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| − | [[ | + | <center><font size= 4> |
| + | '''[[Digital_signal_processing_practice_problems_list|Practice Question on "Digital Signal Processing"]]''' | ||
| + | </font size> | ||
| + | |||
| + | Topic: Computing an inverse z-transform | ||
| + | |||
| + | </center> | ||
---- | ---- | ||
| − | = | + | ==Question== |
| − | 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 === | ||
| − | + | ||
| + | <math> | ||
| + | X(z) = \frac{1}{1+3z} = \left(\frac{1}{3}\right)\left(\frac{1}{1-\frac{-1}{3z}}\right), |z| > \frac{1}{3} | ||
| + | </math> | ||
| + | |||
| + | <math> | ||
| + | |z| > \frac{1}{3} => |\frac{1}{3z}| < 1 | ||
| + | </math> | ||
| + | |||
| + | <math> | ||
| + | \begin{align} | ||
| + | X(z) & = \left(\frac{1}{3}\right)\left(\frac{1}{1-\frac{-1}{3z}}\right) \\ | ||
| + | &= \left(\frac{1}{3z}\right) \sum_{k = 0}^{\infty} \left( -\frac{1}{3z} \right)^k \\ | ||
| + | &= \sum_{k = -\infty}^{\infty} u[k]\left(\frac{1}{3z}\right) \left( -\frac{1}{3z} \right)^k \\ | ||
| + | &= \sum_{k = -\infty}^{\infty} u[k](-1)^k \left( \frac{1}{3z} \right)^{k +1} \\ | ||
| + | &= \sum_{k = -\infty}^{\infty} u[k](-1)^k (3z)^{-(k +1)} z^{-(k+1) } \\ | ||
| + | & \text{let n=k+1} \\ | ||
| + | &= \sum_{k = -\infty}^{\infty} u[k](-1)^{n-1} (3z)^{-n} z^{-n} \\ | ||
| + | &\text{by comparison with } \sum_{n=-\infty}^{\infty}x[n]z^{-n} \\ | ||
| + | x[n] &= (-1)^{n-1} u[n-1] \left(\frac{1}{3}\right)^n | ||
| + | \end{align} | ||
| + | </math> | ||
---- | ---- | ||
| − | [[ | + | |
| + | [[2011 Spring ECE 301 Boutin|Back to ECE301 Spring 2011 Prof. Boutin]] | ||
| + | |||
| + | [[Category:ECE301Spring2011Boutin]] [[Category:Problem_solving]] | ||
Latest revision as of 11:51, 26 November 2013
Practice Question on "Digital Signal Processing"
Topic: Computing an inverse z-transform
Question
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
$ X(z) = \frac{1}{1+3z} = \left(\frac{1}{3}\right)\left(\frac{1}{1-\frac{-1}{3z}}\right), |z| > \frac{1}{3} $
$ |z| > \frac{1}{3} => |\frac{1}{3z}| < 1 $
$ \begin{align} X(z) & = \left(\frac{1}{3}\right)\left(\frac{1}{1-\frac{-1}{3z}}\right) \\ &= \left(\frac{1}{3z}\right) \sum_{k = 0}^{\infty} \left( -\frac{1}{3z} \right)^k \\ &= \sum_{k = -\infty}^{\infty} u[k]\left(\frac{1}{3z}\right) \left( -\frac{1}{3z} \right)^k \\ &= \sum_{k = -\infty}^{\infty} u[k](-1)^k \left( \frac{1}{3z} \right)^{k +1} \\ &= \sum_{k = -\infty}^{\infty} u[k](-1)^k (3z)^{-(k +1)} z^{-(k+1) } \\ & \text{let n=k+1} \\ &= \sum_{k = -\infty}^{\infty} u[k](-1)^{n-1} (3z)^{-n} z^{-n} \\ &\text{by comparison with } \sum_{n=-\infty}^{\infty}x[n]z^{-n} \\ x[n] &= (-1)^{n-1} u[n-1] \left(\frac{1}{3}\right)^n \end{align} $
