Revision as of 21:10, 19 September 2013

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.)

Share your answers below

You will receive feedback from your instructor and TA directly on this page. Other students are welcome to comment/discuss/point out mistakes/ask questions too!

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] $

Back to ECE438 Fall 2013 Prof. Boutin