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How can I find the peak location from the FFT output?

20 vues (au cours des 30 derniers jours)
University
University le 25 Oct 2023
Modifié(e) : Star Strider le 26 Oct 2023
Hi,
Please, how can I find the peak location from the FFT output? I have attached my m file

Réponse acceptée

Star Strider
Star Strider le 25 Oct 2023
Missing data file.
However this is straightforwazrd —
Fs = 1000;
L = 10;
t = linspace(0, L*Fs, L*Fs+1)/Fs;
% [1 2 3]'*t
% return
s = sum(sin([1 100:100:400]'*t*2*pi),1).';
t = t(:);
figure
plot(t, s)
grid
Fn = Fs/2;
L = numel(t);
NFFT = 2^nextpow2(L);
FTs = fft(s.*hann(L), NFFT)/L;
Fv = linspace(0, 1, NFFT/2+1)*Fn;
Iv = 1:numel(Fv);
[pks,locs] = findpeaks(abs(FTs(Iv))*2, 'MinPeakProminence',0.25)
pks = 5×1
0.4825 0.4810 0.4952 0.4952 0.4810
locs = 5×1
17 1639 3278 4916 6555
Results = table(Fv(locs).', pks, 'VariableNames',{'Frequency','Peak Magnitude'})
Results = 5×2 table
Frequency Peak Magnitude _________ ______________ 0.97656 0.48247 99.976 0.48103 200.01 0.49516 299.99 0.49516 400.02 0.48103
figure
plot(Fv, abs(FTs(Iv))*2)
grid
ylim([min(ylim) 1.5*max(ylim)])
text(Results{:,1}, Results{:,2}, compose('\\leftarrow Freq = %.1f\n Mag = %.3f',Results{:,[1 2]}), 'Rotation',45)
That approach should also work for your data.
.
  8 commentaires
Star Strider
Star Strider le 26 Oct 2023
Modifié(e) : Star Strider le 26 Oct 2023
As always, my pleasure!
Sure!
It looks like you allready solved that.
Here, I changed ‘FFT1’ to ‘FFT2’ to return two-sided Fourier transforms —
F = openfig('plot2.fig');
T1 = readtable('Fourier_data101.xlsx')
T1 = 101×2 table
time theta ____ _______ 0 0.77481 0.02 0.35591 0.04 0.18156 0.06 0.21909 0.08 0.21391 0.1 0.24668 0.12 0.29114 0.14 0.31926 0.16 0.3504 0.18 0.37767 0.2 0.39259 0.22 0.39536 0.24 0.38582 0.26 0.36519 0.28 0.33557 0.3 0.29983
t1 = T1{:,1};
s1 = T1{:,2};
T2 = readtable('FFT_spatial.xlsx')
T2 = 857×2 table
x theta ___________ ___________ -1e-05 -1.6396e-19 -1e-05 5.2143e-07 -9.9543e-06 0.0017766 -9.9099e-06 0.003522 -9.9087e-06 0.0035718 -9.9074e-06 0.0036231 -9.863e-06 0.0053857 -9.8196e-06 0.0070574 -9.817e-06 0.0071556 -9.8144e-06 0.0072566 -9.7711e-06 0.0088551 -9.7292e-06 0.010345 -9.7251e-06 0.010482 -9.7212e-06 0.010617 -9.6782e-06 0.012015 -9.6357e-06 0.013347
t2 = T2{:,1};
s2 = T2{:,2};
Fs2 = 1/mean(diff(t2))
Fs2 = 4.2800e+07
[s2r,t2r] = resample(s2, t2, Fs2); % Resample To Constant Sampling Frequency
figure
plot(t2, s2, 'DisplayName','Original Signal')
hold on
plot(t2r, s2r, 'DisplayNAme','Resampled Signal')
hold off
grid
xlabel('Time')
ylabel('Amplitude')
title('FFT\_spatial — Time Domain Plot')
legend('Location','best')
[FTs1,Fv1] = FFT2(s1,t1); % Fourier_data101
[pks1,locs1] = findpeaks(abs(FTs1)*2, 'MinPeakProminence',0.005)
pks1 = 3×1
0.1120 0.0186 0.1120
locs1 = 3×1
59 65 71
Results1 = table(Fv1(locs1).', pks1, 'VariableNames',{'Frequency','Peak Magnitude'})
