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128 Chapter 4 THE z-TRANSFORM<br />

1.5<br />

Magnitude Response<br />

Magnitude<br />

1<br />

0.5<br />

0<br />

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1<br />

frequency in π units<br />

0.5<br />

Phase Response<br />

Phase in π units<br />

0<br />

−0.5<br />

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1<br />

frequency in π units<br />

FIGURE 4.10 Frequency response plots for Example 4.13<br />

4.5 SOLUTIONS OF THE DIFFERENCE EQUATIONS<br />

In Chapter 2 we mentioned two forms for the solution of linear constant<br />

coefficient difference equations. One form involved finding the particular<br />

and the homogeneous solutions, while the other form involved finding<br />

the zero-input (initial condition) and the zero-state responses. Using<br />

z-transforms, we now provide a method for obtaining these forms. In addition,<br />

we will also discuss the transient and the steady-state responses.<br />

In digital signal processing, difference equations generally evolve in the<br />

positive n direction. Therefore our time frame for these solutions will be<br />

n ≥ 0. For this purpose we define a version of the bilateral z-transform<br />

called the one-sided z-transform.<br />

DEFINITION 4<br />

The One-sided z Transform<br />

The one-sided z-transform of a sequence x(n) is given by<br />

∞∑<br />

Z + [x(n)] = △ Z [x(n)u(n)] = △ X + [z] = x(n)z −n (4.25)<br />

n=0<br />

Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).<br />

Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

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