---
title: "DIT-FFT of {1,1,2,0,1,2,0,1} and spectrum"
subject: "Digital Signal Analysis and Processing"
chapter: 7
key: "semester_7/digital_signal_analysis_and_processing:dit-fft-spectrum-71s"
source: https://www.bctengineer.com/q/semester_7/digital_signal_analysis_and_processing/dit-fft-spectrum-71s
site: BCT Engineering
---

# DIT-FFT of {1,1,2,0,1,2,0,1} and spectrum

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

- Asked in: 2071 Shrawan, 6+2 marks
- Video: [8-Point DFT Solved using Radix-2 DIT FFT Algorithm | Digital Signal Processing Problem Tutorial](https://www.youtube.com/watch?v=9BCirp2Ta6g) — EC Academy
- 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/dit-fft-spectrum-71s
- Whole chapter: https://www.bctengineer.com/md/semester_7/digital_signal_analysis_and_processing/chapter-7

## Question

Find DFT for {1, 1, 2, 0, 1, 2, 0, 1} using FFT DIT butterfly algorithm and plot the spectrum.

## Short answer

Use radix-2 DIT FFT. X(k)={8, −j0.5858, −j2, j3.4142, 0, −j3.4142, j2, j0.5858}; magnitude is symmetric about k=4 and phase is antisymmetric.

## Answer

#### Method: radix-2 decimation in time (N = 8, 3 stages)
Input in bit-reversed order: x(0), x(4), x(2), x(6), x(1), x(5), x(3), x(7). Each butterfly: A' = A + W·B, B' = A − W·B, with W₈⁰ = 1, W₈¹ = 0.7071 − j0.7071, W₈² = −j, W₈³ = −0.7071 − j0.7071. Stage 1 uses W₂ (= ±1), stage 2 uses W₄⁰ = 1 and W₄¹ = −j, stage 3 uses W₈⁰ … W₈³.

x(n) = {1, 1, 2, 0, 1, 2, 0, 1}

| Position | Input (bit-reversed) | After stage 1 | After stage 2 | After stage 3 = X(k) |
|---|---|---|---|---|
| 0 | 1 | 2 | 4 | 8 |
| 1 | 1 | 0 | −j2 | −j0.5858 |
| 2 | 2 | 2 | 0 | −j2 |
| 3 | 0 | 2 | j2 | j3.4142 |
| 4 | 1 | 3 | 4 | 0 |
| 5 | 2 | -1 | −1 + j1 | −j3.4142 |
| 6 | 0 | 1 | 2 | j2 |
| 7 | 1 | -1 | −1 − j1 | j0.5858 |


**Result:**

```text
X(0) = 8
X(1) = −j0.5858
X(2) = −j2
X(3) = j3.4142
X(4) = 0
X(5) = −j3.4142
X(6) = j2
X(7) = j0.5858
```

Check: Σx(n) = 8 = X(0); the result equals the direct 8-point DFT.

#### Spectrum

| k | Magnitude | Phase (deg) |
|---|---|---|
| 0 | 8.0000 | 0.00 |
| 1 | 0.5858 | -90.00 |
| 2 | 2.0000 | -90.00 |
| 3 | 3.4142 | 90.00 |
| 4 | 0.0000 | 0.00 |
| 5 | 3.4142 | -90.00 |
| 6 | 2.0000 | 90.00 |
| 7 | 0.5858 | 90.00 |

Magnitude spectrum (stem plot, one # ≈ 0.5):

```text
k=0 |################ 8.00
k=1 |# 0.59
k=2 |#### 2.00
k=3 |####### 3.41
k=4 | 0.00
k=5 |####### 3.41
k=6 |#### 2.00
k=7 |# 0.59
```
The magnitude is symmetric about k = 4 (|X(k)| = |X(8 − k)|) and the phase is anti-symmetric because x(n) is real.

**Answer: X(k) = {8, −j0.5858, −j2, j3.4142, 0, −j3.4142, j2, j0.5858}.**

## Check yourself

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

1. What order is the input placed in for DIT?
   A. Bit-reversed
   B. Natural
   C. Descending
   D. Even only
   Answer: A. x(0), x(4), x(2), x(6), x(1), x(5), x(3), x(7). (Key point: DIT input order)

2. What is W8^1?
   A. -j
   B. 0.7071 - j0.7071
   C. 0.7071 + j0.7071
   D. 1
   Answer: B. W8^1 = e^(-j pi/4). (Key point: Twiddle W8^1)

3. For a real input, what is true of the spectrum?
   A. Magnitude anti-symmetric
   B. Both are zero beyond k = 4
   C. Phase symmetric
   D. Magnitude symmetric about k = 4 and phase anti-symmetric
   Answer: D. |X(k)| = |X(8 - k)| for real x(n). (Key point: Spectrum symmetry)

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