(New page: = Practice Question 2, ECE438 Fall 2010, Prof. Boutin = On Computing the z-tramsfprm of a discrete-time signal. ---- Compute the z-transform of the discrete-time signa...)
 
 
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= Practice Question 2, [[ECE438]] Fall 2010, [[User:Mboutin|Prof. Boutin]] =
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[[Category:problem solving]]
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On Computing the z-tramsfprm of a discrete-time signal.
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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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</font size>
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Topic: Computing a z-transform
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</center>
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( [[:Category:Problem_solving|Practice Question]] 2, [[ECE438]] Fall 2010, [[User:Mboutin|Prof. Boutin]] )
 
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Compute the z-transform of the discrete-time signal  
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==Question==
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<div><span style="color:purple">  Compute the z-transform of the discrete-time signal  
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<math> x[n]= 4^n \left(u[n+3]-u[n-4] \right) </math>.
  
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<math>x[n]= 4^n \left(u[n+3]-u[n-4] \right)</math>.
 
  
 
Note: there are two tricky parts in this problem. Do you know what they are?
 
Note: there are two tricky parts in this problem. Do you know what they are?
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----
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Post Your answer/questions below.
 
Post Your answer/questions below.
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*Answer/question
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</span></div>
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*Answer/question
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*Answer/question
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*Answer/question
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----
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== Solution 1 ==
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<span class="texhtml">''x''[''n''] = 4<sup>''n''</sup>''u''[''n'' + 3] − 4<sup>''n''</sup>''u''[''n'' − 4]</span>
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<math>x[n] = \sum_{n=-\infty}^{\infty} 4^n u[n+3] z^{-n} - \sum_{n=-\infty}^{\infty} 4^n u[n-4] z^{-n}</math>
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<math>{\color{red}\not}x {\color{red}\not}[n]  {\color{red}X(z)} = \sum_{n=3}^{\infty} 4^n z^{-n} - \sum_{n=-\infty}^{4} 4^n z^{-n}</math>
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<math>{\color{red}\not}x {\color{red}\not}[n]  {\color{red}X(z)}= \sum_{n=0}^{\infty} (\frac{4}{z})^n - 1 - 4^1z^{-1} - 4^2z^{-2} - 4^3z^{-3} - \sum_{n=4}^{\infty} (\frac{4}{z})^n</math>
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this is the mistake I made on my exam - could you please clarify my work, professor?
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*<div><span style="color:green"> Certainly! This is a very common mistake: splitting a sum that converges for most z's  into two sums that diverge for most z's.  The key is to notice that the first sum above has a finite number of  terms: so convergence of the entire sum is guaranteed, unless one (or more) of the terms of the sum diverge (for example, 1/z diverges when z=0). Observe that, by splitting the sum this way,  you get an empty ROC. The correct ROC for this z-transform is actually all the finite complex plane except zero. -pm</span></div>
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*<div><span style="color:green">Another thing I see is the manipulation of the sum with negative indices, namely&nbsp;:
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</span></div>
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:<math>{\color{green}\sum_{n=-\infty}^{4} 4^n z^{-n } =  \sum_{n=4}^{\infty}(\frac{4}{z})^n }</math>
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:which is incorrect. The correct way to manipulate it is the following:
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:<math>
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\begin{align}
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\sum_{n=-\infty}^{4} 4^n z^{-n }  &=  \sum_{k=\infty}^{-4} 4^{-k} z^{k }  \text{ (letting }k=-n),  \\
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        &=  \sum_{k=-4}^{\infty} 4^{-k} z^{k }  \text{ (since the order of the terms in the sum does not matter)}, \\
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&= 4^{4}z^{-4}+4^{3}z^{-3}+4^{2}z^{-2}+4^{1}z^{-1}+ \sum_{k=0}^{\infty} 4^{-k} z^{k }
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\end{align}</math>
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:<span style="color:green">Hope that helps! -pm
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</span>
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<math>X(z) =\sum_{n=0}^{\infty} (\frac{4}{z})^n - 1 - 4z - 4^2z^{2} - 4^3z^{3} - (4^{4}z^{-4}+4^{3}z^{-3}+4^{2}z^{-2}+4^{1}z^{-1}+ \sum_{k=0}^{\infty} 4^{-k} z^{k })</math>
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::Or better yet:
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:::<math>X(z) =\sum_{n=-3}^{3} (\frac{4}{z})^n</math> -pm
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<br>
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----
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*Answer/question
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----
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<div><span style="color:blue">  Note: although the signal given looks very similar to
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<math>{\color{blue} x_1[n]= 4^n u[n+3]- 2^n u[n-4]  }</math>.
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and to
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<math>{\color{blue} x_2[n]= 4^n u[n+3]- 2^n u[-n-4]  }</math>.
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the computation of the z-transform is very different. -pm
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</span></div>
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----
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*Comment/answer/question
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*Comment/answer/question
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----
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[[Category:z-transform]]

