CODEWITHSUNDEEP
pythonnumpypython-3.xmatplotlib
Sibbs Gambling
Sibbs Gambling

Reputation: 20345

Python - FFT leads to wrong physical meanings

I am new to Python. I intend to do Fourier Transform to an array of discrete points, (time, acceleration), and plot the result out.

I copy and paste the sample FFT code, and modify accordingly.

Please see codes:

import numpy as np
import matplotlib.pyplot as plt

# Load the .txt file in
myData = np.loadtxt('twenty_z_up.txt')

# Extract the time and acceleration columns
time = copy(myData[:,0])

# Extract the acceleration columns
zAcc = copy(myData[:,3])

t = np.arange(10080)
sp = np.fft.fft(zAcc)
freq = np.fft.fftfreq(t.shape[-1])
plt.plot(freq, sp.real)

myData is a rectangular matrix with 10080 rows and 10 columns.

Thus, zAcc is the row3 extracted from the matrix.

In the plot drawn by Spyder, most of the harmonics concentrated around 0. They are all extremely small.

But my data are actually the accelerations of the phone carried by a walking person (including the gravity). So I expect the most significant harmonic happens around 2Hz.

Why is the graph non-sense?

Thanks in advance!

==============UPDATES: My Graphs====================== Time Domain

After the above-mentioned codes

Zoomed-in

The first time domain one:

x-axis is in millisecond.

y-axis is in m/s^2, due to earth gravity, it has a DC offset of ~10.

Upvotes: 1

Views: 3007

Answers (2)

Henry Gomersall
Henry Gomersall

Reputation: 8692

You do get two spikes at (approximately) 2Hz. Your sampling period is around 2.8 ms (as best as I can infer from your first plot), giving +/-2Hz the normalized frequency of +/-0.056, which is about where your spikes are. fft.fftfreq by default returns the normalized frequency (which scales the sampling period). You can set the d argument to be the sampling period, and you'll get a vector containing the actual frequency.

Your huge spike in the middle is obviously the DC offset (which you can trivially remove by subtracting the mean).

Upvotes: 3

gg349
gg349

Reputation: 22671

As others said, we need to see the data, post it somewhere. Just to check, try first fixing the timestep size in fftfreq, then plot this synthetic signal, and then plot your signal to see how they compare:

timestep=1./50.#Assume sampling at 50Hz. Change this accordingly.
N=10080#the number of samples
T=N*timestep
t = np.linspace(0,T,N)#needed only to generate xAcc_synthetic
freq=2.#peak a frequency at 2Hz
#generate synthetic signal at 2Hz and add some noise to it
xAcc_synthetic = sin((2*np.pi)*freq*t)+np.random.rand(N)*0.2
sp_synthetic = np.fft.fft(xAcc_synthetic)
freq = np.fft.fftfreq(t.size,d=timestep)
print max(abs(freq))==(1/timestep)/2.#simple check highest freq.
plt.plot(freq, abs(sp_synthetic))
xlabel('Hz')

Now, at the x axis equal to 2 you actually have a physical frequency of 2Hz, and you may spot the more pronounced peak you are looking for. Moreover, you may want to have a look also at yAcc and zAcc.

Upvotes: 1

Related Questions