Film vs Digital - Dynamic Range

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<p>Helen: Where do you get this from? I really wish that you would stop arguing against me by fabricating stuff and claiming that I have said it.</p>

 

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<p>By your definition of 127 mV - 896 mV as the range. That's really not the standard way to choose range endpoints.</p>

<p>Here's what you said:</p>

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<p><a href="../photodb/user?user_id=817643">Helen Bach</a> <a href="../member-status-icons"><img title="Subscriber" src="http://static.photo.net/v3graphics/member-status-icons/sub6.gif" alt="" title="Subscriber" /> </a> , Dec 10, 2008; 07:37 p.m.<br>

Every input value at or below 127 mV is output as 0; every input value of 896 mV or above is output as 7. We agree on that. The dynamic range of the ADC is 896/127 = 7.</p>

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<p><strong>How can an image file represent the stops of a scene that occur below the ADC output of 1, specifically addressing the reality that we generally see in dslrs - that is, linear relationship between input and output?</strong></p>

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<p>By mapping the output to 0.</p>

<p>So what's your point after that?</p>

<p>Vijay, I agree with everything you said to me in your last post. But what I want to know is why you are purposely ignoring the issue of stops? I want someone to specifically tell me what is wrong with my example that shows that a 4 bit converter can handle at max 4 stops of exposure accurately. I don't give a fig what it's theoretical dynamic range is. Will you answer my questions?</p>

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<p>By mapping the output to 0.</p>

 

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<p>How is that linear man?</p>

<p>The output of an ADC is <em>never linear</em> by that definition. It is quantized; and as linear as a staircase. We simply call the staircase "linear" if it is not curved, but what you are talking about goes to the root of the issue - a quantization process is not linear, ever. It is exactly the difference between a ramp and stairs. Zero has nothing to do with it either; both a ramp and staircase can start from the same point.</p>

<p>An aside: in electronics the strict use of the word "linear" only means "analog" or continuously varying. For digital systems, we use the word "switching"; this is to highlight the difference with true linearity.</p>

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<p>The output of an ADC is <em>never linear</em> by that definition. It is quantized; and as linear as a staircase. We simply call the staircase "linear" if it is not curved, but what you are talking about goes to the root of the issue - a quantization process is not linear, ever. It is exactly the difference between a ramp and stairs. Zero has nothing to do with it either; both a ramp and staircase can start from the same point.</p>

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<p>What has this got to do with STOPS? Your, and others, argument all along has been that you can record ANY number of stops with ANY number of bits, right? So, take an image generated by my 4bit ADC, and show me on the image where stops 5 to infinity (or the noise floor) exist. You can't because they don't exist. Or perhaps more precisely they do exist, but they are all in level 0 (all stuffed in there together), so how does that translate to more stops in the final image?</p>

<p>Bernie, I want you think a little about the staircase and ramp example for a bit, please. I think it is the most precise analogy I've come up with and may help you understand the concepts better. I'll address your stops thing as well. Please bear with me.</p>

<p>First, say you have two points; one is at ground level and one is 10 feet high. You want to build, next to each other, a staircase and a ramp that go from the ground to 10 feet up. Your dynamic range is the 10 feet, regardless of whether you use the ramp; or whether the staircase has 10 steps or 20. They all start at the same point and end at the same point.</p>

<p>I can't start the next flight of stairs at the start point of the last step; I have to do it after the step, because if I did, I'd get a step of twice the height. Same for the lowest point. If I wanted to start another flight of stairs to go to the basement, I'd have to start at the beginning point of the first step, not where it ends.</p>

<p>That's why my dynamic range measurement has to do with the 10 ft height, and the ground, and I can't choose other points. It is the same points as the ramp begins and ends, obviously.</p>

<p>Your question about stops. Say you have 16 steps. You are at the highest point; then you descend one stop, i.e, come down to step 8; then another stop to step 4; then another to step 2, then another to step 1. You're down 4 stops and on step 1. (Step 0 is groundlevel). If at this point, you descend one step, you hit ground straightaway; you can't have 1/2 step or 1/4 step, but this one step could encompass any number of stops or halvings on the ramp.</p>

