[filmscanners] Re: Density vs Dynamic range

["" <peter@softwell.co.uk>]

Austin,

I think we may have some fundamental problems regarding some of the
terminology. I'll try to explain my understanding of it. As an aside, you
have stated that you design scanners for a living - perhaps you could
explain where in the team you fit as it must be a multi-disciplined
environment. My background is an electronic engineering degree and many
years of analogue and digital circuit design, although I've now moved into
other areas.


> > I think where we differ is the assumption the a 5000:1 dynamic range
> > yields 5000 discrete integer values.
>
> Well, it does, that's in and of what a "5000:1" dynamic range means!

This is not the case, 5000:1 simply says that the highest voltage output
(analogue not digital representation) from the CCD is 5000 times greater
than the lowest value. Go back to your maths books and look up ratios -
the are dimensionless and cannot say anything about number of values.
In this case, if the highest value is one volt then the lowest will be
0.2mV. You can determine this by putting the CCD in a black box to measure
the minimum then gradually increase the illumination until the output
increases no more.  Please note I've said nothing about the number of
discrete values in that range - it is, in fact, a continuous range and
theoretically has an infinite number if infinitesimally small steps.

Converting this to a digital value is where the number of steps is
introduced. Sticking with the range above, if I use an AtoD with a 0.1mV
resolution I'll get 10,000 possible values out of that. If I use one with
1mV resolution I'll get 1000. and 0.01mV would give me 100,000. The
problem is that the smaller the step size the longer the AtoD takes to
reach a stable output value so manufacturers have to make a compromise.

> > Light intensity is a linear value so
> > that 5000:1 range can be divided into as many steps as you'd like to
> > use.
>
> Absolutely not true.  The 1 is the noise floor, and you can not discern
> below the noise floor, and you can only discern in increments OF the
> noise
> floor.  The noise floor IS the increment.

This is where the tricky bit of the conversion comes in - there is an
electronic noise floor (which is why I said Vmin was 0.2mV not 0) and when
you put real film in the light path there is a minimum value you'll get
because film is not totally opaque. Similarly clear film is not totally
clear so the maximum value will never be reached. In practice the
manufacturer will adjust the values that can be read to cover the range of
real films and to keep out of the noisy region. The ends of this range
will normally never be seen.

Incidentally, there are techniques for dragging information out of the
noise - the simplest is combining multiple scans.

>
> > It is in any case only a ratio and therefore has no units or integer
> > values. If we double the number of bits (possible values) that doesn't
> > increase the dynamic range of the scanner, only it's ability to
> > represent
> > accurately the value coming from the CCD.
>
> No, increasing the number of bits does not increase the dynamic range
> of the
> scanner, NOR does it increase the ability to represent accurately the
> value
> coming from the CCD, providing you already have enough bits TO
> represent the
> full dynamic range of the scanner.  Assuming you already have enough
> bits TO
> represent the full dynamic range of the scanner, adding more bits simply
> means more bits are "in the noise".

I never said that increasing the bits increases the dynamic range. The
analogy is on you PC going from 16 to 256 to whatever possible grey-scale
values. The more steps the smoother the appearance.
>
> If you believe what you say in the above paragraph, then I'd have to
> say you
> do not understand what dynamic range is, and what it means.  If you
> don't
> increase the dynamic range out of the scanner, more bits will NOT give
> you
> more usable data...as I said, they will simply be "in the noise".
>

The dynamic range of scanner is determined by the CCD's abilities and the
optics in the scanner, just in the same way some lenses are more contrasty
than others.  All of this has absolutely nothing to do with the digital
electronics.

More bits WILL give you more useable data, provided your PC, software,
monitor, and printer can use it. It will not increase the dynamic range
and it will not be "in the noise" unless the manufacturer has designed it
so.

> > If not, then your statement is still wrong, or at best, an
> > over-simplification.
>
> It's (my initial statement you disagree with) hardly an
> over-simplification,
> in fact, it is about as complete and accurate as you can get.  It is a
> plain
> and simple fact that a dynamic range of 5000:1 requires 13 bits to
> represent
> that full dynamic range.  Again, 1 is the first discernable signal (as
> in
> the noise floor), and since you can only measure in increments OF the
> noise
> floor...a dynamic range of 5000:1 simply means you can have every
> integer
> value from 1 to 5000...and to represent every integer value from 1 to
> 5000,
> you need 13 bits.
>

This is where your understanding falls down - a 5000:1 range does not
require, or limit, the output to 5000 possible digital values - re-read my
second paragraph above if you still do not see this.

> > I'm sure you'll agree that it's the noise floor and the saturation
> > level
> > that define the true dynamic range.
>
> Yes, and if you know that than I am miffed you say that my statement is
> wrong, because it says the EXACT same thing.  You can only differentiate
> values that are different by the level of the noise....and it appears
> you
> somehow believe you can differentiate to a finer degree than the noise,
> which you can not.
>

I think this is the crux of the difficulties. In all the preceding I have
been referring to the noise floor - that is, the small amount of noise
that is present in every circuit due to the moving electrons. This defines
the minimum signal level that can be discerned, but this noise is present
at all times. I'll call this the analogue noise hereafter.

> Simply put, dynamic range is the (max value - min value) divided by
> noise.
> That is the maximum number of differentiable values you can get.  You
> can't
> measure half the noise, because your unit of measure IS noise.
>

Without going deep into the maths of AtoD converters, I agree that there
is a noise level and it's determined by the various things, not least of
which is the step size. If we stick to the example step size of .1mV then
any value from (say) 345.55 to 345.64mV would give the same output
representing 345.6mV. The maximum digital noise is +/-0.05mV - half the
step size. If the analogue noise is less than the digital noise then
decreasing the step size (increasing the number of steps) will improve the
overall noise figure. If the reverse is true then there is nothing more to
be gained. The analogue noise is also a function of temperature (a complex
function - not linear) so any manufacturer's claims for their scanners
should include a temperature value. Perhaps we'll see liquid nitrogen
cooled CCDs sometime in the future (as they do now for astronomy).

Nowhere in the above have I used the term "density range". This only
becomes meaningful when film is introduced into the scanner and, in my
opinion, should only be quoted in conjunction with specific film types.

As we both know, marketing types will always quote whatever numbers they
think sounds best so anything published must be taken with a pinch of
salt.

Peter, Nr Clonakilty, Co Cork, Ireland

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