Spectrogram generation in SciPy - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of spectrogram shape

What is the shape of the spectrogram array generated by this code snippet?

SciPy

import numpy as np
from scipy.signal import spectrogram
fs = 1000
x = np.sin(2 * np.pi * 50 * np.linspace(0, 1, fs))
f, t, Sxx = spectrogram(x, fs)
print(Sxx.shape)

A (129, 10)

B (128, 10)

C (130, 10)

D (129, 9)

💡 Hint

Think about how the spectrogram divides the signal and the default window size.

The default window length is 256 samples, so the frequency bins are 129 (NFFT/2 + 1). The time bins depend on the overlap and signal length, resulting in 10 segments.

❓ data_output

intermediate

Spectrogram frequency bins values

Given this code, what is the first frequency bin value printed?

SciPy

import numpy as np
from scipy.signal import spectrogram
fs = 500
x = np.random.randn(fs)
f, t, Sxx = spectrogram(x, fs, nperseg=100)
print(round(f[0], 2))

A 0.0

B 5.0

C 2.5

D 50.0

💡 Hint

Frequency bins start from zero frequency (DC component).

The first frequency bin is always 0.0 Hz representing the DC component of the signal.

❓ visualization

advanced

Spectrogram visualization output

What will the color intensity represent in the spectrogram plot generated by this code?

SciPy

import numpy as np
import matplotlib.pyplot as plt
from scipy.signal import spectrogram
fs = 1000
t = np.linspace(0, 1, fs)
x = np.sin(2 * np.pi * 100 * t) + np.sin(2 * np.pi * 200 * t)
f, t, Sxx = spectrogram(x, fs)
plt.pcolormesh(t, f, 10 * np.log10(Sxx))
plt.ylabel('Frequency [Hz]')
plt.xlabel('Time [sec]')
plt.title('Spectrogram')
plt.colorbar(label='Intensity [dB]')
plt.show()

A Amplitude of the signal at each frequency and time

B Raw signal values over time

C Phase angle of the signal at each frequency and time

D Power spectral density in decibels at each frequency and time

💡 Hint

The plot uses 10 * log10 of Sxx values.

The spectrogram shows power spectral density (PSD) which is the power of each frequency component over time. The 10 * log10 converts it to decibels (dB) for better visualization.

🔧 Debug

advanced

Identify error in spectrogram call

What error will this code produce and why?

SciPy

import numpy as np
from scipy.signal import spectrogram
fs = 1000
x = np.sin(2 * np.pi * 50 * np.linspace(0, 1, fs))
f, t, Sxx = spectrogram(x, fs, nperseg=2000)
print(Sxx.shape)

A IndexError: index out of range

B TypeError: spectrogram() got an unexpected keyword argument 'nperseg'

C ValueError: nperseg must not be greater than input length

D No error, prints shape (1001, 1)

💡 Hint

Check the length of the input signal vs nperseg parameter.

The input signal length is 1000 samples but nperseg is set to 2000, which is invalid and causes a ValueError.

🚀 Application

expert

Detecting frequency components from spectrogram data

Given a spectrogram output Sxx and frequency array f, which code snippet correctly identifies the dominant frequency at each time slice?

SciPy

import numpy as np
from scipy.signal import spectrogram
fs = 1000
t = np.linspace(0, 1, fs)
x = np.sin(2 * np.pi * 150 * t) + np.sin(2 * np.pi * 300 * t)
f, t, Sxx = spectrogram(x, fs)
# Identify dominant frequency code here

A dominant_freqs = np.max(Sxx, axis=1)

B dominant_freqs = f[np.argmax(Sxx, axis=0)]

C dominant_freqs = t[np.argmax(Sxx, axis=1)]

D dominant_freqs = f[np.argmax(Sxx, axis=1)]

💡 Hint

Find the frequency index with max power for each time column.

Sxx shape is (frequency_bins, time_bins). To find dominant frequency per time, find max along frequency axis (axis=0) for each time slice, then map indices to frequencies.

Practice

1. What does the scipy.signal.spectrogram function primarily compute from a signal?

easy

A. The frequency content of the signal over time

B. The total energy of the signal

C. The average amplitude of the signal

D. The phase shift of the signal

Solution

1. Step 1: Understand the purpose of spectrogram

   A spectrogram shows how frequencies in a signal change over time, so it computes frequency content over time.

