Visualize Fast Fourier Transform¶

Background¶

Visualize Fft

Educational script demonstrating visualize fft concepts.

Code¶

```python """ Visualize Fft

Educational script demonstrating visualize fft concepts. """

============================================================================¶

VISUALIZE_FFT.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) # d = t[1] - t[0]

# 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. A signal is composed of two sinusoids at frequencies 50 Hz and 120 Hz with amplitudes 1 and 0.5. Write the mathematical expression for this signal and compute its FFT analytically.

Solution to Exercise 1

The signal is \(x(t) = \sin(2\pi \cdot 50\,t) + 0.5\sin(2\pi \cdot 120\,t)\).

The FFT of a sinusoid \(\sin(2\pi f_0 t)\) produces peaks at \(\pm f_0\) with magnitude \(N/2\) (where \(N\) is the number of samples). After normalization by \(N\), the magnitude spectrum shows peaks of height \(1/2\) at \(\pm 50\) Hz and \(0.25\) at \(\pm 120\) Hz.

The negative frequencies are a consequence of the real-valued signal: by Euler's formula, \(\sin(\omega t) = (e^{i\omega t} - e^{-i\omega t})/(2i)\), producing symmetric positive and negative frequency components.

Exercise 2. Explain the Nyquist theorem. With a sampling frequency of \(f_s = 1000\) Hz, what is the maximum frequency that can be represented without aliasing?

Solution to Exercise 2

The Nyquist theorem states that a signal can be perfectly reconstructed from its samples if the sampling rate \(f_s\) exceeds twice the highest frequency component: \(f_s > 2f_{\max}\).

The Nyquist frequency is \(f_N = f_s / 2 = 500\) Hz. Any frequency component above 500 Hz would be aliased -- it would appear as a spurious lower-frequency component in the sampled signal. In this example, both signal frequencies (50 Hz and 120 Hz) are well below 500 Hz, so no aliasing occurs.

Exercise 3. The code verifies that np.allclose(signal, signal_reconstructed.real) is True. Explain why IFFT perfectly reconstructs the original signal.

Solution to Exercise 3

The DFT and IDFT are exact inverses: \(\text{IDFT}(\text{DFT}(x)) = x\) up to floating-point precision. This is because the DFT represents the signal as a sum of complex exponentials, and the IDFT reconstructs the signal from these coefficients.

Mathematically: \(X_k = \sum_{n=0}^{N-1} x_n e^{-2\pi i kn/N}\) and \(x_n = \frac{1}{N}\sum_{k=0}^{N-1} X_k e^{2\pi i kn/N}\). The orthogonality of complex exponentials ensures exact inversion. The np.allclose check confirms that numerical round-off errors are negligible (typically on the order of \(10^{-15}\)).

Exercise 4. Why does the magnitude spectrum show both positive and negative frequencies? How do you interpret the negative frequency components for a real-valued signal?

Solution to Exercise 4

The DFT produces coefficients for frequencies \(f_k = k \cdot f_s / N\) for \(k = 0, 1, \ldots, N-1\). Frequencies \(k > N/2\) correspond to negative frequencies (by the periodicity of the DFT, \(f_k = f_k - f_s\) for \(k > N/2\)).

For a real-valued signal, the DFT satisfies the conjugate symmetry property: \(X_{N-k} = X_k^*\). This means positive and negative frequency components have equal magnitudes. The physical content is fully captured by the positive frequencies; the negative frequencies are redundant but necessary for the inverse transform to produce a real-valued result.