Filter data series on amplitude as a function of time

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

Answer 1

[Star Strider]

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     

.

[ignacio bobadilla tapia]

@Star Strider Dear, first thank you for your cooperation. Also indicate that the findpeaks function does not work for me, since I have matlab 2014a, what I need to be more specific is to identify the largest peak (the greatest amplitude) and calculate the time difference (time delta) with respect to the previous peak and later, for which I attach an explanatory image. If you could please help me, thanks in advance.

[Star Strider]

The findpeaks function will work in R2014a, although it may require a different argument list.

This should do what you want —

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',5, 'NPeaks',3);

didx = locs - idx;                                                      % Index Difference
dt = t(locs) - t(idx);                                                  % 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 = 3×5 table
     Time     Peak_Amplitude    Location_Indices    Index_Differences    Time_Differences
    ______    ______________    ________________    _________________    ________________

     149.1        7.2444              150                 -164               -164.11     
    313.22        7.7414              314                    0                     0     
    468.32        7.6384              469                  155                155.11     

.

[ignacio bobadilla tapia]

Dear the part that throws me an error is 'MinPeak Prominence', it is not recognized by matlab, I am attaching a message.

[Star Strider]

I was in the process of respoinding to this when my desktop computer crashed to update some software. I have no idea why it could instead just notify me and allow me to update it later.

That aside, since the onlline documentation no longer includes R2014a, I do not have access to that version of findpeaks, so I suggest instead using 'MinPeakDistance' and 'MinPeakHeight'.

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, 'MinPeakDistance',100, 'MinPeakHeight',4.5, 'NPeaks',3);

didx = locs - idx;                                                      % Index Difference
dt = t(locs) - t(idx);                                                  % 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 = 3×5 table
     Time     Peak_Amplitude    Location_Indices    Index_Differences    Time_Differences
    ______    ______________    ________________    _________________    ________________

     149.1        7.2444              150                 -164               -164.11     
    313.22        7.7414              314                    0                     0     
    468.32        7.6384              469                  155                155.11     

Without access to the R2014a documentation, this is the best I can do. It may be necessary to experiment if these options also fail.

.

[ignacio bobadilla tapia]

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]

‘... 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.

.

Answer 2

[ignacio bobadilla tapia]

"... 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.

[Star Strider]

Either way.

If the centre peak is always the highest, use it as such. If it as not, use it as the reference anyway.