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= Practice Question on Computing the inverse z-transform = | = Practice Question on Computing the inverse z-transform = | ||
| − | Compute the inverse z-transform of the following signal. | + | Compute the inverse z-transform of the following signal. |
| − | <math>X(z)=\frac{1}{1+3z} \mbox{, } \Big|z\Big|<\frac{1}{3}</math> | + | <math>X(z)=\frac{1}{1+3z} \mbox{, } \Big|z\Big|<\frac{1}{3}</math> |
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== Share your answers below == | == Share your answers below == | ||
| − | Prof. Mimi gave me this problem in class on Friday, so I'm posting it and my answer here. | + | |
| − | --[[User:Cmcmican|Cmcmican]] 22:22, 16 April 2011 (UTC) | + | Prof. Mimi gave me this problem in class on Friday, so I'm posting it and my answer here. --[[User:Cmcmican|Cmcmican]] 22:22, 16 April 2011 (UTC) |
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=== Answer 1 === | === Answer 1 === | ||
| − | <math>X(z)=\sum_{k=0}^\infty (-3z)^k=\sum_{k=-\infty}^\infty u[k](-3)^kz^k</math> | + | <math>X(z)=\sum_{k=0}^\infty (-3z)^k=\sum_{k=-\infty}^\infty u[k](-3)^kz^k</math> |
| − | let n=-k | + | let n=-k |
| − | <math>=\sum_{n=-\infty}^\infty u[-n](-3)^{-n} z^{-n}</math> | + | <math>=\sum_{n=-\infty}^\infty u[-n](-3)^{-n} z^{-n}</math> |
| − | By comparison with <math | + | By comparison with <math>\sum_{n=-\infty}^\infty x[n] z^{-n}:</math> |
| − | <math>x[n]=(-3)^{-n}u[-n]\,</math> | + | <math>x[n]=(-3)^{-n}u[-n]\,</math> |
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| + | --[[User:Cmcmican|Cmcmican]] 22:22, 16 April 2011 (UTC) | ||
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| + | :<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> | ||
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=== Answer 2 === | === Answer 2 === | ||
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| + | I agree, but for the missing steps on |z|<1/3, you can say | ||
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| + | Since |z| < 1/3, |3z| < 1 | ||
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| + | Therefore, |-3z| < 1 | ||
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| + | By comparison with the geometric series, where it diverges for |-3z| < 1, you can rewrite the problem as shown in Answer 1. | ||
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| + | --[[User:Kellsper|Kellsper]] 16:12, 21 April 2011 (UTC) | ||
=== Answer 3 === | === Answer 3 === | ||
| − | Write it here. | + | |
| + | Write it here. | ||
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| − | [[ | + | |
| + | [[2011 Spring ECE 301 Boutin|Back to ECE301 Spring 2011 Prof. Boutin]] | ||
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| + | [[Category:ECE301Spring2011Boutin]] [[Category:Problem_solving]] | ||
Revision as of 11:12, 21 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
I agree, but for the missing steps on |z|<1/3, you can say
Since |z| < 1/3, |3z| < 1
Therefore, |-3z| < 1
By comparison with the geometric series, where it diverges for |-3z| < 1, you can rewrite the problem as shown in Answer 1.
--Kellsper 16:12, 21 April 2011 (UTC)
Answer 3
Write it here.