<p>You've lost all of those stops; but this is a loss of "tonal" range, what is specifically called "resolution" of an ADC. But that doesn't mean that the ground isn't several stops away, and those several stops of range are actually there, just engulfed by one step on the staircase.</p>

<p>The fact that these stops are "there" but encompassed by one step has nothing to do with the ramp; if you increased the number of steps, you'd still lose tonal resolution at the lowest step because the ramp still has range that can be halved.</p>

<p>Noise: This acts like a fuzzy carpet on both the stairs and the ramp. It is present on all steps, so it makes it hard to tell exactly where a step starts and ends, and if you make the steps too small, then because of the carpet you can't even tell the steps apart from the carpet, so there is really no point to making steps much smaller than the carpet fuzz. This imposes a practical (not theoretical) consideration on the number of steps, but that doesn't mean you can't do it.</p>

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<p>You've lost all of those stops; but this is a loss of "tonal" range, what is specifically called "resolution" of an ADC.</p>

 

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<p>Alright, I can accept this. But this is truly a case of you 'not seeing the forest for the trees'. Your definitions don't change the substance of what I have been trying to convey all along. The substance is this: 4 stop ADC can only output 4 stops of data, identical to the scene (perhaps I should say identical in a <em>discrete</em> sense) to a digital file. Dynamic range could be 10, 100, a billion or infinity (and I accept that the technical definition of dynamic range is more correct than the photographic 'stops' interpretation, but this is a photography forum after all). All that matters is what we see at the other end. We've heard so many hypotheticals and theoreticals in this thread, that many of you have lost touch with the reality of what actually happens. And this comes into being from people who have too much specialisation (ie. blinkered), and not enough generalised skills (ie. like being able to verify all this with linear raw data).</p>

<p>So i would like to try and come to a consensus here: </p>

<p><strong>1.</strong> We both agree that technically dynamic range of a digital image could be infinite (or at least bloody close to it) in the absence of noise.</p>

<p><strong>2.</strong> We both agree that in a linear 4bit ADC with both the ADC and photodiodes limiting equally, and no noise, you could only <em><strong>identify</strong></em> 4 stops of scene exposure in a digital image file.</p>

<p><strong>3.</strong> Point 2 is what is relevant to photographers who wish to capture and meaninfully display a wider tonal range in their images.</p>

<p>What do you think?</p>

<p>I see the forest <em>and</em> the trees. And you are mostly right; except for point 2 which needs slight modification.</p>

<p>If you map the value 0 to pure black on a print; and the value 15 to pure white and take a spotmeter and read off the densities, what do you expect to see? You'll still see the dynamic range that paper can represent. If you used a 16 bit ADC and mapped 0 to black and 65,525 to white, you'd still see the exact same dynamic range from the paper print.</p>

<p>The only difference is, with the 4 bit ADC, the next tone from pure black would be several stops lighter (value of 0001). Then the next stop (0011) would be separated from the previous stop (0001) by one tone (0010); the next stop (0111) would be separated from the previous stop (0011) by three tones (0100, 0101, 0110) and the final stop (1111) would be separated from its previous stop by seven tones (1000, 1001, 1010, 1011, 1100, 1101, 1110).</p>

<p>Thus at the upper end of the scale, you divide a stop of DR into 8 tones, the next stop into 4 tones, the next into 2, the next into 1 and squash all other tones (that even are several stops darker) into one tone.</p>

<p>But when you print, 0000 is pure black and 1111 is pure white; giving you the max DR of the paper.</p>

<p>You can actually do this experiment: create sixteen tones in photoshop or gimp and assign them values from 0 to FFFFFF in sixteen equally spaced steps, and print it. Take a lightmeter and check dynamic range in stops. You'll see the DR of the paper. Voila.</p>

<p>I'm glad we kinda agree. But you gotta stop talking about pure dynamic range. The relevant term is 'stops'. When you halve scene value, you also halve image value whether that be linear raw print or linear raw screen (gamma 1). gotta go.</p>

