5.6 FREQUENCY [RESPONSE FROM](#page-11-0) POLE-ZERO LOCATIONS

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5.6 FREQUENCY RESPONSE FROM 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 Nth-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 (z1, z2,…,zN) and all poles (γ1, γ2, … , γN) 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 r1, r2, … , rN be the lengths and φ1, φ2, … , φ*^N* be the angles, respectively, of the straight lines connecting z1, z2, … , zN to the point ej. Similarly, let d1, d2, … , dN be the lengths and θ1, θ2, … , θ*^N* be the angles, respectively, of the lines connecting γ1, γ2, … , γ*^N* 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 b⁰ > 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 ej⁰ = 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

^† 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.

^‡ Placing the pole at ^{z =} 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^π = −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 = ωT.

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 = ωT. 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 = ejπ/4. Therefore, to enhance the frequency response at ω = 250π, we place a pole in the vicinity of ejπ/4. Because this is a complex pole, we also need its conjugate near e−jπ/4, as indicated in Fig. 5.21a. Let us choose these poles γ¹ and γ² as

^γ1 = |^{γ |}ejπ/⁴ and ^γ2 = |^{γ |}e−jπ/⁴

^† 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 = ωT = 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 < 1/2fh = 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 = ejπ/² = 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 = ejπ/² = j and its conjugate at z = e−jπ/² = −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+a²)/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.

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.