12.3 Inverse z-transform

When given $$X(z)=\frac{z}{z-\alpha }$$ with an ROC of $$\left|z\right|> \alpha$$ we could determine "by inspection" that $$x(n)=\alpha ^{n}u(n)$$

Find the inverse z-transform of $$X(z)=\frac{1+2z^{(-1)}+z^{-2}}{1-3z^{(-1)}+2z^{-2}}$$ where the ROC is $\left|z\right|> 2$ . In this case $M=N=2$ , so we have to use long division to get $$X(z)=\frac{1}{2}+\frac{\frac{1}{2}+\frac{7}{2}z^{(-1)}}{1-3z^{(-1)}+2z^{-2}}$$ Next factor the denominator. $$X(z)=2+\frac{-1+5z^{(-1)}}{(1-2z^{(-1)})(1-z^{(-1)})}$$ Now do partial-fraction expansion. $$X(z)=\frac{1}{2}+\frac{A_1}{1-2z^{(-1)}}+\frac{A_2}{1-z^{(-1)}}=\frac{1}{2}+\frac{\frac{9}{2}}{1-2z^{(-1)}}+\frac{-4}{1-z^{(-1)}}$$ Now each term can be inverted using the inspection method and the z-transform table. Thus, since the ROC is $\left|z\right|> 2$ , $$x(n)=\frac{1}{2}\delta (n)+\frac{9}{2}2^{n}u(n)-4u(n)$$

Now look at the z-transform of a finite-length sequence .

Suppose $$X(z)=\log_n(1+\alpha z^{(-1)})$$ Noting that $$\log_n(1+x)=\sum_{n=1}$$∞ -1 n 1 x n n Then $$X(z)=\sum_{n=1}$$∞ -1 n 1 α n z n n Therefore $$X(z)=\begin{cases}\frac{-1^{(n+1)}\alpha ^n}{n} & \text{if $n\ge 1$}\\ 0 & \text{if $n\le 0$}\end{cases}()$$