Julian,
> At 19:26 29/08/02, David wrote:
> >But what do you mean when you say that "dynamic range is a range"?
> >
> >Dynamic range is _always_ a ratio. If it's not, it's something different.
>
> Yes, the number is a ratio. As I have said many times, this is because
> this is the ONLY way you can quantify a range in a single number so that
> the number is independent of irrelevant things. But the thing
> the ratio is
> measuring is a RANGE.
It is measuring OVER a range, but it is NOT "a" range in and of it self.
> Just look at the equation, it is the biggest level
> divided by a littlest level. It measures the ratio of these two levels,
> that is it quantifies the range they represent, the range of
> which they are
> the endpoints,
Yes, the two endpoints define the bounds of a range...BUT...that does not
mean the result of this equation is also a "range". The result is a
"characterization" of one or more attributes of the inputs to the equation.
The value of "dynamic range" says something more than the ratio of the two
levels, it ALSO characterizes the number of "discernable" values.
> (I'll try and attach it. If it doesn't get through Tony's ISP's
> filters, go to
> .
> The diagram
> is Fig 5, about a quarter down the page).
>
> You can't get much clearer than this. The dynamic range is the range from
> the peak level (25dB) down to the noise floor (-65dB).
It's simply an illustration showing where dynamic range is measured from.
The RANGE it is showing has the endpoints at +25dB down to -65dB, THAT is
the range, and the dynamic range is the result of those two numbers.
> It shows
> unambiguously the max level at the top of the arrow. It shows
> unambiguously the min discernable signal level at the bottom of the
> arrow. The arrow illustrates the range.
The arrow DOES illustrate a range, and it points to TWO endpoints, but a
range OVER which dynamic range is measured, it does not show dynamic range
as a range. It is simply a diagram showing relative positions of log
values. I have never heard dynamic range ever characterizes as "the dynamic
range". If dynamic range were a "range", as you claim that diagram shows,
then so would headroom and SNR be a "range" and they are not. I believe you
are talking the term "range" here way out of context.
Dynamic range is a single number. Because it is characterized OVER a range,
does not make it A range.
> The
> dynamic range is the ratio of these two levels, or in logs since they have
> already done the work for us, 25-(-65)dB = 90dB. That's all there is to
> it, it is VERY straightforward.
That does not make it a range. Simply saying "90dB" does NOT illustrate a
range. Saying 25dB to -65dB DOES illustrate a range. They are different.
> > >>>>>>>>>>>>>>>
> >These values can be obtained from testing, and the bit-depth/resolution
> >within that range is immaterial to the DEFINITION of DyR. It may
> be material
> >to the values you will measure in testing, but it is immaterial to the
> >definition/formula.
> ><<<<<<<<<<<<<<<
> >
> >Yes, I think, sort of, maybe. If you are looking at the analog
> signal prior
> >to the A/D converter.
>
> The bit depth is one of the two things that might determine the minimum
> discernable signal of the scanner. If the smallest (one-bit) step is
> bigger than the noise level, then it will be the determinant of MDS. If
> not, the noise will be the determinant.
Correct.
> As Todd says, bit depth may be material to the values you will
> measure. But it is not part of the definition.
Who said it was?
> It MIGHT be in
> the formula
> if you know that it dominates the noise level in terms of determining
> MDS. In this case, your dynamic range formula would be DR = max signal /
> smallest step size.
>
> Otherwise the formula would be DR = max signal / noise level
Correct, and I've always said exactly that, even though you seem to ignore
that fact.
> >Austin's explained this: in any dynamic range calculation, the maximum
> >signal level can be seen as corresponding to the range of levels handled,
> >assuming the minimum level is defined. The noise (or minimum recognizable
> >signal level) (and the maximum signal level) defines how many meaningful
> >steps the maximum signal level is from the minimum signal level.
> That's all
> >dynamic range is: the number of meaningful steps from min to max. That's
> >normally expressed as a ratio...
>
> David, can you tell me why you have to include all these words
> like "Can be
> seen as corresponding to ..."? You don't need to be so convoluted, the
> reality is much simpler. Try this in place of your complex
> paragraph - "In
> any dynamic range calculation, the maximum signal is itself. The noise
> defines the minimum recognizable signal level. The ratio of these is the
> dynamic range".
That's true, but that doesn't negate what David said above at all.
> Much simpler, more accurate,
It is not "more accurate", both statements are equally as accurate in and of
themselves.
> and agrees with the definition formulae.
Well, so does David's statement.
> You
> do not need to tie yourself up with complex equivalences and second order
> ideas, just keep in mind that dynamic range is what is says it is and what
> the equation says it is, a range.
NO, the equation DOES NOT SAY IT IS A RANGE. You made that up your self.
Show me one equation that says that dynamic range is A range. None do.
> Very easily measured, very easily
> visualised,
If it's so easy visualized, then why are you having so much trouble
visualizing that it also provides the resolution of a system?
> There is another related concept
> (usually related but not always) that is the resolution. Most often you
> calculate these two different but related things using the same formula,
> because that is what the normal assumptions are. But not necessarily.
How are they different, and when are they not related?
> > >>>>>>>
> > If
> >a 1-bit scanner can assign any range a value that is 50% of its density
> >range, what bit depth scanner is it that will assign a signal the entire
> >scanners density range?
> ><<<<<<<<<<<<<<<<
>
> A 1-bit example is pretty pointless and gets people confused. If you want
> to do it, it is quite simple.
>
> The max signal is 1. The min signal DISCERNABLE SIGNAL is 1. The dynamic
> range = 1/1 = 1 or 0dB.
A 1 bit scanner is useless, as it has no dynamic range in and of it self.
You need three states. One below a given low threshold, one above a given
high threshold, and one between the two thresholds. You can assign any dMin
or dMax to either of the thresholds, and your three states can represent ANY
density range you want...but no matter what density range it represents,
it's dynamic range will always be the same.
> >My point is that a value reported by a scanner corresponds to a range of
> >possible values in the film, and that the size of that range is
> given by the
> >worse of the noise in the electronics or the bit resolution of
> the scanner.
>
> Absolutely. Has there been any argument against this concept?
By claiming that dynamic range does not characterize resolution, yes. But,
it appears in this post that you may have changed your position on this?
Once you understand that a measured "value" is actually a representation of
a "range of values", then you can understand that means you only have so
many discernable values...and therefore denotes a resolution.
Austin
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