Z-transform of e^(−at) sin ωt

The sampled signal is $x(kT) = e^{-akT}\sin\omega kT$. It is transformed below directly with Euler's formula and then checked with the complex translation theorem.

Method 1: Euler's formula

$$e^{-akT}\sin\omega kT = \frac{1}{2j}\left[e^{-(a - j\omega)kT} - e^{-(a + j\omega)kT}\right]$$

Using $\mathcal{Z}\left[e^{-bkT}\right] = \dfrac{1}{1 - e^{-bT}z^{-1}}$:

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

Multiply by $z^2/z^2$:

$$X(z) = \frac{ze^{-aT}\sin\omega T}{z^2 - 2ze^{-aT}\cos\omega T + e^{-2aT}}$$

Method 2: Complex translation theorem (check)

$\mathcal{Z}[\sin\omega t] = \dfrac{z\sin\omega T}{z^2 - 2z\cos\omega T + 1}$. Since $\mathcal{Z}[e^{-at}x(t)] = X(ze^{aT})$:

$$X(z) = \frac{ze^{aT}\sin\omega T}{z^2e^{2aT} - 2ze^{aT}\cos\omega T + 1}$$

Multiplying numerator and denominator by $e^{-2aT}$ gives the same result as Method 1.

Remarks

• The poles are at $z = e^{-aT}e^{\pm j\omega T}$: radius $e^{-aT}$, angle $\pm\omega T$. For $a > 0$ they lie inside the unit circle, giving a decaying oscillation.
• Example with $a = 1$, $\omega = 2$ rad/s, $T = 0.5$ s: $e^{-0.5} = 0.6065$, $\sin 1 = 0.8415$, $\cos 1 = 0.5403$, so $X(z) = \dfrac{0.5104z}{z^2 - 0.6554z + 0.3679}$.

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