Practice Question on "Digital Signal Processing"

Topic: Computing an inverse z-transform


Question

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] $

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

Write it here.

Answer 4

Write it here.



Back to ECE438 Fall 2013 Prof. Boutin

Alumni Liaison

To all math majors: "Mathematics is a wonderfully rich subject."

Dr. Paul Garrett