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| − | [[Category:ECE301Spring2011Boutin]] [[Category:Problem_solving]] | + | [[Category:ECE301Spring2011Boutin]] [[Category:Problem_solving]] [[Category:z-transform]] |
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| − | = Practice Question on Computing | + | <center><font size= 4> |
| + | '''[[Digital_signal_processing_practice_problems_list|Practice Question on "Digital Signal Processing"]]''' | ||
| + | </font size> | ||
| + | |||
| + | Topic: Computing a z-transform | ||
| + | |||
| + | </center> | ||
| + | ---- | ||
| + | ==Question== | ||
Compute the z-transform of the following signal. | Compute the z-transform of the following signal. | ||
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let l = -n | let l = -n | ||
| − | <math>X(z) = \sum_{l=0}^{\infty}z^{ | + | <math>X(z) = \sum_{l=0}^{\infty}z^{l} = \begin{cases}\frac{1}{1-z}, &|z|<1 \\ diverdges, &else \end{cases}</math> |
=== Answer 3 === | === Answer 3 === | ||
Latest revision as of 11:50, 26 November 2013
Practice Question on "Digital Signal Processing"
Topic: Computing a z-transform
Question
Compute the z-transform of the following signal.
$ x[n]=u[-n] $
Prof. Mimi gave me this problem in class on Friday, so I'm posting it and my answer here. --Cmcmican 22:09, 16 April 2011 (UTC)
Answer 1
$ X(z)=\sum_{n=-\infty}^\infty u[-n]z^{-n} $
let k=-n
$ =\sum_{k=0}^\infty z^{k} $
$ X(z)=\frac{1}{1-z} \mbox{, ROC: }\Big|z\Big|<1 $
--Cmcmican 22:09, 16 April 2011 (UTC)
- TA's comment: Correct!
- Instructor's comment: Exactly where do you get that the norm of z must be less than one for convergence? It is important to clearly state it.
Answer 2
$ X(z) = \sum_{n = -\infty}^{\infty} u[-n]z^{-n} $
$ X(z) = \sum_{n = -\infty}^{0} z^{-n} $
let l = -n
$ X(z) = \sum_{l=0}^{\infty}z^{l} = \begin{cases}\frac{1}{1-z}, &|z|<1 \\ diverdges, &else \end{cases} $
Answer 3
Write it here.
