</sup> = *u*, we have
-
-$$
-\mathcal{F}[x(t-t_0)] = \int_{-\infty}^{\infty} x(u)e^{-j\omega(u+t_0)} du = e^{-j\omega t_0} \int_{-\infty}^{\infty} x(u)e^{-j\omega u} du = X(\omega)e^{-j\omega t_0}
-$$
-
-This result shows that *delaying a signal by t*<sup>0</sup> *seconds does not change its amplitude spectrum. The phase spectrum, however, is changed by* −ω*t*0.
-
-### PHYSICAL EXPLANATION OF THE LINEAR PHASE
-
-Time delay in a signal causes a linear phase shift in its spectrum. This result can also be derived by heuristic reasoning. Imagine *x*(*t*) being synthesized by its Fourier components, which are sinusoids of certain amplitudes and phases. The delayed signal *x*(*t* − *t*0) can be synthesized by the same sinusoidal components, each delayed by *t*<sup>0</sup> seconds. The amplitudes of the components remain unchanged. Therefore, the amplitude spectrum of *x*(*t* − *t*0) is identical to that of *x*(*t*). The time delay of *t*<sup>0</sup> in each sinusoid, however, does change the phase of each component. Now, a sinusoid
-
-**Figure 7.22** Physical explanation of the time-shifting property.
-
-cosω*t* delayed by *t*<sup>0</sup> is given by
-
-$$
-\cos \omega (t - t_0) = \cos (\omega t - \omega t_0)
-$$
-
-Therefore a time delay *t*<sup>0</sup> in a sinusoid of frequency ω manifests as a phase delay of ω*t*0. This is a linear function of ω, meaning that higher-frequency components must undergo proportionately higher phase shifts to achieve the same time delay. This effect is depicted in Fig. 7.22 with two sinusoids, the frequency of the lower sinusoid being twice that of the upper. The same time delay *t*<sup>0</sup> amounts to a phase shift of π/2 in the upper sinusoid and a phase shift of π in the lower sinusoid. This verifies the fact that *to achieve the same time delay, higher-frequency sinusoids must undergo proportionately higher phase shifts*. The principle of linear phase shift is very important, and we shall encounter it again in distortionless signal transmission and filtering applications.
-
-### **EXAMPLE 7.13 Fourier Transform Time-Shifting Property**
-
-Use the time-shifting property to find the Fourier transform of *e*−*a*|*t*−*t*0<sup>|</sup> .
-
-This function, shown in Fig. 7.23a, is a time-shifted version of *e*−*a*|*t*<sup>|</sup> (depicted in Fig. 7.21a). From Eqs. (7.28) and (7.29), we have
-
-$$
-e^{-a|t-t_0|} \Longleftrightarrow \frac{2a}{a^2 + \omega^2} e^{-j\omega t_0}
-$$
-
-The spectrum of *e*−*a*|*t*−*t*0<sup>|</sup> (Fig. 7.23b) is the same as that of *e*−*a*|*t*<sup>|</sup> (Fig. 7.21b), except for an added phase shift of −ω*t*0.
-
-### **EXAMPLE 7.14 Fourier Transform of a Time-Shifted Rectangular Pulse**
-
-Find the Fourier transform of the time-shifted rectangular pulse *x*(*t*) illustrated in Fig. 7.24a.
-
-The pulse *x*(*t*) is the gate pulse rect(*t*/τ ) in Fig. 7.10a delayed by 3τ/4 seconds. Hence, according to Eq. (7.29), its Fourier transform is the Fourier transform of rect(*t*/τ ) multiplied by *e*−*j*ω(3τ /4) . Therefore,
-
-$$
-X(\omega) = \tau \operatorname{sinc}\left(\frac{\omega \tau}{2}\right) e^{-j\omega(3\tau/4)}
-$$
-
-The amplitude spectrum |*X*(ω)| (depicted in Fig. 7.24b) of this pulse is the same as that indicated in Fig. 7.10c. But the phase spectrum has an added linear term −3ωτ/4. Hence, the phase spectrum of *x*(*t*) (Fig. 7.24a) is identical to that in Fig. 7.10d plus a linear term −3ωτ/4, as shown in Fig. 7.24c.
-
-### PHASE SPECTRUM USING PRINCIPAL VALUES
-
-There is an alternate way of spectral representation of *X*(ω). The phase angle computed on a calculator or by using a computer subroutine is generally the principal value (modulo 2π value) of the phase angle, which always lies in the range −π to π. For instance, the principal value of angle 3π/2 is −π/2, and so on. The principal value differs from the actual value by ±2π radians (and its integer multiples) in a way that ensures that the principal value remains within −π to π. Thus, the principal value will show jump discontinuities of ±2π whenever the actual
-
-### 710 CHAPTER 7 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM
-
-phase crosses ±π. The phase plot in Fig. 7.24c is redrawn in Fig. 7.24d using the principal value for the phase. This phase pattern, which contains phase discontinuities of magnitudes 2π and π, becomes repetitive at intervals of ω = 8π/τ .
-
-## **DR ILL 7.5 Fourier Transform Time-Shifting Property**
-
-Use pair 18 of Table 7.1 and the time-shifting property to show that the Fourier transform of sinc [ω0(*t* − *T*)] is (π/ω0)rect(ω/2ω0)*e*−*j*ω*<sup>T</sup>* . Sketch the amplitude and phase spectra of the Fourier transform.
-
-### THE FREQUENCY-SHIFTING PROPERTY If
-
-$$
-x(t) \Longleftrightarrow X(\omega)
-$$
-
-then
-
-$$
-x(t)e^{j\omega_0 t} \Longleftrightarrow X(\omega - \omega_0) \tag{7.30}
-$$
-
-**Proof.** By definition,
-
-$$
-\mathcal{F}[x(t)e^{j\omega_0 t}] = \int_{-\infty}^{\infty} x(t)e^{j\omega_0 t}e^{-j\omega t} dt = \int_{-\infty}^{\infty} x(t)e^{-j(\omega - \omega_0)t} dt = X(\omega - \omega_0)
-$$
-
-According to this property, the multiplication of a signal by a factor *ej*<sup>ω</sup>0*<sup>t</sup>* shifts the spectrum of that signal by ω = ω0. Note the duality between the time-shifting and the frequency-shifting properties.
-
-Changing ω<sup>0</sup> to −ω<sup>0</sup> in Eq. (7.30) yields
-
-$$
-x(t)e^{-j\omega_0 t} \Longleftrightarrow X(\omega + \omega_0) \tag{7.31}
-$$
-
-Because *ej*<sup>ω</sup>0*<sup>t</sup>* is not a real function that can be generated, frequency shifting in practice is achieved by multiplying *x*(*t*) by a sinusoid. Observe that
-
-$$
-x(t)\cos \omega_0 t = \frac{1}{2} [x(t)e^{j\omega_0 t} + x(t)e^{-j\omega_0 t}]
-$$
-
-From Eqs. (7.30) and (7.31), it follows that
-
-$$
-x(t)\cos\omega_0 t \Longleftrightarrow \frac{1}{2}[X(\omega - \omega_0) + X(\omega + \omega_0)]\tag{7.32}
-$$
-
-This result shows that the multiplication of a signal *x*(*t*) by a sinusoid of frequency ω<sup>0</sup> shifts the spectrum *X*(ω) by ±ω0, as depicted in Fig. 7.25.
-
-Multiplication of a sinusoid cosω0*t* by *x*(*t*) amounts to modulating the sinusoid amplitude. This type of modulation is known as *amplitude modulation*. The sinusoid cosω0*t* is called the *carrier,* the signal *x*(*t*) is the *modulating signal*, and the signal *x*(*t*) cosω0*t* is the *modulated signal*. Further discussion of modulation and demodulation appears in Sec. 7.7.
-
-To sketch a signal *x*(*t*) cos ω0*t*, we observe that
-
-$$
-x(t)\cos\omega_0 t = \begin{cases} x(t) & \text{when } \cos\omega_0 t = 1\\ -x(t) & \text{when } \cos\omega_0 t = -1 \end{cases}
-$$
-
-Therefore, *x*(*t*) cos ω0*t* touches *x*(*t*) when the sinusoid cos ω0*t* is at its positive peaks and touches −*x*(*t*) when cos ω0*t* is at its negative peaks. This means that *x*(*t*) and −*x*(*t*) act as envelopes for the signal *x*(*t*) cos ω0*t* (see Fig. 7.25). The signal −*x*(*t*) is a mirror image of *x*(*t*) about the horizontal axis. Figure 7.25 shows the signals *x*(*t*) and *x*(*t*) cos ω0*t* and their spectra.
-
-**Figure 7.25** Amplitude modulation of a signal causes spectral shifting.
-
-### **EXAMPLE 7.15 Spectral Shifting by Amplitude Modulation**
-
-Find and sketch the Fourier transform of the modulated signal *x*(*t*) cos 10*t* in which *x*(*t*) is a gate pulse rect(*t*/4), as illustrated in Fig. 7.26a.
-
-From pair 17 of Table 7.1, we find rect(*t*/4) ⇐⇒ 4 sinc (2ω), which is depicted in Fig. 7.26b. From Eq. (7.32) it follows that
-
-$$
-x(t)\cos 10t \Longleftrightarrow \frac{1}{2}[X(\omega+10) + X(\omega-10)]
-$$
-
-In this case, *X*(ω) = 4 sinc (2ω). Therefore,
-
-*x*(*t*) cos 10*t* ⇐⇒ 2 sinc [2(ω +10)] +2 sinc [2(ω −10)]
-
-The spectrum (Fig. 7.26c) of *x*(*t*) cos 10*t* is obtained by shifting *X*(ω) in Fig. 7.26b to the left by 10 and also to the right by 10, and then multiplying it by 0.5, as depicted in Fig. 7.26d.
-
-## **DR ILL 7.6 Fourier Transform of an Amplitude-Modulated Signal**
-
-Sketch signal *e*−|*t*<sup>|</sup> cos 10*t*. Find the Fourier transform of this signal and sketch its spectrum. **Answer:** *<sup>X</sup>*(ω) <sup>=</sup> <sup>1</sup> (ω−10)2+<sup>1</sup> <sup>+</sup> <sup>1</sup> (ω+10)2+<sup>1</sup> . See Fig. 7.21b for the spectrum of *<sup>e</sup>*−*a*|*t*<sup>|</sup> .
-
-## **DR ILL 7.7 Amplitude Modulation Using a Phase-Shifted Carrier**
-
-Show that
-
-```
-x(t) cos(ω0t +θ ) ⇐⇒ 1
-                     2
-
-                       X(ω −ω0)ejθ +X(ω +ω0)e−jθ
-```
-
-### APPLICATIONS OF MODULATION
-
-Modulation is used to shift signal spectra. Some of the situations that call for spectrum shifting are presented next.
-
-1. If several signals, all occupying the same frequency band, are transmitted simultaneously over the same transmission medium, they will all interfere; it will be impossible to separate or retrieve them at a receiver. For example, if all radio stations decide to broadcast audio signals simultaneously, a receiver will not be able to separate them. This problem is solved
-
-Old is gold, but sometimes it is fool's gold.
-
-by using modulation, whereby each radio station is assigned a distinct carrier frequency. Each station transmits a modulated signal. This procedure shifts the signal spectrum to its allocated band, which is not occupied by any other station. A radio receiver can pick up any station by tuning to the band of the desired station. The receiver must now demodulate the received signal (undo the effect of modulation). Demodulation therefore consists of another spectral shift required to restore the signal to its original band. Note that both modulation and demodulation implement spectral shifting; consequently, demodulation operation is similar to modulation (see Sec. 7.7).
-
-This method of transmitting several signals simultaneously over a channel by sharing its frequency band is known as *frequency-division multiplexing (FDM)*.
-
-2. For effective radiation of power over a radio link, the antenna size must be of the order of the wavelength of the signal to be radiated. Audio signal frequencies are so low (wavelengths are so large) that impracticably large antennas would be required for radiation. Here, shifting the spectrum to a higher frequency (a smaller wavelength) by modulation solves the problem.
-
-### CONVOLUTION
-
-The time-convolution property and its dual, the frequency-convolution property, state that if
-
-$$
-x_1(t) \Longleftrightarrow X_1(\omega) \quad \text{and} \quad x_2(t) \Longleftrightarrow X_2(\omega)
-$$
-
-then
-
-$$
-x_1(t) * x_2(t) \Longleftrightarrow X_1(\omega) X_2(\omega) \quad \text{(time convolution)} \tag{7.33}
-$$
-
-### 7.3 Some Properties of the Fourier Transform 715
-
-and
-
-$$
-x_1(t)x_2(t) \Longleftrightarrow \frac{1}{2\pi}X_1(\omega) * X_2(\omega) \quad \text{(frequency convolution)} \tag{7.34}
-$$
-
-**Proof.** By definition,
-
-$$
-\mathcal{F}|x_1(t) * x_2(t)| = \int_{-\infty}^{\infty} e^{-j\omega t} \left[ \int_{-\infty}^{\infty} x_1(\tau) x_2(t-\tau) d\tau \right] dt
-$$
-$$
-= \int_{-\infty}^{\infty} x_1(\tau) \left[ \int_{-\infty}^{\infty} e^{-j\omega t} x_2(t-\tau) d\tau \right] d\tau
-$$
-
-The inner integral is the Fourier transform of *x*2(*t* − τ ), given by [time-shifting property in Eq. (7.29)] *X*2(ω)*e*−*j*ωτ . Hence,
-
-$$
-\mathcal{F}[x_1(t) * x_2(t)] = \int_{-\infty}^{\infty} x_1(\tau) e^{-j\omega \tau} X_2(\omega) d\tau = X_2(\omega) \int_{-\infty}^{\infty} x_1(\tau) e^{-j\omega \tau} d\tau = X_1(\omega) X_2(\omega)
-$$
-
-Let *H*(ω) be the Fourier transform of the unit impulse response *h*(*t*), that is,
-
-*h*(*t*) ⇐⇒ *H*(ω)
-
-Application of the time-convolution property to *y*(*t*) = *x*(*t*) ∗ *h*(*t*) yields [assuming that both *x*(*t*) and *h*(*t*) are Fourier transformable]
-
-$$
-Y(\omega) = X(\omega)H(\omega) \tag{7.35}
-$$
-
-The frequency-convolution property of Eq. (7.34) can be proved in exactly the same way by reversing the roles of *x*(*t*) and *X*(ω).
-
-### **EXAMPLE 7.16 Time-Convolution Property to Show the Time-Integration Property**
-
-Use the time-convolution property to show that if
-
-$$
-x(t) \Longleftrightarrow X(\omega)
-$$
-
-then
-
-$$
-\int_{-\infty}^{t} x(\tau) d\tau \Longleftrightarrow \frac{X(\omega)}{j\omega} + \pi X(0)\delta(\omega)
-$$
-
-Because
-
-$$
-u(t-\tau) = \begin{cases} 1 & \tau \leq t \\ 0 & \tau > t \end{cases}
-$$
-
-it follows that
-
-$$
-x(t) * u(t) = \int_{-\infty}^{\infty} x(\tau)u(t-\tau) d\tau = \int_{-\infty}^{t} x(\tau) d\tau
-$$
-
-#### 716 CHAPTER 7 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM
-
-Now, from the time-convolution property [Eq. (7.33)], it follows that
-
-$$
-x(t) * u(t) = \int_{-\infty}^{t} x(\tau) d\tau \Longleftrightarrow X(\omega) \left[ \frac{1}{j\omega} + \pi \delta(\omega) \right]
-$$
-$$
-= \frac{X(\omega)}{j\omega} + \pi X(0) \delta(\omega)
-$$
-
-In deriving the last result, we used Eq. (1.10).
-
-### **DR ILL 7.8 Fourier Transform Time-Convolution Property**
-
-Use the time-convolution property to show that:
-
-$$
-(a) x(t) * \delta(t) = x(t)
-$$
-
-**(b)** *<sup>e</sup>*−*atu*(*t*)∗*e*−*btu*(*t*) <sup>=</sup> <sup>1</sup> *<sup>b</sup>*−*<sup>a</sup>* [*e*−*at* <sup>−</sup>*e*−*bt*]*u*(*t*)
-
-## TIME DIFFERENTIATION AND TIME INTEGRATION If
-
-$$
-x(t) \Longleftrightarrow X(\omega)
-$$
-
-then†
-
-$$
-\frac{dx(t)}{dt} \Longleftrightarrow j\omega X(\omega) \quad \text{(time differentiation)} \tag{7.36}
-$$
-
-and
-
-$$
-\int_{-\infty}^{t} x(\tau) d\tau \Longleftrightarrow \frac{X(\omega)}{j\omega} + \pi X(0)\delta(\omega) \quad \text{(time integration)} \tag{7.37}
-$$
-
-**Proof.** Differentiation of both sides of Eq. (7.10) yields
-
-$$
-\frac{dx(t)}{dt} = \frac{1}{2\pi} \int_{-\infty}^{\infty} j\omega X(\omega) e^{j\omega t} d\omega
-$$
-
-$$
-\int_{-\infty}^{\infty} \left| \frac{dx(t)}{dt} \right| dt < \infty
-$$
-
-<sup>†</sup> Valid only if the transform of *dx*/*dt* exists. In other words, *dx*/*dt* must satisfy the Dirichlet conditions. The first Dirichlet condition implies
-
-We also require that *x*(*t*) → 0 as *t* → ±∞. Otherwise, *x*(*t*) has a dc component, which gets lost in differentiation, and there is no one-to-one relationship between *x*(*t*) and *dx*/*dt*.
-
-| Operation                       | x(t)                | X(ω)                        |
-|---------------------------------|---------------------|-----------------------------|
-| Scalar multiplication           | kx(t)               | kX(ω)                       |
-| Addition                        | x1(t)+x2(t)         | X1(ω) +X2(ω)                |
-| Conjugation                     | x∗(t)               | X∗(−ω)                      |
-| Duality                         | X(t)                | 2πx(−ω)                     |
-| Scaling (a real)                | x(at)               | 1<br>ω<br><br>X<br> a <br>a |
-| Time shifting                   | x(t −t0)            | X(ω)e−jωt0                  |
-| Frequency shifting (ω0<br>real) | x(t)ejω0t           | X(ω −ω0)                    |
-| Time convolution                | x1(t)∗x2(t)         | X1(ω)X2(ω)                  |
-| Frequency convolution           | x1(t)x2(t)          | 1<br>2π X1(ω)∗X2(ω)         |
-| Time differentiation            | dnx(t)<br>dtn       | (jω)nX(ω)                   |
-| Time integration                | # t<br>x(u)du<br>−∞ | X(ω)<br>+πX(0)δ(ω)<br>jω    |
-
-**TABLE 7.2** Fourier Transform Properties
-
-This result shows that
-
-$$
-\frac{dx(t)}{dt} \Longleftrightarrow j\omega X(\omega)
-$$
-
-Repeated application of this property yields
-
-$$
-\frac{d^n x(t)}{dt^n} \Longleftrightarrow (j\omega)^n X(\omega)
-$$
-
-The time-integration property [Eq. (7.37)] has already been proved in Ex. 7.16. Table 7.2 summarizes the most important properties of the Fourier transform.
-
-### **EXAMPLE 7.17 Fourier Transform Time-Differentiation Property**
-
-Use the time-differentiation property to find the Fourier transform of the triangle pulse (*t*/τ ) illustrated in Fig. 7.27a. Verify the correctness of the spectrum by using it to synthesize a periodic replication of the original time-domain signal with τ = 1.
-
-**Figure 7.27** Finding the Fourier transform of a piecewise-linear signal using the time-differentiation property.
-
-To find the Fourier transform of this pulse, we differentiate the pulse successively, as illustrated in Fig. 7.27b and 7.27c. Because *dx*/*dt* is constant everywhere, its derivative, *d*<sup>2</sup>*x*/*dt*2, is zero everywhere. But *dx*/*dt* has jump discontinuities with a positive jump of 2/τ at *t* = ±τ/2, and a negative jump of 4/τ at *t* = 0. Recall that the derivative of a signal at a jump discontinuity is an impulse at that point of strength equal to the amount of jump. Hence, *d*<sup>2</sup>*x*/*dt*2, the derivative of *dx*/*dt*, consists of a sequence of impulses, as depicted in Fig. 7.27c; that is,
-
-$$
-\frac{d^2x(t)}{dt^2} = \frac{2}{\tau} \left[ \delta \left( t + \frac{\tau}{2} \right) - 2\delta(t) + \delta \left( t - \frac{\tau}{2} \right) \right]
-$$
-
-From the time-differentiation property [Eq. (7.36)],
-
-$$
-\frac{d^2x(t)}{dt^2} \Longleftrightarrow (j\omega)^2 X(\omega) = -\omega^2 X(\omega)
-$$
-
-Also, from the time-shifting property [Eq. (7.29)],
-
-$$
-\delta(t-t_0) \Longleftrightarrow e^{-j\omega t_0}
-$$
-
-Combining these results, we obtain
-
-$$
--\omega^2 X(\omega) = \frac{2}{\tau} \left[ e^{j(\omega \tau/2)} - 2 + e^{-j(\omega \tau/2)} \right] = \frac{4}{\tau} \left( \cos \frac{\omega \tau}{2} - 1 \right) = -\frac{8}{\tau} \sin^2 \left( \frac{\omega \tau}{4} \right)
-$$
-
-and
-
-$$
-X(\omega) = \frac{8}{\omega^2 \tau} \sin^2\left(\frac{\omega \tau}{4}\right) = \frac{\tau}{2} \left[ \frac{\sin\left(\frac{\omega \tau}{4}\right)}{\frac{\omega \tau}{4}} \right]^2 = \frac{\tau}{2} \text{sinc}^2\left(\frac{\omega \tau}{4}\right)
-$$
-
-The spectrum *X*(ω) is depicted in Fig. 7.27d. This procedure of finding the Fourier transform can be applied to any function *x*(*t*) made up of straight-line segments with *x*(*t*) → 0 as |*t*|→∞. The second derivative of such a signal yields a sequence of impulses whose Fourier transform can be found by inspection. This example suggests a numerical method of finding the Fourier transform of an arbitrary signal *x*(*t*) by approximating the signal by straight-line segments.
-
-### SYNTHESIZING A PERIODIC REPLICATION TO VERIFY SPECTRUM CORRECTNESS
-
-While a signal's spectrum *X*(ω) provides useful insight into signal character, it can be difficult to look at *X*(ω) and know that it is correct for a particular signal *x*(*t*). Is it obvious, for example, that *<sup>X</sup>*(ω) <sup>=</sup> <sup>τ</sup> <sup>2</sup> sinc<sup>2</sup> (ωτ/4) is really the spectrum of a τ -duration rectangle function? Or is it possible that a mathematical error was made in the determination of *X*(ω)? It is difficult to be certain by simple inspection of the spectrum.
-
-The same uncertainties exist when we are looking at a periodic signal's Fourier series spectrum. In the Fourier series case, we can verify the correctness of a signal's spectrum by synthesizing *x*(*t*) with a truncated Fourier series; the synthesized signal will match the original only if the computed spectrum is correct. This is exactly the approach that was taken in Ex. 6.11. And since a truncated Fourier series involves a simple sum, tools like MATLAB make waveform synthesis relatively simple, at least in the case of the Fourier series.
-
-In the case of the Fourier transform, however, synthesis of *x*(*t*) using Eq. (7.10) requires integration, a task not well suited to numerical packages such as MATLAB. All is not lost, however. Consider Eq. (7.5). By scaling and sampling the spectrum *X*(ω) of an aperiodic signal *x*(*t*), we obtain the Fourier series coefficient of a signal that is the periodic replication of *x*(*t*). Similar to Ex. 6.11, we can then synthesize a periodic replication of *x*(*t*) with a truncated Fourier series to verify spectrum correctness. Let us demonstrate the idea for the current example with τ = 1.
-
-To begin, we represent *<sup>X</sup>*(ω) <sup>=</sup> <sup>τ</sup> <sup>2</sup> sinc<sup>2</sup> (ωτ/4) using an anonymous function in MATLAB. Since MATLAB computes sinc(x) as (sin(π*x*))/π*x*, we must scale the input by 1/π to match the notation of sinc in this book.
-
->> tau = 1; X = @(omega) tau/2\*(sinc(omega\*tau/(4\*pi))).^2;
-
-For our periodic replication, let us pick *T*<sup>0</sup> = 2, which is comfortably wide enough to accommodate our (τ = 1)-width function without overlap. We use Eq. (7.5) to define the needed Fourier series coefficients *Dn*.
-
->> TO = 2; omega0 = 
-$$
-2*pi/TO
-$$
-; D =  $\mathcal{Q}(n)$  X(n\*omega0)/TO;
-
-Let us use 25 harmonics to synthesize the periodic replication *x*25(*t*) of our triangular signal *x*(*t*). To begin waveform synthesis, we set the dc portion of the signal.
-
->> t = (-T0:.001:T0); x25 = D(0)\*ones(size(t));
-
-To add the desired 25 harmonics, we enter a loop for 1 ≤ *n* ≤ 25 and add in the *Dn* and *D*−*<sup>n</sup>* terms. Although the result should be real, small round-off errors cause the reconstruction to be complex. These small imaginary parts are removed by using the real command.
-
->> for n = 1:25, >> x25 = x25+real(D(n)\*exp(1j\*omega0\*n\*t)+D(-n)\*exp(-1j\*omega0\*n\*t)); >> end
-
-Lastly, we plot the resulting truncated Fourier series synthesis of *x*(*t*).
-
->> plot(t,x25,'k'); xlabel('t'); ylabel('x\_{25}(t)');
-
-Since the synthesized waveform shown in Fig. 7.28 closely matches a 2-periodic replication of the triangle wave in Fig. 7.27a, we have high confidence that both the computed *Dn* and, by extension, the Fourier spectrum *X*(ω) are correct.
-
-**Figure 7.28** Synthesizing a 2-periodic replication of *x*(*t*) using a truncated Fourier series.
-
-### <span id="page-740-0"></span>**DR ILL 7.9 Fourier Transform Time-Differentiation Property**
-
-Use the time-differentiation property to find the Fourier transform of rect(*t*/τ ).

engineering/linear-systems-and-signals-3rd-edition-bp-lathi/092_7.4 SIGNAL TRANSMISSION THROUGH LTIC SYSTEMS.md

@@ -1,282 +0,0 @@
-## **7.4 SIGNAL [TRANSMISSION](#page-12-0) THROUGH LTIC SYSTEMS**
-
-If *x*(*t*) and *y*(*t*) are the input and output of an LTIC system with impulse response *h*(*t*), then, as demonstrated in Eq. (7.35),
-
-$$
-Y(\omega) = H(\omega)X(\omega)
-$$
-
-This equation does not apply to (asymptotically) unstable systems because *h*(*t*) for such systems is not Fourier transformable. It applies to BIBO-stable as well as most of the marginally stable systems.† Similarly, this equation does not apply if *x*(*t*) is not Fourier transformable.
-
-In Ch. 4, we saw that the Laplace transform is more versatile and capable of analyzing all kinds of LTIC systems whether stable, unstable, or marginally stable. Laplace transform can also handle exponentially growing inputs. In comparison to the Laplace transform, the Fourier transform in system analysis is not just clumsier, but also very restrictive. Hence, the Laplace transform is preferable to the Fourier transform in LTIC system analysis. We shall not belabor the application of the Fourier transform to LTIC system analysis. We consider just one example here.
-
-### **EXAMPLE 7.18 Fourier Transform to Determine the Zero-State Response**
-
-Use the Fourier transform to find the zero-state response of a stable LTIC system with frequency response
-
-$$
-H(s) = \frac{1}{s+2}
-$$
-
-and the input is *x*(*t*) = *e*−*<sup>t</sup> u*(*t*). Stability implies that the region of convergence of *H*(*s*) includes the ω axis.
-
-In this case,
-
-$$
-X(\omega) = \frac{1}{j\omega + 1}
-$$
-
-<sup>†</sup> For marginally stable systems, if the input *x*(*t*) contains a finite-amplitude sinusoid of the system's natural frequency, which leads to resonance, the output is not Fourier transformable. It does, however, apply to marginally stable systems if the input does not contain a finite-amplitude sinusoid of the system's natural frequency.
-
-### 722 CHAPTER 7 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM
-
-Moreover, because the system is stable, the frequency response *H*(*j*ω) = *H*(ω). Hence,
-
-$$
-H(\omega) = H(s)|_{s=j\omega} = \frac{1}{j\omega + 2}
-$$
-
-Therefore,
-
-$$
-Y(\omega) = H(\omega)X(\omega) = \frac{1}{(j\omega + 2)(j\omega + 1)}
-$$
-
-Expanding the right-hand side in partial fractions yields
-
-$$
-Y(\omega) = \frac{1}{j\omega + 1} - \frac{1}{j\omega + 2}
-$$
-
-and
-
-$$
-y(t) = (e^{-t} - e^{-2t})u(t)
-$$
-
-## **DR ILL 7.10 Fourier Transform to Determine the Zero-State Response**
-
-For the system in Ex. 7.18, show that the zero-input response to the input *et <sup>u</sup>*(−*t*) is *<sup>y</sup>*(*t*) <sup>=</sup> <sup>1</sup> <sup>3</sup> [*et u*(−*t*)+*e*−2*<sup>t</sup> u*(*t*)]. [*Hint:* Use pair 2 (Table 7.1) to find the Fourier transform of *et u*(−*t*).]
-
-### HEURISTIC UNDERSTANDING OF LINEAR SYSTEM RESPONSE
-
-In finding the linear system response to arbitrary input, the time-domain method uses convolution integral and the frequency-domain method uses the Fourier integral. Despite the apparent dissimilarities of the two methods, their philosophies are amazingly similar. In the time-domain case, we express the input *x*(*t*) as a sum of its impulse components; in the frequency-domain case, the input is expressed as a sum of everlasting exponentials (or sinusoids). In the former case, the response *y*(*t*) obtained by summing the system's responses to impulse components results in the convolution integral; in the latter case, the response obtained by summing the system's response to everlasting exponential components results in the Fourier integral. These ideas can be expressed mathematically as follows:
-
-1. For the time-domain case,
-
-| δ(t)<br>⇒ h(t)                             | shows the system response<br>δ(t)<br>h(t)<br>to<br>is the impulse response               |
-|--------------------------------------------|------------------------------------------------------------------------------------------|
-| = \$ ∞<br>x(t)<br>x(τ )δ(t<br>−τ )dτ<br>−∞ | expresses<br>x(t)<br>as a sum<br>of impulse components                                   |
-| = \$ ∞<br>y(t)<br>x(τ )h(t<br>−τ )dτ<br>−∞ | expresses<br>y(t)<br>as a sum of responses to<br>the impulse components of input<br>x(t) |
-
-<span id="page-742-0"></span>2. For the frequency-domain case,
-
-$$
-e^{j\omega t} \implies H(\omega)e^{j\omega t} \qquad \text{shows the system response}
-$$
-  
-\nto  $e^{j\omega t}$  is  $H(\omega)e^{j\omega t}$   
-\n
-$$
-x(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} X(\omega)e^{j\omega t} d\omega \qquad \text{expresses } x(t) \text{ as a sum}
-$$
-  
-\nof everyday exponential components  
-\n
-$$
-y(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} X(\omega)H(\omega)e^{j\omega t} d\omega \qquad \text{expresses } y(t) \text{ as a sum of responses to}
-$$
-  
-\nthe exponential components of input  $x(t)$ 
-
-The frequency-domain view sees a system in terms of its frequency response (system response to various sinusoidal components). It views a signal as a sum of various sinusoidal components. Transmission of an input signal through a (linear) system is viewed as transmission of various sinusoidal components of the input through the system.
-
-It was not by coincidence that we used the impulse function in time-domain analysis and the exponential *ej*ω*<sup>t</sup>* in studying the frequency domain. The two functions happen to be duals of each other. Thus, the Fourier transform of an impulse δ(*t* − τ ) is *e*−*j*ωτ , and the Fourier transform of *ej*<sup>ω</sup>0*<sup>t</sup>* is an impulse 2πδ(ω − ω0). This *time-frequency duality* is a constant theme in the Fourier transform and linear systems.
-
-### **[7.4-1 Signal Distortion During Transmission](#page-12-0)**
-
-For a system with frequency response *H*(ω), if *X*(ω) and *Y*(ω) are the spectra of the input and the output signals, respectively, then
-
-$$
-Y(\omega) = X(\omega)H(\omega) \tag{7.38}
-$$
-
-The transmission of the input signal *x*(*t*) through the system changes it into the output signal *y*(*t*). Equation (7.38) shows the nature of this change or modification. Here, *X*(ω) and *Y*(ω) are the spectra of the input and the output, respectively. Therefore, *H*(ω) is the spectral response of the system. The output spectrum is obtained by the input spectrum multiplied by the spectral response of the system. Equation (7.38), which clearly brings out the spectral shaping (or modification) of the signal by the system, can be expressed in polar form as
-
-$$
-|Y(\omega)|e^{j\angle Y(\omega)} = |X(\omega)||H(\omega)|e^{j[\angle X(\omega)+\angle H(j\omega)]}
-$$
-
-Therefore,
-
-$$
-|Y(\omega)| = |X(\omega)| |H(\omega)| \quad \text{and} \quad \angle Y(\omega) = \angle X(\omega) + \angle H(\omega)
-$$
-
-During transmission, the input signal amplitude spectrum |*X*(ω)| is changed to |*X*(ω)||*H*(ω)|. Similarly, the input signal phase spectrum *X*(ω) is changed to *X*(ω) + *H*(ω). An input signal spectral component of frequency ω is modified in amplitude by a factor |*H*(ω)| and is shifted in phase by an angle *H*(ω). Clearly, |*H*(ω)| is the amplitude response, and *H*(ω) is the phase response of the system. The plots of |*H*(ω)| and *H*(ω) as functions of ω show at a glance how the system modifies the amplitudes and phases of various sinusoidal inputs. This is the reason why *H*(ω) is also called the *frequency response* of the system. During transmission through the system, some frequency components may be boosted in amplitude, while others may be attenuated. The
-
-### 724 CHAPTER 7 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM
-
-relative phases of the various components also change. In general, the output waveform will be different from the input waveform.
-
-### DISTORTIONLESS TRANSMISSION
-
-In several applications, such as signal amplification or message signal transmission over a communication channel, we require that the output waveform be a replica of the input waveform. In such cases we need to minimize the distortion caused by the amplifier or the communication channel. It is, therefore, of practical interest to determine the characteristics of a system that allows a signal to pass without distortion (*distortionless transmission*).
-
-Transmission is said to be distortionless if the input and the output have identical waveshapes within a multiplicative constant. A delayed output that retains the input waveform is also considered to be distortionless. Thus, in distortionless transmission, the input *x*(*t*) and the output *y*(*t*) satisfy the condition
-
-$$
-y(t) = G_0 x(t - t_d)
-$$
-
-The Fourier transform of this equation yields
-
-$$
-Y(\omega) = G_0 X(\omega) e^{-j\omega t_d}
-$$
-
-But
-
-$$
-Y(\omega) = X(\omega)H(\omega)
-$$
-
-Therefore,
-
-$$
-H(\omega) = G_0 e^{-j\omega t_d}
-$$
-
-This is the frequency response required of a system for distortionless transmission. From this equation, it follows that
-
-$$
-|H(\omega)| = G_0 \quad \text{and} \quad \angle H(\omega) = -\omega t_d \tag{7.39}
-$$
-
-This result shows that for distortionless transmission, the amplitude response |*H*(ω)| must be a constant, and the phase response *H*(ω) must be a linear function of ω with slope −*td*, where *td* is the delay of the output with respect to input (Fig. 7.29).
-
-### MEASURE OF TIME-DELAY VARIATION WITH FREQUENCY
-
-The gain |*H*(ω)| = *G*<sup>0</sup> means that every spectral component is multiplied by a constant *G*0. Also, as seen in connection with Fig. 7.22, a linear phase *H*(ω) = −ω*td* means that every spectral
-
-**Figure 7.29** LTIC system frequency response for distortionless transmission.
-
-component is delayed by *td* seconds. This results in the output equal to *G*<sup>0</sup> times the input delayed by *td* seconds. Because each spectral component is attenuated by the same factor (*G*0) and delayed by exactly the same amount (*td*), the output signal is an exact replica of the input (except for attenuating factor *G*<sup>0</sup> and delay *td*).
-
-For distortionless transmission, we require a *linear phase* characteristic. The phase is not only a linear function of ω, it should also pass through the origin ω = 0. In practice, many systems have a phase characteristic that may be only approximately linear. A convenient way of judging phase linearity is to plot the slope of *H*(ω) as a function of frequency. This slope, which is constant for an ideal linear phase (ILP) system, is a function of ω in the general case and can be expressed as
-
-$$
-t_g(\omega) = -\frac{d}{d\omega} \angle H(\omega)
-$$
-\n(7.40)
-
-If *tg*(ω) is constant, all the components are delayed by the same time interval *tg*. But if the slope is not constant, the time delay *tg* varies with frequency. This variation means that different frequency components undergo different amounts of time delay, and consequently, the output waveform will not be a replica of the input waveform. As we shall see, *tg*(ω) plays an important role in bandpass systems and is called the *group delay* or *envelope* delay. Observe that constant *td* [Eq. (7.39)] implies constant *tg*. Note that *H*(ω) = φ<sup>0</sup> −ω*tg* also has a constant *tg*. Thus, constant group delay is a more relaxed condition.
-
-It is often thought (erroneously) that flatness of amplitude response |*H*(ω)| alone can guarantee signal quality. However, a system that has a flat amplitude response may yet distort a signal beyond recognition if the phase response is not linear (*td* not constant).
-
-### THE NATURE OF DISTORTION IN AUDIO AND VIDEO SIGNALS
-
-Generally speaking, the human ear can readily perceive amplitude distortion but is relatively insensitive to phase distortion. For the phase distortion to become noticeable, the variation in delay [variation in the slope of *H*(ω)] should be comparable to the signal duration (or the physically perceptible duration, in case the signal itself is long). In the case of audio signals, each spoken syllable can be considered to be an individual signal. The average duration of a spoken syllable is of a magnitude of the order of 0.01 to 0.1 second. Audio systems may have nonlinear phases, yet no noticeable signal distortion results because in practical audio systems, maximum variation in the slope of *H*(ω) is only a small fraction of a millisecond. This is the real truth underlying the statement that "the human ear is relatively insensitive to phase distortion" [3]. As a result, the manufacturers of audio equipment make available only |*H*(ω)|, the amplitude response characteristic of their systems.
-
-For video signals, in contrast, the situation is exactly the opposite. The human eye is sensitive to phase distortion but is relatively insensitive to amplitude distortion. Amplitude distortion in television signals manifests itself as a partial destruction of the relative half-tone values of the resulting picture, but this effect is not readily apparent to the human eye. Phase distortion (nonlinear phase), on the other hand, causes different time delays in different picture elements. The result is a smeared picture, and this effect is readily perceived by the human eye. Phase distortion is also very important in digital communication systems because the nonlinear phase characteristic of a channel causes pulse dispersion (spreading out), which in turn causes pulses to interfere with neighboring pulses. Such interference between pulses can cause an error in the pulse amplitude at the receiver: a binary **1** may read as **0**, and vice versa.
-
-### <span id="page-745-0"></span>**[7.4-2 Bandpass Systems and Group Delay](#page-12-0)**
-
-The distortionless transmission conditions [Eq. (7.39)] can be relaxed slightly for bandpass systems. For lowpass systems, the phase characteristics not only should be linear over the band of interest but also should pass through the origin. For bandpass systems, the phase characteristics must be linear over the band of interest but need not pass through the origin.
-
-Consider an LTI system with amplitude and phase characteristics as shown in Fig. 7.30, where the amplitude spectrum is a constant *G*<sup>0</sup> and the phase is φ<sup>0</sup> − ω*tg* over a band 2*W* centered at frequency ω*c*. Over this band, we can describe *H*(ω) as†
-
-$$
-H(\omega) = G_0 e^{j(\phi_0 - \omega t_g)} \qquad \omega \ge 0 \tag{7.41}
-$$
-
-The phase of *H*(ω) in Eq. (7.41), shown dotted in Fig. 7.30b, is linear but does not pass through the origin.
-
-Consider a modulated input signal *z*(*t*) = *x*(*t*) cosω*ct*. This is a bandpass signal, whose spectrum is centered at ω = ω*c*. The signal cosω*ct* is the carrier, and the signal *x*(*t*), which is a lowpass signal of bandwidth *W* (see Fig. 7.25), is the *envelope* of *z*(*t*). ‡ We shall now show that the transmission of *z*(*t*) through *H*(ω) results in distortionless transmission of the envelope *x*(*t*). However, the carrier phase changes by φ0. To show this, consider an input *z*ˆ(*t*) = *x*(*t*)*ej*ω*ct* and the corresponding output *y*ˆ(*t*). From Eq. (7.30), *Z*ˆ(ω) = *X*(ω − ω*c*), and the corresponding output
-
-**Figure 7.30** Generalized linear phase characteristics.
-
-<sup>†</sup> Because the phase function is an odd function of <sup>ω</sup>, if *<sup>H</sup>*(ω) <sup>=</sup> <sup>φ</sup><sup>0</sup> <sup>−</sup> <sup>ω</sup>*tg* for <sup>ω</sup> <sup>≥</sup> 0, over the band 2*<sup>W</sup>* (centered at ω*c*), then *H*(ω) = −φ<sup>0</sup> − ω*tg* for ω < 0 over the band 2*W* (centered at −ω*c*), as shown in Fig. 7.30a.
-
-<sup>‡</sup> The envelope of a bandpass signal is well defined only when the bandwidth of the envelope is well below the carrier ω*<sup>c</sup>* (*W* ω*c*).
-
-spectrum *Y*ˆ(ω) is given by
-
-$$
-\hat{Y}(\omega) = H(\omega)\hat{Z}(\omega) = H(\omega)X(\omega - \omega_c)
-$$
-
-Recall that the bandwidth of *X*(ω) is *W* so that the bandwidth of *X*(ω −ω*c*) is 2*W*, centered at ω*c*. Over this range, *H*(ω) is given by Eq. (7.41). Hence,
-
-$$
-\hat{Y}(\omega) = G_0 X(\omega - \omega_c) e^{j(\phi_0 - \omega t_g)} = G_0 e^{j\phi_0} X(\omega - \omega_c) e^{-\omega t_g}
-$$
-
-Use of Eqs. (7.29) and (7.30) yields *y*ˆ(*t*) as
-
-$$
-\hat{y}(t) = G_0 e^{j\phi_0} x(t - t_g) e^{j\omega_c(t - t_g)} = G_0 x(t - t_g) e^{j[\omega_c(t - t_g) + \phi_0]}
-$$
-
-This is the system response to input *z*ˆ(*t*) = *x*(*t*)*ej*ω*ct* , which is a complex signal. We are really interested in finding the response to the input *z*(*t*) = *x*(*t*) cosω*ct*, which is the real part of *z*ˆ(*t*) = *x*(*t*)*ej*ω*ct* . Hence, we use Eq. (2.31) to obtain *y*(*t*), the system response to the input *z*(*t*) = *x*(*t*) cosω*ct*, as
-
-$$
-y(t) = G_0 x(t - t_g) \cos [\omega_c (t - t_g) + \phi_0)]
-$$
-\n(7.42)
-
-where *tg*, the *group* (or *envelope*) delay, is the negative slope of *H*(ω) at ω*c*. † The output *y*(*t*) is basically the delayed input *z*(*t* − *tg*), except that the output carrier acquires an extra phase φ0. The output envelope *x*(*t* − *tg*) is the delayed version of the input envelope *x*(*t*) and is not affected by extra phase φ<sup>0</sup> of the carrier. In a modulated signal, such as *x*(*t*) cosω*ct*, the information generally resides in the envelope *x*(*t*). Hence, the transmission is considered to be distortionless if the envelope *x*(*t*) remains undistorted.
-
-Most practical systems satisfy Eq. (7.41), at least over a very small band. Figure 7.30b shows a typical case in which this condition is satisfied for a small band *W* centered at frequency ω*c*.
-
-A system in Eq. (7.41) is said to have a *generalized linear phase* (GLP), as illustrated in Fig. 7.30. The ideal linear phase (ILP) characteristics is shown in Fig. 7.29. For distortionless transmission of bandpass signals, the system need satisfy Eq. (7.41) only over the bandwidth of the bandpass signal.
-
-**Caution.** Recall that the phase response associated with the amplitude response may have jump discontinuities when the amplitude response goes negative. Jump discontinuities also arise because of the use of the principal value for phase. Under such conditions, to compute the group delay [Eq. (7.40)], we should ignore the jump discontinuities.
-
-$$
-y(t) = G_o x(t - t_g) \cos \omega_c (t - t_{ph})
-$$
-
-where *t*ph, called the *phase delay* at ω*c*, is given by *t*ph(ω*c*) = (ω*ctg* − φ0)/ω*c*. Generally, *t*ph varies with ω, and we can write
-
-$$
-t_{\rm ph}(\omega) = \frac{\omega t_g - \phi_0}{\omega}
-$$
-
-Recall also that *tg* itself may vary with ω.
-
-<sup>†</sup> Equation (7.42) can also be expressed as
-
-### **EXAMPLE 7.19 Distortionless Bandpass Transmission**
-
-**(a)** A signal *z*(*t*), shown in Fig. 7.31b, is given by
-
-$$
-z(t) = x(t) \cos \omega_c t
-$$
-
-where ω*<sup>c</sup>* = 2000π. The pulse *x*(*t*) (Fig. 7.31a) is a lowpass pulse of duration 0.1 second and has a bandwidth of about 10 Hz. This signal is passed through a filter whose frequency response is shown in Fig. 7.31c (shown only for positive ω). Find and sketch the filter output *y*(*t*).
-
-**(b)** Find the filter response if ω*<sup>c</sup>* = 4000π.
-
-**(a)** The spectrum *Z*(ω) is a narrow band of width 20 Hz, centered at frequency *f*<sup>0</sup> = 1 kHz. The gain at the center frequency (1 kHz) is 2. The group delay, which is the negative of the slope of the phase plot, can be found by drawing tangents at ω*c*, as shown in Fig. 7.31c. The negative of the slope of the tangent represents *tg*, and the intercept along the vertical axis by the tangent represents φ<sup>0</sup> at that frequency. From the tangents at ω*c*, we find *tg*, the group delay, as
-
-$$
-t_g = \frac{2.4\pi - 0.4\pi}{2000\pi} = 10^{-3}
-$$
-
-The vertical axis intercept is φ<sup>0</sup> = −0.4π. Hence, by using Eq. (7.42) with gain *G*<sup>0</sup> = 2, we obtain
-
-$$
-y(t) = 2x(t - t_g)\cos[\omega_c(t - t_g) - 0.4\pi] \qquad \omega_c = 2000\pi \quad t_g = 10^{-3}
-$$
-
-Figure 7.31d shows the output *y*(*t*), which consists of the modulated pulse envelope *x*(*t*) delayed by 1 ms and the phase of the carrier changed by −0.4π. The output shows no distortion of the envelope *x*(*t*), only the delay. The carrier phase change does not affect the shape of envelope. Hence, the transmission is considered distortionless.
-
-**(b)** Figure 7.31c shows that when ω*<sup>c</sup>* = 4000π, the slope of *H*(ω) is zero so that *tg* = 0. Also, the gain *G*<sup>0</sup> = 1.5, and the intercept of the tangent with the vertical axis is φ<sup>0</sup> = −3.1π. Hence,
-
-$$
-y(t) = 1.5x(t)\cos(\omega_c t - 3.1\pi)
-$$
-
-This, too, is a distortionless transmission for the same reasons as for case (a).

