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[[Category:ECE301]]
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[[Category:ECE438]]
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[[Category:ECE438Fall2013Boutin]]
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[[Category:problem solving]]
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[[Category:z-transform]]
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[[Category:inverse z-transform]]
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= [[: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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= [[: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.
 
 
----
 
----
 +
 
Compute the inverse z-transform of  
 
Compute the inverse z-transform of  
  
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(Write enough intermediate steps to fully justify your answer.)  
 
(Write enough intermediate steps to fully justify your answer.)  
 +
 
----
 
----
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==Share your answers below==
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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!
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== 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!  
 +
 
 
----
 
----
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===Answer 1===
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 +
=== Answer 1 ===
  
 
<math>X[z] = \sum_{n=0}^{+\infty} 3^{-1-n} z^{n}</math>  
 
<math>X[z] = \sum_{n=0}^{+\infty} 3^{-1-n} z^{n}</math>  
 +
 
  <math>      = \sum_{n=-\infty}^{+\infty} u[n] 3^{-1-n} z^{n}</math>
 
  <math>      = \sum_{n=-\infty}^{+\infty} u[n] 3^{-1-n} z^{n}</math>
  
 
  NOTE: Let n=-k
 
  NOTE: Let n=-k
  
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<math>    = \sum_{n=-\infty}^{+\infty} u[-k] 3^{-1+k} z^{-k}</math>  
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<math>    = \sum_{n=-\infty}^{+\infty} u[-k] 3^{-1+k} z^{-k}</math> ('''compare with''' <math>\sum_{n=-\infty}^{+\infty} x[n] z^{-k}</math>)  
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('''compare with''' <math>\sum_{n=-\infty}^{+\infty} x[n] z^{-k}</math>)  
+
  
 
<math>    = \sum_{n=-\infty}^{+\infty} u[-k] 3^{-1+k} z^{-k}</math>  
 
<math>    = \sum_{n=-\infty}^{+\infty} u[-k] 3^{-1+k} z^{-k}</math>  
  
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Therefore, <math> x[n]= 3^{-1+n} u[-n] </math>
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Therefore, <span class="texhtml">''x''[''n''] = 3<sup> − 1 + ''n''</sup>''u''[ − ''n'']</span>  
  
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=== Answer 2===
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=== Answer 2 ===
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<math>X(z) = \frac{1}{3-z} = \frac{1}{3} \frac{1}{1-\frac{z}{3}} </math>
+
  
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<math>X(z) = \frac{1}{3} \sum_{n=-\infty}^{+\infty} u[n] (\frac{z}{3})^n </math>
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<math>X(z) = \frac{1}{3-z} = \frac{1}{3} \frac{1}{1-\frac{z}{3}} </math>  
  
−
Let n = -k
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<math>X(z) = \frac{1}{3} \sum_{n=-\infty}^{+\infty} u[n] (\frac{z}{3})^n </math>
  
−
<math>X(z) = \frac{1}{3} \sum_{n=-\infty}^{+\infty} u[-k]3^{k}z^{-k} </math>
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Let n = -k  
  
−
By comparison with the x-transform formula,
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<math>X(z) = \frac{1}{3} \sum_{n=-\infty}^{+\infty} u[-k]3^{k}z^{-k} </math>
  
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<math>x[n] = 3^{n-1} u[-n] </math>
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By comparison with the x-transform formula,
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===Answer 3===
+
 
−
By Yeong Ho Lee
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<span class="texhtml">''x''[''n''] = 3<sup>''n'' − 1</sup>''u''[ − ''n'']</span>  
 +
 
 +
=== Answer 3 ===
 +
 
 +
By Yeong Ho Lee  
  
 
<math> X[z] = \sum_{n=0}^{\infty}z^{n}3^{-n-1} </math>  
 
<math> X[z] = \sum_{n=0}^{\infty}z^{n}3^{-n-1} </math>  
  
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<math>      = \sum_{n=-\infty}^{+\infty} 3^{-n-1}z^{n}u[n] </math>
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<math>      = \sum_{n=-\infty}^{+\infty} 3^{-n-1}z^{n}u[n] </math>  
  
 
  Now, let n = -k
 
  Now, let n = -k
  
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<math>      = \sum_{n=-\infty}^{+\infty} 3^{k-1}z^{-k}u[-k] </math>
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<math>      = \sum_{n=-\infty}^{+\infty} 3^{k-1}z^{-k}u[-k] </math>  
  
−
Using the z-transform formula,  
+
Using the z-transform formula, <span class="texhtml">''x''[''n''] = 3<sup>''n'' − 1</sup>''u''[ − ''n'']</span>  
−
<math> x[n]= 3^{n-1}u[-n] </math>
+
  
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===Answer 4===
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=== Answer 4 ===
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Write it here.
+
  
 +
Write it here.
  
