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======================
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.
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).
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.
Related
I have a vibration data in time domain and want to convert it to frequency domain with fft. However the plot of the FFT only shows a big spike at zero and nothing else.
This is my vibration data: https://pastebin.com/7RK57kJW
My code:
import numpy as np
import matplotlib.pyplot as plt
t = np.arange(3000)
a1_fft= np.fft.fft(a1, axis=0)
freq = np.fft.fftfreq(t.shape[-1])
plt.plot(freq, a1_fft)
My FFT Plot:
What am I doing wrong here? I am pretty sure my data is uniform, which provoces in other cases a similar problem with fft.
The bins of the FFT correspond to the frequencies at 0, df, 2df, 3df, ..., F-2df, F-df, where df is determined by the number of bins and F is 1 cycle per bin.
Notice the zero frequency at the beginning. This is called the DC offset. It's the mean of your data. In the data that you show, the mean is ~1.32, while the amplitude of the sine wave is around 0.04. It's not surprising that you can't see a peak that's 33x smaller than the DC term.
There are some common ways to visualize the data that help you get around this. One common methods is to keep the DC offset but use a log scale, at least for the y-axis:
plt.semilogy(freq, a1_fft)
OR
plt.loglog(freq, a1_fft)
Another thing you can do is zoom in on the bottom 1/33rd or so of the plot. You can do this manually, or by adjusting the span of the displayed Y-axis:
p = np.abs(a1_fft[1:]).max() * [-1.1, 1.1]
plt.ylim(p)
If you are plotting the absolute values already, use
p = np.abs(a1_fft[1:]).max() * [-0.1, 1.1]
Another method is to remove the DC offset. A more elegant way of doing this than what #J. Schmidt suggests is to simply not display the DC term:
plt.plot(freq[1:], a1_fft[1:])
Or for the positive frequencies only:
n = freq.size
plt.plot(freq[1:n//2], a1_fft[1:n//2])
The cutoff at n // 2 is only approximate. The correct cutoff depends on whether the FFT has an even or odd number of elements. For even numbers, the middle bin actual has energy from both sides of the spectrum and often gets special treatment.
The peak at 0 is the DC-gain, which is very high since you didn't normalize your data. Also, the Fourier transform is a complex number, you should plot the absolute value and phase separately. In this code I also plotted only the positive frequencies:
import numpy as np
import matplotlib.pyplot as plt
#Import data
a1 = np.loadtxt('a1.txt')
plt.plot(a1)
#Normalize a1
a1 -= np.mean(a1)
#Your code
t = np.arange(3000)
a1_fft= np.fft.fft(a1, axis=0)
freq = np.fft.fftfreq(t.shape[-1])
#Only plot positive frequencies
plt.figure()
plt.plot(freq[freq>=0], np.abs(a1_fft)[freq>=0])
I am trying to perform the discrete fourier transform using the FFT algorithm from numpy in python to a very large dataset (more than 900k points).
Here is the original function's graph:
And here is the transform after I plot it:
I've tried detrending the data using scipy's detrend() function and also subtracting the average from the data points. But the only difference that happens is that the thin gap at 0 Hz is gone.
I was expecting two peaks which would result from the big spike and then the little bump.
What can be causing the transform to come out this way? I have not checked if my time data points are spaced out evenly - I assumed that. The big spike's values go all the way to infinity and for the purposes of applying the FFT, I replaced those points with sys.maxsize. Could this somehow factor into the resulting waveform? I need to understand what is causing the resulting waveform to come out this way.
Here is my code for performing the transform and plotting it:
channelA and time are numpy arrays.
fY = np.fft.fft(channelA)
freq = np.fft.fftfreq(len(channelA), time[1]-time[0])
plt.title("Transform")
plt.xlabel("Frequency / Hz")
plt.ylabel("Magnitude")
plt.plot(freq, np.abs(fY))
plt.show()
I suspect that there's something I'm missing in my understanding of the Fourier Transform, so I'm looking for some correction (if that's the case). How should I gather peak information from the first plot below?
The dataset is hourly data for 911 calls over the past 17 years (for a particular city).
I've removed the trend from my data, and am now removing the seasonality. When I run the Fourier transform, I get the following plot:
I believe the dataset does have some seasonality to it (looking at weekly data, I have this pattern):
How do I pick out the values of the peaks in the first plot? Presumably for all of the "peaks" under, say 5000 in the first plot, I may ignore the inclusion of that seasonality in my final model, but only at a loss of accuracy, correct?
