The method of how to calculate the FWHM.

I have searched extensively on this website and the internet, but all the methods for calculating the FWHM (Full Width at Half Maximum) seem overly complex. I was wondering: can we simplify this by using a cut_index on the X-axis to determine the corresponding Y-axis values?

My idea is that we could compute the FWHM using the formula:

FWHM=(max_Y_value - min_Y_value)/2

and display the result accordingly in the window.

BTW, my sample is just a very simple Gaussian curve.

Would this approach be possible or effective? I’d like to discuss this idea further with the community. Thank you!

Answer 1

[Torsten]

xmax = 0;
ymax = 4;
x = xmax:0.01:3;
f = -x.^2 + ymax;
FWHM = 2*interp1(f,x,ymax/2)

FWHM = 2.8284

Or directly use the inverse function of your Gaussian curve and evaluate it at ymax/2.

Answer 2

[Star Strider]

A different approach —

x = linspace(0, 10);
y = exp(-(x-5).^2*0.25);

ymin = min(y)

ymin = 0.0019

[ymax,idx] = max(y - ymin)

ymax = 0.9974

idx = 50

halfmax = ymax/2;

for k = 1:2
    idxrng = (1:idx) + idx*(k-1);
    xhalf(k) = interp1(y(idxrng), x(idxrng), halfmax + ymin);
end

xhalf

xhalf = 1×2

3.3364 6.6636

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FWHM = diff(xhalf)

FWHM = 3.3272

figure
plot(x, y)
hold on
plot(xhalf, [1 1]*(halfmax+ymin), 'rs')
hold off
grid
text(mean(xhalf), (halfmax+ymin), sprintf('\\leftarrow FWHM = %.3f \\rightarrow', FWHM), 'Horiz','center')

.