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Nasty Autocorrelation Integral

  1. Jan 5, 2014 #1
    Hi guys,

    Long story short, I need to compute an autocorrelation integral. Here's the problem:

    There are two arbitrary gaussian pulses, one following the other by a fixed distance. By computing the autocorrelation over space(not time) and taking the derivative of the space-shift autocorrelation and setting it equal to zero, important information hopefully could be obtained.

    The mathematics of this would be as followed:
    [itex]\frac{\partial}{\partial\tau}\int_{-\infty}^{\infty}(Ae^{-a(x-c)^2}+Be^{-b(x-d)^2})(Ae^{-a(x-c-\tau)^2}+Be^{-b(x-d-\tau)^2})dx=0[/itex]
    [itex]\int_{-\infty}^{\infty}(Ae^{-a(x-c)^2}+Be^{-b(x-d)^2})(a(x-c-\tau)Ae^{-a(x-c-\tau)^2}+b(x-d-\tau)Be^{-b(x-d-\tau)^2})dx=0[/itex]

    I am NOT asking anyone to do this for me - I'll do it myself but I just need some ideas or directions on how to go about it.
    I have experience in fourier transforms, complex analysis and calculus of course. I've considered doing a complex contour integral but I'm not sure how reasonable that is after seeing how big of a pain the normal gaussian contour integral is. I've considered fourier transforms a little - I didn't immediately see much help due to the fourier transform of a gaussian just being another gaussian. I've thought about parametrization or even centering the integral about the center of the two gaussians but I don't know where to start I guess.

    It's clearly a bound integral but is it just too impossibly hard to try?
     
    Last edited: Jan 5, 2014
  2. jcsd
  3. Jan 5, 2014 #2

    mfb

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    Staff: Mentor

    You can reduce the problem to several integrals of the types ##\int dx e^{-(x+d)^2 - (x-d)^2}##, ##\int dx x e^{-(x+d)^2 - (x-d)^2}## and maybe something I missed and look for solution methods for those integrals.
     
  4. Jan 5, 2014 #3
    Yes, that's definitely one way to do it. However, this may take a couple dozen sheets of paper and a few hours considering the factoring. I guess after seeing so many tricks in math classes I just assumed there might be a quick way around this... But real world problems versus classroom problems aren't a fair comparison I suppose.
     
  5. Jan 5, 2014 #4
    Well I tried it the old fashion way and it turned out better than I thought - but I don't think [itex]\tau[/itex] can be solved for explicitly.

    [itex]\frac{-A^{2}}{2\sqrt{2a}}e^{\frac{-a\tau^{2}}{2}}+\frac{-B^{2}}{2\sqrt{2b}}e^{\frac{-b\tau^{2}}{2}} = \frac{2ABe^{\frac{-ab(j(j-2k)+k^{2})}{a+b}}}{(a+b)\sqrt{a+b}}e^{\frac{-ab\tau^{2}}{a+b}}(\tau cosh(\frac{2ab\tau(j-k)}{a+b})-(j-k)sinh(\frac{2ab\tau(j-k)}{a+b}))[/itex]

    Any other ideas? I'm not an expert on fourier transforms but I'm beginning to think that's the only way because it should be a multi-valued answer - just not sure how to approach it.
     
    Last edited: Jan 5, 2014
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