---
title: "DIF-FFT butterfly; compute X(2) and X(1)"
subject: "Digital Signal Analysis and Processing"
chapter: 7
key: "semester_7/digital_signal_analysis_and_processing:difft-x2-x1-only"
source: https://www.bctengineer.com/q/semester_7/digital_signal_analysis_and_processing/difft-x2-x1-only
site: BCT Engineering
---

# DIF-FFT butterfly; compute X(2) and X(1)

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

- Asked in: 2074 Chaitra, 7 marks
- Video: [Digital Signal Processing-DIF FFT Algorithm](https://www.youtube.com/watch?v=oLHPKyB8wlo) — All Clear Tube
- 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/difft-x2-x1-only
- Whole chapter: https://www.bctengineer.com/md/semester_7/digital_signal_analysis_and_processing/chapter-7

## Question

Draw the butterfly structure to compute the DFT of the following signal using Radix-2 DIFFFT algorithm, and compute X(2) and X(1) only: x[n] = {1.5, −1, 1.8, 0.6, 3, 1.7}.

## Short answer

Zero-pad to N=8 and apply radix-2 DIF butterflies; outputs appear bit-reversed. X(2)=2.7−j0.1 and X(1)=−3.8335−j0.3151.

## Answer

#### Reading of the question
A radix-2 algorithm needs N = 2^m points, but only six values are given. So I pad with two zeros: x[n] = {1.5, −1, 1.8, 0.6, 3, 1.7, 0, 0}, N = 8.

#### Butterfly structure (8-point DIF)
Basic DIF butterfly (A and B are the two inputs):

```text
 A ──────────(+)──────────► A + B
        ╲   ╱
         ╲ ╱
          ╳
         ╱ ╲
        ╱   ╲
 B ──────────(−)──[× W]───► (A − B)·W
```
- Stage 1 butterflies: (x0,x4) with W₈⁰, (x1,x5) with W₈¹, (x2,x6) with W₈², (x3,x7) with W₈³
- Stage 2 butterflies: positions (0,2) with W₄⁰, (1,3) with W₄¹, (4,6) with W₄⁰, (5,7) with W₄¹
- Stage 3 butterflies: positions (0,1), (2,3), (4,5), (6,7), all with W₂⁰ = 1

In equations:

- Stage 1: a(n) = x(n) + x(n + 4), b(n) = [x(n) − x(n + 4)]·W₈ⁿ, n = 0 … 3
- Stage 2: on each half, c(n) = a(n) + a(n + 2), d(n) = [a(n) − a(n + 2)]·W₄ⁿ, n = 0, 1
- Stage 3: adjacent pairs e = c + c', f = c − c'
- The output is in bit-reversed order: positions 0 … 7 hold X(0), X(4), X(2), X(6), X(1), X(5), X(3), X(7).

#### Stage values
| Position | Input | After stage 1 | After stage 2 | After stage 3 |
|---|---|---|---|---|
| 0 | 1.5 | 4.5 | 6.3 | 7.6 |
| 1 | -1 | 0.7 | 1.3 | 5 |
| 2 | 1.8 | 1.8 | 2.7 | 2.7 − j0.1 |
| 3 | 0.6 | 0.6 | −j0.1 | 2.7 + j0.1 |
| 4 | 3 | -1.5 | −1.5 − j1.8 | −3.8335 − j0.3151 |
| 5 | 1.7 | −1.9092 + j1.9092 | −2.3335 + j1.4849 | 0.8335 − j3.2849 |
| 6 | 0 | −j1.8 | −1.5 + j1.8 | 0.8335 + j3.2849 |
| 7 | 0 | −0.4243 − j0.4243 | 2.3335 + j1.4849 | −3.8335 + j0.3151 |

#### Required outputs
- X(2) is at position 2 (the bit-reversal of 2 = 010 is 010): X(2) = **2.7 − j0.1**
- X(1) is at position 4 (the bit-reversal of 1 = 001 is 100): X(1) = **−3.8335 − j0.3151**

Check: X(2) = Σ x(n)(−j)ⁿ = [x(0) − x(2) + x(4) − x(6)] − j[x(1) − x(3) + x(5) − x(7)] = (1.5 − 1.8 + 3 − 0) − j(−1 − 0.6 + 1.7 − 0) = 2.7 − j0.1, the same as above.

**Answer: X(2) = 2.7 − j0.1, X(1) = −3.8335 − j0.3151** (for the 8-point zero-padded sequence).

## Check yourself

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

1. Six samples are given for a radix-2 algorithm. What is done?
   A. Use radix-3
   B. Drop two samples to get N = 4
   C. Repeat the first two samples
   D. Pad with two zeros to N = 8
   Answer: D. Radix-2 needs N = 2^m, so the sequence is padded to 8 points. (Key point: Padding)

2. How many stages does the 8-point radix-2 DIF-FFT have, and what are the stage pairings?
   A. 3 stages; all pairs adjacent
   B. 8 stages
   C. 3 stages; pairs n and n+4, then n and n+2, then adjacent
   D. 2 stages; pairs n and n+4
   Answer: C. log2(8) = 3 stages with spacing 4, 2, then 1. (Key point: Stages)

3. Where does X(1) appear in the DIF output?
   A. Position 7
   B. Position 4 (bit reversal of 001 is 100)
   C. Position 2
   D. Position 1
   Answer: B. Outputs are bit-reversed; 001 reversed is 100 = 4. (Key point: Locating outputs)

4. What is X(2) for the padded sequence?
   A. 2.7 + j0.1
   B. 2.7 - j0.1
   C. 6.3
   D. -3.8335 - j0.3151
   Answer: B. X(2) sits at position 2 (010 reversed is 010) and equals 2.7 - j0.1. (Key point: Result)

## Related in Digital Signal Analysis and Processing

- [8-point DIF-FFT of {1/2,1/2,1/2,1/2,0,0,0,0}](https://www.bctengineer.com/question.md?key=semester_7%2Fdigital_signal_analysis_and_processing%3Adif-fft-half-ones) — Digital Signal Analysis and Processing, chapter 7
- [8-point DIF-FFT of {1,1,2,2,1,1,2,1}](https://www.bctengineer.com/question.md?key=semester_7%2Fdigital_signal_analysis_and_processing%3Adifft-8pt-69c) — Digital Signal Analysis and Processing, chapter 7
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- [How fast is FFT; X(7) by DIT-FFT](https://www.bctengineer.com/question.md?key=semester_7%2Fdigital_signal_analysis_and_processing%3Afft-speed-x7) — Digital Signal Analysis and Processing, chapter 7, asked 2×

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