---
title: "Why do we need FFT"
subject: "Digital Signal Analysis and Processing"
chapter: 7
key: "semester_7/digital_signal_analysis_and_processing:why-need-fft"
source: https://www.bctengineer.com/q/semester_7/digital_signal_analysis_and_processing/why-need-fft
site: BCT Engineering
---

# Why do we need FFT

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

- Asked in: 2073 Chaitra, 2 marks; 2069 Chaitra, 2 marks (asked 2 times)
- Video: [What is a Discrete Fourier Transform (DFT) and an FFT?](https://www.youtube.com/watch?v=BPf_pw_1Mg4) — Iain Explains Signals, Systems, and Digital Comms
- 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/why-need-fft
- Whole chapter: https://www.bctengineer.com/md/semester_7/digital_signal_analysis_and_processing/chapter-7

## Question

Why do we need FFT?

## Short answer

The radix-2 FFT exploits symmetry to reduce DFT multiplications from N² to (N/2)log₂N. For N=8 it needs 12 complex multiplications vs 64 directly (≈5.3× faster), making real-time spectrum analysis and fast convolution practical.

## Answer

Computing an N-point DFT directly needs N² complex multiplications and N(N−1) complex additions. For large N (for example 1024 points: about one million multiplications) this is too slow for real-time processing.

The **FFT** exploits the symmetry and periodicity of the twiddle factor W_N^(kn) to split the DFT into smaller DFTs. It needs only (N/2)log₂N complex multiplications and N log₂N additions. For N = 1024 that is 5120 against 1,048,576 multiplications, about 200 times faster. This makes real-time spectrum analysis, fast convolution and filtering, and communications systems practical.

## Check yourself

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

1. How many complex multiplications does direct N-point DFT need?
   A. N log N
   B. N
   C. N^2
   D. 2N
   Answer: C. Each of N outputs sums N terms. (Key point: Direct cost)

2. How many does the FFT need?
   A. N^2/2
   B. N log N^2
   C. N/2
   D. (N/2) log2 N
   Answer: D. Divide-and-conquer reduces the count. (Key point: FFT cost)

3. What does the FFT exploit?
   A. Symmetry and periodicity of the twiddle factor
   B. Linearity only
   C. Real input only
   D. Sparse input
   Answer: A. Properties of W_N^(kn) allow splitting into smaller DFTs. (Key point: Basis)

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