Choosing one peak out of two in FFT function

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Hussein Kokash 2022-10-5

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评论： William Rose 2022-10-6

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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Answer 1

William Rose 2022-10-5

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@Hussein Kokash,

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.

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William Rose 2022-10-5

在 MATLAB Online 中打开

@Hussein Kokash,

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.

Hussein Kokash 2022-10-5

Hello William,

I appreciate your responces, so I am investigating a flow behind an object, I have a line of points behind this object (in the spanwise direction).

At each time step, I have a file that contains the spanwise (z) velocity for all of the points that makes this line (zmin to zmax), this is a sample:

0 -2.9577131E-12 0.236238

0.002 2.2367207E-10 0.236238

0.004 3.4525567E-10 0.236238

0.006 3.1392004E-10 0.236238

0.008 1.7650122E-10 0.236238

0.01 -7.5099878E-12 0.236238

0.012 -1.8817284E-10 0.236238

0.014 -3.1705425E-10 0.236238

0.016 -3.3854163E-10 0.236238

0.018 -2.1011342E-10 0.236238

0.02 1.8830366E-11 0.236238

0.022 2.4242510E-10 0.236238

0.024 3.5704751E-10 0.236238

0.026 3.1862220E-10 0.236238

0.028 1.7535199E-10 0.236238

0.03 -1.3538867E-11 0.236238

0.032 -1.9900941E-10 0.236238

0.034 -3.3320666E-10 0.236238

0.036 -3.6012169E-10 0.236238

0.038 -2.3574904E-10 0.236238

0.04 -7.6009011E-12 0.236238

0.042 2.1944382E-10 0.236238

0.044 3.4179600E-10 0.236238

0.046 3.1526771E-10 0.236238

0.048 1.8628746E-10 0.236238

0.05 1.0866790E-11 0.236238

0.052 -1.6580260E-10 0.236238

0.054 -2.9918222E-10 0.236238

0.056 -3.3471170E-10 0.236238

0.058 -2.2668630E-10 0.236238

0.06 -1.6748083E-11 0.236238

0.062 1.9794763E-10 0.236238

0.064 3.1805352E-10 0.236238

0.066 2.9838742E-10 0.236238

0.068 1.8193196E-10 0.236238

0.07 2.0337331E-11 0.236238

0.072 -1.4520383E-10 0.236238

0.074 -2.7288414E-10 0.236238

0.076 -3.0918668E-10 0.236238

0.078 -2.0794240E-10 0.236238

0.08 -9.7146442E-12 0.236238

0.082 1.9173136E-10 0.236238

0.084 3.0145340E-10 0.236238

0.086 2.7771161E-10 0.236238

0.088 1.6262074E-10 0.236238

0.09 5.5778040E-12 0.236238

0.092 -1.5427218E-10 0.236238

0.094 -2.7635975E-10 0.236238

0.096 -3.0763103E-10 0.236238

0.098 -2.0193161E-10 0.236238

0.1 3.3841662E-13 0.236238

0.102 2.0144567E-10 0.236238

0.104 3.0403225E-10 0.236238

0.106 2.6902292E-10 0.236238

0.108 1.4431987E-10 0.236238

0.11 -1.6204361E-11 0.236238

0.112 -1.7232844E-10 0.236238

0.114 -2.8591619E-10 0.236238

0.116 -3.0836106E-10 0.236238

0.118 -1.9723919E-10 0.236238

0.12 7.6309077E-12 0.236238

0.122 2.1455048E-10 0.236238

0.124 3.2751216E-10 0.236238

0.126 3.0043147E-10 0.236238

0.128 1.7559393E-10 0.236238

0.13 4.9337891E-12 0.236238

0.132 -1.6878562E-10 0.236238

0.134 -3.0112440E-10 0.236238

0.136 -3.3568558E-10 0.236238

0.138 -2.2463601E-10 0.236238

0.14 -1.0262878E-11 0.236238

0.142 2.1156070E-10 0.236238

0.144 3.3978994E-10 0.236238

0.146 3.2244513E-10 0.236238

0.148 1.9825914E-10 0.236238

0.15 1.9627189E-11 0.236238

0.152 -1.6648793E-10 0.236238

0.154 -3.0945631E-10 0.236238

0.156 -3.4744837E-10 0.236238

0.158 -2.3123037E-10 0.236238

0.16 -9.0491139E-12 0.236238

0.162 2.1339540E-10 0.236238

0.164 3.3424095E-10 0.236238

0.166 3.0936684E-10 0.236238

0.168 1.8137765E-10 0.236238

0.17 3.1284328E-12 0.236238

0.172 -1.8034544E-10 0.236238

0.174 -3.2127409E-10 0.236238

0.176 -3.5975507E-10 0.236238

0.178 -2.4529187E-10 0.236238

0.18 -1.9049244E-11 0.236238

0.182 2.1396086E-10 0.236238

0.184 3.4508396E-10 0.236238

0.186 3.2556036E-10 0.236238

0.188 1.9843574E-10 0.236238

0.19 1.9107855E-11 0.236238

0.192 -1.6491127E-10 0.236238

0.194 -3.0476838E-10 0.236238

0.196 -3.4132776E-10 0.236238

0.198 -2.2626663E-10 0.236238

0.2 -2.9577131E-12 0.236238

1st column (z), 2nd column (coresspoding z-velocity), 3rd colum (corresponding time)

The idea is I have say many files like these, where time step is variant.

The reason that I want to know the frequency at of each peak for each file is to see if there is a change in wavelength in time.

What makes it challenging is there are two main frequencies and if I just used the regular max or findpeaks, it would not be quite accurate which one was captured as a peak/frequency.

Thanks again!

William Rose 2022-10-6

@Hussein Kokash,

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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Choosing one peak out of two in FFT function

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