<p>Try the experiment I said in my last post. You'll see pure dynamic range at work. You're the master at digital imaging, shouldn't be too hard for you.</p>

<p>The halving that you see in practice works because you have a 9 stop dynamic range encoded as 14 stops. Like the carpet fuzz in my stairs example, you can't distinguish the steps from the ramp at that kind of bit depth. Even that halving gives up on the last quantization step before 0, only you can't see it because the engineers that made your camera aren't stupid - they purposely made the quantization step less than the noise floor.</p>

<p>We kinda agree, but I kinda disagree on your point 2 as I already said. But no matter; you can try the simple experiment for yourself.</p>

<p>I think what Bernie is saying is that when multiple stops map to the same step of an ADC then the ability to differentiate among the stops is lost (and they become essentially the same), so as far as the end result is considered those multiply-mapped stops never made it through.</p>

<p>However, as Vijay noted in above posts, when multiple stops map to the same ADC step, that step should be interpreted as the least of the stops mapped to it; thus (in Bernie's example) the 4-bit ADC still provides information pertaining to whole 8 stop input range.</p>

<p>At this point, I'd request some input from Rishi, Helen, Ryan, Max, Rob and anyone else who can and wants to perform the experiment I mentioned (if you need more details let me know), then measure the reflected values from black and white and tell me the range of stops you see. If the DR of the paper seems too close to 4 stops, you can change to eight bands of equally spaced gray values from 0 to FFFFFF (values are: 000000, 1F1F1F, 3F3F3F, 5F5F5F, 7F7F7F, 9F9F9F, BFBFBF, DFDFDF and FFFFFF). If you want to input 16-bit values, do (000000000000, 1FFF1FFF1FFF, 3FFF3FFF3FFF etc).</p>

<p>Heck, you could just create two bands with values 000000 and FFFFFF and tell me what you see.</p>

<p>Thank you in advance if you do this.</p>

<p><em>Rob: I think what Bernie is saying is that when multiple stops map to the same step of an ADC then the ability to differentiate among the stops is lost (and they become essentially the same), so as far as the end result is considered those multiply-mapped stops never made it through.</em></p>

<p>Absolutely true for tonal range. Absolutely false for dynamic range. The point is the tone that is represented by 0 is black and the next tone, represented by 1 is several stops lighter. That doesn't change the ratio of the whitest white to the blackest black (this is the DR of the printer). And I've always said that 4 bits can't provide any more than 4 stops of tonal information and even said that this is absolute, inviolable. But dynamic range is another matter.</p>

<p>Call this loss of resolution, loss of tonal range or compressing many stops worth of tonal range into one value; but this isn't by any stretch of the imagination a loss of <em>dynamic range</em> . In the real world, the output from an older Xerox machine was very high contrast - only black or only white. It had practically no tonal range, but when it output a half black page and a half white page, you got the full dynamic range - several stops worth. You can try that experiment too. Rather easily.</p>

 

<p>B&W film creates tone through what is essentially a halftone process, i.e. silver grains are either there (black) or not (clear base).<br>

Oh...wait...wrong thread ;-)<br>

(Just wanted to say hi to the people in this thread who will get that joke.)</p>

<p>Hey DLT! Good to see you again. Vijay's been up to his old tricks again. Rishi's been no support. So it's just been good ol' me holding the fort for the forces of good. You should pop back into that other thread and see where it's at. I wound up on Vijay's side for a while there, but I'm back in the undecided camp again (I think....).</p>

<p>Rob>>> However, as Vijay noted in above posts, when multiple stops map to the same ADC step, that step should be interpreted as the least of the stops mapped to it; thus (in Bernie's example) the 4-bit ADC still provides information pertaining to whole 8 stop input range.</p>

<p>Thanks for the support (1st paragraph), but I don't agree with this paragraph. This is the reason why: If I handed you an 8 bit print created from an 8bit ADC (with the specs of all my previous examples, including no noise for simplicity), there is no way when you looked at that print that you could tell how many stops where encoded to level 0 (ie. black). The only useful information you could tell from that print is that it had 8 stops of exposure range. It could have been of a 8 stops scene, a 9 stop scene, or a 50 stop scene. There would be no way for you to tell. This is what I have been getting at all along.</p>

