High-precision IFFT

10 ビュー (過去 30 日間)
Demian Augusto Vera
Demian Augusto Vera 2022 年 1 月 24 日
コメント済み: Star Strider 2022 年 1 月 25 日
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 件のコメント
Paul
Paul 2022 年 1 月 24 日
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
Demian Augusto Vera 2022 年 1 月 25 日
Thanks for your answer, Paul. The symbolic array x is just a high-precision complex array of numers; no symbolic expression involved.

サインインしてコメントする。

採用された回答

Star Strider
Star Strider 2022 年 1 月 24 日
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 件のコメント
Demian Augusto Vera
Demian Augusto Vera 2022 年 1 月 25 日
Thanks, Star. Yes, I'll have to make my own ifft function. Thanks for your help!
Star Strider
Star Strider 2022 年 1 月 25 日
As always, my pleasure!

サインインしてコメントする。

その他の回答 (1 件)

Paul
Paul 2022 年 1 月 25 日
編集済み: Paul 2022 年 1 月 25 日
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.
  1 件のコメント
Demian Augusto Vera
Demian Augusto Vera 2022 年 1 月 25 日
Thanks, Paul. The theory indicates that it is conjugate symmetric. Using double precision that's ensured, as I can see. Don't know if using high-precision arithmetic it remains conjugate symmetric, but I think it has to.
Thanks again for your help answers.

サインインしてコメントする。

カテゴリ

Help Center および File ExchangeCalculus についてさらに検索

製品


リリース

R2021a

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by