---
title: "Complex translation theorem"
subject: "Digital Control System"
chapter: 2
key: "semester_6/digital_control_system:complex-translation-theorem"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/complex-translation-theorem
site: BCT Engineering
---

# Complex translation theorem

Digital Control System · Chapter 2: The Z-Transform

- Asked in: 2072 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/complex-translation-theorem
- Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-2

## Question

State Complex Translation theorem. Obtain the z-transform of e⁻ᵃᵗsin ωt by using Complex Translation theorem.

## Short answer

Complex translation theorem: Z[e^{−at}x(t)] = X(ze^{aT}). For sin ωt, Z[sin ωt] = z sin ωT / (z^2−2z cos ωT+1). Applying the theorem gives Z[e^{−at} sin ωt] = ze^{−aT} sin ωT / (z^2−2z e^{−aT} cos ωT + e^{−2aT}).

## Answer

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

## Check yourself

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

1. The complex translation theorem states Z[e^{-at}x(t)] equals:
   A. X(z e^{aT})
   B. X(z e^{-aT})
   C. e^{-aT}X(z)
   D. X(z) e^{aT}
   Answer: A. Multiplying by e^{-at} replaces z by z e^{aT}. (Key point: Statement)

2. When z is replaced by z e^{aT}, z^{-1} becomes:
   A. e^{-aT} z^{-1}
   B. e^{aT} z^{-1}
   C. z^{-1}
   D. e^{-aT} z
   Answer: A. (z e^{aT})^{-1} = e^{-aT} z^{-1}, and z^{-2} becomes e^{-2aT} z^{-2}. (Key point: Substitution effect)

3. Z[e^{-at} sin wt] is:
   A. z e^{-aT} sin wT/(z^2 - 2z e^{-aT} cos wT + e^{-2aT})
   B. z sin wT/(z^2 - 2z e^{-aT} cos wT + e^{-2aT})
   C. z e^{-aT} cos wT/(z^2 - 2z e^{-aT} cos wT + e^{-2aT})
   D. z e^{-aT} sin wT/(z^2 - 2z cos wT + 1)
   Answer: A. Applying the theorem to z sin wT/(z^2 - 2z cos wT + 1) gives this. (Key point: Result)

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