How to get the exact amplitudes of an irregular wave?

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Martin Thomas 2020-8-15

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评论： Star Strider 2020-8-27

I have time and displacement data of an irregular wave, and generated the amplitude spectrum by fft but the spectrum is very peaky it has lot of amplitudes (figure2). I need to extract the exact amplitude from that spectrum. For eg: given fig 1 has two peak amplitudes i need to get those amplitude at one time. If there was only one peak then it is easy to check by giving max command but for multiple peaks i dont know how to get that.

After fft i got the below figure, how to i get the exact amplitude values from the below figure?. Is there is any other way to extract the amplitudes?

The code i used for irregular sinwave to get the amplitude is,

T=time

X=Displacement

L = length(T);

nfft = 2^nextpow2(L);

f = Fs/2*linspace(0,1,nfft/2+1);

xfft = fft(X,nfft)/L;

xfftwant = 2*xfft(1:length(xfft)/2+1);

plot(f,abs(xfftwant ))

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Star Strider 2020-8-15

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The second figure displays broadband noise. The easiest way to deal with that would be to denoise the original time-domain signal, either using the smoothdata function, the Wavelet Toolbox functions, or the Signal Processing Toolbox sgolayfilt function. The reason is that denoising the time-domain signal may be the easiest, since that allows interactive visual determination of the smoothed signal and the original signal. Then take the fft of the denoised signal to get the spectrum you want.

With broadband noise, frequency-selective filters will not work.

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Star Strider 2020-8-18

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I did some processing of your data, and I get significantly different results.

First, it is straightforward to read your data with textscan.

Second, the sampling intervals vary significantly, although they average out to produce a sampling frequency of about 128 Hz. After resampling them to a constant sampling interval (necessary for all signal processing including the fft), the data are much less noisy and relatively easy to interpret.

fidi1 = fopen('stillwaterT223H75.tom','rt');
D1C = textscan(fidi1, '%f%f%f%f', 'Delimiter','\t', 'CollectOutput',1);
D1 = cell2mat(D1C);
fclose(fidi1);
fidi2 = fopen('Te223Hs75.tom','rt');
D2C = textscan(fidi2, '%f%f%f%f', 'Delimiter','\t', 'CollectOutput',1);
D2 = cell2mat(D2C);
fclose(fidi2);
Fs=128;                 %sampling frequency
[D1r,t1r] = resample(D1(:,2:end), D1(:,1), Fs);                     % Resample To Constant Sampling Interval
[D2r,t2r] = resample(D2(:,2:end), D2(:,1), Fs);                     % Resample To Constant Sampling Interval
figure
plot(t1r, D1r)
grid
xlabel('Time')
ylabel('Amplitude')
title('stillwaterT223H75')
figure
plot(t2r, D2r)
grid
xlabel('Time')
ylabel('Amplitude')
title('Te223Hs75')
D2rs = D2r - mean(D1r);                                             % Subtract ‘mean(stillwaterT223H75)’ From ‘Te223Hs75’
D2rsm = mean(D2rs);
figure
plot(t2r, D2rs)
grid
title('Te223Hs75 Zeroed')
D2rss = smoothdata(D2rs, 'gaussian', 1E+3);                         % Choose Smoothing Method & Window
figure
plot(t2r, D2rss)
grid
title('Te223Hs75 Smoothed')
L = numel(t2r);                                                     % Data Length
Fn = Fs/2;                                                          % Nyquist Frequency
FTD2rs = fft(D2rss - mean(D2rss))/L;
Fv = linspace(0, 1, fix(L/2)+1)*Fn;                                 % Frequency Vector
Ix = 1:numel(Fv);                                                   % Index Vector
figure
nsp = size(FTD2rs,2);
for k = 1:nsp
    subplot(nsp,1,k)
    plot(Fv, abs(FTD2rs(Ix,k))*2)
    grid
    xlim([0 0.1])
    ylabel('Amplitude')
    title(sprintf('Column %d',k))
end
xlabel('Frequency (Hz)')

The Fourier transform results (after subtracting the means of each column to eliminiate the D-C offset and make the remaining peaks more apparent) are then:

I found your code somewhat difficult to follow, however I believe I captured the essence of it in my code.

Experiment with my code — especially with the smoothdata arguments — to get the result you want. I have no idea what I am looking at, or what the ‘correct’ result would be.

Martin Thomas 2020-8-27

hey one last doubt, fft result has more values than we need how to trim that values.? how to identify?

Star Strider 2020-8-27

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I am not certain what you want to do. I would not change the fft call to produce fewer points, although that is an option using the optional second argument to the fft function. Depending on what you want, you can just trim the end of the ‘FTD2rs’ matrix, or decimate it, for example as:

FTD2rsd = FTD2rs(1:10:end,:);

to keep only every tenth value. Change the ‘step’ value (10 here) to produce different results and different row-length ‘FTD2rsd’ variables. There are other possibilities as well, such as interpolating it to a new frequency vector, using linspace to define the interopolation vector and interp1 to do the actual interpolation. The choice is yours.

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