I need some help and clarifications for my notations in 3D centered fft. Consider the 1D centered fft of a 2D image of size $(N+1)\times (N+1)$, where $N$ is even, along X axis it can be computed using 1D FFT along each row such that, \begin{eqnarray} F^{}_x (r,c) = \sum\limits_{n=-N/2}^{N/2} f(r,n) e^{-i\frac{2\pi c n}{N+1}}, \nonumber \\ -N/2 \leq r \leq N/2, -N/2 \leq c \leq N/2 \end{eqnarray} Next we operate on each column such that the 2D centered FFT can be written as, \begin{eqnarray} F^{}_{xy}(r,c) = \sum\limits_{n=-N/2}^{N/2}F^{}_x (n,c) e^{-i\frac{2\pi r n}{N+1}},\nonumber \\ -N/2 \leq r \leq N/2 , -N/2 \leq c \leq N/2 \end{eqnarray}
The order of operation is not important. We can even begin with each column operation first then each row operation the calculation would not be affected.
I hope so far my notations are correct. Now comes the confusing 3D part (it has row, column and depth indexes), consider 1D fft operation along each row or X-axis, \begin{eqnarray} & F^{}_x (r,c,d) = \sum\limits_{n=-N/2}^{N/2} f(r,n,d) e^{-j\frac{2\pi c n}{N+1}}, \nonumber \\ & r,c,d = -\frac{N}{2},\cdots, \frac{N}{2} \nonumber \\ %& -N/2 \leq r \leq N/2, -N/2 \leq c \leq N/2, -N/2 \leq d \leq N/2 \nonumber \\ \end{eqnarray}
Next taking 1D fft along each column or Y-axis we have the equation,
\begin{eqnarray} & F^{}_{xy} (r,c,d) = \sum\limits_{n=-N/2}^{N/2} F^{}_x(n,c,d) e^{-j\frac{2\pi r n}{N+1}}, \nonumber \\ & r,c,d = -\frac{N}{2},\cdots, \frac{N}{2} \nonumber \\ %& -N/2 \leq r \leq N/2, -N/2 \leq c \leq N/2, -N/2 \leq d \leq N/2 \nonumber \\ \end{eqnarray}
Now how do I represent the final 1D operation along Z-axis or each depth index ?
Can I write something like this ?
\begin{eqnarray}
&(i) \qquad F^{}_{xyz} (r,c,d) = \sum\limits_{n=-N/2}^{N/2} F^{}_{xy}(n,c,d) e^{-j\frac{2\pi r n}{N+1}}, \qquad \qquad or\\
&(ii)\qquad F^{}_{xyz} (r,c,d) = \sum\limits_{n=-N/2}^{N/2} F^{}_{xy}(r,n,d) e^{-j\frac{2\pi c n}{N+1}}, \nonumber \qquad \qquad or\\
&(iii)\qquad F^{}_{xyz} (r,c,d) = \sum\limits_{n=-N/2}^{N/2} F^{}_{xy}(r,c,n) e^{-j\frac{2\pi d n}{N+1}}, \nonumber \\
& r,c,d = -\frac{N}{2},\cdots, \frac{N}{2} \nonumber \\
%& -N/2 \leq r \leq N/2, -N/2 \leq c \leq N/2, -N/2 \leq d \leq N/2 \nonumber \\
\end{eqnarray}
Are the above equations representation for 1D fft along z-axis correct ? Can someone help to clarify my confusion ?
First I should discuss a little about the reason for such notations that I am looking for. I wish to decompose the 3D operation into a bunch of 1D operations so I can do something in parallel.
The Fast Fourier Transformation (FFT) of three-dimensional (3D) data is of particular importance for many numerical simulations used in High Performance Computing codes. In order to do this for large data it is required to make the code parallelizable.
Hence I wanted clarification for my 3D notations in the form of 1D FFTs. If you come from 2D to 3D this confusion arises, since we index images differently.
After scouring through the internet and looking for texts with similar notation I came at this one, Parallel 3D FFT. With this reference I was able to understand what notation I should finally use.
So now the correct notations are For 1D fft operation along X-axis, \begin{eqnarray} & F^{}_x (r,c,d) = \sum\limits_{n=-N/2}^{N/2} f(n,c,d) e^{-j\frac{2\pi r n}{N+1}}, \nonumber \\ & r,c,d = -\frac{N}{2},\cdots, \frac{N}{2} \nonumber \\ %& -N/2 \leq r \leq N/2, -N/2 \leq c \leq N/2, -N/2 \leq d \leq N/2 \nonumber \\ \end{eqnarray}
Next taking 1D fft along Y-axis we have the equation,
\begin{eqnarray} & F^{}_{xy} (r,c,d) = \sum\limits_{n=-N/2}^{N/2} F^{}_x(r,n,d) e^{-j\frac{2\pi c n}{N+1}}, \nonumber \\ & r,c,d = -\frac{N}{2},\cdots, \frac{N}{2} \nonumber \\ %& -N/2 \leq r \leq N/2, -N/2 \leq c \leq N/2, -N/2 \leq d \leq N/2 \nonumber \\ \end{eqnarray} Finally the 1D fft along Z-axis gives, \begin{eqnarray} &F^{}_{xyz} (r,c,d) = \sum\limits_{n=-N/2}^{N/2} F^{}_{xy}(r,c,n) e^{-j\frac{2\pi d n}{N+1}}, \nonumber \\ & r,c,d = -\frac{N}{2},\cdots, \frac{N}{2} \nonumber \\ %& -N/2 \leq r \leq N/2, -N/2 \leq c \leq N/2, -N/2 \leq d \leq N/2 \nonumber \\ \end{eqnarray}