[filmscanners] Re: Dynamic range

["Julian Robinson" <jrobinso@pcug.org.au>]

Austin,

> > I have never read whatever paper you are talking about, but I
> > GUARANTEE you
> > it does not SAY that dynamic range is a resolution.  I am sure that you,
> > Austin, INTERPRET it to say that, but it will not actually say that.
>
>You probably should have read the paper before commenting...

But no, that is the point. I don't need to because I know that no paper
will say what you believe - you are mistaken in this and still, to this
date, after buckets of wasted electrons and keyboard hours, you have still
not produced a single reference that says what you say.  I gave you this
totally unsupported challenge as a free kick - you had absolutely every
chance to smother me in extracts from this paper that I have  not even
looked at with quotes that agree with you.  The fact that you have not done
so I think proves the point.


...
> > Austin, if you have a scanner with a noise level of 36dB below the max
> > signal (i.e. 3.6D or 1/4096),
>
>No, where did you get 3.6D???  You can't equate DENSITY values with DYNAMIC
>RANGE.  Density values are absolute things, like volts are, though density
>units are expressed in log form.  They are NOT relative to noise.

Good grief Austin, you are playing semantics.  I only included the 3.6D to
stop you from having a go at me for using dB in an area that usually we use
Bels.  I should have said Bels, except no-one understands what they are.
Forget the D, I wasn't referring to density absolute, I was using it as the
Bel version of 36dB.  You have used this semantic and unrelated
approximation of mine as your only argument below. It is not relevant, I do
understand what D is about, I was trying to protect myself and readers from
another of your interminable side branches designed to get you off the topic.

> > I am sure you'd agree that you need a 12 bit
> > downstream system to maximise the utility of this scanner.
> > (because 12 bits
> > digitises to 4096 levels, and one level is then just equal to the noise
> > level of 1/4096 * max signal.  You won't have wasted bits being lost below
> > the noise, and you won't waste good information by failing to digitise the
> > smallest possible discernable signal)
>
>That's correct, but don't confuse density values with dynamic range.
>
> > Call this Case 0.
> > ***The dynamic range is 36dB.  I say that is the RANGE of this
> > scanner, you say it is the RESOLUTION.  In this case it is both.
> > ***So, resolution also = 36dB.
>
>Well, no.  As I've said, you are confusing density values with dynamic
>range.

Now that I've removed the D equivalent, can you make a substantive comment
on the point that was being made?

> > CASE 1
> > ********
> > Now, if this same scanner only had a 10-bit downstream system
> > (such as from
> > the old days when A/D's were incredibly expensive), what is the dynamic
> > range?  The noise level is 1/4096, and the smallest digital
> > non-zero signal
> > is one bit or 1/1024.   Obviously the minimum usable or detectable signal
> > cannot be smaller than either of these, or in other words it is
> > the maximum
> > or the two figures.  In this case it is 1024, and the MDS is determined
> > ONLY by the bit-size.   Noise level is 4 times smaller than this, so is
> > irrelevant.  So DR is 1024:1 or 30dB or D3.0.
>
>The dynamic range IS 1024:1 or 30dB...but that has nothing to do with
>DENSITY values.

I am not discussing density values.  I am discussing dynamic range and only
dynamic range as you know, so please reframe your response accordingly and
stop this disingenuous nonsense.

>This scanner can, technically, still encode ANY range of
>DENSITY values into those 1024 available values.

As you once said, duh!

> > ***In this case DR = 30 dB
> > ***Resolution is still 36dB if you stick with your formula = max/noise, or
> > 30dB as it obviously is in fact, given you have a digital step size of
> > 1/1024 or 30dB.
>
>Well, here you go again, Julian...and this is why I get pissy with you.  You
>take things out of context and apply them to something else.  I NEVER said
>the MDS was ALWAYS noise.  In the case of the ORIGINAL SIGNAL, it is noise,
>in the case of the digitized signal, it is NOT noise.

Well Austin, let me quote one of many interminable exchanges where I was
tearing my hair out because you were insisting that MDS was noise.  Please
note carefully the contradiction, clear and unambiguous between these two
statements:

A)  "I NEVER said the MDS was ALWAYS noise. " - from this post

B1) "The "smallest discernable signal" IS noise." - from post in June

B2) "This is a misunderstanding of the concept of dynamic range. It is ALWAYS
based on noise." - post in June

B3) "Noise and "smallest discernable signal" are EXACTLY the same thing." -
post in June.

I have struggled for months to get you to agree that noise and MDS are not
the same thing, and now you tell me you have always thought this!!  I am
pleased that you are coming round, but flabbergasted at the same time.

Here is the exchange for the record so you don't accuse me of taking you
out of context:
------start of exchange last June--------------
Julian:
 > iii) How can you tell me that "smallest discernable signal" is not the
 > correct term!?

Austin:
It IS the correct term, but you are using the wrong definition for it! The
"smallest discernable signal" IS noise.

