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 22:38, 16 April 2011 (UTC)

$X(z)=\frac{1}{3z}\frac{1}{(1+\frac{1}{3z})}$

since $\Bigg|z\Bigg|>\frac{1}{3} ==>\Bigg|\frac{1}{z}\Bigg|<3 ==>\Bigg|\frac{1}{3}\frac{1}{z}\Bigg|<1$

$X(z)=\sum_{k=0}^\infty \frac{1}{3z}\Bigg(\frac{1}{3z}\Bigg)^k=\sum_{k=-\infty}^\infty u[k] \Bigg(\frac{1}{3}\Bigg)^{k+1}z^{-(k+1)}$

let n=k+1

$=\sum_{n=-\infty}^\infty u[n-1]\Bigg(\frac{1}{3}\Bigg)^{n}z^{-n}$

By comparison with $\sum_{n=-\infty}^\infty x[n] z^{-n}:$

$x[n]=\Bigg(\frac{1}{3}\Bigg)^{n}u[n-1]\,$

--Cmcmican 22:38, 16 April 2011 (UTC)

TA's comment: I think you have a mistake here. You can check that by taking the Z-transform of your answer.

$X(z)=\frac{1}{3z}\frac{1}{(1+\frac{1}{3z})}=\frac{1}{3z}\frac{1}{(1-(-\frac{1}{3z}))}$

since $\Bigg|z\Bigg|>\frac{1}{3} ==>\Bigg|\frac{1}{z}\Bigg|<3 ==>\Bigg|\frac{1}{3}\frac{1}{z}\Bigg|<1$

$X(z)=\sum_{k=0}^\infty \frac{1}{3z}\Bigg(\frac{-1}{3z}\Bigg)^k=\sum_{k=-\infty}^\infty u[k] (-1)^{k}\Bigg(\frac{1}{3}\Bigg)^{k+1}z^{-(k+1)}$

let n=k+1

$=\sum_{n=-\infty}^\infty u[n-1](-1)^{1-n}\Bigg(\frac{1}{3}\Bigg)^{n}z^{-n}$

By comparison with $\sum_{n=-\infty}^\infty x[n] z^{-n}:$

$x[n]=(-1)^{1-n}\Bigg(\frac{1}{3}\Bigg)^{n}u[n-1]\,$

--Srigney 12:32, 21 April 2011 (UTC)

$X(z) = \frac{1}{1+3z} = \left(\frac{1}{3}\right)\left(\frac{1}{1-\frac{-1}{3z}}\right), |z| > \frac{1}{3}$
$|z| > \frac{1}{3} => |\frac{1}{3z}| < 1$
\begin{align} X(z) & = \left(\frac{1}{3}\right)\left(\frac{1}{1-\frac{-1}{3z}}\right) \\ &= \left(\frac{1}{3z}\right) \sum_{k = 0}^{\infty} \left( -\frac{1}{3z} \right)^k \\ &= \sum_{k = -\infty}^{\infty} u[k]\left(\frac{1}{3z}\right) \left( -\frac{1}{3z} \right)^k \\ &= \sum_{k = -\infty}^{\infty} u[k](-1)^k \left( \frac{1}{3z} \right)^{k +1} \\ &= \sum_{k = -\infty}^{\infty} u[k](-1)^k (3z)^{-(k +1)} z^{-(k+1) } \\ & \text{let n=k+1} \\ &= \sum_{k = -\infty}^{\infty} u[k](-1)^{n-1} (3z)^{-n} z^{-n} \\ &\text{by comparison with } \sum_{n=-\infty}^{\infty}x[n]z^{-n} \\ x[n] &= (-1)^{n-1} u[n-1] \left(\frac{1}{3}\right)^n \end{align} 