[ideas wanted]: How to find "peaks" of sloping signal?

Shoutout to you all math wizards! I need to detect peaks of (possibly) sloping signals. i.e sometimes first "peak" is real peak (findpeaks), but other times "peak" is in upward sloping portion of signal. See example figure below: the blue triangle is what findpeaks finds as first peak. But I need to detect that "peak" before it (see red arrow).

All ideas are much appreciated!

data is in attachment (data.mat)

Answer 1

[Star Strider]

The online ability to run scripts seems to be broken.

I used MATLAB Online to devise this —

% LD = load('data.mat')
file = websave('data.mat','https://www.mathworks.com/matlabcentral/answers/uploaded_files/1809778/data.mat');
LD = load(file);
data = LD.data;
x = linspace(0, 5.5, numel(data));                      % Supply Missing Coordinate
Fs = 1/x(2)

[FTs1,Fv] = FFT1(data,x);

figure
plot(Fv, abs(FTs1)*2)
xlim([0 3])
title('Fourier Transform (Informs Filter Design)')

data_filt = highpass(data, 0.0425, Fs, ImpulseResponse='iir');

[pksf,locsf] = findpeaks(data_filt);
[pks,locs] = findpeaks(data);

locs = [locsf(1) locs]
pks = data(locs)

figure
plot(x,  data)
hold on
plot(x(locs), data(locs), 'rv' )
hold off

figure
plot(x,  data_filt)
title('Filtered ‘data’')

function [FTs1,Fv] = FFT1(s,t)
% Arguments:
%    s: Signal Vector Or Matrix
%    t: Associated Time Vector 
    t = t(:);
    L = numel(t);
    if size(s,2) == L
        s = s.';
    end
    Fs = 1/mean(diff(t));
    Fn = Fs/2;
    NFFT = 2^nextpow2(L);
    FTs = fft((s - mean(s)) .* hann(L).*ones(1,size(s,2)), NFFT)/sum(hann(L));
    Fv = Fs*(0:(NFFT/2))/NFFT;
%     Fv = linspace(0, 1, NFFT/2+1)*Fn;
    Iv = 1:numel(Fv);
    Fv = Fv(:);
    FTs1 = FTs(Iv,:);
end

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[Sven Larsen]

elegant solution! thank you so much :)

[Star Strider]

My pleasure!

If my Answer helped you solve your problem, please Accept it!

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[Sven Larsen]

by the way, why have you hard coded value 0.0425 on :

data_filt = highpass(data, 0.0425, Fs, ImpulseResponse='iir');

how did you come up to that particular number? im applying your solution to my overall program, and need to understand the reasoning for that value.

[Star Strider]

That is the frequency that works best. (Signal processinig is frequently experimental.) I used the Fourier transform result to provide the first approximation, then went from there until I got a result that works.

Note that it is based on the ‘x’ vector. Basing it on the index vector is an optiion, and in that instance the sampling frequency is 1, and the cutoff frequency would have to be calculated differently, and likely experimentally, although scaling it as a function of the calculated sampling frequency here (0.0425/Fs) would likely work.

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