RE: filmscanners: Re: So it's the bits?
Austin wrote: Typically, the image data only falls in part of the range of the CCD, and should be more in the middle, not the ends. Well, that's part of my point. You're suggesting treating the CCD non-linearly it appears. No. I'm saying that the signal to noise ratio changes depending on the input voltage. There is a thought to that, but I will say, that you're probably not going to get any better (read as more usable) information from it...would be my first thought. In my experience there if far less noise in mid tones than dark tones of a slide. I believe that doing either multiple exposures and/or multiple input ranges pretty much does the same thing, doesn't it? That's something else. I've tried to aviod the idea of hardware that is variable because I'm just talking about a given piece of hardware - like my LS30 - which can't vary the input to the D/A. Rob Rob Geraghty [EMAIL PROTECTED] http://wordweb.com
RE: filmscanners: Re: So it's the bits?
Typically, the image data only falls in part of the range of the CCD, and should be more in the middle, not the ends. Well, that's part of my point. You're suggesting treating the CCD non-linearly it appears. No. I'm saying that the signal to noise ratio changes depending on the input voltage. I understand that, and I believe if you think it through, you need to treat this nonlinearly in order for it to be meaningful. You are suggesting that the actual dynamic range of the sensor is non-linear, that it is less at the ends of the scale, and higher in the middle. You get less 'usable' bits on the ends. There is a thought to that, but I will say, that you're probably not going to get any better (read as more usable) information from it...would be my first thought. In my experience there if far less noise in mid tones than dark tones of a slide. No doubt, but that doesn't necessarily translate into more usable information, if the system already is designed to take advantage of it. I believe that doing either multiple exposures and/or multiple input ranges pretty much does the same thing, doesn't it? That's something else. But gives the same results. I believe your idea is really limited by the 'resolution' of the CCD, ie, its ability to discriminate...where noise will eventually overcome any advantage additional bits will give you. I would hope that the number of bits from the A/D 'system' is already matched to the CCD in such a way that 'more' bits wouldn't give you any more usable information.
Re: filmscanners: Re: So it's the bits?
On Sat, 13 Jan 2001 13:18:46 +1100 Julian Robinson ([EMAIL PROTECTED]) wrote: http://www.klt.co.jp/Nikon/Press_Release/ls-4000.html ... Density range 4.2 Hmmm. Except the omission of the word 'optical' is slippery, wibbly-wobbly and misleading - and doubtless deliberate. 'Optical' would tie performance to film. By itself 'Density Range' can mean anything - some internal ability of the scanner, real or theoretical. And that's misdirection, getting the punter to see what they want to see instead of what's there. Regards Tony Sleep http://www.halftone.co.uk - Online portfolio exhibit; + film scanner info comparisons
RE: filmscanners: Re: So it's the bits?
On Fri, 12 Jan 2001 18:21:18 -0500 Austin Franklin ([EMAIL PROTECTED]) wrote: The scanner manufacturers use Dmax as a specification item Used by itself like this, it would be a statment of noise level - ie any higher DMax will be lost in noise, it being below the scanners ability to discriminate. But I think Leaf probably meant to say OD range = 3.7, as most mfr's specify this parameter, meaningless though it is without qualification. Regards Tony Sleep http://www.halftone.co.uk - Online portfolio exhibit; + film scanner info comparisons
RE: filmscanners: Re: So it's the bits?
On Sat, 13 Jan 2001 13:19:01 +1100 Julian Robinson ([EMAIL PROTECTED]) wrote: Nikon may argue that their Dmin is measured with the exposure set low, and Dmax with the exposure set as high as possible. This means that they can get up to another 2 to 4 stops(!!!) into their claimed DENSITY RANGE. Which might explain why they use the term Density Range and not Dynamic Range - Dynamic Range certainly means the range that can be covered without changing the setup i.e. the range available at one instant. Hm. Well spotted! Regards Tony Sleep http://www.halftone.co.uk - Online portfolio exhibit; + film scanner info comparisons
RE: filmscanners: Re: So it's the bits?
