---
title: "State and prove circular convolution property"
subject: "Digital Signal Analysis and Processing"
chapter: 7
key: "semester_7/digital_signal_analysis_and_processing:circular-conv-property-prove"
source: https://www.bctengineer.com/q/semester_7/digital_signal_analysis_and_processing/circular-conv-property-prove
site: BCT Engineering
---

# State and prove circular convolution property

Digital Signal Analysis and Processing · Chapter 7: Discrete Fourier Transform

- Asked in: 2070 Chaitra, 2+4 marks
- Video: [Circular Convolution property of DFT in Digital Signal Processing || EC Academy](https://www.youtube.com/watch?v=jEJ-j3ghJtQ) — EC Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/digital_signal_analysis_and_processing/circular-conv-property-prove
- Whole chapter: https://www.bctengineer.com/md/semester_7/digital_signal_analysis_and_processing/chapter-7

## Question

State and prove the circular convolution property of DFT.

## Short answer

If x₁(n) to X₁(k) and x₂(n) to X₂(k) are N-point DFT pairs, then the N-point circular convolution x₁⊛x₂ has DFT X₁(k)X₂(k). The proof substitutes DFT definitions into the inverse DFT of the product, interchanges sums, and uses orthogonality to obtain the circular sum.

## Answer

#### Statement
If x₁(n) ↔ X₁(k) and x₂(n) ↔ X₂(k) are N-point DFT pairs, then

x₁(n) ⊛ x₂(n) ↔ X₁(k) · X₂(k)

where the N-point circular convolution is x₃(n) = Σ (m = 0 to N−1) x₁(m) x₂((n − m) mod N). Multiplying DFTs in the frequency domain is equivalent to circular convolution in the time domain.

#### Proof
Let X₃(k) = X₁(k)X₂(k). The inverse DFT is

x₃(n) = (1/N) Σ (k = 0 to N−1) X₁(k) X₂(k) e^(j2πkn/N)

Substitute the definitions X₁(k) = Σ_m x₁(m) e^(−j2πkm/N) and X₂(k) = Σ_l x₂(l) e^(−j2πkl/N):

x₃(n) = (1/N) Σ_k Σ_m Σ_l x₁(m) x₂(l) e^(j2πk(n − m − l)/N)

Interchange the order of summation:

x₃(n) = Σ_m Σ_l x₁(m) x₂(l) · [ (1/N) Σ_k e^(j2πk(n − m − l)/N) ]

The bracket is a geometric sum:

(1/N) Σ (k = 0 to N−1) e^(j2πk(n − m − l)/N) = 1 if (n − m − l) is a multiple of N, i.e. l = (n − m) mod N; and 0 otherwise.

So only l = (n − m) mod N survives:

x₃(n) = Σ (m = 0 to N−1) x₁(m) x₂((n − m) mod N)

which is the N-point circular convolution of x₁ and x₂. ∎

#### Remarks
- Both sequences must be padded to the same length N.
- To make circular convolution equal to linear convolution, choose N ≥ L₁ + L₂ − 1.
- The dual property: multiplication x₁(n)x₂(n) ↔ (1/N) X₁(k) ⊛ X₂(k).

## Check yourself

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

1. What does the property state?
   A. Multiplication of DFTs equals linear convolution
   B. Multiplication of N-point DFTs equals circular convolution in time
   C. Addition of DFTs equals convolution
   D. Multiplication in time equals convolution of DFTs only
   Answer: B. x1 (*) x2 <-> X1(k) X2(k). (Key point: Statement)

2. How does the proof begin?
   A. Differentiate both sides
   B. Use Parseval
   C. Take the IDFT of X1(k)X2(k) and substitute the DFT definitions
   D. Assume x1 = x2
   Answer: C. Then the order of sums is interchanged. (Key point: Proof start)

3. Which sum evaluates to 1 or 0?
   A. sum of x(n)
   B. sum of X(k)
   C. sum of |X(k)|^2
   D. (1/N) sum_k e^(j2 pi k(n-m-l)/N)
   Answer: D. It is 1 when n - m - l is a multiple of N. (Key point: Key step)

4. What is the dual property?
   A. x1(n)x2(n) <-> (1/N) X1(k) circular-convolved with X2(k)
   B. x1(n)x2(n) <-> X1(k)X2(k)
   C. x1(n)+x2(n) <-> X1(k)X2(k)
   D. x1(n) conv x2(n) <-> (1/N) X1 X2
   Answer: A. Multiplication in time is circular convolution in frequency with 1/N. (Key point: Dual)

## Related in Digital Signal Analysis and Processing

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