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Find the Initial and final Halves, $ x(z)=\frac{1+z^{-1}}{1-0.25 z^{-2}} $
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Solution:

Initial Value,

$ \begin{aligned} x(0) & =\operatorname{lt}_{z \rightarrow \infty} x(z) \\\\ & =\operatorname{lt}_{z \rightarrow \infty} \frac{1+z^{-1}}{1-0.25 z^{-2}}\\ \end{aligned}\\ $

$ =\frac{1+\infty^{-1}}{1-0.250^{-2}}=\frac{1+0}{1-0}=1\\ $

$x(0)=1$

Final Value,

$ \begin{aligned} x(\infty) & =\operatorname{lt}_{z \rightarrow 1}\left(1-z^{-1}\right) \times(z) \\\\ & =\operatorname{lt}_{z \rightarrow 1}\left(1-z^{-1}\right) \frac{\left(1+z^{-1}\right)}{1-0.25 z^{-2}}\\ \end{aligned} $

$ =\operatorname{lt}_{z \rightarrow 1} \frac{1-z^{-2}}{1-0.25 z^{-2}}\\ $

$ =\operatorname{lt}_{z \rightarrow 1} \frac{z^2-1}{z^2-25}\\ $

$ =\frac{1-1}{1-.25}=0\\ $

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