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[[Category:ECE301Spring2011Boutin]] [[Category:Problem_solving]]
 
[[Category:ECE301Spring2011Boutin]] [[Category:Problem_solving]]
 
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= Practice Question on Computing the z-transform  =
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= [[:Category:Problem_solving|Practice Question]] on Computing the z-transform  =
  
 
Compute the z-transform of the following signal.
 
Compute the z-transform of the following signal.

Revision as of 09:31, 11 November 2011


Practice Question on Computing the z-transform

Compute the z-transform of the following signal.

$ x[n]=u[n] $


Share your answers below

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.


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