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 />
X ( z)<br />
=<br />
∞<br />
∑<br />
n=<br />
0<br />
x[<br />
n]<br />
z<br />
= 1+<br />
0.8z<br />
1<br />
=<br />
1−<br />
0.8z<br />
−1<br />
−1<br />
−n<br />
+<br />
= 1+<br />
0.8z<br />
−1<br />
+ 0.8<br />
−1<br />
2<br />
−1<br />
( 0.8z<br />
) + ( 0.8z<br />
)<br />
z<br />
=<br />
z − 0.8<br />
2<br />
z<br />
−2<br />
3<br />
3<br />
+ 0.8 z<br />
+ ...<br />
−3<br />
+ ...<br />
1<br />
b) Reconstruct the signal corresponding <strong>to</strong> the z-transform: X ( z)<br />
= z + 1.2<br />
−1<br />
( = 1 z<br />
−1⎛<br />
1 ⎞<br />
X z)<br />
= = z<br />
⎜<br />
⎟<br />
−1<br />
z + 1.2 1 + 1.2z<br />
⎝1<br />
− (1.2z<br />
−1<br />
) ⎠<br />
Using the series summation formula in Eq.(3.13) (where 1.2z −1 < 1 is required.),<br />
X(<br />
z)<br />
= z<br />
= z<br />
−1<br />
−1<br />
−1<br />
−1<br />
2<br />
−1<br />
3<br />
{ 1+<br />
( −1.2<br />
z ) + ( −1.2<br />
z ) + ( −1.2<br />
z ) + ...}<br />
−1.2z<br />
−2<br />
_1.44z<br />
−3<br />
−1.728z<br />
−4<br />
+ ...<br />
Therefore, the reconstructed original signal can be obtained as<br />
x[n]=[0 1 -1.2 1.44 -1.728 ... ].<br />
↑<br />
The signal is shown in Figure 5.1(b).<br />
(a)<br />
1<br />
x[n]<br />
0.5<br />
0<br />
-2 -1 0 1 2 3 4 5 6 7 8 9<br />
n<br />
(b)<br />
5<br />
x[n]<br />
0<br />
-5<br />
-2 -1 0 1 2 3 4 5 6 7 8 9<br />
n<br />
Figure 5.1 <strong>Signal</strong>s in the examples<br />
61<br />
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