<p>Vijay>> Try the experiment I said in my last post. You'll see pure dynamic range at work.</p>

<p>I don't need to try it, as I am in total agreement with you on this example. But once again, pure dynamic range isn't relevant. 'Stops' are.</p>

<p>Vijay>> If you used a 16 bit ADC and mapped 0 to black and 65,525 to white, you'd still see the exact same dynamic range from the paper print.</p>

<p>Yes, but you'd still only see a max of 16 stops represented. (remember Vijay, STOPS is what is important in this discussion).</p>

<p>Vijay>> You want to build, next to each other, a staircase and a ramp that go from the ground to 10 feet up. Your dynamic range is the 10 feet</p>

<p>No it's not. 10/0 = infinity (or 'not defined'). Not that I want to keep using pure engineering terms, but this was to show you a flaw in your analogy.</p>

<p>Now, I want to throw a bit of noise reality into the mix. It's been grand talking all this theoretical stuff, but let's get real for a second shall we:</p>

<p>8 stop scene, 6bit linear ADC (both ADC and photodiodes max out at same voltage). Noise floor say 32 volts, max voltage say 1024. Expose to the right:</p>

<p>input => output<br>

1024V => 63<br>

512V => 32<br>

256V => 16<br>

128V => 8<br>

64V => 4<br>

32V => 2<br>

if noise wasn't present:<br>

16V => 1<br>

8V => 0<br>

4V => 0</p>

<p>Couple of points:</p>

<p>1. Only 5 stops can be accurately and precisely (more-or-less) recorded (6 max minus 1 stop to noise).</p>

<p>2. Dynamic range of the scene (if it is not considered infinity - topic for another discussion) is 1024/4 = 256 (this would be it's range as recorded by the ADC if noise floor was at 4 volts instead)<br>

3. Dynamic range of the image file is 63/2 = 31.5 </p>

<p>4. Dynamic range of the scene and the image file are not equal. What's the thinking on this paradox? I thought ALL dynamic ranges could be recorded by ALL bit depths.</p>

<p>5. I am happy for this example to be modified to include voltage ranges, like Vijay did in his example, BUT..... as long as the max voltage of one stop and the max voltage of the next stop down map to two integer such that one is half of the other (or close enough when rounding is employed). This last point is something that Vijay's example didn't do, which we know to be not reflected with the reality of most dslr's.</p>

<p>I almost want to recant that last post, as I know it is now going to generate a whole lot more discussion about dynamic range, which once again, isn't the relevant point of this debate. I also see a couple of mistakes in there. Dynamic range of the image file will be 63/0 (i am assuming there could be a zero noise signal in there?) = infinity (or 'not defined').</p>

<p>The point I really wanted to make was that this talk of infinite dynamic ranges is not relevant because the scene as recorded by the ADC will have a noise floor. The important point is how bits and noise interact to represent scene dynamic range. I'm bowing out at this point, as I feel my initial argument has been vindicated. You guys/gals can knock yourselves out on pure DR as much as you want now. Cheers.</p>

<p>Very interesting test Mauro.The comparison of the dynamic range between digitals and analogs often is discussed among the photographers.I believe the best side-by-side test would be to see printed photos from the same professional printer but this cannot be done in the web.Dynamic range is one of the weakness of digital representation.Since the sensors even in the newest 24 Mp cameras just... try to immitate the analog 35mm negatives and slides is enough to add it to the cons in the pros/cons chart.Perhaps a medium or large format digital with the extra terrestial price is closer to REALITY.But it's a matter of technology to increase the sampling frequency of the analog signal and achive better results.If it is easy for you do the same test using the 35mm film of this specific test and one of the flagships of Canon, Nikon and Sony cameras up to 24Mp.The final verdict would be very very handy.Till then i don't think there is any consumer or prosumer camera that can compete films.Regards</p>

