Power spectral density estimation in Signal Processing - Time & Space Complexity

Time Complexity: Power spectral density estimation

O(n log n)

Understanding Time Complexity

When estimating power spectral density, we want to know how the time to compute grows as the signal length increases.

We ask: How does the processing time change when the input signal gets longer?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

import numpy as np

def estimate_psd(signal):
    n = len(signal)
    fft_vals = np.fft.fft(signal)
    psd = (np.abs(fft_vals) ** 2) / n
    return psd[:n // 2]
    

This code computes the power spectral density by applying a Fast Fourier Transform (FFT) on the input signal and then calculating the squared magnitude.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The FFT algorithm processes the entire signal of length n.
• How many times: The FFT internally performs operations proportional to n times log n.

How Execution Grows With Input

As the signal length increases, the number of operations grows a bit faster than just the length itself.

Input Size (n)	Approx. Operations
10	About 33 (10 * log2(10))
100	About 664 (100 * log2(100))
1000	About 9966 (1000 * log2(1000))

Pattern observation: The operations grow faster than the input size but much slower than its square. This means doubling the input length slightly more than doubles the work.

Final Time Complexity

Time Complexity: O(n log n)

This means the time to estimate the power spectral density grows a bit faster than the signal length, but not as fast as checking every pair of points.

Common Mistake

[X] Wrong: "The FFT runs in linear time, so doubling the signal length doubles the time exactly."

[OK] Correct: FFT uses a divide-and-conquer approach, so its time grows like n times log n, which is a bit more than just n.

Interview Connect

Understanding how FFT time grows helps you explain why it is efficient for signal analysis and why it is preferred over slower methods.

Self-Check

"What if we used a simple Discrete Fourier Transform (DFT) instead of FFT? How would the time complexity change?"

Practice

1. What does the Power Spectral Density (PSD) of a signal represent?

easy

A. How the power of a signal is distributed over different frequencies

B. The total energy of the signal in the time domain

C. The phase shift of the signal over time

D. The amplitude of the signal at a single point in time

Solution

1. Step 1: Understand the definition of PSD

   PSD shows how the power of a signal is spread across frequencies, not just total energy or amplitude.

2. Step 2: Compare options with PSD meaning

   Only How the power of a signal is distributed over different frequencies correctly describes PSD as power distribution over frequencies.

3. Final Answer:

   How the power of a signal is distributed over different frequencies -> Option A

4. Quick Check:

   PSD = power distribution over frequency [OK]

Hint: PSD shows power spread by frequency, not time or amplitude [OK]

Common Mistakes:

• Confusing PSD with total signal energy
• Thinking PSD shows signal amplitude at a time
• Mixing PSD with phase information

2. Which Python function from the scipy.signal library is commonly used to estimate the Power Spectral Density using Welch's method?

easy

A. scipy.signal.spectrogram()

B. scipy.signal.fft()

C. scipy.signal.correlate()

D. scipy.signal.welch()

Solution

1. Step 1: Identify Welch's method function

   Welch's method for PSD estimation is implemented in scipy.signal.welch().

2. Step 2: Differentiate from other functions

   fft() computes Fourier transform, correlate() computes correlation, spectrogram() shows time-frequency but not PSD by Welch.

3. Final Answer:

   scipy.signal.welch() -> Option D

4. Quick Check:

   Welch PSD = scipy.signal.welch() [OK]

Hint: Welch PSD uses scipy.signal.welch() function [OK]

Common Mistakes:

• Using fft() instead of welch() for PSD
• Confusing spectrogram with PSD estimation
• Trying correlate() for PSD calculation

3. Given the following Python code, what will be the length of the frequency array f returned by scipy.signal.welch?

import numpy as np
from scipy.signal import welch

fs = 1000  # Sampling frequency
x = np.random.randn(1024)  # Random signal
f, Pxx = welch(x, fs=fs, nperseg=256)

medium

A. 1024

B. 129

C. 512

D. 256

Solution

1. Step 1: Understand nperseg effect on frequency bins

   Welch returns frequency bins of length nperseg/2 + 1 for real signals.

2. Step 2: Calculate frequency array length

   nperseg=256, so length = 256/2 + 1 = 129.

3. Final Answer:

   129 -> Option B

4. Quick Check:

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

Hint: Frequency bins length = nperseg/2 + 1 for real signals [OK]

Common Mistakes:

• Assuming frequency length equals nperseg
• Using full signal length instead of segment length
• Confusing frequency bins with time samples

4. Identify the error in this code snippet estimating PSD with Welch's method:

import numpy as np
from scipy.signal import welch

fs = 500
x = np.sin(2 * np.pi * 50 * np.arange(1000) / fs)
f, Pxx = welch(x, fs=fs, nperseg=2000)

medium

A. Sampling frequency fs is too low

B. Signal x is not a numpy array

C. nperseg is larger than the signal length

D. Missing window parameter in welch()

Solution

1. Step 1: Check nperseg vs signal length

   nperseg=2000 is greater than signal length 1000, which causes a warning or error.

2. Step 2: Verify other parameters

   fs=500 is valid, x is numpy array, window parameter is optional with default.

3. Final Answer:

   nperseg is larger than the signal length -> Option C

4. Quick Check:

   nperseg > signal length causes error [OK]

Hint: nperseg must not exceed signal length [OK]

Common Mistakes:

• Setting nperseg larger than signal length
• Assuming window parameter is mandatory
• Confusing sampling frequency with segment length

5. You have a noisy signal sampled at 2000 Hz. To get a smoother PSD estimate with better frequency resolution using Welch's method, which combination of parameters should you choose?

hard

A. Increase segment length (nperseg) and use overlapping segments

B. Decrease segment length and avoid overlapping

C. Use no window and set nperseg to 1

D. Increase sampling frequency beyond 2000 Hz

Solution

1. Step 1: Understand effect of segment length and overlap

   Longer segments improve frequency resolution; overlapping segments reduce variance (smoother PSD).

2. Step 2: Evaluate options for smoothing and resolution

   Increase segment length (nperseg) and use overlapping segments correctly suggests increasing nperseg and using overlap; others reduce resolution or are impractical.

3. Final Answer:

   Increase segment length (nperseg) and use overlapping segments -> Option A

4. Quick Check:

   Longer segments + overlap = smoother, better PSD [OK]

Hint: Longer segments + overlap = smoother PSD estimate [OK]

Common Mistakes:

• Using very short segments losing frequency detail
• Ignoring overlap reduces averaging benefits
• Thinking sampling rate increase smooths PSD