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.

Practice

1. What is the main purpose of the Short-Time Fourier Transform (STFT) in signal processing?

easy

A. To compress a signal into fewer data points

B. To analyze how frequencies change over time in a signal

C. To remove noise from a signal completely

D. To convert a signal from digital to analog form

Solution

1. Step 1: Understand STFT function

   STFT breaks a signal into small time segments and finds frequency content in each segment.

2. Step 2: Identify main goal

   This helps us see how frequencies change over time, unlike a normal Fourier transform which shows overall frequencies.

3. Final Answer:

   To analyze how frequencies change over time in a signal -> Option B

4. Quick Check:

   STFT = frequency changes over time [OK]

Hint: STFT = frequency analysis in short time windows [OK]

Common Mistakes:

• Confusing STFT with noise removal
• Thinking STFT compresses data
• Mixing STFT with analog conversion

2. Which of the following is the correct Python function call to compute STFT using the scipy library on a signal array named signal with a window size of 256?

easy

A. scipy.stft(signal, nperseg=256)

B. scipy.stft(signal, window=256)

C. scipy.signal.fft(signal, window=256)

D. scipy.signal.stft(signal, nperseg=256)

Solution

1. Step 1: Recall scipy STFT syntax

   The STFT function is in scipy.signal as stft(), and the window size is set by nperseg parameter.

2. Step 2: Match correct syntax

   scipy.signal.stft(signal, nperseg=256) uses scipy.signal.stft(signal, nperseg=256), which is correct syntax.

3. Final Answer:

   scipy.signal.stft(signal, nperseg=256) -> Option D

4. Quick Check:

   Correct function and parameter = scipy.signal.stft(signal, nperseg=256) [OK]

Hint: Use scipy.signal.stft with nperseg for window size [OK]

Common Mistakes:

• Using scipy.stft instead of scipy.signal.stft
• Confusing fft() with stft()
• Passing window size as 'window' instead of 'nperseg'

3. Given the following Python code using scipy.signal.stft, what is the shape of the output array Zxx?

import numpy as np
from scipy.signal import stft
signal = np.sin(2 * np.pi * 5 * np.linspace(0, 1, 100))
f, t, Zxx = stft(signal, nperseg=20)
print(Zxx.shape)

medium

A. (11, 6)

B. (20, 5)

C. (50, 2)

D. (10, 10)

Solution

1. Step 1: Calculate frequency bins

   With nperseg=20, frequency bins = nperseg//2 + 1 = 10 + 1 = 11 rows.

2. Step 2: Calculate time segments

   Signal length=100, window=20, default no overlap means about 6 segments (100/20 rounded up).

3. Final Answer:

   (11, 6) -> Option A

4. Quick Check:

   Zxx shape = (freq bins, time segments) = (11, 6) [OK]

Hint: Output shape = (nperseg/2+1, number of windows) [OK]

Common Mistakes:

• Confusing nperseg with output shape rows
• Miscounting time segments
• Assuming output shape equals input length

4. Identify the error in this STFT code snippet and choose the correct fix:

from scipy.signal import stft
signal = [1, 2, 3, 4, 5]
f, t, Zxx = stft(signal, nperseg=10)
print(Zxx)

medium

A. Error: nperseg is larger than signal length; fix by setting nperseg=5

B. Error: signal must be a numpy array; fix by converting signal to np.array(signal)

C. Error: stft requires window parameter; fix by adding window='hann'

D. No error; code runs fine

Solution

1. Step 1: Check nperseg vs signal length

   Signal length is 5, but nperseg=10 is larger, which causes a warning or error in scipy.

2. Step 2: Fix nperseg value

   Set nperseg to 5 or less to match signal length and avoid error.

3. Final Answer:

   Error: nperseg is larger than signal length; fix by setting nperseg=5 -> Option A

4. Quick Check:

   nperseg ≤ signal length to avoid error [OK]

Hint: nperseg must not exceed signal length [OK]

Common Mistakes:

• Ignoring nperseg size error
• Thinking signal type causes error
• Adding unnecessary window parameter

5. You want to analyze a music clip with fast-changing notes using STFT. Which combination of window size and overlap will give you better time resolution but less frequency resolution?

hard

A. Large window size with high overlap

B. Large window size with low overlap

C. Small window size with high overlap

D. Small window size with low overlap

Solution

1. Step 1: Understand window size effect

   Small window size improves time resolution but reduces frequency resolution.

2. Step 2: Understand overlap effect

   High overlap keeps time segments close, improving time detail further.

3. Final Answer:

   Small window size with high overlap -> Option C

4. Quick Check:

   Better time resolution = small window + high overlap [OK]

Hint: Small window + high overlap = better time detail [OK]

Common Mistakes:

• Choosing large window for better time resolution
• Ignoring overlap effect
• Assuming low overlap improves time resolution