Apache-Talk @lexa.ru 

Inet-Admins @info.east.ru 

Filmscanners @halftone.co.uk 

Security-alerts @yandex-team.ru 

nginx-ru @sysoev.ru 




      :: Filmscanners
Filmscanners mailing list archive (filmscanners@halftone.co.uk)

[Date Prev][Date Next][Thread Prev][Thread Next][Date Index][Thread Index]

[filmscanners] RE: Density vs Dynamic range

Hi Julian,

> I want to take issue with a view that Austin is
> promoting and which I believe is incorrect and greatly confusing the

I snipped a BIG "mine is bigger than yours" but suffice to say, I am an
authority on this subject, and a reasonably well known one within my field.
I am happy to give my qualifications.  What I've said is absolutely correct,
and is the exact same understanding as other industry experts (not Photoshop
gurus, and other authors who have never designed a bit of imaging equipment
in their life, or ever software folks...many of these people also have
proliferated this misunderstanding).  If any of you know Kennedy McEwen (who
is a far more well known expert than I am in this EXACT field), he and I
have had many a discussion on this, and our understandings are precisely the

> At Sun, 9 Jun 2002 23:14:46 -0400 Austin wrote:
> >Dynamic range is, in our case, (dMax - dMin) / noise. Density range is
> >simply dMax - dMin. Dynamic range is the number of discernable values
> >within a density range (in our case). Density range is simply the max
> >density value you can get minus the minimum density value you can get.
> This is a definition of dynamic range that is at variance with the normal
> definition and VERY confusing.

It is EXACTLY in line with the ENGINEERING definition.  What has happened is
a number of very non-technical people (or people whose technical expertise
is outside this area), who believed they understood something they did not,
have mis-used the term over many years, specifically in the imaging
industry.  I have many engineering reference books, and they all coincide
with my use of the term.

> Dynamic range is a RANGE,

No, it is a DYNAMIC range...that's why the word "dynamic" is in there.  Look
up the word:

"Characterized by continuous change, activity"

In other words, the changes over a given range...and change must be
perceived for it to be change.  You must specify the range for dynamic range
to be meaningful.  Obviously, if you double the range you are resolving over
(and keep the resolution the same), you get a higher dynamic range (and a
higher range)...but no better resolution.

Increasing the overall range DOES increase the dynamic range, but that isn't
the only thing...increasing the resolution (or decreasing the noise) within
that range ALSO increases the dynamic range.  Range is an absolute thing,
DYNAMIC range is not, as it is "dynamic"...

If it was simply a range, then it would NOT be called "dynamic" range, but

I hate to bring in web references, as there are far more that have it wrong,
than have it right, but these I believe have it right:


And here is another resource (requires registration):


Which goes on to say:

"The dynamic range of a pixel (vertical resolution) is a function of the
linear well size and noise. A pixel that can hold 500,000 electrons with a
read noise of five electrons will offer a signal-to-noise ratio of 100,000:1
or approximately 17 bits of information (dynamic range = 2n, where n = the
number of bits). To take full advantage of this, the on-board processor of
the chip and the downstream components need to have 16- to 18-bit

> this
> idea that it
> is "divided by noise" is just strange.

But that IS THE definition of it!

  It is NOT "a number of values", it
> is a RANGE, that is, a ratio.

A range is NOT a ratio.  It is a range.  Again, definition:

"The maximum extent or distance limiting operation"

No where in the definition of range will you find ratio.

> And it is nothing *necessarily* to do with
> number of bits, in fact dynamic range existed for a long time before
> digital processing existed.

Of course, but you have to represent a range, and the resolution within that
range digitally...and you NEED a certain number of bits TO represent it.

> Where number of bits comes in is that for a
> given dynamic range there is a particular number of bits which is capable
> of coding all the possible levels in a meaningful way, without throwing
> away significant information.


> Dynamic range is no more or less than the range between ...

Absolutely not.  Simply stating the "range between" does not at all in any
way shape or form tell you how many.

To show the flaw in your argument, say you have a density range of 1.  Tell
me how many levels are there in that range.  You can't, because you need
more information, like the noise...


> The lowest level as has been said many times is determined by the noise
> level, and the highest level is determined by the part of the system which
> first starts limiting or saturating.

