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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> | <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=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>x[n] = \sum_{n=0}^{\infty} (\frac{4}{z})^n - 85 - \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 - 85 - \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? | 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. | ||
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Revision as of 14:26, 19 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
$ 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.
$ 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 - 85 - \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.
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