David J. Littleboy said:
If the limiting resolution for a 1.6:1 target is 63 lp/mm, then the MTF at
100 lp/mm is a big fat zero _for 1.6:1 targets. So the published MTF curve
_doesn't apply to 1.6:1 targets_.
Completely wrong David!
The MTF still exists at 100cy/mm for 1.6:1 targets but the resulting
contrast in the image has fallen below the noise level. That certainly
does not mean that the MTF is zero, it just means that it requires a
higher contrast original source to measure it or a tedious averaging
method to reduce the noise floor of the measurement medium.
Most systems eventually do have a zero in their MTF, such as the
diffractive cut-off of a circularly symmetric optic or the MTF null on a
CCD pixel, but this has nothing whatsoever to do with the contrast of
the target used for test.
MTF is a measurement of the optical component or system, it is
completely independent of the illumination levels used in the test -
although that may well restrict how much of the unit's performance can
be assessed.
Right. The point where the contrast at the next higher measured frequency is
zero.
Note the difference between zero *contrast* in the resulting image and
zero *MTF* for the medium - they are *NOT* the same thing!
??? If the measured contrast is zero, then the MTF is zero. ???
No! No! No!
Absolutely not!
Measuring MTF with a low contrast signal is like measuring the
dimensions of CCD pixels with a foot rule! Sure, you can do it - but
you have to understand the limits of your measurement technique and make
sure that the results you get are not just a consequence of the method.
Your assessment of a 1.6:1 contrast resolution limit as an upper limit
of the MTF curve is just such a limitation - and the Fuji data sheet you
linked to actually says as much! *Nowhere* on that data sheet does it
state that the MTF is measured with a particular contrast signal -
because MTF is *independent* of the source contrast. It is a transfer
function of the medium - that is why it is called Modulation Transfer
Function. The resolution limit of the film is given on the data sheet
for two different contrasts, because these are the contrasts which, when
transformed by the MTF of the film, result in a signal which is
detectable with a defined (but not stated) signal to noise ratio. The
noise in this case being the film grain.
Hardly. The sky was at zone VI, may be VII. The transformers were in full
sun<g>.
And the shadows? That is where the contrast with the fully lit sky was
well in excess of 1000:1. The contrast between the highlights on the
transformer and the sky was certainly much lower, but that wasn't the
parts of the image I suggested were in high contrast.
If none of your images ever exceed 1000:1 contrast, why bother with
quality film? You can probably make reversal emulsions which have a
Dmax of 3 in your kitchen sink, let alone the crap, fogged and badly
stored films you can buy down the average flea market! Your argument
basically boils down to a statement that even the worst film available
commercially has far more latitude and dynamic range than you ever
encounter in you photographic exercises. Sorry, but I don't believe
that you are that limited, David - remember, I have seen some of your
work!
(Many of the results on photodo show better resolution at f/8 than at lower
f stops. Of course, I'm only looking at short tele and shorter lenses. But
that indicates that these lenses are a long way from diffraction at anything
smaller than f/8.)
Indeed that is often the case, but not with high quality optics.
The basic rule for the diffractive optical cut-off of a lens is 1/(W.f)
where W is the mean wavelength of the light (in the case of the visible
spectrum around 0.55um) and f is the f/# of the lens. However it
requires a very precise surface form to achieve the diffraction limit
and this becomes more difficult at large apertures, low f/#s, for two
reasons:
1. The diffractive limit is higher
2. The diameter of the optical aperture and the incident angle due to
the curvature of the lens surfaces are greater over that aperture.
In short, not only do you have to maintain precision over a greater area
of glass, but the level of precision required increases. That is why
you will often find lenses produce a "sweet spot" in resolution terms
around f/4-f/5.6 - where the diffraction limit actually reduces the
resolution more than residual design and manufacturing aberrations. A
good measure of the intrinsic quality of the lens is the f/# that sweet
spot occurs at - the lower the f/# the better the lens. Even budget
lenses manage sweet spots around f/8.
The usual estimate I see is 1600/f for 0% and 800/f for 50%, so that's 50%
at f/8 for the Minolta. Times the 20% for the film is 10%.