Nikon may argue that their Dmin is measured with the exposure set low, and Dmax with the exposure set as high as possible. This means that they can get up to another 2 to 4 stops(!!!) into their claimed DENSITY RANGE. Which might explain why they use the term Density Range and not Dynamic Range - Dynamic Range certainly means the range that can be covered without changing the setup i.e. the range available at one instant. Hm. Well spotted! Tony Sleep But they *do* use Dynamic Range in some of their literature (4.2 in Nikon's product data sheet for the 4000 ED and their model comparison sheet, provided by Nikon at this past week's Mac World).
RE: filmscanners: Re: So it's the bits?
Dynamic Range certainly means the range that can be covered without changing the setup i.e. the range available at one instant. It depends on what you mean by 'changing the setup'. Dynamic range is a system measurement. If the system can provide a particular 'dynamic range' by doing, say, three passes, and varying the input range of the A/D converter therefore taking three measurements, that would certainly expand the dynamic range of the systems output.
Re: filmscanners: Re: So it's the bits?
Hi Austin. Austin Franklin wrote: If you do the math, you'll find that using a 14-bit A/D on most CCD scanners is kind of silly; in such cases, one LSB generally equates to about 10-50 microvolts of signal. How do you work out this figure? I make it more like 170 microvolts, since most CCDs have a saturation voltage in the region of 2.8 volts. I believe both are wrong. The voltage output range of the CCD has to be matched to the input range of the A/D. Of course it does, but the voltage to toggle the LSB of the A/D, *relative to the maximum voltage from the CCD* is one sixteen thousandth of the CCDs maximum voltage, which is about 170 microvolts. It would be wasteful of dynamic range for a scanner designer not to use the CCD close to its saturation voltage. You can NOT associate the volts/bit of the A/D without knowing the input voltage range of the A/D. And you especially can not associate the volts/bit of the output voltage of the CCD without knowing the circuitry between the CCD and the A/D, and then the A/D capture range. I wasn't even thinking of the input of the A/D, just the voltage relative to the CCD maximum output. I've looked at the data sheets of nearly all the currently available CCD linear array sensors, and they really don't vary that much. They saturate at around 2.8 volts and have a dynamic range of around 5000 to 1, whether they're made by Kodak, Sony or NEC. A/D converters vary more, of course, but I don't see what the input voltage has to do with the usability of a 14 bit output. Any A/D converter that wasn't able to 'see' the voltage required to toggle its LSB wouldn't be of much use. The extra bits also give room for the scanner hardware to take advantage of any improvement in sensor technology that may come along, without a major re-build. A bit of 'future proofing' by the circuit designers. Again, I disagree completely, that is not how the system is designed. The CCD has an output voltage range that is then amplified or attenuated and also voltage shifted to match the input voltage range of the A/D. Yes, that's true, but most A/D converters made specifically for scanner use have programmable gain and offset amplifiers built in. Ref: Analog Devices AD9816 and AD9814, and the Texas Instruments series of scanner A to Ds. Anyway, adding a bit of gain is hardly the basis for an entire design philosophy. "Output of CCD = 2.8 volts, input of A/D converter = 4 volts. Ugh! Add gain of 1.4." That's a no-brainer. And doesn't explain why all the scanner manufacturers are now moving to 14 bits. I don't think it's entirely a marketing numbers game. Regards, Pete.
RE: filmscanners: Re: So it's the bits?
