How can I optimize the FFT of very large arrays

Hello,

I am running a code involving the FFT and the IFFT of very big (up to millions of elements in total) 3D arrays currently using fftn and ifftn. Since the operation is carried out several times, it constitutes a big chunck of the calculation time of my code. How can I optimize my code?

• Would the use of fftw library be beneficial?
• The MATLAB guide for fftn also includes the use of thread-based environments, GPU, and distributed arrays to improve the performance of the code. How would these work? I'm sorry, but I'm not familiar to these methodologies.

Thank you for your help in advance

[Bruno Luong]

"Would the use of fftw library be beneficial? "

No matlab FFT has FFTW under the hood, and pretty much the best one can do. (I have tried it). I doubt you can do better since FFTW is natively already multithread.

You might try GPU though. Just transform your data in gpuArray and call fft on it.

[riccardo ozzola]

Thank you for your anwer.

Instead of the GPU way, would having a compiled version of that piece of code help me?

[Bruno Luong]

What piece of the code?

[riccardo ozzola]

The function involving the calculation of the FFTs

[Bruno Luong]

Honestly I don't know what you refer to, beside native functions fft, fft2, fftn

If you want FFTW the source code and compile DLL for windows is here https://www.fftw.org/download.html

[Walter Roberson]

MATLAB fft function implementation is already compiled code that automatically uses multiple cores if available. It is not written in MATLAB itself.

Answer 1

[Star Strider]

One option may be to be certain that the original signal so that its length is zero-padded to a power of 2:

Fs = 256;
t = linspace(0, 1E+8, 1E+8+1)/Fs;
signal = sin(2*pi*t*2);

L = numel(signal)

L = 100000001

tic
FTsignal1 = fft(signal)/L;
toc

Elapsed time is 16.542528 seconds.

NFFT = 2^nextpow2(L)

NFFT = 134217728

tic
FTsignal2 = fft(signal,NFFT)/L;
toc

Elapsed time is 4.468461 seconds.

For the same signal, the calculation time is about 25% that of the un-padded signal.

.