 +
<br>
  
 +
<br>
  
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===Answer 5===
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=== Answer 5 ===
  
−
By Yixiang Liu
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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}{3-Z}</math>  
  
 +
<br> <math>X(z) = \frac{\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>  
  
−
<math>X(z) = \frac{1}{3} * \frac{1} { 1-\frac{Z}{3}} </math>
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by geometric series
  
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by geometric series
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<math>X(z) = \frac{1}{3} \sum_{n=0}^{+\infty} (\frac{z}{3})^n </math>
  
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<math>X(z) = \frac{1}{3} \sum_{n=0}^{+\infty} (\frac{z}{3})^n </math>
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<math>X(z) = \frac{1}{3} \sum_{n=-\infty}^{+\infty} u[n](\frac{z}{3})^n </math>  
  
−
<math>X(z) = \frac{1}{3} \sum_{n=-\infty}^{+\infty} u[n](\frac{z}{3})^n </math>
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<math>X(z) = \frac{1}{3} \sum_{k=-\infty}^{+\infty} u[-k](\frac{z}{3})^{-k} </math>  
  
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<math>X(z) = \frac{1}{3} \sum_{k=-\infty}^{+\infty} u[-k](\frac{z}{3})^{-k} </math>
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<math>X(z) = \frac{1}{3} \sum_{k=-\infty}^{+\infty} u[-k](\frac{1}{3})^{-k} Z^{-k}</math>  
  
−
<math>X(z) = \frac{1}{3} \sum_{k=-\infty}^{+\infty} u[-k](\frac{1}{3})^{-k} Z^{-k}</math>
+
By comparison with the x-transform formula
  
−
By comparison with the x-transform formula
+
<math>x[n] = \frac{1}{3} u[-n](\frac{1}{3})^{-n} </math>
  
−
<math>x[n] = \frac{1}{3} u[-n](\frac{1}{3})^{-n} </math>
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<math>x[n] = (\frac{1}{3})^{-n+1} u[-n] </math>  
  
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<math>x[n] = (\frac{1}{3})^{-n+1} u[-n] </math>
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<span class="texhtml">''x''[''n''] = 3<sup>''n'' − 1</sup>''u''[ − ''n'']</span>  
  
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<math>x[n] = 3^{n-1} u[-n] </math>
 
 
----
 
----
 +
 +
Answer 6 - Ryan Atwell
 +
 +
<math>X(z) =\frac{1}{3-z}, \quad \text{ROC} \quad |z|<3 </math>
 +
 +
<math>X(z) = \frac{1}{3} \frac{1}{1-\frac{z}{3}} </math>
 +
 +
<math>X(z) = \frac{1}{3} \sum_{n=0}^{\infty} (\frac{z}{3})^{n} </math>
 +
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<math>X(z) = \frac{1}{3} \sum_{n=-\infty}^{\infty} u[n](\frac{z}{3})^{n}</math>
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 +
n=-k
 +
 +
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<math>X(z) = \frac{1}{3} \sum_{k=-\infty}^{\infty} u[-k](\frac{z}{3})^{-k}</math>
 +
 +
<math>X(z) = \frac{1}{3} \sum_{k=-\infty}^{\infty} u[-k]{3}^{k}{z}^{-k}</math>
 +
 +
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<math>X(z) = \sum_{k=-\infty}^{\infty} u[-k]{3}^{k-1}{z}^{-k}</math>
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 +
by formula
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<math>X[n]={3}^{n-1}u[-n]</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]]
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</math>
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[[Category:ECE301]] [[Category:ECE438]] [[Category:ECE438Fall2013Boutin]] [[Category:Problem_solving]] [[Category:Z-transform]] [[Category:Inverse_z-transform]]

Revision as of 16:17, 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.)

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] = \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 + nu[ − 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] = 3n − 1u[ − n]

Answer 3

By Yeong Ho Lee

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

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

Now, let n = -k

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

Using the z-transform formula, x[n] = 3n − 1u[ − n]

Answer 4

Write it here.



Answer 5

By Yixiang Liu

$ 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}} $

by geometric series

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

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

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

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

By comparison with the x-transform formula

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

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

x[n] = 3n − 1u[ − n]

Answer 6 - Ryan Atwell

$ X(z) =\frac{1}{3-z}, \quad \text{ROC} \quad |z|<3 $

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

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

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

n=-k


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

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


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

by formula

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

Back to ECE438 Fall 2013 Prof. Boutin </math>

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