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===Answer 1===
 
===Answer 1===
 
<math> X(z) =\frac{1}{\frac{3z}{z} - z} = \frac{-1}{z} \frac{1}{1-\frac{3}{z}} </math>
 
<math> X(z) =\frac{1}{\frac{3z}{z} - z} = \frac{-1}{z} \frac{1}{1-\frac{3}{z}} </math>
  <math>      =\frac{-1}{z} \sum_{n=0}^{+\infty} (\frac{3}{z})^n </math>
+
  <math>      =\frac{-1}{z} \sum_{n=0}^{+\infty} (\frac{3}{z})^n = \sum_{n=0}^{+\infty} (-z^{-1}) (3z^{-1})^n </math>
<math>      =\sum_{n=0}^{+\infty} (-z^{-1}) (3z^{-1})^n </math>
+
 
  NOTE: <math> (3z^{-1})^n = (3^n) (z^{-n}) (-z^{-1}) </math>  
 
  NOTE: <math> (3z^{-1})^n = (3^n) (z^{-n}) (-z^{-1}) </math>  
  

Revision as of 14:01, 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.)


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Answer 1

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

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


Answer 2

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Answer 3

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Answer 4

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