Finally got to the point where no new camera holds my interest(waiting for specific offering)

In article , Eric Stevens
wrote:

I will give it one more try.

stop digging yourself a deeper hole.

All that the ADC is doing is digitizing the level of charge which has
accumulated in the sensor cells. It does not need to know, nor would
it care, how much exposure to light that represents. Hook it up to a
sensor and a 14-bit ADC will give you a 14-bit datastream which will
tell you all about the charge and nothing about the anount of light
which has fallen on the individual cells. As far as the ADC is
concerned, the important thing is whether or not the maximum voltage
of the charges to which it is exposed is within the ADC's input range.
If it is the ADC will digitize it. How much light it required does not
matter. The ADC will as happily digitize 2 levels of charge as it will
20,000 levels of charge. No matter what it digitizes it will output a
14-bit data stream. THat's why it won't care whether it is digitizing
a collection of charges with a dynamic range of 5 or 50. All will be
OK as long as the peak is within the range of the ADC's input.

still not getting it.

On Sun, 13 Jan 2019 10:05:17 -0500, Alan Browne
wrote:

On 2019-01-12 20:18, Eric Stevens wrote:
On Sat, 12 Jan 2019 09:26:28 -0500, Alan Browne
wrote:

On 2019-01-11 18:20, Eric Stevens wrote:
On Fri, 11 Jan 2019 12:51:28 -0500, Alan Browne
wrote:

On 2019-01-11 10:28, nospam wrote:
In article , Eric Stevens
wrote:


The problem is clearly DXO's testing methods. No matter how you look
at this, you have to be able to imagine all kinds of sources of
inaccurate measurements, especially if they are slight. I have to
agree with nospam and Alan. You can't get DR outside of the limits of
the ADC because that is the output you see, but you can certainly get
test results outside of that limit.

But the digital DR of the output of the ADC is not the same as the
analog DR of the sensor. Nor is there any reason why it should be.

nobody said it was, however, it's always going to be limited by the adc.


Got that Eric?

What is 'it'? The DR of the sensor or the DR of the output of the ADC?

Obvious. The ADC is the limiting factor. Always. There is NO WAY for
DxO to probe the sensor directly (and it would be meaningless to
everyone even if they could...)

The ADC won't be the limiting factor if it has a better dynamic range
than the sensor.

The limiting factor is most likely the sensor and the engineers select
the ADC appropriately. They could have sampled at 16 bits - but that
would just be sampling more noise in the bottom 2+ bits. I suppose if
you needed an entropy source that's as good as any if you use a couple
dozen sensor sites ...

But the ADC is digitizing the analog output of one sensor site at a
time. If there is noise it will not become apparent until after the
digital value of the ADC outputs are collated.

I think at this point I should stop and ask you whereabouts in the
pipeline the sensor DR range should be measured. As perhaps an extreme
example, do you think it should be measured by what is output to a
memory card? ... or should it be closer to the sensor? If so, where
(and how)?

Where delivered to the end user: the stored image file (raw). That is
the only thing that counts and is useful to the user of the product.

Nothing else matters.

But the raw file is not the end product. It has to be decoded by the
raw decoder and at that point whatever DR is inherent in the initial
image will be restored.

And if, as you propose, some compression has occurred between the ADC
and the stored value, then the hocus pocus has simply exchanged 1 form
of noise for another.

No, I'm saying some form of compression occurs in the act of
digitizing.

To the engineer deciding on what bit depth to use, intimate knowledge
about the sensors would have him (in the design/systems engineering
phase) do tests on the sensors to determine its statistics and
appropriately decide on the bit depth.

Agreed. There is no point in using 14 bits on a sensor with a dynamic
range of (say) 5.
--

Regards,

Eric Stevens

In article , Eric Stevens
wrote:

But the digital DR of the output of the ADC is not the same as the
analog DR of the sensor. Nor is there any reason why it should be.

nobody said it was, however, it's always going to be limited by the
adc.

Got that Eric?

What is 'it'? The DR of the sensor or the DR of the output of the ADC?

Obvious. The ADC is the limiting factor. Always. There is NO WAY for
DxO to probe the sensor directly (and it would be meaningless to
everyone even if they could...)

The ADC won't be the limiting factor if it has a better dynamic range
than the sensor.

The limiting factor is most likely the sensor and the engineers select
the ADC appropriately. They could have sampled at 16 bits - but that
would just be sampling more noise in the bottom 2+ bits. I suppose if
you needed an entropy source that's as good as any if you use a couple
dozen sensor sites ...

But the ADC is digitizing the analog output of one sensor site at a
time. If there is noise it will not become apparent until after the
digital value of the ADC outputs are collated.

false.

I think at this point I should stop and ask you whereabouts in the
pipeline the sensor DR range should be measured. As perhaps an extreme
example, do you think it should be measured by what is output to a
memory card? ... or should it be closer to the sensor? If so, where
(and how)?

Where delivered to the end user: the stored image file (raw). That is
the only thing that counts and is useful to the user of the product.

Nothing else matters.

But the raw file is not the end product.

it is as far as the camera is concerned, and more importantly, many
photographers.

measuring dynamic range from a jpeg is not a good idea (although some
review sites have done that).

It has to be decoded by the
raw decoder and at that point whatever DR is inherent in the initial
image will be restored.

no.

