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.)

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{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] $

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} $

Answer 3

Write it here.

Answer 4

Write it here.

Back to ECE438 Fall 2013 Prof. Boutin