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System Representation in the z-Domain 123
15
Magnitude Response
Magnitude
10
5
0
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
frequency in π units
0
Phase Response
Phase in π units
−0.1
−0.2
−0.3
−0.4
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
frequency in π units
FIGURE 4.8 Frequency response plots in Example 4.11
of the unit circle begins at ω = π. Toovercome this problem, we will use the
second form of the freqz function as follows.
>> [H,w] = freqz(b,a,200,’whole’);
>> magH = abs(H(1:101)); phaH = angle(H(1:101));
Now the 101st element of the array H will correspond to ω = π. Asimilar
result can be obtained using the third form of the freqz function.
>> w = [0:1:100]*pi/100; H = freqz(b,a,w);
>> magH = abs(H); phaH = angle(H);
In the future we will use any one of these forms, depending on our convenience.
Also note that in the plots we divided the w and phaH arrays by pi
so that the plot axes are in the units of π and easier to read. This practice
is strongly recommended.
c. From the z-transform in Table 4.1
[
]
h(n) =Z −1 1
1 − 0.9z , |z| > 0.9 =(0.9) n u(n) □
−1
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