---
title: "Circular convolution of two sequences"
subject: "Signals and Systems"
chapter: 3
key: "semester_4/signals_and_systems:circular-convolution"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/circular-convolution
site: BCT Engineering
---

# Circular convolution of two sequences

Signals and Systems · Chapter 3: Fourier Transform

- Asked in: 2069 Bhadra, 5 marks
- Video: [Periodic or Circular Convolution](https://www.youtube.com/watch?v=A6b3UkraTgw) — TutorialsPoint
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/circular-convolution
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-3

## Question

Find circular convolution of the signal x[n] = {1, 0, 0, 1} (x[0] = 1, first sample) and y[n] = {2, 0, 2} (y[0] = 2, first sample).

## Short answer

Pad y to N=4: {2,0,2,0}. Circular convolution of {1,0,0,1} and {2,0,2,0} yields z[n] = {2, 2, 2, 2}.

## Answer

$x[n]$ has 4 samples and $y[n]$ has 3, so use $N = 4$ and pad $y[n]$ with one zero:

$$
x[n] = \{1, 0, 0, 1\}, \qquad y[n] = \{2, 0, 2, 0\}
$$

**Formula:**

$$
z[n] = x[n] \circledast y[n] = \sum_{m=0}^{3}x[m]\,y[(n-m)_4], \quad n = 0,1,2,3
$$

**Matrix (circulant) method:** columns of the matrix are circular shifts of $y[n]$:

$$
\begin{bmatrix} z[0] \\ z[1] \\ z[2] \\ z[3] \end{bmatrix}
=
\begin{bmatrix} 2 & 0 & 2 & 0 \\ 0 & 2 & 0 & 2 \\ 2 & 0 & 2 & 0 \\ 0 & 2 & 0 & 2 \end{bmatrix}
\begin{bmatrix} 1 \\ 0 \\ 0 \\ 1 \end{bmatrix}
$$

**Each sample:**

$$
\begin{aligned}
z[0] &= x[0]y[0] + x[1]y[3] + x[2]y[2] + x[3]y[1] = 2 + 0 + 0 + 0 = 2 \\
z[1] &= x[0]y[1] + x[1]y[0] + x[2]y[3] + x[3]y[2] = 0 + 0 + 0 + 2 = 2 \\
z[2] &= x[0]y[2] + x[1]y[1] + x[2]y[0] + x[3]y[3] = 2 + 0 + 0 + 0 = 2 \\
z[3] &= x[0]y[3] + x[1]y[2] + x[2]y[1] + x[3]y[0] = 0 + 0 + 0 + 2 = 2
\end{aligned}
$$

**Answer:** $z[n] = x[n] \circledast y[n] = \{2, 2, 2, 2\}$

**Check with linear convolution:** $x * y = \{2, 0, 2, 2, 0, 2\}$ (length 6). Wrapping samples $n = 4, 5$ onto $n = 0, 1$ (time aliasing for $N = 4$): $\{2+0,\; 0+2,\; 2,\; 2\} = \{2,2,2,2\}$. Also, the sum of $z$ (8) equals sum of $x$ (2) times sum of $y$ (4).

## Check yourself

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

1. x[n] has 4 samples and y[n] has 3. What N is used for the circular convolution and what is done to y[n]?
   A. N = 3; drop one sample of x[n]
   B. N = 4; pad y[n] with one zero
   C. N = 7; pad both signals
   D. N = 6; pad y[n] with three zeros
   Answer: B. N is the larger length (4), so y[n] becomes {2,0,2,0}. (Key point: Choice of N)

2. Which formula gives the N-point circular convolution?
   A. z[n] = sum over m of x[m] y[n-m] with no wrap-around
   B. z[n] = x[n] y[n] sample by sample
   C. z[n] = sum over m of x[m] y[(n-m) mod N]
   D. z[n] = sum over m of x[m] y[(n+m) mod N]
   Answer: C. The index (n-m) is taken modulo N, so the shifted y wraps around. (Key point: Circular formula)

3. What is x[n] circularly convolved with y[n] for x = {1,0,0,1}, y = {2,0,2,0}?
   A. {2, 0, 2, 2}
   B. {2, 0, 2, 2, 0, 2}
   C. {4, 0, 4, 0}
   D. {2, 2, 2, 2}
   Answer: D. z[0]=2, z[1]=2, z[2]=2, z[3]=2 from the circulant matrix product. (Key point: Result)

4. The linear convolution is {2,0,2,2,0,2}. How does it give the N = 4 circular result?
   A. Samples n = 4, 5 wrap onto n = 0, 1 and are added
   B. Samples n = 4, 5 are simply discarded
   C. The sequence is divided by 4
   D. Samples n = 4, 5 are subtracted from n = 2, 3
   Answer: A. Time aliasing adds the tail onto the start: {2+0, 0+2, 2, 2} = {2,2,2,2}. (Key point: Check by aliasing)

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## Related in other subjects

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