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= Practice Question 2, [[ECE438]] Fall 2010, [[User:Mboutin|Prof. Boutin]] = | = Practice Question 2, [[ECE438]] Fall 2010, [[User:Mboutin|Prof. Boutin]] = | ||
| − | On Computing the z-tramsfprm of a discrete-time signal. | + | <span style="color:blue">On Computing the z-tramsfprm of a discrete-time signal.</span> |
---- | ---- | ||
| − | Compute the z-transform of the discrete-time signal | + | <div><span style="color:blue"> Compute the z-transform of the discrete-time signal |
| − | <math>x[n]= 4^n \left(u[n+3]-u[n-4] \right)</math>. | + | <math>{\color{blue} 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> | ||
| + | ---- | ||
| + | ==Solution 1== | ||
<math>x[n] = 4^n u[n+3] - 4^n u[n-4]</math> | <math>x[n] = 4^n u[n+3] - 4^n u[n-4]</math> | ||
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this is the mistake I made on my exam - could you please clarify my work, professor? | 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 | + | *<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> |
| − | *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>\sum_{n=-\infty}^{4} 4^n z^{-n } = \sum_{n=4}^{\infty}(\frac{4}{z})^n </math> | + | :<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: | :which is incorrect. The correct way to manipulate it is the following: | ||
:<math> | :<math> | ||
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\end{align} | \end{align} | ||
</math> | </math> | ||
| − | :Hope that helps! -pm | + | :<span style="color:green">Hope that helps! -pm |
| + | </span> | ||
<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> | <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 | ||
| + | |||
| + | |||
---- | ---- | ||
Revision as of 03:18, 20 October 2010
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 signal
$ {\color{blue} 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] = 4^n u[n+3] - 4^n u[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
- Answer/question
- Answer/question
