Revision as of 12:27, 21 April 2011

Practice Question on Computing the inverse z-transform

Compute the inverse z-transform of the following signal.

$ X(z)=\frac{1}{1+3z} \mbox{, } \Big|z\Big|<\frac{1}{3} $

Share your answers below

Prof. Mimi gave me this problem in class on Friday, so I'm posting it and my answer here. --Cmcmican 22:22, 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.

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