---
title: "4-point DFT of {3, −2, 1}"
subject: "Signals and Systems"
chapter: 3
key: "semester_4/signals_and_systems:dft-three-point-sequence"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/dft-three-point-sequence
site: BCT Engineering
---

# 4-point DFT of {3, −2, 1}

Signals and Systems · Chapter 3: Fourier Transform

- Asked in: 2079 Jestha, 6 marks
- Video: [Computation of 4 point DFT (Example 2)](https://www.youtube.com/watch?v=LzVXaVivwAY) — EnggClasses
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/dft-three-point-sequence
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-3

## Question

Compute the 4-point DFT for x[n] = {3, −2, 1}.

## Short answer

Zero-pad to length 4: {3, −2, 1, 0}. Using the 4-point DFT matrix (W₄ = −j) gives X[0]=2, X[1]=2+2j, X[2]=6, X[3]=2−2j. DFT = {2, 2+2j, 6, 2−2j}.

## Answer

The sequence has 3 samples, so for a 4-point DFT it is padded with one zero: $x[n] = \{3, -2, 1, 0\}$ for $n = 0, 1, 2, 3$.

#### Formula
$$
X[k] = \sum_{n=0}^{N-1}x[n]e^{-j2\pi kn/N}, \qquad N = 4, \quad e^{-j\pi kn/2} = (-j)^{kn}
$$
The twiddle factor is $W_4 = e^{-j\pi/2} = -j$, with $W^0 = 1$, $W^1 = -j$, $W^2 = -1$, $W^3 = j$.

#### Matrix form
$$
\begin{bmatrix}X[0]\\X[1]\\X[2]\\X[3]\end{bmatrix} = \begin{bmatrix}1 & 1 & 1 & 1\\ 1 & -j & -1 & j\\ 1 & -1 & 1 & -1\\ 1 & j & -1 & -j\end{bmatrix}\begin{bmatrix}3\\-2\\1\\0\end{bmatrix}
$$

#### Each value
$$
\begin{aligned}
X[0] &= 3 + (-2) + 1 + 0 = 2 \\
X[1] &= 3 + (-2)(-j) + 1(-1) + 0(j) = 3 + 2j - 1 = 2 + 2j \\
X[2] &= 3 + (-2)(-1) + 1(1) + 0(-1) = 3 + 2 + 1 = 6 \\
X[3] &= 3 + (-2)(j) + 1(-1) + 0(-j) = 3 - 2j - 1 = 2 - 2j
\end{aligned}
$$

#### Result
| k | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| X[k] | 2 | 2 + 2j | 6 | 2 − 2j |
| \|X[k]\| | 2 | 2.828 | 6 | 2.828 |
| ∠X[k] | 0 | 45° | 0 | −45° |

**Answer:** $X[k] = \{2,\ 2 + 2j,\ 6,\ 2 - 2j\}$.

**Checks:**
- $X[3] = X^*[1]$, as required for a real sequence.
- $X[0] = \sum x[n] = 2$.
- Parseval: $\sum|x[n]|^2 = 9 + 4 + 1 = 14$, and $\frac{1}{4}(4 + 8 + 36 + 8) = 14$.
- Inverse DFT: $x[0] = \frac{1}{4}\sum X[k] = \frac{1}{4}(2 + 2 + 2j + 6 + 2 - 2j) = 3$. Correct.

## Check yourself

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

1. How is the 3-sample sequence {3, -2, 1} handled for a 4-point DFT?
   A. Repeated periodically to {3, -2, 1, 3}
   B. Padded with one zero: {3, -2, 1, 0}
   C. Truncated to two samples
   D. Padded with its average
   Answer: B. A sequence shorter than N is zero-padded to N samples. (Key point: Zero padding)

2. What is X[1] for {3, -2, 1, 0}?
   A. 2 - 2j
   B. 2 + 2j
   C. 6
   D. 4 + 2j
   Answer: B. X[1] = 3 + (-2)(-j) + 1(-1) = 2 + 2j. (Key point: X[1])

3. The 4-point DFT of {3, -2, 1, 0} is:
   A. {2, 2 + 2j, 6, 2 - 2j}
   B. {2, 2 - 2j, 6, 2 + 2j}
   C. {6, 2 + 2j, 2, 2 - 2j}
   D. {2, 2 + 2j, 2, 2 - 2j}
   Answer: A. X[0]=2, X[1]=2+2j, X[2]=3+2+1=6, X[3]=2-2j. (Key point: Full result)

4. Which check confirms the result for a real sequence?
   A. X[3] = X*[1] and X[0] = sum of samples
   B. X[2] = 0
   C. |X[k]| is the same for all k
   D. X[1] is real
   Answer: A. Real signals give conjugate-symmetric DFTs, and X[0] equals the sum of samples (2 here). (Key point: Verification)

## Related in Signals and Systems

- [4-point DFT of {0, 1, 2, 3}](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Adft-four-point-ramp) — Signals and Systems, chapter 3
- [4-point DFT of {2, 1+j, 1−j}](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Adft-complex-sequence) — Signals and Systems, chapter 3
- [4-point DFT by linear transformation](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Adft-linear-transformation) — Signals and Systems, chapter 3

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