engineering/linear-systems-and-signals-3rd-edition-bp-lathi/093_7.5 IDEAL AND PRACTICAL FILTERS.md

@@ -1,78 +0,0 @@
-## <span id="page-749-0"></span>**[7.5 IDEAL AND](#page-13-0) PRACTICAL FILTERS**
-
-Ideal filters allow distortionless transmission of a certain band of frequencies and completely suppress the remaining frequencies. The ideal lowpass filter (Fig. 7.32), for example, allows all components below ω = *W* rad/s to pass without distortion and suppresses all components above ω = *W*. Figure 7.33 illustrates ideal highpass and bandpass filter characteristics.
-
-The ideal lowpass filter in Fig. 7.32a has a linear phase of slope −*td*, which results in a time delay of *td* seconds for all its input components of frequencies below *W* rad/s. Therefore, if the input is a signal *x*(*t*) bandlimited to *W* rad/s, the output *y*(*t*) is *x*(*t*) delayed by *td*: that is,
-
-$$
-y(t) = x(t - t_d)
-$$
-
-The signal *x*(*t*) is transmitted by this system without distortion, but with time delay *td*. For this filter, |*H*(ω)| = rect(ω/2*W*) and *H*(ω) = *e*−*j*ω*td* so that
-
-**Figure 7.32** Ideal lowpass filter: **(a)** frequency response and **(b)** impulse response.
-
-**Figure 7.33** Ideal **(a)** highpass and **(b)** bandpass filter frequency responses.
-
-The unit impulse response *h*(*t*) of this filter is obtained from pair 18 (Table 7.1) and the time-shifting property
-
-$$
-h(t) = \mathcal{F}^{-1} \left[ \text{rect}\left(\frac{\omega}{2W}\right) e^{-j\omega t_d} \right] = \frac{W}{\pi} \operatorname{sinc}[W(t - t_d)]
-$$
-
-Recall that *h*(*t*) is the system response to impulse input δ(*t*), which is applied at *t* = 0. Figure 7.32b shows a curious fact: the response *h*(*t*) begins even before the input is applied (at *t* = 0). Clearly, the filter is noncausal and therefore physically unrealizable. Similarly, one can show that other ideal filters (such as the ideal highpass or ideal bandpass filters depicted in Fig. 7.33) are also physically unrealizable.
-
-For a physically realizable system, *h*(*t*) must be causal; that is,
-
-$$
-h(t) = 0 \qquad \text{for } t < 0
-$$
-
-In the frequency domain, this condition is equivalent to the well-known *Paley–Wiener criterion,* which states that the necessary and sufficient condition for the amplitude response |*H*(ω)| to be realizable is†
-
-$$
-\int_{-\infty}^{\infty} \frac{|\ln|H(\omega)|}{1 + \omega^2} d\omega < \infty
-$$
-\n(7.43)
-
-If *H*(ω) does not satisfy this condition, it is unrealizable. Note that if |*H*(ω)| = 0 over any finite band, |ln|*H*(ω)|| = ∞ over that band, and Eq. (7.43) is violated. If, however, *H*(ω) = 0 at a single frequency (or a set of discrete frequencies), the integral in Eq. (7.43) may still be finite even though the integrand is infinite at those discrete frequencies. Therefore, for a physically realizable system, *H*(ω) may be zero at some discrete frequencies, but it cannot be zero over any finite band. In addition, if |*H*(ω)| decays exponentially (or at a higher rate) with ω, the integral in Eq. (7.43) goes to infinity, and |*H*(ω)| cannot be realized. Clearly, |*H*(ω)| cannot decay too fast with ω. According to this criterion, ideal filter characteristics (Figs. 7.32 and 7.33) are unrealizable.
-
-The impulse response *h*(*t*) in Fig. 7.32 is not realizable. One practical approach to filter design is to cut off the tail of *h*(*t*) for *t* < 0. The resulting causal impulse response:*h*(*t*), given by
-
-$$
-h(t) = h(t)u(t)
-$$
-
-is physically realizable because it is causal (Fig. 7.34). If *td* is sufficiently large, :*h*(*t*) will be a close approximation of *h*(*t*), and the resulting filter *H* :(ω) will be a good approximation of an ideal filter. This close realization of the ideal filter is achieved because of the increased value of time delay *td*. This observation means that the price of close realization is higher delay in the output; this situation is common in noncausal systems. Of course, theoretically, a delay *td* = ∞ is needed to realize the ideal characteristics. But a glance at Fig. 7.32b shows that a delay *td* of three or four times <sup>π</sup> *<sup>W</sup>* will make :*h*(*t*) a reasonably close version of *h*(*t* − *td*). For instance, an audio filter is required to handle frequencies of up to 20 kHz (*W* = 40,000π). In this case, a *td* of about 10−<sup>4</sup>
-
-$$
-\int_{-\infty}^{\infty} |H(\omega)|^2 d\omega < \infty
-$$
-
-Note that the Paley–Wiener criterion is a criterion for the realizability of the amplitude response |*H*(ω)|.
-
-<sup>†</sup> We are assuming that <sup>|</sup>*H*(ω)<sup>|</sup> is square integrable, that is,
-
-**Figure 7.34** Approximate realization of an ideal lowpass filter by truncation of its impulse response.
-
-(0.1 ms) would be a reasonable choice. The truncation operation [cutting the tail of *h*(*t*) to make it causal], however, creates some unsuspected problems. We discuss these problems and their cure in Sec. 7.8.
-
-In practice, we can realize a variety of filter characteristics that approach the ideal. Practical (realizable) filter characteristics are gradual, without jump discontinuities in amplitude response.
-
-### **DR ILL 7.11 The Unrealizable Gaussian Response**
-
-Show that a filter with Gaussian frequency response *<sup>H</sup>*(ω) <sup>=</sup> *<sup>e</sup>*−αω<sup>2</sup> is unrealizable. Demonstrate this fact in two ways: first by showing that its impulse response is noncausal, and then by showing that |*H*(ω)| violates the Paley–Wiener criterion. [*Hint:* Use pair 22 in Table 7.1.]
-
-### THINKING IN THE TIME AND FREQUENCY DOMAINS: A TWO-DIMENSIONAL VIEW OF SIGNALS AND SYSTEMS
-
-Both signals and systems have dual personalities, the time domain and the frequency domain. For a deeper perspective, we should examine and understand both these identities because they offer complementary insights. An exponential signal, for instance, can be specified by its time-domain description such as *e*−2*<sup>t</sup> u*(*t*) or by its Fourier transform (its frequency-domain description) 1/(*j*ω +2). The time-domain description depicts the waveform of a signal. The frequency-domain description portrays its spectral composition [relative amplitudes of its sinusoidal (or exponential) components and their phases]. For the signal *e*−2*<sup>t</sup>* , for instance, the time-domain description portrays the exponentially decaying signal with a time constant 0.5. The frequency-domain description characterizes it as a lowpass signal, which can be synthesized by sinusoids with amplitudes decaying with frequency roughly as 1/ω.
-
-An LTIC system can also be described or specified in the time domain by its impulse response *h*(*t*) or in the frequency domain by its frequency response *H*(ω). In Sec. 2.6, we studied intuitive insights in the system behavior offered by the impulse response, which consists of characteristic modes of the system. By purely qualitative reasoning, we saw that the system responds well to signals that are similar to the characteristic modes and responds poorly to signals that are very different from those modes. We also saw that the shape of the impulse response *h*(*t*) determines the system time constant (speed of response), and pulse dispersion (spreading), which, in turn, determines the rate of pulse transmission.
-
-The frequency response *H*(ω) specifies the system response to exponential or sinusoidal input of various frequencies. This is precisely the filtering characteristic of the system.
-
-<span id="page-752-0"></span>Experienced electrical engineers instinctively think in both domains (time and frequency) whenever possible. When they look at a signal, they consider its waveform, the signal width (duration), and the rate at which the waveform decays. This is basically a time-domain perspective. They also think of the signal in terms of its frequency spectrum, that is, in terms of its sinusoidal components and their relative amplitudes and phases, whether the spectrum is lowpass, bandpass, highpass, and so on. This is a frequency-domain perspective. Experienced electrical engineers think of a system in terms of its impulse response *h*(*t*). The width of *h*(*t*) indicates the time constant (response time): that is, how quickly the system is capable of responding to an input, and how much dispersion (spreading) it will cause. This is a time-domain perspective. From the frequency-domain perspective, these engineers view a system as a filter, which selectively transmits certain frequency components and suppresses the others [frequency response *H*(ω)]. Knowing the input signal spectrum and the frequency response of the system, they create a mental image of the output signal spectrum. This concept is precisely expressed by *Y*(ω) = *X*(ω)*H*(ω).
-
-We can analyze LTI systems by time-domain techniques or by frequency-domain techniques. Then why learn both? The reason is that the two domains offer complementary insights into system behavior. Some aspects are easily grasped in one domain; other aspects may be easier to see in the other domain. Both time-domain and frequency-domain methods are as essential for the study of signals and systems as two eyes are essential to a human being for correct visual perception of reality. A person can see with either eye, but for proper perception of three-dimensional reality, both eyes are essential.
-
-It is important to keep the two domains separate, and not to mix the entities in the two domains. If we are using the frequency domain to determine the system response, we must deal with all signals in terms of their spectra (Fourier transforms) and all systems in terms of their frequency responses. For example, to determine the system response *y*(*t*) to an input *x*(*t*), we must first convert the input signal into its frequency-domain description *X*(ω). The system description also must be in the frequency domain, that is, the frequency response *H*(ω). The output signal spectrum *Y*(ω) = *X*(ω)*H*(ω). Thus, the result (output) is also in the frequency domain. To determine the final answer *y*(*t*), we must take the inverse transform of *Y*(ω).

engineering/linear-systems-and-signals-3rd-edition-bp-lathi/094_7.6 SIGNAL ENERGY.md

@@ -1,110 +0,0 @@
-## **[7.6 SIGNAL](#page-13-0) ENERGY**
-
-The signal energy *Ex* of a signal *x*(*t*) was defined in Ch. 1 as
-
-$$
-E_x = \int_{-\infty}^{\infty} |x(t)|^2 dt
-$$
-\n(7.44)
-
-Signal energy can be related to the signal spectrum *X*(ω) by substituting Eq. (7.10) in Eq. (7.44):
-
-$$
-E_x = \int_{-\infty}^{\infty} x(t)x^*(t) dt = \int_{-\infty}^{\infty} x(t) \left[ \frac{1}{2\pi} \int_{-\infty}^{\infty} X^*(\omega) e^{-j\omega t} d\omega \right] dt
-$$
-
-#### 734 CHAPTER 7 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM
-
-Here, we used the fact that *x*∗(*t*), being the conjugate of *x*(*t*), can be expressed as the conjugate of the right-hand side of Eq. (7.10). Now, interchanging the order of integration yields
-
-$$
-E_x = \frac{1}{2\pi} \int_{-\infty}^{\infty} X^*(\omega) \left[ \int_{-\infty}^{\infty} x(t) e^{-j\omega t} dt \right] d\omega
-$$
-  
-=  $\frac{1}{2\pi} \int_{-\infty}^{\infty} X(\omega) X^*(\omega) d\omega = \frac{1}{2\pi} \int_{-\infty}^{\infty} |X(\omega)|^2 d\omega$ 
-
-Consequently,
-
-$$
-E_x = \int_{-\infty}^{\infty} |x(t)|^2 dt = \frac{1}{2\pi} \int_{-\infty}^{\infty} |X(\omega)|^2 d\omega \qquad (7.45)
-$$
-
-This is *Parseval's theorem* (for the Fourier transform). A similar result was obtained in Eqs. (6.26) and (6.27) for a periodic signal and its Fourier series. This result allows us to determine the signal energy from either the time-domain specification *x*(*t*) or the corresponding frequency-domain specification *X*(ω).
-
-The right-hand side of Eq. (7.45) can be interpreted to mean that the energy of a signal *x*(*t*) results from energies contributed by all the spectral components of the signal *x*(*t*). The total signal energy is the area under |*X*(ω)<sup>2</sup>| (divided by 2π). If we consider a small band ω (ω → 0), as illustrated in Fig. 7.35, the energy *Ex* of the spectral components in this band is the area of |*X*(ω)| <sup>2</sup> under this band (divided by 2π):
-
-$$
-\Delta E_x = \frac{1}{2\pi} |X(\omega)|^2 \,\Delta \omega = |X(\omega)|^2 \,\Delta f \qquad \frac{\Delta \omega}{2\pi} = \Delta f \,\mathrm{Hz}
-$$
-
-Therefore, the energy contributed by the components in this band of *f* (in hertz) is |*X*(ω)| <sup>2</sup>*f* . The total signal energy is the sum of energies of all such bands and is indicated by the area under |*X*(ω)| <sup>2</sup> as in Eq. (7.45). Therefore, |*X*(ω)| <sup>2</sup> is the *energy spectral density* (per unit bandwidth in hertz).
-
-For real signals, *X*(ω) and *X*(−ω) are conjugates, and |*X*(ω)| <sup>2</sup> is an even function of ω because
-
-$$
-|X(\omega)|^2 = X(\omega)X^*(\omega) = X(\omega)X(-\omega)
-$$
-
-**Figure 7.35** Interpretation of energy spectral density of a signal.
-
-Therefore, the energy of real signal *x*(*t*) can be expressed as†
-
-$$
-E_x = \frac{1}{\pi} \int_0^\infty |X(\omega)|^2 d\omega \tag{7.46}
-$$
-
-The signal energy *Ex*, which results from contributions from all the frequency components from ω = 0 to ∞, is given by (1/π times) the area under |*X*(ω)| <sup>2</sup> from ω = 0 to ∞. It follows that the energy contributed by spectral components of frequencies between ω<sup>1</sup> and ω<sup>2</sup> is
-
-$$
-\Delta E_x = \frac{1}{\pi} \int_{\omega_1}^{\omega_2} |X(\omega)|^2 d\omega \tag{7.47}
-$$
-
-### **EXAMPLE 7.20 Signal Energy and Parseval's Theorem**
-
-Find the energy of signal *x*(*t*) = *e*−*atu*(*t*). Determine the frequency *W* (rad/s) so that the energy contributed by the spectral components of all the frequencies below *W* is 95% of the signal energy *Ex*.
-
-We have
-
-$$
-E_x = \int_{-\infty}^{\infty} x^2(t) \, dt = \int_0^{\infty} e^{-2at} \, dt = \frac{1}{2a}
-$$
-
-We can verify this result by Parseval's theorem. For this signal,
-
-$$
-X(\omega) = \frac{1}{j\omega + a}
-$$
-
-and
-
-$$
-E_x = \frac{1}{\pi} \int_0^{\infty} |X(\omega)|^2 d\omega = \frac{1}{\pi} \int_0^{\infty} \frac{1}{\omega^2 + a^2} d\omega = \frac{1}{\pi a} \tan^{-1} \frac{\omega}{a} \Big|_0^{\infty} = \frac{1}{2a}
-$$
-
-The band ω = 0 to ω = *W* contains 95% of the signal energy, that is, 0.95/2*a*. Therefore, from Eq. (7.47) with ω<sup>1</sup> = 0 and ω<sup>2</sup> = *W*, we obtain
-
-$$
-\frac{0.95}{2a} = \frac{1}{\pi} \int_0^W \frac{d\omega}{\omega^2 + a^2} = \frac{1}{\pi a} \tan^{-1} \frac{\omega}{a} \Big|_0^W = \frac{1}{\pi a} \tan^{-1} \frac{W}{a}
-$$
-
-or
-
-$$
-\frac{0.95\pi}{2} = \tan^{-1}\frac{W}{a} \implies W = 12.706a \text{ rad/s}
-$$
-
-<sup>†</sup> In Eq. (7.46), it is assumed that *<sup>X</sup>*(ω) does not contain an impulse at <sup>ω</sup> <sup>=</sup> 0. If such an impulse exists, it should be integrated separately with a multiplying factor of 1/2π rather than 1/π.
-
-### <span id="page-755-0"></span>736 CHAPTER 7 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM
-
-This result indicates that the spectral components of *x*(*t*) in the band from 0 (dc) to 12.706*a* rad/s (2.02*a* Hz) contribute 95% of the total signal energy; all the remaining spectral components (in the band from 12.706*a* rad/s to ∞) contribute only 5% of the signal energy.
-
-### **DR ILL 7.12 Signal Energy and Parseval's Theorem**
-
-Use Parseval's theorem to show that the energy of the signal *x*(*t*) = 2*a*/(*t* <sup>2</sup> +*a*2) is 2π/*a*. [*Hint:* Find *X*(ω) using pair 3 of Table 7.1 and the duality property.]
-
-### THE ESSENTIAL BANDWIDTH OF A SIGNAL
-
-The spectra of all practical signals extend to infinity. However, because the energy of any practical signal is finite, the signal spectrum must approach 0 as ω → ∞. Most of the signal energy is contained within a certain band of *B* Hz, and the energy contributed by the components beyond *B* Hz is negligible. We can therefore suppress the signal spectrum beyond *B* Hz with little effect on the signal shape and energy. The bandwidth *B* is called the *essential bandwidth* of the signal. The criterion for selecting *B* depends on the error tolerance in a particular application. We may, for example, select *B* to be that band which contains 95% of the signal energy.† This figure may be higher or lower than 95%, depending on the precision needed. Using such a criterion, we can determine the essential bandwidth of a signal. The essential bandwidth *B* for the signal *e*−*atu*(*t*), using 95% energy criterion, was determined in Ex. 7.20 to be 2.02*a* Hz.
-
-Suppression of all the spectral components of *x*(*t*) beyond the essential bandwidth results in a signal *x*ˆ(*t*), which is a close approximation of *x*(*t*). If we use the 95% criterion for the essential bandwidth, the energy of the error (the difference) *x*(*t*)− ˆ*x*(*t*) is 5% of *Ex*.

engineering/linear-systems-and-signals-3rd-edition-bp-lathi/095_7.7 APPLICATION TO COMMUNICATIONS - AMPLITUDE MODULATION.md

