How to find FWHM from this?

Question

hnn 2022-10-8

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1819950-how-to-find-fwhm-from-this

评论： hnn 2022-10-8

I am very new to Matlab, and am trying to find the FWHM from this curve. I have tried findpeaks, and don't really understand any of it. I have these (time and power) as my variables and have tried but really don't know how to do it.

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Answer 1

Image Analyst 2022-10-8

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https://ww2.mathworks.cn/matlabcentral/answers/1819950-how-to-find-fwhm-from-this#answer_1069475

在 MATLAB Online 中打开

See if this is what you want

halfMaxValue = max(power) / 2; % Find the half max value.
% Find indexes of power where the power first and last is above the half max value.
leftIndex = find(power >= halfMaxValue, 1, 'first');
rightIndex = find(power >= halfMaxValue, 1, 'last');
% Compute the delta time value by using those indexes in the time vector.
fwhm = t(rightIndex) - t(leftIndex)

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hnn 2022-10-8

Awesome thank you! That gave me exactly what I needed

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Answer 2

Chunru 2022-10-8

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链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1819950-how-to-find-fwhm-from-this#answer_1069295

在 MATLAB Online 中打开

Your data has no half power points so you cannot find fwhm.

load(websave("fwhmdata.mat", "https://www.mathworks.com/matlabcentral/answers/uploaded_files/1149260/fwhm%20data.mat"))

whos

Name Size Bytes Class Attributes M 384x2 6144 double cmdout 1x33 66 char data 384x1 3072 double index1 384x1 3072 double index2 384x1 3072 double power 384x1 3072 double time 384x1 3072 double

plot(time, power);

[pmax, imax] = max(power);

hold on

plot(time(imax), pmax, 'ro');

i1 = find(power(imax:-1:1) <= 0.5*pmax, 1, 'first')

i1 = 0×1 empty double column vector

i2 = find(power(imax:end) <= 0.5*pmax, 1, 'first')

i2 = 0×1 empty double column vector

if ~isempty(i1) & ~isempty(i1)

fwhm = tmax(i2+imax-1) - tmin(imax-i1+1);

else

fwhm = nan;

end

fwhm

fwhm = NaN

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hnn 2022-10-8

Your method did not work but I used a different function which gave me the correct FWHM value which also matches my python answer, as well as matching the physical beam size of the telescope used to gather this data. Thanks for your super super super helpful reply! Have a nice day