FFT-based filtering in SciPy - Time & Space Complexity

Time Complexity: FFT-based filtering

O(n log n)

Understanding Time Complexity

When using FFT-based filtering, we want to know how the time to process data changes as the data size grows.

We ask: How does the filtering time increase when the input signal gets longer?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

import numpy as np
from scipy.fft import fft, ifft

def fft_filter(signal, filter_kernel):
    n = len(signal) + len(filter_kernel) - 1
    signal_fft = fft(signal, n)
    filter_fft = fft(filter_kernel, n)
    filtered_fft = signal_fft * filter_fft
    filtered_signal = ifft(filtered_fft)
    return np.real(filtered_signal)

This code filters a signal by transforming both the signal and filter to frequency space, multiplying them, and then transforming back.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The FFT and inverse FFT computations, which internally perform repeated calculations over the input arrays.
• How many times: Each FFT processes an array of size approximately n (sum of input lengths), doing about n log n steps.

How Execution Grows With Input

As the input size n grows, the number of operations grows a bit faster than n but much slower than n squared.

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

Pattern observation: Operations grow faster than the input size but much slower than if we did every pairwise multiplication.

Final Time Complexity

Time Complexity: O(n log n)

This means the time to filter grows a bit faster than the input size but remains efficient even for large inputs.

Common Mistake

[X] Wrong: "FFT-based filtering takes time proportional to n squared because it multiplies all pairs of points."

[OK] Correct: FFT uses a clever method to reduce repeated calculations, so it runs much faster than checking every pair.

Interview Connect

Understanding FFT time complexity shows you can analyze efficient algorithms that handle large data quickly, a useful skill in many data science tasks.

Self-Check

"What if we used a direct convolution instead of FFT-based filtering? How would the time complexity change?"

Practice

1. What is the main purpose of using FFT-based filtering in data analysis?

easy

A. To transform data into frequency domain for easier noise removal

B. To sort data values in ascending order

C. To calculate the average of data points

D. To convert data into text format

Solution

1. Step 1: Understand FFT transformation

   FFT converts data from time domain to frequency domain, revealing frequency components.

2. Step 2: Purpose of filtering in frequency domain

   Filtering in frequency domain allows easy removal of unwanted noise or frequencies.

3. Final Answer:

   To transform data into frequency domain for easier noise removal -> Option A

4. Quick Check:

   FFT filtering = frequency domain noise removal [OK]

Hint: FFT helps remove noise by working in frequency domain [OK]

Common Mistakes:

• Confusing FFT with sorting or averaging
• Thinking FFT changes data to text
• Assuming FFT directly cleans data without filtering

2. Which of the following is the correct way to compute the FFT of a 1D numpy array data using scipy?

easy

A. fft_data = scipy.fft_transform(data)

B. fft_data = scipy.fft(data)

C. fft_data = scipy.fft.fft(data)

D. fft_data = scipy.fft_fft(data)

Solution

1. Step 1: Identify correct scipy FFT function

   The scipy FFT module uses fft() to compute the Fast Fourier Transform.

2. Step 2: Check syntax correctness

   The correct syntax is scipy.fft.fft(data), other options are invalid function names.

3. Final Answer:

   fft_data = scipy.fft.fft(data) -> Option C

4. Quick Check:

   Correct FFT call = scipy.fft.fft(data) [OK]

Hint: Use scipy.fft.fft() to compute FFT of array [OK]

Common Mistakes:

• Using incorrect function names like fft_transform
• Omitting the .fft part after scipy.fft
• Confusing with numpy.fft functions

3. Given the code below, what will be the output of filtered_signal?

import numpy as np
from scipy.fft import fft, ifft

data = np.array([1, 2, 3, 4])
fft_data = fft(data)
mask = np.array([1, 0, 0, 1])
filtered_fft = fft_data * mask
filtered_signal = ifft(filtered_fft).real
print(np.round(filtered_signal, 2))

medium

A. [1. 0. 0. 4.]

B. [2. 2. 3. 3.]

C. [1.5 0. 0. 3.5]

D. [2. 0. 0. 2.]

Solution

1. Step 1: Compute FFT and apply mask

   FFT of data is computed, then multiplied by mask [1,0,0,1], zeroing middle frequencies.

