4.8 FREQUENCY [RESPONSE OF AN](#page-10-0) LTIC SYSTEM

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4.8 FREQUENCY RESPONSE OF AN 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 = jω in this relationship yields

$$e^{j\omega t} \Longrightarrow H(j\omega)e^{j\omega t} \tag{4.41}$$

Noting that cosωt is the real part of ejω*^t* , 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(jω) 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 cosωt 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(ωt +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 = jω 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 = jω [see Eq. (4.10)]. This means that H(s) when s = jω 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(jω)| times the input amplitude, and the phase of the output sinusoid is shifted by H(jω) with respect to the input phase (see later Fig. 4.38 in Ex. 4.27). For instance, a certain system with |H(j10)| = 3 and H(j10) = −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(10t + 50◦) is 3 × 5 cos(10t + 50◦ − 30◦) = 15 cos(10t +20◦).

Clearly |H(jω)| is the amplitude gain of the system, and a plot of |H(jω)| versus ω shows the amplitude gain as a function of frequency ω. We shall call |H(jω)| the amplitude response. It also goes under the name magnitude response. ‡ Similarly, H(jω) is the phase response, and a plot of H(jω) versus ω shows how the system modifies or changes the phase of the input sinusoid. Plots of the magnitude response |H(jω)| and phase response H(jω) show at a glance how a system responds to sinusoids of various frequencies. Observe that H(jω) has the information of |H(jω)| and H(jω) 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 2t

(b) cos(10t −50◦)

In this case,

$$H(j\omega) = \frac{j\omega + 0.1}{j\omega + 5}$$

^† This may also be argued as follows. For BIBO-unstable systems, the zero-input response contains nondecaying natural mode terms of the form cosω0t or eat cosω0t (a > 0). Hence, the response of such a system to a sinusoid cosωt will contain not just the sinusoid of frequency ω, but also nondecaying natural modes, rendering the concept of frequency response meaningless.

^‡ Strictly speaking, ^|H(ω)^| 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)|.

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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 2t, 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 2t is

$$y(t) = 0.372 \cos(2t + 65.3^{\circ})$$

The input cos 2t and the corresponding system response 0.372cos(2t + 65.3◦) are illustrated in Fig. 4.38b.

(b) For the input cos(10t − 50◦), instead of computing the values |H(jω)| and H(jω) 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(10t −50◦) is

$$y(t) = 0.894 \cos (10t - 50^\circ + 26^\circ) = 0.894 \cos (10t - 24^\circ)$$

If the input were sin(10t − 50◦), the response would be 0.894 sin(10t − 50◦ + 26◦) = 0.894 sin(10t −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 jω 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 cosωt 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(ωt − ωT) has a phase shift −ωT with respect to the input cosωt. 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,

^H(jω) ^{= jω = ω}ejπ/²

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 cosωt is passed through an ideal differentiator, the output is −ωsin ωt = ωcos[ωt + (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 cosωt. 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(jω)| = 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 cosωt is passed through an ideal integrator, the output is (1/ω)sin ωt = (1/ω) cos[ωt −(π/2)]. Therefore, the amplitude response is inversely proportional to ω, and the phase response is constant (−π/2)

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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(3t +35◦).

ANSWER

10.23 sin(3t −61.91◦)

4.8-1 Steady-State Response to Causal Sinusoidal Inputs

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 ejω*^t u*(t), which starts at t = 0 rather than at t = −∞. In this case X(s) = 1/(s + jω). Moreover, according to Eq. (4.27), H(s) = P(s)/Q(s), where Q(s) is the

^† 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 ejω*^t* starts at ^t = −∞. 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.

characteristic polynomial given by Q(s) = (s−λ1)(s−λ2)··· (s−λN). † 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 − λN) be k1, k2, … , kN. The coefficient corresponding to the last term (s−jω) is P(s)/Q(s)|^s=j^ω = H(jω). 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^λit decay with time, and, therefore, constitute the so-called transient component of the response. The last term H(jω)ejω*^t* 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 ejω*^t* results in the total response H(jω)ejω*^t* 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(jω)ejω*^t* .

From the argument that led to Eq. (4.43), it follows that for a causal sinusoidal input cosωt, 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(jω)| cos[ωt + H(jω)] is the total response to everlasting sinusoid cosωt. In contrast, it is the steady-state response to the same input applied at t = 0.