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|>3 $.

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

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Answer 1

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

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

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

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

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

Let $ n=k-1 $

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

By observing that $ X(z) =\sum_{n=-\infty}^{+\infty}x[n]z^{-n} $

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

$ =6^{n-1}u[n-1] $

Grader's comment: You should use partial fractions to split up into two parts

Answer 2

alec green

$ X(z) = \frac{1}{(3-z)(2-z)} = \frac{A}{(3-z)} + \frac{B}{(2-z)} = -\frac{1}{(3-z)} + \frac{1}{(2-z)} $

given the ROC, rewrite as:

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

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

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

letting -k = -n-1, and therefore n = k-1:

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

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

finally, by comparison with:

$ X(z) = \sum_{n=-\infty}^{+\infty} x[n]z^{-n} $

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

Grader's comment: Correct Answer

Answer 3

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Answer 4

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