Practice Question, ECE438 Fall 2013, Prof. Boutin

On computing the inverse z-transform of a discrete-time signal.

Compute the inverse z-transform of

$ X(z) =\frac{1}{(3-z)(2-z)}, \quad \text{ROC} \quad |z|<2 $.

(Write enough intermediate steps to fully justify your answer.)

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Answer 1

$ X(z) = \frac{A}{3-z}+\frac{B}{2-z} $

$ = -\frac{1}{3-z}-\frac{1}{2-z} $

$ = -\frac{1}{3}(\frac{1}{1-\frac{z}{3}})-\frac{1}{2}(\frac{1}{1-\frac{z}{2}}) $

$ = -\frac{1}{3}\sum_{n=0}^{+\infty} (\frac{z}{3})^n -\frac{1}{2}\sum_{n=0}^{+\infty} (\frac{z}{2})^n $

$ = \sum_{n=0}^{+\infty}[(-\frac{1}{3}) (\frac{1}{3})^n + (-\frac{1}{2})(\frac{1}{2})^n]z^n $

Let k=-n

$ = \sum_{k=-\infty}^{+\infty}u[-k][(-\frac{1}{3})3^k + (-\frac{1}{2})2^k]z^{-k} $

by comparison with z-transform formula

$ x[n]=u[-n](-3^{n-1}-2^{n-1}) $

Answer 2

Kyungjun Kim Using a partial fraction expansion, we can change the original equation to $ X(z) = \frac{A}{3-z}+\frac{B}{2-z} $ Where A = 1, B = -1, so we get $ = -\frac{1}{3-z}-\frac{1}{2-z} $ By factoring out 1/3 for the first term, and 1/2 for the second term, we can have both terms in form of $ \frac{1}{1-r} $, which is equal to $ sum_{n=0}^{+\infty} (\frac{1}{r})^n $ $ = -\frac{1}{3}(\frac{1}{1-\frac{z}{3}})-\frac{1}{2}(\frac{1}{1-\frac{z}{2}}) $ $ = -\frac{1}{3}\sum_{n=0}^{+\infty} (\frac{z}{3})^n -\frac{1}{2}\sum_{n=0}^{+\infty} (\frac{z}{2})^n $ $ = \sum_{n=0}^{+\infty}[(-\frac{1}{3}) (\frac{1}{3})^n + (-\frac{1}{2})(\frac{1}{2})^n]z^n $ Then let k=-n $ = \sum_{k=-\infty}^{+\infty}u[-k][(-\frac{1}{3})3^k + (-\frac{1}{2})2^k]z^{-k} $ Comparing it with z-transform formula, we can get

$ x[n]=u[-n](-3^{n-1}-2^{n-1}) $

Answer 3

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Answer 4

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