Here's the bit of code I'm working with, currently:
from scipy import fftpack
fft = fftpack.fft(calls_grouped_hour.detrended_residuals - calls_grouped_hour.detrended_residuals.mean())
plt.plot(1./(17*365)*np.arange(len(fft)), np.abs(fft))
plt.xlim([-.1, 23/2]);
EDIT:
After Mark Snider's initial answer, I have the following plot:
Adding code attempt to get peak values from fft:
Do I need to convert the values back using ifft first?
fft_x_y = np.stack((fft.real, fft.imag), -1)
peaks = []
for x, y in np.abs(fft_x_y):
if (y >= 0):
spipeakskes.append(x)
peaks = np.unique(peaks)
print('Length: ', len(peaks))
print('Peak values: ', '\n', np.sort(peaks))
threshold = 5000
fft[np.abs(fft)<threshold] = 0
This'll give you an fft that ignores everything except the peaks. And no, I wouldn't imagine that the "noise" represents actual seasonality. The peak at fft[0] doesn't represent seasonality, either - it's a multiple of the mean of the data, so if you plan on subtracting the ifft of the peaks I wouldn't include fft[0] either unless you want your data to be centered.
If you want just the peak values and not the full fft that you can invert, you can just do this:
peaks = [np.abs(value) for value in fft if np.abs(value)>threshold]
I am analyzing a time-series dataset that I am pretty sure can be broken down using fft. I want to develop a model to estimate the data using a sum of sin/cos but I am having trouble with the syntax to find the frequencies in python
Here is a graph of the data
data graph
And here's a link to the original data: https://drive.google.com/open?id=1mqZtQ-txdd_AFbKGBlbSL6903CK-_kXl
Most of the examples I have seen have multiple samples per second/time period, however the data in this set represent by-minute observations of some metric. Because of this, I've had trouble translating the answers online to this problem
Here's my naive first approach
X = fftpack.fft(data)
freqs = fftpack.fftfreq(len(data))
plt.plot(freqs, np.abs(X))
plt.show()
Instead of peaking at the major frequencies, my plot only has one peak at 0.
result
The FFT you posted has been shifted so that 0 is at the center. Data to the left of the center represents negative frequencies and to the right represents positive frequencies. If you zoom in and look more closely, I think you will see that there are two peaks close to the center that you are interpreting as a single peak at 0. Just looking at the positive side, the location of this peak will tell you which frequency is contributing significant signal power.
Like you said, your x-axis is probably incorrect. scipy.fftpack.fftfreq needs to know the time between samples (in seconds, I think) of your time-domain signal to correctly determine the bandwidth and create the x-axis array in Hz. This should do it:
dt = 60 # 60 seconds between samples
freqs = fftpack.fftfreq(len(data),dt)
Audio processing is pretty new for me. And currently using Python Numpy for processing wave files. After calculating FFT matrix I am getting noisy power values for non-existent frequencies. I am interested in visualizing the data and accuracy is not a high priority. Is there a safe way to calculate the clipping value to remove these values, or should I use all FFT matrices for each sample set to come up with an average number ?
regards
Edit:
from numpy import *
import wave
import pymedia.audio.sound as sound
import time, struct
from pylab import ion, plot, draw, show
fp = wave.open("500-200f.wav", "rb")
sample_rate = fp.getframerate()
total_num_samps = fp.getnframes()
fft_length = 2048.
num_fft = (total_num_samps / fft_length ) - 2
temp = zeros((num_fft,fft_length), float)
for i in range(num_fft):
tempb = fp.readframes(fft_length);
data = struct.unpack("%dH"%(fft_length), tempb)
temp[i,:] = array(data, short)
pts = fft_length/2+1
data = (abs(fft.rfft(temp, fft_length)) / (pts))[:pts]
x_axis = arange(pts)*sample_rate*.5/pts
spec_range = pts
plot(x_axis, data[0])
show()
Here is the plot in non-logarithmic scale, for synthetic wave file containing 500hz(fading out) + 200hz sine wave created using Goldwave.
Simulated waveforms shouldn't show FFTs like your figure, so something is very wrong, and probably not with the FFT, but with the input waveform. The main problem in your plot is not the ripples, but the harmonics around 1000 Hz, and the subharmonic at 500 Hz. A simulated waveform shouldn't show any of this (for example, see my plot below).
First, you probably want to just try plotting out the raw waveform, and this will likely point to an obvious problem. Also, it seems odd to have a wave unpack to unsigned shorts, i.e. "H", and especially after this to not have a large zero-frequency component.
