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| − | = Practice Question 2, [[ECE438]] Fall 2010, [[User:Mboutin|Prof. Boutin]] | + | [[Category:problem solving]] |
| − | + | ||
| + | <center><font size= 4> | ||
| + | '''[[Digital_signal_processing_practice_problems_list|Practice Question on "Digital Signal Processing"]]''' | ||
| + | </font size> | ||
| + | |||
| + | Topic: Computing a z-transform | ||
| + | |||
| + | </center> | ||
| + | ( [[:Category:Problem_solving|Practice Question]] 2, [[ECE438]] Fall 2010, [[User:Mboutin|Prof. Boutin]] ) | ||
---- | ---- | ||
| − | <div><span style="color: | + | ==Question== |
| + | <div><span style="color:purple"> Compute the z-transform of the discrete-time signal | ||
| + | |||
| + | <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? | ||
Post Your answer/questions below. | Post Your answer/questions below. | ||
| − | </span></div> | + | </span></div> |
---- | ---- | ||
| − | ==Solution 1== | + | == Solution 1 == |
| − | < | + | <span class="texhtml">''x''[''n''] = 4<sup>''n''</sup>''u''[''n'' + 3] − 4<sup>''n''</sup>''u''[''n'' − 4]</span> |
| − | <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> | + | <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> |
| − | <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> | + | <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> |
| − | <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> | + | <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> |
| + | |||
| + | this is the mistake I made on my exam - could you please clarify my work, professor? | ||
| − | |||
*<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> | *<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> | ||
| − | *<div><span style="color:green">Another thing I see is the manipulation of the sum with negative indices, namely : | + | *<div><span style="color:green">Another thing I see is the manipulation of the sum with negative indices, namely : |
| − | :<math>{\color{green}\sum_{n=-\infty}^{4} 4^n z^{-n } = \sum_{n=4}^{\infty}(\frac{4}{z})^n }</math> | + | </span></div> |
| − | :which is incorrect. The correct way to manipulate it is the following: | + | |
| + | :<math>{\color{green}\sum_{n=-\infty}^{4} 4^n z^{-n } = \sum_{n=4}^{\infty}(\frac{4}{z})^n }</math> | ||
| + | :which is incorrect. The correct way to manipulate it is the following: | ||
:<math> | :<math> | ||
\begin{align} | \begin{align} | ||
| Line 32: | Line 45: | ||
&= \sum_{k=-4}^{\infty} 4^{-k} z^{k } \text{ (since the order of the terms in the sum does not matter)}, \\ | &= \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 } | &= 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} | + | \end{align}</math> |
| − | </math> | + | |
:<span style="color:green">Hope that helps! -pm | :<span style="color:green">Hope that helps! -pm | ||
</span> | </span> | ||
| − | |||
| − | ::Or better yet: | + | |
| + | <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> | ||
| + | |||
| + | ::Or better yet: | ||
:::<math>X(z) =\sum_{n=-3}^{3} (\frac{4}{z})^n</math> -pm | :::<math>X(z) =\sum_{n=-3}^{3} (\frac{4}{z})^n</math> -pm | ||
| + | <br> | ||
---- | ---- | ||
| − | *Answer/question | + | *Answer/question |
| − | * | + | |
| − | * | + | ---- |
| + | <div><span style="color:blue"> Note: although the signal given looks very similar to | ||
| + | |||
| + | <math>{\color{blue} x_1[n]= 4^n u[n+3]- 2^n u[n-4] }</math>. | ||
| + | |||
| + | and to | ||
| + | |||
| + | <math>{\color{blue} x_2[n]= 4^n u[n+3]- 2^n u[-n-4] }</math>. | ||
| + | |||
| + | the computation of the z-transform is very different. -pm | ||
| + | |||
| + | </span></div> | ||
| + | ---- | ||
| + | *Comment/answer/question | ||
| + | *Comment/answer/question | ||
---- | ---- | ||
| − | |||
| − | [[ | + | [[Practice question 1 eECE439F10|Previous practice problem]] |
| + | [[Practice Question 3 ECE439F10|Next practice problem]] | ||
| − | [[2010 Fall ECE 438 Boutin|Back to 2010 Fall ECE 438 Boutin]] | + | <br> [[2010 Fall ECE 438 Boutin|Back to 2010 Fall ECE 438 Boutin]] |
[[Category:2010_Fall_ECE_438_Boutin]] | [[Category:2010_Fall_ECE_438_Boutin]] | ||
| + | [[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
- Or better yet:
- 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
