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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 />

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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