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78 Chapter 3 THE DISCRETE-TIME FOURIER ANALYSIS
Solution
Rewrite the difference equation as y(n) − 0.8y(n − 1) = x(n).
a. Using (3.21), we obtain
H(e jω )=
1
1 − 0.8e −jω (3.22)
b. In the steady state the input is x(n) =cos(0.05πn) with frequency ω 0 =
0.05π and θ 0 =0 ◦ . The response of the system is
Therefore
H(e j0.05π )=
1
=4.0928e−j0.5377
1 − 0.8e−j0.05π y ss(n) =4.0928 cos(0.05πn − 0.5377) = 4.0928 cos [0.05π(n − 3.42)]
This means that at the output the sinusoid is scaled by 4.0928 and shifted
by 3.42 samples. This can be verified using MATLAB.
>> subplot(1,1,1)
>> b = 1; a = [1,-0.8];
>> n=[0:100];x = cos(0.05*pi*n);
>> y = filter(b,a,x);
>> subplot(2,1,1); stem(n,x);
>> xlabel(’n’); ylabel(’x(n)’); title(’Input sequence’)
>> subplot(2,1,2); stem(n,y);
>> xlabel(’n’); ylabel(’y(n)’); title(’Output sequence’)
From the plots in Figure 3.8, we note that the amplitude of y ss(n) isapproximately
4. To determine the shift in the output sinusoid, we can compare
zero crossings of the input and the output. This is shown in Figure 3.8, from
which the shift is approximately 3.4 samples.
□
In Example 3.15 the system was characterized by a 1st-order difference
equation. It is fairly straightforward to implement (3.22) in
MATLAB as we did in Example 3.13. In practice the difference equations
are of large order and hence we need a compact procedure to implement
the general expression (3.21). This can be done using a simple matrixvector
multiplication. If we evaluate H(e jω )atk =0, 1,...,K equispaced
frequencies over [0,π], then
∑ M
H(e jω k m=0
)=
b m e −jω km
1+ ∑ N
l=1 a , k =0, 1,...,K (3.23)
l e −jω kl
If we let {b m }, {a l } (with a 0 = 1), {m =0,...,M}, {l =0,...,N}, and
{ω k } be arrays (or row vectors), then the numerator and the denominator
of (3.23) become
b exp(−jm T ω); a exp(−jl T ω)
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