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[filmscanners] Re: Density vs Dynamic range



Austin,

I've answered many of these points in a different post, so this will be
just a few brief comments.

> > > > 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.

This is only true for the digital side.  Looking at the overall system the
base, unchangeable for a given temperature, noise figure is the analogue
noise from the CCD.

>
> > 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.

I don't agree, see above. Under certain circumstances it could be, but not
under cost-effective mass production.

>
> > 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.

0.2mV was chosen as it is 1/5000 of the maximum of 1V I used for this
example.

>
> > 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.

I agree, but I did say theoretically :-) Practically, the noise from the
CCD should be very small.

>
> > Converting this to a digital value is where the number of steps is
> > introduced.
>
> No, noise introduces that...and limits that.
>
Do you not agree that an AtoD with (say) a 0.1mV step size introduces
0.1mV of noise into the system?

> > 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.
>
This may be just a terminology issue. Digital resolution is defined by the
step size. Theoretical range is defined by the maximum and minimum values.
Practical range is limited by noise at one end and saturation at the
other. In practice resolution will be limited if the analogue noise in the
CCD exceeds the stepsize-defined digital 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.
>
> No, but you NEED a certain number of bits TO represent a particular
> dynamic
> range.
>

Agreed!

> > More bits WILL give you more useable data,
>
> NO, not if they are in the noise.
>

But if they are not in the noise, then they will - assuming you are
referring to the CCD noise here!

> > 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.
>

This is where I think you are the one who does not understand. More bits
give you more resolution NOT more range.  The range can be whatever you
design it to be.

If I say to you "this is a stream of 8-bit digital words" all you can tell
me is that this represents 256 possible values. I need to tell you what FF
and 00 represent for you to make any judgement. If I add an offset before
the AtoD so that 00 is just above the noise floor of the CCD and adjust
the gain of the analogue circuit so that FF represent the output of the
CCD just as it reaches saturation then I will have extracted the maximum
dynamic range from the CCD with a resolution of 256 steps.
If I then change the AtoD to one giving 16-bits (65536 steps) and change
nothing else I have not changed the dynamic range of the scanner, just
it's resolution. I will have reduced the digital noise in the system.

> > > > 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.

1 is the DIGITAL noise level - halve the step size; halve the DIGITAL
noise.
>
> > > > 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.
>

I think we agree on this.

> > > 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?????
>

It is your failure to understand that increasing the number of bits from
the AtoD does not necessarily increase the dynamic range of the scanner.


Peter, Nr Clonakilty, Co Cork, Ireland

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