@@ -1,242 +0,0 @@
-## **[7.7 APPLICATION TO](#page-13-0) COMMUNICATIONS: AMPLITUDE MODULATION**
-
-*Modulation* causes a spectral shift in a signal and is used to gain certain advantages mentioned in our discussion of the frequency-shifting property. Broadly speaking, there are two classes of modulation: amplitude (linear) modulation and angle (nonlinear) modulation. In this section, we shall discuss some practical forms of amplitude modulation.
-
-<sup>†</sup> For lowpass signals, the essential bandwidth may also be defined as a frequency at which the value of the amplitude spectrum is a small fraction (about 1%) of its peak value. In Ex. 7.20, for instance, the peak value, which occurs at ω = 0, is 1/*a*.
-
-### <span id="page-756-0"></span>**[7.7-1 Double-Sideband, Suppressed-Carrier \(DSB-SC\) Modulation](#page-13-0)**
-
-In amplitude modulation, the amplitude *A* of the carrier *A*cos(ω*ct* + θ*c*) is varied in some manner with the *baseband* (message)† signal *m*(*t*) (known as the *modulating signal*). The frequency ω*<sup>c</sup>* and the phase θ*<sup>c</sup>* are constant. We can assume θ*<sup>c</sup>* = 0 without loss of generality. If the carrier amplitude *A* is made directly proportional to the modulating signal *m*(*t*), the modulated signal is *m*(*t*) cos ω*ct* (Fig. 7.36). As was indicated earlier [Eq. (7.32)], this type of modulation simply shifts the spectrum of *m*(*t*) to the carrier frequency (Fig. 7.36c). Thus, if
-
-**Figure 7.36** DSB-SC modulation.
-
-<sup>†</sup> The term *baseband* is used to designate the band of frequencies of the signal delivered by the source or the input transducer.
-
-then
-
-$$
-m(t)\cos\omega_c t \Longleftrightarrow \frac{1}{2}[M(\omega + \omega_c) + M(\omega - \omega_c)] \tag{7.48}
-$$
-
-Recall that *M*(ω − ω*c*) is *M*(ω)-shifted to the right by ω*<sup>c</sup>* and *M*(ω + ω*c*) is *M*(ω)-shifted to the left by ω*c*. Thus, the process of modulation shifts the spectrum of the modulating signal to the left and the right by ω*c*. Note also that if the bandwidth of *m*(*t*) is *B* Hz, then, as indicated in Fig. 7.36c, the bandwidth of the modulated signal is 2*B* Hz. We also observe that the modulated signal spectrum centered at ω*<sup>c</sup>* is composed of two parts: a portion that lies above ω*c*, known as the *upper sideband (USB)*, and a portion that lies below ω*c*, known as the *lower sideband (LSB)*. Similarly, the spectrum centered at −ω*<sup>c</sup>* has upper and lower sidebands. This form of modulation is called *double sideband (DSB)* modulation for the obvious reason.
-
-The relationship of *B* to ω*<sup>c</sup>* is of interest. Figure 7.36c shows that ω*<sup>c</sup>* ≥ 2π*B* to avoid the overlap of the spectra centered at ±ω*c*. If ω*<sup>c</sup>* < 2π*B*, the spectra overlap and the information of *m*(*t*) are lost in the process of modulation, a loss that makes it impossible to get back *m*(*t*) from the modulated signal *m*(*t*) cos ω*ct*. †
-
-### **EXAMPLE 7.21 Double-Sideband Suppressed-Carrier Modulation**
-
-For a baseband signal *m*(*t*) = cos ω*mt*, find the DSB-SC signal and sketch its spectrum. Identify the upper and lower sidebands.
-
-We shall work this problem in the frequency domain as well as the time domain to clarify the basic concepts of DSB-SC modulation. In the frequency-domain approach, we work with the signal spectra. The spectrum of the baseband signal *m*(*t*) = cos ω*mt* is given by
-
-$$
-M(\omega) = \pi \left[ \delta(\omega - \omega_m) + \delta(\omega + \omega_m) \right]
-$$
-
-The spectrum consists of two impulses located at ±ω*m*, as depicted in Fig. 7.37a.
-
-The DSB-SC (modulated) spectrum, as indicated by Eq. (7.48), is the baseband spectrum in Fig. 7.37a shifted to the right and the left by ω*<sup>c</sup>* (times 0.5), as depicted in Fig. 7.37b. This spectrum consists of impulses at ±(ω*<sup>c</sup>* − ω*m*) and ±(ω*<sup>c</sup>* + ω*m*). The spectrum beyond ω*<sup>c</sup>* is the upper sideband (USB), and the one below ω*<sup>c</sup>* is the lower sideband (LSB). Observe that the DSB-SC spectrum does not have as a component the carrier frequency ω*c*. This is why the term *double-sideband, suppressed carrier* (DSB-SC) is used for this type of modulation.
-
-<sup>†</sup> Practical factors may impose additional restrictions on ω*c*. For instance, in broadcast applications, a radiating antenna can radiate only a narrow band without distortion. This restriction implies that avoiding distortion caused by the radiating antenna calls for ω*c*/2π*B* 1. The broadcast band AM radio, for instance, with *B* = 5 kHz and the band of 550–1600 kHz for carrier frequency gives a ratio of ω*c*/2π*B* roughly in the range of 100–300.
-
-**Figure 7.37** An example of DSB-SC modulation.
-
-In the time-domain approach, we work directly with signals in the time domain. For the baseband signal *m*(*t*) = cos ω*mt*, the DSB-SC signal ϕDSB-SC(*t*) is
-
-$$
-\varphi_{\text{DSB-SC}}(t) = m(t) \cos \omega_c t
-$$
-  
-=  $\cos \omega_m t \cos \omega_c t$   
-=  $\frac{1}{2} [\cos (\omega_c + \omega_m)t + \cos (\omega_c - \omega_m)t]$  (7.49)
-
-This result shows that when the baseband (message) signal is a single sinusoid of frequency ω*m*, the modulated signal consists of two sinusoids: the component of frequency ω*<sup>c</sup>* + ω*<sup>m</sup>* (the upper sideband), and the component of frequency ω*<sup>c</sup>* − ω*<sup>m</sup>* (the lower sideband). Figure 7.37b illustrates precisely the spectrum of ϕDSB-SC(*t*). Thus, each component of frequency ω*<sup>m</sup>* in the modulating signal results in two components of frequencies ω*<sup>c</sup>* + ω*<sup>m</sup>* and ω*<sup>c</sup>* − ω*<sup>m</sup>* in the modulated signal. This being a DSB-SC (suppressed-carrier) modulation, there is no component of the carrier frequency ω*<sup>c</sup>* on the right-hand side of Eq. (7.49).†
-
-### DEMODULATION OF DSB-SC SIGNALS
-
-The DSB-SC modulation translates or shifts the frequency spectrum to the left and the right by ω*<sup>c</sup>* (i.e., at +ω*<sup>c</sup>* and −ω*c*), as seen from Eq. (7.48). To recover the original signal *m*(*t*) from
-
-<sup>†</sup> The term *suppressed carrier* does not necessarily mean absence of the spectrum at the carrier frequency. "Suppressed carrier" merely implies that there is no discrete component of the carrier frequency. Since no discrete component exists, the DSB-SC spectrum does not have impulses at ±ω*c*, which further implies that the modulated signal *m*(*t*) cos ω*ct* does not contain a term of the form *k* cos ω*ct* [assuming that *m*(*t*) has a zero mean value].
-
-**Figure 7.38** Demodulation of DSB-SC: **(a)** demodulator and **(b)** spectrum of *e*(*t*).
-
-the modulated signal, we must retranslate the spectrum to its original position. The process of recovering the signal from the modulated signal (retranslating the spectrum to its original position) is referred to as *demodulation*, or *detection*. Observe that if the modulated signal spectrum in Fig. 7.36c is shifted to the left and to the right by ω*<sup>c</sup>* (and halved), we obtain the spectrum illustrated in Fig. 7.38b, which contains the desired baseband spectrum in addition to an unwanted spectrum at ±2ω*c*. The latter can be suppressed by a lowpass filter. Thus, demodulation, which is almost identical to modulation, consists of multiplication of the incoming modulated signal *m*(*t*) cos ω*ct* by a carrier cos ω*ct* followed by a lowpass filter, as depicted in Fig. 7.38a. We can verify this conclusion directly in the time domain by observing that the signal *e*(*t*) in Fig. 7.38a is
-
-$$
-e(t) = m(t)\cos^2\omega_c t = \frac{1}{2}[m(t) + m(t)\cos 2\omega_c t]
-$$
-
-Therefore, the Fourier transform of the signal *e*(*t*) is
-
-$$
-E(\omega) = \frac{1}{2}M(\omega) + \frac{1}{4}[M(\omega + 2\omega_c) + M(\omega - 2\omega_c)]
-$$
-
-Hence, *e*(*t*) consists of two components (1/2)*m*(*t*) and (1/2)*m*(*t*) cos 2ω*ct*, with their spectra, as illustrated in Fig. 7.38b. The spectrum of the second component, being a modulated signal with carrier frequency 2ω*c*, is centered at ±2ω*c*. Hence, this component is suppressed by the lowpass filter in Fig. 7.38a. The desired component (1/2)*M*(ω), being a lowpass spectrum (centered at ω = 0), passes through the filter unharmed, resulting in the output (1/2)*m*(*t*).
-
-A possible form of lowpass filter characteristics is depicted (dotted) in Fig. 7.38b. In this method of recovering the baseband signal, called *synchronous detection,* or *coherent detection,* we use a carrier of exactly the same frequency (and phase) as the carrier used for modulation. Thus, for demodulation, we need to generate a local carrier at the receiver in frequency and phase coherence (synchronism) with the carrier used at the modulator. We shall demonstrate in Ex. 7.22 that both phase and frequency synchronism are extremely critical.
-
-### **EXAMPLE 7.22 Frequency and Phase Incoherence in DSB-SC**
-
-Discuss the effect of lack of frequency and phase coherence (synchronism) between the carriers at the modulator (transmitter) and the demodulator (receiver) in DSB-SC.
-
-Let the modulator carrier be cos ω*ct* (Fig. 7.36a). For the demodulator in Fig. 7.38a, we shall consider two cases: with carrier cos(ω*ct*+θ ) (phase error of θ) and with carrier cos(ω*c*+ω)*t* (frequency error ω).
-
-**(a)** With the demodulator carrier cos(ω*ct* + θ ) (instead of cos ω*ct*) in Fig. 7.38a, the multiplier output is *e*(*t*) = *m*(*t*) cos ω*ct* cos(ω*ct* + θ ) instead of *m*(*t*) cos2ω*ct*. From the trigonometric identity, we obtain
-
-$$
-e(t) = m(t)\cos\omega_c t \cos(\omega_c t + \theta)
-$$
-  
-=  $\frac{1}{2}m(t)[\cos\theta + \cos(2\omega_c t + \theta)]$ 
-
-The spectrum of the component (1/2)*m*(*t*) cos(2ω*ct* + θ ) is centered at ±2ω*c*. Consequently, it will be filtered out by the lowpass filter at the output. The component (1/2)*m*(*t*) cos θ is the signal *m*(*t*) multiplied by a constant (1/2) cos θ. The spectrum of this component is centered at ω = 0 (lowpass spectrum) and will pass through the lowpass filter at the output, yielding the output (1/2)*m*(*t*) cos θ.
-
-If θ is constant, the phase asynchronism merely yields an output that is attenuated (by a factor cos θ). Unfortunately, in practice, θ is often the phase difference between the carriers generated by two distant generators and varies randomly with time. This variation would result in an output whose gain varies randomly with time.
-
-**(b)** In the case of frequency error, the demodulator carrier is cos(ω*c*+ω)*t*. This situation is very similar to the phase error case in part (a) with θ replaced by (ω)*t*. Following the analysis in part (a), we can express the demodulator product *e*(*t*) as
-
-$$
-e(t) = m(t)\cos\omega_c t \cos(\omega_c + \Delta\omega)t
-$$
-  
-=  $\frac{1}{2}m(t)[\cos(\Delta\omega)t + \cos(2\omega_c + \Delta\omega)t]$ 
-
-The spectrum of the component (1/2)*m*(*t*) cos(2ω*<sup>c</sup>* + ω)*t* is centered at ±(2ω*<sup>c</sup>* + ω). Consequently, this component will be filtered out by the lowpass filter at the output. The component (1/2)*m*(*t*) cos(ω)*t* is the signal *m*(*t*) multiplied by a low-frequency carrier of frequency ω. The spectrum of this component is centered at ±ω. In practice, the frequency error (ω) is usually very small. Hence, the signal (1/2)*m*(*t*) cos(ω)*t* (whose spectrum is centered at ±ω) is a lowpass signal and passes through the lowpass filter at the output, resulting in the output (1/2)*m*(*t*) cos(ω)*t*. The output is the desired signal *m*(*t*) multiplied by a very-low-frequency sinusoid cos(ω)*t*. The output in this case is not merely an attenuated replica of the desired signal *m*(*t*), but represents *m*(*t*) multiplied by a time-varying gain cos(ω)*t*. If, for instance, the transmitter and the receiver carrier frequencies differ just by 1 Hz, the output will be the desired signal *m*(*t*) multiplied by a time-varying signal whose gain goes from the maximum to 0 every half-second. This is like a restless child fiddling with the volume control knob of a receiver, going from maximum volume to zero volume every half-second. This kind of distortion (called the *beat effect*) is beyond repair.
-
-### <span id="page-761-0"></span>**[7.7-2 Amplitude Modulation \(AM\)](#page-13-0)**
-
-For the suppressed-carrier scheme just discussed, a receiver must generate a carrier in frequency and phase synchronism with the carrier at a transmitter that may be located hundreds or thousands of miles away. This situation calls for a sophisticated receiver, which could be quite costly. The other alternative is for the transmitter to transmit a carrier *A* cosω*ct* [along with the modulated signal *m*(*t*) cosω*ct*] so that there is no need to generate a carrier at the receiver. In this case, the transmitter needs to transmit much larger power, a rather expensive procedure. In point-to-point communications, where there is one transmitter for each receiver, substantial complexity in the receiver system can be justified, provided there is a large enough saving in expensive high-power transmitting equipment. On the other hand, for a broadcast system with a multitude of receivers for each transmitter, it is more economical to have one expensive high-power transmitter and simpler, less expensive receivers. The second option (transmitting a carrier along with the modulated signal) is the obvious choice in this case. This is amplitude modulation (AM), in which the transmitted signal ϕAM (*t*) is given by
-
-$$
-\varphi_{AM}(t) = A\cos\omega_c t + m(t)\cos\omega_c t = [A + m(t)]\cos\omega_c t \tag{7.50}
-$$
-
-Recall that the DSB-SC signal is *m*(*t*) cos ω*ct*. From Eq. (7.50) it follows that the AM signal is identical to the DSB-SC signal with *A*+*m*(*t*) as the modulating signal [instead of *m*(*t*)]. Therefore, to sketch ϕAM (*t*), we sketch *A* + *m*(*t*) and −[*A* + *m*(*t*)] as the envelopes and fill in between with the sinusoid of the carrier frequency. Two cases are considered in Fig. 7.39. In the first case, *A* is large enough so that *A* + *m*(*t*) ≥ 0 (is nonnegative) for all values of *t*. In the second case, *A* is not large enough to satisfy this condition. In the first case, the envelope (Fig. 7.39d) has the same shape as *m*(*t*) (although riding on a dc of magnitude *A*). In the second case, the envelope shape is not *m*(*t*), for some parts get rectified (Fig. 7.39e). Thus, we can detect the desired signal *m*(*t*) by detecting the envelope in the first case. In the second case, such a detection is not possible. We shall see that envelope detection is an extremely simple and inexpensive operation, which does not require generation of a local carrier for the demodulation. But as just noted, the envelope of AM has the information about *m*(*t*) only if the AM signal [*A*+*m*(*t*)] cos ω*ct* satisfies the condition *A*+*m*(*t*) > 0 for all *t*. Thus, the condition for envelope detection of an AM signal is
-
-$$
-A + m(t) \ge 0 \qquad \text{for all } t \tag{7.51}
-$$
-
-If *mp* is the peak amplitude (positive or negative) of *m*(*t*), then Eq. (7.51) is equivalent to
-
-$$
-A\geq m_p
-$$
-
-Thus, the minimum carrier amplitude required for the viability of envelope detection is *mp*. This point is clearly illustrated in Fig. 7.39.
-
-We define the *modulation index* μ as
-
-$$
-\mu = \frac{m_p}{A} \tag{7.52}
-$$
-
-where *A* is the carrier amplitude. Note that *mp* is a constant of the signal *m*(*t*). Because *A* ≥ *mp* and because there is no upper bound on *A*, it follows that
-
-0 ≤ μ ≤ 1
-
-**Figure 7.39** An AM signal **(a)** for two values of A **(b, c)** and the respective envelopes **(d, e)**.
-
-as the required condition for the viability of demodulation of AM by an envelope detector.
-
-When *A* < *mp*, Eq. (7.52) shows that μ > 1 (overmodulation, shown in Fig. 7.39e). In this case, the option of envelope detection is no longer viable. We then need to use synchronous demodulation. Note that synchronous demodulation can be used for any value of μ (see Prob. 7.7-7). The envelope detector, which is considerably simpler and less expensive than the synchronous detector, can be used only when μ ≤ 1.
-
-### **EXAMPLE 7.23 Amplitude Modulation**
-
-Sketch ϕAM (*t*) for modulation indices of μ = 0.5 (50% modulation) and μ = 1 (100% modulation), when *m*(*t*) = *B*cos ω*mt*. This case is referred to as *tone modulation* because the modulating signal is a pure sinusoid (or tone).
-
-### 744 CHAPTER 7 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM
-
-In this case, *mp* = *B* and the modulation index according to Eq. (7.52) is
-
-$$
-\mu = \frac{B}{A}
-$$
-
-Hence, *B* = μ*A* and
-
-$$
-m(t) = B\cos\omega_m t = \mu A\cos\omega_m t
-$$
-
-Therefore,
-
-$$
-\varphi_{AM}(t) = [A + m(t)] \cos \omega_c t = A[1 + \mu \cos \omega_m t] \cos \omega_c t
-$$
-
-The modulated signals corresponding to μ = 0.5 and μ = 1 appear in Figs. 7.40a and 7.40b, respectively.
-
-### DEMODULATION OF AM: THE ENVELOPE DETECTOR
-
-The AM signal can be demodulated coherently by a locally generated carrier (see Prob. 7.7-7). Since, however, coherent, or synchronous, demodulation of AM (with μ ≤ 1) will defeat the very purpose of AM, it is rarely used in practice. We shall consider here one of the noncoherent methods of AM demodulation, *envelope detection*. †
-
-In an envelope detector, the output of the detector follows the envelope of the (modulated) input signal. The circuit illustrated in Fig. 7.41a functions as an envelope detector. During the positive cycle of the input signal, the diode conducts and the capacitor *C* charges up to the peak voltage of the input signal (Fig. 7.41b). As the input signal falls below this peak value, the diode is cut off, because the capacitor voltage (which is very nearly the peak voltage) is greater than the input signal voltage, a circumstance causing the diode to open. The capacitor now discharges through the resistor *R* at a slow rate (with a time constant *RC*). During the next positive cycle,
-
-<sup>†</sup> There are also other methods of noncoherent detection. The rectifier detector consists of a rectifier followed by a lowpass filter. This method is also simple and almost as inexpensive as the envelope detector [4]. The nonlinear detector, although simple and inexpensive, results in a distorted output.
-
-**Figure 7.41** Demodulation by means of envelope detector.
-
-the same drama repeats. When the input signal becomes greater than the capacitor voltage, the diode conducts again. The capacitor again charges to the peak value of this (new) cycle. As the input voltage falls below the new peak value, the diode cuts off again and the capacitor discharges slowly during the cutoff period, a process that changes the capacitor voltage very slightly.
-
-In this manner, during each positive cycle, the capacitor charges up to the peak voltage of the input signal and then decays slowly until the next positive cycle. Thus, the output voltage *vC*(*t*) follows closely the envelope of the input. The capacitor discharge between positive peaks, however, causes a ripple signal of frequency ω*<sup>c</sup>* in the output. This ripple can be reduced by increasing the time constant *RC* so that the capacitor discharges very little between the positive peaks (*RC* 1/ω*c*). Making *RC* too large, however, would make it impossible for the capacitor voltage to follow the envelope (see Fig. 7.41b). Thus, *RC* should be large in comparison to 1/ω*<sup>c</sup>* but small in comparison to 1/2π*B*, where *B* is the highest frequency in *m*(*t*). Incidentally, these two conditions also require that ω*<sup>c</sup>* 2π*B*, a condition necessary for a well-defined envelope.
-
-The envelope-detector output *vC*(*t*) is *A* + *m*(*t*) plus a ripple of frequency ω*c*. The dc term *A* can be blocked out by a capacitor or a simple *RC* highpass filter. The ripple is reduced further by
-
-### <span id="page-765-0"></span>746 CHAPTER 7 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM
-
-another (lowpass) *RC* filter. In the case of audio signals, the speakers also act as lowpass filters, which further enhances suppression of the high-frequency ripple.
-
-### **[7.7-3 Single-Sideband Modulation \(SSB\)](#page-13-0)**
-
-Now consider the baseband spectrum *M*(ω) (Fig. 7.42a) and the spectrum of the DSB-SC modulated signal *m*(*t*) cos ω*ct* (Fig. 7.42b). The DSB spectrum in Fig. 7.42b has two sidebands: the upper and the lower (USB and LSB), both containing complete information on *M*(ω) [see Eq. (7.12)]. Clearly, it is redundant to transmit both sidebands, a process that requires twice the bandwidth of the baseband signal. A scheme in which only one sideband is transmitted is known
-
-**Figure 7.42** Spectra for single-sideband transmission: **(a)** baseband, **(b)** DSB, **(c)** USB, **(d)** LSB, and **(e)** synchronously demodulated signal.
-
-as *single-sideband (SSB) transmission,* which requires only half the bandwidth of the DSB signal. Thus, we transmit only the upper sidebands (Fig. 7.42c) or only the lower sidebands (Fig. 7.42d).
-
-An SSB signal can be coherently (synchronously) demodulated. For example, multiplication of a USB signal (Fig. 7.42c) by 2 cos ω*ct* shifts its spectrum to the left and to the right by ω*c*, yielding the spectrum in Fig. 7.42e. Lowpass filtering of this signal yields the desired baseband signal. The case is similar with an LSB signal. Hence, demodulation of SSB signals is identical to that of DSB-SC signals, and the synchronous demodulator in Fig. 7.38a can demodulate SSB signals. Note that we are talking of SSB signals without an additional carrier. Hence, they are suppressed-carrier signals (SSB-SC).
-
-### **EXAMPLE 7.24 Single-Sideband Modulation**
-
-Find the USB (upper sideband) and LSB (lower sideband) signals when *m*(*t*) = cos ω*mt*. Sketch their spectra, and show that these SSB signals can be demodulated using the synchronous demodulator in Fig. 7.38a.
-
-The DSB-SC signal for this case is
-
-$$
-\varphi_{\text{DSB-SC}}(t) = m(t)\cos\omega_c t = \cos\omega_m t \cos\omega_c t \qquad = \frac{1}{2} [\cos(\omega_c - \omega_m)t + \cos(\omega_c + \omega_m)t]
-$$
-
-As pointed out in Ex. 7.21, the terms (1/2) cos(ω*<sup>c</sup>* + ω*m*)*t* and (1/2) cos(ω*<sup>c</sup>* − ω*m*)*t* represent the upper and lower sidebands, respectively. The spectra of the upper and lower sidebands are given in Figs. 7.43a and 7.43b. Observe that these spectra can be obtained from the DSB-SC spectrum in Fig. 7.37b by using a proper filter to suppress the undesired sidebands. For instance, the USB signal in Fig. 7.43a can be obtained by passing the DSB-SC signal (Fig. 7.37b) through a highpass filter of cutoff frequency ω*c*. Similarly, the LSB signal in Fig. 7.43b can be obtained by passing the DSB-SC signal through a lowpass filter of cutoff frequency ω*c*.
-
-If we apply the LSB signal (1/2) cos(ω*<sup>c</sup>* − ω*m*)*t* to the synchronous demodulator in Fig. 7.38a, the multiplier output is
-
-$$
-e(t) = \frac{1}{2}\cos{(\omega_c - \omega_m)t}\cos{\omega_c t} = \frac{1}{4}[\cos{\omega_m t} + \cos{(2\omega_c - \omega_m)t}]
-$$
-
-The term (1/4) cos(2ω*c*−ω*m*)*t* is suppressed by the lowpass filter, producing the desired output (1/4) cos ω*mt* [which is *m*(*t*)/4]. The spectrum of this term is π[δ(ω+ω*m*)+δ(ω−ω*m*)]/4, as depicted in Fig. 7.43c. In the same way, we can show that the USB signal can be demodulated by the synchronous demodulator.
-
-In the frequency domain, demodulation (multiplication by cos ω*ct*) amounts to shifting the LSB spectrum (Fig. 7.43b) to the left and the right by ω*<sup>c</sup>* (times 0.5) and then suppressing the high frequency, as illustrated in Fig. 7.43c. The resulting spectrum represents the desired signal (1/4)*m*(*t*).
-
-### GENERATION OF SSB SIGNALS
-
-Two methods are commonly used to generate SSB signals. The *selective-filtering method* uses sharp cutoff filters to eliminate the undesired sideband, and the second method uses phase-shifting networks to achieve the same goal [4].† We shall consider here only the first method.
-
-Selective filtering is the most commonly used method of generating SSB signals. In this method, a DSB-SC signal is passed through a sharp cutoff filter to eliminate the undesired sideband.
-
-To obtain the USB, the filter should pass all components above ω*<sup>c</sup>* unattenuated and completely suppress all components below ω*c*. Such an operation requires an ideal filter, which is unrealizable. It can, however, be realized closely if there is some separation between the passband and the stopband. Fortunately, the voice signal provides this condition, because its spectrum shows little power content at the origin (Fig. 7.44). Moreover, articulation tests show that for speech signals, frequency components below 300 Hz are not important. In other words, we may suppress all speech components below 300 Hz without appreciably affecting intelligibility.‡ Thus, filtering of the unwanted sideband becomes relatively easy for speech signals because we have a 600 Hz transition region around the cutoff frequency ω*c*. For some signals, which have considerable power
-
-<sup>†</sup> Yet another method, known as Weaver's method, is also used to generate SSB signals.
-
-<sup>‡</sup> Similarly, suppression of components of a speech signal above 3500 Hz causes no appreciable change in intelligibility.
-
-<span id="page-768-0"></span>at low frequencies (around ω = 0), SSB techniques cause considerable distortion. Such is the case with video signals. Consequently, for video signals, instead of SSB, we use another technique, the *vestigial sideband (VSB)*, which is a compromise between SSB and DSB. It inherits the advantages of SSB and DSB but avoids their disadvantages at a cost of slightly increased bandwidth. VSB signals are relatively easy to generate, and their bandwidth is only slightly (typically 25%) greater than that of SSB signals. In VSB signals, instead of rejecting one sideband completely (as in SSB), we accept a gradual cutoff from one sideband [4].
-
-### **[7.7-4 Frequency-Division Multiplexing](#page-13-0)**
-
-Signal multiplexing allows transmission of several signals on the same channel. Later, in Ch. 8 (Sec. 8.2-2), we shall discuss time-division multiplexing (TDM), where several signals time-share the same channel, such as a cable or an optical fiber. In frequency-division multiplexing (FDM), the use of modulation, as illustrated in Fig. 7.45, makes several signals share the band of the same channel. Each signal is modulated by a different carrier frequency. The various carriers are adequately separated to avoid overlap (or interference) between the spectra of various modulated signals. These carriers are referred to as *subcarriers*. Each signal may use a different kind of modulation, for example, DSB-SC, AM, SSB-SC, VSB-SC, or even other forms of modulation, not discussed here [such as FM (frequency modulation) or PM (phase modulation)]. The modulated-signal spectra may be separated by a small guard band to avoid interference and to facilitate signal separation at the receiver.
-
-When all the modulated spectra are added, we have a composite signal that may be considered to be a new baseband signal. Sometimes, this composite baseband signal may be used to further modulate a high-frequency (radio frequency, or RF) carrier for the purpose of transmission.
-
-At the receiver, the incoming signal is first demodulated by the RF carrier to retrieve the composite baseband, which is then bandpass-filtered to separate the modulated signals. Then each modulated signal is individually demodulated by an appropriate subcarrier to obtain all the basic baseband signals.

engineering/linear-systems-and-signals-3rd-edition-bp-lathi/096_7.8 DATA TRUNCATION - WINDOW FUNCTIONS.md

@@ -1,73 +0,0 @@
-## **7.8 DATA [TRUNCATION: WINDOW](#page-13-0) FUNCTIONS**
-
-We often need to truncate data in diverse situations from numerical computations to filter design. For example, if we need to compute numerically the Fourier transform of some signal, say, *e*−*<sup>t</sup> u*(*t*), we will have to truncate the signal *e*−*<sup>t</sup> u*(*t*) beyond a sufficiently large value of *t* (typically five time constants and above). The reason is that in numerical computations, we have to deal with
-
-**Figure 7.45** Frequency-division multiplexing: **(a)** FDM spectrum **(b)** transmitter, and **(c)** receiver.
-
-data of finite duration. Similarly, the impulse response *h*(*t*) of an ideal lowpass filter is noncausal and approaches zero asymptotically as |*t*|→∞. For a practical design, we may want to truncate *h*(*t*) beyond a sufficiently large value of |*t*| to make *h*(*t*) causal and of finite duration. In signal sampling, to eliminate aliasing, we must use an antialiasing filter to truncate the signal spectrum beyond the half-sampling frequency ω*s*/2. Again, we may want to synthesize a periodic signal by adding the first *n* harmonics and truncating all the higher harmonics. These examples show that data truncation can occur in both time and frequency domains. On the surface, truncation appears to be a simple problem of cutting off the data at a point at which values are deemed to be sufficiently small. Unfortunately, this is not the case. Simple truncation can cause some unsuspected problems.
-
-### WINDOW FUNCTIONS
-
-Truncation operation may be regarded as multiplying a signal of a large width by a window function of a smaller (finite) width. Simple truncation amounts to using a *rectangular window wR*(*t*) (shown later in Fig. 7.48a) in which we assign unit weight to all the data within the window width (|*t*| < *T*/2), and assign zero weight to all the data lying outside the window (|*t*| > *T*/2). It is also possible to use a window in which the weight assigned to the data within the window may not be constant. In a *triangular window wT* (*t*), for example, the weight assigned to data decreases linearly over the window width (shown later in Fig. 7.48b).
-
-Consider a signal *x*(*t*) and a window function *w*(*t*). If *x*(*t*) ⇐⇒ *X*(ω) and *w*(*t*) ⇐⇒ *W*(ω), and if the windowed function *xw*(*t*) ⇐⇒ *Xw*(ω), then
-
-$$
-x_w(t) = x(t)w(t)
-$$
- and  $X_w(\omega) = \frac{1}{2\pi}X(\omega)*W(\omega)$ 
-
-According to the width property of convolution, it follows that the width of *Xw*(ω) equals the sum of the widths of *X*(ω) and *W*(ω). Thus, truncation of a signal increases its bandwidth by the amount of bandwidth of *w*(*t*). Clearly, the truncation of a signal causes its spectrum to spread (or smear) by the amount of the bandwidth of *w*(*t*). Recall that the signal bandwidth is inversely proportional to the signal duration (width). Hence, the wider the window, the smaller its bandwidth, and the smaller the *spectral spreading*. This result is predictable because a wider window means that we are accepting more data (closer approximation), which should cause smaller distortion (smaller spectral spreading). Smaller window width (poorer approximation) causes more spectral spreading (more distortion). In addition, since *W*(ω) is really not strictly bandlimited and its spectrum → 0 only asymptotically, the spectrum of *Xw*(ω) → 0 asymptotically also at the same rate as that of *W*(ω), even if *X*(ω) is, in fact, strictly bandlimited. Thus, windowing causes the spectrum of *X*(ω) to spread into the band where it is supposed to be zero. This effect is called *leakage*. The following example clarifies these twin effects of spectral spreading and leakage.
-
-Let us consider *x*(*t*) = cos ω0*t* and a rectangular window *wR*(*t*) = rect(*t*/*T*), illustrated in Fig. 7.46b. The reason for selecting a sinusoid for *x*(*t*) is that its spectrum consists of spectral lines of zero width (Fig. 7.46a). Hence, this choice will make the effect of spectral spreading and leakage easily discernible. The spectrum of the truncated signal *xw*(*t*) is the convolution of the two impulses of *X*(ω) with the sinc spectrum of the window function. Because the convolution of any function with an impulse is the function itself (shifted at the location of the impulse), the resulting spectrum of the truncated signal is 1/2π times the two sinc pulses at ±ω0, as depicted in Fig. 7.46c (also see Fig. 7.26). Comparison of spectra *X*(ω) and *Xw*(ω) reveals the effects of truncation. These are:
-
-1. The spectral lines of *X*(ω) have zero width. But the truncated signal is spread out by 2π/*T* about each spectral line. The amount of spread is equal to the width of the mainlobe of the window spectrum. One effect of this *spectral spreading* (or smearing) is that if *x*(*t*) has two spectral components of frequencies differing by less than 4π/*T* rad/s (2/*T* Hz), they
-
-**Figure 7.46** Windowing and its effects.
-
-will be indistinguishable in the truncated signal. The result is loss of spectral resolution. We would like the spectral spreading [mainlobe width of *W*(ω)] to be as small as possible.
-
-2. In addition to the mainlobe spreading, the truncated signal has sidelobes, which decay slowly with frequency. The spectrum of *x*(*t*) is zero everywhere except at ±ω0. On the other hand, the truncated signal spectrum *Xw*(ω) is zero nowhere because of the sidelobes. These sidelobes decay asymptotically as 1/ω. Thus, the truncation causes spectral *leakage* in the band where the spectrum of the signal *x*(*t*) is zero. The peak *sidelobe* magnitude is 0.217 times the mainlobe magnitude (13.3 dB below the peak mainlobe magnitude). Also, the sidelobes decay at a rate 1/ω, which is −6 dB/octave (or −20 dB/decade). This is the sidelobe's *rolloff rate*. We want smaller sidelobes with a faster rate of decay (high rolloff rate). Figure 7.46d, which plots |*WR*(ω)| as a function of ω, clearly shows the mainlobe and sidelobe features, with the first sidelobe amplitude −13.3 dB below the mainlobe amplitude and the sidelobes decaying at a rate of −6 dB/octave (or −20 dB/decade).
-
-So far, we have discussed the effect on the signal spectrum of signal truncation (truncation in the time domain). Because of the time-frequency duality, the effect of spectral truncation (truncation in frequency domain) on the signal shape is similar.
-
-### REMEDIES FOR SIDE EFFECTS OF TRUNCATION
-
-For better results, we must try to minimize the twin side effects of truncations: spectral spreading (mainlobe width) and leakage (sidelobe). Let us consider each of these ills.
-
-- 1. The spectral spread (mainlobe width) of the truncated signal is equal to the bandwidth of the window function *w*(*t*). We know that the signal bandwidth is inversely proportional to the signal width (duration). Hence, to reduce the spectral spread (mainlobe width), we need to increase the window width.
-- 2. To improve the leakage behavior, we must search for the cause of the slow decay of sidelobes. In Ch. 6, we saw that the Fourier spectrum decays as 1/ω for a signal with jump discontinuity, decays as 1/ω<sup>2</sup> for a continuous signal whose first derivative is discontinuous, and so on.† Smoothness of a signal is measured by the number of continuous derivatives it possesses. The smoother the signal, the faster the decay of its spectrum. Thus, we can achieve a given leakage behavior by selecting a suitably smooth (tapered) window.
-- 3. For a given window width, the remedies for the two effects are incompatible. If we try to improve one, the other deteriorates. For instance, among all the windows of a given width, the rectangular window has the smallest spectral spread (mainlobe width), but its sidelobes have high level and they decay slowly. A tapered (smooth) window of the same width has smaller and faster decaying sidelobes, but it has a wider mainlobe.‡ But we can compensate for the increased mainlobe width by widening the window. Thus, we can remedy both the side effects of truncation by selecting a suitably smooth window of sufficient width.
-
-There are several well-known tapered-window functions, such as Bartlett (triangular), Hanning (von Hann), Hamming, Blackman, and Kaiser, which truncate the data gradually. These
-
-<sup>†</sup> This result was demonstrated for periodic signals. However, it applies to aperiodic signals also. This is because we showed in the beginning of this chapter that if *xT*<sup>0</sup> (*t*) is a periodic signal formed by periodic extension of an aperiodic signal *x*(*t*), then the spectrum of *xT*<sup>0</sup> (*t*) is (1/*T*<sup>0</sup> times) the samples of *X*(ω). Thus,
-
-what is true of the decay rate of the spectrum of *xT*<sup>0</sup> (*t*) is also true of the rate of decay of *<sup>X</sup>*(ω). ‡ A tapered window yields a higher mainlobe width because the effective width of a tapered window is smaller than that of the rectangular window; see Sec. 2.6-2 [Eq. (2.47)] for the definition of effective width. Therefore, from the reciprocity of the signal width and its bandwidth, it follows that the rectangular window mainlobe is narrower than a tapered window.
-
-| No. | Window w(t)                                                         | Mainlobe<br>Width | Rolloff<br>Rate<br>Level (dB) | Peak<br>Sidelobe |
-|-----|---------------------------------------------------------------------|-------------------|-------------------------------|------------------|
-| 1   | t<br>Rectangular: rect<br>T                                         | 4π<br>T           | −6                            | −13.3            |
-| 2   | t<br>Bartlett:<br>2T                                                | 8π<br>T           | −12                           | −26.5            |
-| 3   | 2πt<br>!<br>Hanning: 0.5<br>1+cos<br>T                              | 8π<br>T           | −18                           | −31.5            |
-| 4   | 2πt<br>Hamming: 0.54+0.46 cos<br>T                                  | 8π<br>T           | −6                            | −42.7            |
-| 5   | 2πt<br>4πt<br>Blackman: 0.42+0.5 cos<br>+0.08 cos<br>T<br>T         | 12π<br>T          | −18                           | −58.1            |
-| 6   | <br>2<br>t<br>I0<br>α<br>1−4<br>T<br>Kaiser:<br>0 ≤ α ≤ 10<br>I0(α) | 11.2π<br>T        | −6                            | −59.9            |
-|     |                                                                     |                   | (α = 8.168)                   |                  |
-
-**TABLE 7.3** Some Window Functions and Their Characteristics
-
-**Figure 7.47 (a)** Hanning and **(b)** Hamming windows.
-
-windows offer different trade-offs with respect to spectral spread (mainlobe width), the peak sidelobe magnitude, and the leakage rolloff rate, as indicated in Table 7.3 [5, 6]. Observe that all windows are symmetrical about the origin (i.e., are even functions of *t*). Because of this feature, *W*(ω) is a real function of ω; that is, *W*(ω) is either 0 or π. Hence, the phase function of the truncated signal has a minimal amount of distortion.
-
-Figure 7.47 shows two well-known tapered-window functions, the von Hann (or Hanning) window *w*Han(*x*) and the Hamming window *w*Ham(*x*). We have intentionally used the independent variable *x* because windowing can be performed in the time domain as well as in the frequency domain, so *x* could be *t* or ω, depending on the application.
-
-There are hundreds of windows, all with different characteristics. But the choice depends on a particular application. The rectangular window has the narrowest mainlobe. The Bartlett <span id="page-774-0"></span>(triangle) window (also called the Fejer or Cesaro) is inferior in all respects to the Hanning window. For this reason, it is rarely used in practice. Hanning is preferred over Hamming in spectral analysis because it has faster sidelobe decay. For filtering applications, on the other hand, the Hamming window is chosen because it has the smallest sidelobe magnitude for a given mainlobe width. The Hamming window is the most widely used general-purpose window. The Kaiser window, which uses *I*0(α), the modified zero-order Bessel function, is more versatile and adjustable. Selecting a proper value of α (0 ≤ α ≤ 10) allows the designer to tailor the window to suit a particular application. The parameter α controls the mainlobe-sidelobe trade-off. When α = 0, the Kaiser window is the rectangular window. For α = 5.4414, it is the Hamming window, and when α = 8.885, it is the Blackman window. As α increases, the mainlobe width increases and the sidelobe level decreases.
-
-### **[7.8-1 Using Windows in Filter Design](#page-13-0)**
-
-We shall design an ideal lowpass filter of bandwidth *W* rad/s, with frequency response *H*(ω), as shown in Fig. 7.48e or Fig. 7.48f. For this filter, the impulse response *h*(*t*) = (*W*/π )sinc (*Wt*) (Fig. 7.48c) is noncausal and, therefore, unrealizable. Truncation of *h*(*t*) by a suitable window (Fig. 7.48a) makes it realizable, although the resulting filter is now an approximation to the desired ideal filter.† We shall use a rectangular window *wR*(*t*) and a triangular (Bartlett) window *wT* (*t*) to truncate *h*(*t*), and then examine the resulting filters. The truncated impulse responses *hR*(*t*) = *h*(*t*)*wR*(*t*) and *hT* (*t*) = *h*(*t*)*wT* (*t*) are depicted in Fig. 7.48d. Hence, the windowed filter frequency response is the convolution of *H*(ω) with the Fourier transform of the window, as illustrated in Figs. 7.48e and 7.48f. We make the following observations.
-
-- 1. The windowed filter spectra show *spectral spreading* at the edges, and instead of a sudden switch there is a gradual transition from the passband to the stopband of the filter. The transition band is smaller (2π/*T* rad/s) for the rectangular case than for the triangular case (4π/*T* rad/s).
-- 2. Although *H*(ω) is bandlimited, the windowed filters are not. But the stopband behavior of the triangular case is superior to that of the rectangular case. For the rectangular window, the leakage in the stopband decreases slowly (as 1/ω) in comparison to that of the triangular window (as 1/ω2). Moreover, the rectangular case has a higher peak sidelobe amplitude than that of the triangular window.

engineering/linear-systems-and-signals-3rd-edition-bp-lathi/097_7.9 MATLAB - FOURIER TRANSFORM TOPICS.md

