Introduction to Digital Signal and System Analysis - Tutorsindia

Introduction to Digital Signal and System Analysis
Frequency Domain Analysis
h[n]
0.4
0.3
0.2
0.1
0
-6 -4 -2 0 2 4 6
n
1.5
|H(W)|
1
0.5
0
0 1 2 3 4 5 6
2
-p/2 ≤ Φ ≤ p/2
1
0
-1
-2
0 1 2 3 4 5 6
0 ≤ W ≤ 2p
Figure 4.5 Impulse response and frequency response in Example 4.6 (a)
b) h[n]= 0.5d[n]+ 0.25d[n-1]+ 0.125d[n-2]+...
H(
Ω)
=
∞
n=−∞
= 0.5
h[
n]exp
∞
n=
0
( − jΩn)
{ } n
0.5exp( − jΩ
= 0.5 + 0.25exp( − jΩ)
+ 0.125exp( − j2Ω)
+ ...
i.e.
0.5
H ( W)
=
.
1 − 0.5exp( − jW)
The modulus and phase are shown in Figure 4.6.
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