And if, as you propose, some compression has occurred between the ADC
and the stored value, then the hocus pocus has simply exchanged 1 form
of noise for another.

No, I'm saying some form of compression occurs in the act of
digitizing.

that would be incorrect.

To the engineer deciding on what bit depth to use, intimate knowledge
about the sensors would have him (in the design/systems engineering
phase) do tests on the sensors to determine its statistics and
appropriately decide on the bit depth.

Agreed. There is no point in using 14 bits on a sensor with a dynamic
range of (say) 5.

that part is true.

On Sun, 13 Jan 2019 10:06:13 -0500, Alan Browne
wrote:

On 2019-01-12 21:15, Eric Stevens wrote:
On Sat, 12 Jan 2019 09:25:23 -0500, Alan Browne
wrote:

On 2019-01-11 18:18, Eric Stevens wrote:

The recorded output of the ADC is limited by the capabilities of the
ADC. But these have no effect on the capabilities of the sensor. If

If there is no way to encode the information, then that is the mootest
of moot points.

That may well be but, as I have several times said, it is possible to
scale the dynamic range of the sensor to fit the narrower dynamic
range of the ADC.

To which I've replied numberous times. In a nutshell, you're trading
one form of noise for another.

The 'scaling' is done during the conversion of analog to digital in
the ADC and involves no more noise than is inherent in any analog to
digital conversion.
--

Regards,

Eric Stevens

On Sun, 13 Jan 2019 10:13:45 -0500, Alan Browne
wrote:

Yep. An interesting point: according to DxO, when the test uses a
paper target, some of the noise may actually be the texture of the
paper of the target.

That's not noise. To claim it's noise absurd. It's part of the image
being recorded.

Agreed, but DxO appear to suspect that some testers are making that
error.
--

Regards,

Eric Stevens

In article , Eric Stevens
wrote:

The recorded output of the ADC is limited by the capabilities of the
ADC. But these have no effect on the capabilities of the sensor. If

If there is no way to encode the information, then that is the mootest
of moot points.

That may well be but, as I have several times said, it is possible to
scale the dynamic range of the sensor to fit the narrower dynamic
range of the ADC.

To which I've replied numberous times. In a nutshell, you're trading
one form of noise for another.

The 'scaling' is done during the conversion of analog to digital in
the ADC and involves no more noise than is inherent in any analog to
digital conversion.

no. it's linear. no scaling.

On Sun, 13 Jan 2019 10:15:15 -0500, Alan Browne
wrote:

On 2019-01-12 21:51, Eric Stevens wrote:
On Sat, 12 Jan 2019 12:11:32 -0500, Alan Browne
wrote:

On 2019-01-12 11:22, nospam wrote:
In article , Eric Stevens
wrote:

Now if I said, for example f/11 you would understand it. But that is
not an EV.

as i said, you don't understand it.

here's a hint:
what's the difference between f/8 and f/11 ?

a) f/3
b) 1 stop
c) 1 EV
d) a and b.
e) b and c.

C'mon _E_ric!
I'm rooting for you!

Try (8*f - 11*f)/88 or (-3*f)/88

The correct reply was e).

But thanks for playing.

Seriously this time: b) is always correct. c) is also correct if the
lens is mounted in a camera with (presumably) a shutter speed and ISO
set. But, without a shutter speed and ISO there can be no EV.
--

Regards,

Eric Stevens

On Sun, 13 Jan 2019 10:25:18 -0500, Alan Browne
wrote:

On 2019-01-12 22:39, Eric Stevens wrote:

AS you say, they (probably) have to take it from an image file.
Although they could tether the camera.


Tethering the camera is just getting the image file w/o storing it on
the camera's memory card.

I'm done. Really. This is pointless. You have no knowledge of
electrical engineering from what I can see. I'm not an engineer (but
worked for many and many worked for me) and what I and nospam have been
pointing out to you is pretty basic systems engineering reality.

I may need to produce those diagrams yet. :-(

TANSTAAFL. Look that up. It really embodies the entire thing.

snipped the rest because it's just blather that doesn't address the
issue at all. Icing to disguise a bad cake, really
--

Regards,

Eric Stevens

In article , Eric Stevens
wrote:


AS you say, they (probably) have to take it from an image file.
Although they could tether the camera.


Tethering the camera is just getting the image file w/o storing it on
the camera's memory card.

I'm done. Really. This is pointless. You have no knowledge of
electrical engineering from what I can see. I'm not an engineer (but
worked for many and many worked for me) and what I and nospam have been
pointing out to you is pretty basic systems engineering reality.

I may need to produce those diagrams yet. :-(

for entertainment purposes, by all means do so.

for educational purposes, do not.

In article , Eric Stevens
wrote:

Now if I said, for example f/11 you would understand it. But that is
not an EV.

as i said, you don't understand it.

here's a hint:
what's the difference between f/8 and f/11 ?

a) f/3
b) 1 stop
c) 1 EV
d) a and b.
e) b and c.

C'mon _E_ric!
I'm rooting for you!

Try (8*f - 11*f)/88 or (-3*f)/88

The correct reply was e).

But thanks for playing.

Seriously this time: b) is always correct. c) is also correct if the
lens is mounted in a camera with (presumably) a shutter speed and ISO
set. But, without a shutter speed and ISO there can be no EV.

false.