How to find FWHM from this?

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.

Answer 1

[Image Analyst]

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)

[hnn]

Awesome thank you! That gave me exactly what I needed

Answer 2

[Chunru]

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

[Chunru]

"Which is definitely possible from this data". How definite you are?

[Chunru]

fwhm stands for full width of half maximum. Your data has a maximum of ~7.2e-4. Half of the maximum is ~3.6e-4. You draw a line of y=3.6e-4 on the curve to see what happens.

[Chunru]

Are you sure you have not misunderstood what 7.2e-4 means?

[Chunru]

Are you sure what is half height?

[Chunru]

You are not using FWHM of the data. You are using the width at mid of maximum and minimum.

[Chunru]

https://en.wikipedia.org/wiki/Full_width_at_half_maximum

[hnn]

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

[Chunru]

If you want the width at mid of max and min. Just add:

power = power - min(power);