But the lenses I own
only begin to degrade noticeably at f/22 _in real-world image tests_. f/16
is indstinguishable from f/11, and it's not clear that I'm not just
persuading myself that f/22 is worse. f/16 really is 0%MTF at 100 lp/mm.
You need to measure the performance a lot better - the diffraction
limits of f/16 are clearly visible on most film.
f/22 certainly is worse - I have already had this discussion several
times, including one poster on another forum who went so far as to buy
some Provia and post scans of his results to demonstrate I was wrong,
but then changed his argument to "not too much worse". I don't know if
his images are still available, I'll have a search and if they are I
will post the link in an update, but this was almost a decade ago.
Yep. That's what I'm refering to. Which happens to be the issue that I was
responding to. Since that was the issue at hand, I didn't bring up grain
aliasing in my original note.
You responded to my post about aliasing on scanned film - which
intrinsically includes the aliasing of the emulsions characteristics. In
my subsequent response to that I already agreed that in most, but not
all, cases your comments were correct in terms of the image content but
that was not the whole story because grain plays a significant effect
and its spatial frequencies certainly do alias.
You seem to have a problem understanding that a random granular pattern
still contains spatial frequencies, some of which will alias. I can
sympathise with this view, but it is not correct. When Jean Baptiste
Fourier first proposed to Paris Academy of Science in 1807 that *all*
waveforms were simply the summation of sine waves of specific
frequencies, amplitude and phase, none other than the greatest
mathematician alive at the time, Lagrange, declared that this was
impossible and the Academy consequently refused to publish Fourier's
work - so your view is not without its significant historical
supporters. Nevertheless, Fourier was right, though Lagrange died
without ever admitting it.
Which film and lenses don't have, so the image on the film has smoothed
edges.
Exactly, but that doesn't mean those frequencies which would alias have
been eliminated, merely reduce by the MTF of the total system.
And optics, in fact almost all imaging processes, but again, unless the
MTF is actually zero (which does happen with optics, even ideal optics)
all frequencies are reproduced at some level. If some of these none
zero frequencies are above the Nyquist limit then they will alias - even
though the edge smears over several pixels.
So far so good. What I've been saying all along.
Aliasing occurs when there is a noticeable _in-band response to an out of
band signal_.
Since the out of band signals are attenuated to well under 10% (and that 10%
is a gross overestimate), I just don't see aliasing being an issue.
10% of the *original* contrast (for the image content) and significantly
more that 10% for the 100% contrast of the emulsion grain! With
sufficient original scene contrast, a perfectly adequate signal can be
recorded on many films, particularly single layer minimal scattering b&w
emulsions, at resolutions which will alias with a 4000ppi scanner.
Hmm. Extended edges in the image implies no detail to alias. You are
claiming here that this implication is problematic. I don't see any basis
for this claim. Explain, please.
For an explanation, read the text I wrote again!
This comes down to which spatial frequencies are present in an edge
transition and how well they are reproduced. Unless all of the spatial
frequencies are reproduced with the same response, the edge will change
its shape - reduce only mid spatial frequencies and the edge will ring,
reduce only low spatial frequencies and the edge will overshoot, remove
the fundamental frequency and the edge will have the same intensity on
either side, reducing all spatial frequencies as a function of the
frequency and the edge will smear. A smeared edge does not mean that
all spatial frequencies that could alias have been eliminated, it only
means that they have been attenuated more that the low frequencies.
Unless those high frequencies have been eliminated then they will alias.
They may be less than the low frequencies (and certainly will be) due to
the MTF of the system, resulting in edge smearing, but that does not
mean that they have been eliminated. Unfortunately, high contrast, high
spatial frequency signals occur naturally all over the place - any
specular reflection you have ever seen is an example. A low, but
non-zero, MTF at high frequencies will result in such image content
being reproduced on film and subsequently aliased by the scanning
system.
(On grain aliasing, aliasing is (by definition) an in-band response to an
out of band signal. Random dots don't appear to meet that definition, since
the components of the transform required to produce that response are out of
band.)
Sorry David, but that is simply BS. Random dots, by definition, have
*all* spatial frequencies present and thus contain out of band signals
which certainly do alias.
I suggest you generate some random steps on a single axis (make sure the
dots are larger than the sampling structure otherwise you will be off
into aliasing again) and fourier transform the waveform if you have any
doubt as to its spectral content.