@@ -1,204 +0,0 @@
-## **[7.9 MATLAB: FOURIER](#page-13-0) TRANSFORM TOPICS**
-
-MATLAB is useful for investigating a variety of Fourier transform topics. In this section, a rectangular pulse is used to investigate the scaling property, Parseval's theorem, essential bandwidth, and spectral sampling. Kaiser window functions are also investigated.
-
-<sup>†</sup> In addition to truncation, we need to delay the truncated function by *T*/2 to render it causal. However, the time delay only adds a linear phase to the spectrum without changing the amplitude spectrum. Thus, to simplify our discussion, we shall ignore the delay.
-
-**Figure 7.48** Window-based filter design.
-
-### <span id="page-776-0"></span>**[7.9-1 The Sinc Function and the Scaling Property](#page-13-0)**
-
-As shown in Ex. 7.2, the Fourier transform of *x*(*t*) = rect(*t*/τ ) is *X*(ω) = τ sinc (ωτ/2). To represent *X*(ω) in MATLAB, a sinc function is first required. As an alternative to the signal processing toolbox function sinc, which computes sinc(*x*) as sin(π*x*)/π*x*, we create our own function that follows the conventions of this book and defines sinc(*x*) = sin(*x*)/*x*.
-
-function [y] = CH7MP1(x) % CH7MP1.m : Chapter 7, MATLAB Program 1 % Function M-file computes the sinc function, y = sin(x)/x.
-
-y(x==0) = 1; y(x~=0) = sin(x(x~=0))./x(x~=0);
-
-The computational simplicity of sinc (*x*) = sin(*x*)/*x* is somewhat deceptive: sin(0)/0 results in a divide-by-zero error. Thus, program CH7MP1 assigns sinc (0) = 1 and computes the remaining values according to the definition. Notice that CH7MP1 cannot be directly replaced by an anonymous function. Anonymous functions cannot have multiple lines or contain certain commands such as =, if, or for. M-files, however, can be used to define an anonymous function. For example, we can represent *X*(ω) as an anonymous function that is defined in terms of CH7MP1.
-
-```
->> X = @(omega,tau) tau*CH7MP1(omega*tau/2);
-```
-
-Once we have defined *X*(ω), it is simple to investigate the effects of scaling the pulse width τ . Consider the three cases τ = 1.0, τ = 0.5, and τ = 2.0.
-
-```
->> omega = linspace(-4*pi,4*pi,200);
->> plot(omega,X(omega,1),'k-',omega,X(omega,0.5),'k-.',omega,X(omega,2),'k--');
->> grid; axis tight; xlabel('\omega'); ylabel('X(\omega)');
->> legend('Baseline (\tau = 1)','Compressed (\tau = 0.5)',...
->> 'Expanded (\tau = 2.0)');
-```
-
-Figure 7.49 confirms the reciprocal relationship between signal duration and spectral bandwidth: time compression causes spectral expansion, and time expansion causes spectral compression. Additionally, spectral amplitudes are directly related to signal energy. As a signal is compressed, signal energy and thus spectral magnitude decrease. The opposite effect occurs when the signal is expanded.
-
-**Figure 7.49** Spectra *X*(ω) = τ sinc (ωτ/2) for τ = 1.0, τ = 0.5, and τ = 2.0.
-
-### <span id="page-777-0"></span>**[7.9-2 Parseval's Theorem and Essential Bandwidth](#page-13-0)**
-
-Parseval's theorem concisely relates energy between the time domain and the frequency domain:
-
-$$
-\int_{-\infty}^{\infty} |x(t)|^2 dt = \frac{1}{2\pi} \int_{-\infty}^{\infty} |X(\omega)|^2 d\omega
-$$
-
-This too is easily verified with MATLAB. For example, a unit amplitude pulse *x*(*t*) with duration τ has energy *Ex* = τ . Thus,
-
-$$
-\int_{-\infty}^{\infty} |X(\omega)|^2 \, d\omega = 2\pi \, \tau
-$$
-
-Letting τ = 1, the energy of *X*(ω) is computed by using the quad function.
-
-```
->> X_squared = @(omega, tau) (tau*CH7MP1(omega*tau/2)).^2;
->> quad(X_squared,-1e6,1e6,[],[],1)
-    ans = 6.2817
-```
-
-Although not perfect, the result of the numerical integration is consistent with the expected value of 2π ≈ 6.2832. For quad, the first argument is the function to be integrated, the next two arguments are the limits of integration, the empty square brackets indicate default values for special options, and the last argument is the secondary input τ for the anonymous function X\_squared. Full format details for quad are available from MATLAB's help facilities.
-
-A more interesting problem involves computing a signal's essential bandwidth. Consider, for example, finding the essential bandwidth *W*, in radians per second, that contains fraction β of the energy of the square pulse *x*(*t*). That is, we want to find *W* such that
-
-$$
-\frac{1}{2\pi} \int_{-W}^{W} |X(\omega)|^2 d\omega = \beta \tau
-$$
-
-Program CH7MP2 uses a guess-and-check method to find *W*.
-
-```
-function [W,E_W] = CH7MP2(tau,beta,tol)
-% CH7MP2.m : Chapter 7, MATLAB Program 2
-% Function M-file computes essential bandwidth W for square pulse.
-% INPUTS: tau = pulse width
-% beta = fraction of signal energy desired in W
-% tol = tolerance of relative energy error
-% OUTPUTS: W = essential bandwidth [rad/s]
-% E_W = Energy contained in bandwidth W
-W = 0; step = 2*pi/tau; % Initial guess and step values
-X_squared = @(omega,tau) (tau*CH7MP1(omega*tau/2)).^2;
-E = beta*tau; % Desired energy in W
-relerr = (E-0)/E; % Initial relative error is 100 percent
-while(abs(relerr) > tol),
-   if (relerr>0), % W too small, so...
-       W=W+step; % ... increase W by step
-   elseif (relerr<0), % W too large, so...
-       step = step/2; % ... decrease step and then W
-```
-
-```
-W = W-step;
-    end
-    E_W = 1/(2*pi)*quad(X_squared,-W,W,[],[],tau);
-    relerr = (E - E_W)/E;
-end
-```
-
-Although this guess-and-check method is not the most efficient, it is relatively simple to understand: CH7MP2 sensibly adjusts *W* until the relative error is within tolerance. The number of iterations needed to converge to a solution depends on a variety of factors and is not known beforehand. The while command is ideal for such situations:
-
-```
-while expression,
-```
-
-*statements;*
-
-end
-
-While the *expression* is true, the *statements* are continually repeated.
-
-To demonstrate CH7MP2, consider the 90% essential bandwidth *W* for a pulse of 1 second duration. Typing [W,E\_W]=CH7MP2(1,0.9,0.001) returns an essential bandwidth *W* = 5.3014 that contains 89.97% of the energy. Reducing the error tolerance improves the estimate. CH7MP2(1,0.9,0.00005) returns an essential bandwidth *W* = 5.3321 that contains 90.00% of the energy. These essential bandwidth calculations are consistent with estimates presented after Ex. 7.2.
-
-### **[7.9-3 Spectral Sampling](#page-13-0)**
-
-Consider a signal with finite duration τ . A periodic signal *xT*<sup>0</sup> (*t*) is constructed by repeating *x*(*t*) every *T*<sup>0</sup> seconds, where *T*<sup>0</sup> ≥ τ . From Eq. (7.5), we can write the Fourier series coefficients of *xT*<sup>0</sup> (*t*) as *Dn* = (1/*T*0)*X*(*n*2π/*T*0). Put another way, the Fourier series coefficients are obtained by sampling the spectrum *X*(ω).
-
-By using spectral sampling, it is simple to determine the Fourier series coefficients for an arbitrary duty-cycle, square-pulse periodic signal. The square pulse *x*(*t*) = rect(*t*/τ ) has spectrum *X*(ω) = τ sinc(ωτ /2). Thus, the *n*th Fourier coefficient of the periodic extension *xT*<sup>0</sup> (*t*) is *Dn* = (τ/*T*0)sinc (*n*πτ/*T*0). As in Ex. 6.4, τ = π and *T*<sup>0</sup> = 2π provide a square-pulse periodic signal. The Fourier coefficients are determined by
-
-```
->> tau = pi; T_0 = 2*pi; n = [0:10];
-```
-
->> D\_n = tau/T\_0\*MS7P1(n\*pi\*tau/T\_0);
-
-```
->> stem(n,D_n); xlabel('n'); ylabel('D_n');
-```
-
->> axis([-0.5 10.5 -0.2 0.55]);
-
-The results, shown in Fig. 7.50, agree with Fig. 6.6b. Doubling the period to *T*<sup>0</sup> = 4π effectively doubles the density of spectral samples and halves the spectral amplitude, as shown in Fig. 7.51.
-
-As *T*<sup>0</sup> increases, the spectral sampling becomes progressively finer while the amplitude becomes infinitesimal. An evolution of the Fourier series toward the Fourier integral is seen by allowing the period *T*<sup>0</sup> to become large. Figure 7.52 shows the result for *T*<sup>0</sup> = 40π.
-
-If *T*<sup>0</sup> = τ , the signal *xT*<sup>0</sup> is a constant and the spectrum should concentrate energy at dc. In this case, the sinc function is sampled at the zero crossings and *Dn* = 0 for all *n* not equal to 0. Only the sample corresponding to *n* = 0 is nonzero, indicating a dc signal, as expected. It is a simple matter to modify the previous code to verify this case.
-
-<span id="page-779-0"></span>**Figure 7.50** Fourier spectra for τ = π and *T*<sup>0</sup> = 2π.
-
-**Figure 7.51** Fourier spectra for τ = π and *T*<sup>0</sup> = 4π.
-
-**Figure 7.52** Fourier spectra for τ = π and *T*<sup>0</sup> = 40π.
-
-### **[7.9-4 Kaiser Window Functions](#page-13-0)**
-
-A window function is useful only if it can be easily computed and applied to a signal. The Kaiser window, for example, is flexible but appears rather intimidating:
-
-$$
-w_K(t) = \begin{cases} \frac{I_0(\alpha\sqrt{1 - 4(t/T)^2})}{I_0(\alpha)} & |t| < T/2\\ 0 & \text{otherwise} \end{cases}
-$$
-
-Fortunately, the bark of a Kaiser window is worse than its bite! The function *I*0(*x*), a zero-order modified Bessel function of the first kind, can be computed according to
-
-$$
-I_0(x) = \sum_{k=0}^{\infty} \left(\frac{x^k}{2^k k!}\right)^2
-$$
-
-or, more simply, by using the MATLAB function besseli(0,x). In fact, MATLAB supports a wide range of Bessel functions, including Bessel functions of the first and second kinds (besselj and bessely), modified Bessel functions of the first and second kinds (besseli and besselk), Hankel functions (besselh), and Airy functions (airy).
-
-Program CH7MP3 computes Kaiser windows at times *t* by using parameters *T* and α.
-
-```
-function [w_K] = CH7MP3(t,T,alpha)
-% CH7MP3.m : Chapter 7, MATLAB Program 3
-% Function M-file computes a width-T Kaiser window using parameter alpha.
-% Alpha can also be a string identifier: 'rectangular', 'Hamming', or
-% 'Blackman'.
-% INPUTS: t = independent variable of the window function
-% T = window width
-% alpha = Kaiser parameter or string identifier
-% OUTPUTS: w_K = Kaiser window function
-if strncmpi(alpha,'rectangular',1),
-   alpha = 0;
-elseif strncmpi(alpha,'Hamming',3),
-   alpha = 5.4414;
-elseif strncmpi(alpha,'Blackman',1),
-   alpha = 8.885;
-elseif isa(alpha,'char')
-   disp('Unrecognized string identifier.'); return
-end
-w_K = zeros(size(t)); i = find(abs(t)<T/2);
-w_K(i) = besseli(0,alpha*sqrt(1-4*t(i).^2/(T^2)))/besseli(0,alpha);
-```
-
-Recall that α = 0, α = 5.4414, and α = 8.885 correspond to rectangular, Hamming, and Blackman windows, respectively. CH7MP3 is written to allow these special-case Kaiser windows to be identified by name rather than by α value. While unnecessary, this convenient feature is achieved with the help of the strncmpi command.
-
-The strncmpi(S1,S2,N) command compares the first N characters of strings S1 and S2, ignoring case. More completely, MATLAB has four variants of string comparison: strcmp, strcmpi, strncmp, and strncmpi. Comparisons are restricted to the first *N* characters when n is present; case is ignored when i is present. Thus, CH7MP3 identifies any string alpha that starts with the letter r or R as a rectangular window. To prevent confusion with a Hanning window, the first three characters must match to identify a Hamming window. The isa(alpha,'char') command determines whether alpha is a character string. MATLAB help documents the many other classes that isa can identify. In CH7MP3, isa is used to terminate execution if a string identifier alpha has not been recognized as one of the three special cases.
-
-<span id="page-781-0"></span>**Figure 7.53** Special-case, unit-duration Kaiser windows.
-
-Figure 7.53 shows the three special-case, unit-duration Kaiser windows generated by
-
-```
->> t = [-0.6:.001:0.6]; T = 1;
-```
-
-```
->> plot(t,CH7MP3(t,T,'r'),'k-',t,CH7MP3(t,T,'ham'),'k-.',t,CH7MP3(t,T,'b'),'k--');
-```
-
-```
->> axis([-0.6 0.6 -.1 1.1]); xlabel('t'); ylabel('w_K(t)');
-```
-
-```
->> legend('Rectangular','Hamming','Blackman','Location','EastOutside');
-```

engineering/linear-systems-and-signals-3rd-edition-bp-lathi/098_7.10 SUMMARY.md

@@ -1,15 +0,0 @@
-## **[7.10 SUMMARY](#page-13-0)**
-
-In Ch. 6, we represented periodic signals as a sum of (everlasting) sinusoids or exponentials (Fourier series). In this chapter we extended this result to aperiodic signals, which are represented by the Fourier integral (instead of the Fourier series). An aperiodic signal *x*(*t*) may be regarded as a periodic signal with period *T*<sup>0</sup> → ∞ so that the Fourier integral is basically a Fourier series with a fundamental frequency approaching zero. Therefore, for aperiodic signals, the Fourier spectra are continuous. This continuity means that a signal is represented as a sum of sinusoids (or exponentials) of all frequencies over a continuous frequency interval. The Fourier transform *X*(ω), therefore, is the spectral density (per unit bandwidth in hertz).
-
-An ever-present aspect of the Fourier transform is the duality between time and frequency, which also implies duality between the signal *x*(*t*) and its transform *X*(ω). This duality arises because of near-symmetrical equations for direct and inverse Fourier transforms. The duality principle has far-reaching consequences and yields many valuable insights into signal analysis.
-
-The scaling property of the Fourier transform leads to the conclusion that the signal bandwidth is inversely proportional to signal duration (signal width). Time shifting of a signal does not change its amplitude spectrum, but it does add a linear phase component to its spectrum. Multiplication of a signal by an exponential *ej*<sup>ω</sup>0*<sup>t</sup>* shifts the spectrum to the right by ω0. In practice, spectral shifting is achieved by multiplying a signal by a sinusoid such as cosω0*t* (rather than the exponential *ej*<sup>ω</sup>0*<sup>t</sup>* ). This process is known as amplitude modulation. Multiplication of two signals results in convolution of their spectra, whereas convolution of two signals results in multiplication of their spectra.
-
-For an LTIC system with the frequency response *H*(ω), the input and output spectra *X*(ω) and *Y*(ω) are related by the equation *Y*(ω) = *X*(ω)*H*(ω). This is valid only for asymptotically stable systems. It also applies to marginally stable systems if the input does not contain a finite-amplitude <span id="page-782-0"></span>sinusoid of the natural frequency of the system. For asymptotically unstable systems, the frequency response *H*(ω) does not exist. For distortionless transmission of a signal through an LTIC system, the amplitude response |*H*(ω)| of the system must be constant, and the phase response *H*(ω) should be a linear function of ω over a band of interest. Ideal filters, which allow distortionless transmission of a certain band of frequencies and suppress all the remaining frequencies, are physically unrealizable (noncausal). In fact, it is impossible to build a physical system with zero gain [*H*(ω) = 0] over a finite band of frequencies. Such systems (which include ideal filters) can be realized only with infinite time delay in the response.
-
-The energy of a signal *x*(*t*) is equal to 1/2π times the area under |*X*(ω)<sup>2</sup>| (Parseval's theorem). The energy contributed by spectral components within a band *f* (in hertz) is given by |*X*(ω)| <sup>2</sup>*f* . Therefore, |*X*(ω)| <sup>2</sup> is the energy spectral density per unit bandwidth (in hertz).
-
-The process of modulation shifts the signal spectrum to different frequencies. Modulation is used for many reasons: to transmit several messages simultaneously over the same channel for the sake of utilizing channel's high bandwidth, to effectively radiate power over a radio link, to shift a signal spectrum at higher frequencies to overcome the difficulties associated with signal processing at lower frequencies, and to effect the exchange of transmission bandwidth and transmission power required to transmit data at a certain rate. Broadly speaking, there are two types of modulation, amplitude and angle modulation. Each class has several subclasses.
-
-In practice, we often need to truncate data. Truncating is like viewing data through a window, which permits only certain portions of the data to be seen and hides (suppresses) the remainder. Abrupt truncation of data amounts to a rectangular window, which assigns a unit weight to data seen from the window and zero weight to the remaining data. Tapered windows, on the other hand, reduce the weight gradually from 1 to 0. Data truncation can cause some unsuspected problems. For example, in computation of the Fourier transform, windowing (data truncation) causes spectral spreading (spectral smearing) that is characteristic of the window function used. A rectangular window results in the least spreading, but it does so at the cost of a high and oscillatory spectral leakage outside the signal band, which decays slowly as 1/ω. In comparison to a rectangular window, tapered windows, in general, have larger spectral spreading (smearing), but the spectral leakage is smaller and decays faster with frequency. If we try to reduce spectral leakage by using a smoother window, the spectral spreading increases. Fortunately, spectral spreading can be reduced by increasing the window width. Therefore, we can achieve a given combination of spectral spread (transition bandwidth) and leakage characteristics by choosing a suitable tapered window function of a sufficiently long width *T*.

engineering/linear-systems-and-signals-3rd-edition-bp-lathi/099_REFERENCES.md

@@ -1,8 +0,0 @@
-### **[REFERENCES](#page-13-0)**
-
-- 1. Churchill, R. V., and Brown, J. W. *Fourier Series and Boundary Value Problems,* 3rd ed. McGraw-Hill, New York, 1978.
-- 2. Bracewell, R. N. *Fourier Transform and Its Applications,* rev. 2nd ed. McGraw-Hill, New York, 1986.
-- 3. Guillemin, E. A. *Theory of Linear Physical Systems*. Wiley, New York, 1963.
-- 4. Lathi, B. P. *Modern Digital and Analog Communication Systems,* 3rd ed. Oxford University Press, New York, 1998.
-- 5. Hamming, R. W. *Digital Filters,* 2nd ed. Prentice-Hall, Englewood Cliffs, NJ, 1983.
-- 6. Harris, F. J. On the use of windows for harmonic analysis with the discrete Fourier transform. *Proceedings of the IEEE,* vol. 66, no. 1, pp. 51–83, January 1978.

engineering/linear-systems-and-signals-3rd-edition-bp-lathi/100_PROBLEMS.md

@@ -1,440 +0,0 @@
-## <span id="page-783-0"></span>**[PROBLEMS](#page-13-0)**
-
-- **7.1-1** Suppose signal *x*(*t*) = *t* <sup>2</sup> [*u*(*t*)−*u*(*t* −2)] has Fourier transform *X*(ω). Define a 3-periodic replication of *<sup>x</sup>*(*t*) as *<sup>y</sup>*(*t*) <sup>=</sup> %<sup>∞</sup> *<sup>n</sup>*=−∞ 2*x*(*t* − 1 − 3*n*). Determine *Yk*, the Fourier series of *y*(*t*), in terms of the Fourier transform *X*(·).
-- **7.1-2** Show that for a real *x*(*t*), Eq. (7.10) can be expressed as
-
-$$
-x(t) = \frac{1}{\pi} \int_0^\infty |X(\omega)| \cos[\omega t + \angle X(\omega)] d\omega
-$$
-
-This is the trigonometric form of the Fourier integral. Compare this with the compact trigonometric Fourier series.
-
-**7.1-3** Show that if *x*(*t*) is an even function of *t*, then
-
-$$
-X(\omega) = 2 \int_0^\infty x(t) \cos \omega t \, dt
-$$
-
-and if *x*(*t*) is an odd function of *t*, then
-
-$$
-X(\omega) = -2j \int_0^\infty x(t) \sin \omega t \, dt
-$$
-
-Hence, prove that if *x*(*t*) is a real and even function of *t*, then *X*(ω) is a real and even function of ω. In addition, if *x*(*t*) is a real and odd function of *t*, then *X*(ω) is an imaginary and odd function of ω.
-
-**7.1-4** A signal *x*(*t*) can be expressed as the sum of even and odd components (see Sec. 1.5-2):
-
-*x*(*t*) = *xe*(*t*) +*xo*(*t*)
-
-(a) If *x*(*t*) ⇐⇒ *X*(ω), show that for real *x*(*t*),
-
-*xe*(*t*) ⇐⇒ Re[*X*(ω)]
-
-and
-
-$$
-x_o(t) \Longleftrightarrow j \operatorname{Im}[X(\omega)]
-$$
-
-- (b) Verify these results by finding the Fourier transforms of the even and odd components of the following signals: **(i)** *u*(*t*) and **(ii)** *e*−*atu*(*t*).
-- **7.1-5** Using Eq. (7.9), find the Fourier transforms of the signals *x*(*t*) in Fig. P7.1-5.
-- **7.1-6** Using Eq. (7.9), find the Fourier transforms of the signals depicted in Fig. P7.1-6.
-- **7.1-7** Use Eq. (7.10) to find the inverse Fourier transforms of the spectra in Fig. P7.1-7.
-
-**Figure P7.1-6**
-
-**Figure P7.1-8**
-
-- **7.1-8** Use Eq. (7.10) to find the inverse Fourier transforms of the spectra in Fig. P7.1-8.
-- **7.1-9** If *x*(*t*) ⇐⇒ *X*(ω), then show that
-
-$$
-X(0) = \int_{-\infty}^{\infty} x(t) dt
-$$
-
-and
-
-$$
-x(0) = \frac{1}{2\pi} \int_{-\infty}^{\infty} X(\omega) d\omega
-$$
-
-Also show that
-
-$$
-\int_{-\infty}^{\infty} \operatorname{sinc}(x) dx = \int_{-\infty}^{\infty} \operatorname{sinc}^{2}(x) dx = \pi
-$$
-
-**Figure P7.2-4**
-
-- **7.2-1** Sketch the following functions: (a) rect(*t*/2)
-  - (b) (3ω/100)
-  - (c) rect((*t* −10)/8)
-  - (d) sinc(πω/5)
-  - (e) sinc ((ω/5)−2π)
-  - (f) sinc (*t*/5)rect(*t*/10π )
-- **7.2-2** Using Eq. (7.9), show that the Fourier transform of rect(*<sup>t</sup>* <sup>−</sup> <sup>5</sup>) is sinc(ω/2)*e*−*j*5ω. Sketch the resulting amplitude and phase spectra.
-- **7.2-3** Using Eq. (7.10), show that the inverse Fourier transform of rect((ω <sup>−</sup>10)/2π ) is sinc(π*t*) *ej*10*<sup>t</sup>* .
-- **7.2-4** Find the inverse Fourier transform of *X*(ω) for the spectra illustrated in Fig. P7.2-4. [*Hint*:
-
-*<sup>X</sup>*(ω) = |*X*(ω)|*ej <sup>X</sup>*(ω). This problem illustrates how different phase spectra (both with the same amplitude spectrum) represent entirely different signals.]
-
-- **7.2-5** (a) Can you find the Fourier transform of *eatu*(*t*) when *<sup>a</sup>*>1 by setting *<sup>s</sup>* <sup>=</sup> *<sup>j</sup>*<sup>ω</sup> in the Laplace transform of *eatu*(*t*)? Explain.
-  - (b) Find the Laplace transform of *x*(*t*) shown in Fig. P7.2-5. Can you find the Fourier transform of *x*(*t*) by setting *s* = *j*ω in its Laplace transform? Explain. Verify your answer by finding the Fourier and the Laplace transforms of *x*(*t*).
-
-**Figure P7.2-5**
-
-- **7.3-1** Apply the duality property to the appropriate pair in Table 7.1 to show that
-  - (a) <sup>1</sup> <sup>2</sup> [δ(*t*)+*j*/π*t*] ⇐⇒ *u*(ω)
-  - (b) δ(*t* +*T*) +δ(*t* −*T*) ⇐⇒ 2 cos *T*ω
-  - (c) δ(*t* +*T*)−δ(*t* −*T*) ⇐⇒ 2*j*sin *T*ω
-
-- **7.3-2** A signal *x*(*t*) has Fourier transform *X*(ω). Determine the Fourier transform *Y*(ω) in terms of *X*(ω) for each of the following signals *y*(*t*): (a) *<sup>y</sup>*(*t*) <sup>=</sup> <sup>1</sup> <sup>5</sup> *x*(−2*t* +3) (b) *<sup>y</sup>*(*t*) <sup>=</sup> *ej*2*<sup>t</sup> x*∗(−3*t* −6)
-- **7.3-3** A signal *x*(*t*) has Fourier transform *X*(ω). Determine the inverse Fourier transform *y*(*t*) in terms of *x*(*t*) for each of the following spectra *Y*(ω),
-
-(a) 
-$$
-Y(\omega) = \frac{4}{3}e^{-j2\omega/3}X(-\omega/3)
-$$
-
-(b) 
-$$
-Y(\omega) = \frac{1}{3}e^{j2(\omega - 2)}X^*\left(\frac{\omega - 2}{3}\right)
-$$
-
-**7.3-4** The Fourier transform of the triangular pulse *x*(*t*) in Fig. P7.3-4 is expressed as
-
-$$
-X(\omega) = \frac{1}{\omega^2} (e^{j\omega} - j\omega e^{j\omega} - 1)
-$$
-
-Use this information, and the time-shifting and time-scaling properties, to find the Fourier transforms of the signals *xi*(*t*)(*i* = 1, 2, 3, 4, 5) shown in Fig. P7.3-4.
-
-- **7.3-5** Using only the time-shifting property and Table 7.1, find the Fourier transforms of the signals depicted in Fig. P7.3-5.
-- **7.3-6** Consider the fact that the τ -duration triangle function <sup>ω</sup> τ has inverse Fourier transform
-
-**Figure P7.3-4**
-
-2 0 234 0 -
-
-(a) (b)
-
-4 -3 -
-
-**Figure P7.3-7**
-
--4 -3 -
-
-τ <sup>4</sup><sup>π</sup> sinc2 *<sup>t</sup>*<sup>τ</sup> 4 . Use the duality property to determine the Fourier transform *Y*(ω) of signal *y*(*t*) = (*t*).
-
-**7.3-7** Use the time-shifting property to show that if *x*(*t*) ⇐⇒ *X*(ω), then
-
-$$
-x(t+T) + x(t-T) \Longleftrightarrow 2X(\omega)\cos T\omega
-$$
-
-This is the dual of Eq. (7.32). Use this result and Table 7.1 to find the Fourier transforms of the signals shown in Fig. P7.3-7.
-
-**7.3-8** Prove the following results, which are duals of each other:
-
-$$
-x(t)\sin \omega_0 t \Longleftrightarrow \frac{1}{2j}[X(\omega - \omega_0) - X(\omega + \omega_0)]
-$$
-  
-$$
-\frac{1}{2j}[x(t+T) - x(t-T)] \Longleftrightarrow X(\omega)\sin T\omega
-$$
-
-Use the latter result and Table 7.1 to find the Fourier transform of the signal in Fig. P7.3-8.
-
-2 234
-
-- **7.3-9** The signals in Fig. P7.3-9 are modulated signals with carrier cos 10*t*. Find the Fourier transforms of these signals by using the appropriate properties of the Fourier transform and Table 7.1. Sketch the amplitude and phase spectra for Figs. P7.3-9a and P7.3-9b.
-- **7.3-10** Use the frequency-shifting property and Table 7.1 to find the inverse Fourier transform of the spectra depicted in Fig. P7.3-10.
-- **7.3-11** Let *X*(ω) = rect(ω) be the Fourier transform of a signal *x*(*t*).
-  - (a) For *y*a(*t*) = *x*(*t*) ∗ *x*(*t*), sketch *Y*a(ω).
-  - (b) For *y*b(*t*) = *x*(*t*) ∗ *x*(*t*/2), sketch *Y*b(ω).
-
-(a) (b)
-
-**Figure P7.3-10**
-
-- (c) For *y*c(*t*) = 2*x*(*t*), sketch *Y*c(ω).
-- (d) For *y*d(*t*) = *x*2(*t*), sketch *Y*d(ω).
-- (e) For *y*e(*t*) = 1−*x*2(*t*), sketch *Y*e(ω).
-- **7.3-12** Use the time-convolution property to prove pairs 2, 4, 13, and 14 in Table 2.1 (assume λ < 0 in pair 2, λ<sup>1</sup> and λ<sup>2</sup> < 0 in pair 4, λ<sup>1</sup> < 0 and λ<sup>2</sup> > 0 in pair 13, and λ<sup>1</sup> and λ<sup>2</sup> > 0 in pair 14). These restrictions are placed because of the Fourier transformability issue for the signals concerned. For pair 2, you need to apply the result in Eq. (1.10).
-- **7.3-13** A signal *x*(*t*) is bandlimited to *B* Hz. Show that the signal *xn*(*t*) is bandlimited to *nB* Hz.
-- **7.3-14** Find the Fourier transform of the signal in Fig. P7.3-5a by three different methods:
-  - (a) By direct integration using Eq. (7.9).
-  - (b) Using only pair 17 (Table 7.1) and the time-shifting property.
-  - (c) Using the time-differentiation and time-shifting properties, along with the fact that δ(*t*) ⇐⇒ 1.
-
-**7.3-15** (a) Prove the frequency-differentiation property (dual of the time-differentiation property):
-
-$$
--jtx(t) \Longleftrightarrow \frac{d}{d\omega}X(\omega)
-$$
-
-- (b) Use this property and pair 1 (Table 7.1) to determine the Fourier transform of *te*−*atu*(*t*).
-- **7.3-16** Adapt the method of Ex. 7.17 and use the frequency-differentiation (see Prob. 7.3-15) and other properties to find the inverse Fourier transform *x*(*t*) of the triangular spectrum *X*(ω) = (ω/2).
-- **7.3-17** Adapt the method of Ex. 7.17 and use the frequency-differentiation (see Prob. 7.3-15) and other properties to find the inverse Fourier transform *x*(*t*) of the spectrum *X*(ω) = π ω 2 rect <sup>ω</sup> 4 .
-- **7.4-1** For a stable LTIC system with transfer function
-
-$$
-H(s) = \frac{1}{s+1}
-$$
-
-find the (zero-state) response if the input *x*(*t*) is
-
-- (a) *e*−2*<sup>t</sup> u*(*t*)
-- (b) *e*−*<sup>t</sup> u*(*t*)
-- (c) *et u*(−*t*)
-- (d) *u*(*t*)
-- **7.4-2** A stable LTIC system is specified by the frequency response
-
-$$
-H(\omega) = \frac{-1}{j\omega - 2}
-$$
-
-Find the impulse response of this system and show that this is a noncausal system. Find the (zero-state) response of this system if the input *x*(*t*) is
-
-- (a) *e*−*<sup>t</sup> u*(*t*)
-- (b) *et u*(−*t*)
-- **7.4-3** A periodic signal *x*(*t*) = 1 + 2 cos(5π*t*) + 3 sin(8π*t*) is applied to an LTIC system with impulse response *h*(*t*) = 8sinc(4*t*) cos(2π*t*) to produce output *y*(*t*) = *x*(*t*) ∗ *h*(*t*).
-  - (a) Determine ω0, the fundamental radian frequency of *x*(*t*).
-  - (b) Determine *X*(ω), the Fourier transform of *x*(*t*).
-  - (c) Sketch the system's magnitude response |*H*(ω)| over −10π ≤ ω ≤ 10π.
-  - (d) Is the system *h*(*t*) distortionless? Explain.
-  - (e) Determine *y*(*t*).
-- **7.4-4** A periodic delta train *<sup>x</sup>*(*t*) <sup>=</sup> %<sup>∞</sup> *<sup>n</sup>*=−∞ δ(*t* − π*n*) is applied to an LTIC system with impulse response *<sup>h</sup>*(*t*) <sup>=</sup> sin(3*t*)sinc2 *<sup>t</sup>* π to produce zero-state output *y*(*t*) = *x*(*t*) ∗ *h*(*t*).
-  - (a) Determine ω0, the fundamental radian frequency of *x*(*t*).
-  - (b) Determine *X*(ω), the Fourier transform of *x*(*t*).
-  - (c) Sketch the system's magnitude response |*H*(ω)| over −10π ≤ ω ≤ 10π.
-  - (d) Is the system *h*(*t*) distortionless? Explain.
-  - (e) Determine *y*(*t*).
-- **7.4-5** Signals *x*1(*t*)=104rect(104*t*) and *x*2(*t*) = δ(*t*) are applied at the inputs of the ideal lowpass filters *H*1(ω) = rect(ω/40,000π ) and *H*2(ω) = rect(ω/20,000π ) (Fig. P7.4-5). The outputs *y*1(*t*) and *y*2(*t*) of these filters are multiplied to obtain the signal *y*(*t*) = *y*1(*t*)*y*2(*t*). (a) Sketch *X*1(ω) and *X*2(ω).
-
-- (b) Sketch *H*1(ω) and *H*2(ω).
-- (c) Sketch *Y*1(ω) and *Y*2(ω).
-- (d) Find the bandwidths of *y*1(*t*), *y*2(*t*), and *y*(*t*).
-
-- **7.4-6** A lowpass system time constant is often defined as the width of its unit impulse response *h*(*t*) (see Sec. 2.6-2). An input pulse *p*(*t*) to this system acts like an impulse of strength equal to the area of *p*(*t*) if the width of *p*(*t*) is much smaller than the system time constant, and provided *p*(*t*) is a lowpass pulse, implying that its spectrum is concentrated at low frequencies. Verify this behavior by considering a system whose unit impulse response is *h*(*t*) = rect(*t*/10−3). The input pulse is a triangle pulse *p*(*t*) = (*t*/10−6). Show that the system response to this pulse is very nearly the system response to the input *A*δ(*t*), where *A* is the area under the pulse *p*(*t*).
-- **7.4-7** A lowpass system time constant is often defined as the width of its unit impulse response *h*(*t*) (see Sec. 2.6-2). An input pulse *p*(*t*) to this system passes practically without distortion if the width of *p*(*t*) is much greater than the system time constant, and provided *p*(*t*) is a lowpass pulse, implying that its spectrum is concentrated at low frequencies. Verify this behavior by considering a system whose unit impulse response is *h*(*t*) = rect(*t*/10−3). The input pulse is a triangle pulse *p*(*t*)=(*t*). Show that the system output to this pulse is very nearly *kp*(*t*), where *k* is the system gain to a dc signal, that is, *k* = *H*(0).
-- **7.4-8** A causal signal *h*(*t*) has a Fourier transform *H*(ω). If *R*(ω) and *X*(ω) are the real and the imaginary parts of *H*(ω), that is, *H*(ω) = *R*(ω)+ *jX*(ω), then show that
-
-$$
-R(\omega) = \frac{1}{\pi} \int_{-\infty}^{\infty} \frac{X(\omega)}{\omega - y} d\omega
-$$
-
-and
-
-$$
-X(\omega) = -\frac{1}{\pi} \int_{-\infty}^{\infty} \frac{R(\omega)}{\omega - y} d\omega
-$$
-
-assuming that *h*(*t*) has no impulse at the origin. This pair of integrals defines the *Hilbert transform*. [*Hint:* Let *he*(*t*) and *ho*(*t*) be the even and odd components of *h*(*t*). Use the results in Prob. 7.1-4. See Fig. 1.24 for the relationship between *he*(*t*) and *ho*(*t*).]
-
-This problem states one of the important properties of causal systems: that the real and imaginary parts of the frequency response of a causal system are related. If one specifies the real part, the imaginary part cannot be specified independently. The imaginary part is predetermined by the real part, and vice versa. This result also leads to the conclusion that the magnitude and angle of *H*(ω) are related, provided all the poles and zeros of *H*(ω) lie in the LHP.
-
-**7.5-1** Consider a filter with the frequency response
-
-$$
-H(\omega) = e^{-(k\omega^2 + j\omega t_0)}
-$$
-
-Show that this filter is physically unrealizable by using the time-domain criterion [noncausal *h*(*t*)] and the frequency-domain (Paley–Wiener) criterion. Can this filter be made approximately realizable by choosing *t*<sup>0</sup> sufficiently large? Use your own (reasonable) criterion of approximate realizability to determine *t*0. [*Hint:* Use pair 22 in Table 7.1.]
-
-**7.5-2** Show that a filter with frequency response
-
-$$
-H(\omega) = \frac{2(10^5)}{\omega^2 + 10^{10}} e^{-j\omega t_0}
-$$
-
-is unrealizable. Can this filter be made approximately realizable by choosing a sufficiently large *t*0? Use your own (reasonable) criterion of approximate realizability to determine *t*0.
-
-- **7.5-3** Determine whether the filters with the following frequency response *H*(ω) are physically realizable. If they are not realizable, can they be realized approximately by allowing a finite time delay in the response?
-  - (a) 10−<sup>6</sup> sinc (10−6ω)
-  - (b) 10−<sup>4</sup> (ω/40,000π)
-  - (c) 2π δ(ω)
-
-- **7.5-4** Consider signal *x*1(*t*), its Fourier transform *X*1(*f*), and several other signals, as shown in Fig. P7.5-4. Notice, spectra are drawn as a function of hertzian frequency *f* rather than radian frequency ω.
-  - (a) Accurately sketch *X*2(*f*), the Fourier transform of *x*2(*t*).
-  - (b) Accurately sketch *x*3(*t*), the inverse Fourier transform of *X*3(*f*).
-  - (c) The signal *x*4(*t*) = *x*1(*t*) + *x*2(*t*) is passed through an ideal lowpass filter with 3 Hz cutoff to produce output *y*4(*t*). Accurately sketch *y*4(*t*).
-
-**Figure P7.5-4**
-
-- **7.6-1** Define *<sup>x</sup>*(*t*) <sup>=</sup> <sup>1</sup> <sup>2</sup><sup>π</sup> sinc(*t*/2) with Fourier transform *X*(ω) = rect(ω). Use Parseval's theorem to determine \$ <sup>∞</sup> −∞ sinc2(*<sup>t</sup>* <sup>−</sup>2)*dt*.
-- **7.6-2** Show that the energy of a Gaussian pulse
-
-$$
-x(t) = \frac{1}{\sigma\sqrt{2\pi}}e^{-t^2/2\sigma^2}
-$$
-
-is 1/(2σ <sup>√</sup>π ). Verify this result by using Parseval's theorem to derive the energy *Ex* from *X*(ω). [*Hint:* See pair 22 in Table 7.1. Use the fact that \$ ∞ −∞ *<sup>e</sup>*−*x*2/<sup>2</sup> *dx* <sup>=</sup> <sup>√</sup>2π.]
-
-**7.6-3** Use Parseval's theorem of Eq. (7.45) to show that
-
-$$
-\int_{-\infty}^{\infty} \operatorname{sinc}^2(kx) dx = \frac{\pi}{k}
-$$
-
-- **7.6-4** A lowpass signal *x*(*t*) is applied to a squaring device. The squarer output *x*2(*t*) is applied to a lowpass filter of bandwidth *f* (in hertz) (Fig. P7.6-4). Show that if *f* is very small (*f* → 0), then the filter output is a dc signal *y*(*t*) ≈ 2*Exf* . [*Hint:* If *x*2(*t*) ⇐⇒ *A*(ω), then show that *Y*(ω) ≈ [4π*A*(0)*f*]δ(ω) if *f* → 0. Now, show that *A*(0) = *Ex*.]
-- **7.6-5** Generalize Parseval's theorem to show that for real, Fourier-transformable signals *x*1(*t*) and *x*2(*t*)
-
-$$
-\int_{-\infty}^{\infty} x_1(t) x_2(t) dt
-$$
-  
-=  $\frac{1}{2\pi} \int_{-\infty}^{\infty} X_1(-\omega) X_2(\omega) d\omega$   
-=  $\frac{1}{2\pi} \int_{-\infty}^{\infty} X_1(\omega) X_2(-\omega) d\omega$ 
-
-**7.6-6** Show that
-
-$$
-\int_{-\infty}^{\infty} \operatorname{sinc} (Wt - m\pi) \operatorname{sinc} (Wt - n\pi) dt
-$$
-$$
-= \begin{cases} 0 & m \neq n \\ \frac{\pi}{W} & m = n \end{cases}
-$$
-
-( )2 Lowpass filter *x*(*t*) *x y*(*t*) 2*Exf* 2(*t*)
-
-**Figure P7.6-4**
-
-[*Hint:* Recognize that
-
-$$
-\operatorname{sinc}(Wt - k\pi) = \operatorname{sinc}\left[W\left(t - \frac{k\pi}{W}\right)\right] \\
-\Longleftrightarrow \frac{\pi}{W} \operatorname{rect}\left(\frac{\omega}{2W}\right) e^{-jk\pi\omega/W}
-$$
-
-Use this fact and the result in Prob. 7.6-5.]
-
-- **7.6-7** (a) What does it mean to compute the 95% essential bandwidth *B* of a signal *x*(*t*) with Fourier transform *X*(ω)?
-  - (b) Determine the 95% essential bandwidth *B* of a signal with spectrum *X*(ω) = rect(ω).
-  - (c) Determine the 95% essential bandwidth *B* of a signal with spectrum *X*(ω) = (ω).
-- **7.6-8** Using a 95% energy criterion, determine the essential bandwidth *B* of a signal that has a Fourier transform given by *X*(ω) = *e*−|ω<sup>|</sup> .
-- **7.6-9** For the signal
-
-$$
-x(t) = \frac{2a}{t^2 + a^2}
-$$
-
-determine the essential bandwidth *B* (in hertz) of *x*(*t*) such that the energy contained in the spectral components of *x*(*t*) of frequencies below *B* Hz is 99% of the signal energy *Ex*.
-
-- **7.7-1** For each of the following baseband signals **(i)** *m*(*t*) = cos 1000*t*, **(ii)** *m*(*t*) = 2 cos 1000*t* + cos 2000*t*, and **(iii)** *m*(*t*) = cos 1000*t* cos 3000*t*:
-  - (a) Sketch the spectrum of *m*(*t*).
-  - (b) Sketch the spectrum of the DSB-SC signal *m*(*t*) cos 10,000*t*.
-  - (c) Identify the upper sideband (USB) and the lower sideband (LSB) spectra.
-  - (d) Identify the frequencies in the baseband, and the corresponding frequencies in the DSB-SC, USB, and LSB spectra. Explain the nature of frequency shifting in each case.
-- **7.7-2** A message signal *<sup>m</sup>*(*t*) with spectrum *<sup>M</sup>*(ω) <sup>=</sup> <sup>1</sup> 1000 <sup>ω</sup> 2π6000 is to be transmitted using a communication system. Assume all single-sideband systems have suppressed carriers.
-  - (a) What is the hertzian bandwidth of *m*(*t*)?
-
-- (b) Sketch the spectrum of the transmitted signal if the communication system is DSB-SC with ω*<sup>c</sup>* = 2π 100,000.
-- (c) Sketch the spectrum of the transmitted signal if the communication system is AM with ω*<sup>c</sup>* = 2π 100,000 and a modulation index of μ = 1. What is the corresponding carrier amplitude *A*?
-- (d) Sketch the spectrum of the transmitted signal if the communication system is USB with ω*<sup>c</sup>* = 2π 100,000.
-- (e) Sketch the spectrum of the transmitted signal if the communication system is LSB with ω*<sup>c</sup>* = 2π 100,000.
-- (f) Suppose we want to transmit *m*(*t*) on each of an FDM system's four channels: DSB-SC at carrier ω1, AM (μ = 1) at carrier ω2, USB at carrier ω3, and LSB at carrier ω4. Determine carrier frequencies ω<sup>1</sup> < ω<sup>2</sup> < ω<sup>3</sup> < ω<sup>4</sup> so that the FDM spectrum begins at a frequency of 100,000 Hz with 5,000 Hz deadbands separating adjacent messages. What is the end hertzian frequency of the FDM signal?
-- **7.7-3** You are asked to design a DSB-SC modulator to generate a modulated signal *km*(*t*) cosω*ct*, where *m*(*t*) is a signal bandlimited to *B* Hz (Fig. P7.7-3a). Figure P7.7-3b shows a DSB-SC modulator available in the stockroom. The bandpass filter is tuned to ω*<sup>c</sup>* and has a bandwidth of 2*B* Hz. The carrier generator available generates not cosω*ct*, but cos<sup>3</sup> ω*ct*.
-  - (a) Explain whether you would be able to generate the desired signal using only this equipment. If so, what is the value of *k*?
-
-- (b) Determine the signal spectra at points *b* and *c*, and indicate the frequency bands occupied by these spectra.
-- (c) What is the minimum usable value of ω*c*?
-- (d) Would this scheme work if the carrier generator output were cos2 ω*ct*? Explain.
-- (e) Would this scheme work if the carrier generator output were cos*<sup>n</sup>* ω*ct* for any integer *n* ≥ 2?
-- **7.7-4** In practice, the analog multiplication operation is difficult and expensive. For this reason, in amplitude modulators, it is necessary to find some alternative to multiplication of *m*(*t*) with cosω*ct*. Fortunately, for this purpose, we can replace multiplication with a switching operation. A similar observation applies to demodulators. In the scheme depicted in Fig. P7.7-4a, the period of the rectangular periodic pulse *x*(*t*)
-
-**Figure P7.7-4**
-
-**Figure P7.7-3**
-
-shown in Fig. P7.7-4b is *T*<sup>0</sup> = 2π/ω*c*. The bandpass filter is centered at ±ω*<sup>c</sup>* and has a bandwidth of 2*B* Hz. Note that multiplication by a square periodic pulse *x*(*t*) in Fig. P7.7-4b amounts to periodic on-off switching of *m*(*t*), which is bandlimited to *B* Hz. Such a switching operation is relatively simple and inexpensive. Show that this scheme can generate an amplitude-modulated signal *k* cos ω*ct*. Determine the value of *k*. Show that the same scheme can also be used for demodulation, provided the bandpass filter in Fig. P7.7-4a is replaced by a lowpass (or baseband) filter.
-
-**7.7-5** Figure P7.7-5a shows a scheme to transmit two signals *m*1(*t*) and *m*2(*t*) simultaneously on the same channel (without causing spectral interference). Such a scheme, which transmits more than one signal, is known as signal *multiplexing*. In this case, we transmit multiple signals by sharing an available spectral band on the channel; hence, this is an example of the *frequency-division* multiplexing. The signal at point *b* is the multiplexed signal, which now modulates a carrier of frequency 20,000 rad/s. The modulated signal at point *c* is now transmitted over the channel.
-
-- (a) Sketch the spectra at points *a*, *b*, and *c*.
-- (b) What must be the minimum bandwidth of the channel?
-- (c) Design a receiver to recover signals *m*1(*t*) and *m*2(*t*) from the modulated signal at point *c*.
-- **7.7-6** The system shown in Fig. P7.7-6 is used for scrambling audio signals. The output *y*(*t*) is the scrambled version of the input *m*(*t*).
-
-**Figure P7.7-6**
-
-- (a) Find the spectrum of the scrambled signal *y*(*t*).
-- (b) Suggest a method of descrambling *y*(*t*) to obtain *m*(*t*).
-
-A slightly modified version of this scrambler was first used commercially on the 25-mile radio-telephone circuit connecting Los Angeles and Santa Catalina Island.
-
-- **7.7-7** Figure P7.7-7 presents a scheme for coherent (synchronous) demodulation. Show that this scheme can demodulate the AM signal [*A* + *m*(*t*)] cos ω*ct* regardless of the value of *A*.
-- **7.7-8** Sketch the AM signal [*A* + *m*(*t*)] cos ω*ct* for the periodic triangle signal *m*(*t*) illustrated in Fig. P7.7-8 corresponding to the following modulation indices:
-  - (a) μ = 0.5
-  - (b) μ = 1
-  - (c) μ = 2
-  - (d) μ = ∞
-
-How do you interpret the case μ = ∞?
-
-**7.9-1** Consider the signal *x*(*t*) defined as
-
-$$
-x(t) = \begin{cases} 1 - |t| & -\frac{1}{2} \le t \le \frac{1}{2} \\ 0 & \text{otherwise} \end{cases}
-$$
-
-- (a) Sketch the signal *x*(*t*) over −2 ≤ *t* ≤ 2.
-- (b) Use time-differentiation and other Fourier transform properties to determine *X*(ω). The only integration you should use is to determine the dc component *X*(0).
-
-(c) Using MATLAB, verify the correctness of *X*(ω) by synthesizing a 3-periodic replication of the original time-domain signal *x*(*t*).
-
-[*Hint:* Follow the approach taken in Ex. 7.17.]
-
-- **7.9-2** Consider the signal *<sup>x</sup>*(*t*) = |*t*|rect *<sup>t</sup>*−<sup>1</sup> 3 .
-  - (a) Sketch the signal *x*(*t*) over −5 ≤ *t* ≤ 5.
-  - (b) Use time-differentiation and other Fourier transform properties to determine *X*(ω). The only integration you should use is to determine the dc component *X*(0).
-  - (c) Use MATLAB to plot the magnitude spectrum |*X*(ω)| and the phase spectrum *X*(ω) over suitable ranges of ω.
-  - (d) Using MATLAB, verify the correctness of *X*(ω) by synthesizing a 10-periodic replication of the original time-domain signal *x*(*t*).
-
-[*Hint:* Follow the approach taken in Ex. 7.17.]
-
-- **7.9-3** Consider the continuous-time aperiodic signal *x*(*t*) = rect(*t*) with Fourier transform *X*(ω) = sinc(ω/2). Furthermore, let *y*(*t*) = (1 − |*t* − 1|)(*u*(*t*) − *u*(*t* − 2)) with Fourier transform *Y*(ω).
-  - (a) Express the Fourier transform *Y*(ω) in terms of *X*(ω).
-  - (b) Suppose we create Fourier series coefficients *Vk* by sampling *Y*(ω) according to *Vk* = *Y*(2π*k*/3). Sketch the corresponding time-domain signal *v*(*t*) over a suitable range of time *t*.
-  - (c) Use MATLAB to synthesize and plot *v*(*t*) using the Fourier series coefficients *Vk* =
-
-*Y*(2π*k*/3). Verify that the synthesized waveform matches the result of part (b).
-
-- (d) Suppose we again create Fourier series coefficients *Vk* according to *Vk* = *Y*(2π*k*/3). Next, we upsample *Vk* by factor 2 to create *Wk*. Sketch the time domain signal *p*(*t*) that has Fourier series coefficients *Pk* = *Vk* +*Wk*.
-- (e) Use MATLAB to synthesize and plot *p*(*t*) using the Fourier series coefficients *Pk* = *Vk* + *Wk* defined in part (d). Verify that the synthesized waveform matches the result of part (d).
-- **7.9-4** Consider the signal *<sup>x</sup>*(*t*) <sup>=</sup> *<sup>e</sup>*−*atu*(*t*). Modify CH7MP2 to compute the following essential bandwidths.
-  - (a) Setting *a*=1, determine the essential bandwidth *W*<sup>1</sup> that contains 95% of the signal energy. Compare this value with the theoretical value presented in Ex. 7.20.
-  - (b) Setting *a*=2, determine the essential bandwidth *W*<sup>2</sup> that contains 90% of the signal energy.
-  - (c) Setting *a*=3, determine the essential bandwidth *W*<sup>3</sup> that contains 75% of the signal energy.
-- **7.9-5** A unit amplitude pulse with duration τ is defined as
-
-$$
-x(t) = \begin{cases} 1 & |t| \le \tau/2 \\ 0 & \text{otherwise} \end{cases}
-$$
-
-- (a) Determine the duration τ<sup>1</sup> that results in a 95% essential bandwidth of 5 Hz.
-- (b) Determine the duration τ<sup>2</sup> that results in a 90% essential bandwidth of 10 Hz.
-- (c) Determine the duration τ<sup>3</sup> that results in a 75% essential bandwidth 20 Hz.
-- **7.9-6** Consider the signal *<sup>x</sup>*(*t*) <sup>=</sup> *<sup>e</sup>*−*atu*(*t*).
-
-- (a) Determine the decay parameter *a*<sup>1</sup> that results in a 95% essential bandwidth of 5 Hz.
-- (b) Determine the decay parameter *a*<sup>2</sup> that results in a 90% essential bandwidth of 10 Hz.
-- (c) Determine the decay parameter *a*<sup>3</sup> that results in a 75% essential bandwidth 20 Hz.
-- **7.9-7** Use MATLAB to determine the 95, 90, and 75% essential bandwidths of a one-second triangle function with a peak amplitude of 1. Recall that a triangle function can be constructed by the convolution of two rectangular pulses.
-- **7.9-8** A 1/3 duty-cycle square-pulse *T*0-periodic signal *x*(*t*) is described as
-
-$$
-x(t) = \begin{cases} 1 & -T_0/6 \le t \le T_0/6 \\ 0 & T_0/t \le |t| \le T_0/2 \\ x(t+T_0) & \forall t \end{cases}
-$$
-
-- (a) Use spectral sampling to determine the Fourier series coefficients *Dn* of *x*(*t*) for *T*<sup>0</sup> = 2π. Evaluate and plot *Dn* for (0 ≤ *n* ≤ 10).
-- (b) Use spectral sampling to determine the Fourier series coefficients *Dn* of *x*(*t*) for *T*<sup>0</sup> = π. Evaluate and plot *Dn* for (0 ≤ *n* ≤ 10). How does this result compare with your answer to part (a)? What can be said about the relation of *T*<sup>0</sup> to *Dn* for signal *x*(*t*), which has fixed duty cycle of 1/3?
-- **7.9-9** Determine the Fourier transform of a Gaussian pulse defined as *<sup>x</sup>*(*t*) <sup>=</sup> *<sup>e</sup>*−*<sup>t</sup>* 2 . Plot both *x*(*t*) and *X*(ω). How do the two curves compare? [*Hint:*
-
-$$
-\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-(t-a)^2/2}dt=1
-$$
-
-for any real or imaginary *a*.]

