Spectral leakage concept in Signal Processing - Practice Problems & Coding Challenges

Challenge - 5 Problems

🧠 Conceptual

intermediate

What causes spectral leakage in signal processing?

Which of the following best explains the cause of spectral leakage when analyzing signals using the Fourier Transform?

A Sampling a signal at a rate higher than the Nyquist frequency

B Applying a Fourier Transform to a perfectly periodic infinite signal

C Using a window function that perfectly matches the signal frequency

D Using a finite-length window on a signal that is not periodic within that window

💡 Hint

Think about what happens when you cut a signal into a short segment before transforming it.

Spectral leakage happens because the Fourier Transform assumes the signal is periodic within the window. If the signal is not exactly periodic in the window length, the discontinuities cause energy to spread to other frequencies.

❓ Predict Output

intermediate

Output of FFT magnitude with and without windowing

Given a pure sine wave sampled and transformed using FFT, what is the effect of applying a rectangular window versus a Hann window on the FFT magnitude?

Consider the code below:

import numpy as np
import matplotlib.pyplot as plt

fs = 1000
f = 50
N = 1000

t = np.arange(N) / fs
signal = np.sin(2 * np.pi * f * t)

fft_rect = np.fft.fft(signal)
window = np.hanning(N)
fft_hann = np.fft.fft(signal * window)

mag_rect = np.abs(fft_rect)[:N//2]
mag_hann = np.abs(fft_hann)[:N//2]

peak_rect = np.argmax(mag_rect)
peak_hann = np.argmax(mag_hann)

print(peak_rect, peak_hann)

A Both peak_rect and peak_hann equal 50

B peak_rect equals 50, peak_hann equals 49

C peak_rect equals 50, peak_hann equals 50

D peak_rect equals 49, peak_hann equals 50

💡 Hint

Consider that the FFT bin index corresponds to frequency bins. The window affects leakage but not the peak frequency bin.

The peak frequency bin corresponds to the sine wave frequency. Both rectangular and Hann windows keep the peak at the correct bin (50). The Hann window reduces leakage but does not shift the peak.

❓ data_output

advanced

Number of significant FFT bins after windowing

After applying a rectangular window and a Hann window to the same signal and computing their FFT magnitudes, how does the number of bins with magnitude above 10% of the peak compare?

Signal Processing

import numpy as np
fs = 1000
f = 50
N = 1000
t = np.arange(N) / fs
signal = np.sin(2 * np.pi * f * t)

fft_rect = np.fft.fft(signal)
window = np.hanning(N)
fft_hann = np.fft.fft(signal * window)

mag_rect = np.abs(fft_rect)[:N//2]
mag_hann = np.abs(fft_hann)[:N//2]

threshold_rect = 0.1 * np.max(mag_rect)
threshold_hann = 0.1 * np.max(mag_hann)

count_rect = np.sum(mag_rect > threshold_rect)
count_hann = np.sum(mag_hann > threshold_hann)

print(count_rect, count_hann)

A count_rect equals count_hann

B count_rect is larger than count_hann

C count_rect is smaller than count_hann

D count_rect is zero, count_hann is nonzero

💡 Hint

Think about how windowing affects the spread of energy in the frequency domain.

The rectangular window causes more spectral leakage, spreading energy across more bins above the threshold. The Hann window reduces leakage, so fewer bins exceed the threshold.

🔧 Debug

advanced

Identify the error causing unexpected spectral leakage

Consider this code snippet that applies a window before FFT but shows unexpected spectral leakage:

import numpy as np
fs = 1000
f = 50
N = 1000
t = np.arange(N) / fs
signal = np.sin(2 * np.pi * f * t)
window = np.hanning(N)
fft_result = np.fft.fft(signal)
windowed_fft = fft_result * window
mag = np.abs(windowed_fft)[:N//2]
print(np.argmax(mag))

What is the main error causing incorrect spectral leakage behavior?

A The window is applied after the FFT instead of before

B The frequency f is not an integer

C The signal length N is too small for FFT

D The FFT is computed on the window instead of the signal

💡 Hint

Consider when the window should be applied relative to the FFT.

The window must be applied to the time-domain signal before computing the FFT. Applying it after the FFT multiplies frequency components incorrectly, causing wrong spectral leakage behavior.

🚀 Application

expert

Choosing window length to minimize spectral leakage

You have a signal with a frequency of 123.4 Hz sampled at 1000 Hz. You want to analyze it with FFT to minimize spectral leakage. Which window length (N) choice will best reduce leakage?

A N = 811 samples, which is an integer multiple of the signal period

B N = 1000 samples (1 second), which is not an integer multiple of the signal period

C N = 1000 samples with a rectangular window

D N = 1000 samples with zero padding to 2048 samples

💡 Hint

Think about how window length relates to the signal period and periodicity assumptions in FFT.

Choosing a window length that is an integer multiple of the signal period makes the signal appear periodic within the window, minimizing discontinuities and spectral leakage.