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

By Yeong Ho Lee

First, using partial fraction we get..

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

A(2-z) + B(3-z) = 1

let z=2, then B=1

let z=3, then A=-1

$ = -\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 $

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

now let n = -k

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

by comparison with z-transfrom formula

x[n] = − 3^{n − 1}u[ − n] + 2^{n − 1}u[ − n]

x[n] = ( − 3^{n − 1} + 2^{n − 1})u[ − n]

Answer 4

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

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

Substitute k with -n

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

Look up Z transform equation on RHEA table and see that X(z) becomes...

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

Answer 5

by partical fraction, we get,

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

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

For $ \quad \text{ROC} \quad |z|<2 $

$ X(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 $

assume n=-k.

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

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

So, x[n] = (−3^n-1+2^n-1)u[-n]

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