engineering/linear-systems-and-signals-3rd-edition-bp-lathi/101_8 SAMPLING - THE BRIDGE FROM CONTINUOUS TO DISCRETE.md

@@ -1,5 +0,0 @@
-# **SAMPLING: THE BRIDGE FROM [CONTINUOUS TO](#page-13-0) DISCRETE**
-
-A continuous-time signal can be processed by applying its samples through a discrete-time system. For this purpose, it is important to maintain the signal sampling rate high enough to permit the reconstruction of the original signal from these samples without error (or with an error within a given tolerance). The necessary quantitative framework for this purpose is provided by the sampling theorem derived in Sec. 8.1.
-
-Sampling theory is the bridge between the continuous-time and discrete-time worlds. The information inherent in a sampled continuous-time signal is equivalent to that of a discrete-time signal. A sampled continuous-time signal is a sequence of impulses, while a discrete-time signal presents the same information as a sequence of numbers. These are basically two different ways of presenting the same data. Clearly, all the concepts in the analysis of sampled signals apply to discrete-time signals. We should not be surprised to see that the Fourier spectra of the two kinds of signal are also the same (within a multiplicative constant).

engineering/linear-systems-and-signals-3rd-edition-bp-lathi/102_8.1 THE SAMPLING THEOREM.md

@@ -1,168 +0,0 @@
-## **8.1 THE [SAMPLING](#page-13-0) THEOREM**
-
-We now show that a real signal whose spectrum is bandlimited to *B* Hz [*X*(ω) = 0 for |ω| > 2π*B*] can be reconstructed exactly (without any error) from its samples taken uniformly at a rate *fs* > 2*B* samples per second. In other words, the minimum sampling frequency is *fs* = 2*B* Hz.†
-
-To prove the sampling theorem, consider a signal *x*(*t*) (Fig. 8.1a) whose spectrum is bandlimited to *B* Hz (Fig. 8.1b).‡ For convenience, spectra are shown as functions of ω as well as of *f* (hertz). Sampling *x*(*t*) at a rate of *fs* Hz ( *fs* samples per second) can be accomplished by multiplying *x*(*t*) by an impulse train δ*<sup>T</sup>* (*t*) (Fig. 8.1c), consisting of unit impulses repeating periodically every *T* seconds, where *T* = 1/*fs*. The schematic of a sampler is shown in Fig. 8.1d. The resulting sampled signal *x*(*t*) is shown in Fig. 8.1e. The sampled signal consists of impulses
-
-<span id="page-795-0"></span>CHAPTER
-
-**8**
-
-<sup>†</sup> The theorem stated here (and proved subsequently) applies to lowpass signals. A bandpass signal whose spectrum exists over a frequency band *fc* − (*B*/2) < |*f* | < *fc* + (*B*/2) has a bandwidth of *B* Hz. Such a signal is uniquely determined by 2*B* samples per second. In general, the sampling scheme is a bit more complex in this case. It uses two interlaced sampling trains, each at a rate of *B* samples per second. See, for example, [1]. ‡ The spectrum *X*(ω) in Fig. 8.1b is shown as real, for convenience. However, our arguments are valid for complex *X*(ω) as well.
-
-**Figure 8.1** Sampled signal and its Fourier spectrum.
-
-spaced every *T* seconds (the sampling interval). The *n*th impulse, located at *t* = *nT*, has a strength *x*(*nT*), the value of *x*(*t*) at *t* = *nT*.
-
-$$
-\bar{x}(t) = x(t)\delta_T(t) = \sum_n x(nT)\delta(t - nT)
-$$
-
-Because the impulse train δ*<sup>T</sup>* (*t*) is a periodic signal of period *T*, it can be expressed as a trigonometric Fourier series like that already obtained in Ex. 6.9 [Eq. (6.25)],
-
-$$
-\delta_T(t) = \frac{1}{T} [1 + 2\cos\omega_s t + 2\cos 2\omega_s t + 2\cos 3\omega_s t + \cdots] \qquad \omega_s = \frac{2\pi}{T} = 2\pi f_s
-$$
-
-Therefore,
-
-$$
-\bar{x}(t) = x(t)\delta_T(t) = \frac{1}{T}[x(t) + 2x(t)\cos\omega_s t + 2x(t)\cos 2\omega_s t + 2x(t)\cos 3\omega_s t + \cdots]
-$$
-(8.1)
-
-To find *X*(ω), the Fourier transform of *x*(*t*), we take the Fourier transform of the right-hand side of Eq. (8.1), term by term. The transform of the first term in the brackets is *X*(ω). The transform
-
-#### 778 CHAPTER 8 SAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE
-
-of the second term 2*x*(*t*) cos ω*st* is *X*(ω − ω*s*) + *X*(ω + ω*s*) [see Eq. (7.32)]. This represents spectrum *X*(ω) shifted to ω*<sup>s</sup>* and −ω*s*. Similarly, the transform of the third term 2*x*(*t*) cos 2ω*st* is *X*(ω − 2ω*s*) + *X*(ω + 2ω*s*), which represents the spectrum *X*(ω) shifted to 2ω*<sup>s</sup>* and −2ω*s*, and so on to infinity. This result means that the spectrum *X*(ω) consists of *X*(ω) repeating periodically with period ω*<sup>s</sup>* = 2π/*T* rad/s, or *fs* = 1/*T* Hz, as depicted in Fig. 8.1f. There is also a constant multiplier 1/*T* in Eq. (8.1). Therefore,
-
-$$
-\overline{X}(\omega) = \frac{1}{T} \sum_{n=-\infty}^{\infty} X(\omega - n\omega_s)
-$$
-\n(8.2)
-
-If we are to reconstruct *x*(*t*) from *x*(*t*), we should be able to recover *X*(ω) from *X*(ω). This recovery is possible if there is no overlap between successive cycles of *X*(ω). Figure 8.1f indicates that this requires
-
-$$
-f_s > 2B \tag{8.3}
-$$
-
-Also, the sampling interval *T* = 1/*fs*. Therefore,
-
-$$
-T < \frac{1}{2B}
-$$
-
-Thus, as long as the sampling frequency *fs* is greater than twice the signal bandwidth *B* (in hertz), *X*(ω) consists of nonoverlapping repetitions of *X*(ω). Figure 8.1f shows that the gap between the two adjacent spectral repetitions is *fs* − 2*B* Hz, and *x*(*t*) can be recovered from its samples *x*(*t*) by passing the sampled signal *x*(*t*) through an ideal lowpass filter having a bandwidth of any value between *B* and *fs* − *B* Hz. The minimum sampling rate *fs* = 2*B* required to recover *x*(*t*) from its samples *x*(*t*) is called the *Nyquist rate* for *x*(*t*), and the corresponding sampling interval *T* = 1/2*B* is called the *Nyquist interval* for *x*(*t*). Samples of a signal taken at its Nyquist rate are the *Nyquist samples* of that signal.
-
-We are saying that the Nyquist rate 2*B* Hz is the minimum sampling rate required to preserve the information of *x*(*t*). This contradicts Eq. (8.3), where we showed that to preserve the information of *x*(*t*), the sampling rate *fs* needs to be greater than 2*B* Hz. Strictly speaking, Eq. (8.3) is the correct statement. However, if the spectrum *X*(ω) contains no impulse or its derivatives at the highest frequency *B* Hz, then the minimum sampling rate 2*B* Hz is adequate. In practice, it is rare to observe *X*(ω) with an impulse or its derivatives at the highest frequency. If the contrary situation were to occur, we should use Eq. (8.3).†
-
-<sup>†</sup> An interesting observation is that if the impulse is because of a cosine term, the sampling rate of 2*B* Hz is adequate. However, if the impulse is because of a sine term, then the rate must be greater than 2*B* Hz. This may be seen from the fact that samples of sin 2π*Bt* using *T* = 1/2*B* are all zero because sin 2π*BnT* = sinπ*n* = 0. But, samples of cos 2π*Bt* are cos 2π*BnT* <sup>=</sup> cosπ*<sup>n</sup>* <sup>=</sup> (−1)*n*. We can reconstruct cos 2π*Bt* from these samples. This peculiar behavior occurs because in the sampled signal spectrum corresponding to the signal cos 2π*Bt*, the impulses, which occur at frequencies (2*n* ± 1)*B* Hz (*n* = 0,±1,±2,. . .), interact constructively, whereas in the case of sin 2π*Bt*, the impulses, because of their opposite phases (*e*±*j*π/2), interact destructively and cancel out in the sampled signal spectrum. Hence, sin 2π*Bt* cannot be reconstructed from its samples at a rate 2*B* Hz. A similar situation exists for signal cos(2π*Bt*+θ ), which contains a component of the form sin 2π*Bt*. For this reason, it is advisable to maintain sampling rate above 2*B* Hz if a finite-amplitude component of a sinusoid of frequency *B* Hz is present in the signal.
-
-The sampling theorem proved here uses samples taken at uniform intervals. This condition is not necessary. Samples can be taken arbitrarily at any instants as long as the sampling instants are recorded and there are, on average, 2*B* samples per second [2]. The essence of the sampling theorem was known to mathematicians for a long time in the form of the *interpolation formula* [see later, Eq. (8.6)]. The origin of the sampling theorem was attributed by H. S. Black to Cauchy in 1841. The essential idea of the sampling theorem was rediscovered in the 1920s by Carson, Nyquist, and Hartley.
-
-### **EXAMPLE 8.1 Sampling at, Below, and Above the Nyquist Rate**
-
-In this example, we examine the effects of sampling a signal at the Nyquist rate, below the Nyquist rate (undersampling), and above the Nyquist rate (oversampling). Consider a signal *<sup>x</sup>*(*t*) <sup>=</sup> sinc<sup>2</sup> (5π*t*) (Fig. 8.2a) whose spectrum is *<sup>X</sup>*(ω) <sup>=</sup> 0.2(ω/20π ) (Fig. 8.2b). The bandwidth of this signal is 5 Hz (10π rad/s). Consequently, the Nyquist rate is 10 Hz; that is, we must sample the signal at a rate no less than 10 samples/s. The Nyquist interval is *T* = 1/2*B* = 0.1 second.
-
-Recall that the sampled signal spectrum consists of (1/*T*)*X*(ω) = (0.2/*T*)(ω/20π ) repeating periodically with a period equal to the sampling frequency *fs* Hz. For the three sampling rates *fs* = 5 Hz (undersampling), 10 Hz (Nyquist rate), and 20 Hz (oversampling), we see that
-
-| fs<br>(Hz) | T<br>= 1<br>(s)<br>fs | TX(ω)<br>1     | Comments      |
-|------------|-----------------------|----------------|---------------|
-| 5          | 0.2                   | ω<br><br>20π   | Undersampling |
-| 10         | 0.1                   | 2 ω<br><br>20π | Nyquist rate  |
-| 20         | 0.05                  | 4 ω<br><br>20π | Oversampling  |
-
-In the first case (undersampling), the sampling rate is 5 Hz (5 samples/s), and the spectrum (1/*T*)*X*(ω) repeats every 5 Hz (10π rad/s). The successive spectra overlap, as depicted in Fig. 8.2d, and the spectrum *X*(ω) are not recoverable from *X*(ω); that is, *x*(*t*) cannot be reconstructed from its samples *x*(*t*) in Fig. 8.2c. In the second case, we use the Nyquist sampling rate of 10 Hz (Fig. 8.2e). The spectrum *X*(ω) consists of back-to-back, nonoverlapping repetitions of (1/*T*)*X*(ω) repeating every 10 Hz. Hence, *X*(ω) can be recovered
-
-**Figure 8.2** Effects of undersampling and oversampling.
-
-<span id="page-800-0"></span>from *X*(ω) using an ideal lowpass filter of bandwidth 5 Hz (Fig. 8.2f). Finally, in the last case of oversampling (sampling rate 20 Hz), the spectrum *X*(ω) consists of nonoverlapping repetitions of (1/*T*)*X*(ω) (repeating every 20 Hz) with empty bands between successive cycles (Fig. 8.2h). Hence, *X*(ω) can be recovered from *X*(ω) by using an ideal lowpass filter or even a practical lowpass filter (shown dashed in Fig. 8.2h).†
-
-### **DR ILL 8.1 Nyquist Sampling**
-
-Find the Nyquist rate and the Nyquist sampling interval for the signals sinc(100π*t*) and sinc(100π*t*)+sinc(50π*t*).
-
-### **ANSWERS**
-
-The Nyquist sampling interval is 0.01 s and the Nyquist sampling rate is 100 Hz for both signals.
-
-### FOR SKEPTICS ONLY
-
-Rare is the reader who, at first encounter, is not skeptical of the sampling theorem. It seems impossible that Nyquist samples can define the one and the only signal that passes through those sample values. We can easily picture infinite number of signals passing through a given set of samples. However, among all these (infinite number of) signals, only one has the minimum bandwidth *B* ≤ 1/2*T* Hz, where *T* is the sampling interval. See Prob. 8.2-15.
-
-To summarize, for a given set of samples taken at a rate *fs* Hz, there is only one signal of bandwidth *B* ≤ *fs*/2 that passes through those samples. All other signals that pass through those samples have bandwidth higher than *fs*/2, and the samples are sub-Nyquist rate samples for those signals.
-
-### **[8.1-1 Practical Sampling](#page-13-0)**
-
-In proving the sampling theorem, we assumed ideal samples obtained by multiplying a signal *x*(*t*) by an impulse train that is physically unrealizable. In practice, we multiply a signal *x*(*t*) by a train of pulses of finite width, depicted in Fig. 8.3c. The sampler is shown in Fig. 8.3d. The sampled signal *x*(*t*) is illustrated in Fig. 8.3e. We wonder whether it is possible to recover or reconstruct *x*(*t*) from this *x*(*t*). Surprisingly, the answer is affirmative, provided the sampling rate is not below the Nyquist rate. The signal *x*(*t*) can be recovered by lowpass filtering *x*(*t*) as if it were sampled by impulse train.
-
-<sup>†</sup> The filter should have a constant gain between 0 and 5 Hz and zero gain beyond 10 Hz. In practice, the gain beyond 10 Hz can be made negligibly small, but not zero.
-
-**Figure 8.3** Effect of practical sampling.
-
-The plausibility of this result becomes apparent when we consider the fact that reconstruction of *x*(*t*) requires the knowledge of the Nyquist sample values. This information is available or built into the sampled signal *x*(*t*) in Fig. 8.3e because the *n*th sampled pulse strength is *x*(*nT*). To prove the result analytically, we observe that the sampling pulse train *pT* (*t*) depicted in Fig. 8.3c, being a periodic signal, can be expressed as a trigonometric Fourier series
-
-$$
-p_T(t) = C_0 + \sum_{n=1}^{\infty} C_n \cos(n\omega_s t + \theta_n) \qquad \omega_s = \frac{2\pi}{T}
-$$
-
-Thus,
-
-$$
-\overline{x}(t) = x(t)p_T(t) = x(t)\left[C_0 + \sum_{n=1}^{\infty} C_n \cos(n\omega_s t + \theta_n)\right]
-$$
-$$
-= C_0 x(t) + \sum_{n=1}^{\infty} C_n x(t) \cos(n\omega_s t + \theta_n)
-$$
-
-The sampled signal *x*(*t*) consists of *C*0*x*(*t*), *C*1*x*(*t*) cos(ω*st* + θ1), *C*2*x*(*t*) cos(2ω*st* + θ2), ... . Note that the first term *C*0*x*(*t*) is the desired signal and all the other terms are modulated signals with spectra centered at ±ω*s*,±2ω*s*,±3ω*s*,..., as illustrated in Fig. 8.3f. Clearly the signal *x*(*t*) can be recovered by lowpass filtering of *x*(*t*), as shown in Fig. 8.3d. As before, it is necessary that ω*<sup>s</sup>* > 4π*B* (or *fs* > 2*B*).
-
-### **EXAMPLE 8.2 Practical Sampling**
-
-Demonstrate practical sampling by sampling signal *<sup>x</sup>*(*t*) <sup>=</sup> sinc2 (5π*t*) with the rectangular pulse sequence *pT* (*t*) illustrated in Fig. 8.4c. Sketch the original and sampled signals and their spectra, and discuss recovery of *x*(*t*) from its samples.
-
-**Figure 8.4** An example of practical sampling.
-
-#### 784 CHAPTER 8 SAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE
-
-The signal *x*(*t*) and its spectrum are shown in Figs. 8.4a and 8.4b, respectively.
-
-The period of *pT* (*t*) is 0.1 second so that the fundamental frequency (which is the sampling frequency) is 10 Hz. Hence, ω*<sup>s</sup>* = 20π. The Fourier series for *pT* (*t*) can be expressed as
-
-$$
-p_T(t) = C_0 + \sum_{n=1}^{\infty} C_n \cos n\omega_s t
-$$
-
-Hence,
-
-$$
-\overline{x}(t) = x(t)p_T(t)
-$$
-  
-= C<sub>1</sub>x(t) + C<sub>1</sub>x(t) cos 20 $\pi$ t + C<sub>2</sub>x(t) cos 40 $\pi$ t + C<sub>3</sub>x(t) cos 60 $\pi$ t + ...
-
-Use of Eq. (6.14) yields *<sup>C</sup>*<sup>0</sup> <sup>=</sup> <sup>1</sup> <sup>4</sup> and *Cn* <sup>=</sup> <sup>2</sup> *<sup>n</sup>*<sup>π</sup> sin *<sup>n</sup>*<sup>π</sup> 4 . Consequently, we have
-
-$$
-\overline{x}(t) = x(t)p_T(t)
-$$
-  
-=  $\frac{1}{4}x(t) + C_1x(t) \cos 20\pi t + C_2x(t) \cos 40\pi t + C_3x(t) \cos 60\pi t + \cdots$ 
-
-and
-
-$$
-\overline{X}(\omega) = \frac{1}{4}X(\omega) + \frac{C_1}{2}[X(\omega - 20\pi) + X(\omega + 20\pi)] + \frac{C_2}{2}[X(\omega - 40\pi) + X(\omega + 40\pi)] + \frac{C_3}{2}[X(\omega - 60\pi) + X(\omega + 60\pi)] + \cdots
-$$
-
-where *Cn* = (2/*n*π )sin(*n*π/4). The sampled signal and its spectrum are shown in Figs. 8.4d and 8.4e, respectively.
-
-The spectrum *X*(ω) consists of *X*(ω) repeating periodically at the interval of 20π rad/s (10 Hz). Hence, there is no overlap between cycles, and *X*(ω) can be recovered by using an ideal lowpass filter of bandwidth 5 Hz. An ideal lowpass filter of unit gain (and bandwidth 5 Hz) will allow the first term on the right-hand side of the foregoing equation to pass fully and suppress all the other terms. Hence, the output *y*(*t*) is
-
-> *<sup>y</sup>*(*t*) <sup>=</sup> <sup>1</sup> <sup>4</sup> *x*(*t*)
-
-### **DR ILL 8.2 The Role of Sampling Pulse Area**
-
-Show that the basic pulse *p*(*t*) used in the sampling pulse train in Fig. 8.4c cannot have zero area if we wish to reconstruct *x*(*t*) by lowpass-filtering the sampled signal.

engineering/linear-systems-and-signals-3rd-edition-bp-lathi/103_8.2 SIGNAL RECONSTRUCTION.md