I was able to get a pretty close duplicate to your FFT by applying clipping to the waveform, as was suggested by both the subharmonic and higher harmonics (and Trevor). You could be introducing clipping either in the simulation or the unpacking. Either way, I bypassed this by creating the waveforms in numpy to start with.
Here's what the proper FFT should look like (i.e. basically perfect, except for the broadening of the peaks due to the windowing)
Here's one from a waveform that's been clipped (and is very similar to your FFT, from the subharmonic to the precise pattern of the three higher harmonics around 1000 Hz)
Here's the code I used to generate these
from numpy import *
from pylab import ion, plot, draw, show, xlabel, ylabel, figure
sample_rate = 20000.
times = arange(0, 10., 1./sample_rate)
wfm0 = sin(2*pi*200.*times)
wfm1 = sin(2*pi*500.*times) *(10.-times)/10.
wfm = wfm0+wfm1
# int test
#wfm *= 2**8
#wfm = wfm.astype(int16)
#wfm = wfm.astype(float)
# abs test
#wfm = abs(wfm)
# clip test
#wfm = clip(wfm, -1.2, 1.2)
fft_length = 5*2048.
total_num_samps = len(times)
num_fft = (total_num_samps / fft_length ) - 2
temp = zeros((num_fft,fft_length), float)
for i in range(num_fft):
temp[i,:] = wfm[i*fft_length:(i+1)*fft_length]
pts = fft_length/2+1
data = (abs(fft.rfft(temp, fft_length)) / (pts))[:pts]
x_axis = arange(pts)*sample_rate*.5/pts
spec_range = pts
plot(x_axis, data[2], linewidth=3)
xlabel("freq (Hz)")
ylabel('abs(FFT)')
show()
FFT's because they are windowed and sampled cause aliasing and sampling in the frequency domain as well. Filtering in the time domain is just multiplication in the frequency domain so you may want to just apply a filter which is just multiplying each frequency by a value for the function for the filter you are using. For example multiply by 1 in the passband and by zero every were else. The unexpected values are probably caused by aliasing where higher frequencies are being folded down to the ones you are seeing. The original signal needs to be band limited to half your sampling rate or you will get aliasing. Of more concern is aliasing that is distorting the area of interest because for this band of frequencies you want to know that the frequency is from the expected one.
The other thing to keep in mind is that when you grab a piece of data from a wave file you are mathmatically multiplying it by a square wave. This causes a sinx/x to be convolved with the frequency response to minimize this you can multiply the original windowed signal with something like a Hanning window.
It's worth mentioning for a 1D FFT that the first element (index [0]) contains the DC (zero-frequency) term, the elements [1:N/2] contain the positive frequencies and the elements [N/2+1:N-1] contain the negative frequencies. Since you didn't provide a code sample or additional information about the output of your FFT, I can't rule out the possibility that the "noisy power values at non-existent frequencies" aren't just the negative frequencies of your spectrum.
EDIT: Here is an example of a radix-2 FFT implemented in pure Python with a simple test routine that finds the FFT of a rectangular pulse, [1.,1.,1.,1.,0.,0.,0.,0.]. You can run the example on codepad and see that the FFT of that sequence is
[0j, Negative frequencies
(1+0.414213562373j), ^
0j, |
(1+2.41421356237j), |
(4+0j), <= DC term
(1-2.41421356237j), |
0j, v
(1-0.414213562373j)] Positive frequencies
Note that the code prints out the Fourier coefficients in order of ascending frequency, i.e. from the highest negative frequency up to DC, and then up to the highest positive frequency.
I don't know enough from your question to actually answer anything specific.
But here are a couple of things to try from my own experience writing FFTs:
Make sure you are following Nyquist rule
If you are viewing the linear output of the FFT... you will have trouble seeing your own signal and think everything is broken. Make sure you are looking at the dB of your FFT magnitude. (i.e. "plot(10*log10(abs(fft(x))))" )
Create a unitTest for your FFT() function by feeding generated data like a pure tone. Then feed the same generated data to Matlab's FFT(). Do a absolute value diff between the two output data series and make sure the max absolute value difference is something like 10^-6 (i.e. the only difference is caused by small floating point errors)
Make sure you are windowing your data
If all of those three things work, then your fft is fine. And your input data is probably the issue.
Check the input data to see if there is clipping http://www.users.globalnet.co.uk/~bunce/clip.gif
Time doamin clipping shows up as mirror images of the signal in the frequency domain at specific regular intervals with less amplitude.