Digital BW, The Print

> I give up! You dragged this out, and tested Mike, as though this carried
> great import. He answered it doesn't matter early on.

It matters as far as the question that was being asked, and that was what
causes the bunching up in the histogram.

> I'm still hoping you will explain in a simple coherent way, how
> the scanner
> assigns values to the input data, how that process is affected by the
> scanners bit depth, and how that shows up in the histogram.

I believe I can do that pretty easily.  BUT unfortunately, I have to relate
it to actual hardware...

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.

The A/D measures voltage (in this case).  Say the CCD/front end has an
output range of +-3V, and the A/D input range is also +- 3V.  Say the noise
is .0003 volts.  This is determined by experimentation with the CCD/analog
front end, the noise can be easily measured in a laboratory with the correct
equipment.

That means the number of different meaningful values the CCD can give us is
6 volts divided by .0003 volts or 20,000 different values.  The A/D has to
have the resources to handle 20,000 different values.  We would need 15 bits
(32,768) to be able to hold 20,000 different values, since our range does
not fit in 14 bits (16,384).

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.

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.

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.