High-precision IFFT
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Demian Augusto Vera
am 24 Jan. 2022
Kommentiert: Star Strider
am 25 Jan. 2022
Hi! I need some high precision computations, so I use the function vpa() with about 32 digits. My final result is obtained by making ifft(x, 'symmetric'). The problem is that I need to transform x to double again, loosing my required precision. Is there any way to overcome this? I think the only way is to make my own ifft function, I am correct? Thanks!
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Paul
am 24 Jan. 2022
I'm pretty sure that you are correct that you will have to roll your own, symbolic, ifft() function. Is the symbolic array x all high precision numbers? Or are the elements of x symbolic expressions?
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Star Strider
am 24 Jan. 2022
Since it appears that these calculations are all symbolic, one option would be the ifourier function, or if you are using your own Fourier transform integration code, simply reversing the sign of the argument of the exponent will produce the inverse transform with essentially the same code, although with a different variable of integration (ω rather than t).
That way, everything remains symbolic, with the desired precision.
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Paul
am 25 Jan. 2022
Bearbeitet: Paul
am 25 Jan. 2022
Given that x is an array of high precision numbers, not symbolic expressions, it should be straightforward to implement the ifft sum. It might not be efficient and it might be slow, but it should work. The one thing I'm not sure about is how well this will work to ensure that the time domain sequence, X[n], is real, which I think is the expectation based on the symmetric flag in the call to numerical ifft. If the result does have a small imaginary part, it can always be removed I suppose, but I don't know what that indicates about precision of the solution you're trying to obtain. OTOH, I'm curious if you are ensuring that the sequence x is exactly conjugate symmetric in the first place, to whatever precision you're using. Or the ifft sum can be implemented assuming x is conjugate symmetric.
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