🚀 go-pugleaf

In article <1992Aug27.215818.10407@waikato.ac.nz>, singlis@waikato.ac.nz writes:
|> Thanks to all who replied about FFT segmentation.
|> 
|> How is the 2D case explained? Is the 2D transform a sum of "planar" sin/cos
|> "waves", or is it more of a breakdown of the horizontal groups of vertical
|> transforms?

The two are equivalent, as you can see by looking at the equations.
E.g. define the relationship between the Fourier pair f(x,y), F(u,v) as

	F(u,v) = \int \int f(x,y) \exp (-2 \pi i (ux + vy) )  dx dy     (1)
	f(x,y) = \int \int F(u,v) \exp ( 2 \pi i (ux + vy) )  du dv     (2)

Reorder the integration in (1) to get

	F(u,v) = \int  [ \int f(x,y) \exp (-2\pi i vy) dy ]  \exp (-2\pi i ux) dx

which shows you can think of the 2D transform F(u,v) as the 1D "horizontal"
transform of the function that is the 1D "vertical" transform of f(x,y) for 
each x.

On the other hand, (2) shows that f(x,y) is the sum of functions of the
form \exp(2\pi i (ux + vy)), each weighted by F(u,v) du dv.  Each component is a 
plane complex exponential of frequency \sqrt(u^2 + v^2) and 
orientation \arctan (v/y).

	--Paul