Complex translation theorem

Complex translation theorem

If $\mathcal{Z}[x(t)] = X(z)$, then

$$\mathcal{Z}\left[e^{-at}x(t)\right] = X\!\left(ze^{aT}\right)$$

So multiplying a time function by $e^{-at}$ is the same as replacing $z$ by $ze^{aT}$ in its z-transform.

Proof:

$$\mathcal{Z}\left[e^{-at}x(t)\right] = \sum_{k=0}^{\infty}x(kT)e^{-akT}z^{-k} = \sum_{k=0}^{\infty}x(kT)\left(ze^{aT}\right)^{-k} = X\!\left(ze^{aT}\right)$$

Z-transform of $\sin\omega t$

Write $\sin\omega kT = \dfrac{e^{j\omega kT} - e^{-j\omega kT}}{2j}$:

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

Apply the theorem

Replace $z$ by $ze^{aT}$, so that $z^{-1}$ becomes $e^{-aT}z^{-1}$ and $z^{-2}$ becomes $e^{-2aT}z^{-2}$:

$$\mathcal{Z}\left[e^{-at}\sin\omega t\right] = \frac{e^{-aT}z^{-1}\sin\omega T}{1 - 2e^{-aT}z^{-1}\cos\omega T + e^{-2aT}z^{-2}}$$

In positive powers of $z$:

$$\mathcal{Z}\left[e^{-at}\sin\omega t\right] = \frac{ze^{-aT}\sin\omega T}{z^2 - 2ze^{-aT}\cos\omega T + e^{-2aT}}$$

The poles are at $z = e^{-aT}e^{\pm j\omega T}$, inside the unit circle for $a > 0$.

Answer: $\dfrac{ze^{-aT}\sin\omega T}{z^2 - 2ze^{-aT}\cos\omega T + e^{-2aT}}$.