# Fourier Transform of a Gaussian

1. Aug 18, 2008

### maverick280857

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

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

then

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

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:

$$G(f) = \int_{-\infty}^{\infty}g(t)e^{-j2\pi ft}dt$$

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

$$g(f) = \int_{-\infty}^{\infty}g(t)e^{-j2\pi ft}dt$$

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 $g(t) = e^{-\pi t^{2}}$.

Cheers.

2. Aug 19, 2008

### maze

I would attack the problem by using the Liebniz integral rule to diferentiate under the integral sign.
http://mathworld.wolfram.com/LeibnizIntegralRule.html

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

Once you got the DE, you could separate variables and integrate, and hopefully that will result in a gaussian.

3. Aug 19, 2008

### rbj

$$x(-t) = x(t)$$

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

$$g(t) = x(t) + X(t)$$

where

$$X(f) = \mathcal{F}\{x(t)\}$$

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

example:

$$g(t) = \mathrm{rect}(t) + \mathrm{sinc}(t)$$

Last edited: Aug 19, 2008
4. Aug 19, 2008

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:

$$f(t) = \sum_{i=-\infty}^{\infty} \delta(t-ci) [/itex] 5. Aug 19, 2008 ### maverick280857 Oh yes, I totally forgot about the impulse train. But then, I want the exact same functional form...i.e. G(f) = g(f)..so I should be able to get the Fourier Transform of the signal just by replacing f with t. Thanks for the inputs. Btw maze, when you use Leibniz' rule and differentiate under the integral wrt f, you get [tex]-j2\pi\int_{-\infty}^{\infty}tg(t)e^{-j2\pi t}dt$$

I suppose you missed the 't' term inside, which is the source of the problem.

6. Aug 20, 2008