Digital BW, The Print

Austin,

You're doing a great job, I can sort of follow. BTW, this is a lot of work,
thanks! I guess I wont know if I've got this completely until we move
forward. I'll just ask a couple of questions, but only answer them if you
think they are relevant to building upon. I really don't want to sidetrack
the conversation too much.

Furthermore if you'd be answering any of these questions in the next part
anyway feel free to move on and do so in flow.
 
> CCD give you a range of voltage out of it.  That voltage range, no matter
> what it is, has noise in it.  That noise level will determine just how much
> signal to noise you can get from the CCD/analog "front end" (circuitry
> between the CCD and the A/D).  A CCD that can "look further into the dark
> regions" will have a lower noise level, and therefore a higher signal to
> noise ratio.
> 
> Now we have to match the CCD to the A/D, in two ways, one voltage, which we
> don't care about, it's done in the "front end", and second, the A/Ds ability
> to discern different values all way down to the noise level of the CCD/front
> end.  That's where the number of bits is important.  We need to have enough
> bits to represent from the minimum signal level out of the CCD/front end to
> the highest signal level out of the CCD/front end, and all discernable
> levels in-between that are above noise.

When you speak of noise, this is noise that is a by product of the scanners
circuitry, like hum, which I suppose3 has a certain predictability to it, or
noise like extra data bits that don't hold statistically meaningful data
bits. I guess I mean values which are too closely spaced to be
distinguishable as significantly different from one another. Would this be
non-integer ratio values? I'm a bit over my head here, but Dan Margulis
explained that when the scanner gets into the highest density regions it
doesn't just hit a wall and call everything that dense and denser black (in
the case of a chrome), it tend to sort of guess a value, and there is a
randomness to it. I think this is the noise we generally associate with
"noisy shadows" of a scan. But with negative materials this occurs in the
highlights of the IMAGE, though it's still considered shadow in this
context, correct? In the context of scanners, shadow is the film density
region, yes?
 
 
> The A/D measures the input voltages from the CCD/front end.  -3V would give
> an A/D output of 0, to +3 would give an A/D output of full scale, and for 15
> bits, that would be 32,767 decimal.  That is 32,768 different values.
 
> So, 0 represents as close to no light as the CCD can measure...the CCD is
> putting out no voltage...and 32,767 represents the highest amount of light
> you are calibrated to measure.  I could go into more detail about gain and
> calibration, but I believe that's another discussion.
> 
> The end voltage, and the intermediate voltages out of the CCD/front end are
> obviously also converted in to values between 0 and 32,767, and that is what
> is the raw output out of the A/D.  Those are the values that are assigned to
> the pixels.

Okay, to confirm, whatever tone the scanner measures to be darkest (or in
the case of the random noise I spoke of, assigns to be darkest) is assigned
the tonal value of zero, and each next whole integer value is assigned in
order up the scale? It is the A/D converters function to assign tonal values
to the voltages received by the CCD?
 
> Also, the reason more bit depth gives better dark detail is simple.  Note
> the value of -3 out of the CCD gives us a value of 0 out of the A/D.  Well,
> that means the scanner sees no light.  A voltage of .0003V out of the CCD
> would give us an A/D output of 1.  Remember, that this value of .0003 is the
> noise level of the CCD...and that it is simply because of the noise level
> that we require so many bits.  In order to see something between 0 and 1, we
> have to lower the noise level, say to .00015V.  That would give us 6V
> divided by .00015V number of values, or 40,000...which would require 16 bits
> to represent 40,000 different values.

So, the lower the noise, the more tones the front end can distinguish, and
the higher the bit depth (BD) of the A/D required, to keep up, so to speak?
With more noise you can get away with a lower BD A/D cause anything more is
wasted. Okay, so this is where the manufacturers claims become suspect
because they speak of the bit depth of their scanners relative to their A/Ds
without specifying the S/N ratio of the front end which is the primary
limiting factor?

In the instance where a manufacturer does put a high BD A/D in with a high
noise CCD, I get that you will still have a lot of shadow noise (which again
are tones which have bee spuriously assigned by the scanner, because it
can't truly distinguish them as discreet from one another, or from the noise
of the front end?). But when you get out of that threshold area, do you then
have more useable tones further along the scale than if you used the lower
BD A/D with that CCD?

> This should explain why bit depth limits dynamic range in a scanner (and the
> dark detail, not the light detail), and how the scanner "assigns" values to
> the data.  Let me know if you get this, any of this or none of this... Then
> we can move past it.

Nice job Austin. I think I'm getting it so far (roughly anyway). I look
forward to the next chapter. ;-)

Todd