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 - 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 :
- $ \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 - 4^1z^{-1} - 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 }) $
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