Find the $Z$-transform of $\sin (\alpha k), k \ge0$

Question

Find the $Z$-transform of $\sin (\alpha k), k \ge0$

Solution

Could anyone please explain how we got the second step (in terms of $e$ and $i$) after writing it in basic $Z$-transform notation? And how it was simplified? I can't follow.

Answer 1

This result can be found in any introductory signal processing book, but I'll post the solution of a closely related problem here.

Given $f(k) = e^{i\alpha k}$, its $Z$-transform is $$ \mathcal{Z}\{f(k)\} = \sum_{k=0}^{\infty} f(k)z^{-k} = \sum_{k=0}^{\infty}e^{i\alpha k}z^{-k} = \sum_{k=0}^{\infty}(e^{i\alpha}z^{-1})^k = 1 + e^{i\alpha}z^{-1} + (e^{i\alpha}z^{-1})^2 + \dots = F(z), $$ for $|z| < 1$. To find a closed-form for $F(z)$, multiply it by $e^{i\alpha}z^{-1}$ and note $$ e^{i\alpha}z^{-1}F(z) = e^{i\alpha}z^{-1} + (e^{i\alpha}z^{-1})^2 + (e^{i\alpha}z^{-1})^3 + \dots = F(z) - 1\\ F(z)(1 - e^{i\alpha}z^{-1}) = 1\\ F(z) = {1 \over 1 - e^{i\alpha}z^{-1}}\cdot {1-e^{-i\alpha}z^{-1} \over 1 - e^{-i\alpha}z^{-1}} = {1 - e^{-i\alpha}z^{-1} \over 1 - z^{-1}(e^{i\alpha} + e^{-i\alpha}) + z^{-2}}. $$ Using $e^{ix} = \cos x + i\sin x$, $$ F(z) = {1 - z^{-1}(\cos\alpha - i\sin\alpha) \over 1 - 2z^{-1}\cos\alpha + z^{-2}}.\tag1 $$ The $Z$-transforms of $\cos(\alpha k)$ and $\sin(\alpha k)$ can be trivially found from $(1)$.