FFT of 3D array in MATLAB

I am trying to understand how the FFT of different directions in MATLAB works to reproduce in C/C++ instead.

So far I have the following simple example in MATLAB:

clearvars; clc; close all;
%3D FFT test
Nx = 8;  
Ny = 4; 
Nz= 6;
Lx =16;
Ly = 6;
dx = Lx/Nx; 
dy = Ly/Ny;
%-----------
xi_x =  (2*pi)/Lx;
yi_y =  (2*pi)/Ly;
xi  = ((0:Nx-1)/Nx)*(2*pi);
yi  = ((0:Ny-1)/Ny)*(2*pi);
x =  xi/xi_x;
y =  yi/yi_y;

zlow = 0; %a
zupp =6; %b
Lz = (zupp-zlow);
eta_zgl  = 2/Lz;
[D,zgl] = cheb(Nz);
zgl = (1/2)*(((zupp-zlow)*zgl) + (zupp+zlow));
[X,Z,Y] = meshgrid(x,zgl,y); %this gives 3d grid with z-by-x-by-y size (i.e. ZXY)
%ICs
A = 2*pi / Lx;
B = 2*pi / Ly;
u = (Z-zlow) .* (Z-zupp) .* sin(A*X).* sin(B*Y);
uh1 =(fft(u,[],3));%ZXY
uh2 =(fft(u,[],1));%ZXY
uh3 =(fft(u,[],2));%ZXY

So, in C/C++ I have a 3D tensor with (Nz+1) rows and Nx coumns and Ny matrices and taking the 1D FFT along each "row" of u returns the same results as the following in MATLAB:

uh3 =(fft(u,[],2));%ZXY

While taking the 1D FFT of u along each column of u in C/C++ returns the same result as the following in MATLAB:

uh2 =(fft(u,[],1));%ZXY

Then my question is what does this 1D FFT represent? and how should I represent it in C/C++?

uh1 =(fft(u,[],3));%ZXY

The cheb(N) function is:

function [ D, x ] = cheb ( N )
  if ( N == 0 )
    D = 0.0;
    x = 1.0;
    return
  end
  x = cos ( pi * ( 0 : N ) / N )';
  c = [ 2.0; ones(N-1,1); 2.0 ] .* (-1.0).^(0:N)';
  X = repmat ( x, 1, N + 1 );
  dX = X - X';
  
%  Set the off diagonal entries.
  D =( c * (1.0 ./ c )' ) ./ ( dX + ( eye ( N + 1 ) ) );
%  Diagonal entries.
  D = D - diag ( sum ( D' ) );
 
  return
 end

Answer 1

[Matt J]

Why reinvent the wheel. Why not just use the open source FFTW C/C++ library (which Matlab is based on).

https://www.fftw.org/

In any case, the mathematical interpretation of Matlab's 1D FFT is in the documentation,

https://www.mathworks.com/help/matlab/ref/fft.html#buuutyt-5

[Janee]

I am using the FFTW library in C++ and I have tested Eigen FFT functions as well. FFTW library is not super user-friendly honestly and I have tried to reproduce the FFT MATLAB output to no avail. Thanks.

[Matt J]

Is the difference more than just a global scaling factor? We cannot assess the problem without examples.

[Janee]

There's no difference. It is just that I am not sure what "3" here represents in

uh1 =(fft(u,[],3));%ZXY

Like, I get that it supposed to be along "y" direction, but what does that really mean? for a 3D tensor u with nz+1 rows, nx columns and y matrices what does 1D FFT of u along y mean (i.e. 3)?

[Janee]

As the link you posted explains FFT along each columns is (u,[],1) and FFT along each row is (u,[],2) then what is fft(u,[],3)?

[Matt J]

Since you understand that fft(u,[],1) and fft(u,[],2) are the same operation (along different dimensions), why do you think there is a different meaning to fft(u,[],3)? There isn't.

[Matt J]

It is the same as if you did it in 3 steps:

tmp=shiftdim(u,2);
Tmp=fft(tmp,[],1);
uh3=shiftdim( Tmp  ,  -2);

[Janee]

So, fft(u,[],3) is just 1D fft along each row of u?

[Matt J]

No, it is the fft along each vector u(i,j,:).

[Janee]

This line "fft along each u(i,j,:)" actually solves my issue and makes more sense!