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= [[:Category:Problem_solving|Practice Question]], [[ECE438]] Fall 2013, [[User:Mboutin|Prof. Boutin]] =
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<center><font size= 4>
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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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</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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  Therefore, <math> x[n]= -3^{n-1} u[n-1] </math>
 
  Therefore, <math> x[n]= -3^{n-1} u[n-1] </math>
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 +
:<span style="color:blue"> Grader's comment: Correct Answer </span>
  
 
=== Answer 2===
 
=== Answer 2===
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<math> x[n] = -u[n-1] 3^{n-1} </math>
 
<math> x[n] = -u[n-1] 3^{n-1} </math>
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:<span style="color:blue"> Grader's comment: Correct Answer </span>
  
 
===Answer 3===
 
===Answer 3===
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<math> x[n] = -u[n-1] 3^{n-1} </math>
 
<math> x[n] = -u[n-1] 3^{n-1} </math>
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:<span style="color:blue"> Grader's comment: Correct Answer </span>
  
 
===Answer 4===
 
===Answer 4===
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<math>X(Z) = \frac{-1}{Z} \frac{1}{1-\frac{3}{Z}}</math>
 
<math>X(Z) = \frac{-1}{Z} \frac{1}{1-\frac{3}{Z}}</math>
  
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<math>X(Z) = \frac{-1}{Z} \sum_{n=0}^{+\infty} (\frac{3}{Z})^{n}</math>  
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Since,<math>|3|<Z</math>
  
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Since,<math>|2|<Z<|3|</math>
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<math>\frac{1}{1-\frac{3}{Z}} = \sum_{n=0}^{+\infty} (\frac{3}{Z})^{n}</math>
  
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So,  
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Thus,  
  
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<math>X(Z) = \frac{-1}{Z} \sum_{n=-\infty}^{+\infty} (\frac{3}{Z})^{n}</math>
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<math>X(Z) = \frac{-1}{Z} \sum_{n=0}^{+\infty} (\frac{3}{Z})^{n}</math>
  
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<math>X(Z) = Z^{-1} \sum_{n=-\infty}^{+\infty} (\frac{3}{Z})^{n}</math>
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<math>X(Z) = -Z^{-1} \sum_{n=0}^{+\infty} (\frac{3}{Z})^{n}</math>
  
 
<math>X(Z) = -\sum_{n=-\infty}^{+\infty} u[n] 3^{n}Z^{-n-1}</math>
 
<math>X(Z) = -\sum_{n=-\infty}^{+\infty} u[n] 3^{n}Z^{-n-1}</math>
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Therefore, <math>x(n) = -u[n-1] 3^{n-1}</math>
 
Therefore, <math>x(n) = -u[n-1] 3^{n-1}</math>
  
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:<span style="color:blue"> Grader's comment: Correct Answer </span>
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=== Answer 5 ===
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By Yixiang Liu
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<math>X(z) = \frac{1}{3-Z}</math>
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<math>X(z) = \frac{1}{-z}* \frac{1} {\frac{3}{-Z}+1} </math>  :<span style="color:blue"> Grader's comment: Use x instead of convolution symbol </span>
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<math>X(z) = \frac{1}{-z}* \frac{1} {1-\frac{3}{Z}} </math>
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<math>X(z) = \frac{1}{-Z} \sum_{n=0}^{+\infty} (\frac{3}{3})^n </math>
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<math>X(z) = \sum_{n=0}^{+\infty} 3^{n} Z^{-n-1} </math>
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<math>X(z) = \sum_{-\infty}^{+\infty} u[n] 3^{n} Z^{-n-1} </math>
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Let -k = -n-1, so n = k-1
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<math>X(z) = \sum_{-\infty}^{+\infty} u[k-1] 3^{k-1} Z^{-k} </math>
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By comparison with the x-transform formula
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<math>x[n] = 3^{n-1} u[n-1] </math>
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:<span style="color:blue"> Grader's comment: Forgot the negative sign </span>
  
 
----
 
----
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Answer 6
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<math>X(z) = \frac{1}{3-z}</math>
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we have |z| > 3,
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<math>X(z) = \frac{1}{3} \frac{1}{1-\frac{z}{3}} </math>
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<math>x[n] = (\frac{1}{3})^{-n+1} u[-n] </math>
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:<span style="color:blue"> Grader's comment: Wrong approach </span>
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----
 
