How to calculate the width that contains 60% of the total area of peaks?

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Lope
Lope on 25 Apr 2022
Commented: Lope on 25 Apr 2022
So starting from the center (where the max peak always lies), I want to calculate automatically the width that contains 60% of the total area under multiple peaks. How can I do that? Attached is the normalized intensities and positions.
Below is the code to an attempt to get the contribution of the main peak only from the total area:
TotalArea=trapz(x2,Int);
lims = (x2>= -3.90625E-5 ) & (x2<= 3.90625E-5); %manually selected
MainPeakArea=trapz(x2(lims),Int(lims));
MainLobePower = (MainPeakArea/TotalArea)*100;

Accepted Answer

Chunru
Chunru on 25 Apr 2022
load position
load Intensity
whos
Name Size Bytes Class Attributes Int 1x2048 16384 double ans 1x38 76 char cmdout 1x33 66 char x2 1x2048 16384 double
plot(x2, Int); hold on
% find the max
[maxI, idx] = max(Int);
maxX = x2(idx)
maxX = 0
n = length(x2);
Atotal = trapz(x2, Int);
for w=1:length(x2)
ii = max(idx-w, 1):min(idx+w, n);
A = trapz(x2(ii), Int(ii));
if A/Atotal>=0.6
break
end
end
area(x2(ii), Int(ii));
plot(x2(idx), Int(idx), 'r^');
xlim([-1 1]*1e-3)
title(sprintf('Width: %g - %g', x2(ii(1)), x2(ii(end)) ));

More Answers (1)

Bruno Luong
Bruno Luong on 25 Apr 2022
Edited: Bruno Luong on 25 Apr 2022
load Intensity.mat
load position.mat
ifun=@(x) interp1(x2,Int,x,'linear','extrap');
I0=integral(ifun,x2(1),x2(end))
I0 = 5.6250e-05
x = fzero(@(x) integral(ifun,-x,x)-0.6*I0,0) % the interval is (-x,x) width is 2*x
x = 3.1468e-05

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