| (4 intermediate revisions by 3 users not shown) | |||
| Line 1: | Line 1: | ||
| − | [[ | + | <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]] 22:22, 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) |
| + | |||
---- | ---- | ||
| + | |||
=== 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> |
| + | |||
| + | --[[User:Cmcmican|Cmcmican]] 22:22, 16 April 2011 (UTC) | ||
| + | |||
| + | :<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 === | ||
| − | + | ||
| + | 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. | ||
| + | |||
| + | --[[User:Kellsper|Kellsper]] 16:12, 21 April 2011 (UTC) | ||
| + | :<span style="color: blue">Instructor's comment: Good. You may shorten this explanation a bit when you write it on the exam. Just say </span> | ||
| + | ::<math class="inline"> X(z)=\frac{1}{1+3z}=\frac{1}{1-(-3z)}=\sum_{k=0}^\infty (-3z)^k</math>, since <math>|-3z|=|3z|<1</math> when <math class="inline">|z|<\frac{1}{3}</math>. | ||
| + | ::<span style="color: blue">-pm .</span> | ||
| + | |||
=== 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]] [[Category:inverse z-transform]] | ||
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 22:22, 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)
- Instructor's comment: Good. You may shorten this explanation a bit when you write it on the exam. Just say
- $ X(z)=\frac{1}{1+3z}=\frac{1}{1-(-3z)}=\sum_{k=0}^\infty (-3z)^k $, since $ |-3z|=|3z|<1 $ when $ |z|<\frac{1}{3} $.
- -pm .
Answer 3
Write it here.
