---
title: "CTFT of e^(−2jω₀t)"
subject: "Signals and Systems"
chapter: 3
key: "semester_4/signals_and_systems:ctft-complex-exponential-negative"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/ctft-complex-exponential-negative
site: BCT Engineering
---

# CTFT of e^(−2jω₀t)

Signals and Systems · Chapter 3: Fourier Transform

- Asked in: 2083 Bhadra (new course), 5 marks
- Video: [Fourier Transform of Basic Signals (Complex Exponential Signal)](https://www.youtube.com/watch?v=fDrChXcplcA) — Neso Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/ctft-complex-exponential-negative
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-3

## Question

Let us consider the signal x(t) = e^(−2jω₀t). Find the continuous time Fourier transform of the given signal x(t).

## Short answer

For x(t) = e^{-j2ω₀t}, X(jω) = 2π δ(ω + 2ω₀), a single impulse at ω = –2ω₀.

## Answer

$x(t) = e^{-j2\omega_0 t}$ is a complex exponential of frequency $-2\omega_0$. It is not absolutely integrable ($|x(t)| = 1$ for all $t$), so its FT exists only with an impulse.

**Step 1: FT of a constant.** $1 \leftrightarrow 2\pi\delta(\omega)$, because

$$
\frac{1}{2\pi}\int_{-\infty}^{\infty}2\pi\,\delta(\omega)\,e^{j\omega t}d\omega = 1
$$

**Step 2: frequency shifting property.** $e^{j\beta t}x(t) \leftrightarrow X(j(\omega - \beta))$. Here $\beta = -2\omega_0$:

$$
e^{-j2\omega_0t}\cdot 1 \;\longleftrightarrow\; 2\pi\,\delta(\omega + 2\omega_0)
$$

**Verification by inverse FT:**

$$
\frac{1}{2\pi}\int_{-\infty}^{\infty}2\pi\,\delta(\omega + 2\omega_0)\,e^{j\omega t}d\omega = e^{j(-2\omega_0)t} = e^{-j2\omega_0t}
$$

**Answer:**

$$
X(j\omega) = 2\pi\,\delta(\omega + 2\omega_0)
$$

```text
              X(jw)
   2*pi ^
        |
--------+--------+--------> w
     -2w0        0
```

**Discussion:** The spectrum is a single impulse at $\omega = -2\omega_0$ of strength $2\pi$. A complex exponential contains only one frequency, so the spectrum has no mirror image at $+2\omega_0$ (a real sinusoid would have impulses at both $\pm 2\omega_0$).

## Check yourself

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

1. Why does the FT of e^(-j2w0 t) need an impulse?
   A. It is a real-valued signal
   B. |x(t)| = 1 for all t, so it is not absolutely integrable
   C. It is zero for t < 0
   D. It decays too fast
   Answer: B. The ordinary FT integral does not converge, so an impulse in frequency is used. (Key point: Why impulse)

2. What is the FT of the constant 1?
   A. δ(ω)
   B. 1
   C. 2π δ(ω)
   D. 2π
   Answer: C. Inverse FT of 2π δ(ω) gives (1/2π)(2π)=1. (Key point: FT of constant)

3. Which property is used to get the answer from 1 <-> 2π δ(ω)?
   A. Time shifting
   B. Time scaling
   C. Differentiation in time
   D. Frequency shifting: e^(jβt) x(t) <-> X(j(ω - β))
   Answer: D. Multiplying by e^(jβt) shifts the spectrum by β; here β = -2ω0. (Key point: Frequency shift)

4. What is the FT of x(t) = e^(-j2ω0 t)?
   A. 2π δ(ω + 2ω0)
   B. 2π δ(ω - 2ω0)
   C. π[δ(ω - 2ω0) + δ(ω + 2ω0)]
   D. δ(ω + 2ω0)
   Answer: A. The impulse sits at ω = -2ω0 with strength 2π. (Key point: Result)

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