Results1 = 3×2 table
Frequency Peak Magnitude _________ ______________ -2.3438 0.11201 0 0.018623 2.3438 0.11201
figure
plot(Fv1, abs(FTs1)*2)
grid
ylim([min(ylim) 1.5*max(ylim)])
xlabel('Frequency')
ylabel('Magnitude')
title('Fourier\_data101')
xlim(xlim/3)
text(Results1{:,1}, Results1{:,2}, compose('\\leftarrow Freq = %.2f\n Mag = %.3f',Results1{:,[1 2]}), 'Rotation',45)
[FTs2,Fv2] = FFT2(s2r,t2r); % FFT_spatial
[pks2,locs2] = findpeaks(abs(FTs2)*2, 'MinPeakProminence',0.01)
pks2 = 8×1
0.2116 0.2211 0.1216 0.1947 0.1947 0.1216 0.2211 0.2116
locs2 = 8×1
497 499 502 512 514 524 527 529
Results2 = table(Fv2(locs2).', pks2, 'VariableNames',{'Frequency','Peak Magnitude'})
Results2 = 8×2 table
Frequency Peak Magnitude ___________ ______________ -6.6875e+05 0.21162 -5.8516e+05 0.22111 -4.5977e+05 0.12159 -41797 0.19473 41797 0.19473 4.5977e+05 0.12159 5.8516e+05 0.22111 6.6875e+05 0.21162
figure
plot(Fv2, abs(FTs2)*2)
grid
ylim([min(ylim) 1.5*max(ylim)])
xlabel('Frequency')
ylabel('Magnitude')
title('FFT\_spatial')
xlim(xlim/12)
text(Results2{:,1}, Results2{:,2}, compose('\\leftarrow Freq = %.2f\n Mag = %.3f',Results2{:,[1 2]}), 'Rotation',45)
function [FTs2,Fv] = FFT2(s,t)
s = s(:);
t = t(:);
L = numel(t);
Fs = 1/mean(diff(t));
Fn = Fs/2;
NFFT = 2^nextpow2(L);
FTs = fftshift(fft((s - mean(s)).*hann(L), NFFT)/sum(hann(L)));
Fv = linspace(-Fs/2, Fs/2-Fs/length(FTs), length(FTs));
Iv = 1:numel(Fv);
FTs2 = FTs(Iv);
end
Make approopriate changes to the rest of the code to get the result you want. (I do not recoognise the spectrum in ‘plot2.fig’ so I guess it was not a signal already uploaded here.)
EDIT — (26 Oct 2023 at 11:12)
Resmapling is necessary because the fft function requires evenly-sampled (or reasonably close to evenly-sampled) data. (The nufft function does not, however I prefer to use fft with resampled data, since resampling also permits other signal processing techniques such as filtering, that are otherwise impossible.) Looking at ‘t2’, the mean of the differences (‘diff(t)’) is 2.3364e-8 and the standard deviation of the differences is 1.6398e-8. It should be several orders-of-magnitude less than the mean. The sampling time differences range from 2.1069e-12 to 7.0182e-8. That range is simply too much for fft (or other signal processing functions, all of which require regular sampling) to deal with. Without resampling, the results would be inaccurate. Resampling to a common sampling frequency (chosen here as the inverse of the mean of the differences (using the median would also work), makes the subsequent analyses reliable. The time-domain plots would look substantially the same when plotted (added above).
.
University
University le 26 Oct 2023
Modifié(e) : University le 26 Oct 2023
Thank you so much for your assistance. Please what is the function of "hann function" in Matlab?

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