Inverse Z-Transform

Sources

1. InverseZTransform (Lecture Slides & Annotations)
2. Lecture Notes

Example 1 - Inverse z-transform

Problem 1.a.

Find the inverse z-transform of

$$X(z)=\frac{1}{1-1.5z^{-1}+0.5z^{-2}}$$

a.) $\text{ROC: }|z|>1$

b.) $\text{ROC: }|z|<0.5$

c.) $\text{ROC: }0.5<|z|<1$

Positive power of $z$

$$\begin{array}{rl}X(z)& =\frac{1}{1-1.5z^{-1}+0.5z^{-2}}\cdot \frac{z^2}{z^2}\\ & =\frac{z^2}{z^2-1.5z+0.5}\end{array}$$

Set a side one $z$

$$\begin{array}{rl}\frac{X(z)}{z}& =\frac{z}{z^2-1.5z+0.5}\\ & =\frac{z}{(z-0.5)(z-1)}\\ \frac{X(z)}{z}& =\frac{A_1}{z-0.5}+\frac{A_2}{z-1}\end{array}$$ $$\boxed{A_k=(z-P_k)\frac{X(z)}{z}{|}_{z=P_k}}$$ $$\begin{array}{rl}{A}_1& =(z-0.5)\frac{X(z)}{z}{|}_{z=0.5}\\ & =(z-0.5)\frac{1}{(z-0.5)(z-1)}{|}_{z=0.5}\\ & =\frac{z}{z-1}{|}_{z=0.5}\\ & =-1\\ A_2& =(z-1)\frac{X(z)}{z}{|}_{z=1}\\ & =\frac{z}{z-0.5}{|}_{z=1}\\ & =2\end{array}$$ $$\begin{array}{rl}\frac{X(z)}{z}& =\frac{-1}{z-0.5}+\frac{2}{z-1}\\ X(z)& =\frac{-z}{z-0.5}+\frac{2z}{z-1}\\ X(z)& =\frac{-1}{1-0.5z^{-1}}+\frac{2}{1-z^{-1}}\end{array}$$

a.) ROC: |z| > 1 (Causal)

$$\boxed{\begin{array}{rl}x[n]& =-(0.5{)}^{n}u[n]+2(1{)}^{n}u[n]\\ & =-(0.5{)}^{n}u[n]+2u[n]\end{array}}$$

b.) ROC: |z| < 0.5 (Anticausal)

$$\boxed{x[n]=(0.5{)}^{n}u[-n-1]-2u[-n-1]}$$

c.) ROC: 0.5 < |z| < 1 (Noncausal)

$$\boxed{x[n]=\underset{|z|>0.5}{\underbrace{-(0.5{)}^{n}u[n]}}-\underset{|z|<1}{\underbrace{2u[-n-1]}}}$$