---
title: "Fourier transform of CT periodic signals"
subject: "Signals and Systems"
chapter: 3
key: "semester_4/signals_and_systems:ft-periodic-signal-derivation"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/ft-periodic-signal-derivation
site: BCT Engineering
---

# Fourier transform of CT periodic signals

Signals and Systems · Chapter 3: Fourier Transform

- Asked in: 2072 Magh, 5 marks
- Video: [Fourier Transform of Periodic Signals](https://www.youtube.com/watch?v=zkbFHDmOA1o) — Salaar Khan
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/ft-periodic-signal-derivation
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-3

## Question

Derive the expression for Fourier transform of continuous time periodic signal.

## Short answer

Expand the periodic signal in a Fourier series: x(t)=∑ a_k e^{jkω0t}. Since e^{jkω0t} ↔ 2π δ(ω−kω0), linearity gives X(jω)=∑ 2π a_k δ(ω−kω0). Thus the FT is a train of impulses at harmonic frequencies with area 2π a_k.

## Answer

A periodic signal is not absolutely integrable, so its ordinary Fourier integral does not converge. Its Fourier transform is obtained from its Fourier series using impulses.

**Step 1: Fourier series.** Let $x(t)$ be periodic with period $T_0$, $\omega_0 = 2\pi/T_0$:

$$
x(t) = \sum_{k=-\infty}^{\infty} a_k\,e^{jk\omega_0 t}, \qquad a_k = \frac{1}{T_0}\int_{T_0} x(t)\,e^{-jk\omega_0 t}\,dt
$$

**Step 2: FT of one complex exponential.** Consider $X(j\omega) = 2\pi\,\delta(\omega - \omega_0)$. Its inverse FT is

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

by the sifting property. Hence

$$
e^{j\omega_0 t} \;\longleftrightarrow\; 2\pi\,\delta(\omega - \omega_0), \qquad e^{jk\omega_0 t} \;\longleftrightarrow\; 2\pi\,\delta(\omega - k\omega_0)
$$

**Step 3: apply linearity** to the Fourier series term by term:

$$
X(j\omega) = \sum_{k=-\infty}^{\infty} 2\pi\,a_k\,\delta(\omega - k\omega_0)
$$

**Result:** The FT of a periodic signal is a train of impulses located at the harmonic frequencies $k\omega_0$, with area $2\pi a_k$.

**Link with the FT of one period:** If $X_0(j\omega)$ is the FT of one period of $x(t)$, then $a_k = \frac{1}{T_0}X_0(jk\omega_0)$, so $X(j\omega) = \omega_0\sum_k X_0(jk\omega_0)\delta(\omega-k\omega_0)$.

**Examples:**

- $\cos\omega_0 t$: $a_{\pm1} = \frac12$, so $X(j\omega) = \pi[\delta(\omega-\omega_0) + \delta(\omega+\omega_0)]$.
- $\sin\omega_0 t$: $a_{\pm1} = \pm\frac{1}{2j}$, so $X(j\omega) = \frac{\pi}{j}[\delta(\omega-\omega_0) - \delta(\omega+\omega_0)]$.
- Impulse train $\sum_n\delta(t - nT)$: $a_k = \frac1T$, so $X(j\omega) = \frac{2\pi}{T}\sum_k\delta\left(\omega - \frac{2\pi k}{T}\right)$.

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

## Check yourself

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

1. Why can't a periodic signal's FT be found by the ordinary integral?
   A. It is non-causal
   B. It has no Fourier series
   C. It is always complex
   D. It is not absolutely integrable, so the integral does not converge
   Answer: D. Periodic signals have infinite energy, so impulses are needed. (Key point: Why not Fourier integral)

2. Which pair is used for each harmonic?
   A. e^(jkω0t) <-> 2π δ(ω - kω0)
   B. e^(jkω0t) <-> δ(ω - kω0)
   C. e^(jkω0t) <-> 2π δ(ω + kω0)
   D. e^(jkω0t) <-> π δ(ω - kω0)
   Answer: A. Verified by the sifting property in the inverse FT. (Key point: Exponential pair)

3. What is X(jω) for a periodic signal with coefficients a_k?
   A. Σ a_k δ(ω - kω0)
   B. Σ 2π a_k δ(ω - kω0)
   C. Σ a_k/(2π) δ(ω - kω0)
   D. Σ 2π a_k δ(ω - k)
   Answer: B. Linearity applied term by term to the Fourier series. (Key point: Result)

4. What is the FT of cos ω0 t?
   A. 2π[δ(ω-ω0) + δ(ω+ω0)]
   B. π/2 [δ(ω-ω0) + δ(ω+ω0)]
   C. π[δ(ω-ω0) + δ(ω+ω0)]
   D. π[δ(ω-ω0) - δ(ω+ω0)]
   Answer: C. a_{±1} = 1/2, so each impulse has area 2π(1/2) = π. (Key point: Cosine example)

## Close variants set in other exams

- [FT of periodic signals; e^(−2|t|)sin t](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Aft-periodic-damped-sine) — Signals and Systems, chapter 3, asked 2×
- [FT of periodic signals; pulse example](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Aft-periodic-pulse-example) — Signals and Systems, chapter 3, asked 2×

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