FFT-based filtering in SciPy - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of FFT and inverse FFT on a simple signal

What is the output of the following code snippet that applies FFT and then inverse FFT on a simple signal?

SciPy

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

signal = np.array([1, 2, 3, 4])
fft_signal = fft(signal)
recovered_signal = ifft(fft_signal)
print(np.round(recovered_signal.real, 2))

A [1. 2. 3. 4.]

B [0. 0. 0. 0.]

C [1. 2. 3. 5.]

D [4. 3. 2. 1.]

💡 Hint

FFT followed by inverse FFT should recover the original signal.

FFT transforms the signal to frequency domain and inverse FFT brings it back. The recovered signal matches the original input.

❓ data_output

intermediate

Length of filtered signal after FFT-based low-pass filtering

Given a signal of length 8, after applying FFT-based low-pass filtering by zeroing out frequencies above index 3 and then inverse FFT, what is the length of the resulting filtered signal?

SciPy

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

signal = np.arange(8)
fft_signal = fft(signal)
fft_signal[4:] = 0  # zero out frequencies above index 3
filtered_signal = ifft(fft_signal)
print(len(filtered_signal))

A 0

B 4

C 7

D 8

💡 Hint

FFT and inverse FFT do not change the length of the signal.

The length of the signal remains the same after FFT and inverse FFT, even if some frequency components are zeroed out.

🔧 Debug

advanced

Identify the error in FFT-based filtering code

What error does the following code produce when trying to apply FFT-based filtering?

SciPy

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

signal = np.array([1, 2, 3, 4])
fft_signal = fft(signal)
# Incorrect slicing to zero out high frequencies
fft_signal[4:] = 0
filtered_signal = ifft(fft_signal)
print(filtered_signal)

A ValueError: operands could not be broadcast together with shapes (4,) (3,)

B IndexError: index 4 is out of bounds for axis 0 with size 4

C TypeError: unsupported operand type(s) for +: 'int' and 'complex'

D No error, prints filtered signal

💡 Hint

Check the length of the FFT output and the slice used.

The FFT output length is 4, so trying to access fft_signal[4:] causes an IndexError.

🧠 Conceptual

advanced

Effect of zeroing out frequency components in FFT

What is the effect on the time-domain signal when you zero out the high-frequency components in its FFT representation before applying inverse FFT?

A The signal becomes more noisy with added high-frequency noise

B The signal becomes completely zero

C The signal becomes smoother by removing rapid changes

D The signal length doubles

💡 Hint

High frequencies correspond to rapid changes in the signal.

Removing high-frequency components filters out rapid changes, making the signal smoother.

🚀 Application

expert

Identify the correct code for band-pass filtering using FFT

Which code snippet correctly applies a band-pass filter to keep only frequencies between indices 2 and 5 (inclusive) in the FFT of a signal, then recovers the filtered signal?

A

fft_signal = fft(signal)
filtered_fft = fft_signal.copy()
filtered_fft[:2] = 0
filtered_fft[6:] = 0
filtered_signal = ifft(filtered_fft)

B

fft_signal = fft(signal)
filtered_fft = np.zeros_like(fft_signal)
filtered_fft[2:6] = fft_signal[2:6]
filtered_signal = ifft(filtered_fft)

C

fft_signal = fft(signal)
filtered_fft = fft_signal
filtered_fft[2:5] = 0
filtered_signal = ifft(filtered_fft)

D

fft_signal = fft(signal)
filtered_fft = np.zeros_like(fft_signal)
filtered_fft[3:5] = fft_signal[3:5]
filtered_signal = ifft(filtered_fft)

💡 Hint

Band-pass means keep frequencies in a range and zero out others.

Option A correctly zeros frequencies outside 2 to 5 inclusive and keeps the rest, then applies inverse FFT.