Windowing before FFT in SciPy - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Effect of Windowing on FFT Magnitude

What is the output magnitude array when applying a Hamming window before FFT on a simple sine wave?

SciPy

import numpy as np
from scipy.signal import windows

fs = 1000
f = 50
N = 256

t = np.arange(N) / fs
signal = np.sin(2 * np.pi * f * t)
window = windows.hamming(N)
windowed_signal = signal * window
fft_result = np.fft.fft(windowed_signal)
magnitude = np.abs(fft_result)[:5]
print(np.round(magnitude, 2))

A [32.00 0. 0. 0. 0. ]

B [31.99 0. 0. 0. 0. ]

C [15.99 0. 0. 0. 0. ]

D [16.00 0. 0. 0. 0. ]

💡 Hint

Recall that the Hamming window reduces spectral leakage but scales the amplitude.

The Hamming window scales the amplitude close to 0.54 times the sum of the window. The FFT magnitude peak is about 32 for this sine wave with windowing.

🧠 Conceptual

intermediate

Purpose of Windowing Before FFT

Why do we apply a window function to a signal before computing its FFT?

A To reduce spectral leakage caused by signal discontinuities at the edges.

B To increase the frequency resolution of the FFT.

C To remove noise from the signal before FFT.

D To speed up the FFT computation.

💡 Hint

Think about what happens at the edges of a signal segment when FFT assumes periodicity.

Windowing smooths the edges of the signal segment, reducing abrupt changes that cause spectral leakage in the FFT.

❓ data_output

advanced

Comparing FFT Results With and Without Windowing

Given a noisy signal, which option shows the correct difference in FFT magnitude peak when using a rectangular window vs a Hann window?

SciPy

import numpy as np
from scipy.signal import windows

fs = 500
f = 60
N = 128

t = np.arange(N) / fs
signal = np.sin(2 * np.pi * f * t) + 0.5 * np.random.randn(N)

rect_window = np.ones(N)
hann_window = windows.hann(N)

fft_rect = np.fft.fft(signal * rect_window)
fft_hann = np.fft.fft(signal * hann_window)

peak_rect = np.max(np.abs(fft_rect))
peak_hann = np.max(np.abs(fft_hann))
diff = round(peak_rect - peak_hann, 2)
print(diff)

A Approximately 5.0

B Approximately 0.0

C Approximately 10.0

D Approximately -10.0

💡 Hint

Windowing usually reduces the peak magnitude due to tapering.

The Hann window tapers the signal edges, reducing the peak FFT magnitude compared to the rectangular window, so the difference is negative.

🔧 Debug

advanced

Identify the Error in Windowing Application

What error occurs when this code tries to apply a window before FFT?

SciPy

import numpy as np
from scipy.signal import windows

signal = np.array([1, 2, 3, 4, 5])
window = windows.hann(10)
windowed_signal = signal * window
fft_result = np.fft.fft(windowed_signal)
print(np.round(np.abs(fft_result), 2))

A ValueError due to shape mismatch in multiplication

B TypeError because window is not a numpy array

C No error, prints FFT magnitude

D IndexError accessing window elements

💡 Hint

Check the sizes of the signal and window arrays before multiplying.

The signal has length 5 but the window has length 10, so element-wise multiplication fails with a ValueError.

🚀 Application

expert

Choosing the Best Window for Frequency Resolution

You want to detect two close frequencies in a signal using FFT. Which window choice helps best to resolve them?

A Hamming window for lowest side lobes

B Bartlett window for fastest computation

C Rectangular window for narrow main lobe and high resolution

D Blackman window for widest main lobe

💡 Hint

Frequency resolution depends on main lobe width in window's frequency response.

The rectangular window has the narrowest main lobe, giving the best frequency resolution to distinguish close frequencies, despite higher side lobes.

Practice

1. What is the main purpose of applying a window function to a signal before performing an FFT?

easy

A. To increase the signal amplitude

B. To reduce spectral leakage by smoothing signal edges

C. To convert the signal from time domain to frequency domain

D. To remove noise from the signal

Solution

1. Step 1: Understand spectral leakage in FFT

   Spectral leakage occurs when the signal edges are abrupt, causing spreading of frequency components.

2. Step 2: Role of window functions

   Window functions smooth the edges of the signal to reduce this leakage before FFT.

