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$ \text { Determine inverse } z \text {-transform } x(z)=\frac{1}{1-0.5 z^{-1}+0.5 z^{-2}} $ for (i) $R_{\propto}:|z|>1$ (ii) $R_{0 C}+1<0.5$ (iii) ROC $0.5<|2|<1$
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Solution:

Step 1: Convert $x(z)$ to positive powers of z

$ x(z)=\cdot \frac{z^2}{z^2-1.5 z+0.5}\\ $

$ \begin{aligned} \frac{x(z)}{z} & =\frac{z}{z^2-1.5 z+0.5} \\\\ & =\frac{z}{(z-1)(z-0.5)}\\ \end{aligned}\\ $

Step 2 =: Partal Fraction

$ \frac{x(z)}{z}=\frac{A}{z-1}+\frac{B}{z-0.5} $

$ A=\left.(z-1) \cdot \frac{z}{(z-1)(z-0.5}\right|_{z=1}=\frac{1}{1-0.5}=2\\ $

$ B=\left.\frac{(z-0.5) z}{(z-1)(z-0.5)}\right|_{z=0.5}=\frac{.5}{.5-1}=-1\\ $

$ \frac{x(z)}{z}=\frac{z}{z-1}-\frac{1}{z-0.5} $

$ \begin{aligned} x(z) & =\frac{2 z}{z-1}-\frac{z}{z-0.5} \\\\ & =\frac{2}{1-z^{-1}}-\frac{1}{1-0.5 z^{-1}}\\ \end{aligned}\\ $

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