Latest revision as of 11:48, 26 November 2013


Practice Question on "Digital Signal Processing"

Topic: Computing a z-transform

( Practice Question 2, ECE438 Fall 2010, Prof. Boutin )

Question

Compute the z-transform of the discrete-time signal

$ x[n]= 4^n \left(u[n+3]-u[n-4] \right) $.


Note: there are two tricky parts in this problem. Do you know what they are?

Post Your answer/questions below.

Solution 1

x[n] = 4nu[n + 3] − 4nu[n − 4]

$ x[n] = \sum_{n=-\infty}^{\infty} 4^n u[n+3] z^{-n} - \sum_{n=-\infty}^{\infty} 4^n u[n-4] z^{-n} $

$ {\color{red}\not}x {\color{red}\not}[n] {\color{red}X(z)} = \sum_{n=3}^{\infty} 4^n z^{-n} - \sum_{n=-\infty}^{4} 4^n z^{-n} $

$ {\color{red}\not}x {\color{red}\not}[n] {\color{red}X(z)}= \sum_{n=0}^{\infty} (\frac{4}{z})^n - 1 - 4^1z^{-1} - 4^2z^{-2} - 4^3z^{-3} - \sum_{n=4}^{\infty} (\frac{4}{z})^n $

this is the mistake I made on my exam - could you please clarify my work, professor?

  • Certainly! This is a very common mistake: splitting a sum that converges for most z's into two sums that diverge for most z's. The key is to notice that the first sum above has a finite number of terms: so convergence of the entire sum is guaranteed, unless one (or more) of the terms of the sum diverge (for example, 1/z diverges when z=0). Observe that, by splitting the sum this way, you get an empty ROC. The correct ROC for this z-transform is actually all the finite complex plane except zero. -pm
  • Another thing I see is the manipulation of the sum with negative indices, namely :
$ {\color{green}\sum_{n=-\infty}^{4} 4^n z^{-n } = \sum_{n=4}^{\infty}(\frac{4}{z})^n } $
which is incorrect. The correct way to manipulate it is the following:
$ \begin{align} \sum_{n=-\infty}^{4} 4^n z^{-n } &= \sum_{k=\infty}^{-4} 4^{-k} z^{k } \text{ (letting }k=-n), \\ &= \sum_{k=-4}^{\infty} 4^{-k} z^{k } \text{ (since the order of the terms in the sum does not matter)}, \\ &= 4^{4}z^{-4}+4^{3}z^{-3}+4^{2}z^{-2}+4^{1}z^{-1}+ \sum_{k=0}^{\infty} 4^{-k} z^{k } \end{align} $
Hope that helps! -pm


$ X(z) =\sum_{n=0}^{\infty} (\frac{4}{z})^n - 1 - 4z - 4^2z^{2} - 4^3z^{3} - (4^{4}z^{-4}+4^{3}z^{-3}+4^{2}z^{-2}+4^{1}z^{-1}+ \sum_{k=0}^{\infty} 4^{-k} z^{k }) $

Or better yet:
$ X(z) =\sum_{n=-3}^{3} (\frac{4}{z})^n $ -pm


  • Answer/question
Note: although the signal given looks very similar to

$ {\color{blue} x_1[n]= 4^n u[n+3]- 2^n u[n-4] } $.

and to

$ {\color{blue} x_2[n]= 4^n u[n+3]- 2^n u[-n-4] } $.

the computation of the z-transform is very different. -pm

  • Comment/answer/question
  • Comment/answer/question

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