Revision as of 17:32, 19 September 2013 by Chen653 (Talk | contribs)


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 2<|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

Ruofei

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

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

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

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

Since $ |2|<Z<|3| $

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

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

Thus,

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

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

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

In $ -\frac{1}{3} \sum_{n=-\infty}^{+\infty} u[n] (\frac{Z}{3})^{n} $, Let k=-n, then -k=n

In $ \frac{-1}{Z} \sum_{n=-\infty}^{+\infty} u[n] (\frac{2}{Z})^{n} $, Let i=n+1, then n=i-1

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

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

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


Answer 2

Answer 3

Write it here.

Answer 4

Write it here.



Back to ECE438 Fall 2013 Prof. Boutin

Alumni Liaison

Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale.

Dr. Paul Garrett