Linking 1-D and 2-D FFT's

Taking the one-dimensional Z transform of ${\bf b}$ in the helical coordinate system gives  

$$B(Z_h) = \sum_{p_h=0}^{N_x N_y -1} b_{p_h} Z_h^{p_h}.$$ (1)

Here, Z_h represents the unit delay operator in the sampled (helical) coordinate system. The summation in equation (1) can be split into two components,

$$\begin{eqnarray} B(Z_h) & = & \sum_{p_y=0}^{N_y-1} \; \sum_{p_x=0}^{N_x-1} b_{p... ...; \sum_{p_x=0}^{N_x-1} b_{p_x,p_y} \; Z_h^{p_x} \; Z_h^{N_x p_y}.\end{eqnarray}$$ (2) (3)

Ignoring boundary effects, a single unit delay in the helical coordinate system is equivalent to a single unit delay on the x-axis; similarly, but irrespective of boundary conditions, N_x unit delays in the helical coordinate system are equivalent to a single delay on the y-axis. This leads to the following definitions of Z_h and Z_h^N_x in terms of delay operators, Z_x and Z_y, or wavenumbers, k_x and k_y:

$$\begin{eqnarray} Z_h & \approx & Z_x \; = \; e^{i k_x \Delta x}, \ Z_h^{N_x} & = & Z_y \; = \; e^{i k_y \Delta y},\end{eqnarray}$$ (4) (5)

where $\Delta x$ and $\Delta y$ define the grid-spacings along the x and y-axis respectively.

Substituting equations (4) and (5) into equation (3) leaves

$$\begin{eqnarray} B(k_x,k_y) \; = \; B(Z_h) & = & \sum_{p_y=0}^{N_y-1} \; \sum_{... ... b_{p_x,p_y} \; e^{i k_x \Delta x p_x} \; e^{i k_y \Delta y p_y}.\end{eqnarray}$$ (6) (7)

Equation (7) implies that, if we ignore boundary effects, the one-dimensional FFT of ${\bf b}(x,y)$ in helical coordinates is equivalent to its two-dimensional Fourier transform.