Visualize Fast Fourier Transform Low Pass Filter¶
Background¶
Visualize Fft Low Pass
Educational script demonstrating visualize fft low pass concepts.
Code¶
```python """ Visualize Fft Low Pass
Educational script demonstrating visualize fft low pass concepts. """
============================================================================¶
VISUALIZE_FFT_LOW_PASS_FILTER.py¶
============================================================================¶
import matplotlib.pyplot as plt import matplotlib.gridspec as gridspec import numpy as np
Time domain: sample 1 second of signal at 1000 Hz¶
if name == "main": fs = 1000 # Sampling frequency (samples per second) T = 1 # Total time in seconds N = fs * T # Total number of samples t = np.linspace(0, T, N, endpoint=False) # Time array
# Signal: combination of 50 Hz and 120 Hz sinusoids
freqs = (50, 120)
mags = (1, 0.5)
signal = np.zeros_like(t)
for mag, freq in zip(mags, freqs):
signal += mag * np.sin(2 * np.pi * freq * t)
# FFT and frequency bins
signal_fft = np.fft.fft(signal)
frequencies = np.fft.fftfreq(N, d=1/fs)
low_pass_filter = np.exp( - 4 * np.pi**2 * frequencies**2 * 0.00001 )
signal_fft *= low_pass_filter
# IFFT reconstruction
signal_reconstructed = np.fft.ifft(signal_fft)
print("Reconstruction matches original:", np.allclose(signal, signal_reconstructed.real))
# Set up custom layout: 2 rows, 2 columns
fig = plt.figure(figsize=(14, 6))
gs = gridspec.GridSpec(2, 2, height_ratios=[1, 1])
# First row: original vs reconstructed
ax0 = fig.add_subplot(gs[0, :])
ax0.plot(t, signal, label='Original Signal', alpha=0.8)
ax0.plot(t, signal_reconstructed.real, '--', label='Reconstructed Signal (IFFT)', alpha=0.8)
ax0.set_title("Original vs Reconstructed Signal")
ax0.set_xlabel("Time (s)")
ax0.set_ylabel("Amplitude")
ax0.legend()
ax0.grid(True)
# Second row, left: positive frequencies
half_N = N // 2
ax1 = fig.add_subplot(gs[1, 0])
ax1.plot(frequencies[:half_N], np.abs(signal_fft[:half_N]) / N)
ax1.set_title("Magnitude Spectrum (Positive Frequencies)")
ax1.set_xlabel("Frequency (Hz)")
ax1.set_ylabel("Magnitude")
ax1.grid(True)
# Second row, right: full spectrum
ax2 = fig.add_subplot(gs[1, 1])
ax2.plot(frequencies, np.abs(signal_fft) / N)
ax2.set_title("Full Magnitude Spectrum (Including Negative Frequencies)")
ax2.set_xlabel("Frequency (Hz)")
ax2.set_ylabel("Magnitude")
ax2.grid(True)
plt.tight_layout()
plt.show()
```
Exercises¶
Exercise 1. The low-pass filter used is \(H(f) = \exp(-4\pi^2 f^2 \cdot \alpha)\) with \(\alpha = 10^{-5}\). Explain why this is called a Gaussian low-pass filter and what determines the cutoff frequency.
Solution to Exercise 1
The filter \(H(f) = \exp(-4\pi^2 \alpha f^2)\) is a Gaussian function of frequency \(f\). It attenuates high frequencies more than low frequencies because \(|H(f)|\) decreases with \(|f|\).
The effective cutoff frequency \(f_c\) is where the filter magnitude drops to \(e^{-1} \approx 0.37\): \(4\pi^2 \alpha f_c^2 = 1\), giving \(f_c = 1/(2\pi\sqrt{\alpha})\). For \(\alpha = 10^{-5}\): \(f_c = 1/(2\pi \cdot 0.00316) \approx 50.3\) Hz. This means the 50 Hz component is barely attenuated while the 120 Hz component is significantly reduced.
Exercise 2. After applying the low-pass filter, the reconstruction no longer matches the original signal. Compute the expected attenuation of the 120 Hz component with \(\alpha = 10^{-5}\).
Solution to Exercise 2
The attenuation at 120 Hz is
So the 120 Hz component is reduced to about 0.34% of its original amplitude -- essentially removed.
For comparison, at 50 Hz: \(H(50) = \exp(-4\pi^2 \cdot 10^{-5} \cdot 2500) = \exp(-0.987) \approx 0.373\). The 50 Hz component is attenuated to about 37%.
Exercise 3. Explain the connection between this Gaussian low-pass filter and the heat equation. What role does the parameter \(\alpha\) play?
Solution to Exercise 3
The Gaussian filter \(\exp(-4\pi^2 D k^2 t)\) is exactly the frequency-domain response of the heat equation \(u_t = Du_{xx}\). Here \(\alpha\) plays the role of \(Dt\) (diffusivity times time).
Applying the filter to a signal is equivalent to evolving it under the heat equation for time \(t = \alpha/D\). Larger \(\alpha\) means more diffusion (stronger smoothing). This connection explains why the heat equation is a natural model for smoothing and denoising: it exponentially damps high-frequency components while preserving low-frequency structure.
Exercise 4. Design a band-pass filter (using the FFT) that passes only frequencies between 100 Hz and 150 Hz. Write the filter function \(H(f)\).
Solution to Exercise 4
A sharp band-pass filter is
A smoother version using Gaussians:
This is centered at 125 Hz with bandwidth controlled by the denominator. In code: H = np.exp(-((np.abs(frequencies) - 125)**2) / (2 * 25**2)). Applying this filter to the test signal would retain only the 120 Hz component and suppress the 50 Hz component.