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

# CTFT of e^(jω₀t)

Signals and Systems · Chapter 3: Fourier Transform

- Asked in: 2083 Baisakh (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
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-3

## Question

Find the continuous time Fourier transform of the signal x(t) = e^(jω₀t). Discuss the result.

## Short answer

e^{jω₀t} ↔ 2π δ(ω – ω₀), a single impulse at ω₀ with area 2π. This represents all energy at one frequency and serves as the basis for FT of periodic signals.

## Answer

$x(t) = e^{j\omega_0t}$ has $|x(t)| = 1$ for all $t$, so it is not absolutely integrable and the FT integral does not converge in the ordinary sense. We use an impulse in frequency.

**Step 1: guess and verify.** Take $X(j\omega) = 2\pi\,\delta(\omega - \omega_0)$ and find its inverse FT:

$$
\begin{aligned}
x(t) &= \frac{1}{2\pi}\int_{-\infty}^{\infty}2\pi\,\delta(\omega - \omega_0)\,e^{j\omega t}\,d\omega \\
&= e^{j\omega_0t}
\end{aligned}
$$

by the sifting property. Since the FT is unique,

$$
e^{j\omega_0t} \;\longleftrightarrow\; 2\pi\,\delta(\omega - \omega_0)
$$

(Equivalently: $1 \leftrightarrow 2\pi\delta(\omega)$, then frequency shifting by $\omega_0$.)

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

**Discussion:**

- The spectrum is a single impulse at $\omega = \omega_0$ with area $2\pi$: all the signal content is at one frequency.
- The spectrum is not symmetric about $\omega = 0$, because $x(t)$ is complex.
- The magnitude is constant forever, so the signal has infinite energy but finite power (1). It is a power signal, which is why an impulse appears instead of an ordinary function.
- This pair is the basis for the FT of all periodic signals: from $x(t) = \sum a_ke^{jk\omega_0t}$ we get $X(j\omega) = \sum 2\pi a_k\delta(\omega - k\omega_0)$. For example, $\cos\omega_0t \leftrightarrow \pi[\delta(\omega-\omega_0) + \delta(\omega+\omega_0)]$.

## Check yourself

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

1. Why is the FT of e^(jω0 t) an impulse rather than an ordinary function?
   A. The signal is causal
   B. The signal has constant magnitude, is not absolutely integrable, and is a power signal
   C. The signal has finite energy
   D. The signal is real and even
   Answer: B. |x(t)|=1 forever, so the integral does not converge normally. (Key point: Why impulse)

2. How is the FT pair found in the answer?
   A. Direct integration of e^(jω0 t) e^(-jωt) over all t giving a finite value
   B. Using the Laplace transform at s = 0
   C. Guess X = 2π δ(ω - ω0) and check its inverse FT using sifting
   D. Using the Parseval relation
   Answer: C. Inverse FT of 2π δ(ω-ω0) gives e^(jω0 t); FT uniqueness finishes the proof. (Key point: Method)

3. What is the FT of e^(jω0 t)?
   A. δ(ω - ω0)
   B. 2π δ(ω + ω0)
   C. π[δ(ω - ω0) + δ(ω + ω0)]
   D. 2π δ(ω - ω0)
   Answer: D. A single impulse at ω0 with area 2π. (Key point: Result)

4. Why is the spectrum of e^(jω0 t) not symmetric about ω = 0?
   A. Because x(t) is complex
   B. Because x(t) is odd
   C. Because x(t) is periodic
   D. Because the impulse has area 2π
   Answer: A. Only real signals have conjugate-symmetric spectra. (Key point: Discussion)

## Related in Signals and Systems

- [Fourier transform of a CT signal](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Aft-unspecified-ct-signal) — Signals and Systems, chapter 3
- [Time shifting property of CTFT](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Actft-time-shifting) — Signals and Systems, chapter 3, asked 5×
- [CTFT linearity and time shifting](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Actft-linearity-time-shifting) — Signals and Systems, chapter 3
- [Fourier transform of unit step](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Aft-unit-step) — Signals and Systems, chapter 3, asked 3×
- [Fourier transform of CT periodic signals](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Aft-periodic-signal-derivation) — Signals and Systems, chapter 3, asked 3×
- [Derive the CT Fourier transform pair](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Actft-derivation) — Signals and Systems, chapter 3, asked 3×

## Related in other subjects

- [Z-transform of ramp-and-hold continuous signal](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Awaveform-ramp-to-1-at-6) — Digital Control System, chapter 2
- [Fourier transform of a DC signal](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Aft-dc-signal) — Data Communication, chapter 3
- [Properties of continuous-time Fourier series](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Afourier-series-properties) — Data Communication, chapter 3
- [Fourier transform property: Time shifting](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aft-property-time-shifting) — Communication Systems, chapter 1