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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}, \quad \text{ROC} \quad |z|<3 $.

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

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Answer 1

$ X[z] = \sum_{n=0}^{+\infty} 3^{-1-n} z^{n} $

$       = \sum_{n=-\infty}^{+\infty} u[n] 3^{-1-n} z^{n} $

NOTE: Let n=-k

$ = \sum_{n=-\infty}^{+\infty} u[-k] 3^{-1+k} z^{-k} $ (compare with $ \sum_{n=-\infty}^{+\infty} x[n] z^{-k} $)

$ = \sum_{n=-\infty}^{+\infty} u[-k] 3^{-1+k} z^{-k} $

Therefore, $ x[n]= 3^{-1+n} u[-n] $

Answer 2

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

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

Let n = -k

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

By comparison with the x-transform formula,

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

Answer 3

Write it here.

Answer 4

Write it here.

Answer 5

By Yixiang Liu

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

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

$ X(z) = \frac{1}{3} * \frac{1} { 1-\frac{Z}{3}} $

by geometric series

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

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

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

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

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

$ x[n] = (\frac{1}{3})^{-n+1} u[-n] $

$ x[n] = \3^{n-1} u[-n] $

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