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[filmscanners] RE: Density vs Dynamic range




Hi Peter,

> 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.

Aside from designing the overall system, I typically design all the
electronics...from CCD to SCSI out, as well as the DSP processing algorithms
that happen on board.  I do not design the optics.  As well as an EE/CE
degree, I am an ME, so I do participate in quite a lot of the mechanical
design, at least the specifications thereof.  Over the past 15 years, I am
typically the CTO of the company, or the Chief Engineer of the project.
Currently I own the company.

> 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 felt that was your background.

>
> > > 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.

Not when referring to dynamic range...as when referring to dynamic range the
lowest value (divisor) is considered noise.  That IS the definition of
dynamic range.

> Go back to your maths books and look up ratios -

He he...go (back) to signal processing books and look up dynamic range ;-)

> the are dimensionless and cannot say anything about number of values.

Except in the case of dynamic range...and once you agree that the "1" is the
noise level, you will understand why.

> In this case, if the highest value is one volt then the lowest will be
> 0.2mV.

Why is the lowest value 0.2mV?????  Where did that number come from?  Again,
read the CCD specs that I referenced....it CLEARLY says exactly what I've
been saying.

> 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.

NO, absolutely not true.  You can only measure to the level of noise!  It's
a misconception that analog is infinitely continuous...it is always limited
by noise.

> Converting this to a digital value is where the number of steps is
> introduced.

No, noise introduces that...and limits that.

> 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.

Exactly...and with respect to dynamic range that low value is...NOISE!!!
BTW, not always is the low value noise in a simple range...as you can have
some offset in the given range...but in the case of dynamic range, the
divisor IS noise, and again, that IS part and parcel of the definition OF
dynamic range.

> I never said that increasing the bits increases the dynamic range.

No, but you said increasing the number of bits increases the
resolution...and it does not...as the noise is the same, so your lower bits
are "in the noise".  Once you agree that dynamic range is based on noise,
you will understand why increasing the bits does not increase the resolution
either.

> 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.

No, but you NEED a certain number of bits TO represent a particular dynamic
range.

> More bits WILL give you more useable data,

NO, not if they are in the noise.

> It will not increase the dynamic range
> and it will not be "in the noise" unless the manufacturer has designed it
> so.

Well, since the conversation was about an "optimal" number of bits, and it
was given that that number of bits being used WAS the optimal, adding more
bits DOES put these added bits "in the noise".  You can NOT effectively
increase the number of bits (meaning get useful data) without increasing the
dynamic range.

> > > 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

The 1 is the noise, and YOU CAN NOT RESOLVE BEYOND THE LEVEL OF NOISE,
PERIOD.  Therefore, 5000:1 DOES REQUIRE AND LIMIT the output to 5000 USABLE
values.

> > > 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.

Yes, and EVERY READING YOU TAKE HAS A RANGE WITHIN IT SELF AS TO WHAT IT
REPRESENTS.  If you have a reading of .100, and your noise level is .001,
your reading only gives you an actual reading that is within the range of
.999-.101.

> > 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,

(which I have designed...and one converter company uses my
sigma-delta/delta-sigma designs)...

> 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.

Correct, and if you understand that, then you agree with what I have said.

> 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.

Well, the conversation started with a given that the optimal number of bits
for the analog noise was what was being increased, and increasing it does
NOT increase the resolution or dynamic range, once you already have that
optimal number of bits.

> If the reverse is true then there is nothing more to
> be gained.

Exactly what I've been saying....

> 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).

Agreed.

> 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.

So, now that at the bottom here, you have completely agreed with what I've
said, where is the misunderstanding?????

Regards,

Austin

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