Luminous Landscape Forum
Equipment & Techniques => Digital Cameras & Shooting Techniques => Topic started by: Henry Goh on November 29, 2007, 01:01:54 pm
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What does it take to make 16-bit DSLRs? How much more would it cost? What are the restrictions if any?
Most people do not need to go beyond 10/12 Mp captures but they do need/want better renditions of tones and gradations. I think I would love to have a 16-bit Nikon D3xx. I'm not technically familiar as to what needs to go into the making of such a camera. Anyone better equipped can share their knowledge?
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Most people do not need to go beyond 10/12 Mp captures but they do need/want better renditions of tones and gradations. I think I would love to have a 16-bit Nikon D3xx
Please explain, how you would utilize the 16-bit range, or, turning it around, when - in which situations - do you experience a lack of tones/graduations, with particular attention to the dynamic range of the envisioned camera.
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I wouldn't say there's any problem with the tones and graduations on current cameras. Dynamic range can always be improved, and without such an improvement, a move to 16bit is not justified.
I've not seen an MF cameras properly tested for DR, but I highly doubt them capable of true 16bit performance.
Graeme
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Some brief comments:
- Dynamic Range: 16-bit RAW files by itlsef won't improve the dynamic range of a camera. DR is today limited by noise in the shadows, so if there is no noise improvement at the same time as upgrading from 12 to 14 or 16 bits, there will be no dynamic range expansion at all.
- Tonal quality (gradations): sometimes we forget most camera sensors have a Bayer pattern distribution, where 2/3 (i.e. 66.6%) of levels of a developed 12 bit RAW file, are interpolated straight in the 16 bit range. That doesn't mean they have more precision than the 12 bit samples (since they are interpolated from them) but it means they will take many more values in the huge 0..65535 range. This will make an upgrade from 12 or 14 to 16 bit RAW much les noticeable in terms of softer gradations.
- Photoshop is 15 bit, so if you feel happy using Photoshop and buy a 16 bit camera in the future, you must think PS will steal one bit from your image so you will be only 1 bit ahead of the recentrly introduced 14 bit cameras.
- Everytime more and more, storing RAW files is becoming a big problem. And since an increasing number of Mpx is the common trend, file sizes will get larger and larger. To store 16 bit images instead of 12 or 14 bit, with no apparent advantage in terms of image quality will not be a good business.
After saying all that, I am sure there will be 16-bit DSLR cameras in the near future
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Some brief comments:
- Dynamic Range: 16-bit RAW files by itlsef won't improve the dynamic range of a camera. DR is today limited by noise in the shadows, so if there is no noise improvement at the same time as upgrading from 12 to 14 or 16 bits, there will be no dynamic range expansion at all.
If you say that, lots of people around here will argue that. I will not, I totally agree. The more I read, the more this gets reinforced in my mind (speaking of which, this is discussed as we both believe in a new book called The HDRI Handbook by Christian Bloch. Just started it, but its quite good and in the first chapter, he backs up this point).
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- Photoshop is 15 bit, so if you feel happy using Photoshop and buy a 16 bit camera in the future, you must think PS will steal one bit from your image so you will be only 1 bit ahead of the recentrly introduced 14 bit cameras.
[a href=\"index.php?act=findpost&pid=157071\"][{POST_SNAPBACK}][/a]
Actually, it's 15 full bits plus one level, which is important (and why the engineers did it) relative to processing and so there would be an exact middle level.
I've seen and heard of scanners and cameras "claiming full 16 bit" and I think that's bogus. They may have something north of 14 bit but I think it would depend on what one would consider noise and signal. 15 bits of signal is more than plenty...
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What does it take to make 16-bit DSLRs? How much more would it cost? What are the restrictions if any?
[a href=\"index.php?act=findpost&pid=157007\"][{POST_SNAPBACK}][/a]
Henry - are you asking about whether 16-bit dynamic range from the sensor could be achieved, aka up to 16 stop range? This would require the ability to read out very low photon counts from the deep shadows , adequately distinguished from read-out noise, from a well in a pixel on the sensor (conceptually a photon bucket) with a capacity of 2^15 above that - that's a very big bucket, and the sensor manufacturers don't seem to be there yet. Some read-out noise can be decreased by cooling the sensor (astro cameras do that) and well size can be increased by making bigger pixels. I'm sure the practical aspects of sensor fabrication/design to improve DR are carefully guarded and proprietary !
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What does it take to make 16-bit DSLRs? How much more would it cost? What are the restrictions if any?
Most people do not need to go beyond 10/12 Mp captures but they do need/want better renditions of tones and gradations. I think I would love to have a 16-bit Nikon D3xx. I'm not technically familiar as to what needs to go into the making of such a camera. Anyone better equipped can share their knowledge?
[a href=\"index.php?act=findpost&pid=157007\"][{POST_SNAPBACK}][/a]
A DSLR with 16 useful bits would be wonderful. The problem seems to be that manufacturers don't seem to have any way of reading out the full range of the sensor (IOW, at the lowest ISO) without introducing a significant amount of electrical noise, so anything more than 12 bits is recording mostly noise at this point in technology. Some readout technologies, like those used by Canon DSLRs and some of the newer Nikons can read out a lower range of sensor charges (IOW, high ISO) with less noise, relative to absolute signal. The quality of 12-bit ISO 1600 RAW data from these cameras would do excellent as the lowest 12 bits of a 16-bit ISO 100; they have as little as about 15% of the noise added in the readout/digitization process at ISO 100, relative to absolute signal.
Canon and Nikon have just raised their bit depths to 14, and if you take these RAWs and replace the lowest two bits (which can be either '00', '01', '10', or '11')with '10', and perform torture tests on the data, and they don't seem to be any the worse for it. The introduction of 14 bits is simply coinciding with other improvements in image quality, and gets undeserved credit, IMO.
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If you say that, lots of people around here will argue that. I will not, I totally agree. The more I read, the more this gets reinforced in my mind (speaking of which, this is discussed as we both believe in a new book called The HDRI Handbook by Christian Bloch. Just started it, but its quite good and in the first chapter, he backs up this point).
[a href=\"index.php?act=findpost&pid=157073\"][{POST_SNAPBACK}][/a]
Why would someone argue with this?
The definition of DR depends very much on the way noise in the shadows is measured right?
Bit depth has nothing to do with DR. You can capture an amazing B&W 50 stops perfect gradient with an 8 bits sampling, the transitions will just not be smooth if you have an image with more than 255 pixels. You will get banding.
Cheers,
Bernard
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Why would someone argue with this?
http://luminous-landscape.com/forum/index....27&hl=staircase (http://luminous-landscape.com/forum/index.php?showtopic=19927&hl=staircase)
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Why would someone argue with this?
The definition of DR depends very much on the way noise in the shadows is measured right?
Bit depth has nothing to do with DR. You can capture an amazing B&W 50 stops perfect gradient with an 8 bits sampling, the transitions will just not be smooth if you have an image with more than 255 pixels. You will get banding.
Cheers,
Bernard
[{POST_SNAPBACK}][/a] (http://index.php?act=findpost&pid=157105\")
Bernard,
With linear integer encoding, dynamic range is limited by the bit depth as shown by the [a href=\"http://www.normankoren.com/digital_tonality.html]table[/url] on Norman Koren's web site. The table shows that 10 bits can encode only 10 stops, and that allows only one level in the darkest f/stop. You would really need more to avoid banding.
The trouble with linear integer encoding is that it devotes more bits than needed to the brighter stops than to the darker stops. Gamma encoding at 2.2 improves the distribution somewhat.
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Tests of the human eye's ability to distinct between shades show, that one can distinguish between two shades if their lightness differes at least 1% (I guess this is a rounded value.)
Consequently, we can distinguish between at least 70 levels within one stop dynamic range (2^70 = 2.00...). In order to eliminate visible transitions, the number of levels needs to be higher. I don't know how high and I am cautios to accept anything, which is not based on tests of a wide range; I remember to have been told, that 24 frames per second are all needed, for we don't "see" anything above it.
Anyway, let's accept this 70 as an initial value. Select the one stop wide range of the lowest lightness, which needs "full service", i.e. all levels. From here above every further stop requires twice as many levels because of the linearity of the sensor.
Accordingly, 10 stop "clean" dynamic range requires 70*2^9 = 35840 levels. Add to this the number of levels, which are required for the "dirty" stops (there is not much point to distinguish between 70 levels of a noisy stop).
This range requires over 15 bits, and the dynamic range is still not very large.
However, what does one do with such an image file?
The consideration, how many bits Photoshop can handle it totally irrelevant. Not only, because Photoshop is not the center of the universe, but because the software will support much greater bit depth as soon as it becomes actual.
