如何使用FFT查找方波的频率

Question

In the FFT (second) plot, I am expecting a bigger peak at frequency = 1.0, compared to other frequencies, since it is a 1 Hz Square Wave signal sampled at 5Hz.

I am a beginner at this, possibly missing something silly here Here's what I have done:

import numpy as np
from matplotlib import pyplot as plt
from scipy import signal
t500 = np.linspace(0,5,500,endpoint=False)
s1t500 = signal.square(2*np.pi*1.0*t500)

First plot shows 1 Hz Square Wave sampled at 5Hz for 5 seconds:

t5 = np.linspace(0,5,25,endpoint=False)
t5 = t5 + 1e-14
s1t5 = signal.square(2.0*np.pi*1.0*t5)
plt.ylim(-2,2); plt.plot(t500,s1t500,'k',t5,s1t5,'b',t5,s1t5,'bo'); plt.show()

Here in the Second plot, I am expecting the magnitude at f=1 Hz to be more than at f=2. Am I missing something ?

y1t5 = np.fft.fft(s1t5)
ff1t5 = np.fft.fftfreq(25,d=0.2)
plt.plot(ff1t5,y1t5); plt.show()

Answer 1

It seems you missed the fact that Fourier transform produces functions (or sequences of numbers in case of DFT/FFT) in complex space:

>>> np.fft.fft(s1t5)
[ 5. +0.j          0. +0.j          0. +0.j          0. +0.j          0. +0.j
  5.-15.38841769j  0. +0.j          0. +0.j          0. +0.j          0. +0.j
  5. +3.63271264j  0. +0.j          0. +0.j          0. +0.j          0. +0.j
# and so on

In order to see the amplitude spectrum on your plot, apply np.absolute or abs :

>>> np.absolute(np.fft.fft(s1t5))
[  5.           0.           0.           0.           0.          16.18033989
   0.           0.           0.           0.           6.18033989   0.           0.
   0.           0.           6.18033989   0.           0.           0.           0.
   16.18033989  0.           0.           0.           0.        ]

Otherwise only real part will be shown.