Z-transform of sin ωt

Sampling with period $T$: $x(kT) = \sin\omega kT$ for $k \ge 0$. Use Euler's formula:

$$\sin\omega kT = \frac{e^{j\omega kT} - e^{-j\omega kT}}{2j}$$

Derivation

Since $\mathcal{Z}[a^k] = \sum_{k=0}^{\infty} a^k z^{-k} = \frac{1}{1-az^{-1}}$ for $|z| > |a|$:

$$\begin{aligned} X(z) &= \frac{1}{2j}\left[\frac{1}{1-e^{j\omega T}z^{-1}} - \frac{1}{1-e^{-j\omega T}z^{-1}}\right] \\ &= \frac{1}{2j}\cdot\frac{(e^{j\omega T} - e^{-j\omega T})z^{-1}}{1 - (e^{j\omega T}+e^{-j\omega T})z^{-1} + z^{-2}} \\ &= \frac{z^{-1}\sin\omega T}{1 - 2z^{-1}\cos\omega T + z^{-2}} \end{aligned}$$

using $e^{j\omega T} - e^{-j\omega T} = 2j\sin\omega T$ and $e^{j\omega T} + e^{-j\omega T} = 2\cos\omega T$.

Multiplying numerator and denominator by $z^2$:

$$X(z) = \frac{z\sin\omega T}{z^2 - 2z\cos\omega T + 1}, \quad |z| > 1$$

Answer: $\mathcal{Z}[\sin\omega t] = \dfrac{z\sin\omega T}{z^2 - 2z\cos\omega T + 1}$. The poles $z = e^{\pm j\omega T}$ lie on the unit circle, as expected for an undamped sinusoid.