But how does one present such image? I don't know what is on the horizon of printer development, but the monitors are not far away from reproducing not 12 but 16 stops. Some consumer grade monitors offer already contrast ratio 3000:1 (ok, that's a claim,. but I guess that is at least 2000:1), and new development promises tens of thousands of contrast ratio. These monitors (even those with 3000:1) are not for computer work but video (the resolution is not high enough for example for image processing), but that too is only a question of demand - and anyway, why could not images be presented on TV monitors?
The new JPEG standard will handle such images, so we only need cameras with higher dynamic range.
As the dynamic range is not limited by the noise in shadow but by the well capacity (the former is the consequence of the latter), we only need less pixels on the same sensor with the newest technology. Unfortunately, misguided consumer considerations force manufacturers to cram more and more pixels on the sensor.
Any time I hear people talking about digital cameras, the first question is always "how many megapixels". We should start an action to disspell the myth, that the measure of camera's quality is the number of pixels.
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Some brief comments:
- Dynamic Range: 16-bit RAW files by itlsef won't improve the dynamic range of a camera. DR is today limited by noise in the shadows, so if there is no noise improvement at the same time as upgrading from 12 to 14 or 16 bits, there will be no dynamic range expansion at all.
[{POST_SNAPBACK}][/a] (http://index.php?act=findpost&pid=157071\")
Data are starting to dribble out on the performance of the new Canon and Nikon sensors with 14 bit ADCs.
[a href=\"http://theory.uchicago.edu/~ejm/pix/20d/posts/tests/D300_40D_tests/]Emil Martinec[/url], professor of physics at the University of Chicago, has published data for the Nikon D300 and Canon 40D. He uses the engineering definition of DR (dynamic range = raw saturation level divided by read noise). At base ISO both cameras have a DR of about 11.3 stops, and gain about 1/3 of a stop when going from 12 bit to 14 bit mode. It would appear that 12 bits would be sufficient for these cameras.
EJ Martin, also at the U of C, reported results for the Nikon D3 (http://forums.dpreview.com/forums/read.asp?forum=1021&message=25627719). He found a DR of 11.7 f/stops. Since the D3 pixel size is considerably larger than that of the D300 or 40D, one would expect a better DR with the D3 if read noise is held to the same level, since DR = full well capacity/read noise.
These gains in DR are rather modest when compared with Roger Clark's prediction of a 2 stop gain for a properly implemented 14 bit camera as compared to 12 bits.
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Thanks guys. You make such a wonderful resource center
So at the end of the day, will I get an image with better tonal range if they made the camera I'm asking for here? Is it coming anytime soon?
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Tests of the human eye's ability to distinct between shades show, that one can distinguish between two shades if their lightness differes at least 1% (I guess this is a rounded value.)
Consequently, we can distinguish between at least 70 levels within one stop dynamic range (2^70 = 2.00...). In order to eliminate visible transitions, the number of levels needs to be higher.
Anyway, let's accept this 70 as an initial value. Select the one stop wide range of the lowest lightness, which needs "full service", i.e. all levels. From here above every further stop requires twice as many levels because of the linearity of the sensor.
Accordingly, 10 stop "clean" dynamic range requires 70*2^9 = 35840 levels. Add to this the number of levels, which are required for the "dirty" stops (there is not much point to distinguish between 70 levels of a noisy stop).
This range requires over 15 bits, and the dynamic range is still not very large.
[{POST_SNAPBACK}][/a] (http://index.php?act=findpost&pid=157126\")
The above line of reasoning is similar to that reported by [a href=\"http://www.normankoren.com/digital_tonality.html]Norman Koren[/url]; however, he states that fewer levels are needed in the darker zones, where the eye is less sensitive. Norman bases his calculations on the Weber-Fechner law, which dates to the mid 19th century, and has been replaced by the Steven's law (http://www.neuro.uu.se/fysiologi/gu/nbb/lectures/WebFech.html). It would be interesting to repeat the calculations using the newer law.
However, what does one do with such an image file?
The consideration, how many bits Photoshop can handle it totally irrelevant. Not only, because Photoshop is not the center of the universe, but because the software will support much greater bit depth as soon as it becomes actual.
But how does one present such image? I don't know what is on the horizon of printer development, but the monitors are not far away from reproducing not 12 but 16 stops. Some consumer grade monitors offer already contrast ratio 3000:1 (ok, that's a claim,. but I guess that is at least 2000:1), and new development promises tens of thousands of contrast ratio. These monitors (even those with 3000:1) are not for computer work but video (the resolution is not high enough for example for image processing), but that too is only a question of demand - and anyway, why could not images be presented on TV monitors?
[{POST_SNAPBACK}][/a] (http://index.php?act=findpost&pid=157126\")
[a href=\"http://www.dolby.com/promo/hdr/technology.html]BrightSide[/url] produces a monitor with a claimed 200,000:1 contrast ratio (17.6 f/stops) at a resolution of 1,920 x 1,080 on a 37 inch screen.
Photoshop can handle such images using 32 bit floating point representation (HDR). However, rendering such a contrast ratio into a printer space is problematic, and the full range of Photoshop tools are not available with 32 bit floating point representation.
Bill
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BrightSide produces a monitor with a claimed 200,000:1 contrast ratio (17.6 f/stops) at a resolution of 1,920 x 1,080 on a 37 inch screen
With the high Canadian dollar, this a consumer grade monitor (only US$45,000); however, only 1080pix high sucks, so I rather stick to the Samsung 24" 244T, contrast ratio 1000:1 (claimed).
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I've seen and heard of scanners and cameras "claiming full 16 bit" and I think that's bogus. They may have something north of 14 bit but I think it would depend on what one would consider noise and signal. 15 bits of signal is more than plenty...
[{POST_SNAPBACK}][/a] (http://index.php?act=findpost&pid=157077\")
[a href=\"http://www.dalsa.com/sensors/products/productdetails.asp?productID=FTF5066c]Dalsa[/url] and Kodak (http://www.kodak.com/US/en/dpq/site/SENSORS/name/KAF-39000_product/show/KAF-39000_productSpecifications) both publish the DR of their sensors used in medium format backs, and the typical DR is 70-72 db, or about 12 f/stops. Even though they may have 16 bit ADCs, the least significant bits most likely consist largely of noise, so 15 bits should be more than adequate, as you say. I have no data on scanners or scanning backs.
Bill
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Bernard,
With linear integer encoding, dynamic range is limited by the bit depth as shown by the table (http://www.normankoren.com/digital_tonality.html) on Norman Koren's web site. The table shows that 10 bits can encode only 10 stops, and that allows only one level in the darkest f/stop. You would really need more to avoid banding.
[a href=\"index.php?act=findpost&pid=157114\"][{POST_SNAPBACK}][/a]
Yes, of course, but this only has to do wit banding.
I am not saying that you can cover a 50 stops DR with 8 bits for practical applications, but my example is just about showing that there is no relationship between DR and bit depth.
Cheers,
Bernard
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Yes, of course, but this only has to do wit banding.
I am not saying that you can cover a 50 stops DR with 8 bits for practical applications, but my example is just about showing that there is no relationship between DR and bit depth.
Cheers,
Bernard
[a href=\"index.php?act=findpost&pid=157151\"][{POST_SNAPBACK}][/a]
Bernard,
No, you can not cover 50 stops with 8 bit linear integer coding at all. The first stop would have 128 levels, the 2nd 64, the 3rd 32... down to the 8th where you would have one level. You can't go any darker because you've run out of values. The DR of 8 bit linear encoding is 8 f/stops. If you use gamma 2.2 encoding, then you can do better. Did you look at Norman's table?
Regards
Bill
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there is no relationship between DR and bit depth
Sometimes I state in a discussions, that a digital image does not have any dynamic range; some people are outraged to hear that.
you can not cover 50 stops with 8 bit linear integer coding at all. The first stop would have 128 levels, the 2nd 64, the 3rd 32... down to the 8th where you would have one level. You can't go any darker because you've run out of values. The DR of 8 bit linear encoding is 8 f/stops
You can cover a range of 50 stops even with one bit. Of course some stops will not be distinguishable from others (a kind of "posterization"), but so what? The stop is an artificial unit. What is the "natural meaning" of a stop in conjunction with scanned data or with any non-photographic digital image?
There is an uncountable number of intensities within any given stop, which are not distinguishable from each other. The "stop" is only a point on that scale.
For example a monitor's dynamic range will not be measured in stops but in the ratio of intensity between the brightest and darkest levels. The contrast ratio of 1000:1 does not imply any number of required levels.
