Re: Michael Reichmann reasoning for AA filters?
- From: Kennedy McEwen <rkm@xxxxxxxxxxxxxxxxxx>
- Date: Sat, 20 Oct 2007 14:55:49 +0100
In article <1192887107.282629.95650@xxxxxxxxxxxxxxxxxxxxxxxxxxx>, acl <achilleaslazarides@xxxxxxxxxxx> writes
On Oct 20, 1:24 pm, Kennedy McEwen <r...@xxxxxxxxxxxxxxxxxx> wrote:I think you have gone wrong before the integral...In article <_o-dnbPbVMtI2YTanZ2dnUVZ_judn...@xxxxxxxxxxxx>, David J.
Littleboy <davi...@xxxxxxx> writes
>That doesn't work very well because diffraction causes more loss in the pass
>band than a good AA filter (the MTF curve for diffraction is a straight line
>from 100% at 0 lp/mm to 0% at 1600/(f number), but what you want for an AA
>filter is a step function).
Actually, the diffraction MTF for a circular aperture is not quite a
straight line, it flattens off significantly towards the cut-off. The
diffraction MTF is the autocorrelation of the pupil.
Is that so? Then if I take the pupil to be described by a step
function between 0 and 1 (radially), the MTF would be $(1-f)^2$ (err..
I hope, at least) with f in the inverse of the length units in which
the radius is 1. But the inverse fourier transform of that isn't an
airy function, so this can't be the mtf... Or I misunderstood what you
said. Or screwed up some integral.
Remember you are calculating the common area between two overlapped circles.
Without going round the houses, the function you are looking for is:
MTF(u)=(2.a - sin(2.a))/pi where
cos(a) = w.u.f
w = wavelength
u = spatial frequency
f = f/#
--
Kennedy
Yes, Socrates himself is particularly missed;
A lovely little thinker, but a bugger when he's pissed.
Python Philosophers (replace 'nospam' with 'kennedym' when replying)
.
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