Filmscanners mailing list archive (filmscanners@halftone.co.uk)
[Date Prev][Date Next][Thread Prev][Thread Next][Date Index][Thread Index]
[filmscanners] RE: Dynamic range
Julian,
> >Come on, Todd, the paper CLEARLY says dynamic range is a
> resolution. Why on
> >earth do they say so many times that you need so many bits to represent a
> >particular dynamic range? Forget the diagrams, you are confused
> by them, as
> >they clearly represent two different things.
>
> 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...
> You ask: "Why on earth do they say ...that you need so many bits to
> represent a particular dynamic range?"
>
> Because, Austin, bit numbers put a limit to the Minimum Discernable Signal
> (MDS) and hence DR, that's why.
Not necessarily true. That is only ONE of the two terms used to determine
dynamic range. The other is the overall signal range, so decreasing the
number of bits can ALSO simply decrease the overall range you are "capturing
over", while keeping the same minimum discernable signal.
> 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.
> 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 or 3.6D. 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.
> 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. This scanner can, technically, still encode ANY range of
DENSITY values into those 1024 available values.
> Remember... the DR in this
> case is NOT determined here by noise level.
> Please note this, it is
> determined by MDS as it always is, and in this case MDS is determined by
> number of bits, not noise.
Of course, I've always said exactly that.
> ***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.
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.
> 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...
> : 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. 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.
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...
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. 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.
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.
> 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.
> 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.
> It is important to understand this accurately,
I do, and always have. I HAVE designed scanners, I do not believe YOU have.
> 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....
> 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
----------------------------------------------------------------------------------------
Unsubscribe by mail to listserver@halftone.co.uk, with 'unsubscribe
filmscanners'
or 'unsubscribe filmscanners_digest' (as appropriate) in the message title or
body
|
