How to calculate FWHM from gaussian histogram fit?

Hello

I was just trying to get the FWHM value for my histrogram with 8 bins for X = [0.6268 0.5372 0.5382 0.4272 0.4635 0.5422 0.5869 0.5308 0.5188 0.4759 0.6365 0.5532 0.5753 0.5734 0.5734 0.5465];

It will be really heplfull if someone guided with small portion of code to fit the gauassian distribution line and related mean, variance and fwhm.

thanks in advance!!

Answer 1

[Star Strider]

Thje easiest way is to use histfit and then findpeaks with the name-value pair appropriate arguments —

X = [0.6268    0.5372    0.5382 0.4272    0.4635    0.5422  0.5869    0.5308    0.5188 0.4759    0.6365    0.5532 0.5753    0.5734    0.5734  0.5465]; 

figure
h = histfit(X)

h = 
  2×1 graphics array:

  Bar
  Line

line_x = h(2).XData;
line_y = h(2).YData;

[pk,loc,w] = findpeaks(line_y, 'WidthReference','halfheight')

pk = 6.9351

loc = 51

w = 38.8687

FWHM = w

FWHM = 38.8687

x_peak = line_x(loc)

x_peak = 0.5458

pd = fitdist(X(:),'Normal')

pd = 
  NormalDistribution

  Normal distribution
       mu =  0.544113   [0.514699, 0.573526]
    sigma = 0.0551991   [0.0407758, 0.0854312]

To get the parameters of the distribution, use the fitdist function. (This is the function histfit uses.)

.

[Somnath Kale]

Thank You @Star Strider

It really hard to belive that the extrated value FWHM = 38.86 is resonable.

Just by looking at the fitted distribution line it seems that the FWHM should be arround 0.12 - 0.13

Now im not sure that where the things went wrong!

[Star Strider]

I usually use findpeaks with only one numeric argument because I want the index values in the ‘locs’ output.

Adding ‘line_x’ as the second argument produces —

X = [0.6268    0.5372    0.5382 0.4272    0.4635    0.5422  0.5869    0.5308    0.5188 0.4759    0.6365    0.5532 0.5753    0.5734    0.5734  0.5465]; 

figure
h = histfit(X)

h = 
  2×1 graphics array:

  Bar
  Line

line_x = h(2).XData;
line_y = h(2).YData;

[pk,loc,w] = findpeaks(line_y, line_x, 'WidthReference','halfheight')

pk = 6.9351

loc = 0.5458

w = 0.1300

FWHM = w

FWHM = 0.1300

x_peak = loc

x_peak = 0.5458

pd = fitdist(X(:),'Normal')

pd = 
  NormalDistribution

  Normal distribution
       mu =  0.544113   [0.514699, 0.573526]
    sigma = 0.0551991   [0.0407758, 0.0854312]

This is likely closer to what you expect.

.

[Somnath Kale]

Yupp!

Thank you very much for your time efforts and suggestions!

looking fourword to your kind help in future as well!!

[Star Strider]

As always, my pleasure!

I will help as much I am able!