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[filmscanners] RE: Density vs Dynamic range




Hi Julian,

> One of my points is that these two ideas/equations, both from you, are NOT
> consistent.

Ah, but they simply ARE!!!

> Making the assumption that min value = noise, and translating (2) into the
> same simple but meaningful jargon as (1) and forgetting the logs, and
> putting (1) and (2) side by side, we have:
>
> DR = (max - noise) / noise ...(1)
>
> DR = max / noise ...(2)
>
> These are NOT the same!

Well, yes they are.  When using "max" in the first, it's the top value, not
the range, and "noise" is the bottom value...let's assign values, 3 for max
here, and 1 for noise here...and you get a (max - noise) of 2.

Same holds true for DR = (max - min) / noise (preferred
equation)...obviously, if min is 1 and noise is 1, but say max was 4, min 2,
and noise 1...same overall range (2), and 2/1 is simply 2.

In the second "max", that would simply be 2, as it's the overall swing of
the signal.

> But as for "dynamic" and what that means, we have totally different ideas
> and I absolutely do not agree with your interpretation.

But it's not an interpretation, it's simply as it's defined in engineering
resources, and what the concept of DYNAMIC range means.

> I went to some trouble to explain what is dynamic range where I come from,
> but you haven't responded to that.  Dynamic range is a range, like a red
> house is a house. It is not a number of steps.

Sure I've responded...and I've said that's wrong, or it would simply be
called a range, without the word dynamic.  Dynamic has a very standard
meaning...and it means, in a nut shell, change.  You can look at "dynamic
range" to mean "the dynamics OVER a range"...

> Even better, here is a definition of dynamic range, the first I found...

Well, it's not necessarily better...

> dynamic range: 1. In a system or device, the ratio of (a) a specified
> maximum level of a parameter, such as power, current, voltage, or
> frequency
> to (b) the minimum detectable value of that parameter.

That is correct, and is EXACTLY the same equation I've said.

> There it is, a ratio.

It can be expressed as a ratio, and I've never said differently.

> Nothing to do with resolution at all,

It absolutely does have to do with resolution, and for some reason, you're
not understanding that.  The ratio is overall range to noise, fine.  The
noise IS the resolution, you can only resolve to the level of noise, that's
plain and simple and indisputable.  If you have a range of 1 and noise of
1/4, you can only get four different steps within that range.  You can NOT
get 8 discernable values, because your tolerance is simply 1/4...which is
the finest you can resolve to.

> although it is
> true that for those situations where the minimum level is determined by
> self-sourced noise, then that lower level = noise = the resolution.

But it doesn't matter, the noise is ALWAYS the resolution, whether it's the
same as the low signal level or not.

> Later you said:
>
> >And this I completely disagree with.  Dynamic range defines the
> resolution
> >within a range, and a "range" simply specifies the endpoints, and says
> >nothing about the resolution within that range.
> >
> >Here's an easy example.  You measure one foot in 1" increments.
> That gives
> >you a min of 1 and a max of 12.  Dynamic range = max/min, or 12 in this
> >case.  Range is still 12.  Now change your increment to 1/4".
> Dynamic range
> >= min/max, or 12/.25 and the result is 48.  Hum.  Over the EXACT
> same range,
> >we now have two different DYNAMIC ranges.
>
> But no!  It is NOT the exact same range.

Well, yes they are...you are making an assumption that the noise is the
lower signal level (start of the range), and that's not what I specified.  I
specified 1.

> In the first case, the
> range is 1
> to 12 i.e delta = 11 or ratio = 12.

> In the second case the range is 0.25
> to 12 i.e. delta = 11.75 or ratio = 48.  That is ----  two different
> ranges, or dynamic ranges, same thing in this example.

I never said that noise was the lower limit...I said that the lower limit
was 1.  You can start a range ANYWHERE, and I chose to start it at 1.  But,
I should have said 0...but it doesn't matter.  Like I said THE RANGES ARE
EXACTLY THE SAME, it's merely the noise that's different.  If you re-do your
"thinking" to accommodate that:

Surely, you can understand that you can have two exact same ranges, with
different noise?  That can't be hard to understand?

Yours (and mine) "equations" corrected:

Range of 1-12...noise of 1 = 11/1 or 11
Range of 1-12...noise of 1/4 = 44/1 or 44

four times the resolution...

  My interpretation
> follows if you mean that the minimum you could *measure* was 1" and 1/4"
> because there is some kind of measurement noise operating. If
> not, then you
> can measure 0" perfectly well in both cases, and the range in this case is
> in fact 12" in both cases.  There is no meaningful ratio to be
> made, so any
> attempt to do so is invalid.

Sigh...the ratio is to the NOISE, NOT to the lower limit.  Like I said, the
lower limit of the range can start anywhere.  That is something that I think
you need to understand...it's quite important, and basic to understanding
this.

> The resolution is increased in the second
> example, but this has NOTHING to do with the range.

Correct, increasing the resolution has nothing to do with the range, per se,
except minutely at the ends...possibly may be but depends on the
system...but it doesn't matter, because the data you get won't be any more
accurate anyway.

> The ruler example is
> dangerous because it is not clear what the limitation or "minimum signal"
> is, in turn because it is resolution limiting you here, not noise.

I believe it's a perfect example, and illustrates the concept well.

> I won't carry on any more about this, so this is forewarning that I won't
> write again (on this topic anyway!).  I don't mean this rudely and I hope
> you *do* respond to this, but I really don't think we are going
> to converge
> and my efforts are not likely to help anyone else.  I now think I
> understand what you mean by dynamic range, but I don't agree and I thought
> (maybe I was wrong) that  your definition was causing confusion.  As I am
> absolutely sure that I am not improving comprehension for anyone else,
> there is no  usefulness in continuation except for our intellectual
> satisfaction and I am happy to agree to differ.

I HOPE you take the time to try to read what I am saying, because I believe
you aren't.  I am repeating the same things...as I believe you are...and I
feel like you're reading something else and not what I'm writing...  It's
REALLY very simple, and for some reason, the concept isn't getting across.

Regards,

Austin

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