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--[[User:Cmcmican|Cmcmican]] 22:22, 16 April 2011 (UTC) | --[[User:Cmcmican|Cmcmican]] 22:22, 16 April 2011 (UTC) | ||
:<span style="color:green">TA's comment: Good Job!</span> | :<span style="color:green">TA's comment: Good Job!</span> | ||
| + | :<span style="color:blue">Instructor's comment: You may want to mention where you use the fact that |z|<1/3.</span> | ||
=== Answer 2 === | === Answer 2 === | ||
Write it here. | Write it here. | ||
Revision as of 03:04, 18 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 22:22, 16 April 2011 (UTC)
Answer 1
$ X(z)=\sum_{k=0}^\infty (-3z)^k=\sum_{k=-\infty}^\infty u[k](-3)^kz^k $
let n=-k
$ =\sum_{n=-\infty}^\infty u[-n](-3)^{-n} z^{-n} $
By comparison with $ \sum_{n=-\infty}^\infty x[n] z^{-n}: $
$ x[n]=(-3)^{-n}u[-n]\, $
--Cmcmican 22:22, 16 April 2011 (UTC)
- TA's comment: Good Job!
- Instructor's comment: You may want to mention where you use the fact that |z|<1/3.
Answer 2
Write it here.
Answer 3
Write it here.
