---
title: "Circular convolution of {1,2} and u[n]−u[n−4]"
subject: "Digital Signal Analysis and Processing"
chapter: 7
key: "semester_7/digital_signal_analysis_and_processing:circular-conv-1-2-step"
source: https://www.bctengineer.com/q/semester_7/digital_signal_analysis_and_processing/circular-conv-1-2-step
site: BCT Engineering
---

# Circular convolution of {1,2} and u[n]−u[n−4]

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

- Asked in: 2071 Chaitra, 5 marks; 2070 Asar, 5 marks (asked 2 times)
- Video: [Q2.a Circular Convolution using Time domain (Example 4) | DSP](https://www.youtube.com/watch?v=SU5Q-QT0-Eo) — EnggClasses
- 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-1-2-step
- Whole chapter: https://www.bctengineer.com/md/semester_7/digital_signal_analysis_and_processing/chapter-7

## Question

Find circular convolution between x[n] = {1, 2} and y[n] = u[n] − u[n−4].

## Short answer

Pad x[n]={1,2} to length N=4 with zeros and y[n]=[1,1,1,1]. Since y[n] is all ones, each output sample equals the sum of x: 1+2+0+0=3. The 4-point circular convolution is {3,3,3,3}.

## Answer

y[n] = u[n] − u[n−4] = {1, 1, 1, 1} (n = 0 … 3) and x[n] = {1, 2}. The longer sequence has 4 points, so take the circular convolution with **N = 4**; pad x[n] to {1, 2, 0, 0}.

x₃[n] = Σ (m = 0 to 3) x[m] y[(n − m) mod 4]

Since y[n] = 1 for all four indices, y[(n − m) mod 4] = 1 for every n and m. Hence

x₃[n] = x[0] + x[1] + x[2] + x[3] = 1 + 2 + 0 + 0 = 3, for every n.

Matrix method (circulant of x, columns shifted):

```text
 | 1 0 0 2 |   | 1 |   | 3 |
 | 2 1 0 0 | · | 1 | = | 3 |
 | 0 2 1 0 |   | 1 |   | 3 |
 | 0 0 2 1 |   | 1 |   | 3 |
```
Check by DFT: Y(k) = 4δ(k) (the sum of y), X(0) = 3; so X(k)Y(k) = 12 δ(k), whose inverse DFT is 12/4 = 3 for all n.

**Answer: x₃[n] = {3, 3, 3, 3}** (4-point circular convolution).

(The linear convolution is {1, 3, 3, 3, 2}; the wrap-around of the last sample, 2, onto n = 0 gives 1 + 2 = 3.)

## Check yourself

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

1. What is y[n] = u[n] - u[n-4]?
   A. {1, 0, 0, 0}
   B. {1, 1, 1, 1}
   C. {1, 2, 3, 4}
   D. {0, 1, 1, 1}
   Answer: B. A rectangular pulse of length 4. (Key point: Sequence y)

2. What N is used and how is x padded?
   A. N = 2
   B. N = 5
   C. N = 4, x = {1, 2, 0, 0}
   D. N = 3, no padding
   Answer: C. Both sequences must have the same length N. (Key point: Length)

3. Why is every output equal?
   A. x is symmetric
   B. The DFT is 1
   C. y is zero
   D. y is 1 at all indices, so each output is the sum of x = 3
   Answer: D. Circular shift of an all-ones sequence is unchanged. (Key point: Why constant)

4. What is the 4-point circular convolution?
   A. {3, 3, 3, 3}
   B. {1, 3, 3, 3, 2}
   C. {3, 3, 3, 2}
   D. {1, 2, 0, 0}
   Answer: A. The last linear sample 2 wraps onto n = 0 giving 1 + 2 = 3. (Key point: Result)

## Related in Digital Signal Analysis and Processing

- [State and prove circular convolution property](https://www.bctengineer.com/question.md?key=semester_7%2Fdigital_signal_analysis_and_processing%3Acircular-conv-property-prove) — Digital Signal Analysis and Processing, chapter 7
- [Circular convolution of h and x, 5 and 4 points](https://www.bctengineer.com/question.md?key=semester_7%2Fdigital_signal_analysis_and_processing%3Acircular-conv-5-and-4) — Digital Signal Analysis and Processing, chapter 7
- [Linear convolution via circular convolution](https://www.bctengineer.com/question.md?key=semester_7%2Fdigital_signal_analysis_and_processing%3Alinear-conv-via-circular) — Digital Signal Analysis and Processing, chapter 7

## Related in other subjects

- [Circular convolution of two sequences](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Acircular-convolution) — Signals and Systems, chapter 3