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Re: I Miss my Viewfinder !

 
 
Whisky-dave
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Posts: n/a
 
      06-23-2011
On Jun 23, 1:30*am, Mxsmanic <(E-Mail Removed)> wrote:
> Whisky-dave writes:
> >http://www.sequoia.co.uk/components/...1&f=91&p=683&f...
> > SiI9127A HDMI 1.3 Receiver with Deep Colour outputs
> > 2-Port HDMI 1.3, HDCP 1.3, and DVI 1.0 compliant Receiver
> > Integrated TMDS core running at 25-225 MHz
> > 36-bit digital video interface supports video processors:
> > xvYCC to extended RGB
> > 36-bit RGB / YCbCr 4:4:4
> > 16/20/24-bit YCbCr 4:2:2
> > 8/10/12-bit YCbCr 4:2:2 (ITU BT.656)
> > Colour Space Conversion for both RGB-to- YCbCr

>
> So what is your point? Measure it. It's surprising how many high-end displays
> can't display simple test patterns correctly.
>
> > Well done, people also view pictures and watch TV don;t they.

>
> Yes. All analog.


So no one has a digital TV, or are you saying it's all just
advertising or marketing.


> > So don't use it. As they say **** in **** out.
> > if the digital part can';t be precise or accurate then neither can the
> > analogue
> > it's taken from so best to filter it out.

>
> So why record 48 bits of color?


How about for astronomy you view a star NOT just in visible light,
but there's infra red, ultra violet, gamma, X-rays.
This data isn't ignored just because the human eye or a monitor can't
see it.
All this is valuable information.


>
> > >There's no point in having 32-bit color if your camera
> > > can only resolve 16 bits.

>
> > it does make processing faster though, which is an advantage.

>
> Granted.


 
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nospam
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      06-23-2011
In article <(E-Mail Removed)>, Mxsmanic
<(E-Mail Removed)> wrote:

> > not when the sensor is capable of capturing it.

>
> No sensor is capturing 48 bits of data.


many sensors do. stop buying crappy cameras.

> > you do for compositing.

>
> Displays don't do compositing.


the operating system that puts the images on the display does.
 
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Whisky-dave
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      06-23-2011
On Jun 23, 3:25*pm, (E-Mail Removed) (Floyd L. Davidson) wrote:
> Whisky-dave <(E-Mail Removed)> wrote:


>
> >> You think so? *Are you drunk?

>
> >> >So if you switch from 5V to 0V or 0V to 5V *what is the rise time ?

>
> >> It depends on the circuit. *How much reactance, how much resistance....

>
> >yes as I said, it depends on more than just the switch.
> >Turning a switch off means the voltage will go from 5V to 0V
> >it won't have to go through every voltage from 5-0 unless something is
> >storing that voltage.

>
> And there is *always* something "storing that voltage".


Such as what...... and how is it relevant .
If a sensor captures the light from a scene and measures the
amount of light on it, you're saying it's wrong because that value
gets stored
so it can't update, because it has to go through all the voltages.

> It *necessarily* goes through every voltage, absolutely every time.
> It is *impossible* to do otherwise.


So you can't actually take a valid reading of the voltage then can
you.
Is that what you're saying.


>
> If you every take the time to find out what a "potential difference" is
> that will be perfectly clear. *Until you do...


yep, I know you don't



>
> >> It is not a "physical quantity", is is not a constant. *It depends
> >> of the rise time, and you have not provided the information required
> >> to calculate rise time.

>
> >So why don;t you provide it you're the one that thinks it is
> >important.

>
> I don't have that information.


What information do you need.
You have 5V switched to 0V or 0 to 5 up to you.
shown the calculation for rise time.

>
> >The rise time is a human invention a figure like bandwidth
> >a mathematical invention so people can refer to analogue things
> >in discrete terms that make sense.

>
> Are you still drunk? *That's nonsensical gibberish.


You do know what bandwidth is don;t you .
You do know about the -3dB points don;t you which actually defines
such things,
it's a measure of when the power falls by 50% or half .


>
> >> >When you insert a battery into a device does that device see a higher
> >> >voltage than
> >> >the voltage supplied by the battery, if so then how.

>
> >> It often does.

>
> >No it doesn't

>
> You've never looked at a battery lead with something like a spectrum
> analyzer! *It often does!


it doesn't .... Batteries are DC not AC
Where talking about the battery not the lead anyway.

