[CST-2] Inf. Th. & Cod.

[Anthony Jones]

On Wed, May 29, 2002 at 12:51:46AM +0100, George van den Driessche wrote:
> AFAICT, the PSD is the Fourier transform of the square of the signal. How is
> that "flat", if the signal is Gaussian? Surely the square of the Gaussian is
> another Gaussian, in which case the Fourier transform is yet another
> Gaussian?

Not quite:

http://www.ee.nmt.edu/~elosery/ee446/power_spectral_density.pdf

(FT of the signal)^2 rather than (FT of the signal^2) by the looks of this.

Not that that makes *any* difference to your reasoning if it's a Gaussian ;)

> Or are we talking about the PSD of white noise with Gaussian distribution?
> If so, is it true to say that it is the whiteness of the noise that
> guarantees the flatness of the PSD, which is independent of the Gaussian?

The question I think Mat was referring to is 8b in the Learning guide,
which *is* talking about white noise with Gaussian distribution. 

Presumably there are infinitely many functions that satisfy F(k)^2 =
constant (infinitely many because F(k) can be either -constant or +constant
for any k), and presumably these reverse transform to signals with lots of
different distributions -- as a trivial example, (IIRC) F(k) = +constant
reverse transforms to the Dirac delta function...

Anyway, I'd guess that you're right -- it's the whiteness of the noise that
makes it flat, I think it being Gaussian distributed just maximises the
power level...

Ant