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Perhaps it would be helpful to distinguish between linearly-encoded sensor image data (which does have a fairly fixed relationship between bit depth and maximum encodable DR) and other image encoding schemes with non-linear tone curves (HDR, ProPhoto, sRGB, etc), where the DR limit is flexible and limited only by the number of levels one wishes to devote per stop. Such encoding schemes do not have a fixed limit to the DR they can contain.
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Why would someone argue with this?
[a href=\"index.php?act=findpost&pid=157105\"][{POST_SNAPBACK}][/a]
See, what did I tell you? <G>
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With the high Canadian dollar, this a consumer grade monitor (only US$45,000); however, only 1080pix high sucks, so I rather stick to the Samsung 24" 244T, contrast ratio 1000:1 (claimed).
[a href=\"index.php?act=findpost&pid=157140\"][{POST_SNAPBACK}][/a]
Well considering that IF your goal is imaging to print, and the best you'll get (again up to debate) is maybe 350:1 contrast ratio on paper, what's the big deal about a 1000:1 contrast ratio? You watching movies? Well displays designed for that use produce serious issues for those of us working with still imagery. The contrast ratio is about the least useful spec for displays I can think of. And the ambient light where the display resides makes a pretty huge difference. Keep the room dim, you don't need a huge range here.
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There is an uncountable number of intensities within any given stop, which are not distinguishable from each other. The "stop" is only a point on that scale.
For example a monitor's dynamic range will not be measured in stops but in the ratio of intensity between the brightest and darkest levels. The contrast ratio of 1000:1 does not imply any number of required levels.
[a href=\"index.php?act=findpost&pid=157163\"][{POST_SNAPBACK}][/a]
With linear integer encoding and a limited bit depth, there are a finite number of intensities within any given f/stop, and your statement of an infinite number is incorrect.
The DR limitations we have been discussing for digital sensors are for linear capture, gamma = 1. The monitor space may have a gamma of 2.2 and coding is more efficient in the shadows. Still a certain number of bits are needed for that 1000:1 ratio, and 8 bits may not be sufficient. A contrast ration of 1000:1 is 3 orders of magnitude (log base 10). It could also be expressed in f/stops (log base 2), and 1000:1 is 9.97 f/stops.
Bill
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Data are starting to dribble out on the performance of the new Canon and Nikon sensors with 14 bit ADCs.
Emil Martinec (http://theory.uchicago.edu/~ejm/pix/20d/posts/tests/D300_40D_tests/), professor of physics at the University of Chicago, has published data for the Nikon D300 and Canon 40D. He uses the engineering definition of DR (dynamic range = raw saturation level divided by read noise). At base ISO both cameras have a DR of about 11.3 stops, and gain about 1/3 of a stop when going from 12 bit to 14 bit mode. It would appear that 12 bits would be sufficient for these cameras.
EJ Martin, also at the U of C, reported results for the Nikon D3 (http://forums.dpreview.com/forums/read.asp?forum=1021&message=25627719). He found a DR of 11.7 f/stops. Since the D3 pixel size is considerably larger than that of the D300 or 40D, one would expect a better DR with the D3 if read noise is held to the same level, since DR = full well capacity/read noise.
These gains in DR are rather modest when compared with Roger Clark's prediction of a 2 stop gain for a properly implemented 14 bit camera as compared to 12 bits.
[a href=\"index.php?act=findpost&pid=157129\"][{POST_SNAPBACK}][/a]
I read some of Roger Clark's site the other day and I don't think he was claiming that moving from a 12 bit ADC to a 14 bit ADC would give 2 extra stops of dynamic range. He did suggest that the lower read noise possible with a 14bit (or 16 bit) device could benefit some current camera's, e.g. canon 5D and presumably Nikon D3 etc, giving a higher achievable dynamic range, but not a full 2 stops higher. Part of the issue is that the DR is limited by full well capcity and the noise floor, which the ADC contributes to. hence a 'quieter ADC' should improve the lowest readable levels.
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(...) He uses the engineering definition of DR (dynamic range = raw saturation level divided by read noise). At base ISO both cameras have a DR of about 11.3 stops, and gain about 1/3 of a stop when going from 12 bit to 14 bit mode. It would appear that 12 bits would be sufficient for these cameras.
Bill, it's nice to check in objective figures what one suspected.
Moreover, I would say that this DR figure in terms of Sat_Level/Read_Noise (11.7 f-stops) is too optimistic when compared to what it's usable in real photography.
I shot a high dynamic range scene with my Canon EOS 350D at ISO100 to make a subjective but valuable calculation of the DR:
(http://www.guillermoluijk.com/article/digitalp02/altorango.jpg)
(http://www.guillermoluijk.com/article/digitalp02/altorangohis.gif)
The histogram shows the scene contained information along around 12 f-stops.
I developed it linear and plotted its distribution in f-stops:
(http://www.guillermoluijk.com/article/digitalp02/zonas.jpg)
Looking at the texture detail in each f-stop:
(http://www.guillermoluijk.com/article/digitalp02/zonasecuencia.jpg)
Beyond f-stop -7EV there is nothing but noise grain. We can conclude that 8 f-stops (0EV to -7EV) of usable dynamic range would be already quite optimistic for this camera. By repeating this test on the new 14-bit Mark III I don't think at all that bit depth will reveal as being determinant in defining the new achievable dynamic range; IMO still noise will set the low end of the DR and probably the influence of the 2 extra bits will be almost negligible for DR. Allowing 1 f-stop or 1.5 f-stops of improvement in noise for the new sensors (which is not a joke), we would estimate the final USABLE DR in around 9 or 9.5 f-stops, still recordable using a 12-bit linear A/D. The improvement would thus be thanks to noise optimisation, not to higher bit depth.
Panopeeper, bit depth doesn't guarantee achievable DR, but sets a physical limit to it if linear RAW encoding is used (and this is the case of most sensors, not Leica's M8 for instance). To have N bits of bit depth is not a sufficient condition, but a necessary condition to capture N f-stops of DR. And that assuming that the lower f-stops would be poorly represented, so actually you will need more than N bits for properly recording N f-stops of DR (remember we are talking about capture, not about image post process encoding).
A 12-bit RAW has the following levels for each f-stop:
0EV: 2048 levels, 2048..4095
-1EV: 1024 levels, 1024..2047
-2EV: 512 levels, 512..1023
-3EV: 256 levels, 256..511
-4EV: 128 levels, 128..255
-5EV: 64 levels, 64..127
-6EV: 32 levels, 32..63
-7EV: 16 levels, 16..31
-8EV: 8 levels, 8..15
-9EV: 4 levels, 4..7
-10EV: 2 levels, 2..3
-11EV: 1 level, 1
As you can see, the lowest f-stops have a strong lack of tonal variety. There is a direct link between a maximum achievable DR and the minimum number of bits required for it.
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Bill, it's nice to check in objective figures what one suspected.
Moreover, I would say that this DR figure in terms of Sat_Level/Read_Noise (11.7 f-stops) is too optimistic when compared to what it's usable in real photography.
As you can see, the lowest f-stops have a strong lack of tonal variety. There is a direct link between a maximum achievable DR and the minimum number of bits required for it.
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Guillermo,
I used different methodology in [a href=\"http://luminous-landscape.com/forum/index.php?showtopic=21300&view=findpost&p=157253]another thread[/url] but arrived at similar conclusions. The engineering definition of DR is not that useful for routine photography and over estimates practical DR.
Bill
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What does it take to make 16-bit DSLRs? How much more would it cost? What are the restrictions if any?
Most people do not need to go beyond 10/12 Mp captures but they do need/want better renditions of tones and gradations. I think I would love to have a 16-bit Nikon D3xx. I'm not technically familiar as to what needs to go into the making of such a camera. Anyone better equipped can share their knowledge?
[a href=\"index.php?act=findpost&pid=157007\"][{POST_SNAPBACK}][/a]
Going back to your original question (not whether it is worth doing or not). You need a 16-bit analogue to digital converter. There are multiple design issues that make increasing the number of bits difficult and expensive. From memory these tend to be:
1/ Number of conversions that can be achieved per second - each time a new pixel is read from the array then it takes time for the signal to settle at the input to the Analogue to Digital Converter. This limits either the maximum number of pixels in the sensor array, the maximum number of frames per second or requires more ADCs in parallel which pushes up cost.
2/ The ADC needs a stable reference voltage or current against which to compare the incoming signal so that you produce the same digital output for the same analogue input. The more bits in the conversion the greater accuracy you need for the reference. NB this is a battery driven device, the reference needs to be constant irrespective of the actual battery voltage.
3/ Noise in the system becomes a limiting factor (even if there was a noise free source/sensor driving the ADC) there is still noise generated by the reference and also within the circuitry of the ADC itself. Eliminating this noise becomes harder the more bits of precision and the faster the conversion rate.
As with most things there already exist 16-bit plus ADCs and ADCs operating at higher frequencies. The problem is getting the combination of bits resolution, performance and acceptable price all at the same time.
[From a personal perspective I would welcome additional bit depth to provide greater colour accuracy in the darker parts of the image. For some of my images it is noticeable that there is a lack of colour information following post processing.]
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With linear integer encoding and a limited bit depth, there are a finite number of intensities within any given f/stop, and your statement of an infinite number is incorrect
1. The concept of dynamic range has nothing to do with encoding.
2. As long as the individual photons can not be reliably counted, lightness remains an analog phenomenon for our purpose. As such, there is no fixed number of levels, there is no fixed spacing, there are no fixed locations of the levels.
3. Dynamic range is an arbitrary measurement of relative intensities. Even if one uses this measurement, it does not pose any requirements regarding the spacing and location of levels.
The DR limitations we have been discussing for digital sensors are for linear capture, gamma = 1
The starting point was Bernard's
I am not saying that you can cover a 50 stops DR with 8 bits for practical applications
This did not relate to sensors, etc. but to DR per se.
Still a certain number of bits are needed for that 1000:1 ratio, and 8 bits may not be sufficient
Not sufficient for what? For preventing posterisation? What about applications, which don't use continuous colors, like drawings, artificial coloring, like in astrophysics, etc?
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bit depth doesn't guarantee achievable DR, but sets a physical limit to it if linear RAW encoding is used (and this is the case of most sensors, not Leica's M8 for instance). To have N bits of bit depth is not a sufficient condition, but a necessary condition to capture N f-stops of DR
You too are mixing up the concept of dynamic range with its measurement.
The meaning of "DR of ten stops" is, that the highest intensity is 1024 times higher than the lowest measurable intensity. It means nothing else. The unit of "f-stop' is not godgiven, it is arbitrary. You could measure it in decibel, or in 5%, or you could make uo your own measurement.
There is no basis to regard the measurement in stops as a basic characteristic of DR. What you are doing is something like
I measure the length in metri units, therefor the length of everything has to be expressable in meters or centimeters
However, if I measured the length in imperial unites, everything looks differently.
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1. The concept of dynamic range has nothing to do with encoding.
You too are mixing up the concept of dynamic range with its measurement.
The meaning of "DR of ten stops" is, that the highest intensity is 1024 times higher than the lowest measurable intensity. It means nothing else. The unit of "f-stop' is not godgiven, it is arbitrary. You could measure it in decibel, or in 5%, or you could make uo your own measurement.
There is no basis to regard the measurement in stops as a basic characteristic of DR. What you are doing is something like
I measure the length in metri units, therefor the length of everything has to be expressable in meters or centimeters
However, if I measured the length in imperial unites, everything looks differently.
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I have to say, Panopeeper makes total sense to me. It seems again that those who keep going back to some specific articles that link dynamic range and bit depth are making the wrong assumption based on encoding. I'll keep an open mind but so far, everything seems to point to my original beliefs about the two being separate, and Panopeeper expresses why this is so. Dynamic range and bit depth are two separate spec's.
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I have to say, Panopeeper makes total sense to me. It seems again that those who keep going back to some specific articles that link dynamic range and bit depth are making the wrong assumption based on encoding. I'll keep an open mind but so far, everything seems to point to my original beliefs about the two being separate, and Panopeeper expresses why this is so. Dynamic range and bit depth are two separate spec's.
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They are only separated by analog noise. In the absence of analog noise, bit depth would strictly define possible DR. 16 linear bits would have *exactly* 4x the dynamic range of 14 linear bits, etc, regardless of what the standard was for minimum usable signal, as long as the standard was consistent.
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Going back to your original question (not whether it is worth doing or not). You need a 16-bit analogue to digital converter. There are multiple design issues that make increasing the number of bits difficult and expensive. From memory these tend to be:
Thank you. That looks like the technical limitations I was seeking to learn.
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They are only separated by analog noise. In the absence of analog noise, bit depth would strictly define possible DR. 16 linear bits would have *exactly* 4x the dynamic range of 14 linear bits, etc, regardless of what the standard was for minimum usable signal, as long as the standard was consistent.
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I'm having trouble conceptualising the difference between these factors associated with DR. I can't think of anything clearer than the staircase analogy.
DR can be represented by the total height of a staircase. The higher the staircase, the greater the dynamic range.
However, the number of steps in that staircase will be determined by bit depth. 16 bit linear equates to a huge number of tiny steps. 8 bit linear equates to a relatively small number of very large steps. The height of the staircase (the dynamic range) can be the same in both cases and will be determined by sensor full-well capacity minus signal-obscuring noise. The ground is the noise and the first step in the staircase is the darkest, usable signal value above the noise floor.
Any flaws in that analogy?
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The Leica DMR back for the R8 and R9 is a 16 bit camera. It has excellent dynamic range and gradation and can retain much more shadow detail than a 12- or 14-bit camera. Those who have used both the DMR and 5D (for example) in RAW mode overwhelmingly prefer the DMR's image quality.
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In the absence of analog noise, bit depth would strictly define possible DR. 16 linear bits would have *exactly* 4x the dynamic range of 14 linear bits, etc
With linear encoding, the proportions between bit depths reflect the proportions between the dynamic ranges, like in the above example. However, this does not define the magnitude of the DR; that depends on the scale.
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With linear encoding, the proportions between bit depths reflect the proportions between the dynamic ranges, like in the above example. However, this does not define the magnitude of the DR; that depends on the scale.
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DR is a relative metric, not an absolute. Scale is totally irrelevant to DR.
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DR is a relative metric, not an absolute. Scale is totally irrelevant to DR.
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I don't see how this can be true. The smallest possible step in a dynamic range is limited by the effect of a single photon. If photodetectors on sensors could actually count photons, which I believe they can't, but lets assume that they can without any interference from noise of any description, then the dynamic range is determined by the maximum number of photons that a sensor can 'process' during any exposure.
This maximum is determined by the size of the sensor, all else being equal, or if you like, the size of individual photodetectors, all else being equal.
In this sense, size or scale is very relevant to DR.
In practice, you have to either subtract or cancel all sources of noise from this maximum signal capacity of the sensor in order to determine a useful DR.
If DR is not 'scale' dependent as you suggest then you could claim, if it were possible to design a completely noise free tiny sensor containing say 100 photodiodes just 1 micron in diameter, that such a tiny sensor could have the same dynamic range as, say a 5D or P45+.
This would be clearly ridiculous. Reductio absurdum!
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DR is a relative metric, not an absolute
That's right. However, the numerical representation does not have to follow that one-to-one.
Example: if the DR is only one stop and the values are stored in 256 levels, then the highest value is 256 times as high as the lowest value - but that represents only twice the lightness.
Likewise, the numerical values from 0 to 7 can represent leves of a dynamic range of 1000:1.
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That's right. However, the numerical representation does not have to follow that one-to-one.
Example: if the DR is only one stop and the values are stored in 256 levels, then the highest value is 256 times as high as the lowest value - but that represents only twice the lightness.
Likewise, the numerical values from 0 to 7 can represent leves of a dynamic range of 1000:1.
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Exactly! So you agree with my staircase analogy, Panopeeper?
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So you agree with my staircase analogy, Panopeeper?
Yes, I do, though I have a small problem with it: one could come to the idea, that "floors" represent "stops", therefor the levels are alway a fraction of a stop. It does not have to be that way: a step could be for example 3 stops high.
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Yes, I do, though I have a small problem with it: one could come to the idea, that "floors" represent "stops", therefor the levels are alway a fraction of a stop. It does not have to be that way: a step could be for example 3 stops high.
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Yes, a step could be 3 stops high if you had a low enough bit depth like 4 bit encoding. But that's going to extremes. However, using f stops to describe DR is just one particular convention, I agree.
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I don't see how this can be true. The smallest possible step in a dynamic range is limited by the effect of a single photon. If photodetectors on sensors could actually count photons, which I believe they can't, but lets assume that they can without any interference from noise of any description, then the dynamic range is determined by the maximum number of photons that a sensor can 'process' during any exposure.
The context in which I was writing was one without noise, in which bit depth is the limiter. DR applies as well to any computer-generated image, in which case you could double the DR with each extra bit of linear depth. I was think of shot noise as a noise in this context.
Of course, with a pure photon-counting situation, the maximum number of photons would determine DR. That is just as hypothetical as my noiseless image (for photography; not for computer-generated images).
This maximum is determined by the size of the sensor, all else being equal, or if you like, the size of individual photodetectors, all else being equal.
In this sense, size or scale is very relevant to DR.
In practice, you have to either subtract or cancel all sources of noise from this maximum signal capacity of the sensor in order to determine a useful DR.
There's where your analogy doesn't work; the maximum number of photons minus the "noise floor" does not relate to DR at all, unless you are still thinking about pure photon counting noise, where the lowest usable standard (by any arbitrary but consistent standard) is always the same number of photons, so there is a unique mapping between max minus noise floor and max divided by noise floor. The division is the most useful and straightforward, though. "Differences" just result in a curve that needs to be converted back to a ratio for any useful calculations. As soon as you start doing anything else besides counting photons/electrons, like introducing read noise, the absolute difference between the noise floor and the max has no direct relationship to DR. A 16-bit camera with a noise floor of 32K ADU would have a traditional DR of only 2x, or one stop, although the difference would be 32K ADU or somewhere from 16K to 400K electrons. An 8-bit camera with a noise floor of 1.5 ADU or 0.75 to 30 electrons out of a max of 30K electrons would have much more DR, even though the difference between the noise floor and max signal is much less, both in ADUs and electrons.
You seem to have a literal conception of "noise floor". It's a poorly chosen term, IMO, which gives off false connotations. It is not the bottom of anything. Signal always exists below the noise floor and is not totally obscured by the noise. It's just an important turning point for SNRs in the deep shadows. Don't forget, most of this discussion is pixel-centric, and that's fine, as long as we understand what that means. The DR we usually speak of is that of the pixel, but the pixel does not determine the image, and depending upon the pixel frequency of the detail we are interested in capturing, you can get usable signal well below the noise floor. You can record a white fat letter that almost fills the frame, on a black background, in a clean DSLR, where the level for the white letter is a small fraction of a single photon. In the same way, as we use more and more pixels in our images, the pixel noise, and pixel DR, become less of an issue to image noise and DR. Never forget, the real world of light is individual photons, and any illusion of smooth levels is achieved by mechanical binning and inability to resolve individual photons.
If DR is not 'scale' dependent as you suggest then you could claim, if it were possible to design a completely noise free tiny sensor containing say 100 photodiodes just 1 micron in diameter, that such a tiny sensor could have the same dynamic range as, say a 5D or P45+.
This would be clearly ridiculous. Reductio absurdum!
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Yes it would be, but as I said before, my context was one where DR isn't separable
from bit depth, and such scenarios can exist, if not in a digital capture. You can do a 3D ray-tracing with the output as a linear DNG that looks like a camera capture, but with the only noise/distortion as quantization. In that case, DR would be directly proportional to bit depth.
I was just trying to give some balance to the idea that DR has nothing to do with bit depth.
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Yes, a step could be 3 stops high if you had a low enough bit depth like 4 bit encoding. But that's going to extremes. However, using f stops to describe DR is just one particular convention, I agree.
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If the noise is high enough, though, 4 bits linear is all you need, and a gamma-encoded 4 bits if the noise is even lower. You can barely see quantization effects with a D2X ISO 1600 RAW quantized to 6 linear bits. I'm sure some of the earliest CCD attempts at 1600 would suffice with 4 bits (although a positive blackpoint offset in the RAW data would be very useful at that extreme).
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That's right. However, the numerical representation does not have to follow that one-to-one.
Example: if the DR is only one stop and the values are stored in 256 levels, then the highest value is 256 times as high as the lowest value - but that represents only twice the lightness.
If you are talking about 256 linear levels (8 bits), then a DR of 2.0 would mean that the noise floor was at level 127.5. If you are going to use 0 through 255 to represent what I would think of as 127.5 to 255, then you are clipping away shadows. You can't clip at the noise floor, without damaging the capture (not that it is a good one anyway, but further damage is further damage).
Likewise, the numerical values from 0 to 7 can represent leves of a dynamic range of 1000:1.
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Yes, but you might want to add some noise to that, if enough weren't there already, unless you're looking to use quantization as a special effect.
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I read some of Roger Clark's site the other day and I don't think he was claiming that moving from a 12 bit ADC to a 14 bit ADC would give 2 extra stops of dynamic range. He did suggest that the lower read noise possible with a 14bit (or 16 bit) device could benefit some current camera's, e.g. canon 5D and presumably Nikon D3 etc, giving a higher achievable dynamic range, but not a full 2 stops higher. Part of the issue is that the DR is limited by full well capcity and the noise floor, which the ADC contributes to. hence a 'quieter ADC' should improve the lowest readable levels.
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I think that the ADCs in the high end camera already aren't introducing the majority of read/blackframe noise. I'm talking about the ones that have ISO 100 blackframe noises of 1.22 to 1.4 12-bit ADUs (1D* series, D3, 40D). The ones that have 2.0 to 4.0 ADU are possibly using inferior ADCs, but that can also be from the photosite read noise.
I think Roger overestimates the role of ADC noise in current DSLRs, especially in Canons. I think that the difference in read noise, in electrons at different ISOs in Canons occurs right at the photosite readout. Roger was expecting an increase in DR the 14-bit DSLRs that were announced, but the benefit never materialized. The fact that ISO 100 read noise stayed at almost exactly the same level between the 1Dmk2 and 1Dmk3 suggests that the ADC is probably not a major contributor to noise, IMO.
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The Leica DMR back for the R8 and R9 is a 16 bit camera. It has excellent dynamic range and gradation and can retain much more shadow detail than a 12- or 14-bit camera. Those who have used both the DMR and 5D (for example) in RAW mode overwhelmingly prefer the DMR's image quality.
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Is this with the same absolute exposure (IOW, same scene with the same lighting, same Av/Tv combo or equivalent)? Differences could be in absolute exposure, otherwise. What a camera says its ISO is isn't very exact. The 5D meters for about 1.2x the ISO it says it is at, for example.
Also, how deep into the shadows are you talking? A MF back could have low shot noise, and high read noise, compared to a DSLR, and look better in the shadows down to a point, and then break up faster as it goes deeper into the shadows, so often, it depends on how deep you go. Anyway, the 5D is not spectacular at ISO 100. It has 1/2 stop more read noise than the 1D* cameras at ISO 100. The extra pixels make up for it a little, compared to a 1Dmk2 or 3, but not enough to equal them in the deep shadows.
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The context in which I was writing was one without noise, in which bit depth is the limiter. DR applies as well to any computer-generated image, in which case you could double the DR with each extra bit of linear depth. I was think of shot noise as a noise in this context.
John,
If you have already defined a situation where bit depth is the limiter to DR, then you might be right that DR is not dependent upon scale. However, it's difficult to follow your reasoning. I would have thought the DR of a computer-generated image would be limited by the contrast ratio of the computer monitor. However, the number of shades between the darkest point and the brightest point would depend on the bit depth. The staircase analogy still applies, does it not?
Of course, with a pure photon-counting situation, the maximum number of photons would determine DR. That is just as hypothetical as my noiseless image (for photography; not for computer-generated images).
Well, it's not entirely hypothetical but rather a matter of accuracy. Current DSLRs produce a DR which is roughly related to a photon count at base ISO, do they not?
If you concede that a futuristic camera that was so accurate and sophisticated it could serve a dual purpose as a photon counter, would have a dynamic range based on the maximum number of photons it could count during a full exposure, ie a DR based on scale, then why should a less accurate camera not have a DR based on scale. Has the nature and definition of DR changed because our capturing device is less accurate and can only give a rough approximation of photon count?
There's where your analogy doesn't work; the maximum number of photons minus the "noise floor" does not relate to DR at all, unless you are still thinking about pure photon counting noise, where the lowest usable standard (by any arbitrary but consistent standard) is always the same number of photons, so there is a unique mapping between max minus noise floor and max divided by noise floor.
Again, I can't follow your reasoning. If I say, 'the maximum number of photons minus the noise floor = DR', then I must be thinking about this ultra-accurate, pure photon counting situation. I don't see how you can move from a position of conceding that at a level of great accuracy DR does relate to scale, but at a level of less accuracy DR has nothing to do with scale.
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That's right. However, the numerical representation does not have to follow that one-to-one.
Example: if the DR is only one stop and the values are stored in 256 levels, then the highest value is 256 times as high as the lowest value - but that represents only twice the lightness.
Likewise, the numerical values from 0 to 7 can represent leves of a dynamic range of 1000:1.
I have to agree with John Sheehy. Panopeeper, as I already said in a post on this thread, we are talking about sensors, i.e. about CAPTURE of dynamic range, not about STORING or PROCESSING. And as long as sensors are LINEAR devices, they simply cannot register more DR than the number of bits used for the LINEAR A/D encoding.
Your example is perfectly right to demonstrate that more DR than N f-stops can be coded into a properly designed NON-LINEAR N-bit encoding system, but as far as I know only Leica's M8 RAW files are non-linearly encoded, and after all they are only 8-bit.
All Canons, Nikons, etc... use linear ADC encoding, and because of this they can achieve to store a maximum of N f-stops of DR in N bits. So for those cameras bit depth is a physical limit to DR.
Just an example: if N=16, your LINEAR range is 0..65535
- Min amount of light we can represent: 1
- Max amount of light we can represent: 65535
- Max contrast we are capable of capturing: 65535/1
What is the corresponding DR to it? DR=log(65535/1)/log(2)=16 f-stops=N
You could store 256 f-stops of DR in 8-bit, just assigning 1 level of the 0..255 range to each f-stop, that is OK. But I insist: there is no capture device (sensor) than works in such a non-linear way today. Surely tomorrow there will be (Leica's M8 achieves nearly the same DR with 8-bit than a Canon 5D with its linear 12-bit RAW files, and it's all thanks to a very intelligent non-linear encoding).
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You could store 256 f-stops of DR in 8-bit, just assigning 1 level of the 0..255 range to each f-stop, that is OK. But I insist: there is no capture device (sensor) than works in such a non-linear way today. Surely tomorrow there will be (Leica's M8 achieves nearly the same DR with 8-bit than a Canon 5D with its linear 12-bit RAW files, and it's all thanks to a very intelligent non-linear encoding).
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Ah! Here we have the explanation for this current difference of opinion. DR can be limited by bit depth in the linear method of encoding used in most digital cameras, whereby each doubling of light values is represented by a doubling of encoded levels; but it doesn't have to be.
This point was made earlier in the thread, a few times, but seems to have got overlooked or forgotten, at least by me.
If you were to put an 8 bit A/D converter in a P45 DB, then you could truly say the dynamic range was limited by bit depth (using the linear method of encoding). However, if you put a 16 bit A/D converter in a current Canon or Olympus DSLR with their current noise levels, then I think it would be true to say that DR was limited by photodiode size.
On the other hand, it might just be limited by noise. Either way, in this situation DR is scale limited; ie, limited by the scale of the noise or the scale of the photoreceptor.
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we are talking about sensors, i.e. about CAPTURE of dynamic range, not about STORING or PROCESSING
We are talking all the time (like yourself just in this opost) about the numerical representation of the values. Raw data is not about the analog values.
Just an example: if N=16, your LINEAR range is 0..65535
The numerical range of the stored data is 1-65535. This does not mean anything on its own.
Think of following: you can measure length in inches or in millimeters. If you measure it in inches, then you can express much larger lengths in the same numerical range. Both are linear measurements, with different scales.
The meaning of linear in the present context is, that every increase of the numerical value by one represents a certain fixed increase of the light intensity. It is not relevant from this point, how much that intensity difference is.
The problem is, that you are concentrating on "stop" and "bit", instead of seeing the dynamic range simply as "proportion". Who said, that it has to be measured on a 2-based log scale? Why not 3, 10 or 1.5? You are mixing the number representation (binary) with your favourite scale of the dynamic range. What about a computer, which can work only with decimal numbers? (There was such.) What about tri-state harware architecture ("trits" instead of "bits")?
Imagine a line with a point on it, let's name it a. Draw markers on top of the line with a fix distance of d, i.e. at a+d, a+2d, a+3d, a+4d, a+5d, etc. These are the "levels", stored as the raw values.
Draw markers at the underside of the line too; pick the first one anywhere to the right of a nd name that a*p, then the next one a*p^2, a*p^3, a*p^4, a*p^5, etc (p > 1). These represent the steps (not necessarily stops) of the dynamic range. If p is 2, then we measure it in stops, but it does not have to be.
There is no fixed relationship between the two row of markers. The linear markers can be much closer or much farther than the proportional markers, up to a certain point. Because p > 1, the proportional scale will "outpace" the linear scale at some point, and that point may fall within the range of our interest, or it may not.
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If you were to put an 8 bit A/D converter in a P45 DB, then you could truly say the dynamic range was limited by bit depth (using the linear method of encoding
You could predict, that the result will be horrendeously posterized, but you could not say anything about the captured dynamic range.
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You could predict, that the result will be horrendeously posterized, but you could not say anything about the captured dynamic range.
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You could say something about the captured dynamic range if the encoding is the same as it currently is in most cameras, ie. 256, 128, 64 , 32, 16, 8, 4, 2, 1, which represents 8 intervals or just 8 stops of DR. The posterization would be horrendous in the lower mid-tones, producing a practically useful DR of considerably less than the theoretical maximum of 8 stops.
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We are talking all the time (like yourself just in this opost) about the numerical representation of the values. Raw data is not about the analog values.
The numerical range of the stored data is 1-65535. This does not mean anything on its own.
Think of following: you can measure length in inches or in millimeters. If you measure it in inches, then you can express much larger lengths in the same numerical range. Both are linear measurements, with different scales.
The meaning of linear in the present context is, that every increase of the numerical value by one represents a certain fixed increase of the light intensity. It is not relevant from this point, how much that intensity difference is.
The problem is, that you are concentrating on "stop" and "bit", instead of seeing the dynamic range simply as "proportion". Who said, that it has to be measured on a 2-based log scale? Why not 3, 10 or 1.5? You are mixing the number representation (binary) with your favourite scale of the dynamic range. What about a computer, which can work only with decimal numbers? (There was such.) What about tri-state harware architecture ("trits" instead of "bits")?
Imagine a line with a point on it, let's name it a. Draw markers on top of the line with a fix distance of d, i.e. at a+d, a+2d, a+3d, a+4d, a+5d, etc. These are the "levels", stored as the raw values.
Draw markers at the underside of the line too; pick the first one anywhere to the right of a nd name that a*p, then the next one a*p^2, a*p^3, a*p^4, a*p^5, etc (p > 1). These represent the steps (not necessarily stops) of the dynamic range. If p is 2, then we measure it in stops, but it does not have to be.
There is no fixed relationship between the two row of markers. The linear markers can be much closer or much farther than the proportional markers, up to a certain point. Because p > 1, the proportional scale will "outpace" the linear scale at some point, and that point may fall within the range of our interest, or it may not.
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Panopeeper, the point of talking about doubling and halving levels, has no relation with bits. It is simply the definition of "f-stop": relative range of lightness where the brightest value is double as the lowest light value.
Forget now about bits or trits, think that you just have an encoding system which can record light levels from 0 to 65535. And you also know this system responds linearly to light, i.e. if a light source generates a level X on that range, double the light would have generated 2*X and half the light would have generated X/2.
With such a linear response system, can you imagine of any level distribution that would allow to capture more than: log(65535)/log(2)=16 f-stops of dynamic range? You can't. It's a numerical limit for such a linear capture system.
The new 14-bit Canon and Nikon cameras have linear encoding, so if they were hypothetically free of noise they could only distinguish a maximum of 14 f-stops. The practical dynamic range will be quite lower that this figure due to noise since the lowest f-stops will be plenty of noise making them unusable from a photographic point of view.
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Forget now about bits or trits, think that you just have an encoding system which can record light levels from 0 to 65535. And you also know this system responds linearly to light, i.e. if a light source generates a level X on that range, double the light would have generated 2*X and half the light would have generated X/2.
With such a linear response system, can you imagine of any level distribution that would allow to capture more than: log(65535)/log(2)=16 f-stops of dynamic range? You can't. It's a numerical limit for such a linear capture system.
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This has been gone over ad naseum times and reasoning such as Guillermo's has always prevailed. Further discussion is futile.
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You can cover a range of 50 stops even with one bit. [a href=\"index.php?act=findpost&pid=157163\"][{POST_SNAPBACK}][/a]
Yes, but you have to use extreme oversampling; otherwise, all you have is a threshold and the concept of "DR" or even "stops" is meaningless.
Real light is always "1-bit", but the receptors in our eyes and the sensor wells and film grain all work towards binning multiple single-bit values into a wider range of shades besides "photon or no photon".
A 10MP 1-bit camera would be of limited usefulness; a 500 GP 1-bit camera would be quite usable, if handling the data were practical.
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You could predict, that the result will be horrendeously posterized, but you could not say anything about the captured dynamic range.
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Again, it depends on what kind of resolution you're looking for. For each pixel, as an independent measuring device, the 8-bit linear reduces DR to slightly less than 8 stops, without any noise, and even less, with noise. When you start looking at groups of pixels, however, then their combined values start creating more levels, and potentially more DR, and here a little bit of linearly distributed noise before digitization can actually help if there is not enough existing noise.
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Yes, but you have to use extreme oversampling; otherwise, all you have is a threshold and the concept of "DR" or even "stops" is meaningless
The subject was not the practicability but the principle. I could have said "20 stops in ten bits", which is still impractical, but not as extreme as "50 stops with one bit".
The concept of "stop" is relevant mainly for photographers, it plays no role whatsoever in the principle.
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With such a linear response system, can you imagine of any level distribution that would allow to capture more than: log(65535)/log(2)=16 f-stops of dynamic range? You can't. It's a numerical limit for such a linear capture system
I just tried to make it understandable, that there is no such connection. You are still concentrating on stops and bits instead of concentrating on the underlying math. The only meaning of the formula log(65535)/log(2) is, that 16 divided by 1 is 16.
Let's see it a different way.
There is a given scenery, a given lens and a given sensor. Assume we can count the photons individually; the numerically expressed intensity is deducted directly from the number of captured photons.
In one case the resulting number representing the intensity equals to the number of photons.
In another case only every 15th photon counts.
In another case only every 200th photon counts, but we can go even further: only every 500th photon counts (i.e. the numerical intensity is the number of photons divided by 500). The 60000 photons of a well would be converted in the numerical range of 0-120, which requires just 7 bits.
The numerical representation is linear in all these cases, but the number of levels covering the entire range is vastly different, and so is the number of required bits to store the data.
In all these cases the covered dynamic range is the same.
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Let's see it a different way....
In all these cases the covered dynamic range is the same.
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The point you are missing is that the dynamic range is NOT the same. Dynamic range by definition is the ratio between maximum signal and minimum signal, not simply the value of the maximum signal. (We are of course still igoring noise, otherwise the definition would be maximum signal divided by noise.)
In your example, the maximum signal is the same in all cases (60000 photons), but the minimum signal varies with the method of photon counting. If you only count one photon in 500, then the minumum signal you can count is 500 photons and the dynamic range is 60000/500 = 120, capturable as you state in 7 bits. But if every photon is counted, then the dynamic range increases to 60000/1 and requires 16 bit to capture. This is a real increase in dynamic range, by any reasonable definition.
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If you only count one photon in 500, then the minumum signal you can count is 500 photons and the dynamic range is 60000/500 = 120, capturable as you state in 7 bits. But if every photon is counted, then the dynamic range increases to 60000/1 and requires 16 bit to capture. This is a real increase in dynamic range, by any reasonable definition.
These are independent issues. I can start counting the photons with 10, then 510, 1010, 1510, etc. Then the dynamic range is 60000/10.
But I can start counting with photon 1, then 501, 1001, etc.
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These are independent issues. I can start counting the photons with 10, then 510, 1010, 1510, etc. Then the dynamic range is 60000/10.
But I can start counting with photon 1, then 501, 1001, etc.
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Do you really think you can increase the dynamic range of a sensor by a factor of ten simply by counting {1, 11, 21, ...} instead of {0, 10, 20, ...}? Doesn't that sound ridiculous?
We all agree that it is possible to concoct non-linear counting schemes which provide extended dynamic range with fewer bits. But such a scheme is mathematically impossible for a linear sensor (with an offset of zero).
Here is a thought exercise which might prove interesting. Imagine an array of 120 light bulbs which I want to photograph. The light bulbs have different brightness, with a linear progression. The first light bulb is turned off. The second light bulb emits 500 photons/second into the aperture of my camera. The third light bulb emits 1000 photons/second.... The last light bulb emits 60000 photons/second.
Now I photograph this array of light bulbs with two different cameras in three different scenarios:
1) Camera full well 60000 photons, linear scale counting one photon out of 500, exposure 1 second. This camera images the light bulbs with readings {0, 1, 2, ... 120).
2) Camera full well 60000 photons, linear scale counting single photons, exposure 1 second. This camera images the light bulbs with readings {0, 500, 1000, ... 60000}.
3) Camera full well 60000 photons, linear scale counting single photons, exposure 1/500 second. This camera images the light bulbs with readings {0, 1, 2, ... 120}.
Would you agree with the following statements?
1) The first camera uses its entire dynamic range to capture the scene. If the scene had brighter light bulbs, this camera could not capture them. If the scene had light bulbs with intermediate brightness, this camera could not resolve them.
2) The second camera, with the same exposure, captures the scene with resolution to spare. The captured dynamic range appears to be the same (because the dimmest bulb still has brightness 500 photons/second), but there is plenty of unused space to capture light bulbs with intermediate brightness.
3) The second camera, with reduced exposure, captures the scene with the same resolution (in fact the same numerical output values), but with dynamic range to spare. This camera could capture a larger range of light bulb brightness.
Clearly the two cameras do not have the same dynamic range!
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You could say something about the captured dynamic range if the encoding is the same as it currently is in most cameras, ie. 256, 128, 64 , 32, 16, 8, 4, 2, 1, which represents 8 intervals or just 8 stops of DR. The posterization would be horrendous in the lower mid-tones, producing a practically useful DR of considerably less than the theoretical maximum of 8 stops.
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8 bits linear is actually sufficient for most cameras (except for the D3 and Canon DSLRS) at ISO 1600, even ISO 800 for some. Again, you never (visibly) see this RAW quantization in practice, because of noise.
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The subject was not the practicability but the principle. I could have said "20 stops in ten bits", which is still impractical, but not as extreme as "50 stops with one bit".
The concept of "stop" is relevant mainly for photographers, it plays no role whatsoever in the principle.
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It's an arbitrary factor, but one that is familiar to photographers, so why not use it? Now decibels, that is something foreign to the photographer.
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In another case only every 200th photon counts, but we can go even further: only every 500th photon counts (i.e. the numerical intensity is the number of photons divided by 500). The 60000 photons of a well would be converted in the numerical range of 0-120, which requires just 7 bits.
That would have less than 7 stops of DR in the absence of analog noise at the pixel level, and more at the image level with a little noise, but greater limitations with ample noise.
The numerical representation is linear in all these cases, but the number of levels covering the entire range is vastly different, and so is the number of required bits to store the data.
In all these cases the covered dynamic range is the same.
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That's only true if the analog noise is the bottleneck to DR. Otherwise, your 200 levels becoming 1 level in the deepest shadows throws away DR.
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These are independent issues. I can start counting the photons with 10, then 510, 1010, 1510, etc. Then the dynamic range is 60000/10.
But I can start counting with photon 1, then 501, 1001, etc.
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Your "counting" has no effect on DR. Bottom line is that without noise, your bottom 500 electrons are squished into 1 level, at the image level, and noise won't help at all at the pixel level.
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Do you really think you can increase the dynamic range of a sensor by a factor of ten simply by counting {1, 11, 21, ...} instead of {0, 10, 20, ...}? Doesn't that sound ridiculous?
That would be truly a ridiculous claim. However, the subject is not the dynamic range of the sensor but the recording of the sensed data. As I posted above:
a given scenery, a given lens and a given sensor
so that only the numeric representation is variable.
But such a scheme is mathematically impossible for a linear sensor (with an offset of zero).
That's your belief.
an array of 120 light bulbs which I want to photograph. The light bulbs have different brightness, with a linear progression. The first light bulb is turned off. The second light bulb emits 500 photons/second into the aperture of my camera. The third light bulb emits 1000 photons/second.... The last light bulb emits 60000 photons/second
The demonstration-idea is excellent. Let's see, what it demosntrates:
3) Camera full well 60000 photons, linear scale counting single photons, exposure 1/500 second. This camera images the light bulbs with readings {0, 1, 2, ... 120}
...
3) The second camera, with reduced exposure, captures the scene with the same resolution (in fact the same numerical output values), but with dynamic range to spare. This camera could capture a larger range of light bulb brightness
Right. So, you have demonstrated, that the second camera can capture the entire original dynamic range in seven bits (just like the first camera), and it can capture even more, in more bits.
In effect you have shown, that if a camera counts only every xth photon, it requires x times higher exposure in order to gain the same number of levels, than another camera, which counts every photon.
No objection here, but this has nothing to do with the numerical representation (bit depth vs. DR).
However, if you are saying, that the first camera is "throwing away" large part of its capability, then consider following:
1. nobody disputed, that greater bit depth results in more levels,
2. in 1/500 sec the first 499 measurable levels are eliminated. They won't be captured with the first camera either, because you conveniently decided, that it counts 0, 500, 1000, etc. (your proof requires this). However, if the first camera counts 5, 505, 1005, etc. then everything looks differently.
The demostration with discrete values leaves much leeway, which can be used so or so.
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However, if the first camera counts 5, 505, 1005, etc. then everything looks differently.
Panopeeper a sensor that assigns the first non-zero level (level 1) after counting 5 photons, and next level (level 2) when the count reaches 505 photons is simply NOT LINEAR; and present cameras are linear so they have the bit depth limitation for DR.
I think we all agree that the limitation of DR with bit depth is only a matter of the encoding system used, and as long as this is linear, N bits can encode a maximum of N f-stops of DR.
In fact we already discussed that typical 12-bit linear sensors loose around the first 256 levels because of the black offset level due to sensor electronics. That means an even more reduced limit to the DR of:
log(4096-256)/log(2)=11,9 f-stops
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Panopeeper a sensor that assigns the first non-zero level (level 1) after counting 5 photons, and next level (level 2) when the count reaches 505 photons is simply NOT LINEAR
So, you think if you measure something with a measure tape and start on the 50cm mark, then the measurement is not linear?
Beside, you can start at 1, then 501, etc., it is really irrelevant.
I made a simple drawing showing the relationship (or the lack thereof) between DR and bit depth:
(http://www.panopeeper.com/Demo/DRvsBitDepth.gif)
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So, you think if you measure something with a measure tape and start on the 50cm mark, then the measurement is not linear?
Yes, unless you have a completely new definition of "linear". If 0 = 0 and 1 = 5 and 2 = 505, then the encoding function is by definition not linear. Or did you fail algebra?
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Just to shift the topic a bit, from counting photons, defining levels, etc ...
The dynamic range is ultimately limited by the noise floor - has anyone made any effort to lower the noise floor by cooling the sensor and other analog electronics? Pelletier devices are quite small (albeit rather power hungry), and would probably fit into a DSLR.
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That is commonly done with cameras designed for astronomy, but the cost/benefit ratio isn't high enough for mainstream DSLR manufacturers to add it as a design feature.
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Even cooling the sensor will not remove photonic shot noise which is numerically the square root of the total number of photons the sensor receives, although it's not exactly clear to me how this noise is distributed.
I imagine if there's one part of the image in deep shadow which receives on average 25 photons per pixel, then noise will be on average 5 photons per pixel or 20% for the total area in deep shadow.
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Even cooling the sensor will not remove photonic shot noise which is numerically the square root of the total number of photons the sensor receives, although it's not exactly clear to me how this noise is distributed.
I imagine if there's one part of the image in deep shadow which receives on average 25 photons per pixel, then noise will be on average 5 photons per pixel or 20% for the total area in deep shadow.
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Here's what you get with a population of 100, averaging 25 electrons:
28 17 22 25 26 35 21 31 36 19
25 36 19 26 27 23 25 13 22 30
21 18 22 17 26 24 23 27 30 23
24 20 18 38 39 22 22 25 22 26
32 18 20 26 17 19 21 20 25 23
21 28 25 29 19 38 20 29 24 30
25 19 21 28 19 25 36 22 31 21
26 25 30 34 33 27 26 29 28 25
25 29 34 38 22 22 30 26 23 37
13 26 31 25 35 23 30 26 26 24
Minimum of 13 and maximum of 39 in this sample, a spread of 26 photons.
With a population of 20 million, I get 4 to 56.
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Even cooling the sensor will not remove photonic shot noise which is numerically the square root of the total number of photons the sensor receives, although it's not exactly clear to me how this noise is distributed.
I imagine if there's one part of the image in deep shadow which receives on average 25 photons per pixel, then noise will be on average 5 photons per pixel or 20% for the total area in deep shadow.
Yes, this is all correct. To be precise, the statistics apply to electrons produced by interacting photons, since that is what the sensor actually "counts".
Shot noise has a Poisson distribution, with fluctuations (standard deviation) equal to the square root of the mean. So in your example, a light level which produces 25 electrons on average will have statistical noise of 5 electrons. You can generate the full distribution if you have Excel handy (look up the Poisson function), but it is essentially a bell shaped curve with a peak at 25 and a width of 5.
Most sensors are dominated by electronics readout noise at very low light levels so shot noise is rarely a concern. For example, if a sensor has readout noise of 20 electrons, shot noise will be smaller until the signal reaches 400 electrons, and by then this noise is only 5% of the signal.
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Here's what you get with a population of 100, averaging 25 electrons:
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That should be, "what you might get"!
You probably won't get exactly this.
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That should be, "what you might get"!
You probably won't get exactly this.
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John,
Just to be sure I've understood the significance of these figures, are you saying that the above variation amongst a group of 100 pixels might apply if we were photographing a dark grey patch where, ideally, each pixel would receive 25 photons if it were free of shot noise? (and hypothetically free of all noise from other sources).
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Most sensors are dominated by electronics readout noise at very low light levels so shot noise is rarely a concern. For example, if a sensor has readout noise of 20 electrons, shot noise will be smaller until the signal reaches 400 electrons, and by then this noise is only 5% of the signal.
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So the implication here is that both shot noise and read noise will not be reduced by cooling, just thermal noise, or is read noise partly thermal?
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Yes, unless you have a completely new definition of "linear". If 0 = 0 and 1 = 5 and 2 = 505, then the encoding function is by definition not linear. Or did you fail algebra?
Do you mind pointing out, where I posted 0 = 0 and 1 = 5 and 2 = 505? Or did you fail reading comprehension?
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John,
Just to be sure I've understood the significance of these figures, are you saying that the above variation amongst a group of 100 pixels might apply if we were photographing a dark grey patch where, ideally, each pixel would receive 25 photons if it were free of shot noise? (and hypothetically free of all noise from other sources).
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Yes. I just wanted to demonstrate how far off values are. The "standard deviation" can seem a bit more conservative than what it really means. The majority of values are often in a spread about 3x as great as the standard deviation.
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Do you mind pointing out, where I posted 0 = 0 and 1 = 5 and 2 = 505? Or did you fail reading comprehension?
Panopeeper, this is getting ridiculous. One last post and then I at least am done with this.
Let's go back to the basics. Several of us claimed that a noiseless camera with a linear encoding scheme necessarily has dynamic range equal to the number of encoded values. Going back to the ladder analogy, the height of the ladder divided by the step size is equal to the number of rungs. (Is this a clear enough mathematical proof?) We neglected to specify that the first rung of the ladder must be placed one step size above ground level.
You concocted a scheme which gets around this simple claim, by defining dynamic range as the height of the ladder divided by the height of the lowest rung above ground, then chopping off the bottom of the ladder so that the first rung is very close to the ground. I will grudgingly admit that this ladder could still be called linear and that with this contorted definition of dynamic range, you have found a way around the proof.
Will you in turn admit that this extension of dynamic range is just a game to win an argument and has no practical application in photography? The more you extend dynamic range this way, the fewer pixels get to participate in the extended dynamic range. Your camere with infinite dynamic range and my camera with limited dynamic range take indistinguishable pictures.
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Do you mind pointing out, where I posted 0 = 0 and 1 = 5 and 2 = 505? Or did you fail reading comprehension?
Does this sound familiar?
These are independent issues. I can start counting the photons with 10, then 510, 1010, 1510, etc. Then the dynamic range is 60000/10.
But I can start counting with photon 1, then 501, 1001, etc.
First increment of photon counter happens at 10, second at 510, or first counter increment at 1, second at 501, third at 1001, sounds pretty non-linear to me. "Linear" requires a reasonably fixed ratio between photons and ADU output values, which is not what you are describing in your December 4 quote. Clipping the black point (which is the net result of the process you describe) does not increase DR, it merely disguises noise by clipping it to black or near-black. The reduction of apparent noise is accompanied by a similar level of shadow detail loss.
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First increment of photon counter happens at 10, second at 510, or first counter increment at 1, second at 501, third at 1001, sounds pretty non-linear to me
The increase is constant. That makes it linear, if the original values are linear.
"Linear" requires a reasonably fixed ratio between photons and ADU output values
There are no photons and no ADU at the stage we are talking about (i.e. numerical representation). There are some measured, linear values, no matter how they have been arrived at. Now the question is, how they can be recorded.
I nowhere suggested to measure the light intensity differently from how it is measured now; "counting the photons" is only symbolic, as there is no such thing on the sensor, AFAIK. I am suggesting, that given a set of measured values, they can be recorded different ways, by varying the correlation between the measured and recorded values. (Note, that this correlation is not only not god-given, it is not fix even with the same camera.)
Furthermore, "0 = 0" is not correct. The measured/recorded value 0 represents all values under some limit. Likewise, 16383 (or whatever the clipping point is) represents all values above that as well. Therefor linearity does not include the two extreme values. In fact, non-linearity can include a range of values. This is the nature of the data, it has nothing to do with the numerical representation.
(Guillermo posted a raw file from the Nikon D3. I found, that the non-linearity of the green values streches from 15750 to 16060, while all red and all blue pixels clip "at once", though red and blue not together.)