>
> >When you put a battery in your camera the voltage doesn't increase
> >when you put the battery in or when you switch on.
> >otherwise it'd be better than perpetual motion.

>
> Nonsense. *The reactance of the leads within the camera can
> often lead to spikes that are significantly higher than the
> battery voltage.


Where do you get that idea from.
The physical properties are what make batteries the voltages they are.



>
> >> A battery is a very good filter, which means that it
> >> will absorb almost all of the high frequency components
> >> resulting from switching voltages on and off. *But it
> >> doesn't get all of them.

>
> >That is not the use of a switch, what high frequencies 'components'
> >do you get when you insert a battery into say a torch.
> >Where do these high frequencies come from the bulb !

>
> From the switching transients due to circuit reactance.
> Here, I've already explained it:


There's no circuit it's a switch.

of course if you add a capacitor or any other component then that
changes thins because it depends
on the component.


>
> >> (Yes, I'm aware that you are going to be totally confused
> >> by the above. *It's something you'd have to study in the
> >> first two or three days of any course on circuit analyzis.

>
> >Something that was done years ago.
> >Please cite what you believe then.
> >How much do you know about ohms law and kerchoffs ?
> >The sum of the voltages should all add up,
> >what you say is that 5V will become higher or is it lower if it's a
> >negative
> >overshoot, or is that a positive undershoot.

>
> Transient response, due to reactance.


The reactance of the switch or the meter ?



> >> >you take say a PP3 battery of 9 volts and you connect a paper clip to
> >> >the + terminal.
> >> >That paper clip will have a potential difference which can be measure
> >> >by using probes
> >> >on it and the -Ve terminal.
> >> > Have you never used one of those screwdrivers (AC) with a neon in it..

>
> >> The paper clip will also have both inductive reactance
> >> and capacitive reactance, as well as resistance.
> >> Therefore the voltage at the distant tip of the clip
> >> will not reach the peak as fast as the voltage at the
> >> end you connect to the battery.

>
> >So now you've dropped this down to speed,

>
> That is exactly what "rise time" is. *Slew rate... remember?


They are not the same.
Rise time is used mainly in digital circuitry
slew rate is used with AC signals which tell you how fast
an output can change for that component typically measured in volts
per microsecond.


>
> >which is slightly less than the speed of light
> >which is the charge propagation speed.

>
> >> Saying that you "have 5 volts and apply it to a piece of
> >> metal" is nonsense. *What is the voltage between your
> >> paper clip and the kitchen stove, or the waste basket in
> >> the bathroom, or the springs in your bed, or the the
> >> quarter in your pocket?

>
> >In your world it can be anything it can be higher than any battery,
> >higher than any AC supply can't it.
> >So why don't you measure it if you want to know.

>
> I have measured it thousands of times on a variety of circuits.


How and what have you measured it with.


> >> You think not... try switching your 5 volts at a 100GHz rate, and
> >> you'll find out that not much voltage ever even reaches the end of
> >> your paper clip! *Too much reactance...

>
> >What to you mean by 100GHz rate ?

>
> Exactly that.


Rate of what exactly is this a sine wave or a square wave what about
the m/s ratio
Are you talking about the peak voltage or the average or the RMS of
this 100GHz rate.

What is 100GHz rate are you sure rate isn't your word for frequency.
And will you measure this frequency, if the voltage goes through every
level
you won;t be able to tell the start from end point.



>
> >> >But where do theses harmonics come from we are talking DC here not AC..
> >> >If you're using a battery you won;t get harmonics.
> >> >DC hasn't harmonics only AC

>
> >> When you switch the voltage... bingo, you've got AC.

>
> >NO AC goes between + 0 and - * * *Plus zero/gnd and negative.
> >Also in case you've missed it AC is alternating current.
> >Going from 0 to + and from + to 0 is NOT AC, that is DC

>
> Sure, plus or minus... *compared to what?*


Compared to your reference point which is normally 0V or GND.

>*You don't seem to
> realize that switching between 5 volts and 0 volts is a 2.5 volt


No ) to 5 is a potential difference of 5V.

> AC signal with plus or mine voltages compared to 2.5.


That makes no sense even if mine=minus

> *It's also
> loaded with harmonics.


A pure sine wave does not contain harmonics.



