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Fourier Transform of a Gaussian

  1. Aug 18, 2008 #1
    Hi everyone

    I am trying to prove that if a signal g(t) is its own Fourier Transform (so that G(f) = g(f), i.e. they have the same functional form), then g(t) must be a Gaussian. I know that the Fourier Transform of a Gaussian is a Gaussian, so thats not the point of the exercise.

    Simon Haykin's book on Communication Systems, 2nd edition gives the following 'proof':

    \frac{d}{dt}g(t) \rightarrow j2\pi fG(f)

    -j2\pi tg(t) \rightarrow \frac{d}{df}G(f)

    Now, from equations 1 and 2, if

    \frac{d}{dt}g(t) = -2\pi tg(t)


    \frac{d}{df}G(f) = -2\pi fG(f)

    which means that the signal and its FT are the same function.

    The problem here is the assumption of that differential equation, which isn't intuitively obvious at first sight. Of course if I stare hard enough at it, it makes sense, but I am trying to prove it from first principles using the definition of the Fourier Transform:

    [tex]G(f) = \int_{-\infty}^{\infty}g(t)e^{-j2\pi ft}dt[/tex]

    as G(f) = g(f), this becomes

    [tex]g(f) = \int_{-\infty}^{\infty}g(t)e^{-j2\pi ft}dt[/tex]

    Now, I tried a couple of things with this..like differentiating wrt t, and using g'(f) = G'(f) and so on, but nothing worked out. I'd be grateful if someone could point me in the right direction with this approach. I am looking to get a differential equation in g(t) which has the solution [itex]g(t) = e^{-\pi t^{2}}[/itex].

    Thanks in advance.
  2. jcsd
  3. Aug 19, 2008 #2
    I would attack the problem by using the Liebniz integral rule to diferentiate under the integral sign.

    The steps would go something as follows:
    [tex]\frac{d}{df}G(f) = \frac{d}{df} \int_{-\infty}^{\infty}g(t)e^{-j2\pi ft}dt = ... liebniz rule ... = -2 \pi f \int_{-\infty}^{\infty}g(t)e^{-j2\pi ft}dt = -2 \pi f G(f)[/tex]

    Once you got the DE, you could separate variables and integrate, and hopefully that will result in a gaussian.
  4. Aug 19, 2008 #3


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    start with any even-symmetry signal:

    [tex] x(-t) = x(t) [/tex]

    consider adding the F.T. of this even-symmetry signal to itself (after replacing f with t).

    [tex] g(t) = x(t) + X(t) [/tex]


    [tex] X(f) = \mathcal{F}\{x(t)\} [/tex]

    you will see that there are an infinite number of theoretical signals that are the F.T. of themselves.


    [tex] g(t) = \mathrm{rect}(t) + \mathrm{sinc}(t) [/tex]
    Last edited: Aug 19, 2008
  5. Aug 19, 2008 #4
    That's pretty neat; I'd never seen that before. The other famous example of a signal that is its own transform is the impulse train:

    f(t) = \sum_{i=-\infty}^{\infty} \delta(t-ci)
  6. Aug 19, 2008 #5
    Oh yes, I totally forgot about the impulse train. But then, I want the exact same functional form...i.e. G(f) = g(f)..so I should be able to get the Fourier Transform of the signal just by replacing f with t.

    Thanks for the inputs. Btw maze, when you use Leibniz' rule and differentiate under the integral wrt f, you get

    [tex]-j2\pi\int_{-\infty}^{\infty}tg(t)e^{-j2\pi t}dt[/tex]

    I suppose you missed the 't' term inside, which is the source of the problem.
  7. Aug 20, 2008 #6
    So, since it isn't the case that a function must be Gaussian to be its own transform, you aren't going to be able to get such a differential equation without invoking further assumptions. Haykin seems to just point out that that diff.eq. is sufficient and move on. As far as basing the proof in the transform definitions, I think you can see that application of Leibniz's rule will get you from the definitions up to the transform properties (eqs. 1 and 2 in the original post). The question then is: what is the minimal set of additional assumptions on f(t) such that you get the differential equation in question. Obviously continuity will rule out the impulse train, and finite energy will get rid of some more, but rbj's examples are not so easily dispensed with. Perhaps the assumption of an exponential family distribution will do it?
  8. Aug 20, 2008 #7


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    FYI, the hyperbolic secant sech(t) transforms into a hyperbolic secant.
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