2. Step 2: Compare with other options

   Total energy, average amplitude, and phase shift are different signal properties, not what spectrogram calculates.

3. Final Answer:

   The frequency content of the signal over time -> Option A

4. Quick Check:

   Spectrogram = frequency over time [OK]

Hint: Spectrogram = frequency changes over time [OK]

Common Mistakes:

• Confusing spectrogram with total energy calculation
• Thinking it returns average amplitude only
• Assuming it gives phase information

2. Which of the following is the correct way to import the spectrogram function from scipy?

easy

A. from scipy import spectrogram

B. import scipy.spectrogram

C. from scipy.signal import spectrogram

D. import spectrogram from scipy.signal

Solution

1. Step 1: Recall correct import syntax for functions

   Functions inside submodules are imported using 'from module.submodule import function'.

2. Step 2: Check each option

   from scipy.signal import spectrogram matches this syntax exactly. Options B, C, and D are invalid Python import statements.

3. Final Answer:

   from scipy.signal import spectrogram -> Option C

4. Quick Check:

   Correct import syntax = from scipy.signal import spectrogram [OK]

Hint: Use 'from scipy.signal import spectrogram' [OK]

Common Mistakes:

• Using 'import scipy.spectrogram' which is invalid
• Trying 'from scipy import spectrogram' ignoring submodule
• Incorrect import order like 'import spectrogram from ...'

3. What will be the shape of the frequency array returned by f, t, Sxx = spectrogram(signal, fs=1000, nperseg=256) if signal length is 1024?

medium

A. (256,)

B. (129,)

C. (512,)

D. (1024,)

Solution

1. Step 1: Understand frequency bins from nperseg

   The frequency array length is nperseg/2 + 1 due to FFT symmetry, so 256/2 + 1 = 129.

2. Step 2: Confirm other options

   256 is the segment length, not frequency bins. 512 and 1024 are unrelated to frequency array size here.

3. Final Answer:

   (129,) -> Option B

4. Quick Check:

   Frequency bins = nperseg/2 + 1 = 129 [OK]

Hint: Frequency bins = nperseg/2 + 1 [OK]

Common Mistakes:

• Assuming frequency array length equals nperseg
• Confusing signal length with frequency bins
• Ignoring FFT symmetry in frequency output

4. Given this code snippet, what is the error?

from scipy.signal import spectrogram
signal = [1, 2, 3, 4, 5]
fs = 100
f, t, Sxx = spectrogram(signal, nperseg=256, fs=fs)

medium

A. nperseg is larger than signal length causing a warning or error

B. fs parameter is missing

C. spectrogram function is not imported correctly

D. signal must be a numpy array, not a list

Solution

1. Step 1: Check parameters and signal length

   Signal length is 5, but nperseg is 256, which is larger than signal length.

2. Step 2: Understand spectrogram behavior

   When nperseg > signal length, scipy issues a warning or error because segment size can't exceed signal length.

3. Final Answer:

   nperseg is larger than signal length causing a warning or error -> Option A

4. Quick Check:

   nperseg > signal length causes error [OK]

Hint: nperseg must be ≤ signal length [OK]

Common Mistakes:

• Ignoring nperseg size relative to signal length
• Thinking fs is missing when it is provided
• Assuming list input causes error (it works but numpy array preferred)

5. You want to generate a spectrogram for a noisy signal sampled at 2000 Hz. To improve time resolution, which parameter should you decrease in scipy.signal.spectrogram?

hard

A. window

B. fs

C. noverlap

D. nperseg

Solution

1. Step 1: Understand time vs frequency resolution trade-off

   Decreasing nperseg (segment length) improves time resolution but reduces frequency resolution.

2. Step 2: Check other parameters

   fs is sample rate, noverlap controls overlap, window shapes segments but doesn't directly improve time resolution.

3. Final Answer:

   nperseg -> Option D

4. Quick Check:

   Decrease nperseg for better time resolution [OK]

Hint: Lower nperseg for better time resolution [OK]

Common Mistakes:

• Trying to change fs which is fixed sample rate
• Confusing noverlap with time resolution
• Changing window type expecting time resolution change