>
> >> >WRONG.
> >> >DC does not contain harmonics that is the FACT.
> >> >http://en.wikipedia.org/wiki/Harmonic

>
> >> I hate to tell you, but there are virtually always transient
> >> spikes on a "DC" voltage bus. *

>
> >Not on a true DC bus.

>
> Look, don't be an idiot. *There is no such thing as "a true DC bus"#


yes I know.

> in that sense. *If nothing else it will have 50/60Hz induction from
> nearby power lines.


Batteries don't.

>
> >But of course now you're moving the goal post.

>
> Nope. *That's just the way it actually is.
>
> >Why are you claiming that a batteries voltage fluctuates up and down
> >as it sites there on the table, unused.

>
> Because it does.


In your dreams .
Batteries sitting on a bench being used the voltage does not increase
with time
it decreases.

>
> >>That is why there are always
> >> many many little bypass capacitors... to aborb them.

>
> >That';s because of other elements in the circuit.

>
> That's an hilarious statement. *You are clueless.


You haven't a clue have you.
De-coupling as it's called is needed due to either excessive demands
on power or
electromagnetic interference from other sources.



> >> >Are you getting slew rates confused with propagation delay.

>
> >> Interesting you bring that up. *In fact it is related to propagation
> >> delay. *Actually to something known as "group delay". *Or "phase delay".
> >> Some frequencies have a different propagation delay than others,

>
> >it's not the frequencies that dictate that though.
> >it's the amount of components that the signal passes through.

>
> It's the frequencies.


No, propagation delay is that delay that a signal has from the input
of the device to the output frequency is irrelevant.


> > >and
> >> the result is an attenuation of higher frequencies, which means the
> >> slew rate is lower...

>
> >No it's not.
> >What do you mean by lower is that quicker or slower ?

>
> The frequency is lower. *1 Hz is a lower frequency that 10 Hz.


Yes, so what about the slew rate of the device.

>
> >Are you saying a hight or low slew rate is better do you even know ?

>
> I'm not saying either. *It depends on the application.


fast or high slew rates are usually better unless you want a slow rate
of course which is unlikely .
Is 0.5V us better than 13V us ? (us= microseconds)


>



 
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Martin Brown
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      06-23-2011
On 23/06/2011 11:21, Floyd L. Davidson wrote:
> Martin Brown<|||newspam|||@nezumi.demon.co.uk> wrote:
>> On 23/06/2011 09:11, Floyd L. Davidson wrote:
>>> Mxsmanic<(E-Mail Removed)> wrote:
>>>> nospam writes:
>>>>
>>>>> not when the sensor is capable of capturing it.
>>>>
>>>> No sensor is capturing 48 bits of data.
>>>
>>> Virtually *all* modern "sensors" capture far in excess
>>> of 48 bits per pixel.


You are double counting here. There is only one pixels worth of real
data sampled for each sensor site - everything else is interpolated.
>>
>> This is a clueless gibberish interpretation of bits per pixel.

>
> I've spent several *decades* working with digital image
> data, and have an extremely good understanding of
> exactly how it works.


And then you go on to prove conclusively that you do not understand it.
>
>> A single photosite on a CCD at room temperature is dynamic range limited
>> by readout noise (~10e) at one end and full well overflow at the other
>> (~100,000e). The intrinsic dynamic range at the time of measurement is
>> typically 1:10000 or 13bits. Put a two stage Peltier cooler on it and
>> you can get nearly 16bits out of the better ones.

>
> You are discussing the dynamic range of luminance values.


Which is the limiting factor for determining colour values - all the
interpolation ever does is compute a decent guess at the colour value
for each of the missing colours at every site. The colour image result
is then 3 x 13 bits of significant data and usually worse than that. The
blue channel is usually somewhat noiser than the others.
>
> The discussion was about *color values*, not exclusively
> luminance.
>
>> Don't take my word for it!
>> Here is what one of the top scientific CCD sensor manufacturers has to
>> say on the subject of dynamic range and bits per pixel:
>>
>> http://learn.hamamatsu.com/articles/dynamicrange.html

>
> Nothing on that site contradicts what I described. It's
> all about grey scale luminance range.
>
>> The Bayer CCD matrix captures one colour at each photo site and
>> recombines them by interpolation to make an RGB image - allowing for the

>
> That is correct.
>
>> colour filtration and the subsampling the total information captured is
>> significantly *less* than with a monochrome CCD.

