High-precision IFFT
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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!
2 个评论
Paul
2022-1-24
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?
Demian Augusto Vera
2022-1-25
采纳的回答
Star Strider
2022-1-24
1 个投票
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.
4 个评论
Demian Augusto Vera
2022-1-25
Star Strider
2022-1-25
My pleasure!
The 'symmetric' flag refers to the result being conjugate symmetric as discussed in Algorithms, however subscript references do not always work the same way with symbolic vectors as with numeric vectors. I am not certain what you are doing, so I cannot offer specific help with it.
One approach might be to integrate symmetrically so that the inverse transform is integrated from a negative angle to the same value positive angle. That is possible using int because it offers that option. I am not certain that ifourier offers the same option.
Demian Augusto Vera
2022-1-25
Star Strider
2022-1-25
As always, my pleasure!
更多回答（1 个）
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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