spectral derivative using fft()

[Ben Abbott]

On Dec 18, 2007, at 8:01 AM, Ben Abbott wrote:

>
> On Dec 17, 2007, at 5:39 PM, Gastón Araguás wrote:
>>
>> i hope someone can help, thanks in advanced
>> Gastón
>
> At 1st glance, I noticed a problem. You used the following to
> represent the derivative of "f" (I've modified it slightly).
>
> 	fp = ifft ( ( 2i * pi  * m ) * fft( f ) );
>
> which is correct for continuous signals with infinite limits (Fourier
> Transform). However, for FFT's the signals are discrete & periodic
> (Discrete Fourier Transforms). So the derivative must be periodic as
> well. I haven't gone through all the details, but the result should
> look something like
>
> 	fp = 1i *  ifft ( sin ( 2 * pi * m ) .* fft ( f ) );
>
> I'm likely off by some scale factor, but you get the idea, yes?
>
> Ben

Ozzy is correct, a sawtooth is needed.

See this link http://web.cecs.pdx.edu/~mperkows/CAPSTONES/Fourier/Lecture22a.ppt

Here's some sample code for evaluating the spectral derivative, "fp",  
of a vector "f", using the FFT

	N = numel(f);
	n = 0:(N-1);
	Df = 1i * fftshift (n-N/2); % discrete differential operator for "f"
	fp = ifft ( D.*fft (f));

Ben
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