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Homework Help: Fourier Transform of a Gaussian With Non-Zero Mean

  1. Nov 30, 2012 #1
    1. The problem statement, all variables and given/known data
    I am looking at finding the fourier transform of:
    [tex] f(t)=\exp \left[ \frac{-(t-m)^2}{2 \sigma^2}\right] [/tex]

    2. Relevant equations

    [tex] \hat{f}(t)=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} f(t) e^{-i \omega t} dt[/tex]

    3. The attempt at a solution
    I did it a little differently that my professor suggested. He said use taylor series but when I do that I end up getting a triple sum (2 from the exponentials and 1 coming from an (x+m)^j; so I used binomial expansion) so I did it the traditional way you deal with gaussians: complete the square.

    I combined the exponentials, completed the square and end up with an answer that looks like:
    [tex] \hat{f}(t)=\frac{1}{2 \sigma}e^{-(\frac{\sigma^2 \omega^2}{2}+i \sigma \omega)}[/tex]
    But wolfram has this:

    With [itex] r= 2 \sigma^2[/itex]

    I was just wondering if anyone could double check me against wolfram? I can type out steps if needed. This isn't a homework or anything, he just brought it up and I suggested completing the square instead. Apparently this means I have to present it to the class on Monday -.-
    Last edited: Nov 30, 2012
  2. jcsd
  3. Nov 30, 2012 #2


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    Comparing your expression for [itex]f(t)[/itex] with the input to the Fourier transform at the Wolfram link, I'm not sure why you put [itex]r = 2\sigma^2[/itex] instead of [itex]r = \sigma[/itex].

    Also, check Wolfram's definition of the Fourier transform. It might not include your [itex]1/\sqrt{2\pi}[/itex] factor. There is no universal consensus regarding this. [edit]: You can see their definition by clicking on the "definition" link. Indeed, their definition omits the [itex]1/\sqrt{2\pi}[/itex]
  4. Nov 30, 2012 #3
    That's what I meant [tex] r=\sigma[/tex]. I typed it in two different ways and I had them mixed up :/

    And omitting the 1/sqrt(2 pi) gives me:
    [tex] \sqrt{\frac{\pi}{2 \sigma^2}}e^{-(\frac{\sigma^2 \omega^2}{2}+\sigma^2 i \omega)}[/tex]
    They have the sigma in the numerator some how.
  5. Nov 30, 2012 #4
    Ahhhh, nevermind I see it!
  6. Nov 30, 2012 #5


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    By the way, you can also solve this problem by using the shift property of the Fourier transform: if [itex]\mathcal{F}[f(t)] = \hat{f}(\omega)[/itex], then [itex]\mathcal{F}[f(t-m)] = \hat{f}(\omega)\exp(-im\omega)[/itex].
  7. Nov 30, 2012 #6
    Thank you. I always forget about these nice shifting formulae.
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