Of course it does, but the voltage to toggle the LSB of the A/D, *relative to the maximum voltage from the CCD* When talking about number of volts/bit (technically, volts/code) the measurement is *USUALLY* done relative to the A/D input voltage range...but if you want to reference to the CCD output voltage range, well, OK...as long as that is stated. A/D converters vary more, of course, but I don't see what the input voltage has to do with the usability of a 14 bit output. Everything. If the input voltage to the converter isn't matched to the converter, you will not get the full dynamic range of the converter. Anyway, adding a bit of gain is hardly the basis for an entire design philosophy. "Output of CCD = 2.8 volts, input of A/D converter = 4 volts. Ugh! Add gain of 1.4." That's a no-brainer. Well, it's possibly a little more than that. If the CCD output is 0-2.8V, and the A/D input is -3V to +3V (typically, A/Ds take +- voltage swing), you need to level shift it (negative offset of 1.4V), then apply gain...I don't know the chips you are referring to, but perhaps they handle it effortlessly... Obviously, you have to also consider how much distortion these circuits introduce into the system, they may or may not be very good... It all really depends on how good a system you want to design. And doesn't explain why all the scanner manufacturers are now moving to 14 bits. Possibly because of cost. The lower bit converters aren't available any more, and 14 bits are cheap... Sometimes this happens, where higher spec parts are cheaper than older technology lower spec parts.
RE: filmscanners: Re: So it's the bits?
I don't think anyone commented on my suggestion that a 14 bit A/D still gives more detail in the middle part of the range of values (where colour neg film generally is) precisely because the noise is lowest there? I understand what you are saying, but I don't believe that's what is done, or that it really helps in the way I believe you mean. The A/D will convert the same delta V over the entire capture range. Typically, the image data only falls in part of the range of the CCD, and should be more in the middle, not the ends. Typically, setpoints are applied to the high bit data, and the usable image data is taken only from that range. Typically, also, that usable image data is then mapped into 8 bit data. It is better to apply tonal curves to the high bit data, so you avoid replication of codes, which will manifest themselves as the 'comb effect' seen in the histogram...
RE: filmscanners: Re: So it's the bits?
At 04:44 15/01/01, : [EMAIL PROTECTED] wrote: .Which might explain why they use the term Density Range and not Dynamic Range - Dynamic Range certainly means the range that can be covered without changing the setup i.e. the range available at one instant. Hm. Well spotted! Tony Sleep But they *do* use Dynamic Range in some of their literature (4.2 in Nikon's product data sheet for the 4000 ED and their model comparison sheet, provided by Nikon at this past week's Mac World). Yes I noticed this about 5 stupid minutes after I wrote the first comment! The truth as usual might be more stupidity than conspiracy. Probably there is some serious thinking about spec presentation by technical people arguing with sales people as to what they can get away with, resulting in a finely balanced agreement as to how to phrase this specification. Then somewhere downstream other sales people mess it all up by not appreciating the niceties of what was agreed elsewhere and plonk in the new figure with what they think is a "synonymous" name. Cheers Julian Julian Robinson in usually sunny, smog free Canberra, Australia
RE: filmscanners: Re: So it's the bits?
Austin wrote: I don't think anyone commented on my suggestion that a 14 bit A/D still gives more detail in the middle part of the range of values (where colour neg film generally is) precisely because the noise is lowest there? I understand what you are saying, but I don't believe that's what is done, or that it really helps in the way I believe you mean. The A/D will convert the same delta V over the entire capture range. I understand that the delta V is the same. The point is that the comparison between the signal voltage and delta V is much more favourable in the middle to upper areas of the capture range. Typically, the image data only falls in part of the range of the CCD, and should be more in the middle, not the ends. Well, that's part of my point. Assuming CCD output voltage rises with light intensity, a really dark section of a slide will produce the lowest voltage, which may get lost in the thermal noise of the CCD. The signal to noise ratio should improve as the voltage increases, so that the middle to higher light intensities should produce much more accurate samples than really dark areas. In the case of a neg, the darkest areas (actually the brightest areas of the original scene) are still nowhere as dark as a slide. So the majority of the actual image information of a neg, or the midtones of a slide should be in the best part of the CCD response and have a really good signal to noise ratio - so 14 bit accuracy is actually *useful* for these areas but less so for the more dense areas of the film. In the case of a neg where you want to expand the subtle range of tonal shifts in the neg, surely the more bits the merrier? Rob Rob Geraghty [EMAIL PROTECTED] http://wordweb.com
RE: filmscanners: Re: So it's the bits?
