Choosing one peak out of two in FFT function
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Hello,
I have a question about how to choose only one out the two peaks that are produced after applying FFT function.
I have multiple text files where each one has "y" values, each file is at a certain time step so the peak of the FFT function is different for every file. Each one produces two peaks the same way in the image, is there a way to take just one of the highest two peaks for each file and store them to either plot or post process.
I tried findpeaks, but the problem is since it is varying peaks, it is difficult to set a parameter such as 'MinPeakDistance' or other.
I use the following:
A = fft(y);
B = abs(A);
figure()
hold on
semilogy(B)
hold off
even if I use C = max(abs(fft(y))); it would not decide which max to choose
Note: the first image is for one file, the second image is for multiple files
Thank you!
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William Rose
2022년 10월 5일
0 개 추천
The output of y=fft(x) is symmetric about the middle of the array y.* This is due to the mathematics of the discrete Fourier transform. So when identifying peaks, you can ignore those that occur above element N/2, where N is the length of x and of y.
*When the input sequence,x, is real numbers.
댓글 수: 7
Hussein Kokash
2022년 10월 5일
Any way to save every peak in different array? say peak1 and peak2 for each file, save in seperate arrays.
something like, P1 (saving the 1st peak for all files), P2 (saving the 2nd peak for all files).
The only reason is to extract the corresponding two x-positions (frequency) for each file
Thanks again William!
Here is an example using a waveform whose Fourier transform is similar to the one you showed. I am using your notation for y and B.
N=101; %number of points in signal
dt=1; %sampling interval (s)
t=(0:N-1)*dt; %time values
y=sawtooth(2*pi*10*t/N,.3); %triangle wave
B=abs(fft(y)); %discrete Fourier transform
subplot(211); plot(y) %plot y=original waveform
title('y')
subplot(212); plot(B) %plot B=abs(fft(y))
title('B')
Since you want to find the peaks, I assume you also want to find the frequencies of the peaks. If so, you need to know the frequencies that go with the fft. Given sampling interval dt, and signal duration N (in points), the frequencies associated with fft(y) are
f=(0:N-1)/(N*dt); %frequencies (in Hz, if dt is in seconds)
With this information, you can plot B versus frequency.
figure;
subplot(211); plot(t,y);
title('y'); xlabel('Time (s)')
subplot(212); plot(f,B);
title('B'); xlabel('Frequency (Hz)'); grid on
Since B is symmetric about its middle, let's keep only the lower half of B (and f):
N2=floor(N/2)+1; %number of points in first half of B, plus 1
B=B(1:N2); f=f(1:N2);
Now do findpeaks on B to find the peak frequencies.
[pks,freqs] = findpeaks(B,f);
[pk1,idx1]=max(pks); %biggest peak and its index
f1=freqs(idx1); %frequency of biggest peak
hold on %hold existing plot
plot(freqs,pks,'g+') %plot + at every peak
plot(f1,pk1,'r*') %plot * at biggest peak
Try it.
Image Analyst
2022년 10월 5일
William Rose
2022년 10월 5일
@Hussein Kokash, I had not read your comment when I posted my comment. Let me think about your question...
The code I gave in my comment makes a vector freqs, which has the frequencies of all the peaks in the lower half of the fft. And it makes a scalar f1 with the frequency of the biggest peak. Due to the mathematics of the discrete Fourier transform, the frequencies of the upper half are the mirror image of the lower half frequencies. The middle frequency is called the Nyquist frequency. It is exactly half the sampling frequency:
where dt is the sampling interval, in units of time. Therefore the frequencies of the upper half peaks are
fs=1/dt; %dt was specified earlier in my code
freqs_up=fs-freqs;
f1_up=fs-f1;
Good luck.
William Rose
2022년 10월 5일
@Hussein Kokash, Why do you want to save the frequencies of the peaks in the upper half of the FFT? The reason I ask is that the upper half frequencies do not contain any additional useful informaiton.
Analogy: If you know and have saved p=probability of heads, for a particular coin, there is no need to save q=1-p=probability of tails.
Hussein Kokash
2022년 10월 5일
William Rose
2022년 10월 6일
You're welcome.
You have an interesting problem and an interesting experiment.
The code I provided gives f1, the frequency with the most power in the signal. Obviously the associated wavelength is lambda1=c/f1. The frequencies of the peaks in the "top half" of the spectrum (the part above the Nyquist frequency) have no additional value for your analysis.
Good luck with your work!
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