R help - Apr 2009 - FFT function

Hi,

A very simple question.

I know that the Fourier transform of a Laplace distribution, with zero
mean and variance 2b^2, is equal to 1/(1+(tb)^2). Therefore, the Fourier
transform is a positive function (actually is always between 0 and 1).

BUT, if I use the fft(alpha) function of R, where alpha is a vector
containing 128 values of the probability density function from -2 to 2, I
get negative values as well.

Why?


Thanks,

AA.

--

On 10/04/2009, at 11:44 PM, Achilleas Achilleos wrote:
> Hi,
>
> A very simple question.
>
> I know that the Fourier transform of a Laplace distribution, with zero
> mean and variance 2b^2, is equal to 1/(1+(tb)^2). Therefore, the  
> Fourier
> transform is a positive function (actually is always between 0 and 1).
>
> BUT, if I use the fft(alpha) function of R, where alpha is a vector
> containing 128 values of the probability density function from -2  
> to 2, I
> get negative values as well.
>
> Why?

Basically because there is a (considerable!) difference between the  
continuous
Fourier transform and the discrete Fourier transform, the latter  
being what
is calculated by fft().  (NOTE:  ***NOT*** ``FFT'' --- R is case  
sensitive.)

	cheers,

		Rolf Turner

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