Introduction to Digital Signal and System Analysis - Tutorsindia
Introduction to Digital Signal and System Analysis - Tutorsindia
Introduction to Digital Signal and System Analysis - Tutorsindia
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<strong>Introduction</strong> <strong>to</strong> <strong>Digital</strong> <strong>Signal</strong> <strong>and</strong> <strong>System</strong> <strong>Analysis</strong><br />
Z Domain <strong>Analysis</strong><br />
Low pass 0 < α < 1<br />
High pass -1 < α < 0<br />
1<br />
1<br />
0<br />
0<br />
-1<br />
0 5 10<br />
n<br />
-1<br />
0 5 10<br />
n<br />
5<br />
5<br />
|H(W)|<br />
-pi/2 ≤ phase ≤ p/2<br />
|H(W)|<br />
0<br />
0 1 2 3<br />
0 ≤ W ≤ 2p<br />
1<br />
0<br />
-1<br />
0 1 2 3<br />
0 ≤ W ≤ 2p<br />
-pi/2 ≤ phase ≤ p/2<br />
0<br />
0 1 2 3<br />
0 ≤ W ≤ 2p<br />
1<br />
0<br />
-1<br />
0 1 2 3<br />
0 ≤ W ≤ 2p<br />
1.5<br />
z-plane<br />
1.5<br />
z-plane<br />
1<br />
1<br />
0.5<br />
0.5<br />
0<br />
-0.5<br />
W=p<br />
α<br />
W=0<br />
0<br />
-0.5<br />
W=p<br />
α<br />
W=0<br />
-1<br />
-1<br />
-1.5<br />
-1 0 1<br />
Low pass<br />
-1.5<br />
-1 0 1<br />
High pass<br />
Figure 5.9 The frequency response, poles <strong>and</strong> zeros of low <strong>and</strong> high pass filters<br />
For a second-order system, the transfer function in polar form is<br />
H<br />
2<br />
( z)<br />
=<br />
z<br />
2<br />
2<br />
2<br />
( z − r exp( jq<br />
)( z − r exp( − jq<br />
)) z − 2rz<br />
cosq<br />
+ r<br />
=<br />
z<br />
2<br />
Its frequency response is<br />
79<br />
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