Leonard Evens said:
I always find you discussions very informative, and this was no
exception. I believe I followed your discussion of grids and square
waves below. But now I'm confused.
I've made several partial attempts at mastering digital sampling
theory, with only partial success. I thought I had the basics, but
now I'm not sure. How does the Nyquist limit come into what you say?
The Epson 2450 provides you with 2400 samples per inch, or about 94
samples per mm. If I understand Nyquist properly, half that or 47
lp/mm is the highest spatial frequency that it can resolve, other
factors aside. I do understand that higher frequencies can be detected
as aliasing, but I'm now confused from you discussion about just what
the Nyquist limit tells you. Could you elaborate further?
More or less exactly as you described it in your question Leonard, so
you probably aren't as confused as you fear you may be. ;-)
2400samples per inch gives a Nyquist limit of 1200cy/in or, as you
convert it, 47lp/mm. That is the hard limit that the scanner can
unambiguously resolve - it only says what the maximum optical resolution
of the scanner can be. However, the resolution of each element in the
CCD is much higher than this, which is why the CCD as a whole will
produce a response to higher spatial frequencies. That response is,
however, aliased by the sampling density around the Nyquist limit. The
Nyquist limit is simply where the onset of aliasing occurs - if higher
spatial frequencies get into the sampling system then they will not be
reproduced faithfully.
The spatial frequency response (ie. effectively the MTF) of a
rectangular pixel is simply sin(pi.a.f)/(pi.a.f) where a is the width of
the pixel, f is the spatial frequency it is viewing and pi is the
obvious greek constant. This is the amplitude of the output that the
single pixel would produce if it was scanned at infinitely fine steps
across the test pattern of spatial frequency f. Similar to taking one
2cm square in my example and plotting the output it would produce as it
was scanned smoothly accross each of the test patterns in my previous
example.
One way of considering how this relates to a CCD, is that the elements
only exist at certain positions, not at every infinitesimal position
scanned across it, so the MTF is effectively sampled by a series of
delta functions, one at each real position of an element in the CCD,
which is where the familiar effects of sampling, such as Nyquist limits
etc. come in. By considering the CCD this way it is easy to see that
the resolution of the CCD and the Nyquist limits of the sampling density
are completely separable - they can be related to each other but they
need not be so.
For example, if the CCD has 100% fill factor then the sampling pitch, p,
is exactly the same as the element width, a. Consequently, at the
Nyquist limit of f=1/2p = 1/2a, the MTF of an ideal CCD is
sin(pi/2)/(pi/2), or 63.66%. However, if the CCD has only 64% fill
factor elements by area then a=0.8p, and the MTF at the Nyquist limit is
now sin(0.4pi)/0.4pi), which is 75.7%. Similarly, if the CCD is a
staggered 2 row device, with the 64% fill factor by area, then the MTF
at the Nyquist limit would be sin(0.8pi)/(0.8pi), which is 23.4%.
The Nyquist limit says nothing about the MTF of the sensor at the onset
of aliasing. As you can see from the above paragraph, a simple linear,
high fill factor CCD will have a very significant MTF at this limit,
meaning that the resolution is artificially truncated, resulting in all
of the familiar aliasing and jagged artefacts.
