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FT of the Gaussian function

  1. Dec 10, 2008 #1
    The other day I was playing around with gaussian functions and I noticed that the fourier transform of a gaussian function looked an awful lot like another gaussian function. I managed to find a single blurb about this fact in the Wikipedia article, and indeed, my hunch was correct. However, the article gave no clue as to how to demonstrate this is true. I've been playing around with the integral for a day and I haven't gotten anywhere with it!

    Can someone help me along with a proof that the FT of a gaussian function is another with the parameters slightly tweaked?

    Additionally, the article said this was only true for when the function is symmetric about the y-axis. However, it seems natural enough to assume that shifted gaussian functions would have a similarly regular FT.
  2. jcsd
  3. Dec 10, 2008 #2
    Your aim to to compute the integral:

    [tex] \int e^{-ikx} e^{-x^2}\, \mathrm{d} x [/tex]

    (or something very similar). Do you know any complex analysis? No problem if not - there's more than one way to skin a cat.
  4. Dec 10, 2008 #3
    I've never studied complex analysis, but I'm sure I could manage.
  5. Dec 10, 2008 #4
    Well it's probably not best (or efficient) for me to give you an introductory course on complex analysis so as to answer this one question! We'll go about things slightly differently. We will consider the Fourier transform of the function [tex]f(x)=\exp[-x^2][/tex] (you can generalise for a general Gaussian yourself). Clearly we have:

    [tex] \frac{ \mathrm{d} f}{\mathrm{d} x} = -2xf(x)[/tex]

    Can you apply the Fourier transform to this equation, and use some standard tricks?
  6. Dec 11, 2008 #5
    So I haven't figured it out, but this is what I've come up with.

    The function [tex]f(x) = e^{-x^2}[/tex] approaches 0 at either infinity. That allows us to use Really Easy integration by parts:

    [tex]F(f') = \int f'(x) e^{-ikx} dx = -\int f(x) \frac{d}{dx} e^{-ikx} dx = ik \int f(x) d^{-ikx}dx = ik F(f)[/tex]

    So [tex]F(-2x e^{-x^2}) = ik F(e^{-x^2})[/tex]. Divide by ik and we have the fourier transform isolated on on side:

    [tex]F(e^{-x^2}) = \frac{1}{ik } F(-2x e^{-x^2})[/tex]

    Expanding the other side, we get

    [tex]F(e^{-x^2}) = \frac{1}{ik } \int -2x e^{-x^2} e^{-ikx} dx
    = \frac{1}{ik } \int -2x e^{-x^2 - ikx} dx[/tex]

    Then I'm kind of stuck.

    Another approach I used was to take:

    [tex]f'(x) - ik f(x) = (-2x - ik)f(x)[/tex]

    Taking the fourier trans. of both sides results in

    [tex]F(f') - ik F(f) = \int (-2x - ik) e^{-x^2} e^{ikx} dx = \int (-2x - ik) e^{-x^2} e^{-ikx} dx = \int \frac{du}{dx} e^u dx = \int e^u du[/tex]

    But the problem with this approach is the integral [tex]\int^\infty_{-\infty} e^u du[/tex] is pretty darn infinite.

    So I'm pretty much lost X-D
  7. Dec 11, 2008 #6

    And I pretty much cheated. The integral is 0, because the function inside the integral is odd, and it reduces to the identity I used in the first part of my post.
  8. Dec 11, 2008 #7
    Ok, we need a couple of lemmas. See if you can prove the following:

    [tex] \mathcal{F}\left[ xf(x)\right] = i \frac{ \mathrm{d} \hat{f}}{\mathrm d k}, \qquad \mathcal{F} \left[ f'(x)\right] = ik \hat{f}(k) [/tex]

    assuming [tex]f[/tex] has sufficient decay at [tex]|x|\rightarrow \infty[/tex]. Hint: integration by parts!!!
  9. Dec 11, 2008 #8
    This gets confusing quickly =-P

    But I narrowed it down to being able to prove that:

    [tex]\lim_{h \to 0} \frac{cos(xh)}{h} = \frac{1}{h}[/tex] and
    [tex]\lim_{h \to 0} \frac{sin(xh)}{h} = x[/tex]

    Which seems to check out.

    Using the two identities, I can see that both f and [tex]\hat{f}[/tex] satisfy the same first-order differential equation. I'm guessing you would solve this with the initial value at k=0 to prove the relation?

    Then, I'm guessing you can use the frequency shift relationship [tex]F(e^{ 2\pi iax} f(x))\ = \hat{f} \left(\xi - a\right)[/tex] to work with gaussians which aren't centered at x=0, and that they simply rotate in the complex plane as you shift across the x-axis.
  10. Dec 12, 2008 #9
    I think you're going a little off track. Did you use my hint? I'll prove them for you. For the first:

    [tex] \mathcal{F}\left[x f(x)\right] = \int xe^{-ikx}f(x)\, \mathrm{d} x = \int i\left(\frac{ \mathrm{d}}{\mathrm{d} k} e^{-ikx}\right)f(x)\, \mathrm{d} x = i \frac{ \mathrm{d}}{\mathrm{d} k} \int e^{-ikx} f(x)\, \mathrm{d} x = i \frac{\mathrm{d} \hat{f}}{\mathrm d k} [/tex]

    And for the second:

    [tex] \mathcal {F}\left[ f'(x)\right] = \int e^{-ikx} \frac{\mathrm{d} f}{\mathrm{d} x}\, \mathrm{d} x = \int \left(\frac{\mathrm{d}}{\mathrm{d} x} \left[ e^{-ikx}f(x)\right] - \left[ \frac{\mathrm{d}}{\mathrm{d} x} e^{-ikx}\right] f(x) \right)\, \mathrm{d} x = ik \int e^{-ikx} f(x)\, \mathrm{d} x = ik \hat{f}(k) [/tex]

    where we've used the fact that [tex]f[/tex] necessarily tends to zero at [tex]|x|\rightarrow \infty[/tex]. So if we apply the Fourier transform to the differential equation:

    [tex] \frac{ \mathrm{d} f}{\mathrm{d} x} = -2xf(x)[/tex]

    and use the lemmas we've just proved, we get:

    [tex] ik \hat{f} = -2i \frac{ \mathrm{d} \hat{f}}{\mathrm{d} k}, \quad \mathrm{i.e} \quad \frac{ \mathrm{d} \hat{f}}{\mathrm{d} k} = - \tfrac{1}{2} k \hat{f}.[/tex]

    Can you solve this differential equation for [tex]\hat{f}[/tex]?
  11. Dec 12, 2008 #10
    I did no such thing =-P I just didn't post my work.

    I took [tex]\hat{f}(k)[/tex] and took the derivative using the standard definition of derivative:

    [tex]\hat{f}'(k) = \lim_{h \to 0} \frac{\hat{f}(k + h) - \hat{f}(k)}{h}[/tex]

    What you end up with is an expression:

    [tex]\hat{f}'(k) = \lim_{h \to 0} \frac{1}{h} \int f(x) e^{-ixk} (\frac{e^{-ixh} - 1}{h})dx[/tex]

    So what I rambled about above was showing that [tex]\lim_{h \to 0} \frac{e^{-ixh} - 1}{h} = -ix[/tex]. Then you can pull out the -i factor and you're left with an x inside the integral, which you then reword as [tex]-iF(x f(x))[/tex].
  12. Dec 12, 2008 #11
    Well, you know best.
  13. Dec 12, 2008 #12
    That's hardly the case.

    Either way, I've played with it enough that I think I get the gist of what's going on. Thanks a lot for your help =-)
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