 > I don't say it IS determined by noise, I say that most of the time it
 > is. Because MOST of the time, the "smallest discernable signal" is
 > determined by noise, so MOST of the time dynamic range is determined by
 > noise.

This is a misunderstanding of the concept of dynamic range. It is ALWAYS
based on noise.

 > The importance of this semantic juggling is twofold, first, it is
 > important to understand the DEFINITION of dynamic range, and the fact that
 > it is NOT defined in terms of noise, it IS defined in terms of "smallest
 > discernable signal".

Noise and "smallest discernable signal" are EXACTLY the same thing.

 > Second, on those odd occasions when "smallest
 > discernable signal" is NOT determined by noise, then you need to make sure
 > that noise is NOT in the equation! (which is one reason why your equation
 > has a problem).

So, you are saying that my reference material is entirely incorrect? I KNOW
that isn't the case.
-----------end of exchange----------------------------

IN fact, noise and digitised step size are the two things that limit MDS in
a scanner, as I have always said.  Whichever one predominates determines
MDS and thus the bottom half of DyR..

>First off, you using the term resolution ambiguously.  The "resolution" of
>the actual signal is still 36dB, the resolution of the digitized signal is
>NOW 30dB.

Austin, we are discussing, this whole discussion, is about the dynamic
range (and resolution) of a SCANNER.  I am talking about the resolution of
the scanner.  This is not ambiguous, DyR is set in this case by the digital
step size.  It is 30dB.

> > CASE 2
> > ********
> > If this same scanner had a 14-bit downstream system, what is the
> > DR?  It is
> > (yawn)
>
>I assume this "yawn" is simply your tiring of all your reams of prolifery?
>I know I'm getting tired on this one post, and you've managed to post about
>6 more...

It was recognition of the fact that I have now written "max signal / MDS"
1000 times since this discussion started.  That I am still struggling to
get you to accept this most basic premise.  DR is not defined as
(max-min)/noise, it is defined as max signal / MDS.

> > : max signal / MDS. This time, MDS is determined by the noise level,
> > because noise level is higher (4 times higher) than the bit size.  MDS =
> > noise level = 1/4096. So the DR of this scanner is 36dB again.  You could
> > have any number of bits over 12, and it would not change the dynamic range
> > one iota.
> >
> > ***In this case DR = 36dB.
> > ***Resolution is --- 36dB by your formula = max/noise (correct this time),
> > or 42 dB if you just consider digital bit numbers and step size.
>
>I really don't know what your point is here.

My point is to demonstrate in agonising detail that your unambiguously
applied formula for DR (of the system) as "something/noise" is not always
correct, and your unambiguously applied formula that DR is determined by
the number of bits is not always correct.  We may even agree on this, but
in past discussions you have blasted people with these as absolute truths
when they are not.  The ONLY absolute in dynamic range that is true all the
time is: DR =  max signal / MDS.  These are what determines it, and it pays
to keep this in mind when making absolute statements like the erroneous
ones of yours I quoted above.  It is precisely these kinds of statements
that cause you to muddy the waters and confuse people.

>If your signal only requires
>12 bits, and you use 14, so what?  There are two wasted bits.  Of course the
>original signal only has a dynamic range of 36dB.  The dynamic range of 14
>bits is still 42dB, 14 bits is 14 bits.  That has nothing to do with the
>dynamic range of the SYSTEM though.  The SYSTEM dynamic range will certainly
>not all of a sudden increase to 42, but be 36 or less.
>
> > CRUNCHY BIT
> > ****************
> > Here is the crunchy part - the dynamic range is NOT determined by the
> > number of bits, it is LIMITED by the number of bits.  The dynamic range is
> > NOT determined by the noise level, it is LIMITED by the noise level.
>
>It depends on your context.  I have always said exactly what you say above,
>but there is more to the story.  Obviously, the dynamic range equation
>contains noise, and therefore, it is one of the DETERMINING factors in
>calculating dynamic range, and therefore dynamic range IS determined by
>noise...as well as overall range.

1.  No it is NOT determined by overall range.  Why do you fail to see this
basic point? It is determined - at the upper end - by the maximum signal, a
VALUE, not a range.  Overall range is a meaningless statement when you are
discussing DR.  The ONLY range that is discussed, or measurable, is what
turns out to be the dynamic range.  A range is defined by a top and a
bottom.  The top of the DR is the maximum signal, and the bottom is the
MDS.  It is, once again, simple.  What are the top and bottom of this
"overall range"?  Please tell me? I am sure you think you know, you have a
feeling about it, but let me know where and how you measure this "overall
range".  why do you introduce this "overall range".  It is an irrelevant
and confusing construct of yours. It is not necessary. It confuses yourself
and your explanations.

2.  YOur discussion of "noise" in this para is exactly why your previous
unqualified statements (that noise is on the bottom of the definition
equation for DR) are dangerous. In the above paragraph of yours you have
mixed up two different "noises".

