Filter data series on amplitude as a function of time

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Dear,
Along with greeting, I need to filter a time series, specifically I need to obtain the maximum peak (maximum amplitude that is on the vertical axis), and calculate the time difference (plotted on the horizontal axis) between this peak and the previous peak, in addition to obtain the time difference between the maximum peak and the next peak.
Beforehand thank you very much.
I am attaching code that I have in the meantime and file with the data.
close all, clear all, clc
data=load('data.txt');
t=0:10:4*3600;
plot(t,data)
xlabel('Time (s)')
ylabel('Amplitude (m)')
grid on

Accepted Answer

Star Strider
Star Strider on 27 May 2021
Try this —
s = readmatrix('https://www.mathworks.com/matlabcentral/answers/uploaded_files/632075/data.txt');
t = linspace(0,numel(s),numel(s));
[maxpk,idx] = max(s)
maxpk = 7.7414
idx = 314
[pks,locs] = findpeaks(s, 'MinPeakProminence',0.05);
didx = idx - locs; % Index Difference
dt = t(idx) - t(locs); % Time Difference
figure
plot(t,s)
hold on
plot(t(locs), pks, '^r')
hold off
grid
PeakTable = table(t(locs).',pks(:),locs(:),didx(:),dt(:), 'VariableNames',{'Time','Peak_Amplitude','Location_Indices','Index_Differences','Time_Differences'})
PeakTable = 40×5 table
Time Peak_Amplitude Location_Indices Index_Differences Time_Differences ______ ______________ ________________ _________________ ________________ 143.1 7.1592 144 170 170.12 149.1 7.2444 150 164 164.11 302.21 7.0417 303 11 11.008 313.22 7.7414 314 0 0 320.22 6.4249 321 -7 -7.0049 439.3 0.23134 440 -126 -126.09 451.31 7.2826 452 -138 -138.1 456.32 7.2931 457 -143 -143.1 461.32 6.8597 462 -148 -148.1 468.32 7.6384 469 -155 -155.11 471.33 7.1554 472 -158 -158.11 491.34 7.0425 492 -178 -178.12 506.35 6.2797 507 -193 -193.13 510.35 6.4715 511 -197 -197.14 573.4 -0.52506 574 -260 -260.18 664.46 -1.5985 665 -351 -351.24
.
  6 Comments
ignacio bobadilla tapia
ignacio bobadilla tapia on 28 May 2021
Dear, thank you, this code works for me. My query, how can I always make sure that among the 3 selected peaks, the one with the highest peak is in the center of the 3, since I have to vary the data, and I am interested that the maximum amplitude is always the central value, I stay tuned Regards.
Star Strider
Star Strider on 28 May 2021
‘... how can I always make sure that among the 3 selected peaks, the one with the highest peak is in the center of the 3 ...’
I am not certain that is possible. However I do not know what you are doing, so it may not be a problem.
In any event, if the three highest peaks are in roughly the same positions, it might be best to consider (or simply define) the centre peak the reference regardless of its relative amplitude, and just use it that way.
.

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More Answers (1)

ignacio bobadilla tapia
ignacio bobadilla tapia on 28 May 2021
"... it would be better to consider (or just define) the central peak as a reference, regardless of its relative amplitude, and use it that way."
This as it would be programmed, or as it would look like in a code to be able to execute it, thanks.
  1 Comment
Star Strider
Star Strider on 28 May 2021
Either way.
If the centre peak is always the highest, use it as such. If it as not, use it as the reference anyway.

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