I've built a function that deals with plotting FFT of real signals. The extra bonus in my function relative to the previous answers is that you get the actual amplitude of the signal.
Also, because of the assumption of a real signal, the FFT is symmetric, so we can plot only the positive side of the x-axis:
import matplotlib.pyplot as plt
import numpy as np
import warnings
def fftPlot(sig, dt=None, plot=True):
# Here it's assumes analytic signal (real signal...) - so only half of the axis is required
if dt is None:
dt = 1
t = np.arange(0, sig.shape[-1])
xLabel = 'samples'
else:
t = np.arange(0, sig.shape[-1]) * dt
xLabel = 'freq [Hz]'
if sig.shape[0] % 2 != 0:
warnings.warn("signal preferred to be even in size, autoFixing it...")
t = t[0:-1]
sig = sig[0:-1]
sigFFT = np.fft.fft(sig) / t.shape[0] # Divided by size t for coherent magnitude
freq = np.fft.fftfreq(t.shape[0], d=dt)
# Plot analytic signal - right half of frequence axis needed only...
firstNegInd = np.argmax(freq < 0)
freqAxisPos = freq[0:firstNegInd]
sigFFTPos = 2 * sigFFT[0:firstNegInd] # *2 because of magnitude of analytic signal
if plot:
plt.figure()
plt.plot(freqAxisPos, np.abs(sigFFTPos))
plt.xlabel(xLabel)
plt.ylabel('mag')
plt.title('Analytic FFT plot')
plt.show()
return sigFFTPos, freqAxisPos
if __name__ == "__main__":
dt = 1 / 1000
# Build a signal within Nyquist - the result will be the positive FFT with actual magnitude
f0 = 200 # [Hz]
t = np.arange(0, 1 + dt, dt)
sig = 1 * np.sin(2 * np.pi * f0 * t) + \
10 * np.sin(2 * np.pi * f0 / 2 * t) + \
3 * np.sin(2 * np.pi * f0 / 4 * t) +\
7.5 * np.sin(2 * np.pi * f0 / 5 * t)
# Result in frequencies
fftPlot(sig, dt=dt)
# Result in samples (if the frequencies axis is unknown)
fftPlot(sig)
