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Thread: Full Width Half Max question??

  1. #1
    May 2010

    Full Width Half Max question??

    I have a plotted a graph of:
    $\displaystyle \frac{I(t)}{|E_p|^2}=e^{(-2\alpha t^2)}$

    Now I understand that the maximum is when t=0, however what about the full width half max?? Why is the half height 1/e??

    Can someone help me obtain the full width half max of this?

    Last edited by chutsu; May 16th 2010 at 01:39 AM.
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  2. #2
    MHF Contributor

    Apr 2005
    I had to google "full width half max". Apparently, it is the distance between $\displaystyle x_1$ and $\displaystyle x_2$ for points such that $\displaystyle f(x_1)= f(x-2) = $ 1/2 the maximum value of the function.

    You have correctly calculated that the maximum value of this function occurs at x= 0 and, so, is e. But the "half max" is not 1/e, it is, as you should guess, half of e: e/2.

    You need to solve $\displaystyle f(x)= e^{-2\alpha x^2}= \frac{e}{2}$. Since f is symmetric about x= 0, the "full width at half maximum" is twice that x value.
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