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===Answer 5===
 
===Answer 5===
 
<math>X(z) = \frac{1}{3-Z}</math>
 
<math>X(z) = \frac{1}{3-Z}</math>
<math>X(z) = \[[\frac{1}{3}]/[1-[\frac{Z}{3}]]]</math>
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 +
 
 +
<math>X(z) = \frac{\frac{1}{3}} { 1-\frac{Z}{3}} </math>
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 +
<math>X(z) = \frac{1}{3} * \frac{1} { 1-\frac{Z}{3}} </math>
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[[2013_Fall_ECE_438_Boutin|Back to ECE438 Fall 2013 Prof. Boutin]]
 
[[2013_Fall_ECE_438_Boutin|Back to ECE438 Fall 2013 Prof. Boutin]]

Revision as of 15:02, 19 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.)


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Answer 1

$ X[z] = \sum_{n=0}^{+\infty} 3^{-1-n} z^{n} $

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

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

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

Let n = -k

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

By comparison with the x-transform formula,

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

Answer 3

Write it here.

Answer 4

Write it here.



Answer 5

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


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

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




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