2. Step 2: Compute inverse FFT and round

   Inverse FFT of masked data gives filtered signal; rounding to 2 decimals yields [2. 0. 0. 2.].

3. Final Answer:

   [2. 0. 0. 2.] -> Option D

4. Quick Check:

   Mask zeros middle frequencies, inverse FFT smooths signal [OK]

Hint: Mask zeros frequencies, inverse FFT returns filtered signal [OK]

Common Mistakes:

• Forgetting to multiply FFT by mask
• Not taking real part after inverse FFT
• Rounding incorrectly or not at all

4. Identify the error in the following FFT-based filtering code snippet:

import numpy as np
from scipy.fft import fft, ifft

data = np.array([1, 2, 3, 4])
fft_data = fft(data)
mask = np.array([1, 0, 0])
filtered_fft = fft_data * mask
filtered_signal = ifft(filtered_fft).real
print(filtered_signal)

medium

A. Data array should be complex type

B. Mask length does not match FFT data length

C. Using ifft instead of fft

D. Missing import for numpy

Solution

1. Step 1: Check mask and FFT data lengths

   FFT of data has length 4, but mask length is 3, causing shape mismatch error.

2. Step 2: Identify error cause

   Multiplying arrays of different lengths raises an error; mask must match FFT length.

3. Final Answer:

   Mask length does not match FFT data length -> Option B

4. Quick Check:

   Mask length must equal FFT length [OK]

Hint: Mask length must match FFT output length [OK]

Common Mistakes:

• Using mask shorter than FFT result
• Confusing ifft with fft usage
• Assuming data must be complex initially

5. You have a noisy signal sampled at 1000 Hz stored in signal. You want to remove frequencies above 100 Hz using FFT-based filtering. Which code snippet correctly implements this filter?

hard

A. import numpy as np
from scipy.fft import fft, ifft
N = len(signal)
freqs = np.fft.fftfreq(N, 1/1000)
fft_signal = fft(signal)
mask = np.abs(freqs) <= 100
filtered_fft = fft_signal * mask
filtered_signal = ifft(filtered_fft).real

B. import numpy as np
from scipy.fft import fft, ifft
N = len(signal)
freqs = np.fft.fftfreq(N)
fft_signal = fft(signal)
mask = freqs < 100
filtered_fft = fft_signal * mask
filtered_signal = ifft(filtered_fft).real

C. import numpy as np
from scipy.fft import fft, ifft
N = len(signal)
freqs = np.fft.fftfreq(N, 1/1000)
fft_signal = fft(signal)
mask = freqs > 100
filtered_fft = fft_signal * mask
filtered_signal = ifft(filtered_fft).real

D. import numpy as np
from scipy.fft import fft, ifft
N = len(signal)
freqs = np.fft.fftfreq(N, 1/1000)
fft_signal = fft(signal)
mask = np.abs(freqs) >= 100
filtered_fft = fft_signal * mask
filtered_signal = ifft(filtered_fft).real

Solution

1. Step 1: Calculate frequency bins correctly

   Use np.fft.fftfreq(N, 1/1000) to get frequencies in Hz for sampling rate 1000 Hz.

2. Step 2: Create mask to keep frequencies ≤ 100 Hz

   Mask should be np.abs(freqs) <= 100 to keep low frequencies and remove higher ones.

3. Step 3: Apply mask and inverse FFT

   Multiply FFT by mask, then apply inverse FFT and take real part for filtered signal.

4. Final Answer:

   mask = np.abs(freqs) <= 100 with freqs = np.fft.fftfreq(N, 1/1000) -> Option A

5. Quick Check:

   Mask keeps frequencies ≤ 100 Hz [OK]

Hint: Use np.abs(freqs) ≤ cutoff for low-pass mask [OK]

Common Mistakes:

• Ignoring sampling rate in frequency calculation
• Using wrong comparison operators in mask
• Not taking absolute value of frequencies