Typically, the image data only falls in part of the range of the CCD, and should be more in the middle, not the ends. Well, that's part of my point. You're suggesting treating the CCD non-linearly it appears. There is a thought to that, but I will say, that you're probably not going to get any better (read as more usable) information from it...would be my first thought. I believe that doing either multiple exposures and/or multiple input ranges pretty much does the same thing, doesn't it?
Re: filmscanners: Re: So it's the bits?
No, they are claiming even more specifically ... and I quote from http://www.klt.co.jp/Nikon/Press_Release/ls-4000.html ... Density range 4.2 Interesting. A Nikon product data sheet for the 4000 ED and model comparison sheet, provided by Nikon at this past week's Mac World, both use the phrase "Dynamic Range" (and give 4.2). However the 8000 ED product sheet reads "Density range 4.2" (though the product comparison sheet that includes the 8000 ED states "Dynamic Range"). -- Bob Shomler http://www.shomler.com/gallery.htm
RE: filmscanners: Re: So it's the bits?
Rafe wrote: Not quite. There's no point going for extra bits, without a corresponding decrease in overall system noise. If the noise is equal to one LSB at 8 bits, then it's 2 LSBs at 9 bits, 4 LSBs at 10 bits, etc. I take your point Rafe, *but* most of the noise in the CCD is when the voltage is lowest. AFAIK there is much more thermal noise and other factors when the input is very dark than at average light intensities. So having more bits to play with over most of the useful range *is* actually useful because it reduces the size of the steps and therefore increases the accuracy of the scan. This is true of most A/D situations - the noise is much more of a problem when the signal is small but *not* when there's a good signal level. So I agree and disagree - you're right in that there will be a lot of dark noise unless more is done to eliminate that noise before it gets to the A/D, but I disagree that more bits are useless for most of the image - quite the contrary. Rob Rob Geraghty [EMAIL PROTECTED] http://wordweb.com
Re: filmscanners: Re: So it's the bits?
Ray wrote: Is there anyone out there other than the participants who has any idea what they are saying? Sorry if the techno-speak is losing people, Ray. If you want a short summary of the most important point it's "Optical Density as quoted by the manufacturer is probably meaningless as a way of choosing a scanner. Testing it yourself is really the best way to know whether it can do what you want it to." Given that, it's a shame that no retailer I'm aware of has a variety of scanners set up so that they *can* be tested by potential purchasers. :( (not in my part of the world anyway!) Rob Rob Geraghty [EMAIL PROTECTED] http://wordweb.com
SV: filmscanners: Re: So it's the bits?
I'm not sure which thread or topic this really is, but since I really need help, I'll ask from my confused heart: Julian wrote: Now, let's unplug the 8 bit D/A, and plug in a 12 bit D/A instead, to the same circuit. My point is - NOTHING CHANGES. Then Ray Amos wrote: Is there anyone out there other than the participants who has any idea what they are saying? For us who think we understand some of it, but lately have started to doubt our earlier pictures of bits and stuff, is there any good place on the web where someone gives a thorough explanation of how D/A converters work - preferably with graphics? Like: what does "linearity" really mean in this particular context? Do more bits really mean finer discrimination between levels? What I think I need is a place that starts from basics but keeps on going to qualified definitions. Not yet hopeless, but lost Ingemar Lindahl
RE: filmscanners: Re: So it's the bits?
