(New page: Category:ECE301Spring2011Boutin Category:Problem_solving ---- = Practice Question on Computing the z-transform = Compute the z-transform of the following signal. <math>x[n]=u[n]...) |
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| − | [[Category:ECE301Spring2011Boutin]] [[Category:Problem_solving]] | + | [[Category:ECE301Spring2011Boutin]] |
| + | [[Category:Problem_solving]] | ||
| + | [[Category:z-transform]] | ||
| + | |||
| + | ---- | ||
| + | <center><font size= 4> | ||
| + | '''[[Digital_signal_processing_practice_problems_list|Practice Question on "Digital Signal Processing"]]''' | ||
| + | </font size> | ||
| + | |||
| + | Topic: Computing a z-transform | ||
| + | |||
| + | </center> | ||
---- | ---- | ||
| − | = | + | ==Question== |
Compute the z-transform of the following signal. | Compute the z-transform of the following signal. | ||
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--[[User:Cmcmican|Cmcmican]] 22:05, 16 April 2011 (UTC) | --[[User:Cmcmican|Cmcmican]] 22:05, 16 April 2011 (UTC) | ||
| + | :<span style="color:green">TA's comment: Correct!</span> | ||
| + | :<span style="color:blue">Instructor's comment: Exactly where do you get that the norm of z must be greater than one for convergence? It is important to clearly state it.</span> | ||
| + | ---- | ||
=== Answer 2 === | === Answer 2 === | ||
| − | + | <math> \begin{align} | |
| + | X(z) &= \sum_{n=-\infty}^{\infty}u[n]z^{-n} \\ | ||
| + | &= \sum_{n=0}^{\infty}z^{-n} = \sum_{n=0}^{\infty} \left( \frac{1}{z} \right)^n \\ | ||
| + | &= \begin{cases} \frac{1}{1-\frac{1}{z}}, & |z| > 1 \\ diverges, & else \end{cases} | ||
| + | \end{align}</math> | ||
| + | If <math class="inline">z \leq 1</math> then <math class="inline">\frac{1}{z} \geq 1</math>, then the sum would diverge. | ||
| + | |||
| + | |||
| + | :<span style="color:blue">Instructor's comment: You forgot the norm around z in the last line. The statement z<1 does not make any sense in the complex plane, so you would lose points for that. </span> | ||
| + | ---- | ||
=== Answer 3 === | === Answer 3 === | ||
| − | + | <math> X(z) = \sum_{n=-\infty}^{\infty}u[n]z^{-n} = \sum_{n=0}^{\infty}z^{-n} = \sum_{n=0}^{\infty} \left( \frac{1}{z} \right)^n </math> | |
| + | |||
| + | So if <math>|\frac{1}{z}| \leq 1</math>, then by the [[More_on_geometric_series|geometric series]] formula we have <math> X(z)=\frac{1}{1-\frac{1}{z}}</math>. | ||
| + | |||
| + | On the other hand, if <math>|\frac{1}{z}| > 1</math>, then X(z) diverges. | ||
| + | |||
| + | :<span style="color:blue"> Instructor's comment: The series does not converge when <math>|\frac{1}{z}| =1</math>. </span> | ||
| + | :<span style="color:blue">Another Instructor's comment: Can you state your answer in simplified form? </span> | ||
| + | ---- | ||
| + | ===Answer 4=== | ||
| + | <math> \begin{align} | ||
| + | X(z) &= \sum_{n=-\infty}^{\infty}u[n]z^{-n} \\ | ||
| + | &= \sum_{n=0}^{\infty}z^{-n} = \sum_{n=0}^{\infty} \left( \frac{1}{z} \right)^n \\ | ||
| + | &= \begin{cases} \frac{1}{1-\frac{1}{z}}, & |z| > 1 \\ diverges, & else \end{cases} | ||
| + | \end{align}</math> | ||
| + | |||
| + | So <math> X(z)= \frac{1}{1-\frac{1}{z}} </math> with ROC <math> |z|>1 </math> | ||
| + | |||
| + | :<span style="color:blue">Instructor's comment: Looks good! </span> | ||
---- | ---- | ||
[[2011_Spring_ECE_301_Boutin|Back to ECE301 Spring 2011 Prof. Boutin]] | [[2011_Spring_ECE_301_Boutin|Back to ECE301 Spring 2011 Prof. Boutin]] | ||
Latest revision as of 08:09, 4 March 2015
Practice Question on "Digital Signal Processing"
Topic: Computing a z-transform
Question
Compute the z-transform of the following signal.
$ x[n]=u[n] $
Prof. Mimi gave me this problem in class on Friday, so I'm posting it and my answer here. --Cmcmican 22:05, 16 April 2011 (UTC)
Answer 1
$ X(z)=\sum_{n=-\infty}^\infty u[n]z^{-n}=\sum_{n=0}^\infty z^{-n} $
$ X(z)=\frac{z}{z-1} \mbox{, ROC: }\Big|z\Big|>1 $
--Cmcmican 22:05, 16 April 2011 (UTC)
- TA's comment: Correct!
- Instructor's comment: Exactly where do you get that the norm of z must be greater than one for convergence? It is important to clearly state it.
Answer 2
$ \begin{align} X(z) &= \sum_{n=-\infty}^{\infty}u[n]z^{-n} \\ &= \sum_{n=0}^{\infty}z^{-n} = \sum_{n=0}^{\infty} \left( \frac{1}{z} \right)^n \\ &= \begin{cases} \frac{1}{1-\frac{1}{z}}, & |z| > 1 \\ diverges, & else \end{cases} \end{align} $
If $ z \leq 1 $ then $ \frac{1}{z} \geq 1 $, then the sum would diverge.
- Instructor's comment: You forgot the norm around z in the last line. The statement z<1 does not make any sense in the complex plane, so you would lose points for that.
Answer 3
$ X(z) = \sum_{n=-\infty}^{\infty}u[n]z^{-n} = \sum_{n=0}^{\infty}z^{-n} = \sum_{n=0}^{\infty} \left( \frac{1}{z} \right)^n $
So if $ |\frac{1}{z}| \leq 1 $, then by the geometric series formula we have $ X(z)=\frac{1}{1-\frac{1}{z}} $.
On the other hand, if $ |\frac{1}{z}| > 1 $, then X(z) diverges.
- Instructor's comment: The series does not converge when $ |\frac{1}{z}| =1 $.
- Another Instructor's comment: Can you state your answer in simplified form?
Answer 4
$ \begin{align} X(z) &= \sum_{n=-\infty}^{\infty}u[n]z^{-n} \\ &= \sum_{n=0}^{\infty}z^{-n} = \sum_{n=0}^{\infty} \left( \frac{1}{z} \right)^n \\ &= \begin{cases} \frac{1}{1-\frac{1}{z}}, & |z| > 1 \\ diverges, & else \end{cases} \end{align} $
So $ X(z)= \frac{1}{1-\frac{1}{z}} $ with ROC $ |z|>1 $
- Instructor's comment: Looks good!
