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Can I find the inverse $\mathcal Z$-transform of this transfer function: $$H(z)=\frac{1}{1-\alpha z^{-8}}$$ in a way other than contour integration and finding the residues of the 8 poles? If so, how?

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Applying the geometric series directly

$$ H(z) = \frac{1}{1-\alpha z^{-8}} = \sum_{n=0}^{\infty}(\alpha z^{-8})^n = \sum_{n=0}^{\infty}\alpha^n z^{-8n} $$ If $h_n = \mathcal Z^{-1}\{H(z)\}$, then $$ h_n = \begin{cases} a^{n/8} \; , \; \text{n is multiple of 8} \\ 0 \; , \; \text{otherwise} \end{cases} $$

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