3. Final Answer:

   To reduce spectral leakage by smoothing signal edges -> Option B

4. Quick Check:

   Windowing reduces leakage = B [OK]

Hint: Windowing smooths edges to reduce leakage before FFT [OK]

Common Mistakes:

• Thinking windowing increases amplitude
• Confusing windowing with noise removal
• Believing windowing converts domains

2. Which of the following is the correct way to create a Hann window of length 256 using scipy?

easy

A. window = scipy.signal.hann(256)

B. window = scipy.get_window('hann', 256)

C. window = scipy.signal.window('hann', 256)

D. window = scipy.signal.get_window('hann', 256)

Solution

1. Step 1: Identify the correct scipy module and function

   The function to create windows is get_window inside scipy.signal.

2. Step 2: Check syntax for get_window

   Correct syntax is scipy.signal.get_window(window_name, length).

3. Final Answer:

   window = scipy.signal.get_window('hann', 256) -> Option D

4. Quick Check:

   Use scipy.signal.get_window('hann', length) = D [OK]

Hint: Use scipy.signal.get_window('hann', length) to create Hann window [OK]

Common Mistakes:

• Using scipy.get_window instead of scipy.signal.get_window
• Calling nonexistent scipy.signal.window function
• Using scipy.signal.hann which does not exist

3. What is the output of the following code snippet?

import numpy as np
from scipy.signal import get_window

signal = np.ones(4)
window = get_window('hann', 4)
windowed_signal = signal * window
print(np.round(windowed_signal, 3))

medium

A. [0. 0.75 0.75 0. ]

B. [0. 0.5 0.5 0. ]

C. [0. 1. 1. 0. ]

D. [1. 1. 1. 1.]

Solution

1. Step 1: Calculate the Hann window values for length 4

   The Hann window for length 4 is [0, 0.75, 0.75, 0].

2. Step 2: Multiply the signal by the window element-wise

   Signal is all ones, so windowed_signal equals the window values.

3. Final Answer:

   [0. 0.75 0.75 0. ] -> Option A

4. Quick Check:

   Hann window at length 4 = [0, 0.75, 0.75, 0] [OK]

Hint: Hann window starts and ends at 0, peaks near middle [OK]

Common Mistakes:

• Assuming window values are 0.5 instead of 0.75
• Forgetting to multiply element-wise
• Confusing Hann with rectangular window

4. Identify the error in the following code that applies a Hamming window before FFT:

import numpy as np
from scipy.signal import get_window

signal = np.array([1, 2, 3, 4, 5])
window = get_window('hamming', 5)
windowed_signal = signal + window
fft_result = np.fft.fft(windowed_signal)
print(np.round(np.abs(fft_result), 2))

medium

A. Window should be multiplied, not added, to the signal

B. The window length must be larger than the signal length

C. np.fft.fft cannot be used on windowed signals

D. Hamming window cannot be created with length 5

Solution

1. Step 1: Check how window is applied to signal

   The code adds the window to the signal instead of multiplying element-wise.

2. Step 2: Correct application of window

   Windowing requires element-wise multiplication to smooth edges before FFT.

3. Final Answer:

   Window should be multiplied, not added, to the signal -> Option A

4. Quick Check:

   Windowing = element-wise multiply, not add [OK]

Hint: Always multiply signal by window, never add [OK]

Common Mistakes:

• Adding window instead of multiplying
• Assuming window length must exceed signal length
• Thinking FFT can't be applied after windowing

5. You have a noisy signal sampled at 1000 Hz with 1024 points. You want to analyze its frequency content using FFT with minimal spectral leakage. Which steps correctly apply windowing before FFT using scipy?

hard

A. Create a Hamming window of length 512, multiply it with the first half of the signal, then compute FFT

B. Create a rectangular window, add it to the signal, then compute FFT

C. Create a Hann window of length 1024, multiply it element-wise with the signal, then compute FFT

D. Compute FFT directly without windowing to preserve original signal

Solution

1. Step 1: Choose appropriate window and length

   Hann window is common for reducing spectral leakage; length must match signal length (1024).

2. Step 2: Apply window and compute FFT

   Multiply signal element-wise by window, then compute FFT on the windowed signal.

3. Final Answer:

   Create a Hann window of length 1024, multiply it element-wise with the signal, then compute FFT -> Option C

4. Quick Check:

   Window length = signal length, multiply before FFT = A [OK]

Hint: Match window length to signal, multiply before FFT [OK]

Common Mistakes:

• Adding window instead of multiplying
• Using wrong window length
• Skipping windowing and accepting leakage