That is not entirely true.  Your "range" can start well above the noise
level, and can stop well before saturation, it depends on what you are
trying to do.

> In my view, despite your (Austin's) statement to the contrary, the dynamic
> range is PRECISELY the measurable density range, in our context.

You are absolutely, unquestionably wrong.


> There is a kind of analogous situation in our context.  In a scanner the
> DYNAMIC RANGE would be the range between the lightest part of an
> image that
> can be scanned meaningfully and the darkest  part.

Again, absolutely WRONG.  That is merely the density range.  This IS the
misconception, misunderstanding that people have.

Show me ONE technical engineering reference book that supports your claim.
Photoshop books to NOT count, nor do the "guides".  Specifically an
ENGINEERING technical book.  I suggest you take a look at one of the
"Bibles" of digital signal processing:

"Digital Signal Processing in VLSI" by Richard Higgins.

It clearly agrees with the definition and formula and understanding I have
outlined for dynamic range.

<more snipping>

> Now back to number of bits.  The number of bits can be related to a given
> dynamic range because, as has been said, you don't want to waste either
> bits on the one hand, or resolution on the other hand.  There is
> nothing to
> stop you using more bits if you want, but it is largely wasted because
> every level you are measuring / coding will be jumping up and
> down by about
> the noise level.

You are correct...and you just countered your whole argument.  What you just
said, was the noise level limits the dynamic range...but noise does NOT
limit the overall range (except minutely at the endpoints, technically).
So, how can dynamic range simply be range?  How does noise limit the overall
range?  It does not.

> So you choose your coding so that the distance between
> successive measurement levels (i.e one bit) is the same as the noise
> level.


> If you use more bits, then you will get coded levels out
> which flip
> between 2 values even when the actual input signal doesn't
> change, so there
> is just no point.  If you use less than optimum bits, you will be losing
> some of your possible resolution.


> So the optimum number of bits is decided by:
> num coded levels =  max level divided by optimum digital step size
>                  = max level divided by noise level.


Above, you said dynamic range was limited by noise.  Here you say number of
levels equals max level (which is dMax - dMin...look familiar, like exactly
like the DENSITY range) divided by noise level.  Now you're contradicting
your self.  Here is what you said (and I quote):

"Dynamic range is a RANGE, this idea that it
is "divided by noise" is just strange."

But yet here, you are DIVIDING IT BY NOISE.  What am I missing here?

> This gives you the number of levels.  The number of bits is then
> determined
> from that by: num bits = log (base2) num levels.
> Note that num levels = max level divided by noise level = dynamic
> range.

Gee, that looks like an awful familiar equation....now where have I seen
that before...

> This is NOT saying that num levels is the same thing as
> the dynamic
> range. They are two totally different things.

That was not the discussion.  The discussion was, how the dynamic range of
the CCD/analog front end is determined, and that is simply, as you said
above, max/noise...and then, how many bits are NEEDED/REQUIRED to represent
THAT GIVEN dynamic range...

>  It is only saying that the
> OPTIMUM number of levels is the same number as the dynamic range expressed
> as a ratio.

Well, duh.  That is PRECISELY what I've been saying.  I think you haven't
been reading my writing very carefully.

> OK I need to eat tonight so I have to stop this.  To reiterate - my aim in
> writing this was to point out that I don't agree with the definition that
> is being used in this discussion.

Yes, but you JUST DID agree with it!

> The problem being that this faulty
> definition is partly responsible for the discussion failing to
> converge.

Well, my definition is absolutely NOT faulty.  It IS the definition of
dynamic range, as stated in all my ENGINEERING technical books, and as
understood by industry experts, of which I am one of them.

> I doubt I have reduced the confusion , and guess I have
> probably started yet another argument, but that was *not* my aim!

But now I don't understand YOUR argument.  You have just agreed with what I
said...but somehow you believe you disagree.  Here is what I've said:

dynamic range (in our discussion of scanners) = density range divided by

Where is this incorrect?


Unsubscribe by mail to listserver@halftone.co.uk, with 'unsubscribe 
or 'unsubscribe filmscanners_digest' (as appropriate) in the message title or 


Copyright © Lexa Software, 1996-2009.