Re: [APML] Fw: images

[Chuck Vaughn <aa6g@aa6g.org>]

Rick,

> Chuck,
>
> I objected to you being surprised at peoples confusion, when I've seen 
> many
> an expert get confused when talking to another expert, mainly because 
> they
> use different definitions or misunderstand each other.  S/N is a very
> confusing subject, mostly because one persons S is anothers N, and so 
> many
> noise sources are lumped together. Some use full scale signal to RMS 
> noise,
> others use RMS Signal to RMS noise.  I see nothing but confusion when 
> this
> topic comes up.

If we're not using the same definitions then we have a problem.

>>
>> You are confusing bit depth with dynamic range. The two have 
>> absolutely
>> nothing to do with each other.
>>
> Says who?
>
> Again, definitions,  (I generally work with linear systems), when I 
> refer to
> how many bits are needed to record a signal,  I generally mean 
> recording the
> signal (without cliping), down to the noise floor (in a linear space). 
> Max
> code large enough to hold the signal, LSB significantly smaller than 
> RMS
> noise level (lower level somewhat vague, depends on the nature of the
> signals, and how linear you A/D is).

You're describing a specific situation which from below it looks we 
agree on. In general there's nothing that says it has to be this way. 
That's why I say they're not related unless you apply a bunch of 
definitions to it.

>> How many steps do you propose to be in the grayscale?
>>
> No steps, continuous from black to white, using a linear scale.  If you
> think of the A/D code as representing a fraction from 0 to 1, then make
> black 0 and white 1.  Now increase the number of bits from 1 to N, stop
> increasing N when the lsb is random for like samples, at the same 
> point on
> the ramp.  That is how many bits I would say the film has.  You can 
> argue
> how correlated the LSB needs  to be considered repeatable.

I'll buy this and I'll bet for a typical astrophoto it will be no more 
than 6 bits.

>> It seems like what you're saying here is that dynamic range implies
>> some bit depth. It does not. One bit can represent Dmax and Dmin,
>> 1=white, 0= black.
>
> It does if you don't want to add  noise by the digitizing process. The
> classic noise added by digitization is LSB level, squared/12 (again 
> linear
> signals, also assumes signal is random).  So you want there to be 
> enough
> bits in the digitization such that the noise level of the film is 
> digitized.

I still think you're going to be around 6 bits. For TP, the hyper fog + 
sky fog is often around 1.5d. The Log of 64 (levels) would give you a 
density range of 1.8 if linear. You add those together and you're at 
3.3 which would probably be around the shoulder of the TP curve. It 
makes some sense that 6 bits is enough to do it. Color film has the 
problem of low density. Few astronomical objects will push the density 
to the shoulder of the curve. Less density range implies fewer bits 
necessary to cover the range.

> Moreover, you want this noise level to come from the photons 
> themselves, not
> from your recording/digitizing of them.   (Note:  since the film 
> density is
> already a non-linear process, it is probably a bit silly to talk in 
> such
> terms.)

Perhaps, but we do digitize the things. ;-)

> Now if your negative is exposed as far as you can without foging the 
> film,
> the system noise is comming from the random variations of skyglow, LP 
> and
> the signal itself, and should reduce the number of bits needed to 
> digitize
> the film, but that does not mean that the inherent film can't record 
> more
> bits than this.

I've tried to clarify that I'm only talking about astrophotos and not 
daylight photos.

> Besides, Chuck, don't you use D19 on your negatives, which is almost a 
> 1/0
> developer anyway <g>.

I know you're joking but D19 just increases the slope of the 
characteristic curve (or contrast index as Kodak calls it) so it 
reaches Dmax faster with low light.

>>
>> The implication of what you said is that the S/N is different 
>> depending
>> on whether the data is stored as linear or log. Of course this is not
>> correct. The only difference between linear and log is that the
>> absolute amplitudes are compressed, i.e. the steps are smaller. The 
>> S/N
>> remains unchanged and so does the bit depth required to represent that
>> S/N.
>>
>
> S/N is not independant of bit depth unless you have enough bits that 
> the
> noise added by digitization is insignificant (compared to some other 
> system
> noise).  Thus without enough bits, the noise added by digitization it 
> will
> be noticed.

I think we said two completely different things here and I don't think 
either is wrong. ;-)

Chuck

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