Area under each peak
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Hi, I have a TGA derivative plot with three peaks (two prominant peaks and one shoulder) and I want to get individual gaussian functions to get the area under each peak (which tells how much mass is reduces during heating), could you please tell me how get area under each curve. and also how to get full width half maximum, FWHM and peak center values too.
Thanks
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Star Strider
2016-7-8
This was certainly an interesting challenge!
The loop first selects a range of x and y values to make the fit easier, then uses fminsearch to do the fit, calculates the area under the respective Gaussian using trapz and saves it to ‘AUC’, then uses fzero to calculate the full-width-half-maximum value and saves it to ‘FWHM’. The plot simply shows the fitted Gaussians, not ‘AUC’ or ‘FWHM’.
The code:
[d,s,r] = xlsread('chamila De Silva Q16N2.csv');
x = d(:,1);
y = d(:,2);
gausfcn = @(b,x) b(1) .* exp(-((x-b(2)).^2)./b(3)); % Gaussian Function
SSECF = @(b,x,y) sum((y - gausfcn(b,x)).^2); % Sum-Squared-Error Cost Function
[pks,locs] = findpeaks(y, 'MinPeakDist',30, 'MinPeakHeight',0.05); % Find Centres
q = x(locs);
for k1 = 1:size(pks,1)
idxrng = locs(k1)-25 : locs(k1)+25;
[Parms(:,k1), SSE(k1)] = fminsearch(@(b)SSECF(b,x(idxrng),y(idxrng)), [pks(k1); x(locs(k1)); 1]);
AUC(k1) = trapz(x, gausfcn(Parms(:,k1),x));
FWHM(k1) = 2*(x(locs(k1)) - fzero(@(x) gausfcn(Parms(:,k1),x) - pks(k1)/2, x(locs(k1))-5));
end
figure(1)
plot(x, y, 'LineWidth',1.5)
hold on
plot(x(locs), pks, '^r')
for k1 = 1:size(pks,1)
plot(x, gausfcn(Parms(:,k1),x), 'LineWidth',1)
end
hold off
grid
The plot:
7 个评论
camila
2016-7-13
Star Strider
2016-7-13
My pleasure.
I would use the detrend function first for baseline correction. If that doesn’t work satisfactorily, we can try other approaches, including a highpass or bandpass filter. (Those require the Signal Processing Toolbox, but are fairly easy to implement.)
Eric Patterson
2018-7-24
How can you determine what parameters you have put in place for the idxrng variable. I am having trouble completely fitting the curve under the peak with this code.
Star Strider
2018-7-24
I’m guessing that you’re referring to the ‘25’ values.
I would tweak the code to something like this:
lims = 25; % Put This Before The Loop
then in the loop:
idxrng = locs(k1)-lims : locs(k1)+lims;
and experiment with the value of ‘lims’ to get the result you want. (My code was designed for this specific Question. It wasn’t intended to generalise automatically to every situation.)
Matin Salehi
2019-8-23
Can you help me regarding a problem like this. I have a 28649x750 data, which are the intensity that is recorded by spectrometer.
each column has 310 peaks that I am interested on it (750x310). I could find the peaks, but now I want to know the the area under each peak. I also have the width, height and location of the peaks.
can you help me how can I write that.
Thor
2021-2-20
Hi star strider, could you please define the meaning of lims? I have tried to applied your code to my data. However, the error " Array indices must be positive integers or logical values." appears. Many thanks @Star Strider
Star Strider
2021-2-21
The ‘lims’ constant simply defines the range (±frame length) of ‘idxrng’.
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