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I wonder whether my calculation of inverse z transform are correct. My IIR system is described as follows in Z-domain $H(z) = \frac{z^{-2}}{1-0.5z^{-2}}$

After using partial fraction decomposition I obtained $H(z)$ in Z-domain as follows $H(z)=\frac{\frac{1}{2}}{1-\sqrt{\frac{1}{2}}z^{-1}} + \frac{\frac{1}{2}}{1+\sqrt{\frac{1}{2}}z^{-1}}$.

After applying inverse Z transform I obtained such form: $h[n]=\frac{1}{2}(\sqrt{\frac{1}{2}}^{n}-\sqrt{\frac{1}{2}}^{n})u[n]$. What confused me is that this is 0 as the insides of parentheses cancel out. Maybe this is some dumb mistake I am making (it is quite late)

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HINT:

Write $H(z)$ as

$$H(z)=G(z^2)\tag{1}$$

with $$G(z)=\frac{z^{-1}}{1-0.5z^{-1}}\tag{2}$$

Determine the inverse transform $g[n]$ of $G(z)$. Figure out what replacing $z$ by $z^2$ means in the time domain. From this, the inverse transform of $H(z)$ is easily obtained from $g[n]$.

EDIT: Your partial fraction expansion lacks the factor $z^{-2}$, and the major mistake is that you transform both terms to the same sequence, even though the poles have opposite signs. If you do it right, the two terms of the sequence only cancel each other for odd $n$, and you can combine them for even $n$. The method shown as a hint in this answer is probably more straightforward because it avoids partial fraction expansion.

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You did make small mistakes, the inverse z transfom for$\frac 1{1+az^{-1}} is {(-a)}^nu(n)$ and in partial fractions you have fallow the rule that numerator degree should be less than the denomination degree. so your question becomes $\frac {z^{-2}} {1-0.5z^{-2}}=-2+\frac 2 {1-0.5z^{-2}} = -2+\frac 1 {1-\frac1 {√2}z^{-1}}+\frac 1 {1+\frac1 {√2}z^{-1}}.$so after applying inverse z transfom it becomes $ -2\delta(n)+{(\frac1 {√2})}^nu(n)+{(-\frac1 {√2})}^nu(n)$.

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