Geometric Evaluation of the z-Transform

4.5 Geometric Evaluation of the z-Transform

In this section we discuss a geometrical method to evaluate a rational function in z at any point in the z-plane, particularly on the unit circle z = e jΩ for obtaining the

frequency response G(Ω) from the pole-zero plot of the system function G[z]. Let us consider a system function G[z] given in a rational form as

G j =1 (z − z j ) (z − z 1 )(z − z 2 ) · · · (z − z M ) [z] = K 1 N

=K

i =1 (z − p i ) (z − p 1 )(z − p 2 ) · · · (z − p N )

232 4 The z-Transform where z j ’s and p i ’s are finite zeros and poles of G[z], respectively. The value of

G [z] at some point z = z 0 in the z-plane is a complex number that can be expressed in the polar form as

(4.5.2) where

G [z 0 ] = |G[z 0 ]|∠G[z 0 ]

j =1 |z 0 −z j |

Magnitude : |G[z 0 ]| = |K | 1 N

(4.5.3a)

i =1 |z 0 −p i |

Phase : ∠G[z 0 ]=

(4.5.3b) (z 0 −z j )’s and (z 0 −p i )’s in the above equations are complex numbers, each of

with ± π only for K < 0

which can be represented by a vector in the z-plane from z j or p i to z 0 . They can

be easily constructed from the pole/zero plot where |z 0 −z j | and |z 0 −p i | are the distances, while ∠z 0 −z j and ∠z 0 −p i are the phase angles.

As mentioned in Remark 4.5(1), the frequency response G(Ω) can be obtained by evaluating the system function G[z] on the unit circle if it exists. Keeping this in mind, we can use the pole-zero plot to get the approximate shape of the magnitude response |G(Ω)| without computing the frequency response from Eq. (4.5.2). Here is the overall feature of frequency response related with the pole-zero pattern:

Remark 4.7 Frequency (Magnitude/Phase) Response and Pole-Zero Pattern (1) For a pole/zero near the unit circle, the magnitude response curve tends to have

a sharp peak/valley and the phase changes rapidly by about 180 ◦ at the corre- sponding frequency. As the pole/zero moves to the unit circle, the peak/valley becomes sharper. On the other hand, as the pole/zero moves to the origin, the peak/valley becomes smoother. Such a tendency can be observed from Fig. 3.2 where the frequency responses of G[z] = z/(z −a) are depicted for a = 0.8 and

0.5. Also, the phase jump of π [rad] occurring at the frequency corresponding to a zero on the unit circle is illustrated by Figs. 3.1, 3.7, 3.11, and 4.8. (2) Generally speaking, if a singularity (pole/zero) is located close to the unit circle, it will dominate the frequency response in the frequency range adjacent to that location. This idea of dominant singularity is helpful not only for getting an approximate frequency response, but also for the pole/zero placement design to achieve a desired frequency response. For example, the magnitude response (Fig. 4.7(b)) of the system described by (E4.11.1) has the maximum around √

Ω = ±π/4 due to the pole at z = 0.5 2e ± jπ/4 and becomes zero at Ω = π because of the zero at z = −1. In contrast, the zero at z = 0 (which is far from

the unit circle) has no influence on the magnitude response.

4.5 Geometric Evaluation of the z-Transform 233 Example 4.11 Pole-Zero Pattern and Frequency Response

For the system function z (z + 1)

G [z] = (z + 1) (E4.11.1)

z 2 − z + 0.5 = (z − 0.5 − j0.5)(z − 0.5 + j0.5)

we have the pole-zero plot and the frequency response magnitude and phase curves depicted in Fig. 4.7(a), (b), and (c), respectively. As shown in Fig. 4.7(b), the mag- nitude of G(Ω) = G[e jΩ ] becomes zero at Ω = π corresponding to the zero z = −1, i.e.,

|G[e jΩ ]|

Ω =π = |G[z]|| z=−1 =0 (E4.11.2)

1 G [ z ] = B z (z+1) z 2 A P –z+0.5

z (z+1) (z –0.5–j –0.5+j 0.5) (z 0.5)

θ 2 θ 1 C Ω=0

Ω=− π Z 2 Re z Z1 } 1 { Ω=2 π

Ω=− π /4 (a) The pole–zero plot of G [ z ] = z (z+1)

z 2 –z+0.5

0 Ω P = 0.68 π

2 π Ω (b) The magnitude curve of the frequency response G ( Ω ) = G [ e j Ω ]

(c) The phase curve of the frequency response G ( Ω ) = G [ e j Ω ] Fig. 4.7 The pole–zero plot and frequency response for Example 4.11

234 4 The z-Transform and it reaches the peak around Ω = ±π/4 adjacent to the phases of the poles

z = 0.5 ± j0.5 = 0.5 2e ± jπ/4

(E4.11.3) (a) Let us find the response of the system G[z] to a sinusoidal input x 1 [n] =

sin(Ω p n ) with Ω p = 0.68.

z sin Ω p

(z + 1)

1 [z] = X 1 [z]G[z] =

2 2 (E4.11.4)

p (z − 0.5) + 0.5 −3.60 × z(z − cos Ω p ) + 3.89 × z sin Ω p

(z − cos Ω p + sin

(z − cos Ω p ) 2 + sin 2 Ω p

3.60 × z(z − R cos Ω s ) − 1.65 × z R sin Ω s

+ s ) 2 s ) (z − R cos Ω 2 + (R sin Ω

R= √ 1 2 ,Ω s = π 4

y 1 [n] = Z −1 {Y 1 [z] } = −3.6 cos(Ω p n ) + 3.89 sin(Ω p n )

√ −n + 2 (3.6 cos(nπ/4) − 1.65 sin(nπ/4))

√ −n

= (5.3 sin(Ω p n − 0.7457) + 2 (3.6 cos(nπ/4) − 1.65 sin(nπ/4)))u s [n]

(E4.11.5) Note that the sinusoidal steady-state response y 1,ss [n] = 5.3 sin(Ω p n − 0.7457)

has the same frequency with the input x 1 [n], but its amplitude is |G(Ω p )| = 5.3 times that of the input and its phase is ∠G(Ω p ) = −0.7457 ≃ −1.1Ω p plus that of the input, as can be seen from Fig. 4.8(a).

