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System Representation in the z-Domain 127
we have
1
H(z) =1.2346 − 0.1173
1+0.9z − 0.1173 1
, |z| > 0.9
−1 1 − 0.9z−1 or from Table 4.1
h(n) =1.2346δ(n) − 0.1173 {1+(−1) n } (0.9) n u(n)
c. From Table 4.1 Z[u(n)] = U(z) =
or
V (z) =H(z)U(z)
[
Finally,
=
=
(1 + z −1 )(1 − z −1 )
(1 + 0.9z −1 )(1− 0.9z −1 )
1
, |z| > 1. Hence
1 − z−1 ] [
1
1 − z −1 ]
, |z| > 0.9 ∩|z| > 1
1+z −1
, |z| > 0.9
(1+0.9z −1 )(1− 0.9z −1 )
1
V (z) =1.0556
1 − 0.9z − 0.0556 1
, |z| > 0.9
−1 1+0.9z−1 v(n) =[1.0556(0.9) n − 0.0556 (−0.9) n ] u(n)
Note that in the calculation of V (z) there is a pole-zero cancellation at z =1.
This has two implications. First, the ROC of V (z) isstill {|z| > 0.9} and not
{|z| > 0.9 ∩|z| > 1=|z| > 1}. Second, the step response v(n) contains no
steady-state term u(n).
d. Substituting z = e jω in H(z),
H(e jω )=
1 − e−j2ω
1 − 0.81e −j2ω
We will use the MATLAB script to compute and plot responses.
>> w = [0:1:500]*pi/500; H = freqz(b,a,w);
>> magH = abs(H); phaH = angle(H);
>> subplot(2,1,1); plot(w/pi,magH); grid
>> xlabel(’frequency in pi units’); ylabel(’Magnitude’)
>> title(’Magnitude Response’)
>> subplot(2,1,2); plot(w/pi,phaH/pi); grid
>> xlabel(’frequency in pi units’); ylabel(’Phase in pi units’)
>> title(’Phase Response’)
The frequency response plots are shown in Figure 4.10.
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