@@ -1,259 +0,0 @@
-## <span id="page-804-0"></span>**8.2 SIGNAL [RECONSTRUCTION](#page-13-0)**
-
-The process of reconstructing a continuous-time signal *x*(*t*) from its samples is also known as *interpolation*. In Sec. 8.1, we saw that a signal *x*(*t*) bandlimited to *B* Hz can be reconstructed (interpolated) exactly from its samples if the sampling frequency *fs* exceeds 2*B* Hz or the sampling interval *T* is less than 1/2*B*. This reconstruction is accomplished by passing the sampled signal through an ideal lowpass filter of gain *T* and having a bandwidth of any value between *B* and *fs* −*B* Hz. From a practical viewpoint, a good choice is the middle value *fs*/2 = 1/2*T* Hz or π/*T* rad/s. This value allows for small deviations in the ideal filter characteristics on either side of the cutoff frequency. With this choice of cutoff frequency and gain *T*, the ideal lowpass filter required for signal reconstruction (or interpolation) is
-
-$$
-H(\omega) = T \operatorname{rect}\left(\frac{\omega}{2\pi f_s}\right) = T \operatorname{rect}\left(\frac{\omega T}{2\pi}\right)
-$$
-\n(8.4)
-
-The interpolation process here is expressed in the frequency domain as a filtering operation. Now we shall examine this process from the time-domain viewpoint.
-
-### TIME-DOMAIN VIEW:ASIMPLE INTERPOLATION
-
-Consider the interpolation system shown in Fig. 8.5a. We start with a very simple interpolating filter, whose impulse response is rect(*t*/*T*), depicted in Fig. 8.5b. This is a gate pulse centered at the origin, having unit height, and width *T* (the sampling interval). We shall find the output of this filter when the input is the sampled signal *x*(*t*) consisting of an impulse train with the *n*th impulse at *t* = *nT* with strength *x*(*nT*). Each sample in *x*(*t*), being an impulse, produces at the output a gate pulse of height equal to the strength of the sample. For instance, the *n*th sample is an impulse of strength *x*(*nT*) located at *t* = *nT* and can be expressed as *x*(*nT*)δ(*t*−*nT*). When this impulse passes through the filter, it produces at the output a gate pulse of height *x*(*nT*), centered at *t* = *nT* (shaded in Fig. 8.5c). Each sample in *x*(*t*) will generate a corresponding gate pulse, resulting in the filter output that is a staircase approximation of *x*(*t*), shown dotted in Fig. 8.5c. This filter thus gives a crude form of interpolation.
-
-The frequency response of this filter *H*(ω) is the Fourier transform of the impulse response rect(*t*/*T*). Thus,
-
-$$
-h(t) = \text{rect}\left(\frac{t}{T}\right) \qquad \text{and} \qquad H(\omega) = T \text{sinc}\left(\frac{\omega T}{2}\right) \tag{8.5}
-$$
-
-The amplitude response |*H*(ω)| for this filter, illustrated in Fig. 8.5d, explains the reason for the crudeness of this interpolation. This filter, also known as the *zero-order hold* (ZOH) filter, is a poor form of the ideal lowpass filter (shaded in Fig. 8.5d) required for exact interpolation.†
-
-We can improve on the ZOH filter by using a *first-order hold* filter, which results in a linear interpolation instead of a staircase interpolation. A linear interpolator, whose impulse response is a triangle pulse (*t*/2*T*), results in an interpolation in which successive sample tops are connected by straight-line segments (see Prob. 8.2-3).
-
-<sup>†</sup> Figure 8.5b shows that the impulse response of this filter is noncausal, and this filter is not realizable. In practice, we make it realizable by delaying the impulse response by *T*/2. This merely delays the output of the filter by *T*/2.
-
-**Figure 8.5** Simple interpolation by means of a zero-order hold (ZOH) circuit. **(a)** ZOH interpolator. **(b)** Impulse response of a ZOH circuit. **(c)** Signal reconstruction by ZOH, as viewed in the time domain. **(d)** Frequency response of a ZOH.
-
-### TIME-DOMAIN VIEW: AN IDEAL INTERPOLATION
-
-The ideal interpolation filter frequency response obtained in Eq. (8.4) is illustrated in Fig. 8.6a. The impulse response of this filter, the inverse Fourier transform of *H*(ω) is
-
-$$
-h(t) = \text{sinc}\left(\frac{\pi t}{T}\right)
-$$
-
-For the Nyquist sampling rate, *T* = 1/2*B*, and
-
-$$
-h(t) = \text{sinc}\left(2\pi Bt\right)
-$$
-
-This *h*(*t*) is depicted in Fig. 8.6b. Observe the interesting fact that *h*(*t*) = 0 at all Nyquist sampling instants (*t* = ±*n*/2*B*) except at *t* = 0. When the sampled signal *x*¯(*t*) is applied at the input of this filter, the output is *x*(*t*). Each sample in *x*(*t*), being an impulse, generates a sinc pulse of height equal to the strength of the sample, as illustrated in Fig. 8.6c. The process is identical to that depicted in Fig. 8.5c, except that *h*(*t*) is a sinc pulse instead of a gate pulse. Addition of the sinc
-
-**Figure 8.6** Ideal interpolation for Nyquist sampling rate.
-
-pulses generated by all the samples results in *x*(*t*). The *n*th sample of the input *x*(*t*) is the impulse *x*(*nT*)δ(*t* −*nT*); the filter output of this impulse is *x*(*nT*)*h*(*t* −*nT*). Hence, the filter output to *x*(*t*), which is *x*(*t*), can now be expressed as a sum
-
-$$
-x(t) = \sum_{n} x(nT)h(t - nT) = \sum_{n} x(nT)\operatorname{sinc}\left[\frac{\pi}{T}(t - nT)\right]
-$$
-
-For the case of Nyquist sampling rate, *T* = 1/2*B*, this expression simplifies to
-
-$$
-x(t) = \sum_{n} x(nT) \operatorname{sinc}(2\pi Bt - n\pi)
-$$
-\n(8.6)
-
-Equation (8.6) is the *interpolation formula*, which yields values of *x*(*t*) between samples as a weighted sum of all the sample values.
-
-## **EXAMPLE 8.3 Bandlimited Interpolation of the Kronecker Delta Function**
-
-Find a signal *x*(*t*) that is bandlimited to *B* Hz, and whose samples are
-
-$$
-x(0) = 1
-$$
- and  $x(\pm T) = x(\pm 2T) = x(\pm 3T) = \cdots = 0$ 
-
-### <span id="page-807-0"></span>788 CHAPTER 8 SAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE
-
-where the sampling interval *T* is the Nyquist interval for *x*(*t*), that is, *T* = 1/2*B*.
-
-Because we are given the Nyquist sample values, we use the interpolation formula of Eq. (8.6) to construct *x*(*t*) from its samples. Since all but one of the Nyquist samples are zero, only one term (corresponding to *n* = 0) in the summation on the right-hand side of Eq. (8.6) survives. Thus,
-
-$$
-x(t) = \operatorname{sinc}\left(2\pi Bt\right)
-$$
-
-This signal is illustrated in Fig. 8.6b. Observe that this is the only signal that has a bandwidth *B* Hz and the sample values *x*(0) = 1 and *x*(*nT*) = 0(*n* = 0). No other signal satisfies these conditions.
-
-### **[8.2-1 Practical Difficulties in Signal Reconstruction](#page-13-0)**
-
-Consider the signal reconstruction procedure illustrated in Fig. 8.7a. If *x*(*t*) is sampled at the Nyquist rate *fs* = 2*B* Hz, the spectrum *X*(ω) consists of repetitions of *X*(ω) without any gap between successive cycles, as depicted in Fig. 8.7b. To recover *x*(*t*) from *x*(*t*), we need to pass the sampled signal *x*(*t*) through an ideal lowpass filter, shown dotted in Fig. 8.7b. As seen in Sec. 7.5, such a filter is unrealizable; it can be closely approximated only with infinite time delay in the response. In other words, we can recover the signal *x*(*t*) from its samples with infinite time delay. A practical solution to this problem is to sample the signal at a rate higher than the Nyquist rate (*fs* > 2*B* or ω*<sup>s</sup>* > 4π*B*). The result is *X*(ω), consisting of repetitions of *X*(ω) with a finite bandgap between successive cycles, as illustrated in Fig. 8.7c. Now, we can recover *X*(ω) from *X*(ω) using a lowpass filter with a gradual cutoff characteristic, shown dotted in Fig. 8.7c. But even in this case, if the unwanted spectrum is to be suppressed, the filter gain must be zero beyond some frequency (see Fig. 8.7c). According to the Paley–Wiener criterion [Eq. (7.43)], it is impossible to realize even this filter. The only advantage in this case is that the required filter can be closely approximated with a smaller time delay. All this means that it is impossible in practice to recover a bandlimited signal *x*(*t*) exactly from its samples, even if the sampling rate is higher than the Nyquist rate. However, as the sampling rate increases, the recovered signal approaches the desired signal more closely.
-
-### THE TREACHERY OF ALIASING
-
-There is another fundamental practical difficulty in reconstructing a signal from its samples. The sampling theorem was proved on the assumption that the signal *x*(*t*) is bandlimited. *All practical signals are timelimited*; that is, they are of finite duration or width. We can demonstrate (see Prob. 8.2-20) that a signal cannot be timelimited and bandlimited simultaneously. If a signal is timelimited, it cannot be bandlimited, and vice versa (but it can be simultaneously nontimelimited and nonbandlimited). Clearly, all practical signals, which are necessarily timelimited, are nonbandlimited, as shown in Fig. 8.8a; they have infinite bandwidth, and the spectrum *X*(ω) consists of overlapping cycles of *X*(ω) repeating every *fs* Hz (the sampling frequency), as
-
-**Figure 8.7 (a)** Signal reconstruction from its samples. **(b)** Spectrum of a signal sampled at the Nyquist rate. **(c)** Spectrum of a signal sampled above the Nyquist rate.
-
-illustrated in Fig. 8.8b.† Because of infinite bandwidth in this case, the spectral overlap is unavoidable, regardless of the sampling rate. Sampling at a higher rate reduces but does not eliminate overlapping between repeating spectral cycles. Because of the overlapping tails, *X*(ω) no longer has complete information about *X*(ω), and it is no longer possible, even theoretically, to recover *x*(*t*) exactly from the sampled signal *x*(*t*). If the sampled signal is passed through an ideal lowpass filter of cutoff frequency *fs*/2 Hz, the output is not *X*(ω) but *Xa*(ω) (Fig. 8.8c), which is a version of *X*(ω) distorted as a result of two separate causes:
-
-- 1. The loss of the tail of *X*(ω) beyond |*f* | > *fs*/2 Hz.
-- 2. The reappearance of this tail inverted or folded onto the spectrum. Note that the spectra cross at frequency *fs*/2 = 1/2*T* Hz. This frequency is called the *folding* frequency.
-
-<sup>†</sup> Figure 8.8b shows that from the infinite number of repeating cycles, only the neighboring spectral cycles overlap. This is a somewhat simplified picture. In reality, all the cycles overlap and interact with every other cycle because of the infinite width of all practical signal spectra. Fortunately, all practical spectra also must decay at higher frequencies. This results in insignificant amount of interference from cycles other than the immediate neighbors. When such an assumption is not justified, aliasing computations become little more involved.
-
-**Figure 8.8** Aliasing effect. **(a)** Spectrum of a practical signal *x*(*t*). **(b)** Spectrum of sampled *x*(*t*). **(c)** Reconstructed signal spectrum. **(d)** Sampling scheme using anti-aliasing filter. **(e)** Sampled signal spectrum (dotted) and the reconstructed signal spectrum (solid) when anti-aliasing filter is used.
-
-The spectrum may be viewed as if the lost tail is folding back onto itself at the folding frequency. For instance, a component of frequency (*fs*/2) + *fz* shows up as or "impersonates" a component of lower frequency (*fs*/2) − *fz* in the reconstructed signal. Thus, the components of frequencies above *fs*/2 reappear as components of frequencies below *fs*/2. This tail inversion, known as *spectral folding* or *aliasing,* is shown shaded in Fig. 8.8b and also in Fig. 8.8c. In the process of aliasing, not only are we losing all the components of frequencies above the folding frequency *fs*/2 Hz, but these very components reappear (aliased) as lower-frequency components, as shown in Figs. 8.8b and 8.8c. Such aliasing destroys the integrity of the frequency components below the folding frequency *fs*/2, as depicted in Fig. 8.8c.
-
-The aliasing problem is analogous to that of an army with a platoon that has secretly defected to the enemy side. The platoon is, however, ostensibly loyal to the army. The army is in double jeopardy. First, the army has lost this platoon as a fighting force. In addition, during actual fighting, the army will have to contend with sabotage by the defectors and will have to find another loyal platoon to neutralize the defectors. Thus, the army has lost two platoons in nonproductive activity.
-
-### DEFECTORS ELIMINATED: THE ANTI-ALIASING FILTER
-
-If you were the commander of the betrayed army, the solution to the problem would be obvious. As soon as the commander got wind of the defection, he would incapacitate, by whatever means, the defecting platoon *before the fighting begins*. This way he loses only one (the defecting) platoon. This is a partial solution to the double jeopardy of betrayal and sabotage, a solution that partly rectifies the problem and cuts the losses to half.
-
-We follow exactly the same procedure. The potential defectors are all the frequency components beyond the folding frequency *fs*/2 = 1/2*T* Hz. We should eliminate (suppress) these components from *x*(*t*) *before sampling x*(*t*). Such suppression of higher frequencies can be accomplished by an ideal lowpass filter of cutoff *fs*/2 Hz, as shown in Fig. 8.8d. This is called the *anti-aliasing filter*. Figure 8.8d also shows that anti-aliasing filtering is performed before sampling. Figure 8.8e shows the sampled signal spectrum (dotted) and the reconstructed signal *Xaa*(ω) when an anti-aliasing scheme is used. An anti-aliasing filter essentially bandlimits the signal *x*(*t*) to *fs*/2 Hz. This way, we lose only the components beyond the folding frequency *fs*/2 Hz. These suppressed components now cannot reappear to corrupt the components of frequencies below the folding frequency. Clearly, use of an anti-aliasing filter results in the reconstructed signal spectrum *Xaa*(ω) = *X*(ω) for |*f* | < *fs*/2. Thus, although we lost the spectrum beyond *fs*/2 Hz, the spectrum for all the frequencies below *fs*/2 remains intact. The effective aliasing distortion is cut in half owing to elimination of folding. We stress again that the anti-aliasing operation must be performed *before the signal is sampled*.
-
-An anti-aliasing filter also helps to reduce noise. Noise, generally, has a wideband spectrum, and without anti-aliasing, the aliasing phenomenon itself will cause the noise lying outside the desired band to appear in the signal band. Anti-aliasing suppresses the entire noise spectrum beyond frequency *fs*/2.
-
-The anti-aliasing filter, being an ideal filter, is unrealizable. In practice, we use a steep cutoff filter, which leaves a sharply attenuated spectrum beyond the folding frequency *fs*/2.
-
-### SAMPLING FORCES NONBANDLIMITED SIGNALS TO APPEAR BANDLIMITED
-
-Figure 8.8b shows that the spectrum of a signal *x*(*t*) consists of overlapping cycles of *X*(ω). This means that *x*(*t*) are sub-Nyquist samples of *x*(*t*). However, we may also view the spectrum in Fig. 8.8b as the spectrum *Xa*(ω) (Fig. 8.8c), repeating periodically every *fs* Hz without overlap. The spectrum *Xa*(ω) is bandlimited to *fs*/2 Hz. Hence, these (sub-Nyquist) samples of *x*(*t*) are actually the Nyquist samples for signal *xa*(*t*). In conclusion, sampling a nonbandlimited signal *x*(*t*) at a rate *fs* Hz makes the samples appear to be the Nyquist samples of some signal *xa*(*t*), bandlimited to *fs*/2 Hz. In other words, sampling makes a nonbandlimited signal appear to be a bandlimited signal *xa*(*t*) with bandwidth *fs*/2 Hz. A similar conclusion applies if *x*(*t*) is bandlimited but sampled at a sub-Nyquist rate.
-
-### VERIFICATION OF ALIASING IN SINUSOIDS
-
-We showed in Fig. 8.8b how sampling a signal below the Nyquist rate causes aliasing, which makes a signal of higher frequency (*fs*/2) + *fz* Hz masquerade as a signal of lower frequency (*fs*/2) − *fz* Hz. Figure 8.8b demonstrates this result in the frequency domain. Let us now verify it in the time domain to gain a deeper appreciation of aliasing.
-
-We can prove our proposition by showing that samples of sinusoids of frequencies (ω*s*/2)+ω*<sup>z</sup>* and (ω*s*/2)−ω*<sup>z</sup>* are identical when the sampling frequency is *fs* = ω*s*/2π Hz.
-
-For a sinusoid *x*(*t*) = cos ω*t*, sampled at intervals of *T* seconds, *x*(*nT*), its *n*th sample (at *t* = *nT*) is
-
-*x*(*nT*) = cosω*nT n* integer
-
-Hence, samples of sinusoids of frequency ω = (ω*s*/2)±ω*<sup>z</sup>* are†
-
-$$
-x(nT) = \cos\left(\frac{\omega_s}{2} \pm \omega_z\right) nT = \cos\left(\frac{\omega_s}{2}\right) nT \cos\omega_z nT \mp \sin\left(\frac{\omega_s}{2}\right) nT \sin\omega_z nT
-$$
-
-Recognizing that ω*sT* = 2π*fsT* = 2π, and sin(ω*s*/2)*nT* = sinπ*n* = 0 for all integer *n*, we obtain
-
-$$
-x(nT) = \cos\left(\frac{\omega_s}{2}\right) nT \cos\omega_z nT
-$$
-
-Clearly, the samples of a sinusoid of frequency (*fs*/2)+*fz* are identical to the samples of a sinusoid (*fs*/2)−*fz*. ‡ For instance, when a sinusoid of frequency 100 Hz is sampled at a rate of 120 Hz, the apparent frequency of the sinusoid that results from reconstruction of the samples is 20 Hz. This follows from the fact that here, 100 = (*fs*/2)+*fz* = 60+*fz* so that *fz* = 40. Hence, (*fs*/2)−*fz* = 20. Such would precisely be the conclusion arrived at from Fig. 8.8b.
-
-<sup>†</sup> Here we have ignored the phase aspect of the sinusoid. Sampled versions of a sinusoid *<sup>x</sup>*(*t*) <sup>=</sup> cos(ω*<sup>t</sup>* <sup>+</sup> θ ) with two different frequencies (ω*s*/2) ± ω*<sup>z</sup>* have identical frequency, but the phase signs may be reversed
-
-depending on the value of <sup>ω</sup>*z*. ‡ The reader is encouraged to verify this result graphically by plotting the spectrum of a sinusoid of frequency (ω*s*/2) + ω*<sup>z</sup>* (impulses at ±[(ω*s*/2) + ω*z*]) and its periodic repetition at intervals ω*s*. Although the result is valid for all values of ω*z*, consider the case of ω*<sup>z</sup>* < ω*s*/2 to simplify the graphics.
-
-This discussion again shows that sampling a sinusoid of frequency *f* aliasing can be avoided if the sampling rate *fs* > 2*f* Hz.
-
-$$
-0 \le f < \frac{f_s}{2} \qquad \text{or} \qquad 0 \le \omega < \frac{\pi}{T}
-$$
-
-Violating this condition leads to aliasing, implying that the samples appear to be those of a lower-frequency signal. Because of this loss of identity, it is impossible to reconstruct the signal faithfully from its samples.
-
-### GENERAL CONDITION FOR ALIASING IN SINUSOIDS
-
-We can generalize the foregoing result by showing that samples of a sinusoid of frequency *f*<sup>0</sup> are identical to those of a sinusoid of frequency *f*<sup>0</sup> + *mfs* Hz (integer *m*), where *fs* is the sampling frequency. The samples of cos 2π(*f*<sup>0</sup> +*mfs*)*t* are
-
-$$
-\cos 2\pi (f_0 + m f_s) nT = \cos (2\pi f_0 nT + 2\pi mn) = \cos 2\pi f_0 nT
-$$
-
-The result follows because *mn* is an integer and *fsT* = 1. This result shows that sinusoids of frequencies that differ by an integer multiple of *fs* result in identical set of samples. In other words, samples of sinusoids separated by frequency *fs* Hz are identical. This implies that samples of sinusoids in any frequency band of *fs* Hz are unique; that is, no two sinusoids in that band have the same samples (when sampled at a rate *fs* Hz). For instance, frequencies in the band from −*fs*/2 to *fs*/2 have unique samples (at the sampling rate *fs*). This band is called the *fundamental band*. Recall also that *fs*/2 is the folding frequency.
-
-From the discussion thus far, we conclude that if a continuous-time sinusoid of frequency *f* Hz is sampled at a rate of *fs* Hz (samples/s), the resulting samples would appear as samples of a continuous-time sinusoid of frequency *fa* in the fundamental band, where
-
-$$
-f_a = f - mf_s \qquad -\frac{f_s}{2} \le f_a < \frac{f_s}{2} \qquad m \text{ an integer} \tag{8.7}
-$$
-
-The frequency *fa* lies in the fundamental band from −*fs*/2 to *fs*/2. Figure 8.9a shows the plot of *fa* versus *f* , where *f* is the actual frequency and *fa* is the corresponding fundamental band frequency, whose samples are identical to those of the sinusoid of frequency *f* , when the sampling rate is *fs* Hz.
-
-Recall, however, that the sign change of a frequency does not alter the actual frequency of the waveform. This is because
-
-$$
-\cos(-\omega_a t + \theta) = \cos(\omega_a t - \theta)
-$$
-
-Clearly the *apparent frequency* of a sinusoid of frequency −*fa* is also *fa*. However, its phase undergoes a sign change. This means the apparent frequency of any sampled sinusoid lies in the range from 0 to *fs*/2 Hz. To summarize, if a continuous-time sinusoid of frequency *f* Hz is sampled at a rate of *fs* Hz (samples/second), the resulting samples would appear as samples of a continuous-time sinusoid of frequency |*fa*| that lies in the band from 0 to *fs*/2. According to Eq. (8.7),
-
-$$
-|f_a| = |f - mf_s| \qquad |f_a| \le \frac{f_s}{2} \qquad m \text{ an integer}
-$$
-
-**Figure 8.9** Apparent frequencies of a sampled sinusoid: **(a)** *fa* versus *f* and **(b)** |*fa*| versus *f* .
-
-The plot of the apparent frequency |*fa*| versus *f* is shown in Fig. 8.9b.† As expected, the apparent frequency |*fa*| of any sampled sinusoid, regardless of its frequency, is always in the range of 0 to *fs*/2 Hz. However, when *fa* is negative, the phase of the apparent sinusoid undergoes a sign change. The frequency belts in which such phase changes occur are shown shaded in Fig. 8.9b.
-
-Consider, for example, a sinusoid cos(2π*ft* +θ ) with *f* =8000 Hz sampled at a rate *fs* = 3000 Hz. Using Eq. (8.7), we obtain *fa* = 8000 − 3 × 3000 = −1000. Hence, |*fa*| = 1000. The samples would appear to have come from a sinusoid cos(2000π*t* − θ ). Observe the sign change of the phase because *fa* is negative.‡
-
-In the light of the foregoing development, let us consider a sinusoid of frequency *f* = (*fs*/2)+*fz*, sampled at a rate of *fs* Hz. According to Eq. (8.7),
-
-$$
-f_a = \frac{f_s}{2} + f_z - (1 \times f_s) = -\frac{f_s}{2} + f_z
-$$
-
-Hence, the apparent frequency is |*fa*| = (*fs*/2) − *fz*, confirming our earlier result. However, the phase of the sinusoid will suffer a sign change because *fa* is negative.
-
-Figure 8.10 shows how samples of sinusoids of two different frequencies (sampled at the same rate) generate identical sets of samples. Both the sinusoids are sampled at a rate *fs* = 5 Hz (*T* = 0.2 second). The frequencies of the two sinusoids, 1 Hz (period 1) and 6 Hz (period 1/6), differ by *fs* = 5 Hz.
-
-<sup>†</sup> The plots in Figs. 8.9 and 5.17 are identical. This is because a sampled sinusoid is basically a discrete-time sinusoid.
-
-<sup>‡</sup> For phase sign change, we are assuming that the signal has the form cos(2π*ft* <sup>+</sup> θ ). If the form is sin(2π*ft* + θ ), the rule changes slightly. It is left as an exercise for the reader to show that when *fa* < 0, this sinusoid appears as −sin(2π|*fa*|*t* − θ ). Thus, in addition to phase change, the amplitude also changes sign.
-
-**Figure 8.10** Demonstration of aliasing.
-
-The reason for aliasing can be clearly seen in Fig. 8.10. The root of the problem is the sampling rate, which may be adequate for the lower-frequency sinusoid but is clearly inadequate for the higher-frequency sinusoid. The figure clearly shows that between the successive samples of the higher-frequency sinusoid, there are wiggles, which are bypassed or ignored, and are unrepresented in the samples, indicating a sub-Nyquist rate of sampling. The frequency of the apparent signal *xa*(*t*) is always the lowest possible frequency that lies within the band |*f* | ≤ *fs*/2. Thus, the apparent frequency of the samples in this example is 1 Hz. If these samples are chosen to reconstruct a signal using a lowpass filter of bandwidth *fs*/2, we shall obtain a sinusoid of frequency 1 Hz.
-
-### **EXAMPLE 8.4 Apparent Frequency of Sampled Sinusoids**
-
-A continuous-time sinusoid cos(2π*ft* + θ ) is sampled at a rate *fs* = 1000 Hz. Determine the apparent (aliased) sinusoid of the resulting samples if the input signal frequency *f* is **(a)** 400 Hz, **(b)** 600 Hz, **(c)** 1000 Hz, and **(d)** 2400 Hz.
-
-The folding frequency is *fs*/2 = 500. Hence, sinusoids below 500 Hz (frequency within the fundamental band) will not be aliased and sinusoids of frequency above 500 Hz will be aliased.
-
-**(c)** Since *f* = 1000 Hz can be expressed as 1000 = 0 + 1000, we see that *fa* = 0. Hence, the aliased frequency is 0 Hz (dc), and there is no phase sign change. The apparent sinusoid is *y*(*t*) = cos(0π*t* ±θ ) = cos(θ ). This is a dc signal with constant sample values for all *n*.
-
-**<sup>(</sup>a)** Since *f* = 400 Hz is less than 500 Hz, there is no aliasing. The apparent sinusoid is cos(2π*ft* +θ ) with *f* = 400.
-
-**<sup>(</sup>b)** Since *f* = 600 Hz can be expressed as 600 = −400 + 1000, we see that *fa* = −400. Hence, the aliased frequency is 400 Hz and the phase changes sign. The apparent (aliased) sinusoid is cos(2π*ft* −θ ) with *f* = 400.
-
-### <span id="page-815-0"></span>796 CHAPTER 8 SAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE
-
-**(d)** Here, *f* = 2400 Hz can be expressed as 2400 = 400 + (2 × 1000) so that *fa* = 400. Hence, the aliased frequency is 400 Hz and there is no sign change for the phase. The apparent sinusoid is cos(2π*ft* +θ ) with *f* = 400.
-
-We could have found these answers directly from Fig. 8.9b. For example, for case (b), we read |*fa*| = 400 corresponding to *f* = 600. Moreover, *f* = 600 lies in the shaded belt. Hence, there is a phase sign change.
-
-## **DR ILL 8.3 A Case of Identical Sampled Sinusoids**
-
-Show that samples of 90 Hz and 110 Hz sinusoids of the form cosω*t* are identical when sampled at a rate 200 Hz.
-
-### **DR ILL 8.4 Apparent Frequency of Sampled Sinusoids**
-
-A sinusoid of frequency *f*<sup>0</sup> Hz is sampled at a rate of 100 Hz. Determine the apparent frequency of the samples if *f*<sup>0</sup> is **(a)** 40 Hz, **(b)** 60 Hz, **(c)** 140 Hz, and **(d)** 160 Hz.
-
-### **ANSWERS**
-
-All four cases have an apparent frequency of 40 Hz.
-
-### **[8.2-2 Some Applications of the Sampling Theorem](#page-13-0)**
-
-The sampling theorem is very important in signal analysis, processing, and transmission because it allows us to replace a continuous-time signal with a discrete sequence of numbers. Processing a continuous-time signal is therefore equivalent to processing a discrete sequence of numbers. Such processing leads us directly into the area of digital filtering. In the field of communication, the transmission of a continuous-time message reduces to the transmission of a sequence of numbers by means of pulse trains. The continuous-time signal *x*(*t*) is sampled, and sample values are used to modify certain parameters of a periodic pulse train. We may vary the amplitudes (Fig. 8.11b), widths (Fig. 8.11c), or positions (Fig. 8.11d) of the pulses in proportion to the sample values of the signal *x*(*t*). Accordingly, we may have *pulse-amplitude modulation* (PAM), *pulse-width modulation* (PWM), or *pulse-position modulation* (PPM). The most important form of pulse modulation today is *pulse-code modulation* (PCM), discussed in Sec. 8.3 in connection with Fig. 8.14b. In all these cases, instead of transmitting *x*(*t*), we transmit the corresponding pulse-modulated signal. At the receiver, we read the information of the pulse-modulated signal and reconstruct the analog signal *x*(*t*).
-
-**Figure 8.11** Pulse-modulated signals. **(a)** The signal. **(b)** The PAM signal. **(c)** The PWM (PDM) signal. **(d)** The PAM signal.
-
-One advantage of using pulse modulation is that it permits the simultaneous transmission of several signals on a time-sharing basis—*time-division multiplexing* (TDM). Because a pulse-modulated signal occupies only a part of the channel time, we can transmit several pulse-modulated signals on the same channel by interweaving them. Figure 8.12 shows the TDM of two PAM signals. In this manner, we can multiplex several signals on the same channel by reducing pulse widths.†
-
-Digital signals also offer an advantage in the area of communications, where signals must travel over distances. Transmission of digital signals is more rugged than that of analog signals because digital signals can withstand channel noise and distortion much better as long as the noise
-
-<sup>†</sup> Another method of transmitting several baseband signals simultaneously is frequency-division multiplexing (FDM) discussed in Sec. 7.7-4. In FDM, various signals are multiplexed by sharing the channel bandwidth. The spectrum of each message is shifted to a specific band not occupied by any other signal. The information of various signals is located in nonoverlapping frequency bands of the channel (Fig. 7.45). In a way, TDM and FDM are duals of each other.
-
-**Figure 8.12** Time-division multiplexing of two signals.
-
-**Figure 8.13** Digital signal transmission: **(a)** at the transmitter, **(b)** received distorted signal (without noise), **(c)** received distorted signal (with noise), and **(d)** regenerated signal at the receiver.
-
-and the distortion are within limits. An analog signal can be converted to digital binary form through sampling and quantization (rounding off), as explained in the next section. The digital (binary) message in Fig. 8.13a is distorted by the channel, as illustrated in Fig. 8.13b. Yet if the distortion remains within a limit, we can recover the data without error because we need only make a simple binary decision: Is the received pulse positive or negative? Figure 8.13c shows the same data with channel distortion and noise. Here again, the data can be recovered correctly as long as the distortion and the noise are within limits. Such is not the case with analog messages. Any distortion or noise, no matter how small, will distort the received signal.
-
-The greatest advantage of digital communication over the analog counterpart, however, is the viability of regenerative repeaters in the former. In an analog transmission system, a message signal grows progressively weaker as it travels along the channel (transmission path), whereas the channel noise and the signal distortion, being cumulative, become progressively stronger. Ultimately, the signal, overwhelmed by noise and distortion, is mutilated. Amplification is of little help because it enhances the signal and the noise in the same proportion. Consequently, the distance over which an analog message can be transmitted is limited by the transmitted power. If a transmission path is long enough, the channel distortion and noise will accumulate sufficiently to overwhelm even a <span id="page-818-0"></span>digital signal. The trick is to set up repeaters along the transmission path at distances short enough to permit detection of signal pulses before the noise and distortion have a chance to accumulate sufficiently. At each repeater, the pulses are detected, and new, clean pulses are transmitted to the next repeater, which, in turn, duplicates the same process. If the noise and distortion remain within limits (which is possible because of the closely spaced repeaters), pulses can be detected correctly.† This way the digital messages can be transmitted over longer distances with greater reliability. In contrast, analog messages cannot be cleaned up periodically, and their transmission is therefore less reliable. The most significant error in digitized signals comes from quantizing (rounding off). This error, discussed in Sec. 8.3, can be reduced as much as desired by increasing the number of quantization levels, at the cost of an increased bandwidth of the transmission medium (channel).

engineering/linear-systems-and-signals-3rd-edition-bp-lathi/104_8.3 ANALOG-TO-DIGITAL (A - D) CONVERSION.md

@@ -1,77 +0,0 @@
-## **[8.3 ANALOG-TO-DIGITAL](#page-13-0) (A/D) CONVERSION**
-
-The amplitude of an *analog* signal can take on any value over a continuous range. Hence, analog signal amplitude can take on an infinite number of values. In contrast, a *digital* signal amplitude can take on only a finite number of values. An analog signal can be converted into a digital signal by means of sampling and *quantizing* (rounding off). Sampling an analog signal alone will not yield a digital signal because a sample of analog signal can still take on any value in a continuous range. It is digitized by rounding off its value to one of the closest permissible numbers (or *quantized levels*), as illustrated in Fig. 8.14a, which represents one possible quantizing scheme. The amplitudes of the analog signal *x*(*t*) lie in the range (−*V*,*V*). This range is partitioned into *L* subintervals, each of magnitude = 2*V*/*L*. Next, each sample amplitude is approximated by the midpoint value of the subinterval in which the sample falls (see Fig. 8.14a for *L* = 16). It is clear that each sample is approximated to one of the *L* numbers. Thus, the signal is digitized with quantized samples taking on any one of the *L* values. This is an *L*-ary digital signal (see Sec. 1.3-2). Each sample can now be represented by one of *L* distinct pulses.
-
-From a practical viewpoint, dealing with a large number of distinct pulses is difficult. We prefer to use the smallest possible number of distinct pulses, the very smallest number being 2. A digital signal using only two symbols or values is the binary signal. A binary digital signal (a signal that can take on only two values) is very desirable because of its simplicity, economy, and ease of engineering. We can convert an *L*-ary signal into a binary signal by using pulse coding. Figure 8.14b shows one such code for the case of *L* = 16. This code, formed by binary representation of the 16 decimal digits from 0 to 15, is known as the *natural binary code (NBC)*. For *L* quantization levels, we need a minimum of *b* binary code digits, where 2*<sup>b</sup>* = *L* or *b* = log2 *L*.
-
-Each of the 16 levels is assigned one binary code word of four digits. Thus, each sample in this example is encoded by four binary digits. To transmit or digitally process the binary data, we need to assign a distinct electrical pulse to each of the two binary states. One possible way is to assign a negative pulse to a binary **0** and a positive pulse to a binary **1** so that each sample is now represented by a group of four binary pulses (pulse code), as depicted in Fig. 8.14b. The resulting binary signal is a digital signal obtained from the analog signal *x*(*t*) through A/D conversion. In communications jargon, such a signal is known as a pulse-code-modulated (PCM) signal.
-
-<sup>†</sup> The error in pulse detection can be made negligible.
-
-| (a) |
-|-----|
-|     |
-|     |
-|     |
-
-| Digit | Binary equivalent | Pulse code waveform |
-|-------|-------------------|---------------------|
-| 0     | 0000              |                     |
-| 1     | 0001              |                     |
-| 2     | 0010              |                     |
-| 3     | 0011              |                     |
-| 4     | 0100              |                     |
-| 5     | 0101              |                     |
-| 6     | 0110              |                     |
-| 7     | 0111              |                     |
-| 8     | 1000              |                     |
-| 9     | 1001              |                     |
-| 10    | 1010              |                     |
-| 11    | 1011              |                     |
-| 12    | 1100              |                     |
-| 13    | 1101              |                     |
-| 14    | 1110              |                     |
-| 15    | 1111              |                     |
-
-(b)
-
-**Figure 8.14** Analog-to-digital (A/D) conversion of a signal: **(a)** quantizing and **(b)** pulse coding.
-
-The convenient contraction of "*b*inary digi*t*" to *bit* has become an industry standard abbreviation.
-
-The audio signal bandwidth is about 15 kHz, but subjective tests show that signal articulation (intelligibility) is not affected if all the components above 3400 Hz are suppressed [3]. Since the objective in telephone communication is intelligibility rather than high fidelity, the components above 3400 Hz are eliminated by a lowpass filter.† The resulting signal is then sampled at a rate of 8000 samples/s (8 kHz). This rate is intentionally kept higher than the Nyquist sampling rate of 6.8 kHz to avoid unrealizable filters required for signal reconstruction. Each sample is finally quantized into 256 levels (*L* = 256), which requires a group of eight binary pulses to encode each sample (28 = 256). Thus, a digitized telephone signal consists of data amounting to 8 × 8000 = 64,000 or 64 kbit/s, requiring 64,000 binary pulses per second for its transmission.
-
-The compact disc (CD), a high-fidelity application of A/D conversion, requires the audio signal bandwidth of 20 kHz. Although the Nyquist sampling rate is only 40 kHz, an actual sampling rate of 44.1 kHz is used for the reason mentioned earlier. The signal is quantized into a rather large number of levels (*L* = 65,536) to reduce quantizing error. The binary-coded samples are now recorded on the CD.
-
-### A HISTORICAL NOTE
-
-The binary system of representing any number by using **1**s and **0**s was invented in India by Pingala (ca. 200 BCE). It was again worked out independently in the West by Gottfried Wilhelm Leibniz (1646–1716). He felt a spiritual significance in this discovery, reasoning that **1** representing unity was clearly a symbol for God, while **0** represented the nothingness. He reasoned that if all numbers can be represented merely by the use of **1** and **0**, this surely proves that God created the universe out of nothing!
-
-### **EXAMPLE 8.5 ADC Bit Number and Bit Rate**
-
-A signal *x*(*t*) bandlimited to 3 kHz is sampled at a rate 33<sup>1</sup> <sup>3</sup>% higher than the Nyquist rate. The maximum acceptable error in the sample amplitude (the maximum error due to quantization) is 0.5% of the peak amplitude *V*. The quantized samples are binary-coded. Find the required sampling rate, the number of bits required to encode each sample, and the bit rate of the resulting PCM signal.
-
-The Nyquist sampling rate is *f*Nyq = 2×3000 = 6000 Hz (samples/s). The actual sampling rate is *fA* <sup>=</sup> <sup>6000</sup>×(1<sup>1</sup> <sup>3</sup> ) = 8000 Hz.
-
-The quantization step is , and the maximum quantization error is ±/2, where = 2*V*/*L*. The maximum error due to quantization, /2, should be no greater than 0.5% of the
-
-<sup>†</sup>Components below 300 Hz may also be suppressed without affecting the articulation.
-
-### 802 CHAPTER 8 SAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE
-
-signal peak amplitude *V*. Therefore,
-
-$$
-\frac{\Delta}{2} = \frac{V}{L} = \frac{0.5}{100}V \implies L = 200
-$$
-
-For binary coding, *L* must be a power of 2. Hence, the next higher value of *L* that is a power of 2 is *L* = 256. Because log2 256 = 8, we need 8 bits to encode each sample. Therefore the bit rate of the PCM signal is
-
-8000×8 = 64,000 bits/s
-
-### **DR ILL 8.5 Bit Number and Bit Rate for ASCII**
-
-The American Standard Code for Information Interchange (ASCII) has 128 characters, which are binary-coded. A certain computer generates 100,000 characters per second. Show that
-
-- **(a)** 7 bits (binary digits) are required to encode each character
-- **(b)** 700,000 bits/s are required to transmit the computer output.

engineering/linear-systems-and-signals-3rd-edition-bp-lathi/105_8.4 DUAL OF TIME SAMPLING - SPECTRAL SAMPLING.md