Julian wrote: there is a definite limit to dynamic range prescribed by the number of bits. An 8 bit scanner can never do better than a "Dmax" or ~dynamic range of log10 2^8 = 2.4. This is because the lowest usable level "step" has to be around one LSB to be meaningful. OK, I understand what you're saying and the reason for confusion is thinking in linear terms not logarithmic terms. It *is* possible to make a scanner with any dynamic range you want, but if it doesn't have enough bits to represent the values or if it has too much noise, the amount of *useful* information it produces will be minimal. It's possible to work around the 8 bit limitation in an A/D by switching the input, but that's not the point of this discussion. So OK, the number of bits determines the *useful* resolution. Maybe the most useful thing which could be concluded from all this (which I don't think anyone has stated - although I think Julian may have said something similar): The number of bits in the output from a scanner determines the maximum *theoretical* dynamic range of the scanner. Therefore if the manufacturer claims a DR greater than the A/D can produce (of frankly even equal to it), that claim should be considered with a very large grain of salt. So for instance if Nikon claim the LS2000 has 3.6 or more (log(2^12)=3.6), it's dubious. The LS30 may be a special case since the A/D is actually 12 bit but the BIOS drops the 2 least significant bits. Am I making sense now? :) Rob PS My apologies to those increasingly disinterested in this thread - I was hoping I could find a useful, practical conclusion at the end of it. Now that I have one, I will try not to pursue the pedantic details. :) Rob Geraghty [EMAIL PROTECTED] http://wordweb.com
Re: filmscanners: Re: So it's the bits?
There is a legal term for lies in advertising: It's called 'puffery'. In principle, I believe it means that if a claim is one that most people would recognize as nonsense, then it is not a 'crime' or 'tort', for which redress could be obtained in court. If it is a lie (or omission) that most people would NOT recognize immediately, then it is subject to a suit. As Charles Dickens said in Pickwick Papers, "The law sir, is a ass!" At 04:24 PM 01/12/2001 +1000, you wrote: Julian wrote: Actually in thinking about it, it is worse than that - they should be open to consumer legal action. If they state an unqualified figure for Dmax, then when measured by some "reasonable" process it should meet that figure. With the likelihood that it will not, this would mean that they are just plain lying, and therefore should be open to action through consumer protection laws. By the same logic, Microsoft should have been out of business years ago. :) I vote for changing the old saw "Lies, damn lies and statistics" to read "Lies, damn lies, and advertising". :) Read some Dilbert sometime - advertising claims seldom match reality. Rob Rob Geraghty [EMAIL PROTECTED] http://wordweb.com
Re: filmscanners: Re: So it's the bits?
On Thu, 11 Jan 2001 23:27:33 -0500 Ray Amos ([EMAIL PROTECTED]) wrote: Is there anyone out there other than the participants who has any idea what they are saying? :-)) Regards Tony Sleep http://www.halftone.co.uk - Online portfolio exhibit; + film scanner info comparisons
Re: filmscanners: Re: So it's the bits?
On Fri, 12 Jan 2001 12:52:47 +1100 Julian Robinson ([EMAIL PROTECTED]) wrote: If they state an unqualified figure for Dmax, then when measured by some "reasonable" process it should meet that figure. With the likelihood that it will not, this would mean that they are just plain lying, and therefore should be open to action through consumer protection laws. But they aren't AFAIK claiming a DMax figure, nor even an OD range (DMax-DMin), but a wibbly-wobbly bit of slipperiness called 'dynamic range'. Really this is all horribly reminiscent of output power specs for HiFi amps - 'RMS', 'Music Power', 'Peak' and so on, all gibberish without qualifying terms. Caveat emptor! Regards Tony Sleep http://www.halftone.co.uk - Online portfolio exhibit; + film scanner info comparisons
RE: filmscanners: Re: So it's the bits?
