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:<span style="color:blue">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.</span> | :<span style="color:blue">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.</span> | ||
=== Answer 2 === | === Answer 2 === | ||
| − | + | <math>X(z) = \sum_{n = -\infty}^{\infty} u[-n]z^{-n}</math> | |
| + | |||
| + | <math>X(z) = \sum_{n = -\infty}^{0} z^{-n}</math> | ||
| + | |||
| + | let l = -n | ||
| + | |||
| + | <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 === | ||
Revision as of 11:44, 21 April 2011
Contents
Practice Question on Computing the z-transform
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.
