Hi everyone(adsbygoogle = window.adsbygoogle || []).push({});

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':

[tex]

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

[/tex]

[tex]

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

[/tex]

Now, from equations 1 and 2, if

[tex]

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

[/tex]

then

[tex]

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

[/tex]

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.

Cheers.

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# Fourier Transform of a Gaussian

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