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)

$X(z)=\sum_{k=0}^\infty (-3z)^k=\sum_{k=-\infty}^\infty u[k](-3)^kz^k$

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.

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 .