<p>Vijay,</p>

<p><em>"By your definition of 127 mV - 896 mV as the range. That's really not the standard way to choose range endpoints."</em></p>

<p>I got that from your table. Then I used your data to illustrate a point and you told me that my data was wrong! AAAAAAAAAGH. What's the point?</p>

<p>Best,<br>

Helen</p>

<p>Bernie,</p>

<p>I don't think I could catch up with either thread at this point, so you're going to have to hold the fort against Vijay alone ;-)</p>

<p>Though it would be interesting to try and catch up. More interesting than my work at the moment, but I'm afraid I just can't push the work off.</p>

<p>Vijay,</p>

<p><em>"And I've always said that 4 bits can't provide any more than 4 stops of tonal information and even said that this is absolute, inviolable."</em></p>

<p>There's no such discrete quantity as 'four stops of tonal information' other than a dynamic range. Four stops is a 16:1 ratio - there are an infinite number of possible 'tones' in that ratio, unless you decide to use human perception, when it will depend on the luminance of them, the human etc etc. What you mean is that 4 bits can provide 16 tones. 2 bits will get you a shave and a haircut.</p>

<p>Best,<br>

Helen</p>

<p>Vijay,</p>

<p><em>"First, say you have two points; one is at ground level and one is 10 feet high. You want to build, next to each other, a staircase and a ramp that go from the ground to 10 feet up. Your dynamic range is the 10 feet, regardless of whether you use the ramp; or whether the staircase has 10 steps or 20. They all start at the same point and end at the same point.</em><br>

<em>I can't start the next flight of stairs at the start point of the last step; I have to do it after the step, because if I did, I'd get a step of twice the height. Same for the lowest point. If I wanted to start another flight of stairs to go to the basement, I'd have to start at the beginning point of the first step, not where it ends.</em><br>

<em>That's why my dynamic range measurement has to do with the 10 ft height, and the ground, and I can't choose other points. It is the same points as the ramp begins and ends, obviously.</em><br>

<em>Your question about stops. Say you have 16 steps. You are at the highest point; then you descend one stop, i.e, come down to step 8; then another stop to step 4; then another to step 2, then another to step 1. You're down 4 stops and on step 1. (Step 0 is groundlevel). If at this point, you descend one step, you hit ground straightaway; you can't have 1/2 step or 1/4 step, but this one step could encompass any number of stops or halvings on the ramp."</em></p>

<p>That's a fair analogy, but there's a big difference between a set of stairs and a scene - you are referring to a finite range. You know that the distance from the ground to the top step is 10 ft. Now what happens if the next gap you have to climb is 100 ft? Your metaphorical staircase has no bottom tread and no top tread - they are there thanks to the fact that you know that one is at 0 ft and the other is at 10 ft.</p>

<p>That is why you can't count the upper and lower limits of the input as the limits of the representable dynamic range - they cannot both be known by the ADC. The ADC may only know the top of the lowest riser and the bottom of the top riser. I've already explained why an ADC (CCD-PGA-ADC system) that goes out of range would not work in a photographic application.</p>

<p>Remember that we are talking about a linear system here - in this case the meaning is that the steps are all equal.</p>

<p>Best,<br>

Helen</p>

<p><em>Bernie wrote: If I handed you an 8 bit print created from an 8bit ADC (with the specs of all my previous examples, including no noise for simplicity), there is no way when you looked at that print that you could tell how many stops where encoded to level 0 (ie. black). The only useful information you could tell from that print is that it had 8 stops of exposure range. It could have been of a 8 stops scene, a 9 stop scene, or a 50 stop scene. There would be no way for you to tell. This is what I have been getting at all along.</em></p>

<p>I think this is wrong (and this gets to the heart of the matter). The number of stops I could distinguish is determined by the range that the 8-bit print is capable of displaying. If the 8-bit ADC is fronted by an imaging sensor capable of recording a 50 stop scene and the printer is capable of printing over 50 stops, then the max value that travels through this system will represent a tone that is 50 stops greater than the tone represented by the minimum value, and this will be expressed in the print (regardless of the number of bits in the ADC).</p>