Short-Time Fourier Transform (STFT) in Signal Processing - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of STFT Magnitude Shape

Given a 1-second sine wave sampled at 1024 Hz, what is the shape of the magnitude array returned by the STFT with a window size of 256 and hop length of 128?

Signal Processing

import numpy as np
from scipy.signal import stft
fs = 1024
T = 1.0
t = np.linspace(0, T, int(T*fs), endpoint=False)
x = np.sin(2*np.pi*50*t)
f, t_seg, Zxx = stft(x, fs=fs, nperseg=256, noverlap=128)
result = Zxx.shape

A (128, 8)

B (128, 7)

C (129, 8)

D (129, 7)

💡 Hint

Remember that the number of frequency bins is nperseg//2 + 1 and the number of time segments depends on the signal length, window size, and overlap.

The STFT returns frequency bins equal to nperseg//2 + 1 = 129. The number of time segments is calculated as floor((len(x) - noverlap) / (nperseg - noverlap)) = 7. So the shape is (129, 7).

❓ data_output

intermediate

STFT Frequency Bin Values

What is the frequency value of the 10th frequency bin returned by STFT with sampling frequency 8000 Hz and window size 512?

Signal Processing

import numpy as np
from scipy.signal import stft
fs = 8000
nperseg = 512
freqs = np.fft.rfftfreq(nperseg, 1/fs)
result = freqs[9]

A 140.625

B 156.25

C 125.0

D 160.0

💡 Hint

Frequency bins are spaced by fs/nperseg.

The frequency resolution is fs/nperseg = 8000/512 = 15.625 Hz. The 10th bin (index 9) corresponds to 9 * 15.625 = 140.625 Hz.

❓ visualization

advanced

STFT Spectrogram Interpretation

You compute the STFT of a signal containing two sine waves at 100 Hz and 300 Hz. Which option best describes the expected spectrogram visualization?

Signal Processing

import numpy as np
import matplotlib.pyplot as plt
from scipy.signal import stft
fs = 1000
t = np.linspace(0, 1, fs, endpoint=False)
x = np.sin(2*np.pi*100*t) + np.sin(2*np.pi*300*t)
f, t_seg, Zxx = stft(x, fs=fs, nperseg=256)
plt.pcolormesh(t_seg, f, np.abs(Zxx), shading='gouraud')
plt.ylabel('Frequency [Hz]')
plt.xlabel('Time [sec]')
plt.title('STFT Magnitude')
plt.show()

A A single bright line at 200 Hz representing the average frequency

B Random scattered bright spots with no clear frequency lines

C Two bright horizontal lines at 100 Hz and 300 Hz across all time segments

D Bright vertical lines at 100 Hz and 300 Hz only at the start of the signal

💡 Hint

Each sine wave frequency appears as a horizontal line in the spectrogram.

The STFT spectrogram shows energy at frequencies present in the signal over time. Two sine waves produce two horizontal bright lines at their frequencies.

🧠 Conceptual

advanced

Effect of Window Overlap on STFT

What is the main effect of increasing the overlap between windows in STFT computation?

A Decreases frequency resolution and reduces computational cost

B Improves time resolution but increases computational cost

C Improves frequency resolution but decreases time resolution

D Increases both time and frequency resolution without cost

💡 Hint

More overlap means windows slide less between segments.

Increasing overlap means more windows cover the signal, improving time resolution but requiring more computations.

🔧 Debug

expert

Identifying Error in STFT Parameters

What error will occur if you call STFT with nperseg=1024 on a signal of length 500 samples without zero-padding?

Signal Processing

from scipy.signal import stft
import numpy as np
x = np.random.rand(500)
f, t, Zxx = stft(x, nperseg=1024, padded=False)

A ValueError: nperseg must not be greater than input length

B TypeError: unsupported operand type(s) for -: 'int' and 'NoneType'

C No error, STFT will zero-pad automatically

D IndexError: index out of range

💡 Hint

Check the input length compared to window size.

STFT requires nperseg to be less than or equal to the input length. Otherwise, it raises a ValueError.