Fast Fourier Transform (FFT) algorithm in Signal Processing - Time & Space Complexity

Time Complexity: Fast Fourier Transform (FFT) algorithm

O(n log n)

Understanding Time Complexity

We want to understand how the time needed to run the FFT algorithm changes as the input signal gets bigger.

Specifically, how does the number of steps grow when we increase the size of the data?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function FFT(x) {
  const N = x.length;
  if (N <= 1) return x;

  const even = FFT(x.filter((_, i) => i % 2 === 0));
  const odd = FFT(x.filter((_, i) => i % 2 === 1));

  const T = [];
  for (let k = 0; k < N / 2; k++) {
    const t = Math.exp(-2 * Math.PI * 1j * k / N) * odd[k];
    T[k] = even[k] + t;
    T[k + N / 2] = even[k] - t;
  }
  return T;
}
    

This code splits the input into even and odd parts, recursively processes them, and combines results to compute the FFT.

Identify Repeating Operations

• Primary operation: Recursive calls splitting the input in half each time.
• How many times: The recursion happens about log₂(n) times because the input size halves each call.
• Loop inside each recursion: A loop runs n/2 times to combine results.

How Execution Grows With Input

Each time we double the input size, the number of steps grows a bit more than double but much less than square.

Input Size (n)	Approx. Operations
10	About 30 steps
100	About 700 steps
1000	About 10,000 steps

Pattern observation: The steps grow roughly like n times log n, which is faster than n squared but slower than just n.

Final Time Complexity

Time Complexity: O(n log n)

This means if you double the input size, the time needed grows a bit more than double but much less than four times.

Common Mistake

[X] Wrong: "FFT runs in linear time O(n) because it just processes the input once."

[OK] Correct: FFT uses recursion and combines results multiple times, so it does more work than just one pass over the data.

Interview Connect

Understanding FFT's time complexity shows you can analyze algorithms that use divide-and-conquer and recursion, a useful skill in many data science and signal processing tasks.

Self-Check

"What if we changed the recursion to split the input into three parts instead of two? How would the time complexity change?"

Practice

1. What is the main purpose of the Fast Fourier Transform (FFT) algorithm in signal processing?

easy

A. To compress the signal data for storage

B. To convert a signal from time domain to frequency domain

C. To filter out noise from the signal

D. To increase the amplitude of the signal

Solution

1. Step 1: Understand FFT role and compare options

   FFT transforms time-domain signal to frequency domain. A matches exactly; B is compression, C filtering, D amplification.

2. Final Answer:

   To convert a signal from time domain to frequency domain -> Option B

3. Quick Check:

   FFT = Time to Frequency conversion [OK]

Hint: FFT changes signal view from time to frequency fast [OK]

Common Mistakes:

• Confusing FFT with noise filtering
• Thinking FFT compresses data
• Assuming FFT amplifies signals

2. Which of the following is the correct way to import the FFT function from NumPy in Python?

easy

A. from numpy import fft

B. import numpy.fft as fft

C. import fft from numpy

D. from numpy.fft import fft

Solution

1. Step 1: Recall and match Python import syntax

   Correct for specific function: 'from numpy.fft import fft' (C). A imports submodule, B aliases submodule (requires fft.fft()), D invalid.

2. Final Answer:

   from numpy.fft import fft -> Option D

3. Quick Check:

   Correct import syntax = from numpy.fft import fft [OK]

Hint: Use 'from module import function' to import FFT [OK]

Common Mistakes:

• Using 'import fft from numpy' which is invalid syntax
• Importing whole numpy.fft as fft causing confusion
• Trying 'from numpy import fft' which imports the module, not function

3. Given the Python code:

import numpy as np
signal = np.array([1, 0, -1, 0])
fft_result = np.fft.fft(signal)
print(np.round(fft_result, 2))

What is the output?

medium

A. [0.+0.j 0.+2.j 0.+0.j 0.-2.j]

B. [0.+0.j 0.+0.j 0.+0.j 0.+0.j]

C. [0.+0.j 2.+0.j 0.+0.j 2.+0.j]

D. [0.+0.j 1.+0.j 0.+0.j 1.+0.j]

Solution

1. Step 1: Compute and verify FFT output

   np.fft.fft([1,0,-1,0]) corresponds to a sine wave component, resulting in imaginary parts at k=1 and k=3. The output is [0.+0.j 0.+2.j 0.+0.j 0.-2.j], matching [0.+0.j 0.+2.j 0.+0.j 0.-2.j].

2. Final Answer:

   [0.+0.j 0.+2.j 0.+0.j 0.-2.j] -> Option A

3. Quick Check:

   FFT sine wave = imaginary components at k=1,3 [OK]

Hint: Sine-like signals yield imaginary FFT components [OK]

Common Mistakes:

• Confusing cosine signal with sine (imaginary output)
• Confusing zeros with non-zero frequency components
• Rounding incorrectly or not at all

4. Consider this Python code snippet:

import numpy as np
signal = [1, 2, 3, 4]
fft_result = fft(signal)
print(fft_result.real)

It raises an error. What is the most likely cause?

medium

A. signal should be a NumPy array, not a list

B. np.fft.fft does not have a real attribute

C. Missing import statement for fft

D. print statement syntax is incorrect

Solution

1. Step 1: Trace code execution for error

   fft(signal) raises NameError: 'fft' undefined. Import is numpy as np (use np.fft.fft); no direct fft import (e.g., from numpy.fft import fft). A, B, D unrelated.

2. Final Answer:

   Missing import statement for fft -> Option C

3. Quick Check:

   Undefined fft = missing import [OK]

Hint: Import fft properly: from numpy.fft import fft [OK]

Common Mistakes:

• Not importing the fft function when using fft()
• Thinking .real is invalid on FFT output
• Confusing print syntax errors

5. You have a noisy signal sampled at 1000 Hz stored in a NumPy array data. You want to remove frequencies above 100 Hz using FFT. Which approach correctly achieves this?

hard

A. Compute FFT, zero out components with frequency > 100 Hz, then inverse FFT

B. Compute FFT, keep only first 100 components, then inverse FFT

C. Apply FFT and multiply all values by 0.1 to reduce noise

D. Use FFT to find max frequency and subtract it from signal

Solution

1. Step 1: Outline correct FFT low-pass filtering

   FFT to freq domain (use fftfreq for bins), zero coeffs where |freq| > 100 (fs=1000, cutoff index ~100/500=0.2*N), ifft back. Matches B precisely; A ignores freq mapping, C/D ineffective.

2. Final Answer:

   Compute FFT, zero out components with frequency > 100 Hz, then inverse FFT -> Option A

3. Quick Check:

   Zero high freq + ifft = low-pass filter [OK]

Hint: Zero unwanted FFT frequencies, then inverse transform [OK]

Common Mistakes:

• Confusing component count with frequency
• Scaling FFT values instead of zeroing
• Misusing FFT to subtract frequencies directly