First and second derivative of the function using fft?

2011 12 7
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David Young
David Young 2011-12-7

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See my answer to this question.

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Feruz
Feruz 2011-12-7
Thank you for your answer, but the idea is to compute approximation to first and second derivative of function (any) with *fft*...
David Young
David Young 2011-12-7
The fft is always applied to numerical data, never to a function proper, so if you are asking about finding a derivative of a function using the fft, you must mean numerically estimating the derivative from data generated by calling that function.
The answer I referred you to does exactly that. The example code is for 2-D data, but the simplification to 1-D is trivial (and in fact you can use the exact same code, just on a vector). Getting the second derivative is simply a matter of multiplying again.
The key thing you need to know is how to write the derivative operator in the frequency domain. The answer I referred you has it, as ftdiff.
Feruz
Feruz 2011-12-7
I see, second derivative you mention multiply again?!
i.e.
g = ifft(bsxfun(@times, fft(f), ftdiff))
gg = ifft(bsxfun(@times, fft(g), ftdiff))
Is that so?
David Young
David Young 2011-12-7
Yes, you can do that. But you don't need the bsxfun if you're working in 1-D - you just multiply. In addition, you can do both multiplications before transforming back. So something like this:
F = fft(f);
FD = F .* ftdiff;
FDD = FD .* ftdiff;
fd = ifft(FD);
fdd = ifft(FDD);
where uppercase variables are in the frequency domain, and d or D means differentiated.

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Dr. Seis
Dr. Seis 2011-12-7

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I have another example, here
In that case I am converting acceleration data to displacement, which is (double) integration in the time domain and division in the frequency domain. In the frequency domain, this is essentially done by dividing the complex acceleration amplitude at frequency 'f' by (sqrt(-1)*2*pi*f)^2.
To go from displacement to velocity, you would multiply the displacement amplitude at frequency 'f' by sqrt(-1)*2*pi*f
To go from displacement to acceleration you would multiply the displacement amplitude at frequency 'f' by (sqrt(-1)*2*pi*f)^2
This is what David was referring to when he says the second derivative is "simply a matter of multiplying again."

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Feruz
Feruz 2011-12-7
I see, but I am doing numerical approximation, and exact, as I understood, my numerical approx is by fft, so in a sense multiplying again, doesn't mean multiply as I wrote???

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