(New page: Category:ECE301Spring2011Boutin Category:Problem_solving ---- = Practice Question on Computing the inverse z-transform = Compute the inverse z-transform of the following signal. ...)
 
 
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[[Category:ECE301Spring2011Boutin]] [[Category:Problem_solving]]
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<center><font size= 4>
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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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</center>
 
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= Practice Question on Computing the inverse z-transform  =
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==Question==
  
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Compute the inverse z-transform of the following signal.
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Compute the inverse z-transform of the following signal.  
  
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<math>X(z)=\frac{1}{1+3z} \mbox{, } \Big|z\Big|>\frac{1}{3}</math>
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<math>X(z)=\frac{1}{1+3z} \mbox{, } \Big|z\Big|>\frac{1}{3}</math>  
  
 
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== Share your answers below  ==
 
== Share your answers below  ==
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Prof. Mimi gave me this problem in class on Friday, so I'm posting it and my answer here.
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--[[User:Cmcmican|Cmcmican]] 22:38, 16 April 2011 (UTC)
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Prof. Mimi gave me this problem in class on Friday, so I'm posting it and my answer here. --[[User:Cmcmican|Cmcmican]] 22:38, 16 April 2011 (UTC)  
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=== Answer 1  ===
 
=== Answer 1  ===
  
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<math>X(z)=\frac{1}{3z}\frac{1}{(1+\frac{1}{3z})}</math>
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<math>X(z)=\frac{1}{3z}\frac{1}{(1+\frac{1}{3z})}</math>  
  
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since <math class="inline">\Bigg|z\Bigg|>\frac{1}{3} ==>\Bigg|\frac{1}{z}\Bigg|<3 ==>\Bigg|\frac{1}{3}\frac{1}{z}\Bigg|<1</math>
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since <math>\Bigg|z\Bigg|>\frac{1}{3} ==>\Bigg|\frac{1}{z}\Bigg|<3 ==>\Bigg|\frac{1}{3}\frac{1}{z}\Bigg|<1</math>  
  
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<math>X(z)=\sum_{k=0}^\infty \frac{1}{3z}\Bigg(\frac{1}{3z}\Bigg)^k=\sum_{k=-\infty}^\infty u[k] \Bigg(\frac{1}{3}\Bigg)^{k+1}z^{-(k+1)}</math>
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<math>X(z)=\sum_{k=0}^\infty \frac{1}{3z}\Bigg(\frac{1}{3z}\Bigg)^k=\sum_{k=-\infty}^\infty u[k] \Bigg(\frac{1}{3}\Bigg)^{k+1}z^{-(k+1)}</math>  
  
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let n=k+1
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let n=k+1  
  
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<math>=\sum_{n=-\infty}^\infty u[n-1]\Bigg(\frac{1}{3}\Bigg)^{n}z^{-n}</math>
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<math>=\sum_{n=-\infty}^\infty u[n-1]\Bigg(\frac{1}{3}\Bigg)^{n}z^{-n}</math>  
  
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By comparison with <math class="inline">\sum_{n=-\infty}^\infty x[n] z^{-n}:</math>
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By comparison with <math>\sum_{n=-\infty}^\infty x[n] z^{-n}:</math>  
  
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<math>x[n]=\Bigg(\frac{1}{3}\Bigg)^{n}u[n-1]\,</math>
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<math>x[n]=\Bigg(\frac{1}{3}\Bigg)^{n}u[n-1]\,</math>  
  
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--[[User:Cmcmican|Cmcmican]] 22:38, 16 April 2011 (UTC)
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--[[User:Cmcmican|Cmcmican]] 22:38, 16 April 2011 (UTC)  
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:TA's comment: I think you have a mistake here. You can check that by taking the Z-transform of your answer.
  
 
=== Answer 2  ===
 
=== Answer 2  ===
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Write it here.
 
  
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<math>X(z)=\frac{1}{3z}\frac{1}{(1+\frac{1}{3z})}=\frac{1}{3z}\frac{1}{(1-(-\frac{1}{3z}))}</math>
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since <math>\Bigg|z\Bigg|>\frac{1}{3} ==>\Bigg|\frac{1}{z}\Bigg|<3 ==>\Bigg|\frac{1}{3}\frac{1}{z}\Bigg|<1</math>
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<math>X(z)=\sum_{k=0}^\infty \frac{1}{3z}\Bigg(\frac{-1}{3z}\Bigg)^k=\sum_{k=-\infty}^\infty u[k] (-1)^{k}\Bigg(\frac{1}{3}\Bigg)^{k+1}z^{-(k+1)}</math>
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let n=k+1
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<math>=\sum_{n=-\infty}^\infty u[n-1](-1)^{1-n}\Bigg(\frac{1}{3}\Bigg)^{n}z^{-n}</math>
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By comparison with <math>\sum_{n=-\infty}^\infty x[n] z^{-n}:</math>
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<math>x[n]=(-1)^{1-n}\Bigg(\frac{1}{3}\Bigg)^{n}u[n-1]\,</math>
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--[[User:Srigney|Srigney]] 12:32, 21 April 2011 (UTC)
 
