---
title: "Find x3[n] from X3=X1X2 (1,2,4)"
subject: "Digital Signal Analysis and Processing"
chapter: 7
key: "semester_7/digital_signal_analysis_and_processing:dft-product-x3-72k"
source: https://www.bctengineer.com/q/semester_7/digital_signal_analysis_and_processing/dft-product-x3-72k
site: BCT Engineering
---

# Find x3[n] from X3=X1X2 (1,2,4)

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

- Asked in: 2072 Kartik, 6 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-x3-72k
- Whole chapter: https://www.bctengineer.com/md/semester_7/digital_signal_analysis_and_processing/chapter-7

## Question

If X1(k) and X2(k) are DFT of sequences x1[n] = {1, 2, 4} and x2[n] = {−1, 2, 3, 1} respectively, then find the sequence x3[n], if DFT of x3[n] is given by X3(k) = X1(k) X2(k).

## Short answer

Use 4-point DFTs, padding x1 with one zero; multiplication corresponds to 4-point circular convolution. The result is x3[n] = {13, 4, 3, 15}.

## Answer

The DFT length is not stated; the smallest size that holds both sequences is N = 4 (the length of x2), so x1[n] is padded with one zero and 4-point DFTs are used. (If a 6-point DFT were used, the result would equal the linear convolution {-1, 0, 3, 15, 14, 4}.) Since X3(k) = X1(k)·X2(k), x3[n] is the 4-point **circular convolution** of x1[n] and x2[n] (multiplication of DFTs ⇔ circular convolution). Pad the shorter sequence(s) with zeros to N = 4:

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

x3[n] = Σ (m = 0 to 3) x1[m] · x2[(n − m) mod 4]

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

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

Row by row:

- x3[0] = (1)(-1) + (0)(2) + (4)(3) + (2)(1) = **13**
- x3[1] = (2)(-1) + (1)(2) + (0)(3) + (4)(1) = **4**
- x3[2] = (4)(-1) + (2)(2) + (1)(3) + (0)(1) = **3**
- x3[3] = (0)(-1) + (4)(2) + (2)(3) + (1)(1) = **15**

Check using DFTs: X1(k)·X2(k) for k = 0 … 3 has the inverse DFT {13, 4, 3, 15} (Σx3 = 35 = X1(0)·X2(0) = 7×5 = 35).

**Answer: x3[n] = {13, 4, 3, 15}, n = 0 … 3.**

## Check yourself

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

1. The DFT length is not stated for x1 (3 points) and x2 (4 points). Which N is chosen?
   A. N = 3
   B. N = 7
   C. N = 4, the smallest size holding both
   D. N = 12
   Answer: C. The smallest size that holds both sequences is the longer length, 4. (Key point: Choice of N)

2. If a 6-point DFT were used instead, what would x3[n] be?
   A. The linear convolution {-1, 0, 3, 15, 14, 4}
   B. Only the first 4 samples of x2
   C. The same as the 4-point result
   D. All zeros
   Answer: A. N = 6 is at least 3 + 4 - 1, so there is no aliasing and the circular result equals the linear convolution. (Key point: Effect of N)

3. How is x1 prepared for the 4-point circular convolution?
   A. Left as 3 points
   B. Repeated to {1,2,4,1}
   C. Truncated to {1, 2}
   D. Padded to {1, 2, 4, 0}
   Answer: D. One zero is appended so both have 4 points. (Key point: Zero padding)

4. Which is the 4-point result x3[n]?
   A. {13, 4, 3, 15}
   B. {-1, 0, 3, 15}
   C. {13, 4, 3, 15, 14, 4}
   D. {4, 13, 15, 3}
   Answer: A. Computed from the circulant: 13, 4, 3, 15; their sum 35 = 7 x 5. (Key point: Result)

## Close variants set in other exams

- [Find x3[n] from X3=X1X2 (5-point)](https://www.bctengineer.com/question.md?key=semester_7%2Fdigital_signal_analysis_and_processing%3Adft-product-x3-74a) — Digital Signal Analysis and Processing, chapter 7, asked 2×

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