Finding area under multiple curves with different y references

Hello all,

I need to find the sum of the area under each peak (see 5 minute skin conductance sample attached). The y axis must correspond to the lowest point of each peak to the peak itself. Therefore, the y axis changes for each peak. I have attached an image of the areas I would like help with.

I have tried for several days with no prevail, using past forums as guides, using functions such as trapz, cumtrapz, findpeaks, etc.

Any help would be greatly appreciated!

Jacob

Answer 1

[Robin Kirsch]

you could try to get the peaks with the findpeaks method from Matlab, and then you for-loop backwards over the data array from the x value as the index and compare the data, as long as it is decreasing. I have made you an example for the first peak.

I attached you an example. Does this make sense? I get as a result 24.6050 for the first low to the first peak.

Of course you must for loop over all peaks and extend the thing! The example is only for the first peak.

My example gets you from that first detected peak to the bottom.

and this is the variable z, so the extracted data from the first peak in the figure above to the bottom. Then simply get the area underneath the data with the trapz method.

If you want to ignore this little peaks and only head for the bigger ones, I think you can add a parameter to the findpeaks method.

data = [3.42
3.42
3.41
3.4
3.4
3.4
3.4
3.4
3.4
3.39
3.39
3.39
3.39
3.39
3.37
3.37
3.37
3.37
3.37
3.37
3.37
3.37
3.36
3.35
3.35
3.35
3.34
3.36
3.37
3.38
3.43
3.54
3.61
3.7
3.79
3.77
3.84
3.9
3.99
4.03
4.07
4.11
4.05
4
3.95
3.9
3.86
3.85
3.84
3.89
3.92
3.9
3.86
3.83
3.83
3.83
3.83
3.83
3.83
3.83
3.84
3.85
3.85
3.91
3.96
3.97
3.96
3.95
3.96
4.01
4.03
4
3.99
3.99
3.97
3.97
3.97
3.97
3.99
3.97
3.97
3.99
4
4
3.99
3.97
3.97
4
3.99
3.97
3.97
3.97
3.97
3.97
3.95
3.95
3.95
3.94
3.92
3.92
3.91
3.9
3.89
3.87
3.86
3.86
3.86
3.89
3.9
3.87
3.85
3.85
3.84
3.83
3.8
3.8
3.84
3.85
3.85
3.85
3.83
3.83
3.8
3.81
3.83
3.8
3.78
3.78
3.78
3.78
3.79
3.8
3.79
3.77
3.76
3.75
3.76
3.85
3.85
3.83
3.8
3.78
3.78
3.9
3.96
3.94
3.89
4.04
4.04
4.03
4.01
4
4.14
4.18
4.16
4.14
4.11
4.11
4.08
4.05
4.04
4.05
4.05
4.07
4.14
4.14
4.14
4.11
4.08
4.08
4.08
4.08
4.08
4.08
4.21
4.29
4.26
4.22
4.21
4.19
4.16
4.16
4.16
4.15
4.14
4.14
4.14
4.14
4.12
4.11
4.11
4.11
4.11
4.11
4.11
4.11
4.11
4.11
4.21
4.32
4.31
4.28
4.25
4.19
4.19
4.16
4.19
4.21
4.19
4.19
4.16
4.16
4.28
4.32
4.31
4.39
4.4
4.34
4.28
4.25
4.25
4.29
4.31
4.28
4.25
4.25
4.22
4.22
4.19
4.19
4.21
4.19
4.22
4.29
4.28
4.25
4.25
4.23
4.26
4.25
4.22
4.22
4.21
4.19
4.19
4.19
4.19
4.21
4.19
4.19
4.19
4.19
4.18
4.26
4.32
4.4
4.4
4.37
4.37
4.37
4.31
4.28
4.28
4.25
4.25
4.25
4.25
4.22
4.22
4.21
4.19
4.19
4.25
4.28
4.37
4.45
4.43
4.4
4.37
4.34
4.39
4.4
4.37
4.34
4.31
4.28
4.34
4.4
4.4
4.37
4.34
4.31
4.31
4.31
4.31
4.32
4.28
4.29
4.32
4.31
4.28
4.28
4.28
4.25
4.25
4.25
4.25
4.23
4.22
4.22
4.22]

[x,y] = findpeaks(data);

%%take first peak
j = 1;
for i=y(1):-1:2
    if data(i-1) < data(i)
        z(j) = data(i);
        j = j+1;
    else
        break;
    end
end

z = flip(z);
area = trapz(z);

[Robin Kirsch]

I just noticed you have to add the data point of the bottom, cause that will be left out, for example in the else method before the break.

like:

if ...

else

z(j) = data(i);

break;

[Jacob Merriman]

Hello,

thank you ever so much for your prompt response, it is greatly apreciated.

When I run this myself within a function I get 65.85, not sure why our answers are different?

Your answer makes sense, though in hindsight I only need to work out the displacement from the highest peak and the lowest peak, which I have done now.

I tried to create a for loop which expanded further but I kept getting strange results.

all the best,

Jacob