>
> That is not a valid statement. Color is encoded in the
> Bayer pattern, separate from the bit depth of the RAW
> sensor data, but it is still information that is
> captured, and contributes a rather large addition to the
> information from "a monochrome CCD". (If it wasn't captured,
> we wouldn't have color images...)


It is true that without the Bayer GRGB matrix you would not collect the
information needed for computing a colour image. But it is also true
that the coloured filters necessarily lose more than half the luminance
information to achieve that result. (A CMYG matrix does slightly better)
>
> The obvious fact that that is true is clear when you
> simply consider that when a RAW file (with color
> information encoded in the Bayer filter pattern) is
> converted to an RGB encoding the size of the data file
> is increased by generally 5 or 6 *times* the size of the
> RAW sensor data file.


But that is empty data. The orginal RAW sensor file at 13bits per pixel
is both necessary and sufficient to reconstruct a Bayer sampled image.
All that interpolating up to RGB24 or RGB48 does is make it possible to
display on suitable hardware. Some in camera JPEG implementations skip
the RGB stage and go straight to Y Cr Cb in a 2x1 subsampled colour.
>
>>> The Bayer pattern encodes the color separate from the 14
>>> bit-depth RAW sensor data data. When a typical raw
>>> converter interpolates the raw data it uses a 3x3
>>> or larger matrix to decode the color for each image
>>> pixel. Each matrix location is a 14-bit byte of data.

>>
>> And when you combine them you get another 14 bit number stored in a
>> 16bit register. There is no free lunch here.

>
> That is absolutely *not* true. When they are combined it is
> an integer the size of the machine's native integer. That is
> generally either a 32 or 64 bit integer.


You really do not have the first clue do you?

Manipulating large images goes memory bandwidth limited whenever you do
anything serious. No-one that understands what they are doing *ever*
makes the working image buffer larger than is absolutely necessary.

>>> Each *pixel* is therefore generated from 9 * 14 bits,
>>> or 126 bits *minimum*.

>>
>> That is a perverse and clueless way of looking at it.

>
> It is precisely *accurate*.


It is clueless and utterly misleading to boot.

The RGB pixel data is generated from a bunch of 9 13bit numbers but the
best that this could ever give in terms of additional signal to noise is
at most 1 additional bit. Conservatively you need at most 3 x 14 = 42
bits per pixel to store any digicam image in RGB format.

Amusing that the answer is 42. Deep thought would be pleased.

>>> On a 64 bit CPU the resulting calculations provide a 3
>>> channel RGB data set that is 3 * 64 bits, or as much as
>>> 192 bits (which is excess for a 3x3 matrix, but not when
>>> a 4x4 matrix with 224 bits per pixel is used).

>>
>> No decent 64 bit CPU software would ever extend image data to full width
>> 64bit. It is sometimes extended to 32bit reals which has enough mantissa
>> to hold a 16bit measurement for nonlinear transforms.

>
> The *only* way any decent 64 bit CPU would calculate the
> matrix is with a full 64 bit value.


The main use for 64 bit data paths is for SIMD instructions to work on
smaller chunks of data in parallel. The data is not promoted to 64bit
integer values wasting 48 bits for every 16bit word processed.

4x 16 bit words or 8x 8 bit bytes are processed simultaneously if a
64bit mode is used.
>
>>> On a 32 bit CPU the calculations generate a 3 channel
>>> RGB data set that is 3 * 32 bits, or 96 bits per pixel.

>>
>> Rubbish. Most packages still hold the image buffer as either 8bit byte
>> or 16bit word except for a handful of specialised ones that keep a
>> working copy in 32 bit floating point for more exotic calculations.

>
> Only *after* the calculation is complete, and only
> *then* is it moved to a 16 or 8 bit value.


You really do not have the first clue what you are talking about.
>
>>> That is only reduced to 48 bits per pixel when the data is
>>> transformed to something like a 16 bit per channel TIFF format.

>>
>> If you are looking at it on the screen it has to transform back to a
>> recognisable OS bitmap buffer from whatever internal storage is used.

>
> Viewing an image is one thing, interpolating raw sensor
> data is entirely different.


Interpolating RAW sensor data does not give you something for nothing.
>
>>> Clearly the point at which the image date is *reduced*
>>> to only 48 or fewer bits per pixel is when it is encoded
>>> into a standard image format. The sensor data has
>>> significantly more data.

