Finding the Full width half maximum of a function I am a bit confused as to how we calculate Full-Width-Half-maxiumums (FWHM) of a function.
Consider a function as $$f(x) = {\frac{\sin(ax/2)^2}{(ax/2)^2}} \cos(bx/2)^2 $$
where b>>a.
It looks something like this:

How would I deduce the FWHM  of both the envelope as well as the FWHM of the enclosed fringe of such a function?
 A: Equate the function to half its peak value and solve.
For eample, if $b\gg a$, then the envelope of the overall function is defined by the sinc function, which has a maximum of 1, so it reaches the FWHM points at
$$ {\rm sinc}^2 \left( \frac{ax}{2} \right) = \frac{1}{2}$$
This must be solved numerically. The solution is $ax/2 \simeq \pm 1.392$.
Thus the FWHM is $\simeq 5.57/a$.
If $b \gg a$ then you can then assume that the sinc function hardly varies whilst the cosine oscillates. In which case the $\cos^2(bx/2)$ term has a maximum of 1 at $x=0$ and its FWHM occurs at values of $x$ where
$$ \cos^2 \left(\frac{bx}{2}\right) = \frac{1}{2} $$
$$ x = \pm \frac{2}{b}\cos^{-1}\left( \frac{1}{\sqrt{2}} \right) = \pm \frac{\pi}{2b}$$
So the FWHM is $\pi/ b$.
A: You would do just as the name implies. For example, the envelop has a height of about 1. So half max is 0.5. The width of the envelop at 0.5 appears to be 0.25 - -0.25 which equals 0.5.
You could also use the function to determine where the maximum occured and how wide it is at half that max. Since you've plotted the data you should have all the info that you need.  Look at the data to determine the values you need.
