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[filmscanners] Re: Density vs Dynamic range
On 6/9/02 11:02 PM, "Austin Franklin" <darkroom@ix.netcom.com> wrote:
> It's (my initial statement you disagree with) hardly an oversimplification,
> 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.
Perhaps this can be made clearer by bringing back a concrete example
(similar to one which I think Austin used earlier in this thread).
Suppose that the output of the CCD analog stage swings from 2.5 to +2.5
volts, and that there is 0.001 volts of noise. If you install an
analogtodigital convertor to sample signals from this system, there is a
lower limit to the step size that you should choose. There is no advantage
to choosing a step size smaller than the noise level of 0.001 volt. Why?
Because when you measure the voltage and get a result x, any value in the
range, you really don't know the actual value. You measured the signal plus
noise. The signal could have been anything between (x + 0.0005) and (x +
0.0005). So the optimum "granularity" is a sampling system with (2.5 
(2.5))/0.001 = 5000 steps. You could use a system with fewer steps, but it
would not be optimal. It would not encode all the information. If you used
a system with more than 5000 steps you would not be measuring with more
accuracy, you would just be building in excess complexity and expense.
The dynamic range of this system is 5000:1. You would need a 13 bit
converter to represent 5000 discrete steps, since 2^13 = 8192 (more than
enough) but 2^12 = 4096 (not enough).

Julian Vrieslander <julianv@mindspring.com>

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