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Re: spectral derivative using fft()

From: Ben Abbott
Subject: Re: spectral derivative using fft()
Date: Tue, 18 Dec 2007 09:09:20 +0800

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

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?


Ozzy is correct, a sawtooth is needed.

See this link

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));


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