Colour could be relevent if the sensor has poor sensitivity to a particular frequency range, or produces more noise in that range (eg. blue, which I often hear contains more noise than other channels :). Very true, but you have to believe the manufacturer is going to use CCDs that don't have these problems. No, that's wrong (and you were doing so well ;-). Dynamic range IS resolution over ANY range, and 8 bits won't give you a DMax of 4. In fact, 8 bits is 48db, or a DMax of log10(2^8) or 2.4. Perhaps you are confusing the meaning of DMax? Maybe I am. I thought that the optical density range was a logarithmic conversion of light intensity. If it is, I don't see any reason why an 8 bit system can't represent an optical density of 4 using a value of 255. Or zero, and represent an OD of 0 with 255. They're arbitrary numbers representing something in the analogue realm. All the number of bits determines is the number of steps between the maximum and minimum values converted. I fail to see why the bit depth is necessarily related directly to the analogue realm - it depends entirely on how one is mapped to the other. Yes, it appears you are confused about what DMax is. There are two 'properties' to the system we are discussing. One is the voltage range, and as you say, that, technically, could be represented by any number of bits. Second is the ability to discriminate within that voltage range, which is 'resolution'and that is what DMax is. DMax is relative in and of itself. A voltage range of +1V and -1V (2V range) with 2 bits (4 values) has the exact same DMax as a voltage range of +20V and -20V with 2 bits, yet their respective ranges are different, and each bit of each range represents a completely different voltage. In one it is .5V, and in the other it is 5V. The output data from the scanner does not care a wit about voltages. What it cares about is the ability to 'discriminate' as many values as possible (DMax), and they are all relative.
Re: filmscanners: Re: So it's the bits?
- Original Message - From: "Austin Franklin" [EMAIL PROTECTED] To: [EMAIL PROTECTED] Sent: Friday, January 12, 2001 8:10 AM Subject: RE: filmscanners: Re: So it's the bits? Austin: I Have been following this thread with some intrest and your example below ties the thread up into a nice clean package. Thanks for the use of your knowledge and expertise. John Horton snip Yes, it appears you are confused about what DMax is. There are two 'properties' to the system we are discussing. One is the voltage range, and as you say, that, technically, could be represented by any number of bits. Second is the ability to discriminate within that voltage range, which is 'resolution'and that is what DMax is. DMax is relative in and of itself. A voltage range of +1V and -1V (2V range) with 2 bits (4 values) has the exact same DMax as a voltage range of +20V and -20V with 2 bits, yet their respective ranges are different, and each bit of each range represents a completely different voltage. In one it is .5V, and in the other it is 5V. The output data from the scanner does not care a wit about voltages. What it cares about is the ability to 'discriminate' as many values as possible (DMax), and they are all relative.
RE: filmscanners: Re: So it's the bits?
Austin writes ... Yes, it appears you are confused about what DMax is. ... Second is the ability to discriminate within that voltage range, which is 'resolution'and that is what DMax is. DMax is relative in and of itself. ... And you have always been such a stickler for definitions. When did Dmax (and Dmin) leave its proper context ... a property of the film only? shAf :o)
Re: filmscanners: Re: So it's the bits?
On Fri, 12 Jan 2001 12:52:47 +1100 Julian Robinson ([EMAIL PROTECTED]) wrote: If they state an unqualified figure for Dmax, then when measured by some "reasonable" process it should meet that figure. With the likelihood that it will not, this would mean that they are just plain lying, and therefore should be open to action through consumer protection laws. I may be wrong, but isn't this a figure of merit that existed for drum scanners, even before there were scanners at all for consumers? I know that books compare this figure of merit for commercial drum scanners vs. home and office ones with this being one of the biggie differences. Further, I think it's a reasonable named spec, even if odd, when taking into account the name of a spec should be shorter than an explanatory paragraph. Talk about lawsuits over un-understandable technical gibberish by the mfgr (in U.S., consumers can't even understand how to work a punchcard ballot and will sue as a result). The spec spec's (AFAIK) resolution of density (whether linear or not...). The term "resolution" is already used (and has the similar faults: "input?", "output?", "optical?" etc.), so something else is needed to eliminate (ha!) confusion. :-) Although the spec dynamic-range/Dmax spec may be foo-foo'd by some here, I'll bet everyone here wants their scanner to have this number be high on their unit! I know I do. Mike K.
RE: filmscanners: Re: So it's the bits?
Austin writes ... Yes, it appears you are confused about what DMax is. ... Second is the ability to discriminate within that voltage range, which is 'resolution'and that is what DMax is. DMax is relative in and of itself. ... And you have always been such a stickler for definitions. When did Dmax (and Dmin) leave its proper context ... a property of the film only? Scanners use DMax as part of their specs, which is what this whole discussion has been about. What's your point?