>>
>> Rubbish. You are confusing and conflating the number of measurements
>> that go to make up one full colour pixel with bit depth.

>
> Except that *is* the way it is done.


You are completely wrong and have no understanding of the subject.

>> can gain a little bit of extra SNR by binning pixels 2x2 gives a 2x
>> improvement and binning NxN gives an Nx improvement but you have to
>> trade resolution for greater *real* bit depth in the image.

>
> That is irrelevant too.


Actually it is precisely relevant to the case of the green channel
interpolation to red or blue sites on a standard Bayer mask. It is a
sparse 2x2 binning by default unless the luminance edge discontinuity
heuristic is invoked.

Regards,
Martin Brown
 
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Wolfgang Weisselberg
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Posts: n/a
 
      06-23-2011
Mxsmanic <(E-Mail Removed)> wrote:
> Wolfgang Weisselberg writes:


>> Cheap TN dispays probably cannot. People who use them for colour
>> critical stuff are stupid.


> Even expensive displays generally cannot, and this is often visually obvious
> if you display the right test pattern.


That very much depends on the displays.
Especially as you use an absolute "expensive" whereas I use a
relative "cheap".

>> CRTs can (they are analog and thus can display infinite colours),
>> better flatscreen panels can, etc.


> CRTs can do better (in theory, sometimes in practice), but they are limited,
> too.


Analog systems always beat digital system. --Mxsmanic


>> [...] Which is why your "most displays cannot" is a singulary stupid
>> argument.


>> But he can tell the difference between 00..001111 and 00..0010000,
>> especially when it's digitally boosted.


> Which is really the difference between 0.01 and 0.00.


Really? By which broken math do you make 16 and 15 into 0.01 and 0.00?

> Which is one part in
> four, hardly worth crowing about.


one part in four is 0.25.

BTW, the difference between $0.01 and $0.00 per kilobyte transmitted
is a slight bit more than "hardly worth crowing about".


What have you done to the real Mxsmanic?

-Wolfgang
 
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nospam
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Posts: n/a
 
      06-23-2011
In article <(E-Mail Removed)>, Mxsmanic
<(E-Mail Removed)> wrote:

> > Virtually *all* modern "sensors" capture far in excess
> > of 48 bits per pixel.

>
> No, none of them are doing that. As I've said, that's an accuracy of one part
> in about 280 trillion, and there is no ADC anywhere that can come close to
> that kind of resolution. So most of the bits are wasted, since they are
> essentially random.


high end sensors have 16 bit a/d converters, for each of red, green and
blue, for a total of 48 bits per pixel.

most cameras have 12 or 14 bit a/d converters, so it's 36 or 42 bits
total.
 
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nospam
Guest
Posts: n/a
 
      06-23-2011
In article <(E-Mail Removed)>, Mxsmanic
<(E-Mail Removed)> wrote:

> > many sensors do. stop buying crappy cameras.

>
> Even helium-cooled sensors do not capture 48 bits of data.


well, that sucks, since you don't need helium cooled sensors to obtain
48 bit results.

if i spent the money for a helium cooled sensor, i'd want it to be much
better than an off the shelf dslr.

hopefully those who were considering buying a helium cooled sensor will
reevaluate their purchase decisions, now that you've brought it to
their attention.

> > the operating system that puts the images on the display does.

>
> The operating system is not a display device.


a display device has nothing to display unless there's an operating
system sending data to it. windows 7 and os x have compositing window
managers.
 
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James Silverton
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Posts: n/a
 
      06-23-2011
On 6/23/2011 2:23 PM, Wolfgang Weisselberg wrote:
> Mxsmanic<(E-Mail Removed)> wrote:
>> Wolfgang Weisselberg writes:

>
>>> Cheap TN dispays probably cannot. People who use them for colour
>>> critical stuff are stupid.

>
>> Even expensive displays generally cannot, and this is often visually obvious
>> if you display the right test pattern.

>
> That very much depends on the displays.
> Especially as you use an absolute "expensive" whereas I use a
> relative "cheap".
>
>>> CRTs can (they are analog and thus can display infinite colours),
>>> better flatscreen panels can, etc.

>
>> CRTs can do better (in theory, sometimes in practice), but they are limited,
>> too.

