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--[[User:Cmcmican|Cmcmican]] 22:38, 16 April 2011 (UTC) | --[[User:Cmcmican|Cmcmican]] 22:38, 16 April 2011 (UTC) | ||
| − | + | :TA's comment: I think you have a mistake here. You can check that by taking the Z-transform of your answer. | |
=== Answer 2 === | === Answer 2 === | ||
Write it here. | Write it here. | ||
Revision as of 08:18, 17 April 2011
Contents
Practice Question on Computing the inverse z-transform
Compute the inverse z-transform of the following signal.
$ X(z)=\frac{1}{1+3z} \mbox{, } \Big|z\Big|>\frac{1}{3} $
Prof. Mimi gave me this problem in class on Friday, so I'm posting it and my answer here. --Cmcmican 22:38, 16 April 2011 (UTC)
Answer 1
$ X(z)=\frac{1}{3z}\frac{1}{(1+\frac{1}{3z})} $
since $ \Bigg|z\Bigg|>\frac{1}{3} ==>\Bigg|\frac{1}{z}\Bigg|<3 ==>\Bigg|\frac{1}{3}\frac{1}{z}\Bigg|<1 $
$ X(z)=\sum_{k=0}^\infty \frac{1}{3z}\Bigg(\frac{1}{3z}\Bigg)^k=\sum_{k=-\infty}^\infty u[k] \Bigg(\frac{1}{3}\Bigg)^{k+1}z^{-(k+1)} $
let n=k+1
$ =\sum_{n=-\infty}^\infty u[n-1]\Bigg(\frac{1}{3}\Bigg)^{n}z^{-n} $
By comparison with $ \sum_{n=-\infty}^\infty x[n] z^{-n}: $
$ x[n]=\Bigg(\frac{1}{3}\Bigg)^{n}u[n-1]\, $
--Cmcmican 22:38, 16 April 2011 (UTC)
- TA's comment: I think you have a mistake here. You can check that by taking the Z-transform of your answer.
Answer 2
Write it here.
Answer 3
Write it here.