RE: filmscanners: Re: So it's the bits?
Austin writes ... Scanners use DMax as part of their specs, which is what this whole discussion has been about. What's your point? I've never seen any scanner use Dmax as a spec ... they often quote "dynamic range" or "optical density" but Dmax and Dmin are absolute terms for the densest and thinnest areas of film. "DR" and "OD", the ability of measuring Dmax - Dmin, are also absolute terms. Dmax, historically, has always been an absolute term ... my point being claiming it is a relative term, no matter the context, is a mis-use and confusing application of the term. shAf :o)
RE: filmscanners: Re: So it's the bits?
I've never seen any scanner use Dmax as a spec ... The Leafscan 45 has listed right in their brochure, and I quote: "Dynamic range: 5000:1 or 3.7 Dmax" So, obviously at least one manufacturer does use Dmax as a spec, and I am sure others do also. the ability of measuring Dmax - Dmin, are also absolute terms. Dmax, historically, has always been an absolute term ... my point being claiming it is a relative term, no matter the context, is a mis-use and confusing application of the term. The pixel values (for which the range of is the theoretically highest Dmax for the scanner) are relative to each other, not absolute, in the systems we are talking about. You can not say one pixel value is differentiated from another by N volts without knowing what the voltage range is. The voltage ranges in different systems vary, and yet they can have the same Dmax. In fact, the voltage range is irrelevant, and the only thing that is relevant is the Dmax, or dynamic range or what ever you want to call it...
RE: filmscanners: Re: So it's the bits?
Austin writes ... The pixel values (for which the range of is the theoretically highest Dmax for the scanner) are relative to each other, not absolute, ... Correct ... the "pixel values" associated with measuring Dmax may be relative ... but "Dmax" is a measured value, is absolute, and belongs to film. Small point, but let's not confuse terms. shAf :o)
Re: filmscanners: Re: So it's the bits?
Hi Rafe. rafeb wrote: The only reason I can see that a greater number of bits would help is that when you are at the extremities of the CCD's range, more bits should help resolve meaningful data from noise, or by reducing the size of the steps, reduce the loss of image information which lies between the steps at a lower bit depth. Not quite. There's no point going for extra bits, without a corresponding decrease in overall system noise. If the noise is equal to one LSB at 8 bits, then it's 2 LSBs at 9 bits, 4 LSBs at 10 bits, etc. Ditto, not quite. You're assuming that all the noise is generated in the CCD and analogue stages. In fact, bit dither in the A/D conversion can make a significant contribution to the overall noise figure. Techniques like correlated double sampling can significantly reduce the contribution of analogue noise, whereas the digital noise, or uncertainty, is always going to be in the region of plus or minus 1 LSB. Each extra bit takes the digital noise down by -6dB. A smaller voltage step between bit levels also means that the analogue noise isn't excessively amplified by causing spurious digital bit changes. For instance, noise that causes an oscillation between bit levels in an 8 bit converter will always look as if it has an ampitude of 1/256th of the maximum voltage input. The same noise in a 14 bit converter *might* look 64 times smaller. It will certainly look closer to its real value. If you do the math, you'll find that using a 14-bit A/D on most CCD scanners is kind of silly; in such cases, one LSB generally equates to about 10-50 microvolts of signal. How do you work out this figure? I make it more like 170 microvolts, since most CCDs have a saturation voltage in the region of 2.8 volts. Anyway, microvolt signal levels are no big deal as long as the source impedance is kept low. I'm not saying that 14 bit A/Ds can be used to their full advantage by any means, but their use isn't entirely wasted. The range of the signal from a CCD amounts to about 12 bits, so the last useable 12 dB (0.6D) causes a change of 16 levels, not just 4, as would be the case with a 12 bit A/D. The extra bits also give room for the scanner hardware to take advantage of any improvement in sensor technology that may come along, without a major re-build. A bit of 'future proofing' by the circuit designers. I think I might detect the signs of such a change in the air. Sony have recently cut their range of linear CCD sensors quite drastically, perhaps to make room for something better? And nearly all the new scanners are coming out with 14 bit A/D converters. Maybe I'm just putting 2 and 2 together and making 5. Regards, Pete.