>
> Analog systems always beat digital system. --Mxsmanic
>
>
>>> [...] Which is why your "most displays cannot" is a singulary stupid
>>> argument.

>
>>> But he can tell the difference between 00..001111 and 00..0010000,
>>> especially when it's digitally boosted.

>
>> Which is really the difference between 0.01 and 0.00.

>
> Really? By which broken math do you make 16 and 15 into 0.01 and 0.00?
>
>> Which is one part in
>> four, hardly worth crowing about.

>
> one part in four is 0.25.
>
> BTW, the difference between $0.01 and $0.00 per kilobyte transmitted
> is a slight bit more than "hardly worth crowing about".
>

It's maybe not on topic for slr but I wonder what the effect is going to
be of the new Lytro focus fixing camera? You won't need autofocus and a
wide aperture, uniform but low quality lens might allow taking pictures
in dim light.
--


James Silverton, Potomac

I'm *not* http://www.velocityreviews.com/forums/(E-Mail Removed)
 
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Martin Brown
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Posts: n/a
 
      06-24-2011
On 23/06/2011 21:56, nospam wrote:
> In article<(E-Mail Removed) >, Mxsmanic
> <(E-Mail Removed)> wrote:
>
>>> many sensors do. stop buying crappy cameras.

>>
>> Even helium-cooled sensors do not capture 48 bits of data.

>
> well, that sucks, since you don't need helium cooled sensors to obtain
> 48 bit results.


True enough. A double stage Peltier cooled CCD will do it at about -50C
or so. But you do not find these in consumer digicams it takes far too
much power. Astronomical cameras routinely operate at 16bits and the
high end ones have fractional least significant bit readout noise.

The CCD thermal noise floor at room temperature corresponds to a dynamic
range at the point of capture of 10,000:1 or 13bits per channel.
Anything beyond 14bits ADC is for bragging rights only.

It is amazing just how clueless some people here are!

You have captured one colour at every pixel on the CCD at 13bit depth.
You can estimate what the other colours were from surrounding pixels
using Bayer demosaic. The best that could ever give you in the green
channel is one more significant bit from noise averaging over 4 cells.
The resulting colour image has at most only 40bits of equivalent data
content per pixel when demosaiced (and can be exactly represented albeit
somewhat inconveniently by 13bits of data per photosite).

Not all demosaics are created equal - some of them are positively flaky
so you really do not have anything like the ideal result at pixel level.
Practical example of their shortcomings for high contrast black text on
a white ground is demonstrated at sixbats with various decoders:

http://sixbats.com/RawCompare.html

It isn't hard to see the various systematic errors introduced by the
Bayer sampling under these tricky circumstances.

Where has this daft idea come from that consumer digicams can produce
meaningful 48bit results?

They might produce raw images that are best handled as 16bit data in a
computer but there are still only 13bits of useful data in each channel.

Regards,
Martin Brown
 
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Martin Brown
Guest
Posts: n/a
 
      06-24-2011
On 23/06/2011 21:55, nospam wrote:
> In article<(E-Mail Removed) >, Mxsmanic
> <(E-Mail Removed)> wrote:
>
>>> Virtually *all* modern "sensors" capture far in excess
>>> of 48 bits per pixel.

>>
>> No, none of them are doing that. As I've said, that's an accuracy of one part
>> in about 280 trillion, and there is no ADC anywhere that can come close to
>> that kind of resolution. So most of the bits are wasted, since they are
>> essentially random.

>
> high end sensors have 16 bit a/d converters, for each of red, green and
> blue, for a total of 48 bits per pixel.


Unless they are actively cooled they are capturing 3 bits of thermal
random noise in the least significant bits. And all that does is
increase file sizes by about 20% without any benefit to the user.

> most cameras have 12 or 14 bit a/d converters, so it's 36 or 42 bits
> total.


Every pixel measures just one colour channel. You have around 13 bits of
independent data for each pixel on the sensor. When it is processed then
you end up with a conventional RGB16 image that needs to be stored in 3x
16bit integers to accommodate the full linear dynamic range.

The total evidence you have for this image is still the original dataset
of NxM 13bit values and it can be compressed on that basis.

Foveon sensors really do measure all three colours at every pixel but
they have significant other failings so unless you really spend your
time photographing test charts in complementary colours their notional
advantage over Bayer sampled images is largely illusory.

You never get something for nothing. What you measured is all you have.

Regards,
Martin Brown
 
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