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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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