@@ -1,82 +0,0 @@
-## **8.4 DUAL OF TIME [SAMPLING: SPECTRAL](#page-13-0) SAMPLING**
-
-As in other cases, the sampling theorem has its dual. In Sec. 8.1, we discussed the time-sampling theorem and showed that a signal bandlimited to *B* Hz can be reconstructed from the signal samples taken at a rate of *fs* > 2*B* samples/s. Note that the signal spectrum exists over the frequency range (in hertz) of −*B* to *B*. Therefore, 2*B* is the spectral width (not the bandwidth, which is *B*) of the signal. This fact means that a signal *x*(*t*) can be reconstructed from samples taken at a rate *fs* > the spectral width of *X*(ω) in hertz ( *fs* > 2*B*).
-
-We now prove the dual of the time-sampling theorem. This is the *spectral sampling theorem*, which applies to timelimited signals (the dual of bandlimited signals). A timelimited signal *x*(*t*) exists only over a finite interval of τ seconds, as shown in Fig. 8.15a. Generally, a timelimited signal is characterized by *x*(*t*) = 0 for *t* < *T*<sup>1</sup> and *t* > *T*<sup>2</sup> (assuming *T*<sup>2</sup> > *T*1). The signal width or duration is τ = *T*<sup>2</sup> −*T*<sup>1</sup> seconds.
-
-The spectral sampling theorem states that the spectrum *X*(ω) of a signal *x*(*t*) timelimited to a duration of τ seconds can be reconstructed from the samples of *X*(ω) taken at a rate *R* samples/Hz, where *R* > τ (the signal width or duration) in seconds.
-
-Figure 8.15a shows a timelimited signal *x*(*t*) and its Fourier transform *X*(ω). Although *X*(ω) is complex in general, it is adequate for our line of reasoning to show *X*(ω) as a real function.
-
-$$
-X(\omega) = \int_{-\infty}^{\infty} x(t)e^{-j\omega t}dt = \int_{0}^{\tau} x(t)e^{-j\omega t}dt
-$$
-\n(8.8)
-
-<span id="page-821-0"></span>
-
-**Figure 8.15** Periodic repetition of a signal amounts to sampling its spectrum.
-
-We now construct *xT*<sup>0</sup> (*t*), a periodic signal formed by repeating *x*(*t*) every *T*<sup>0</sup> seconds (*T*<sup>0</sup> > τ ), as depicted in Fig. 8.15b. This periodic signal can be expressed by the exponential Fourier series
-
-$$
-x_{T_0}(t) = \sum_{n = -\infty}^{\infty} D_n e^{jn\omega_0 t} \qquad \omega_0 = \frac{2\pi}{T_0}
-$$
-
-where (assuming *T*<sup>0</sup> > τ )
-
-$$
-D_n = \frac{1}{T_0} \int_0^{T_0} x(t) e^{-jn\omega_0 t} dt = \frac{1}{T_0} \int_0^{\tau} x(t) e^{-jn\omega_0 t} dt
-$$
-
-From Eq. (8.8), it follows that
-
-$$
-D_n = \frac{1}{T_0} X(n\omega_0)
-$$
-
-This result indicates that the coefficients of the Fourier series for *xT*<sup>0</sup> (*t*) are (1/*T*0) times the sample values of the spectrum *X*(ω) taken at intervals of ω0. This means that the spectrum of the periodic signal *xT*<sup>0</sup> (*t*) is the sampled spectrum *X*(ω), as illustrated in Fig. 8.15b. Now as long as *T*<sup>0</sup> > τ , the successive cycles of *x*(*t*) appearing in *xT*<sup>0</sup> (*t*) do not overlap, and *x*(*t*) can be recovered from *xT*<sup>0</sup> (*t*). Such recovery implies indirectly that *X*(ω) can be reconstructed from its samples. These samples are separated by the fundamental frequency *f*<sup>0</sup> = 1/*T*<sup>0</sup> Hz of the periodic signal *xT*<sup>0</sup> (*t*). Hence, the condition for recovery is *T*<sup>0</sup> > τ ; that is,
-
-$$
-f_0 < \frac{1}{\tau} \, \mathrm{Hz}
-$$
-
-Therefore, to be able to reconstruct the spectrum *X*(ω) from the samples of *X*(ω), the samples should be taken at frequency intervals *f*<sup>0</sup> < 1/τ Hz. If *R* is the sampling rate (samples/Hz), then
-
-$$
-R = \frac{1}{f_0} > \tau \text{ samples/Hz}
-$$
-
-#### 804 CHAPTER 8 SAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE
-
-### SPECTRAL INTERPOLATION
-
-Consider a signal timelimited to τ seconds and centered at *Tc*. We now show that the spectrum *X*(ω) of *x*(*t*) can be reconstructed from the samples of *X*(ω). For this case, using the dual of the approach employed to derive the signal interpolation formula in Eq. (8.6), we obtain the spectral interpolation formula†
-
-$$
-X(\omega) = \sum_{n=-\infty}^{\infty} X(n\omega_0) \operatorname{sinc}\left(\frac{\omega T_0}{2} - n\pi\right) e^{-j(\omega - n\omega_0)T_c} \qquad \omega_0 = \frac{2\pi}{T_0} \qquad T_0 > \tau \tag{8.9}
-$$
-
-For the case in Fig. 8.15, *Tc* = *T*0/2. If the pulse *x*(*t*) were to be centered at the origin, then *Tc* = 0, and the exponential term at the extreme right in Eq. (8.9) would vanish. In such a case, Eq. (8.9) would be the exact dual of Eq. (8.6).
-
-### **EXAMPLE 8.6 Spectral Sampling and Interpolation**
-
-The spectrum *X*(ω) of a unit-duration signal *x*(*t*), centered at the origin, is sampled at the intervals of 1 Hz or 2π rad/s (the Nyquist rate). The samples are
-
-$$
-X(0) = 1
-$$
- and  $X(\pm 2\pi n) = 0$   $n = 1, 2, 3, ...$ 
-
-Find *x*(*t*).
-
-We use the interpolation formula Eq. (8.9) (with *Tc* = 0) to construct *X*(ω) from its samples. Since all but one of the Nyquist samples are zero, only one term (corresponding to *n* = 0) in the summation on the right-hand side of Eq. (8.9) survives. Thus, with *X*(0) = 1 and τ = *T*<sup>0</sup> = 1, we obtain
-
-$$
-X(\omega) = \operatorname{sinc}\left(\frac{\omega}{2}\right)
-$$
- and  $x(t) = \operatorname{rect}(t)$ 
-
-For a signal of unit duration, this is the only spectrum with the sample values *X*(0) = 1 and *X*(2π*n*) = 0(*n* = 0). No other spectrum satisfies these conditions.
-
-<sup>†</sup> This can be obtained by observing that the Fourier transform of *xT*<sup>0</sup> (*t*) is 2<sup>π</sup> % *<sup>n</sup> Dn*δ(ω − *n*ω0) [see Eq. (7.22)]. We can recover *x*(*t*) from *xT*<sup>0</sup> (*t*) by multiplying the latter with rect(*t* − *Tc*)/*T*0, whose Fourier transform is *T*<sup>0</sup> sinc(ω*T*0/2)*e*−*j*ω*Tc* . Hence, *X*(ω) is 1/2π times the convolution of these two Fourier transforms, which yields Eq. (8.9).

engineering/linear-systems-and-signals-3rd-edition-bp-lathi/106_8.5 NUMERICAL COMPUTATION OF THE FOURIER TRANSFORM - THE DISCRETE FOURIER TRANSFORM.md

@@ -1,552 +0,0 @@
-## <span id="page-824-0"></span>**8.5 NUMERICAL [COMPUTATION OF THE](#page-13-0) FOURIER TRANSFORM: THE DISCRETE FOURIER TRANSFORM**
-
-Numerical computation of the Fourier transform of *x*(*t*) requires sample values of *x*(*t*) because a digital computer can work only with discrete data (sequence of numbers). Moreover, a computer can compute *X*(ω) only at some discrete values of ω [samples of *X*(ω)]. We therefore need to relate the samples of *X*(ω) to samples of *x*(*t*). This task can be accomplished by using the results of the two sampling theorems developed in Secs. 8.1 and 8.4.
-
-We begin with a timelimited signal *x*(*t*) (Fig. 8.16a) and its spectrum *X*(ω) (Fig. 8.16b). Since *x*(*t*) is timelimited, *X*(ω) is nonbandlimited. For convenience, we shall show all spectra as functions of the frequency variable *f* (in hertz) rather than ω. According to the sampling theorem, the spectrum *X*(ω) of the sampled signal *x*(*t*) consists of *X*(ω) repeating every *fs* Hz, where *fs* = 1/*T*, as depicted in Fig. 8.16d.† In the next step, the sampled signal in Fig. 8.16c is repeated periodically every *T*<sup>0</sup> seconds, as illustrated in Fig. 8.16e. According to the spectral sampling theorem, such an operation results in sampling the spectrum at a rate of *T*<sup>0</sup> samples/Hz. This sampling rate means that the samples are spaced at *f*<sup>0</sup> = 1/*T*<sup>0</sup> Hz, as depicted in Fig. 8.16f.
-
-The foregoing discussion shows that when a signal *x*(*t*) is sampled and then periodically repeated, the corresponding spectrum is also sampled and periodically repeated. Our goal is to relate the samples of *x*(*t*) to the samples of *X*(ω).
-
-### NUMBER OF SAMPLES
-
-One interesting observation from Figs. 8.16e and 8.16f is that *N*0, the number of samples of the signal in Fig. 8.16e in one period *T*0, is identical to *N* <sup>0</sup>, the number of samples of the spectrum in Fig. 8.16f in one period *fs*. To see this, we notice that
-
-$$
-N_0 = \frac{T_0}{T} \quad N'_0 = \frac{f_s}{f_0} \quad f_s = \frac{1}{T} \quad \text{and} \quad f_0 = \frac{1}{T_0}
-$$
-(8.10)
-
-Using these relations, we see that
-
-$$
-N_0 = \frac{T_0}{T} = \frac{f_s}{f_0} = N'_0
-$$
-
-### ALIASING AND LEAKAGE IN NUMERICAL COMPUTATION
-
-Figure 8.16f shows the presence of aliasing in the samples of the spectrum *X*(ω). This aliasing error can be reduced as much as desired by increasing the sampling frequency *fs* (decreasing the sampling interval *T* = 1/*fs*). The aliasing can never be eliminated for timelimited *x*(*t*), however, because its spectrum *X*(ω) is nonbandlimited. Had we started with a signal having a bandlimited spectrum *X*(ω), there would be no aliasing in the spectrum in Fig. 8.16f. Unfortunately, such a signal is nontimelimited, and its repetition (in Fig. 8.16e) would result in signal overlapping (aliasing in the time domain). In this case, we shall have to contend with errors in signal
-
-<sup>†</sup> There is a multiplying constant 1/*T* for the spectrum in Fig. 8.16d [see Eq. (8.2)], but this is irrelevant to our discussion here.
-
-samples. In other words, in computing the direct or inverse Fourier transform numerically, we can reduce the error as much as we wish, but the error can never be eliminated. This is true of numerical computation of the direct and inverse Fourier transforms, regardless of the method used. For example, if we determine the Fourier transform by direct integration numerically, by using Eq. (7.9), there will be an error because the interval of integration *t* can never be made zero. Similar remarks apply to numerical computation of the inverse transform. Therefore, we should always keep in mind the nature of this error in our results. In our discussion (Fig. 8.16), we assumed *x*(*t*) to be a timelimited signal. If *x*(*t*) were not timelimited, we would need to timelimit it because numerical computations can work only with finite data. Furthermore, this data truncation causes error because of spectral spreading (smearing) and leakage, as discussed in Sec. 7.8. The leakage also causes aliasing. Leakage can be reduced by using a tapered window for signal truncation. But this choice increases spectral spreading or smearing. Spectral spreading can be reduced by increasing the window width (i.e., more data), which increases *T*0, and reduces *f*<sup>0</sup> (increases *spectral* or *frequency resolution*).
-
-### PICKET FENCE EFFECT
-
-The numerical computation method yields only the uniform sample values of *X*(ω). The major peaks or valleys of *X*(ω) can lie between two samples and may remain hidden, giving a false picture of reality. Viewing samples is like viewing the signal and its spectrum through a "picket fence" with upright posts that are very wide and placed close together. What is hidden behind the pickets is much more than what we can see. Such misleading results can be avoided by using a sufficiently large *N*0, the number of samples, to increase resolution. We can also use zero padding (discussed later) or the spectral interpolation formula [Eq. (8.9)] to determine the values of *X*(ω) between samples.
-
-### POINTS OF DISCONTINUITY
-
-If *x*(*t*) or *X*(ω) has a jump discontinuity at a sampling point, the sample value should be taken as the average of the values on the two sides of the discontinuity because the Fourier representation at a point of discontinuity converges to the average value.
-
-### DERIVATION OF THE DISCRETE FOURIER TRANSFORM (DFT)
-
-If *x*(*nT*) and *X*(*r*ω0) are the *n*th and *r*th samples of *x*(*t*) and *X*(ω), respectively, then we define new variables *xn* and *Xr* as
-
-$$
-x_n = Tx(nT) = \frac{T_0}{N_0}x(nT)
-$$
-\n(8.11)
-
-and
-
-$$
-X_r = X(r\omega_0)
-$$
-
-where
-
-$$
-\omega_0 = 2\pi f_0 = \frac{2\pi}{T_0}
-$$
-
-We shall now show that *xn* and *Xr* are related by the following equations† :
-
-$$
-X_r = \sum_{n=0}^{N_0 - 1} x_n e^{-jr\Omega_0 n}
-$$
- (8.12)
-
-and
-
-$$
-x_n = \frac{1}{N_0} \sum_{r=0}^{N_0 - 1} X_r e^{ir\Omega_0 n}
-$$
-\n(8.13)
-
-where
-
-$$
-\Omega_0 = \omega_0 T = \frac{2\pi}{N_0}
-$$
-
-These equations define the direct and the inverse *discrete Fourier transforms,* with *Xr* the direct discrete Fourier transform (DFT) of *xn*, and *xn* the inverse discrete Fourier transform (IDFT) of *Xr*. The notation
-
-$$
-x_n \Longleftrightarrow X_r
-$$
-
-is also used to indicate that *xn* and *Xr* are a DFT pair. Remember that *xn* is *T*0/*N*<sup>0</sup> times the *n*th sample of *x*(*t*) and *Xr* is the *r*th sample of *X*(ω). Knowing the sample values of *x*(*t*), we can use the DFT to compute the sample values of *X*(ω)—and vice versa. Note, however, that *xn* is a function of *n* (*n* = 0, 1, 2,...,*N*<sup>0</sup> − 1) rather than of *t* and that *Xr* is a function of *r* (*r* = 0, 1, 2,...,*N*<sup>0</sup> − 1) rather than of ω. Moreover, both *xn* and *Xr* are periodic sequences of period *N*<sup>0</sup> (Figs. 8.16e, 8.16f). Such sequences are called *N*0*-periodic sequences*. The proof of the DFT relationships in Eqs. (8.12) and (8.13) follows directly from the results of the sampling theorem. The sampled signal *x*(*t*) (Fig. 8.16c) can be expressed as
-
-$$
-\overline{x}(t) = \sum_{n=0}^{N_0 - 1} x(nT)\delta(t - nT)
-$$
-
-Since δ (*t* −*nT*) ⇐⇒ *e*−*jn*ω*<sup>T</sup>* , applying the Fourier transform yields
-
-$$
-\overline{X}(\omega) = \sum_{n=0}^{N_0 - 1} x(nT) e^{-jn\omega T}
-$$
-
-But from Fig. 8.1f [or Eq. (8.2)], it is clear that over the interval |ω| ≤ ω*s*/2, *X*(ω), the Fourier transform of *x*(*t*) is *X*(ω)/*T*, assuming negligible aliasing. Hence,
-
-$$
-X(\omega) = T\overline{X}(\omega) = T\sum_{n=0}^{N_0 - 1} x(nT)e^{-jn\omega T} \qquad |\omega| \le \frac{\omega_s}{2}
-$$
-
-<sup>†</sup> In Eqs. (8.12) and (8.13), the summation is performed from 0 to *<sup>N</sup>*<sup>0</sup> <sup>−</sup>1. It is shown in Sec. 9.1-2 [Eqs. (9.6) and (9.7)] that the summation may be performed over any successive *N*<sup>0</sup> values of *n* or *r*.
-
-and
-
-$$
-X_r = X(r\omega_0) = T \sum_{n=0}^{N_0 - 1} x(nT)e^{-nkr\omega_0 T}
-$$
-\n(8.14)
-
-If we let ω0*T* = 0, then from Eq. (8.10),
-
-$$
-\Omega_0 = \omega_0 T = 2\pi f_0 T = \frac{2\pi}{N_0}
-$$
-
-Also, from Eq. (8.11),
-
-$$
-Tx(nT) = x_n
-$$
-
-Therefore, Eq. (8.14) becomes
-
-$$
-X_r = \sum_{n=0}^{N_0 - 1} x_n e^{-jr\Omega_0 n} \qquad \Omega_0 = \frac{2\pi}{N_0}
-$$
-
-This is Eq. (8.12), which we set to prove.
-
-The inverse transform relationship of Eq. (8.13) can be derived by using a similar procedure with the roles of *t* and ω reversed, but here we shall use a more direct proof. To prove Eq. (8.13), we multiply both sides of Eq. (8.12) by *ejm*0*<sup>r</sup>* and sum over *r* as
-
-$$
-\sum_{r=0}^{N_0-1} X_r e^{jm\Omega_0 r} = \sum_{r=0}^{N_0-1} \left[ \sum_{n=0}^{N_0-1} x_n e^{-jr\Omega_0 n} \right] e^{jm\Omega_0 r}
-$$
-
-By interchanging the order of summation on the right-hand side, we have
-
-$$
-\sum_{r=0}^{N_0-1} X_r e^{jm\Omega_0 r} = \sum_{n=0}^{N_0-1} x_n \left[ \sum_{r=0}^{N_0-1} e^{j(m-n)\Omega_0 r} \right]
-$$
-
-As the footnote below readily shows, the inner sum on the right-hand side is zero for *n* = *m* and is *N*<sup>0</sup> when *n* = *m*. † Thus, the outer sum will have only one nonzero term when *n* = *m*, and it is
-
-† We show that *<sup>N</sup>*
-
-$$
-\sum_{n=0}^{N_0-1} e^{jk\Omega_0 n} = \begin{cases} N_0 & k = 0, \pm N_0, \pm 2N_0, \dots \\ 0 & \text{otherwise} \end{cases}
-$$
- (8.15)
-
-Recall that 0*N*<sup>0</sup> <sup>=</sup> <sup>2</sup>π. So *ejk*0*<sup>n</sup>* <sup>=</sup> 1 when *<sup>k</sup>* <sup>=</sup> 0,±*N*0,±2*N*0,.... Hence, the sum on the left-hand side of Eq. (8.15) is *N*0. To compute the sum for other values of *k*, we note that the sum on the left-hand side of Eq. (8.15) is a geometric progression with common ratio <sup>α</sup> <sup>=</sup> *<sup>e</sup>jk*<sup>0</sup> . Therefore, (see Sec. B.8-3)
-
-$$
-\sum_{n=0}^{N_0-1} e^{jk\Omega_0 n} = \frac{e^{jk\Omega_0 N_0} - 1}{e^{jk\Omega_0} - 1} = 0 \qquad (e^{jk\Omega_0 N_0} = e^{j2\pi m} = 1)
-$$
-
-#### 810 CHAPTER 8 SAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE
-
-*N*0*xn* = *N*0*xm*. Therefore,
-
-$$
-x_m = \frac{1}{N_0} \sum_{r=0}^{N_0 - 1} X_r e^{im\Omega_0 r} \qquad \Omega_0 = \frac{2\pi}{N_0}
-$$
-
-Because *Xr* is *N*<sup>0</sup> periodic, we need to determine the values of *Xr* over any one period. It is customary to determine *Xr* over the range (0, *N*<sup>0</sup> − 1), rather than over the range (−*N*0/2,(*N*0/2)−1).†
-
-## CHOICE OF *T* AND *T*<sup>0</sup>
-
-In DFT computation, we first need to select suitable values for *N*<sup>0</sup> and *T* or *T*0. For this purpose, we begin by deciding on *B*, the essential bandwidth (in hertz) of the signal. The sampling frequency *fs* must be at least 2*B*, that is,
-
-$$
-\frac{f_s}{2} \ge B
-$$
-
-Moreover, the sampling interval *T* = 1/*fs* [Eq. (8.10)], and
-
-$$
-T \le \frac{1}{2B} \tag{8.16}
-$$
-
-Once we pick *B*, we can choose *T* according to Eq. (8.16). Also,
-
-$$
-f_0 = \frac{1}{T_0} \tag{8.17}
-$$
-
-where *f*<sup>0</sup> is the *frequency resolution* [separation between samples of *X*(ω)]. Hence, if *f*<sup>0</sup> is given, we can pick *T*<sup>0</sup> according to Eq. (8.17). Knowing *T*<sup>0</sup> and *T*, we determine *N*<sup>0</sup> from
-
-$$
-N_0 = \frac{T_0}{T}
-$$
-
-### ZERO PADDING
-
-Recall that observing *Xr* is like observing the spectrum *X*(ω) through a picket fence. If the frequency sampling interval *f*<sup>0</sup> is not sufficiently small, we could miss out on some significant details and obtain a misleading picture. To obtain a higher number of samples, we need to reduce *f*0. Because *f*<sup>0</sup> =1/*T*0, a higher number of samples requires us to increase the value of *T*0, the period of repetition for *x*(*t*). This option increases *N*0, the number of samples of *x*(*t*), by adding dummy samples of 0 value. This addition of dummy samples is known as *zero padding*. Thus, zero padding increases the number of samples and may help in getting a better idea of the spectrum *X*(ω) from its samples *Xr*. To continue with our picket fence analogy, zero padding is like using more, and narrower, pickets.
-
-<sup>†</sup> The DFT of Eq. (8.12) and the IDFT of Eq. (8.13) represent a transform in their own right, and they are exact. There is no approximation. However, *xn* and *Xr*, thus obtained, are only approximations to the actual samples of a signal *x*(*t*) and of its Fourier transform *X*(ω).
-
-### ZERO PADDING DOES NOT IMPROVE ACCURACY OR RESOLUTION
-
-Actually, we are not observing *X*(ω) through a picket fence. We are observing a distorted version of *X*(ω) resulting from the truncation of *x*(*t*). Hence, we should keep in mind that even if the fence were transparent, we would see a reality distorted by aliasing. Seeing through the picket fence just gives us an imperfect view of the imperfectly represented reality. Zero padding only allows us to look at more samples of that imperfect reality. It can never reduce the imperfection in what is behind the fence. The imperfection, which is caused by aliasing, can be lessened only by reducing the sampling interval *T*. Observe that reducing *T* also increases *N*0, the number of samples, and is like increasing the number of pickets while reducing their width. But in this case, the reality behind the fence is also better dressed and we see more of it.
-
-### **EXAMPLE 8.7 Number of Samples and Frequency Resolution**
-
-A signal *x*(*t*) has a duration of 2 ms and an essential bandwidth of 10 kHz. It is desirable to have a frequency resolution of 100 Hz in the DFT (*f*<sup>0</sup> = 100). Determine *N*0.
-
-To have *f*<sup>0</sup> = 100 Hz, the effective signal duration *T*<sup>0</sup> must be
-
-$$
-T_0 = \frac{1}{f_0} = \frac{1}{100} = 10
-$$
- ms
-
-Since the signal duration is only 2 ms, we need zero padding over 8 ms. Also, *B* = 10,000. Hence, *fs* = 2*B* = 20,000 and *T* = 1/*fs* = 50 µs. Furthermore,
-
-$$
-N_0 = \frac{f_s}{f_0} = \frac{20,000}{100} = 200
-$$
-
-The *fast Fourier transform* (FFT) algorithm (discussed later; see Sec. 8.6) is used to compute DFT, where it proves convenient (although not necessary) to select *N*<sup>0</sup> as a power of 2; that is, *N*<sup>0</sup> = 2*<sup>n</sup>* (*n*, integer). Let us choose *N*<sup>0</sup> = 256. Increasing *N*<sup>0</sup> from 200 to 256 can be used to reduce aliasing error (by reducing *T*), to improve resolution (by increasing *T*<sup>0</sup> using zero padding), or a combination of both.
-
-**Reducing Aliasing Error.** We maintain the same *T*<sup>0</sup> so that *f*<sup>0</sup> = 100. Hence,
-
-$$
-f_s = N_0 f_0 = 256 \times 100 = 25,600
-$$
- and  $T = \frac{1}{f_s} = 39 \,\mu s$ 
-
-Thus, increasing *N*<sup>0</sup> from 200 to 256 permits us to reduce the sampling interval *T* from 50 µs to 39 µs while maintaining the same frequency resolution (*f*<sup>0</sup> = 100).
-
-**Improving Resolution.** Here, we maintain the same *T* = 50 µs, which yields
-
-$$
-T_0 = N_0 T = 256(50 \times 10^{-6}) = 12.8
-$$
- ms and  $f_0 = \frac{1}{T_0} = 78.125$  Hz
-
-### 812 CHAPTER 8 SAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE
-
-Thus, increasing *N*<sup>0</sup> from 200 to 256 can improve the frequency resolution from 100 to 78.125 Hz while maintaining the same aliasing error (*T* = 50 µs).
-
-**Combination of Reducing Aliasing Error and Improving Resolution.** To simultaneously reduce alias error and improve resolution, we could choose *T* = 45 µs and *T*<sup>0</sup> = 11.5 ms so that *f*<sup>0</sup> = 86.96 Hz. Many other combinations exist as well.
-
-### **EXAMPLE 8.8 DFT to Compute the Fourier Transform of an Exponential**
-
-Use the DFT to compute (samples of) the Fourier transform of *e*−2*<sup>t</sup> u*(*t*). Plot the resulting Fourier spectra.
-
-We first determine *T* and *T*0. The Fourier transform of *e*−2*<sup>t</sup> u*(*t*) is 1/(*j*ω + 2). This lowpass signal is not bandlimited. In Sec. 7.6, we used the energy criterion to compute the essential bandwidth of a signal. Here, we shall present a simpler, but workable alternative to the energy criterion. The essential bandwidth of a signal will be taken as the frequency at which |*X*(ω)| drops to 1% of its peak value (see the footnote on page 736). In this case, the peak value occurs at ω = 0, where |*X*(0)| = 0.5. Observe that
-
-$$
-|X(\omega)| = \frac{1}{\sqrt{\omega^2 + 4}} \approx \frac{1}{\omega} \qquad \omega \gg 2
-$$
-
-Also, 1% of the peak value is 0.01 × 0.5 = 0.005. Hence, the essential bandwidth *B* is at ω = 2π*B*, where
-
-$$
-|X(\omega)| \approx \frac{1}{2\pi B} = 0.005 \quad \Rightarrow \quad B = \frac{100}{\pi} \text{ Hz}
-$$
-
-and from Eq. (8.16),
-
-$$
-T \le \frac{1}{2B} = \frac{\pi}{200} = 0.015708
-$$
-
-Had we used 1% energy criterion to determine the essential bandwidth, following the procedure in Ex. 7.20, we would have obtained *B* = 20.26 Hz, which is somewhat smaller than the value just obtained by using the 1% amplitude criterion.
-
-The second issue is to determine *T*0. Because the signal is not timelimited, we have to truncate it at *T*<sup>0</sup> such that *x*(*T*0) 1. A reasonable choice would be *T*<sup>0</sup> = 4 because *x*(4) = *e*−<sup>8</sup> = 0.0003351. The result is *N*<sup>0</sup> = *T*0/*T* = 254.6, which is not a power of 2. Hence, we choose *T*<sup>0</sup> = 4, and *T* = 0.015625 = 1/64, yielding *N*<sup>0</sup> = 256, which is a power of 2.
-
-Note that there is a great deal of flexibility in determining *T* and *T*0, depending on the accuracy desired and the computational capacity available. We could just as well have chosen *T* = 0.03125, yielding *N*<sup>0</sup> = 128, although this choice would have given a slightly higher aliasing error.
-
-Because the signal has a jump discontinuity at *t* = 0, the first sample (at *t* = 0) is 0.5, the averages of the values on the two sides of the discontinuity. We compute *Xr* (the DFT) from the samples of *e*−2*<sup>t</sup> u*(*t*) according to Eq. (8.12). Note that *Xr* is the *r*th sample of *X*(ω), and these samples are spaced at *f*<sup>0</sup> = 1/*T*<sup>0</sup> = 0.25 Hz (ω<sup>0</sup> = π/2 rad/s).
-
-Because *Xr* is *N*<sup>0</sup> periodic, *Xr* = *X*(*r*+256) so that *X*<sup>256</sup> = *X*0. Hence, we need to plot *Xr* over the range *r* = 0 to 255 (not 256). Moreover, because of this periodicity, *X*−*<sup>r</sup>* = *X*(−*r*+256), and the values of *Xr* over the range *r* = −127 to −1 are identical to those over the range *r* = 129 to 255. Thus, *X*−<sup>127</sup> = *X*129, *X*−<sup>126</sup> = *X*130,...,*X*−<sup>1</sup> = *X*255. In addition, because of the property of conjugate symmetry of the Fourier transform, *X*−*<sup>r</sup>* = *X*<sup>∗</sup> *<sup>r</sup>* , it follows that *X*−<sup>1</sup> = *X*<sup>∗</sup> <sup>1</sup> , *X*−<sup>2</sup> = *X*<sup>∗</sup> <sup>2</sup> ,...,*X*−<sup>128</sup> = *X*<sup>∗</sup> <sup>128</sup>. Thus, we need *Xr* only over the range *r* = 0 to *N*0/2 (128 in this case).
-
-Figure 8.17 shows the computed plots of |*Xr*| and *Xr*. The exact spectra are depicted by continuous curves for comparison. Note the nearly perfect agreement between the two sets of spectra. We have depicted the plot of only the first 28 points rather than all 128 points, which would have made the figure very crowded, resulting in loss of clarity. The points are at the intervals of 1/*T*<sup>0</sup> = 1/4 Hz or ω<sup>0</sup> = 1.5708 rad/s. The 28 samples, therefore, exhibit the plots over the range ω = 0 to ω = 28(1.5708) ≈ 44 rad/s or 7 Hz.
-
-**Figure 8.17** Discrete Fourier transform of an exponential signal *e*−2*<sup>t</sup> u*(*t*).
-
-### 814 CHAPTER 8 SAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE
-
-In this example, we knew *X*(ω) beforehand; hence we could make intelligent choices for *B* (or the sampling frequency *fs*). In practice, we generally do not know *X*(ω) beforehand. In fact, that is the very thing we are trying to determine. In such a case, we must make an intelligent guess for *B* or *fs* from circumstantial evidence. We should then continue reducing the value of *T* and recomputing the transform until the result stabilizes within the desired number of significant digits.
-
-### USING MATLAB TO COMPUTE AND PLOT THE RESULTS
-
-Let us now use MATLAB to confirm the results of this example. First, parameters are defined and MATLAB's fft command is used to compute the DFT.
-
-```
->> T_0 = 4; N_0 = 256; T = T_0/N_0; t = (0:T:T*(N_0-1))';
->> x = T*exp(-2*t); x(1) = x(1)/2;
->> X_r = fft(x); r = [-N_0/2:N_0/2-1]'; omega_r = r*2*pi/T_0;
-```
-
-The true Fourier transform is also computed for comparison.
-
->> omega = linspace(-pi/T,pi/T,5001); X = 1./(j\*omega+2);
-
-For clarity, we display spectrum over a restricted frequency range.
-
-```
->> subplot(1,2,1); stem(omega_r,fftshift(abs(X_r)),'k.');
-```
-
-```
->> line(omega,abs(X),'color',[0 0 0]); axis([-0.01 44 -0.01 0.51]);
-```
-
-- >> xlabel('\omega'); ylabel('|X(\omega)|');
-- >> subplot(1,2,2); stem(omega\_r,fftshift(angle(X\_r)),'k.');
-- >> line(omega,angle(X),'color',[0 0 0]); axis([-0.01 44 -pi/2-0.01 0.01]);
-- >> xlabel('\omega'); ylabel('\angle X(\omega)');
-
-The results, shown in Fig. 8.18, match the earlier results shown in Fig. 8.17.
-
-### **EXAMPLE 8.9 DFT to Compute the Fourier Transform of a Rectangular Pulse**
-
-Use the DFT to compute the Fourier transform of 8 rect(*t*).
-
-This gate function and its Fourier transform are illustrated in Figs. 8.19a and 8.19b. To determine the value of the sampling interval *T*, we must first decide on the essential bandwidth *B*. In Fig. 8.19b, we see that *X*(ω) decays rather slowly with ω. Hence, the essential bandwidth *B* is rather large. For instance, at *B* = 15.5 Hz (97.39 rad/s), *X*(ω) = −0.1643, which is about 2% of the peak at *X*(0). Hence, the essential bandwidth is well above 16 Hz if we use the 1% of the peak amplitude criterion for computing the essential bandwidth. However, we shall deliberately take *B* = 4 for two reasons: to show the effect of aliasing and because the use of *B* > 4 would give an enormous number of samples, which could not be conveniently displayed on the page without losing sight of the essentials. Thus, we shall intentionally accept approximation to graphically clarify the concepts of the DFT.
-
-The choice of *B* = 4 results in the sampling interval *T* = 1/2*B* = 1/8. Looking again at the spectrum in Fig. 8.19b, we see that the choice of the frequency resolution *f*<sup>0</sup> = 1/4 Hz is reasonable. Such a choice gives us four samples in each lobe of *X*(ω). In this case *T*<sup>0</sup> = 1/*f*<sup>0</sup> = 4 seconds and *N*<sup>0</sup> = *T*0/*T* = 32. The duration of *x*(*t*) is only 1 second. We must repeat it every 4 seconds (*T*<sup>0</sup> = 4), as depicted in Fig. 8.19c, and take samples every 1/8 second. This choice yields 32 samples (*N*<sup>0</sup> = 32). Also,
-
-$$
-x_n = Tx(nT) = \frac{1}{8}x(nT)
-$$
-
-Since *x*(*t*) = 8 rect(*t*), the values of *xn* are 1, 0, or 0.5 (at the points of discontinuity), as illustrated in Fig. 8.19c, where *xn* is depicted as a function of *t* as well as *n*, for convenience.
-
-In the derivation of the DFT, we assumed that *x*(*t*) begins at *t* = 0 (Fig. 8.16a), and then took *N*<sup>0</sup> samples over the interval (0, *T*0). In the present case, however, *x*(*t*) begins at −1/2. This difficulty is easily resolved when we realize that the DFT obtained by this procedure is actually the DFT of *xn* repeating periodically every *T*<sup>0</sup> seconds. Figure 8.19c clearly indicates that periodic repeating the segment of *xn* over the interval from −2 to 2 seconds yields the same signal as the periodic repeating the segment of *xn* over the interval from 0 to 4 seconds. Hence, the DFT of the samples taken from −2 to 2 seconds is the same as that of the samples taken from 0 to 4 seconds. Therefore, regardless of where *x*(*t*) starts, we can always take the samples of *x*(*t*) and its periodic extension over the interval from 0 to *T*0. In the present example, the 32 sample values are
-
-$$
-x_n = \begin{cases} 1 & 0 \le n \le 3 \text{ and } 29 \le n \le 31 \\ 0 & 5 \le n \le 27 \\ 0.5 & n = 4,28 \end{cases}
-$$
-
-816 CHAPTER 8 SAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE
-
-**Figure 8.19** Discrete Fourier transform of a gate pulse.
-
-Observe that the last sample is at *t* = 31/8, not at 4, because the signal repetition starts at *t* = 4, and the sample at *t* = 4 is the same as the sample at *t* = 0. Now, *N*<sup>0</sup> = 32 and <sup>0</sup> = 2π/32 = π/16. Therefore [see Eq. (8.12)],
-
-$$
-X_r = \sum_{n=0}^{31} x_n e^{-jr(\pi/16)n}
-$$
-
-Values of *Xr* are computed according to this equation and plotted in Fig. 8.19d.
-
-The samples *Xr* are separated by *f*<sup>0</sup> = 1/*T*<sup>0</sup> Hz. In this case *T*<sup>0</sup> = 4, so the frequency resolution *f*<sup>0</sup> is 1/4 Hz, as desired. The folding frequency *fs*/2 = *B* = 4 Hz corresponds to *r* = *N*0/2 = 16. Because *Xr* is *N*<sup>0</sup> periodic (*N*<sup>0</sup> = 32), the values of *Xr* for *r* = −16 to *n* = −1 are the same as those for *r* = 16 to *n* = 31. For instance, *X*<sup>17</sup> = *X*−15, *X*<sup>18</sup> = *X*−14, and so on. The DFT gives us the samples of the spectrum *X*(ω).
-
-For the sake of comparison, Fig. 8.19d also shows the shaded curve 8 sinc(ω/2), which is the Fourier transform of 8 rect(*t*). The values of *Xr* computed from the DFT equation show aliasing error, which is clearly seen by comparing the two superimposed plots. The error in *X*<sup>2</sup> is just about 1.3%. However, the aliasing error increases rapidly with *r*. For instance, the error in *X*<sup>6</sup> is about 12%, and the error in *X*<sup>10</sup> is 33%. The error in *X*<sup>14</sup> is a whopping 72%. The percent error increases rapidly near the folding frequency (*r* = 16) because *x*(*t*) has a jump discontinuity, which makes *X*(ω) decay slowly as 1/ω. Hence, near the folding frequency, the inverted tail (due to aliasing) is very nearly equal to *X*(ω) itself. Moreover, the final values are the difference between the exact and the folded values (which are very close to the exact values). Hence, the percent error near the folding frequency (*r* = 16 in this case) is very high, although the absolute error is very small. Clearly, for signals with jump discontinuities, the aliasing error near the folding frequency will always be high (in percentage terms), regardless of the choice of *N*0. To ensure a negligible aliasing error at any value *r*, we must make sure that *N*<sup>0</sup> *r*. This observation is valid for all signals with jump discontinuities.
-
-### USING MATLAB TO COMPUTE AND PLOT THE RESULTS
-
-Once again, MATLAB lets us easily confirm the results of this example. First, parameters are defined and MATLAB's fft command is used to compute the DFT.
-
->> T\_0 = 4; N\_0 = 32; T = T\_0/N\_0; >> x\_n = [ones(1,4) 0.5 zeros(1,23) 0.5 ones(1,3)]'; >> X\_r = fft(x\_n); r = [-N\_0/2:N\_0/2-1]'; omega\_r = r\*2\*pi/T\_0;
-
-The true Fourier transform is also computed for comparison.
-
-```
->> omega = linspace(-pi/T,pi/T,5001); X = 8*sinc(omega/(2*pi));
-```
-
-Since it is real, we can display the resulting spectrum using a single plot.
-
->> clf; stem(omega\_r,fftshift(real(X\_r)),'k.'); >> line(omega,X,'color',[0 0 0]);
-
-### <span id="page-837-0"></span>818 CHAPTER 8 SAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE
-
->> xlabel('\omega'); ylabel('X(\omega)'); axis tight
-
-The result, shown in Fig. 8.20, matches the earlier result shown in Fig. 8.19d. The DFT approximation does not perfectly follow the true Fourier transform, especially at high frequencies, because the parameter *B* is deliberately set too small.
-
-### **[8.5-1 Some Properties of the DFT](#page-13-0)**
-
-The discrete Fourier transform is basically the Fourier transform of a sampled signal repeated periodically. Hence, the properties derived earlier for the Fourier transform apply to the DFT as well.
-
-LINEARITY If *xn* ⇐⇒ *Xr* and *gn* ⇐⇒ *Gr*, then
-
-$$
-a_1x_n + a_2g_n \Longleftrightarrow a_1X_r + a_2G_r
-$$
-
-The proof is trivial.
-
-### CONJUGATE SYMMETRY
-
-From the conjugation property *x*∗(*t*) ⇐⇒ *X*∗(−ω), we have
-
-*x*∗ *<sup>n</sup>* ←→ *X*<sup>∗</sup> −*r*
-
-From this equation and the time-reversal property, we obtain
-
-$$
-x_{-n}^* \longleftrightarrow X_r^*
-$$
-
-When *x*(*t*) is real, then the conjugate-symmetry property states that *X*∗(ω) = *X*(−ω). Hence, for real *xn*,
-
-$$
-X_r^* = X_{-r}
-$$
-
-Moreover, *Xr* is *N*<sup>0</sup> periodic. Hence,
-
-$$
-X_r^* = X_{N_0-r}
-$$
-
-Because of this property, we need compute only half the DFTs for real *xn*. The other half are the conjugates.
-
-### TIME SHIFTING
-
-The time-shifting (circular shifting) property states†
-
-$$
-x_{n-k} \Longleftrightarrow X_r e^{-jr\Omega_0 k}
-$$
-
-**Proof.** We use Eq. (8.13) to find the inverse DFT of *Xre*−*jr*0*<sup>k</sup>* as
-
-$$
-\frac{1}{N_0} \sum_{r=0}^{N_0-1} X_r e^{-jr\Omega_0 k} e^{jr\omega_0 n} = \frac{1}{N_0} \sum_{r=0}^{N_0-1} X_r e^{jr\Omega_0 (n-k)} = x_{n-k}
-$$
-
-### FREQUENCY SHIFTING
-
-A dual of the time-shifting property, the frequency-shifting property states
-
-$$
-x_n e^{jn\Omega_0 m} \Longleftrightarrow X_{r-m}
-$$
-
-**Proof.** This proof is identical to that of the time-shifting property except that we start with Eq. (8.12).
-
-### CIRCULAR CONVOLUTION
-
-The circular (or periodic) convolution property states
-
-$$
-x_n \circledast g_n \Longleftrightarrow X_r G_r \tag{8.18}
-$$
-
-and
-
-$$
-x_n g_n \Longleftrightarrow \frac{1}{N_0} X_r \circledast G_r \tag{8.19}
-$$
-
-For two *N*0-periodic sequences *xn* and *gn*, circular (or periodic) convolution is defined by
-
-$$
-x_n \circledast g_n = \sum_{k=0}^{N_0 - 1} x_k g_{n-k} = \sum_{k=0}^{N_0 - 1} g_k x_{n-k}
-$$
-\n(8.20)
-
-<sup>†</sup> Time shifting is also known as *circular shifting* because such a shift can be interpreted as a circular shift of the *N*<sup>0</sup> samples in the first cycle 0 ≤ *n* ≤ *N*<sup>0</sup> −1.
-
-<span id="page-839-0"></span>**Figure 8.21** Graphical depictions of circular convolution.
-
-To prove Eq. (8.18), we find the DFT of the circular convolution *xn*-∗ *gn* as
-
-$$
-\sum_{n=0}^{N_0-1} \left( \sum_{k=0}^{N_0-1} x_k g_{n-k} \right) e^{-j r \omega_0 n} = \sum_{k=0}^{N_0-1} x_k \left( \sum_{n=0}^{N_0-1} g_{n-k} e^{-j r \omega_0 n} \right)
-$$
-$$
-= \sum_{k=0}^{N_0-1} x_k (G_r e^{-j r \Omega_0 k}) = X_r G_r
-$$
-
-Equation (8.19) can be proved in the same way.
-
-For periodic sequences, the convolution can be visualized in terms of two sequences, with one sequence fixed and the other inverted and moved past the fixed sequence, one digit at a time. If the two sequences are *N*<sup>0</sup> periodic, the same configuration will repeat after *N*<sup>0</sup> shifts of the sequence. Clearly the convolution *xn*-∗ *gn* becomes *N*<sup>0</sup> periodic. Such convolution can be conveniently visualized in terms of *N*<sup>0</sup> sequences, as illustrated in Fig. 8.21, for the case of *N*<sup>0</sup> = 4. The inner *N*0-point sequence *xn* is clockwise and fixed. The outer *N*0-point sequence *gn* is inverted so that it becomes counterclockwise. This sequence is now rotated clockwise 1 unit at a time. We multiply the overlapping numbers and add. For example, the value of *xn*-∗ *gn* at *n* = 0 (Fig. 8.21) is
-
-$$
-x_0g_0 + x_1g_3 + x_2g_2 + x_3g_1
-$$
-
-and the value of *xn*-∗ *gn* at *n* = 1 is (Fig. 8.21)
-
-$$
-x_0g_1 + x_1g_0 + x_2g_3 + x_3g_2
-$$
-
-and so on.
-
-### **[8.5-2 Some Applications of the DFT](#page-13-0)**
-
-The DFT is useful not only in the computation of direct and inverse Fourier transforms, but also in other applications such as convolution, correlation, and filtering. Use of the efficient FFT algorithm, discussed shortly (Sec. 8.6), makes it particularly appealing.
-
-### LINEAR CONVOLUTION
-
-Let *x*(*t*) and *g*(*t*) be the two signals to be convolved. In general, these signals may have different time durations. To convolve them by using their samples, they must be sampled at the same rate (not below the Nyquist rate of either signal). Let *xn* (0 ≤ *n* ≤ *N*<sup>1</sup> − 1) and *gn* (0 ≤ *n* ≤ *N*<sup>2</sup> − 1) be the corresponding discrete sequences representing these samples. Now,
-
-$$
-c(t) = x(t) * g(t)
-$$
-
-and if we define three sequences as *xn* = *Tx*(*nT*), *gn* = *Tg*(*nT*), and *cn* = *Tc*(*nT*), then†
-
-$$
-c_n = x_n * g_n
-$$
-
-where we define the linear convolution sum of two discrete sequences *xn* and *gn* as
-
-$$
-c_n = x_n * g_n = \sum_{k=-\infty}^{\infty} x_k g_{n-k}
-$$
-
-Because of the width property of the convolution, *cn* exists for 0≤ *n*≤ *N*1+*N*2−1. To be able to use the DFT circular convolution technique, we must make sure that the circular convolution will yield the same result as does linear convolution. In other words, the signal resulting from the circular convolution must have the same length (*N*<sup>1</sup> + *N*<sup>2</sup> − 1) as that of the signal resulting from linear convolution. This step can be accomplished by adding *N*<sup>2</sup> − 1 dummy samples of zero value to *xn* and *N*<sup>1</sup> −1 dummy samples of zero value to *gn* (zero padding). This procedure changes the length of both *xn* and *gn* to *N*1+*N*<sup>2</sup> −1. The circular convolution now is identical to the linear convolution except that it repeats periodically with period *N*<sup>1</sup> +*N*<sup>2</sup> −1. A little reflection will show that in such a case the circular convolution procedure in Fig. 8.21 over one cycle (0 ≤ *n* ≤ *N*<sup>1</sup> + *N*<sup>2</sup> − 1) is identical to the linear convolution of the two sequences *xn* and *gn*. We can use the DFT to find the convolution *xn* ∗ *gn* in three steps, as follows:
-
-- 1. Find the DFTs *Xr* and *Gr* corresponding to suitably padded *xn* and *gn*.
-- 2. Multiply *Xr* by *Gr*.
-- 3. Find the IDFT of *XrGr*. This procedure of convolution, when implemented by the fast Fourier transform algorithm (discussed later), is known as *fast convolution*.
-
-### FILTERING
-
-We generally think of filtering in terms of a hardware-oriented solution (e.g., building a circuit with *RLC* components and operational amplifiers). However, filtering also has a software-oriented solution [a computer algorithm that yields the filtered output *y*(*t*) for a given input *x*(*t*)]. This goal can be conveniently accomplished by using the DFT. If *x*(*t*) is the signal to be filtered, then *Xr*, the DFT of *xn*, is found. The spectrum *Xr* is then shaped (filtered) as desired by multiplying *Xr* by *Hr*, where *Hr* are the samples of *H*(ω) for the filter [*Hr* = *H*(*r*ω0)]. Finally, we take the IDFT of *XrHr* to obtain the filtered output *yn*[*yn* =*Ty*(*nT*)]. This procedure is demonstrated in the following example.
-
-<sup>†</sup> We can show that *cn* <sup>=</sup> lim*T*→<sup>0</sup> *xn* <sup>∗</sup> *gn*; [see 4]. Error is inherent in any numerical method used to compute convolution of continuous-time signals; since *T* = 0 in practice, there will be some error in this equation.
-
-The signal *x*(*t*) in Fig. 8.22a is passed through an ideal lowpass filter of frequency response *H*(ω) depicted in Fig. 8.22b. Use the DFT to find the sampled version of the filter output.
-
-**Figure 8.22** DFT solution for filtering *x*(*t*) through *H*(ω).
-
-We have already found the 32-point DFT of *x*(*t*) (see Fig. 8.19d). Next we multiply *Xr* by *Hr*. To find *Hr*, we recall using *f*<sup>0</sup> = 1/4 in computing the 32-point DFT of *x*(*t*). Because *Xr* is 32-periodic, *Hr* must also be 32-periodic with samples separated by 1/4 Hz. This fact means that *Hr* must be repeated every 8 Hz or 16π rad/s (see Fig. 8.22c). The resulting 32 samples of *Hr* over (0 ≤ ω ≤ 16π ) are as follows:
-
-$$
-H_r = \begin{cases} 1 & 0 \le r \le 7 \\ 0 & 9 \le r \le 23 \\ 0.5 & r = 8,24 \end{cases} \text{ and } 25 \le r \le 31
-$$
-
-We multiply *Xr* with *Hr*. The desired output signal samples *yn* are found by taking the inverse DFT of *XrHr*. The resulting output signal is illustrated in Fig. 8.22d.
-
-It is quite simple to verify the results of this filtering example using MATLAB. First, parameters are defined, and MATLAB's fft command is used to compute the DFT of *xn*.
-
->> T\_0 = 4; N\_0 = 32; T = T\_0/N\_0; n = (0:N\_0-1); r = n; >> x\_n = [ones(1,4) 0.5 zeros(1,23) 0.5 ones(1,3)]'; X\_r = fft(x\_n);
-
-The DFT of the filter's output is just the product of the filter response *Hr* and the input DFT *Xr*. The output *yn* is obtained using the ifft command and then plotted.
-
-```
->> H_r = [ones(1,8) 0.5 zeros(1,15) 0.5 ones(1,7)]';
->> Y_r = H_r.*X_r; y_n = ifft(Y_r);
->> clf; stem(n,real(y_n),'k.');
->> xlabel('n'); ylabel('y_n'); axis([0 31 -.1 1.1]);
-```
-
-The result, shown in Fig. 8.23, matches the earlier result shown in Fig. 8.22d. Recall, this DFT-based approach shows the samples *yn* of the filter output *y*(*t*) (sampled in this case at a rate *<sup>T</sup>* <sup>=</sup> <sup>1</sup> <sup>8</sup> ) over 0 ≤ *n* ≤ *N*<sup>0</sup> − 1 = 31 when the input pulse *x*(*t*) is periodically replicated to form samples *xn* (see Fig. 8.19c).
-
-**Figure 8.23** Using MATLAB and the DFT to determine filter output.

