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[filmscanners] RE: Density vs Dynamic range
Hi Peter,
> 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.
Not so. There is no noise, per se, in the "digital side", it simply "1"
(what ever your number of bits is, you can only resolve to half of the LSB).
No, everything I have been talking about is purely analog...and I've
discussed translating that from analog to digital.
> Looking at the overall system the
> base, unchangeable for a given temperature, noise figure is the analogue
> noise from the CCD.
I don't see what you're talking about here. The simple dynamic range
equation is for a static condition...of course, any change in noise or range
will change the 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.
>
> I don't agree, see above. Under certain circumstances it could be, but not
> under costeffective mass production.
I have no idea what you're talking about here...it certainly isn't in
response to the above conversation...which was about a dynamic range "ratio"
as some use, say, 5000:1, where "1" IS the base line increment value or the
"ratio" value, and it's always 1. It equates to the noise. If the noise
were half that, relatively, then the dynamic range would be stated as
10,000:1, not 5000:1/2...
> > > 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.
"very small" is not very quantitative. Noise is what it is, small or
large...and if it's small relative to the overall range of the system, your
dynamic range will be high. The higher the noise, given the same overall
range, the lower the dynamic range, the lower the noise, the higher the
dynamic range.
> > > 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?
A/Ds don't typically specify "step sizes". They have an overall range, and
a number of bits within that range...but none the less, if your "step" is
0.1mV, then that is your minimum resolvable value, and the data will be +
0.05mV. But you have to match that to the analog noise. If your analog
noise is also + 0.05mV...the system noise IS +0.05mV. The digital "noise"
doesn't get added to the analog noise...as they both are equal, for the most
part ;)
> 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.
Basically, that's true...BUT...I disagree that in all cases that the limits
are noise and saturation. You can have an offset in the system that is far
greater than the noise...and therefore the min and the noise level aren't
the same.
> In practice resolution will be limited if the analogue noise in the
> CCD exceeds the stepsizedefined digital noise.
OK, but it'll be limited by the analog noise. It's simply a matching
game...you match the output of the CCD to the input of the A/D voltage
wise...and you match the number of bits in the A/D to the dynamic range
presented to the A/D, and you have an optimal system.
> > > 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!
But that was the original point...that was disagreed with...
> > > 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
Agreed.
 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.
Yes, that's EXACTLY what I've said all along. Of course more bits doesn't
give you more range, the range is fixed!!! In fact, I've only talked about
resolution, and not overall range. Go reread my posts...because it appears
you missed what I wrote.
> If I say to you "this is a stream of 8bit 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.
I don't agree with that. If the noise floor (and the minimum discernable
value) was .002V and the range was 6V...you set 00 to .002V and FF to
6V...and you used an 8 bit converter, you would NOT be resolving all the
analog data. You can put an 8 bit converter on ANY analog front end...but
that doesn't mean it's matched to the front end.
> If I then change the AtoD to one giving 16bits (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.
Since I believe your paragraph before this is wrong, I believe this is wrong
too...
> > > 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 was not talking about digital noise levels...it is an analog noise level
simply specified as a ratio to 1. It's the exact same thing as saying a
noise level of .001V and a 5V range. That is a 5000:1 dynamic range.
> > 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.
That's good ;)
> 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.
I've NEVER said that. NEVER. Of course increasing the number of bits does
NOT NECESSARILY increase the dynamic range. I have NEVER said differently.
Of course I understand that. WHERE do you believe I said differently?
PLEASE go reread my posts, as they CLEARLY say that I know that increasing
the number of bits does NOT necessarily increase the dynamic range. In
fact, I said the following (and I quote):
"That is correct....providing the system can actually take advantage of
those bits. If you have a 24 bit converter, and 12 bits of it is useless
(noise), then what good are the 24 bits?"
"Again, bit depth is the limiting factor OF dynamic range, but does not
assure you that the dynamic range of the scanner is equal to the bit
depth...as in the 24 bit, w/ 12 bits of noise, example."
"Yes it can (as I've said many times, if YOU read my posts), but that has
nothing to do with what I said. READ IT AGAIN. "To extract the FULL
dynamic range"... NOTE THE WORD FULL. There IS FOR A FACT an optimal
number of bits, as I said, as Kodak also says. If you use less, you will
NOT get the FULL dynamic range, and if you use more, you will not get any
better data."
"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"."
So, where on earth did you get the idea that I believe that I fail to
understand that "increasing the number of bits from the AtoD does not
necessarily increase the dynamic range of the scanner"???????? OBVIOUSLY
from my four clips above, and ESPECIALLY the last one it should be bloody
obvious that I completely understand what you believe I do not.
Austin

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