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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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<center><font size= 4>
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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</font size>
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Topic: Computing an inverse z-transform
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</center>
 
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==Question==
 
Compute the inverse z-transform of  
 
Compute the inverse z-transform of  
  
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:<span style="color:blue"> Grader's comment: Made a mistake in the last step . It should be 2 instead of 3 </span>
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===Answer 2===
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Li-Pang Mo
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<math>X(z) =\frac{1}{(3-z)(2-z)} </math>
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<math>X(z) =\frac{-1}{3-z} + \frac{1}{2-z} </math>
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<math>X(z) =(\frac{-1}{3})(\frac{1}{1-\frac{z}{3}}) + (\frac{-1}{z})(\frac{1}{1-\frac{2}{z}}) </math>
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<math> |2|<Z<|3| </math>, which makes <math> \frac{z}{3}<1, \frac{2}{z}<1 </math>
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Use geometric series:
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<math>X(z) =\frac{-1}{3} \sum_{n=0}^{+\infty} (\frac{z}{3})^{n} + \frac{-1}{z} \sum_{n=0}^{+\infty} (\frac{2}{z})^{n} </math>
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<math>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} </math>
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<math>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} </math>
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<math>X(z) = -\sum_{n=-\infty}^{+\infty} u[n] (z)^{n} (\frac{1}{3})^{n+1} - \sum_{n=-\infty}^{+\infty} u[n] (2)^{n} (z)^{-n-1} </math>
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<math>let p = -n ,  n = -p,  q = n+1 ,  n = q-1 </math>
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<math>X(z) = -\sum_{n=-\infty}^{+\infty} u[-p] (z)^{-p} (\frac{1}{3})^{-p+1} - \sum_{n=-\infty}^{+\infty} u[q-1] (2)^{q-1} (z)^{q-1} </math>
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By observation:
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<math>x(n) = -u[-n] (\frac{1}{3})^{-n+1} - u[n-1](2)^{n-1} </math>
  
=== Answer 2===
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:<span style="color:blue"> Grader's comment: Answer is Correct </span>
  
 
===Answer 3===
 
===Answer 3===

Latest revision as of 12:55, 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 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} $


Grader's comment: Made a mistake in the last step . It should be 2 instead of 3


Answer 2

Li-Pang Mo

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

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

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

$ |2|<Z<|3| $, which makes $ \frac{z}{3}<1, \frac{2}{z}<1 $


Use geometric series:

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

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

$ let p = -n , n = -p, q = n+1 , n = q-1 $


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

By observation:

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

Grader's comment: Answer is Correct

Answer 3

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Answer 4

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