Line 20: Line 20:
 
----
 
----
 
===Answer 1===
 
===Answer 1===
Write it here.
+
<math>X(z) = \sum_{n=-\0}^{+\infty} 3^{-1-n} z^{n}</math>
 +
<math>    = \sum_{n=-\infty}^{+\infty} u[n] 3^{-1-n} z^{n}</math>
 +
Let n=-k
 +
<math>    = \sum_{n=-\infty}^{+\infty} u[-k] 3^{-1+k} z^{-k}</math>
 +
compare with <math>\sum_{n=-\infty}^{+\infty} x[n] z^{-k}</math>
 +
<math>    = \sum_{n=-\infty}^{+\infty} u[-k] 3^{-1+k} z^{-k}</math>
 +
 
 +
Therefore, <math> x[n]= 3^{-1+n} u[-n] </math>
 +
 
 
=== Answer 2===
 
=== Answer 2===
 
Write it here.
 
Write it here.

Revision as of 14:32, 18 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}, \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) = \sum_{n=-\0}^{+\infty} 3^{-1-n} z^{n} $ $ = \sum_{n=-\infty}^{+\infty} u[n] 3^{-1-n} z^{n} $ Let n=-k $ = \sum_{n=-\infty}^{+\infty} u[-k] 3^{-1+k} z^{-k} $ compare with $ \sum_{n=-\infty}^{+\infty} x[n] z^{-k} $ $ = \sum_{n=-\infty}^{+\infty} u[-k] 3^{-1+k} z^{-k} $

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

Answer 2

Write it here.

Answer 3

Write it here.

Answer 4

Write it here.



Back to ECE438 Fall 2013 Prof. Boutin

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood