dynamic range and A/D conversion

David J Taylor schrieb:
> Johannes,
>
> If the digitising is only eight bits linear, then a lot of dynamic range
> will be lost. As one f/stop is half the light, full scale on 8 bits is
> 256, 1 stop down is 128 and so forth, 2 stops is 64, so the total range is
> only 8 stops. There is no tonal contrast in the lowest level, it is
> either on or off, so the dynamic range may be considered as well less than
> 8 stops *depending how you define dynamic range).
>
> With the JPEG images, a non-linear encoding is used so that the digital
> value is more nearly the square-root of the light level: 1, sqrt (0.5),
> sqrt (0.25), i.e. 255, 181, 128 etc., so that there is a greater dynamic
> range possible, at the expense of a less accurate representation of a
> particular light level (in the highlights).
> Does that help?


Dear David,

As I´m more deeply involved into analog photography than into digital
I define dynamic range as the range of f-stops between the brightest
and the darkest area that still shows perceptable details. As far ist
this definition is concerned I fully understand your explanation: There
is no visible detail in the lowest level and in the highest level
(which are white and black) - that´s a true loss of valuable
information - and so the dynamic range is remarkable decreased after 8
bit conversion. That helped a lot!

Johannes
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Johannes wrote:
> Hi,
>
> Can anybody explain to me if there is a connection between the dynamic
> range of a digital camera/CCD-/CMOS-Sensor and the bit width of the
> analog-to-digital converter - does a 12 bit converter lead to a larger
> dynamic range than a 8 bit converter - and if so why?
>
> Thanks a lot for your input!
> Johannes

You might find these articles and their references interesting reading:

http://www.clarkvision.com/imagedetail/dynamicrange/index.html

http://www.clarkvision.com/imagedetail/digital.signal.to.noise/

David
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Johannes wrote:
> David J Taylor schrieb:
>
>> You might find these articles and their references interesting
>> reading:
>> http://www.clarkvision.com/imagedetail/dynamicrange/index.html
>> http://www.clarkvision.com/imagedetail/digital.signal.to.noise/
>> David
>
> Thanks for the reference but my english is not good enough for this
> kind of technical matter. My question is, to be more precise, if I
> loose dynamic range when the information containing, say, 12 stops of
> range is digitized by a converter that uses only 8 bit.
>
> Best regards!
> Johannes

Johannes,

If the digitising is only eight bits linear, then a lot of dynamic range
will be lost. As one f/stop is half the light, full scale on 8 bits is
256, 1 stop down is 128 and so forth, 2 stops is 64, so the total range is
only 8 stops. There is no tonal contrast in the lowest level, it is
either on or off, so the dynamic range may be considered as well less than
8 stops *depending how you define dynamic range).

With the JPEG images, a non-linear encoding is used so that the digital
value is more nearly the square-root of the light level: 1, sqrt (0.5),
sqrt (0.25), i.e. 255, 181, 128 etc., so that there is a greater dynamic
range possible, at the expense of a less accurate representation of a
particular light level (in the highlights).

Does that help?

Cheers,
David
 >> Stay informed about: dynamic range and A/D conversion 

David J Taylor wrote:
> Johannes wrote:
>> David J Taylor schrieb:
>>
>>> You might find these articles and their references interesting
>>> reading:
>>> http://www.clarkvision.com/imagedetail/dynamicrange/index.html
>>> http://www.clarkvision.com/imagedetail/digital.signal.to.noise/
>>> David
>>
>> Thanks for the reference but my english is not good enough for this
>> kind of technical matter. My question is, to be more precise, if I
>> loose dynamic range when the information containing, say, 12 stops of
>> range is digitized by a converter that uses only 8 bit.
>>
>> Best regards!
>> Johannes
>
> Johannes,
>
> If the digitising is only eight bits linear, then a lot of dynamic
> range will be lost. As one f/stop is half the light, full scale on 8
> bits is 256, 1 stop down is 128 and so forth, 2 stops is 64, so the
> total range is only 8 stops. There is no tonal contrast in the
> lowest level, it is either on or off, so the dynamic range may be
> considered as well less than 8 stops *depending how you define
> dynamic range).
> With the JPEG images, a non-linear encoding is used so that the
> digital value is more nearly the square-root of the light level: 1,
> sqrt (0.5), sqrt (0.25), i.e. 255, 181, 128 etc., so that there is a
> greater dynamic range possible, at the expense of a less accurate
> representation of a particular light level (in the highlights).
>
> Does that help?
>
> Cheers,
> David

I should also have added that the typical ADC is 12-bit accuracy, followed
by the non-linear encoding to 8-bit JPEG.

David
 >> Stay informed about: dynamic range and A/D conversion 

Johannes wrote:
[]
> Dear David,
>
> As I´m more deeply involved into analog photography than into digital
> I define dynamic range as the range of f-stops between the brightest
> and the darkest area that still shows perceptable details. As far ist
> this definition is concerned I fully understand your explanation:
> There is no visible detail in the lowest level and in the highest
> level (which are white and black) - that´s a true loss of valuable
> information - and so the dynamic range is remarkable decreased after 8
> bit conversion. That helped a lot!
>
> Johannes

Johannes,

I'm glad that helped. Certainly 8-bit linear conversion would produce an
image with notably degraded dynamic range, but 8-bit gamma 2.2 corrected
encoded data (such as found in JPEGs or analog/digital/analog TV) can
produce greyscale images which exceed the limits of the human eye (in
terms of dynamic range, under typical display conditions). Once you get
into colour, there is some evidence that 24-bit RGB images can be seen to
limit the displayed colour differences, but these effects are quite
subtle.

For processing, of course, you may want 12-bit raw data if you are going
to alter the levels appreciably. The very best DSLRs are now showing that
12-bit ADCs are not quite enough, and 14 bits would be required to capture
the full range available from the sensor.

Cheers,
David
 >> Stay informed about: dynamic range and A/D conversion 

Johannes wrote:
> Hi,
>
> Can anybody explain to me if there is a connection between the dynamic
> range of a digital camera/CCD-/CMOS-Sensor and the bit width of the
> analog-to-digital converter - does a 12 bit converter lead to a larger
> dynamic range than a 8 bit converter - and if so why?
>
> Thanks a lot for your input!
> Johannes

There are many factors that in total determine dynamic range. Bit
width is ONE of them. The quantization (number of bits) sets a hard
limit- the dynamic range cannot be greater than the range allowed by
the quantization steps of the number of bits.

However, signal-to-noise-ratio is another big factor. If the S/N is
less than appropriate for the quantization steps, then you will merely
be digitizing the noise in greater detail

The S/N, on the other hand, depends on the chip, the quality of the
preamp electronics and the quality of the A/D chip. It will also vary
with exposure.
 >> Stay informed about: dynamic range and A/D conversion 

Johannes wrote:
> David J Taylor schrieb:
snip
>
> Dear David,
>
> As I´m more deeply involved into analog photography than into digital
> I define dynamic range as the range of f-stops between the brightest
> and the darkest area that still shows perceptable details. As far ist
> this definition is concerned I fully understand your explanation: There
> is no visible detail in the lowest level and in the highest level
> (which are white and black) - that´s a true loss of valuable
> information - and so the dynamic range is remarkable decreased after 8
> bit conversion. That helped a lot!
>
> Johannes

There is another type of dynamic range that is sometimes called the
small-signal dynamic range. Sometimes there is a big floor to the
black level, which would seem to be a limit on DR. However, if the
noise is low, that floor can be subtracted out. What is more
bothersome is the noise in medium exposures- variations in level in
medium intensity. So one definition is the ratio of that noise in the
middle of the exposure range to the numerical value of the maximum
possible signal.
 >> Stay informed about: dynamic range and A/D conversion 

David J Taylor wrote:
> Johannes,
>
> If the digitising is only eight bits linear, then a lot of dynamic range
> will be lost. As one f/stop is half the light, full scale on 8 bits is
> 256, 1 stop down is 128 and so forth, 2 stops is 64, so the total range is
> only 8 stops. There is no tonal contrast in the lowest level, it is
> either on or off, so the dynamic range may be considered as well less than
> 8 stops *depending how you define dynamic range).
>
> With the JPEG images, a non-linear encoding is used so that the digital
> value is more nearly the square-root of the light level: 1, sqrt (0.5),
> sqrt (0.25), i.e. 255, 181, 128 etc., so that there is a greater dynamic
> range possible, at the expense of a less accurate representation of a
> particular light level (in the highlights).

Even if all the encoders were linear, surely there's a simple
interplay between dynamic range and accuracy?

A single bit (potential values 0 and 1!) can encode
a massive dynamic range (e.g. 10 f-stops worth).
It just won't have much accuracy within that range

The same logic applies to 8 bit versus 12 bit.

There's no reason they shouldn't have the same dynamic range,
with different accuracy.

Of course, one could use 8bit to have the same accuracy
as 12 bit, but a smaller range.

Or any compromise between these 2 extremes.

(non-linear encoding improves perceievd accuracy,
since (IIRC) the eye is a log (ish) device,
but I believe this is a separable issue)

BugBear
 >> Stay informed about: dynamic range and A/D conversion 

bugbear wrote:
[]
> Even if all the encoders were linear, surely there's a simple
> interplay between dynamic range and accuracy?
>
> A single bit (potential values 0 and 1!) can encode
> a massive dynamic range (e.g. 10 f-stops worth).
> It just won't have much accuracy within that range
>
> The same logic applies to 8 bit versus 12 bit.
>
> There's no reason they shouldn't have the same dynamic range,
> with different accuracy.
>
> Of course, one could use 8bit to have the same accuracy
> as 12 bit, but a smaller range.
>
> Or any compromise between these 2 extremes.
>
> (non-linear encoding improves perceievd accuracy,
> since (IIRC) the eye is a log (ish) device,
> but I believe this is a separable issue)
>
> BugBear

Yes, if you talk about different encoding choices, there are many
different possibilities. That's why I suggested that you needed a
definition of dynamic range to discuss this in a quantitative way.

I don't think that you should ignore any of the components of the vision
chain, in particular the characteristics of the display device and the
human eye, when choosing an optimum encoding for a given number of bits on
the storage or transmission channel.

Cheers,
David
 >> Stay informed about: dynamic range and A/D conversion 

David J Taylor wrote:

> Johannes,
> If the digitising is only eight bits linear, then a lot of dynamic range
> will be lost. As one f/stop is half the light, full scale on 8 bits is
> 256, 1 stop down is 128 and so forth, 2 stops is 64, so the total range is
> only 8 stops. There is no tonal contrast in the lowest level, it is
> either on or off, so the dynamic range may be considered as well less than
> 8 stops *depending how you define dynamic range).

Well, that´s kind of interesting.
I always thought the tonal values stay the same after a/d conversion
and they´re only spread out on a smaller strap. Isn´t it the case
that, when doing a/d conversion with 8 bit or whatever bit width, the
lowest level holds the darkest tonal level that the image contains and
the highest level the brightest and that only the sampling is
closer/smaller? What you imply is a lowest level as total black and a
highest as total white. Is that really so?

> With the JPEG images, a non-linear encoding is used so that the digital
> value is more nearly the square-root of the light level: 1, sqrt (0.5),
> sqrt (0.25), i.e. 255, 181, 128 etc., so that there is a greater dynamic
> range possible, at the expense of a less accurate representation of a
> particular light level (in the highlights).

Maybe I haven´t recognized any loss due to that gamma conversion.

Greetings!
Mr.Adams
 >> Stay informed about: dynamic range and A/D conversion 

mr.adams.TakeThisOut@gmx.net wrote:
> David J Taylor wrote:
>
>> Johannes,
>> If the digitising is only eight bits linear, then a lot of dynamic
>> range will be lost. As one f/stop is half the light, full scale on
>> 8 bits is 256, 1 stop down is 128 and so forth, 2 stops is 64, so
>> the total range is only 8 stops. There is no tonal contrast in the
>> lowest level, it is either on or off, so the dynamic range may be
>> considered as well less than 8 stops *depending how you define
>> dynamic range).
>
> Well, that´s kind of interesting.
> I always thought the tonal values stay the same after a/d conversion
> and they´re only spread out on a smaller strap. Isn´t it the case
> that, when doing a/d conversion with 8 bit or whatever bit width, the
> lowest level holds the darkest tonal level that the image contains and
> the highest level the brightest and that only the sampling is
> closer/smaller? What you imply is a lowest level as total black and a
> highest as total white. Is that really so?
>
>> With the JPEG images, a non-linear encoding is used so that the
>> digital value is more nearly the square-root of the light level: 1,
>> sqrt (0.5), sqrt (0.25), i.e. 255, 181, 128 etc., so that there is a
>> greater dynamic range possible, at the expense of a less accurate
>> representation of a particular light level (in the highlights).
>
> Maybe I haven´t recognized any loss due to that gamma conversion.
>
> Greetings!
> Mr.Adams

The ADC should linearly convert light levels from the sensor to digital
values, yes, and the accuracy depends on the number of bits. However
general photographic usage talks in terms of "stops" which are not a
linear scale, but a logarithmic one (corresponding more closely to how the
eye responds to light). Alternatively, you could look on it as a
percantage accuracy. With a linear conversion, the quantisation step
expressed as a fraction of the signal in a 256-level signal is 1/256 at
the white end, but "100%" at the black end (1/1). With a gamma-corrected
JPEG, the percentage steps are more equal across the whole range.

Does that help?

David
 >> Stay informed about: dynamic range and A/D conversion 

David J Taylor schrieb:
> The ADC should linearly convert light levels from the sensor to digital
> values, yes, and the accuracy depends on the number of bits. However
> general photographic usage talks in terms of "stops" which are not a
> linear scale, but a logarithmic one (corresponding more closely to how the
> eye responds to light). Alternatively, you could look on it as a
> percantage accuracy. With a linear conversion, the quantisation step
> expressed as a fraction of the signal in a 256-level signal is 1/256 at
> the white end, but "100%" at the black end (1/1). With a gamma-corrected
> JPEG, the percentage steps are more equal across the whole range.
> Does that help?

Not really because I still do not see why it should compress the
dynamic range in the picture. In a 256-level signal the spot at 1/256
represents the brightest part of the image and 1/1 the darkest. The
logarithmic range in between can well be 10 stops, it´s only that I
have just 8 different values to represent them. But: The spot at 1/256
is still 10 stops brighter than at 1/1 or am I on the totally wrong
track here?

Greetings!
Tony Adams
 >> Stay informed about: dynamic range and A/D conversion 

mr.adams.RemoveThis@gmx.net wrote:
>David J Taylor schrieb:
>> The ADC should linearly convert light levels from the sensor to digital
>> values, yes, and the accuracy depends on the number of bits. However
>> general photographic usage talks in terms of "stops" which are not a
>> linear scale, but a logarithmic one (corresponding more closely to how the
>> eye responds to light). Alternatively, you could look on it as a
>> percantage accuracy. With a linear conversion, the quantisation step
>> expressed as a fraction of the signal in a 256-level signal is 1/256 at
>> the white end, but "100%" at the black end (1/1). With a gamma-corrected
>> JPEG, the percentage steps are more equal across the whole range.
>> Does that help?
>
>Not really because I still do not see why it should compress the
>dynamic range in the picture. In a 256-level signal the spot at 1/256
>represents the brightest part of the image and 1/1 the darkest. The
>logarithmic range in between can well be 10 stops, it´s only that I
>have just 8 different values to represent them. But: The spot at 1/256
>is still 10 stops brighter than at 1/1 or am I on the totally wrong
>track here?

With a "linear" conversion, the number of bits determines the
ratio of the maximum level to the minimum level that can be
recorded, and that is of course the dynamic range.

Your statement is true for what is called a "non-linear" A-D
conversion. The 8-bit representation used by JPEG is an example
of exactly that. The non-linear transfer function determines
the "dynamic range".

A significant point to remember though, is that "dynamic range"
does not necessarily equate to the same thing as "useful dynamic
range". Hence while it is true that JPEG technically can record
a greater dynamic range in 8 bits that a linear 12 bit file can,
there is a very significant difference between them, and the
_useful_ dynamic range of the 12 bit linear file is greater.

Here is a chart showing the "normalized" values for 12 bit linear
data:

| Number of | |
| Levels | |
Fstop | 12 bit | Normalized Pixel |
Range | Linear | value |
------|------------|------------------|
1 | 2048 | 1.0 |
2 | 1024 | 0.5 |
3 | 512 | 0.25 |
4 | 256 | 0.125 |
5 | 128 | 0.0625 |
6 | 64 | 0.03125 |
7 | 32 | 0.015625 |
8 | 16 | 0.007812 |
9 | 8 | 0.003906 |
10 | 4 | 0.001953 |
11 | 2 | 0.0009765 |
12 | 1 | 0.0004883 |


The brightest level would be 1.0, represented by a binary value
of 1111 1111 1111. Half that brightness would be 0.5,
represented by a binary value of 0111 1111 1111. One quarter of
the maximum brightness, 0.25, would be represented by 0011 1111
1111.

The range above 0.5 all the way to 1.0 is divided into 0111 1111
1111 equal parts, which in decimal is 2048 levels. Hence the
accuracy with which a brightness value in the highest f/stop can
be recorded is extremely good. (Probably more than 10 or 20
times better than the eye can distinquish, and hence using 2048
levels to represent only a 1 f/stop range is a real waste.)

Down at the 9th f/stop, the range is divided into 0000 0000 0111
values (decimal , which is not particularly very good accuracy.

As it happens, 8 levels in an entire f/stop is just about the
minimum that is useful. Fewer than that and the image will
appear to be "posterized".

Hence even though a 12 bit file technically has a dynamic range
of 12 f/stops, the actualy useful dynamic range is about 9
f/stops.

However, with JPEG the increments are not linear (and there is
no waste of 2048 values just for 1 f/stop), and instead a "gamma
correction" curve is applied (basically that is a lookup table
which quantizes a larger number of values into a smaller number,
in a non-linear way). Here is the same chart as above, with
some extra columns showing JPEG data too:

+------- 12 bit linear -------+ +----------- 8 bit JPEG ------------+
/ \/ \
| No. Levels | Normalized Pixel | 8 bit | No. Levels |
Fstop | 12 bit | value | Gam.Cor. | 8 Bit JPEG |
Range | Linear | Linear * Gam.Cor. | Value | Gamma Corr |
------|------------|-------------*--------------|----------|------------|
1 | 2048 | 1.0 * 1.0 | 255 | 69 |
2 | 1024 | 0.5 * 0.72974 | 186 | 50 |
3 | 512 | 0.25 * 0.53252 | 136 | 37 |
4 | 256 | 0.125 * 0.38860 | 99 | 27 |
5 | 128 | 0.0625 * 0.28358 | 72 | 20 |
6 | 64 | 0.03125 * 0.20694 | 53 | 14 |
7 | 32 | 0.015625 * 0.15101 | 38 | 10 |
8 | 16 | 0.007812 * 0.11020 | 28 | 8 |
9 | 8 | 0.003906 * 0.08042 | 21 | 6 |
10 | 4 | 0.001953 * 0.05868 | 15 | 4 |
11 | 2 | 0.0009765 * 0.04282 | 11 | 3 |
12 | 1 | 0.0004883 * 0.03125 | 8 | 2 |
13 | | 0.0002415 * 0.02269 | 6 | 2 |
14 | | 0.0001207 * 0.01656 | 4 | 1 |
15 | | 0.0000604 * 0.01208 | 3 | 1 |
16 | | 0.0000302 * 0.00882 | 2 | 0 |
17 | | 0.0000151 * 0.06434 | 2 | 1 |
18 | | 0.0000075 * 0.00470 | 1 | 1 |
19 | | 0.0000038 * 0.00343 | 1 | 0 |

Some want to claim that the above means that 8 bit gamma
corrected JPEG data can represent a dynamic range of 19 f/stops!
Technically that appears to be true, but obviously none of the
f/stops below the 8th one are of any usefulness, since those all
have fewer than 8 values per fstop.

For all practical purposes, the lowest 15 values in a 12 bit
linear file are of no significance and the lowest 21 values in a
JPEG file are equally useless. A 12-bit linear file can retain
a 9 f/stop dynamic range. A JPEG gamma correctd file can retain
8 f/stops of dynamic range.

--
Floyd L. Davidson <http://www.apaflo.com/floyd_davidson>
Ukpeagvik (Barrow, Alaska) floyd.RemoveThis@apaflo.com
 >> Stay informed about: dynamic range and A/D conversion