----
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Answer 7
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<math>X(z) = \frac{1}{3-z}</math>
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<math>X(z) = \frac{1}{\frac{3}{z} z -z}</math>
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<math>X(z) = \frac{1}{z} \frac{1}{\frac{3}{z} - 1}</math>
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<math>X(z) = \frac{-1}{z} \frac{1}{1-\frac{3}{z}}</math>
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<math>X(z) = \frac{-1}{z} \frac{1-\left( \frac{3}{z} \right) ^{n}}{1-\frac{3}{z}}, \quad \text{As n goes to} \quad \infty \text{  since ROC} \quad |z|>3</math>
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<math>X(z) = \frac{-1}{z} \sum_{n=0}^{\infty} \left( \frac{3}{z} \right)^{n}</math>
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<math>X(z) = \frac{-1}{z} \sum_{n=0}^{\infty} 3^n z^{-n}</math>
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<math>X(z) = -\sum_{n=0}^{\infty} 3^n z^{-\left(n+1 \right)}, \quad \text{Replace n+1 with k}</math>
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<math>X(z) = -\sum_{k=1}^{\infty} 3^{k-1} z^{-k}</math>
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<math>X(z) = -\sum_{-\infty}^{\infty} 3^{k-1} u[k-1] z^{-k}, \quad \text{By comparing with DTFT equation we get}</math>
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<math>x[n] = -3^{\left( n-1 \right) } u[n-1]</math>
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:<span style="color:blue"> Grader's comment: Correct Answer </span>
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=== Answer 5 ===
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<math>X(z) = \frac{1}{\frac{3z}{z}-z}</math>
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<math> = -(\frac{1}{z})(\frac{1}{1-\frac{3}{z}})</math>
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Using the geometric series formula,
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<math> = -(\frac{1}{z})\sum_{n=0}^{\infty}(\frac{3}{z})^n</math>
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<math> = -(\frac{1}{z})\sum_{n=0}^{\infty}3^nz^{-n}</math>
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<math> = \sum_{n=-\infty}^{\infty}-u[n]3^nz^{-n-1}</math>
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Let k = n+1,
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<math> = \sum_{k=-\infty}^{\infty}-u[k-1]3^{k-1}z^{-k}</math>
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By comparison with the Z-transform formula,
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x[n] = -u[n-1]3<sup>n-1</sup>
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:<span style="color:blue"> Grader's comment: Correct Answer </span>
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=== Answer 9 ===
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<math>X(z)=\frac{1}{3-Z}</math>
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<math> = \frac{-1}{z}*\frac{1}{1-\frac{3}{z}}</math>
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<math> = \frac{-1}{z}\sum_{n=0}^{\infty}(\frac{3}{z})^n </math>
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<math> = \sum_{n=0}^{\infty}(-z^{-1})(3z^{-1})^n </math>
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<math> = \sum_{n=0}^{\infty}-3^n z^{-n-1}</math>
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<math> = \sum_{n=-\infty}^{\infty}-3^n u[n] z^{-n-1}</math>
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substituting n+1 with k:
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<math> = \sum_{n=-\infty}^{\infty}-u[k-1]3^{k-1}z^{-k}</math>
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Comparing to common Z-transform tables:
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x[n] = (−3^(n−1))u[n−1]
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:<span style="color:blue"> Grader's comment: Correct Answer </span>
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--------------------------------------------------------------------------------------
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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]]

Latest revision as of 11:54, 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}, \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{-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})  $ 
$  = \sum_{n=0}^{+\infty} -3^n z^{-n-1} = \sum_{n=-\infty}^{+\infty} -3^n u[n] z^{-n-1}  $
NOTE: Let k=n+1

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

Therefore, $  x[n]= -3^{n-1} u[n-1]  $
Grader's comment: Correct Answer

Answer 2

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

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

Let k = n+1, so n = k-1

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

By comparison with the z-transform equation

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

Grader's comment: Correct Answer

Answer 3

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

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

Let k = n+1 then n = k-1

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

By comparison,

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

Grader's comment: Correct Answer

Answer 4

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

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

Since,$ |3|<Z $

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

Thus,

$ X(Z) = \frac{-1}{Z} \sum_{n=0}^{+\infty} (\frac{3}{Z})^{n} $

$ X(Z) = -Z^{-1} \sum_{n=0}^{+\infty} (\frac{3}{Z})^{n} $

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

Let k=n+1, then -k=-n-1,n=k-1

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

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

Grader's comment: Correct Answer

Answer 5

By Yixiang Liu

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


$ X(z) = \frac{1}{-z}* \frac{1} {\frac{3}{-Z}+1} $  : Grader's comment: Use x instead of convolution symbol

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

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

$ X(z) = \sum_{n=0}^{+\infty} 3^{n} Z^{-n-1} $

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

Let -k = -n-1, so n = k-1

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

By comparison with the x-transform formula

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

Grader's comment: Forgot the negative sign

Answer 6

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

we have |z| > 3,

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


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

Grader's comment: Wrong approach


Answer 7


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

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

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

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

$ X(z) = \frac{-1}{z} \frac{1-\left( \frac{3}{z} \right) ^{n}}{1-\frac{3}{z}}, \quad \text{As n goes to} \quad \infty \text{ since ROC} \quad |z|>3 $

$ X(z) = \frac{-1}{z} \sum_{n=0}^{\infty} \left( \frac{3}{z} \right)^{n} $

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

$ X(z) = -\sum_{n=0}^{\infty} 3^n z^{-\left(n+1 \right)}, \quad \text{Replace n+1 with k} $

$ X(z) = -\sum_{k=1}^{\infty} 3^{k-1} z^{-k} $

$ X(z) = -\sum_{-\infty}^{\infty} 3^{k-1} u[k-1] z^{-k}, \quad \text{By comparing with DTFT equation we get} $

$ x[n] = -3^{\left( n-1 \right) } u[n-1] $

Grader's comment: Correct Answer

Answer 5

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

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

Using the geometric series formula,

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

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

$ = \sum_{n=-\infty}^{\infty}-u[n]3^nz^{-n-1} $

Let k = n+1,

$ = \sum_{k=-\infty}^{\infty}-u[k-1]3^{k-1}z^{-k} $

By comparison with the Z-transform formula,

x[n] = -u[n-1]3n-1

Grader's comment: Correct Answer


Answer 9

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

$ = \sum_{n=0}^{\infty}-3^n z^{-n-1} $

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

substituting n+1 with k:

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

Comparing to common Z-transform tables:

x[n] = (−3^(n−1))u[n−1]

Grader's comment: Correct Answer


Back to ECE438 Fall 2013 Prof. Boutin

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