Power spectral density estimation in Signal Processing - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of PSD estimation using Welch's method

What is the shape of the output array psd after running Welch's method on a 1D signal of length 1024 with nperseg=1024?

Signal Processing

import numpy as np
from scipy.signal import welch

fs = 1000  # Sampling frequency
signal = np.random.randn(1024)
frequencies, psd = welch(signal, fs=fs, nperseg=1024)
print(psd.shape)

A (513,)

B (512,)

C (1024,)

D (256,)

💡 Hint

Welch's method returns half the FFT length plus one for the frequency bins.

Welch's method with nperseg=1024 uses an FFT length of 1024 (same as signal length). The output PSD array length is nfft/2 + 1 = 512 + 1 = 513.

❓ data_output

intermediate

Number of frequency bins in periodogram

Given a signal of length 2048 sampled at 500 Hz, what is the number of frequency bins returned by the periodogram method with default FFT length?

Signal Processing

import numpy as np
from scipy.signal import periodogram

fs = 500
signal = np.random.randn(2048)
frequencies, psd = periodogram(signal, fs=fs)
print(len(frequencies))

A 2048

B 2049

C 1025

D 1024

💡 Hint

Periodogram returns nfft/2 + 1 frequency bins for real signals.

The default FFT length is the signal length (2048). The number of frequency bins is nfft/2 + 1 = 1024 + 1 = 1025.

❓ visualization

advanced

Identifying the correct PSD plot

Which plot correctly shows the power spectral density of a noisy sine wave at 50 Hz using Welch's method?

Signal Processing

import numpy as np
import matplotlib.pyplot as plt
from scipy.signal import welch

fs = 500
T = 2
t = np.linspace(0, T, int(T*fs), endpoint=False)
signal = np.sin(2*np.pi*50*t) + 0.5*np.random.randn(len(t))
frequencies, psd = welch(signal, fs=fs, nperseg=256)

plt.figure(figsize=(8,4))
plt.semilogy(frequencies, psd)
plt.xlabel('Frequency (Hz)')
plt.ylabel('PSD (V**2/Hz)')
plt.title('Power Spectral Density using Welch')
plt.grid(True)
plt.show()

A A plot with a peak near 100 Hz instead of 50 Hz

B A flat line with no peaks

C Multiple peaks at random frequencies with no dominant peak

D A plot with a clear peak near 50 Hz and noise floor elsewhere

💡 Hint

The sine wave frequency should appear as a peak in the PSD.

The noisy sine wave at 50 Hz produces a PSD with a clear peak near 50 Hz. Noise contributes to a flat noise floor.

🧠 Conceptual

advanced

Effect of segment length on PSD resolution

How does increasing the segment length (nperseg) in Welch's method affect the frequency resolution and variance of the PSD estimate?

A Increases frequency resolution but increases variance

B Decreases frequency resolution and decreases variance

C Increases frequency resolution and decreases variance

D Decreases frequency resolution but increases variance

💡 Hint

Longer segments give finer frequency bins but fewer averages.

Longer segments increase frequency resolution (smaller frequency bins) but reduce the number of segments averaged, increasing variance.

🔧 Debug

expert

Identifying error in PSD calculation code

What error will this code raise when calculating PSD using Welch's method?

import numpy as np
from scipy.signal import welch

fs = 1000
signal = np.random.randn(500)
frequencies, psd = welch(signal, fs=fs, nperseg=1024)
print(psd)

A No error, prints PSD array

B ValueError: nperseg must not be greater than input length

C IndexError: index out of range

D TypeError: unsupported operand type(s) for /: 'int' and 'str'

💡 Hint

Check if segment length is valid for the signal length.

Welch's method requires nperseg to be less than or equal to the signal length. Here, nperseg=1024 > 500 causes ValueError.