=== Answer 3  ===
 
=== Answer 3  ===
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Write it here.
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<math>
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X(z) = \frac{1}{1+3z} = \left(\frac{1}{3}\right)\left(\frac{1}{1-\frac{-1}{3z}}\right), |z| > \frac{1}{3}
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</math>
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<math>
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|z| > \frac{1}{3} => |\frac{1}{3z}| < 1
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</math>
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<math>
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\begin{align}
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X(z) & = \left(\frac{1}{3}\right)\left(\frac{1}{1-\frac{-1}{3z}}\right) \\
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&= \left(\frac{1}{3z}\right) \sum_{k = 0}^{\infty} \left( -\frac{1}{3z} \right)^k \\
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&= \sum_{k = -\infty}^{\infty} u[k]\left(\frac{1}{3z}\right)  \left( -\frac{1}{3z} \right)^k \\
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&= \sum_{k = -\infty}^{\infty} u[k](-1)^k \left( \frac{1}{3z} \right)^{k +1} \\
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&= \sum_{k = -\infty}^{\infty} u[k](-1)^k (3z)^{-(k +1)} z^{-(k+1) } \\
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& \text{let n=k+1} \\
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&= \sum_{k = -\infty}^{\infty} u[k](-1)^{n-1} (3z)^{-n} z^{-n} \\
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&\text{by comparison with } \sum_{n=-\infty}^{\infty}x[n]z^{-n} \\
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x[n] &= (-1)^{n-1} u[n-1] \left(\frac{1}{3}\right)^n
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\end{align}
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</math>
 
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[[2011_Spring_ECE_301_Boutin|Back to ECE301 Spring 2011 Prof. Boutin]]
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[[2011 Spring ECE 301 Boutin|Back to ECE301 Spring 2011 Prof. Boutin]]
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[[Category:ECE301Spring2011Boutin]] [[Category:Problem_solving]]

Latest revision as of 11:51, 26 November 2013

Practice Question on "Digital Signal Processing"

Topic: Computing an inverse z-transform

Question

Compute the inverse z-transform of the following signal.

$ X(z)=\frac{1}{1+3z} \mbox{, } \Big|z\Big|>\frac{1}{3} $

Share your answers below

Prof. Mimi gave me this problem in class on Friday, so I'm posting it and my answer here. --Cmcmican 22:38, 16 April 2011 (UTC)

Answer 1

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

since $ \Bigg|z\Bigg|>\frac{1}{3} ==>\Bigg|\frac{1}{z}\Bigg|<3 ==>\Bigg|\frac{1}{3}\frac{1}{z}\Bigg|<1 $

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

let n=k+1

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

By comparison with $ \sum_{n=-\infty}^\infty x[n] z^{-n}: $

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

--Cmcmican 22:38, 16 April 2011 (UTC)

TA's comment: I think you have a mistake here. You can check that by taking the Z-transform of your answer.

Answer 2

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

since $ \Bigg|z\Bigg|>\frac{1}{3} ==>\Bigg|\frac{1}{z}\Bigg|<3 ==>\Bigg|\frac{1}{3}\frac{1}{z}\Bigg|<1 $

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

let n=k+1

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

By comparison with $ \sum_{n=-\infty}^\infty x[n] z^{-n}: $

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

--Srigney 12:32, 21 April 2011 (UTC)

Answer 3

$ X(z) = \frac{1}{1+3z} = \left(\frac{1}{3}\right)\left(\frac{1}{1-\frac{-1}{3z}}\right), |z| > \frac{1}{3} $

$ |z| > \frac{1}{3} => |\frac{1}{3z}| < 1 $

$ \begin{align} X(z) & = \left(\frac{1}{3}\right)\left(\frac{1}{1-\frac{-1}{3z}}\right) \\ &= \left(\frac{1}{3z}\right) \sum_{k = 0}^{\infty} \left( -\frac{1}{3z} \right)^k \\ &= \sum_{k = -\infty}^{\infty} u[k]\left(\frac{1}{3z}\right) \left( -\frac{1}{3z} \right)^k \\ &= \sum_{k = -\infty}^{\infty} u[k](-1)^k \left( \frac{1}{3z} \right)^{k +1} \\ &= \sum_{k = -\infty}^{\infty} u[k](-1)^k (3z)^{-(k +1)} z^{-(k+1) } \\ & \text{let n=k+1} \\ &= \sum_{k = -\infty}^{\infty} u[k](-1)^{n-1} (3z)^{-n} z^{-n} \\ &\text{by comparison with } \sum_{n=-\infty}^{\infty}x[n]z^{-n} \\ x[n] &= (-1)^{n-1} u[n-1] \left(\frac{1}{3}\right)^n \end{align} $

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