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

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University 2023-10-25

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编辑： Star Strider 2023-10-26

采纳的回答： Star Strider

calfft.m

Hi,

Please, how can I find the peak location from the FFT output? I have attached my m file

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Answer 1

Star Strider 2023-10-25

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https://ww2.mathworks.cn/matlabcentral/answers/2038441-how-can-i-find-the-peak-location-from-the-fft-output#answer_1340506

在 MATLAB Online 中打开

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.

.

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Star Strider 2023-10-25

编辑：Star Strider 2023-10-25

在 MATLAB Online 中打开

As always, my pleasure!

The Fourier transform will work regardless of the independnet variable units. If theat is time, the frequency units are in cycles/time unit. If spatial, cycles/spatial unit (for example cycles/millimetre).

Try this —

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

[FTs1,Fv1] = FFT1(s1,t1); % Fourier_data101

[pks1,locs1] = findpeaks(abs(FTs1)*2, 'MinPeakProminence',0.05)

pks1 = 0.1120

locs1 = 7

Results1 = table(Fv1(locs1).', pks1, 'VariableNames',{'Frequency','Peak Magnitude'})

Results1 = 1×2 table

Frequency Peak Magnitude _________ ______________ 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')

text(Results1{:,1}, Results1{:,2}, compose('\\leftarrow Freq = %.1f\n Mag = %.3f',Results1{:,[1 2]}), 'Rotation',45)

[FTs2,Fv2] = FFT1(s2r,t2r); % FFT_spatial

[pks2,locs2] = findpeaks(abs(FTs2)*2, 'MinPeakProminence',0.1)

pks2 = 0.2211

locs2 = 15

Results2 = table(Fv2(locs2).', pks2, 'VariableNames',{'Frequency','Peak Magnitude'})

Results2 = 1×2 table

Frequency Peak Magnitude __________ ______________ 5.8516e+05 0.22111

figure

plot(Fv2, abs(FTs2)*2)

grid

ylim([min(ylim) 1.5*max(ylim)])

xlabel('Frequency')

ylabel('Magnitude')

title('FFT\_spatial')

text(Results2{:,1}, Results2{:,2}, compose('\\leftarrow Freq = %.1f\n Mag = %.3f',Results2{:,[1 2]}), 'Rotation',45)

function [FTs1,Fv] = FFT1(s,t)

s = s(:);

t = t(:);

L = numel(t);

Fs = 1/mean(diff(t));

Fn = Fs/2;

NFFT = 2^nextpow2(L);

FTs = fft((s - mean(s)).*hann(L), NFFT)/sum(hann(L));

Fv = linspace(0, 1, NFFT/2+1)*Fn;

Iv = 1:numel(Fv);

FTs1 = FTs(Iv);

end

That seems appropriate to me, however if you want to identify more peaks, decrese the 'MinPeakProminence' value, or eliminate it altogether if desired. You can also use xlim to decrease the width of the plots, to show more detail (especially at the lower frequency values).

My apologies for not seeing ‘Fourier_data101.xlsx’ earlier. I was off doing other things for a while.

.

Star Strider 2023-10-26

编辑：Star Strider 2023-10-26

在 MATLAB Online 中打开

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 2023-10-26

编辑：University 2023-10-26

Thank you so much for your assistance. Please what is the function of "hann function" in Matlab?

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

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

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