How can I extract the width of the following graph.

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I have made a farfield profile from the Phased Array and want to find how the width of the beam changes with the reduction of the number of emitting elements. I extract this data to matlab. And find the HWFM but when doing so I only get the same width for all the elements. The width doesn't change with the number of elements. But in my simulations the width is changing. This is the code that use for extracting the width. I only need the width of the main lobe of this plot. Which is the part of the highest intensity.
E_intensity = abs(E2);
% Find the half max value.
halfMax = (min(E2) + max(E2)) / 2;
% Find where the data first drops below half the max.
index1 = find(E2 >= halfMax, 1, 'first');
% Find where the data last rises above half the max.
index2 = find(E2 >= halfMax, 1, 'last');
fwhm = index2-index1 + 1; % FWHM in indexes.
% OR, if you have an x vector
fwhmx = E2(index2) - E2(index1);
%I have found this code in the forum.
%to do check this if this works.
peakHeight = max(E2);
x1 = min(E2);
x2 = find(E2 <= peakHeight, 1, 'last');
theWidth = x2-x1;
Mathieu NOE
Mathieu NOE on 24 Nov 2020
you're welcome
may I ask you to accept my answer ?
I put it in the answer again so you can accept it

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Accepted Answer

Mathieu NOE
Mathieu NOE on 24 Nov 2020
this piece of code will give you the angle . Demo based on a half sinus wave.
you can easily apply it to your case
x= linspace(0,pi,100);
y = sin(x);
minus3dBpoint = 1/sqrt(2); % -3 dB
z = abs(y-minus3dBpoint);
[p,loc] = findpeaks(-z);
% angle = half of difference between the 2 elements of x(loc)
angle = diff(x(loc))/2
of course , my example (sinus) has a peak amplitude of 1 , which is not your case
you'll figure out how to adapt the code....

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