(b) Let us find the response of the system G[z] to a unit step (DC) input x 2 [n] = u s [n].

0 5 m 0 : Input

11 20 30 : Output

(a) The output of the system of Example 4.11 to a sinusoidal input sin (Ω pn )

0 0 10 20 30 n (b) The output of the system of Example 4.11 to a unit step (DC) input

Fig. 4.8 The output of the system of Example 4.11 to sinusoidal/DC inputs

4.5 Geometric Evaluation of the z-Transform 235

(z + 1)

2 [z] = X 2 [z]G[z] =

z−1 (z − 0.5) 2 + 0.5 2

1 × z0.5 =

4×z

−3 × z(z − 0.5)

2 2 (E4.11.6) z−1

(z − 0.5) + 0.5 where the partial fraction expansion is found as follows:

(z − 0.5) + 0.5

×z

K 2 × z(z − 0.5)

K 3 × z0.5

2 [z] = + z−1

2 + 0.5 2 (z − 0.5) + (z − 0.5) 2 + 0.5 2

(z − 0.5) + 0.5 z=1

K 1 +K 2 = 1, −K 1 − 0.5K 2 + 0.5K 3 = 1, 0.5K 1 + 0.5K 2 − 0.5K 3 =0 K 2 =1−K 1 = 1 − 4 = −3, K 3 =K 1 +K 2 =4−3=1

Thus the response to x 2 [n] = u s [n] can be obtained from the inverse z- transform as

√ −n

2 [n] = Z {Y 2 [z] }= 4+ 2 (−3 cos(nπ/4) + sin(nπ/4)) u s [n] (E4.11.7)

Note that the DC steady-state response y 2,ss [n] = 4 is the DC gain |G(0)| = 4 times that of the input, as can be seen from Fig. 4.8(b).

In fact, the inverse z-transform of Y 1 [z] as well as Figs. 4.7 and 4.8 is obtained by running the following MATLAB program “sig04e11.m”, where we managed to get the coefficients of the partial fraction expansion using residue() since iztrans() does not work properly for this case.

%sig04e11.m clear, clf B=[1 1 0]; A=[1 -1 0.5]; %numerator/denominator of system function (E4.11.1) figure(1), zplane(roots(B),roots(A)) % pole-zero plot in Fig. 4.7(a) % To get the frequency response N=360; dW=2*pi/N; k0=-200; W=[k0:500]*dW; GW= freqz(B,A,W); % frequency response GW mag= abs(GW); GW ph= angle(GW); % magnitude/phase of frequency response [GW mag peak,i]= max(GW mag(-k0+[1:N/2])) % peak frequency response magnitude ip=-k0+i-1; Wp=W(ip); GW ph peak=GW ph(ip); % peak frequency GW dc= GW(-k0+1); % DC gain figure(2), subplot(411), plot(W,GW mag) % Fig. 4.7(b) subplot(412), plot(W,GW ph)

% Fig. 4.7(c)

% To get the time response from filtering of a sine-wave input signal nn=[0:30]; % time index vector xn= sin(Wp*nn); % A sinusoidal input of peak frequency yn= filter(B,A,xn); % With zero initial condition by default % plot the time response subplot(413), stem(nn,xn,’Markersize’,5), hold on

236 4 The z-Transform

stem(nn,yn,’x’,’Markersize’,5) % Try with the inverse z-transform syms z y1=iztrans(sin(Wp)*zˆ2*(z+1)/((z-cos(Wp))ˆ2+sin(Wp)ˆ2)/((z-0.5)ˆ2+0.5ˆ2)) [r,p,k]=residue(sin(Wp)*[1 1 0],conv([1 -2*cos(Wp) 1],[1 -1 0.5])) R = sqrt(real(p(3)*p(4))); Ws= angle(p(3)); % numerator of the 1st&2nd terms reduced to a common denominator n1= r(1)*[1 -p(2)]+r(2)*[1 -p(1)]; K1= n1(1); K2= (n1(2)+K1*cos(Wp))/sin(Wp); % numerator of the 3rd&4th terms reduced to a common denominator n2= r(3)*[1 -p(4)]+r(4)*[1 -p(3)]; K3= n2(1); K4= (n2(2)+K3*R*cos(Ws))/(R*sin(Ws)); y1n= K1*cos(Wp*nn) + K2*sin(Wp*nn) + R.ˆnn.*(K3*cos(Ws*nn)+K4*sin(Ws*nn)); stem(nn,y1n,’rx’,’Markersize’,5) %filtering of a DC input signal nn=[0:30]; xn= ones(size(nn)); % A DC input signal (of zero frequency) yn DC= filter(B,A,xn); % With zero initial condition by default subplot(414) stem(nn,xn,’Markersize’,5) % plot the time response together with the input hold on, stem(nn,yn DC,’m’,’Markersize’,5) % Try with the inverse z-transform y2=iztrans(zˆ2*(z+1)/(z-1)/((z-0.5)ˆ2+0.5ˆ2)) % A nice alternative for the case of all simple poles [r,p,k]=residue([1 1 0],conv([1 -1],[1 -1 0.5])) y2n= 0; for i=1:length(r)

y2n= y2n + r(i)*p(i).ˆnn; end

stem(nn,real(y2n),’mˆ’,’Markersize’,5)