---
title: "State and prove Fourier convolution theorem"
subject: "Applied Mathematics"
chapter: 4
key: "semester_4/applied_mathematics:convolution-theorem-fourier-prove"
source: https://www.bctengineer.com/q/semester_4/applied_mathematics/convolution-theorem-fourier-prove
site: BCT Engineering
---

# State and prove Fourier convolution theorem

Applied Mathematics · Chapter 4: Fourier Transform

- Asked in: 2070 Bhadra, 5 marks; 2068 Magh, 5 marks (asked 2 times)
- Video: [Convolution theorem using fourier transform](https://www.youtube.com/watch?v=dSttSBySnJE) — Engineering Mathematics
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/applied_mathematics/convolution-theorem-fourier-prove
- Whole chapter: https://www.bctengineer.com/md/semester_4/applied_mathematics/chapter-4

## Question

State and prove convolution theorem of Fourier transform.

## Short answer

Convolution theorem: F{f*g} = F(s)G(s) where F(s),G(s) are Fourier transforms of f,g. Proof: F{f*g}=∫∫ f(u)g(x-u) e^{-isx} du dx; change variable w=x-u gives inner integral e^{-isu}G(s); integrating over u yields F(s)G(s).

## Answer

#### Definitions

Fourier transform: F{f}(s) = F(s) = ∫_{−∞}^{∞} f(x) e^{−isx} dx.

Convolution: (f * g)(x) = ∫_{−∞}^{∞} f(u) g(x − u) du.

#### Statement

If F(s) and G(s) are the Fourier transforms of f(x) and g(x), then the Fourier transform of their convolution is the product of the transforms:

**F{f * g} = F(s) · G(s)**

Equivalently, f * g = F⁻¹{F(s)G(s)}. (With the symmetric convention F(s) = (1/√(2π)) ∫ f e^{−isx} dx, the result reads F{f * g} = √(2π) F(s)G(s).)

#### Proof

```text
F{f * g} = ∫ e^{−isx} [ ∫ f(u) g(x − u) du ] dx
```

Interchange the order of integration (f and g are absolutely integrable):

```text
F{f * g} = ∫ f(u) [ ∫ g(x − u) e^{−isx} dx ] du
```

In the inner integral put x − u = w, so x = w + u and dx = dw:

```text
∫ g(w) e^{−is(w + u)} dw = e^{−isu} ∫ g(w) e^{−isw} dw = e^{−isu} G(s)
```

Hence

```text
F{f * g} = ∫ f(u) e^{−isu} G(s) du
         = G(s) ∫ f(u) e^{−isu} du
         = F(s) · G(s)      ∎
```

#### Consequence

The inverse form is ∫_{−∞}^{∞} F(s)G(s) e^{isx} ds = 2π (f * g)(x). It lets us compute a convolution from a simple product.

## Check yourself

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

1. What does the convolution theorem state?
   A. F{f*g} = F(s) + G(s)
   B. F{f*g} = F(s) G(s)
   C. F{f*g} = F(s) * G(s) (convolution)
   D. F{fg} = F(s) G(s)
   Answer: B. Convolution in x becomes multiplication in the transform domain. (Key point: Statement)

2. What is the key move in the proof?
   A. Differentiate under the integral
   B. Integrate by parts
   C. Interchange the order of integration
   D. Use Parseval's identity
   Answer: C. This lets the inner integral be done over x first. (Key point: Proof first step)

3. What substitution is used in the inner integral?
   A. w = x + u
   B. w = u - x only with sign change of limits
   C. w = xu
   D. w = x - u
   Answer: D. Then x = w + u, and the exponential splits into e^{-isu} and e^{-isw}. (Key point: Substitution)

## Close variants set in other exams

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