---
title: "Convolution theorem for FT with importance"
subject: "Applied Mathematics"
chapter: 4
key: "semester_4/applied_mathematics:convolution-theorem-fourier-importance"
source: https://www.bctengineer.com/q/semester_4/applied_mathematics/convolution-theorem-fourier-importance
site: BCT Engineering
---

# Convolution theorem for FT with importance

Applied Mathematics · Chapter 4: Fourier Transform

- Asked in: 2073 Bhadra, 2+3 marks
- Video: [Convolution and the Fourier Transform explained visually](https://www.youtube.com/watch?v=9i6aDdQ9FTQ) — Mark Newman
- 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-importance
- Whole chapter: https://www.bctengineer.com/md/semester_4/applied_mathematics/chapter-4

## Question

State convolution theorem for Fourier transform. Give its importance with suitable example.

## Short answer

If F(s) and G(s) are transforms, F{f*g}=F(s)G(s). It turns convolution into multiplication, key for linear systems filtering and fast FFT computation. Example: for f=g=e^{-x²}, both sides give π e^{-s²/2}, showing convolution of two Gaussians is a Gaussian.

## Answer

#### Statement

If F(s) and G(s) are the Fourier transforms of f(x) and g(x), then the Fourier transform of the convolution (f * g)(x) = ∫ f(u) g(x − u) du is the product:

**F{f * g} = F(s) · G(s)**

(With the 1/√(2π) normalisation the right side carries an extra factor √(2π).)

#### Importance

1. **Convolution becomes multiplication.** Direct convolution needs an integral for every x. In the frequency domain it needs only a product, which is far easier.
2. **Linear systems and filters.** The output of a linear time-invariant system is y = h * x (h is the impulse response). By the theorem Y(s) = H(s)X(s), so the system is described by its frequency response H(s). Filtering in signal processing is done this way.
3. **Fast computation.** With the FFT, convolution of long sequences costs O(N log N) instead of O(N²).
4. **Solving equations.** Integral equations of convolution type and some differential equations are reduced to algebraic ones.
5. **Inverse transforms.** The inverse of a product F(s)G(s) can be obtained as the convolution f * g.

#### Example

Let f(x) = g(x) = e^{−x²}. Then F(s) = G(s) = √π e^{−s²/4}, so

F(s)G(s) = π e^{−s²/2}

The inverse transform of this product is f * g = √(π/2) e^{−x²/2}. The same result follows from the direct integral ∫ e^{−u²} e^{−(x−u)²} du, but the frequency-domain route is much shorter. It also shows that the convolution of two Gaussians is again a Gaussian.

## Check yourself

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

1. Why is the theorem useful?
   A. Multiplication becomes convolution
   B. It removes the need for transforms
   C. Convolution in x becomes multiplication in the frequency domain
   D. It makes functions periodic
   Answer: C. A product is easier than an integral. (Key point: Main idea)

2. In an LTI system y = h * x. What is Y(s)?
   A. H(s) + X(s)
   B. H(s)/X(s)
   C. H(s) * X(s)
   D. H(s) X(s)
   Answer: D. The system is described by frequency response. (Key point: LTI systems)

3. How does FFT help?
   A. Convolution costs O(N log N) instead of O(N^2)
   B. O(N) instead of O(N^2)
   C. O(N^3) instead of O(N)
   D. No change
   Answer: A. Via FFT, multiply, inverse FFT. (Key point: Fast computation)

4. What is the convolution of e^{-x^2} with itself?
   A. e^{-x^2}
   B. sqrt(pi/2) e^{-x^2/2}
   C. pi e^{-x^2/2}
   D. sqrt(pi) e^{-x^2/4}
   Answer: B. Convolution of Gaussians is a Gaussian. (Key point: Example)

## Close variants set in other exams

- [State and prove Fourier convolution theorem](https://www.bctengineer.com/question.md?key=semester_4%2Fapplied_mathematics%3Aconvolution-theorem-fourier-prove) — Applied Mathematics, chapter 4, asked 3×

## Related in Applied Mathematics

- [Convolution theorem for z-transform](https://www.bctengineer.com/question.md?key=semester_4%2Fapplied_mathematics%3Aconvolution-theorem-z-transform) — Applied Mathematics, chapter 2
- [Verify convolution theorem, f = g = e^{-x²}](https://www.bctengineer.com/question.md?key=semester_4%2Fapplied_mathematics%3Aconvolution-fourier-verify-e-x2) — Applied Mathematics, chapter 4
- [Complex Fourier series; verify convolution](https://www.bctengineer.com/question.md?key=semester_4%2Fapplied_mathematics%3Afourier-series-complex-convolution) — Applied Mathematics, chapter 4
- [Fourier transform of 1-x²; evaluate integral](https://www.bctengineer.com/question.md?key=semester_4%2Fapplied_mathematics%3Afourier-transform-1-x2-evaluate) — Applied Mathematics, chapter 4, asked 2×
- [Define z-transform; differ from Fourier transform](https://www.bctengineer.com/question.md?key=semester_4%2Fapplied_mathematics%3Az-transform-vs-fourier) — Applied Mathematics, chapter 2
- [Fourier transform vs Fourier series](https://www.bctengineer.com/question.md?key=semester_4%2Fapplied_mathematics%3Afourier-transform-vs-series) — Applied Mathematics, chapter 4

## Related in other subjects

- [Fourier transform property: Convolution](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aft-property-convolution) — Communication Systems, chapter 1
- [Fourier transform properties; Rayleigh energy theorem](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Afourier-properties-rayleigh) — Communication Systems, chapter 1
- [Convolution theorem and inverse Laplace](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Aconvolution-theorem-and-inverse-laplace) — Mathematics III, chapter 3, asked 3×
- [Convolution theorem and inverse Laplace](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Aconvolution-theorem-and-inverse-laplace-two) — Mathematics III, chapter 3, asked 3×