> > Hi Todd,
> >
> >> This paper appears to speak to many of the issues discussed in
> >> this thread:
> >>
> >> <>
> >
> > I had a chance to look over that paper. The diagram you
> mention (I believe
> > you were referring to the sinusoidal wave +-5V signal...) that
> looked like
> > the Higgins one, though is similar, is being used to show an
> example of 2's
> > compliment arithmetic. Nothing really to do with dynamic
> range, as far as
> > our discussion that is.
>
> No man, I said: sec. 3.1, fig. 5, which illustrates DyR as a range
> between the "noise floor" and "distortion region".
I disagree that it shows it "as a range". That's like pointing to the
window on a picture of the Space Shuttle and saying that the whole thing is
a window... That is an illustration, that is in dB, and merely graphs the
region over which dynamic range is characterized.
> As per the rest, I'm really trying hard to move this AWAY from
> this being a
> conversation of bit depth because I think bit depth is particular
> to digital
> devices, which DyR is not.
That is absolutely correct. It's Roy, who tried to claim that bit depth
didn't have anything to do with dynamic range, and it does.
> I'm trying to discuss the CONCEPT of DyR,
> specifically whether is a resolution or a range.
Well, I think the equation is self explanatory...it's a ratio of the largest
signal to the noise, period. Whether you express it in log form, or
non-log form, it is not a range, but it is measured OVER a particular range.
Each value you get from the A/D is not precise, it represents a range of
possible values, that because of noise, are not distinguishable, and there
are only so many of these "range of values" that are usably
distinguishable...
If dynamic range wasn't the same as resolution, then WHY do all these web
sites say you need so many bits to represent a dynamic range of, say, 30dB,
when you could just simply use 8 bits to represent a dynamic range of 30dB?
It's because you also need to represent ALL the intermediate ratio values.
30dB is 1000:1, and you need to represent 999:1, 998:1...2:1 and 1:1...so
you would need 10 bits. 1000 is the largest signal, and 1 is the minimum
discernable signal.
One of the problems I believe you're having seeing this is the distinction
between log and non-log numbers. The non-log numbers are what is really
important (as that's what the scanner actually does, it doesn't do any log
math internally, there's no need), as far as understanding this. They ARE
the ratio, and they are :1 ("to 1"), such as 5000:1 (the dynamic range of
the Leafscan ;-).
Austin
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