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76 Chapter 3 THE DISCRETE-TIME FOURIER ANALYSIS
Hence
√
|H(e jω 1
)| =
(1 − 0.9 cos ω) 2 +(0.9 sin ω) = 1
√ 2 1.81 − 1.8 cos ω
and
[
]
H(e jω 0.9 sin ω
)=− arctan
1 − 0.9 cos ω
To plot these responses, we can either implement the |H(e jω )| and ̸ H(e jω )
functions or the frequency response H(e jω ) and then compute its magnitude
and phase. The latter approach is more useful from a practical viewpoint [as
shown in (3.18)].
>> w = [0:1:500]*pi/500; % [0, pi] axis divided into 501 points.
>> H = exp(j*w) ./ (exp(j*w) - 0.9*ones(1,501));
>> magH = abs(H); angH = angle(H);
>> subplot(2,1,1); plot(w/pi,magH); grid;
>> xlabel(’frequency in pi units’); ylabel(’|H|’);
>> title(’Magnitude Response’);
>> subplot(2,1,2); plot(w/pi,angH/pi); grid
>> xlabel(’frequency in pi units’); ylabel(’Phase in pi Radians’);
>> title(’Phase Response’);
The plots are shown in Figure 3.7.
□
□ EXAMPLE 3.14 Let an input to the system in Example 3.13 be 0.1u(n). Determine the steadystate
response y ss(n).
Solution
Since the input is not absolutely summable, the discrete-time Fourier transform
is not particularly useful in computing the complete response. However, it can
be used to compute the steady-state response. In the steady state (i.e., n →∞),
the input is a constant sequence (or a sinusoid with ω 0 = θ 0 = 0). Then the
output is
y ss(n) =0.1 × H(e j0 )=0.1 × 10 = 1
where the gain of the system at ω =0(also called the DC gain) is H(e j0 )=10,
which is obtained from Figure 3.7.
□
3.3.4 FREQUENCY RESPONSE FUNCTION FROM DIFFERENCE EQUA-
TIONS
When an LTI system is represented by the difference equation
y(n)+
N∑
a l y(n − l) =
l=1
M∑
b m x(n − m) (3.20)
m=0
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