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:<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> | :<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. | ||
| + | |||
| + | |||
=== Answer 3 === | === Answer 3 === | ||
Revision as of 12:54, 21 April 2011
Contents
Practice Question on Computing the z-transform
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.
Answer 3
Write it here.
