Re: What can we expect by taking Fourier Transform of noise or random samples?

From: Han de Bruijn (Han.deBruijn_at_DTO.TUDelft.NL)
Date: 11/29/04 

Date: Mon, 29 Nov 2004 13:50:26 +0100

lucy wrote:

> Hi all,
>
> I am trying to understand using Fourier Transform to do denoising.
[ ... ]
> Any thoughts?

It all depends upon what your "real" problem is. But ...
You could consider denoising _without_ using a Fourier transform.

Basically, what you do then is taking the convolution of your signal
with a "filter" function, for example a bell shaped (Gauss) function:

Z(t) = norm . integral_(-oo)^(+oo) exp(-(x-t)^2/sigma^2/2) z(x) dx

Where norm = 1/(sigma*sqrt(2.pi))
The "bandwidth" sigma must be chosen properly.

Due to properties of the 'exp', the (numerical) domain of integration
is rather limited, say (t-2.pi.sigma) < x < (t+2.pi.sigma), which makes
the method feasible.

BTW. It is noted that the above convolution integral becomes a product
of functions in the Fourier domain: exp(-(omega*sigma)^2/2) * Y(omega)
Where Y(omega) = Fourier transform of z(t) .

IF your decision is in favour of the Fourier approach, nevertheless,
be well aware then that there exist subtle, but nasty, differences
between the so-called FFT (discrete version) and the continuous FT.

Han de Bruijn

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