---
title: "Find x3[n] from X3=X1X2, x1 5-point, x2 4-point"
subject: "Digital Signal Analysis and Processing"
chapter: 7
key: "semester_7/digital_signal_analysis_and_processing:dft-product-x3-73s"
source: https://www.bctengineer.com/q/semester_7/digital_signal_analysis_and_processing/dft-product-x3-73s
site: BCT Engineering
---

# Find x3[n] from X3=X1X2, x1 5-point, x2 4-point

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

- Asked in: 2073 Shrawan, 7 marks
- Video: [Compute the circular convolution using DFT and IDFT method](https://www.youtube.com/watch?v=7aFkjvC3R0M) — Engg-Course-Made-Easy
- 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/dft-product-x3-73s
- Whole chapter: https://www.bctengineer.com/md/semester_7/digital_signal_analysis_and_processing/chapter-7

## Question

Find x3[n] if DFT of x3[n] is given by X3(k) = X1(k) X2(k) where X1(k) and X2(k) are 5-point DFT of x1[n] = {1, −2, 2, 1, 4} and x2[n] = {2, 1, −3, −1} respectively.

## Short answer

Pad x2 to five points, then take the 5-point circular convolution of x1 and x2. The result is x3[n] = {1, -16, -5, 9, 5}.

## Answer

x2[n] has only 4 points; since the DFTs are 5-point, it is padded with one zero. Because X3(k) = X1(k)·X2(k), x3[n] is the 5-point **circular convolution** of x1[n] and x2[n] (multiplication of DFTs ⇔ circular convolution):

- x1[n] = {1, −2, 2, 1, 4}
- x2[n] = {2, 1, −3, −1, 0}

x3[n] = Σ (m = 0 to 4) x1[m] · x2[(n − m) mod 5]

Written as a matrix product (the first column of the circulant is x1, and each later column is the previous one shifted down circularly):

```text
|    1    4    1    2   -2 |   |  2 |
|   -2    1    4    1    2 |   |  1 |
|    2   -2    1    4    1 |   | -3 |
|    1    2   -2    1    4 |   | -1 |
|    4    1    2   -2    1 |   |  0 |
```

Row by row:

- x3[0] = (1)(2) + (4)(1) + (1)(−3) + (2)(−1) + (−2)(0) = **1**
- x3[1] = (−2)(2) + (1)(1) + (4)(−3) + (1)(−1) + (2)(0) = **−16**
- x3[2] = (2)(2) + (−2)(1) + (1)(−3) + (4)(−1) + (1)(0) = **−5**
- x3[3] = (1)(2) + (2)(1) + (−2)(−3) + (1)(−1) + (4)(0) = **9**
- x3[4] = (4)(2) + (1)(1) + (2)(−3) + (−2)(−1) + (1)(0) = **5**

Check using DFTs: the inverse DFT of X1(k)·X2(k), k = 0 … 4, is {1, −16, −5, 9, 5} (Σx3 = −6 = X1(0)·X2(0) = 6 × (−1)).

**Answer: x3[n] = {1, −16, −5, 9, 5}, n = 0 … 4.**

## Check yourself

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

1. X3(k) = X1(k)·X2(k) means x3[n] is which operation on x1 and x2?
   A. Cross-correlation
   B. N-point circular convolution
   C. Linear convolution of unlimited length
   D. Sample-by-sample product
   Answer: B. Multiplication of DFTs in frequency corresponds to circular convolution in time. (Key point: DFT product property)

2. x2[n] has 4 points but the DFTs are 5-point. What is done?
   A. x1 is cut to 4 points
   B. x2 is repeated periodically
   C. N is changed to 4
   D. x2 is padded with one zero to {2,1,-3,-1,0}
   Answer: D. Both sequences must have length N = 5, so one zero is appended to x2. (Key point: Zero padding)

3. Which quick check should hold for the result x3[n]?
   A. x3[0] equals X1(0)
   B. Sum of x3 equals X1(0)·X2(0), the product of the two sequence sums
   C. Number of samples equals the sum of lengths
   D. Sum of x3 is zero
   Answer: B. X3(0) = sum of x3, and X3(0) = X1(0)X2(0). Here the sum is -6. (Key point: Check)

## Close variants set in other exams

- [Find x3[n] from X3=X1X2 (1,-2,5,1,2)](https://www.bctengineer.com/question.md?key=semester_7%2Fdigital_signal_analysis_and_processing%3Adft-product-x3-73c) — Digital Signal Analysis and Processing, chapter 7, asked 3×
- [Find x3[n] from X3=X1X2, 4-point](https://www.bctengineer.com/question.md?key=semester_7%2Fdigital_signal_analysis_and_processing%3Adft-product-x3-69c) — Digital Signal Analysis and Processing, chapter 7, asked 3×

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