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[[Category:inverse z-transform]]
 
[[Category:inverse z-transform]]
  
= [[:Category:Problem_solving|Practice Question]], [[ECE438]] Fall 2013, [[User:Mboutin|Prof. Boutin]] =
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On computing the inverse z-transform of a discrete-time signal.
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'''[[Digital_signal_processing_practice_problems_list|Practice Question on "Digital Signal Processing"]]'''
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Topic: Computing an inverse z-transform
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==Question==
 
Compute the inverse z-transform of  
 
Compute the inverse z-transform of  
  
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<math> =6^{n-1}u[n-1] </math>
 
<math> =6^{n-1}u[n-1] </math>
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:<span style="color:blue"> Grader's comment: You should use partial fractions to split up into two parts </span>
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=== Answer 2===
 
=== Answer 2===
  
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<math>= \sum_{k=-\infty}^{+\infty}u[k-1](3^{k-1} - 2^{k-1})z^{-k}</math>
 
<math>= \sum_{k=-\infty}^{+\infty}u[k-1](3^{k-1} - 2^{k-1})z^{-k}</math>
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finally, by comparison with:
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<math>X(z) = \sum_{n=-\infty}^{+\infty} x[n]z^{-n}</math>
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<math>x[n] = u[n-1](3^{n-1} - 2^{n-1})</math>
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:<span style="color:blue"> Grader's comment: Correct Answer </span>
  
 
===Answer 3===
 
===Answer 3===

Latest revision as of 12:54, 26 November 2013


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.



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