engineering/linear-systems-and-signals-3rd-edition-bp-lathi/107_8.6 THE FAST FOURIER TRANSFORM (FFT).md

@@ -1,99 +0,0 @@
-## <span id="page-843-0"></span>**8.6 THE FAST FOURIER [TRANSFORM](#page-13-0) (FFT)**
-
-The number of computations required in performing the DFT was dramatically reduced by an algorithm developed by Cooley and Tukey in 1965 [5]. This algorithm, known as the *fast Fourier transform* (FFT), reduces the number of computations from something on the order of *N*<sup>2</sup> <sup>0</sup> to *N*<sup>0</sup> log*N*0. To compute one sample *Xr* from Eq. (8.12), we require *N*<sup>0</sup> complex multiplications and *N*<sup>0</sup> −1 complex additions. To compute *N*<sup>0</sup> such values (*Xr* for *r* = 0, 1,...,*N*<sup>0</sup> −1), we require a total of *N*<sup>2</sup> <sup>0</sup> complex multiplications and *N*0(*N*<sup>0</sup> − 1) complex additions. For a large *N*0, these computations can be prohibitively time-consuming, even for a high-speed computer. The FFT algorithm is what made the use of Fourier transform accessible for digital signal processing.
-
-### HOW DOES THE FFT REDUCE THE NUMBER OF COMPUTATIONS?
-
-It is easy to understand the magic of the FFT. The secret is in the linearity of the Fourier transform and also of the DFT. Because of linearity, we can compute the Fourier transform of a signal *x*(*t*) as a sum of the Fourier transforms of segments of *x*(*t*) of shorter duration. The same principle applies to the computation of the DFT. Consider a signal of length *N*<sup>0</sup> = 16 samples. As seen earlier, DFT computation of this sequence requires *N*<sup>2</sup> <sup>0</sup> = 256 multiplications and *N*0(*N*<sup>0</sup> − 1) = 240 additions. We can split this sequence into two shorter sequences, each of length 8. To compute DFT of each of these segments, we need 64 multiplications and 56 additions. Thus, we need a total of 128 multiplications and 112 additions. Suppose, we split the original sequence in four segments of length 4 each. To compute the DFT of each segment, we require 16 multiplications and 12 additions. Hence, we need a total of 64 multiplications and 48 additions. If we split the sequence in eight segments of length 2 each, we need 4 multiplications and 2 additions for each segment, resulting in a total of 32 multiplications and 8 additions. Thus, we have been able to reduce the number of multiplications from 256 to 32 and the number of additions from 240 to 8. Moreover, some of these multiplications turn out to be multiplications by 1 or −1. All this fantastic economy in the number of computations is realized by the FFT without any approximation! The values obtained by the FFT are identical to those obtained by the DFT. In this example, we considered a relatively small value of *N*<sup>0</sup> = 16. The reduction in the number of computations is much more dramatic for higher values of *N*0.
-
-The FFT algorithm is simplified if we choose *N*<sup>0</sup> to be a power of 2, although such a choice is not essential. For convenience, we define
-
-$$
-W_{N_0} = e^{-(j2\pi/N_0)} = e^{-j\Omega_0}
-$$
-
-so that
-
-$$
-X_r = \sum_{n=0}^{N_0 - 1} x_n W_{N_0}^{nr} \qquad 0 \le r \le N_0 - 1 \tag{8.21}
-$$
-
-and
-
-$$
-x_n = \frac{1}{N_0} \sum_{r=0}^{N_0 - 1} X_r W_{N_0}^{-nr} \qquad 0 \le n \le N_0 - 1 \tag{8.22}
-$$
-
-Although there are many variations of the Tukey–Cooley algorithm, these can be grouped into two basic types: *decimation in time* and *decimation in frequency*.
-
-### THE DECIMATION-IN-TIME ALGORITHM
-
-Here we divide the *N*0-point data sequence *xn* into two (*N*0/2)-point sequences consisting of evenand odd-numbered samples, respectively, as follows:
-
-$$
-\underbrace{x_0, x_2, x_4, \ldots, x_{N_0-2}}_{\text{sequence } g_n}, \underbrace{x_1, x_3, x_5, \ldots, x_{N_0-1}}_{\text{sequence } h_n}
-$$
-
-Then, from Eq. (8.21),
-
-$$
-X_r = \sum_{n=0}^{(N_0/2)-1} x_{2n} W_{N_0}^{2n} + \sum_{n=0}^{(N_0/2)-1} x_{2n+1} W_{N_0}^{(2n+1)r}
-$$
-
-Also, since
-
-$$
-W_{N_0/2} = W_{N_0}^2
-$$
-
-we have
-
-$$
-X_r = \sum_{n=0}^{(N_0/2)-1} x_{2n} W_{N_0/2}^{nr} + W_{N_0}^r \sum_{n=0}^{(N_0/2)-1} x_{2n+1} W_{N_0/2}^{nr}
-$$
-  
-=  $G_r + W_{N_0}^r H_r$  0 \le  $r \le N_0 - 1$  (8.23)
-
-where *Gr* and *Hr* are the (*N*0/2)-point DFTs of the even- and odd-numbered sequences, *gn* and *hn*, respectively. Also, *Gr* and *Hr*, being the (*N*0/2)-point DFTs, are (*N*0/2) periodic. Hence,
-
-$$
-G_{r+(N_0/2)} = G_r \qquad \text{and} \qquad H_{r+(N_0/2)} = H_r \tag{8.24}
-$$
-
-Moreover,
-
-$$
-W_{N_0}^{r + (N_0/2)} = W_{N_0}^{N_0/2} W_{N_0}^r = e^{-j\pi} W_{N_0}^r = -W_{N_0}^r
-$$
-\n(8.25)
-
-From Eqs. (8.23), (8.24), and (8.25), we obtain
-
-$$
-X_{r+(N_0/2)} = G_r - W_{N_0}^r H_r \tag{8.26}
-$$
-
-This property can be used to reduce the number of computations. We can compute the first *N*0/2 points (0 ≤ *n* ≤ (*N*0/2) − 1) of *Xr* by using Eq. (8.23) and the last *N*0/2 points by using Eq. (8.26) as
-
-$$
-X_r = G_r + W_{N_0}^r H_r \qquad 0 \le r \le \frac{N_0}{2} - 1
-$$
-  
-$$
-X_{r + (N_0/2)} = G_r - W_{N_0}^r H_r \qquad 0 \le r \le \frac{N_0}{2} - 1
-$$
- (8.27)
-
-**Figure 8.24** Butterfly signal flow graph.
-
-**Figure 8.25** Successive steps in an 8-point FFT.
-
-<span id="page-846-0"></span>Thus, an *N*0-point DFT can be computed by combining the two (*N*0/2)-point DFTs, as in Eq. (8.27). These equations can be represented conveniently by the *signal flow* graph depicted in Fig. 8.24. This structure is known as a *butterfly*. Figure 8.25a shows the implementation of Eq. (8.24) for the case of *N*<sup>0</sup> = 8.
-
-The next step is to compute the (*N*0/2)-point DFTs *Gr* and *Hr*. We repeat the same procedure by dividing *gn* and *hn* into two (*N*0/4)-point sequences corresponding to the even- and odd-numbered samples. Then we continue this process until we reach the one-point DFT. These steps for the case of *N*<sup>0</sup> = 8 are shown in Figs. 8.25a, 8.25b, and 8.25c. Figure 8.25c shows that the two-point DFTs require no multiplication.
-
-To count the number of computations required in the first step, assume that *Gr* and *Hr* are known. Equation (8.27) clearly shows that to compute all the *N*<sup>0</sup> points of the *Xr*, we require *N*<sup>0</sup> complex additions and *N*0/2 complex multiplications† (corresponding to *W<sup>r</sup> N*0 *Hr*).
-
-In the second step, to compute the (*N*0/2)-point DFT *Gr* from the (*N*0/4)-point DFT, we require *N*0/2 complex additions and *N*0/4 complex multiplications. We require an equal number of computations for *Hr*. Hence, in the second step, there are *N*<sup>0</sup> complex additions and *N*0/2 complex multiplications. The number of computations required remains the same in each step. Since a total of log2*N*<sup>0</sup> steps is needed to arrive at a one-point DFT, we require, conservatively, a total of *N*<sup>0</sup> log2*N*<sup>0</sup> complex additions and (*N*0/2)log2*N*<sup>0</sup> complex multiplications, to compute the *N*0-point DFT. Actually, as Fig. 8.25c shows, many multiplications are multiplications by 1 or −1, which further reduces the number of computations.
-
-The procedure for obtaining IDFT is identical to that used to obtain the DFT except that *WN*<sup>0</sup> = *ej*(2π/*N*0) instead of *e*−*j*(2π/*N*0) (in addition to the multiplier 1/*N*0). Another FFT algorithm, the *decimation-in-frequency* algorithm, is similar to the decimation-in-time algorithm. The only difference is that instead of dividing *xn* into two sequences of even- and odd-numbered samples, we divide *xn* into two sequences formed by the first *N*0/2 and the last *N*0/2 samples, proceeding in the same way until a single-point DFT is reached in log2*N*<sup>0</sup> steps. The total number of computations in this algorithm is the same as that in the decimation-in-time algorithm.

engineering/linear-systems-and-signals-3rd-edition-bp-lathi/108_8.7 MATLAB - THE DISCRETE FOURIER TRANSFORM.md

@@ -1,216 +0,0 @@
-## **[8.7 MATLAB: THE](#page-13-0) DISCRETE FOURIER TRANSFORM**
-
-As an idea, the discrete Fourier transform (DFT) has been known for hundreds of years. Practical computing devices, however, are responsible for bringing the DFT into common use. MATLAB is capable of DFT computations that would have been impractical just a few decades ago.
-
-### **[8.7-1 Computing the Discrete Fourier Transform](#page-13-0)**
-
-The MATLAB command fft(x) computes the DFT of a vector x that is defined over (0 ≤ *n* ≤ *N*<sup>0</sup> −1) (Problem 8.7-1 considers how to scale the DFT to accommodate signals that do not begin at *n* = 0.) As its name suggests, the function fft uses the computationally more efficient fast Fourier transform algorithm when it is appropriate to do so. The inverse DFT is easily computed by using the ifft function.
-
-<sup>†</sup> Actually, *N*0/2 is a conservative figure because some multiplications corresponding to the cases of *W<sup>r</sup> <sup>N</sup>*<sup>0</sup> = 1,*j*, and so on, are eliminated.
-
-To illustrate MATLAB's DFT capabilities, consider 50 points of a 10 Hz sinusoid sampled at *fs* = 50 Hz and scaled by *T* = 1/*fs*.
-
-```
->> T = 1/50; N_0 = 50; n = (0:N_0-1);
->> x = T*cos(2*pi*10*n*T);
-```
-
-In this case, the vector x contains exactly 10 cycles of the sinusoid. The fft command computes the DFT.
-
->> X = fft(x);
-
-Since the DFT is both discrete and periodic, fft needs to return only the *N*<sup>0</sup> discrete values contained in the single period (0 ≤ *f* < *fs*).
-
-While *Xr* can be plotted as a function of *r*, it is more convenient to plot the DFT as a function of frequency *f* . A frequency vector, in hertz, is created by using *N*<sup>0</sup> and *T*.
-
->> f = (0:N\_0-1)/(T\*N\_0); stem(f,abs(X),'k.'); >> axis([0 50 -0.05 0.55]); xlabel('f [Hz]'); ylabel('|X(f)|');
-
-As expected, Fig. 8.26 shows content at a frequency of 10 Hz. Since the time-domain signal is real, *X*(*f*) is conjugate symmetric. Thus, content at 10 Hz implies equal content at −10 Hz. The content visible at 40 Hz is an alias of the −10 Hz content.
-
-Often, it is preferred to plot a DFT over the principal frequency range (−*fs*/2 ≤ *f* < *fs*/2). The MATLAB function fftshift properly rearranges the output of fft to accomplish this task.
-
-```
->> stem(f-1/(T*2),fftshift(abs(X)),'k.');
->> axis([-25 25 -0.05 0.55]); xlabel('f [Hz]'); ylabel('|X(f)|');
-```
-
-When we use fftshift, the conjugate symmetry that accompanies the DFT of a real signal becomes apparent, as shown in Fig. 8.27.
-
-Since DFTs are generally complex-valued, the magnitude plots of Figs. 8.26 and 8.27 offer only half the picture; the signal's phase spectrum, shown in Fig. 8.28, completes it.
-
-```
->> stem(f-1/(T*2),fftshift(angle(X)),'k.');
->> axis([-25 25 -1.1*pi 1.1*pi]); xlabel('f [Hz]'); ylabel('\angle X(f)');
-```
-
-**Figure 8.26** |*X*(*f*)| computed over (0 ≤ *f* < 50) by using fft.
-
-<span id="page-848-0"></span>**Figure 8.27** |*X*(*f*)| displayed over (−25 ≤ *f* < 25) by using fftshift.
-
-**Figure 8.28** *X*(*f*) displayed over (−25≤*f* <25).
-
-Since the signal is real, the phase spectrum necessarily has odd symmetry. Additionally, the phase at ±10 Hz is zero, as expected for a zero-phase cosine function. More interesting, however, are the phase values found at the remaining frequencies. Does a simple cosine really have such complicated phase characteristics? The answer, of course, is no. The magnitude plot of Fig. 8.27 helps identify the problem: there is zero content at frequencies other than ±10 Hz. Phase computations are not reliable at points where the magnitude response is zero. One way to remedy this problem is to assign a phase of zero when the magnitude response is near or at zero.
-
-### **[8.7-2 Improving the Picture with Zero Padding](#page-13-0)**
-
-DFT magnitude and phase plots paint a picture of a signal's spectrum. At times, however, the picture can be somewhat misleading. Given a sampling frequency *fs* = 50 Hz and a sampling interval *T* = 1/*fs*, consider the signal
-
-$$
-y[n] = Te^{j2\pi \left(10\frac{1}{3}\right)n}
-$$
-
-This complex-valued, periodic signal contains a single positive frequency at 10<sup>1</sup> <sup>3</sup> Hz. Let us compute the signal's DFT using 50 samples.
-
-```
->> y = T*exp(j*2*pi*(10+1/3)*n*T); Y = fft(y);
->> stem(f-25,fftshift(abs(Y)),'k.');
->> axis([-25 25 -0.05 1.05]); xlabel('f [Hz]'); ylabel('|Y(f)|');
-```
-
-**Figure 8.29** |*Y*(*f*)| using 50 data points.
-
-**Figure 8.30** |*Yzp*(*f*)| over 5 ≤ *f* ≤ 15 using 50 data points padded with 550 zeros.
-
-In this case, the vector y contains a noninteger number of cycles. Figure 8.29 shows the significant frequency leakage that results. Also notice that since *y*[*n*] is not real, the DFT is not conjugate symmetric.
-
-In this example, the discrete DFT frequencies do not include the actual 10<sup>1</sup> <sup>3</sup> Hz frequency of the signal. Thus, it is difficult to determine the signal's frequency from Fig. 8.29. To improve the picture, the signal is zero-padded to 12 times its original length.
-
-```
->> y_zp = [y,zeros(1,11*length(y))]; Y_zp = fft(y_zp);
->> f_zp = (0:12*N_0-1)/(T*12*N_0);
->> stem(f_zp-25,fftshift(abs(Y_zp)),'k.');
->> axis([-25 25 -0.05 1.05]); xlabel('f [Hz]'); ylabel('|Y_{zp}(f)|');
-```
-
-Figure 8.30, zoomed in to 5 <sup>≤</sup> *<sup>f</sup>* <sup>≤</sup> 15, correctly shows the peak frequency at 10<sup>1</sup> <sup>3</sup> Hz and better represents the signal's spectrum.
-
-It is important to keep in mind that zero padding does not increase the resolution or accuracy of the DFT. To return to the picket fence analogy, zero padding increases the number of pickets in our fence but cannot change what is behind the fence. More formally, the characteristics of the sinc function, such as main beam width and sidelobe levels, depend on the fixed width of the pulse, not on the number of zeros that follow. Adding zeros cannot change the characteristics of the sinc function and thus cannot change the resolution or accuracy of the DFT. Adding zeros simply allows the sinc function to be sampled more finely.
-
-### <span id="page-850-0"></span>**[8.7-3 Quantization](#page-13-0)**
-
-A *B*-bit analog-to-digital converter (ADC) samples an analog signal and quantizes amplitudes by using 2*<sup>B</sup>* discrete levels. This quantization results in signal distortion that is particularly noticeable for small *B*. Typically, quantization is classified as symmetric or asymmetric and as either rounding or truncating. Let us investigate rounding-type quantizers.
-
-The quantized output *x*<sup>q</sup> of an asymmetric rounding converter is given as†
-
-$$
-x_{\rm q} = \frac{x_{\rm max}}{2^{B-1}}\lfloor \frac{x}{x_{\rm max}}2^{B-1} + \frac{1}{2}\rfloor
-$$
-
-The quantized output *x*<sup>q</sup> of a symmetric rounding converter is given as
-
-$$
-x_{\rm q} = \frac{x_{\rm max}}{2^{B-1}}\left(\lfloor \frac{x}{x_{\rm max}}2^{B-1}\rfloor + \frac{1}{2}\right)
-$$
-
-Program CH8MP1 quantizes a signal using one of these two rounding quantizer rules and also ensures no more than 2*<sup>B</sup>* output levels.
-
-```
-function [xq] = CH8MP1(x,xmax,B,method)
-% CH8MP1.m : Chapter 8, MATLAB Program 1
-% Function M-file quantizes x over (-xmax,xmax) using 2^b levels.
-% Uses rounding rule, supports symmetric and asymmetric quantization
-% INPUTS: x = input signal
-% xmax = maximum magnitude of signal to be quantized
-% B = number of quantization bits
-% method = default 'sym' for symmetrical, 'asym' for asymmetrical
-% OUTPUTS: xq = quantized signal
-if (nargin<3),
-   disp('Insufficient number of inputs.'); return
-elseif (nargin==3),
-   method = 'sym';
-elseif (nargin>4),
-   disp('Too many inputs.'); return
-end
-x(abs(x)>xmax)=xmax*sign(x(abs(x)>xmax)); % Limit amplitude to xmax
-switch lower(method)
-   case 'asym'
-       xq = xmax/(2^(B-1))*floor(x*2^(B-1)/xmax+1/2);
-       xq(xq>=xmax)=xmax*(1-2^(1-B)); % Ensure only 2^B levels
-   case 'sym'
-       xq = xmax/(2^(B-1))*(floor(x*2^(B-1)/xmax)+1/2);
-       xq(xq>=xmax)=xmax*(1-2^(1-B)/2); % Ensure only 2^B levels
-```
-
-<sup>†</sup> Large values of *x* may return quantized values *x*<sup>q</sup> outside the 2*<sup>B</sup>* allowable levels. In such cases, *x*<sup>q</sup> should be clamped to the nearest permitted level.
-
-```
-otherwise
-        disp('Unrecognized quantization method.'); return
-end
-```
-
-Several MATLAB commands require discussion. First, the nargin function returns the number of input arguments. In this program, nargin is used to ensure that a correct number of inputs is supplied. If the number of inputs supplied is incorrect, an error message is displayed and the function terminates. If only three input arguments are detected, the quantization type is not explicitly specified and the program assigns the default symmetric method.
-
-As with many high-level languages such as C, MATLAB supports general switch/case structures† :
-
-```
-switch switch_expr,
-case case_expr,
-      statements;
-...
-otherwise,
-      statements;
-```
-
-end
-
-CH8MP1 switches among cases of the string method. In this way, method-specific parameters are easily set. The command lower is used to convert a string to all lowercase characters. In this way, strings such as SYM, Sym, and sym are all indistinguishable. Similar to lower, the MATLAB command upper converts a string to all uppercase.
-
-The floor command rounds input values to the nearest integer toward minus infinity. Mathematically, it computes ·. To accommodate different types of rounding, MATLAB supplies three other rounding commands: ceil, round, and fix. The ceil command rounds input values to the nearest integers toward infinity, ( ·"); the round command rounds input values toward the nearest integer; the fix command rounds input values to the nearest integer toward zero. For example, if x = [-0.5 0.5];, floor(x) yields [-1 0], ceil(x) yields [0 1], round(x) yields [-1 1], and fix(x) yields [0 0]. Finally, CH8MP1 checks and, if necessary, corrects large values of *x*<sup>q</sup> that may be outside the allowable 2*<sup>B</sup>* levels.
-
-To verify operation, CH8MP1 is used to determine the transfer characteristics of a symmetric 3-bit quantizer operating over (−10,10).
-
-```
->> x = (-10:.0001:10); xsq = CH8MP1(x,10,3,'sym');
-```
-
-```
->> plot(x,xsq,'k'); axis([-10 10 -10.5 10.5]); grid on;
-```
-
-```
->> xlabel('Quantizer input'); ylabel('Quantizer output');
-```
-
-Figure 8.31 shows the results. Clearly, the quantized output is limited to 2*<sup>B</sup>* = 8 levels. Zero is not a quantization level for symmetric quantizers, so half of the levels occur above zero and half of the levels occur below zero. In fact, *symmetric quantizers* get their name from the symmetry in quantization levels above and below zero.
-
-By changing the method in CH8MP1 from 'sym' to 'asym', we obtain the transfer characteristics of an asymmetric 3-bit quantizer, as shown in Fig. 8.32. Again, the quantized output is limited to 2*<sup>B</sup>* = 8 levels, and zero is now one of the included levels. With zero as a quantization
-
-<sup>†</sup> A functionally equivalent structure can be written by using if, elseif, and else statements.
-
-**Figure 8.31** Transfer characteristics of a symmetric 3-bit quantizer.
-
-**Figure 8.32** Transfer characteristics of an asymmetric 3-bit quantizer.
-
-level, we need one fewer quantization level above zero than there are levels below. Not surprisingly, *asymmetric quantizers* get their name from the asymmetry in quantization levels above and below zero.
-
-There is no doubt that quantization can change a signal. It follows that the spectrum of a quantized signal can also change. While these changes are difficult to characterize mathematically, they are easy to investigate by using MATLAB. Consider a 1 Hz cosine sampled at *fs* = 50 Hz over 1 second.
-
->> x = cos(2\*pi\*n\*T); X = fft(x); T = 1/50; N\_0 = 50; n = (0:N\_0-1);
-
-Upon quantizing by means of a 2-bit asymmetric rounding quantizer, both the signal and spectrum are substantially changed.
-
-```
->> xaq = CH8MP1(x,1,2,'asym'); Xaq = fft(xaq);
->> subplot(2,2,1); stem(n,x,'k'); axis([0 49 -1.1 1.1]);
->> xlabel('n');ylabel('x[n]');
->> subplot(2,2,2); stem(f-25,fftshift(abs(X)),'k'); axis([-25,25 -1 26])
->> xlabel('f');ylabel('|X(f)|');
->> subplot(2,2,3); stem(n,xaq,'k');axis([0 49 -1.1 1.1]);
-```
-
-<span id="page-853-0"></span>**Figure 8.33** Signal and spectrum effects of quantization.
-
-```
->> xlabel('n');ylabel('x_{aq}[n]');
-```
-
-```
->> subplot(2,2,4); stem(f-25,fftshift(abs(fft(xaq))),'k'); axis([-25,25 -1 26]);
-```
-
-```
->> xlabel('f');ylabel('|X_{aq}(f)|');
-```
-
-The results are shown in Fig. 8.33. The original signal *x*[*n*] appears sinusoidal and has pure spectral content at ±1 Hz. The asymmetrically quantized signal *xaq*[*n*] is significantly distorted. The corresponding magnitude spectrum |*Xaq*(*f*)| is spread over a broad range of frequencies.

engineering/linear-systems-and-signals-3rd-edition-bp-lathi/109_8.8 SUMMARY.md

@@ -1,9 +0,0 @@
-## **[8.8 SUMMARY](#page-13-0)**
-
-A signal bandlimited to *B* Hz can be reconstructed exactly from its samples if the sampling rate *fs* > 2*B* Hz (the sampling theorem). Such a reconstruction, although possible theoretically, poses practical problems such as the need for ideal filters, which are unrealizable or are realizable only with infinite delay. Therefore, in practice, there is always an error in reconstructing a signal from its samples. Moreover, practical signals are not bandlimited, which causes an additional error (aliasing error) in signal reconstruction from its samples. When a signal is sampled at a frequency *fs* Hz, samples of a sinusoid of frequency (*fs*/2) + *x* Hz appear as samples of a lower frequency (*fs*/2) − *x* Hz. This phenomenon, in which higher frequencies appear as lower frequencies, is known as aliasing. Aliasing error can be reduced by bandlimiting a signal to *fs*/2 Hz (half the sampling frequency). Such bandlimiting, done prior to sampling, is accomplished by an anti-aliasing filter that is an ideal lowpass filter of cutoff frequency *fs*/2 Hz.
-
-The sampling theorem is very important in signal analysis, processing, and transmission because it allows us to replace a continuous-time signal with a discrete sequence of numbers. Processing a continuous-time signal is therefore equivalent to processing a discrete sequence of numbers. This leads us directly into the area of digital filtering (discrete-time systems). In the field <span id="page-854-0"></span>of communication, the transmission of a continuous-time message reduces to the transmission of a sequence of numbers. This opens doors to many new techniques of communicating continuous-time signals by pulse trains.
-
-The dual of the sampling theorem states that for a signal timelimited to τ seconds, its spectrum *X*(ω) can be reconstructed from the samples of *X*(ω) taken at uniform intervals not greater than 1/τ Hz. In other words, the spectrum should be sampled at a rate not less than τ samples/Hz.
-
-To compute the direct or the inverse Fourier transform numerically, we need a relationship between the samples of *x*(*t*) and *X*(ω). The sampling theorem and its dual provide such a quantitative relationship in the form of a discrete Fourier transform (DFT). The DFT computations are greatly facilitated by a fast Fourier transform (FFT) algorithm, which reduces the number of computations from something on the order of *N*<sup>2</sup> <sup>0</sup> to *N*<sup>0</sup> log*N*0.