| Line 1: | Line 1: | ||
| − | |||
---- | ---- | ||
| + | |||
= 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> |
---- | ---- | ||
== 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]] 22:38, 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]] 22:38, 16 April 2011 (UTC) |
| + | |||
---- | ---- | ||
| + | |||
=== Answer 1 === | === Answer 1 === | ||
| − | <math>X(z)=\frac{1}{3z}\frac{1}{(1+\frac{1}{3z})}</math> | + | <math>X(z)=\frac{1}{3z}\frac{1}{(1+\frac{1}{3z})}</math> |
| − | since <math | + | since <math>\Bigg|z\Bigg|>\frac{1}{3} ==>\Bigg|\frac{1}{z}\Bigg|<3 ==>\Bigg|\frac{1}{3}\frac{1}{z}\Bigg|<1</math> |
| − | <math>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)}</math> | + | <math>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)}</math> |
| − | let n=k+1 | + | let n=k+1 |
| − | <math>=\sum_{n=-\infty}^\infty u[n-1]\Bigg(\frac{1}{3}\Bigg)^{n}z^{-n}</math> | + | <math>=\sum_{n=-\infty}^\infty u[n-1]\Bigg(\frac{1}{3}\Bigg)^{n}z^{-n}</math> |
| − | By comparison with <math | + | By comparison with <math>\sum_{n=-\infty}^\infty x[n] z^{-n}:</math> |
| − | <math>x[n]=\Bigg(\frac{1}{3}\Bigg)^{n}u[n-1]\,</math> | + | <math>x[n]=\Bigg(\frac{1}{3}\Bigg)^{n}u[n-1]\,</math> |
| + | |||
| + | --[[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. | :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 === | ||
| − | |||
| + | <math>X(z)=\frac{1}{3z}\frac{1}{(1+\frac{1}{3z})}=\frac{1}{3z}\frac{1}{(1-(-\frac{1}{3z}))}</math> | ||
| + | |||
| + | since <math>\Bigg|z\Bigg|>\frac{1}{3} ==>\Bigg|\frac{1}{z}\Bigg|<3 ==>\Bigg|\frac{1}{3}\frac{1}{z}\Bigg|<1</math> | ||
| + | |||
| + | <math>X(z)=\sum_{k=0}^\infty \frac{1}{3z}\Bigg(\frac{-1}{3z}\Bigg)^k=\sum_{k=-\infty}^\infty u[k] (-1)^{k}\Bigg(\frac{1}{3}\Bigg)^{k+1}z^{-(k+1)}</math> | ||
| + | |||
| + | let n=k+1 | ||
| + | |||
| + | <math>=\sum_{n=-\infty}^\infty u[n-1](-1)^{1-n}\Bigg(\frac{1}{3}\Bigg)^{n}z^{-n}</math> | ||
| + | |||
| + | By comparison with <math>\sum_{n=-\infty}^\infty x[n] z^{-n}:</math> | ||
| + | |||
| + | <math>x[n]=(-1)^{1-n}\Bigg(\frac{1}{3}\Bigg)^{n}u[n-1]\,</math> | ||
| + | |||
| + | --[[User:Srigney|Srigney]] 12:32, 21 April 2011 (UTC) | ||
=== Answer 3 === | === Answer 3 === | ||
| − | Write it here. | + | |
| + | Write it here. | ||
| + | |||
---- | ---- | ||
| − | [[ | + | |
| + | [[2011 Spring ECE 301 Boutin|Back to ECE301 Spring 2011 Prof. Boutin]] | ||
| + | |||
| + | [[Category:ECE301Spring2011Boutin]] [[Category:Problem_solving]] | ||
Revision as of 08:32, 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 22:38, 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
$ X(z)=\frac{1}{3z}\frac{1}{(1+\frac{1}{3z})}=\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] (-1)^{k}\Bigg(\frac{1}{3}\Bigg)^{k+1}z^{-(k+1)} $
let n=k+1
$ =\sum_{n=-\infty}^\infty u[n-1](-1)^{1-n}\Bigg(\frac{1}{3}\Bigg)^{n}z^{-n} $
By comparison with $ \sum_{n=-\infty}^\infty x[n] z^{-n}: $
$ x[n]=(-1)^{1-n}\Bigg(\frac{1}{3}\Bigg)^{n}u[n-1]\, $
--Srigney 12:32, 21 April 2011 (UTC)
Answer 3
Write it here.