"The dynamic range equation contains noise" - which equation? If you mean
the equation in the ISO standard, you will remember that noise is only
there at one level removed from the actual equation - signal to noise is
part of a definition of MDS.  In this case, this term "noise" is talking
about all the things in the scanner that limit the MDS.  It INCLUDES
therefore the limitations on signal visibility caused by your
digitisation.  So it includes the effects of digitisation step size.  It is
not the same thing as the analog signal noise that occurs up to the
digitisation, the noise that we have discussed everywhere else in this
argument.

All this is transparent and obvious if you stick with the basic and correct
definition of DR.

>Of course, the number of bits LIMITS the dynamic range, I've always said
>that...but BTW, that contradicts Roy's last round, as he claims that 8 bits
>has the same dynamic range as 16 bits...

Yes, I don't agree with Roy on this.

>I don't get your point here...
>
> > The dynamic range IS determined by the Minimum Detectable Signal,
> > which may
> > be determined by the noise  level OR the number of bits.  Neither of your
> > continued assertions are true :
> >
> > A) It is NOT true to say that DR = max signal / noise.  Case 1 shows this.
> > B) It is NOT true to say that number of bits determines DR, as you see in
> > Case 2.
>
>Ah, now I see what you're doing.  You are making things up in, yet, another
>attempt to claim I'm wrong.

No Austin, I have no malevolent plan here, I am not making anything
up.  Please curb your paranoia.  I am simply trying to prevent inaccurate
information from confusing people and thus avoid stifling what could be
interesting and useful discussions.

>  Well, the two above statements are take OUT OF
>CONTEXT, and, as you've done before, it's completely dishonest of you to do
>so.
>
>A) I KNOW that MDS is not always based on noise, I have always said that it
>is not, and that it can be based on other things.  It is a LIE for you to
>claim I said otherwise.

Well please see my above quotes, maybe you will see why I believed that you
thought this. To remind you, here is a quote again:   "This is a
misunderstanding of the concept of dynamic range. It is ALWAYS
based on noise."  And - "Noise and "smallest discernable signal" are
EXACTLY the same thing".

Can you see the problem?

>B) It is a FACT that the number of bits DOES determine the dynamic range
>that that number of bits can hold.  That is a plain and simple fact, and
>I've said nothing more than that.
>
>
> > You are continually confused by the fact that there is an OPTIMUM
> > number of
> > bits for downstream processing,
>
>I'm not confused at all by this.  If you believe I am, then you simply don't
>read what I write.
>
> > and it is determined by the noise
> > level.  No argument here.
>
>Then why are you arguing at all?  This IS what you've been arguing
>about...and all I've ever said.


No Austin, you didn't read what I said. I said that OPTIMIUM number of bits
is determined by noise level.  Nothing to do with dynamic range.  I have
been discussing dynamic range.


> > BUT you continually go on from this ASSUMPTION based on good
> > engineering to
> > say that both A and B are true.
>
>No, I absolutely do NOT.  YOU ARE MAKING THAT UP.  I NEVER SAID THAT.

See above quotes.

> > In any other case, EVERY case where the number of bits is not
> > exactly equal
> > to the inverse of the noise level, then either A) or B) must not be
> > true.
>
>I've NEVER said otherwise.
>
> > Since the number of bits is limited in practice to powers of 2, and
> > since manufacturers commonly put more bits into their machines than noise
> > justifies, then almost always the scanner will NOT be optimally engineered
> > and one of A) or B) will not be true.
>
>That's not true.  Most manufacturers DO pick A/Ds that are matched to the
>CCDs etc.

Yes of course they try to, but since A/D's come in 2-bit increments in
practice, obviously in practice they will choose the number of bits to
exceed the noise level limit.  Rarely will these exactly coincide.  This is
not important.

> > It is important to understand this accurately,
>
>I do, and always have.  I HAVE designed scanners, I do not believe YOU have.

Correct on the second count but not relevant.  What scanner have you designed?

> > because to discuss any of
> > this topic properly you have to understand WHY dynamic range
> > might or might
> > not be equal to either of the figures as calculated under A or B.
>
>Yes, and I do understand completely, and have NEVER said otherwise.  It's
>only when YOU make things up about what I said, apparently, so you can claim
>I'm wrong, when I'm not....

Despite your remarkable ability to avoid answering questions you don't
like, and to change the subject at other times, and to skip around
unnervingly in your arguments, and to change your stance without appearing
to, I work hard at not quoting anyone, especially you, out of context.  So
if you give me some examples (not just *stating* that I did it, show me the
quote and it's context) it will help me greatly and reinforce your
credibility in this argument. But when you do this, would you also like to
answer the points in my post of 23 August?

> > Or WHY
> > manufacturers adding extra bits does not necessarily change their
> > scanner's
> > dynamic range at all.
>
>And...I've always said that too.
>
> > Without an accurate understanding of the DEFINITION
> > of dynamic range you won't know what it is measuring and you won't be able
> > to properly answer questions on the topic.
>
>Of course.  You really have to stop making things up that I simply did not
>say.  You are either taking them out of context, or doing it purposely to
>say I'm wrong, when I'm simply not.
>
>Austin


Julian

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