---
title: "Z-transform of e^(−at) sin ωt"
subject: "Digital Control System"
chapter: 2
key: "semester_6/digital_control_system:z-transform-exp-sin"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/z-transform-exp-sin
site: BCT Engineering
---

# Z-transform of e^(−at) sin ωt

Digital Control System · Chapter 2: The Z-Transform

- Asked in: 2070 Magh, 4 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/digital_control_system/z-transform-exp-sin
- Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-2

## Question

Find the z-transform of x(t)=e⁻ᵃᵗsin ωt.

## Short answer

Using Euler's formula or the complex translation theorem: Z[e^{−at} sin ωt] = ze^{−aT} sin ωT / (z^2 − 2z e^{−aT} cos ωT + e^{−2aT}).

## Answer

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

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. e^{-akT} sin(wkT) is written as:
   A. (1/2j)[e^{-(a-jw)kT} - e^{-(a+jw)kT}]
   B. (1/2)[e^{-(a-jw)kT} + e^{-(a+jw)kT}]
   C. (1/2j)[e^{-(a-jw)kT} + e^{-(a+jw)kT}]
   D. e^{-(a-jw)kT} - e^{-(a+jw)kT}
   Answer: A. Sine has the 1/(2j) factor and a minus sign between the exponentials. (Key point: Euler decomposition)

2. The poles of the result are at:
   A. z = e^{-aT} e^{±jwT}
   B. z = e^{aT} e^{±jwT}
   C. z = ±jw
   D. z = e^{-aT}
   Answer: A. Radius e^{-aT}, angle ±wT; inside the unit circle for a > 0. (Key point: Pole positions)

3. Which is Z[e^{-akT} sin wkT]?
   A. z e^{-aT} sin wT/(z^2 - 2z e^{-aT} cos wT + e^{-2aT})
   B. z e^{-aT} sin wT/(z^2 - 2z e^{-aT} sin wT + e^{-2aT})
   C. z sin wT/(z^2 - 2z e^{-aT} cos wT + 1)
   D. z e^{-aT} cos wT/(z^2 - 2z cos wT + e^{-2aT})
   Answer: A. Combining the two fractions gives sin wT in the numerator and 2e^{-aT} cos wT in the denominator. (Key point: Final transform)

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