RE: filmscanners: Re: So it's the bits?
The pixel values (for which the range of is the theoretically highest Dmax for the scanner) are relative to each other, not absolute, ... Correct ... the "pixel values" associated with measuring Dmax may be relative ... but "Dmax" is a measured value, is absolute, and belongs to film. Small point, but let's not confuse terms. The scanner manufacturers use Dmax as a specification item, which you said they didn't, but they do. We were talking about that, not a wit about film. We were talking about how many bits correspond to the different values of Dmax (amongst many other things), and that is NOT measured. I don't know what your point is, but I suggest you go read the thread, and if you take issue with the term "Dmax" used with a scanner, than take it up with the scanner manufacturers.
Re: filmscanners: Re: So it's the bits?
At 11:19 PM 1/12/01 +, Pete wrote: I'm not saying that 14 bit A/Ds can be used to their full advantage by any means, but their use isn't entirely wasted. The range of the signal from a CCD amounts to about 12 bits, so the last useable 12 dB (0.6D) causes a change of 16 levels, not just 4, as would be the case with a 12 bit A/D. The extra bits also give room for the scanner hardware to take advantage of any improvement in sensor technology that may come along, without a major re-build. A bit of 'future proofing' by the circuit designers. My primary peeve is that the number-of-bits in the A/D converter is used as a selling point by the manufacturers, and folks feel smug about the fact that they're sending 42-bit images to Photoshop, without really thinking things through. (Without considering, for example, that the four or five LS bits may be pure noise.) The extra bits generally don't hurt anything, aside from consuming bandwidth, power, memory, CPU cycles, etc. But they don't necessarily help much either, if the noise is well in excess of 1 LSB. Of course, it may be possible to trade bandwidth for lower noise, and the extra bits may help with that. I suspect the mfgrs are using 14-bit converters these days mostly because they've gotten very inexpensive. And "14-bits" looks good on the side of the box and in the specs. I hope you're right about pending improvements in sensor technology. That would be nice. Though I wonder if the bulk of research now isn't in area sensors (for digicams) rather than the linear sensors used in film scanners. The digicam market is much larger, I think. PS: My figure of 10-50 uV for a 14-bit step was based on a recent measurement from a cheap flatbed scanner at work. rafe b.
RE: filmscanners: Re: So it's the bits?
At 10:21 13/01/01, Austin wrote: The pixel values (for which the range of is the theoretically highest Dmax for the scanner) are relative to each other, not absolute, ... Correct ... the "pixel values" associated with measuring Dmax may be relative ... but "Dmax" is a measured value, is absolute, and belongs to film. Small point, but let's not confuse terms. The scanner manufacturers use Dmax as a specification item, which you said they didn't, but they do. We were talking about that, not a wit about film. We were talking about how many bits correspond to the different values of Dmax (amongst many other things), and that is NOT measured. Like most specification stuff, nothing is clear cut and manufacturers adopt shorthand methods of describing things - which is fine if everyone understands and agrees. In a former life I wrote specs for radars and processing systems, and wrote and assessed tenders for same so I have participated at first hand in the gamesmanship of manipulating specs. This explains why I am in my element here, and apologies to those who are not. Scanner Dmax, for better or worse, is often used as a shorthand for "Density Range" or "Dynamic Range". This doesn't seem too incomprehensible or even reprehensible to me, since the figures must be close, because Dmin is pretty close to "no film at all" .I mean... Dynamic range (or density range) = Dmax-Dmin where Dmax is the maximum film density that can be measured by the scanner, Dmin the minimum Since clear film (fully exposed slide) is almost transparent, Dmin is close to zero, so making an assumption that the scanner is set so that it can just record Dmin (by adjusting exposure), then density range = Dmax-Dmin ~= Dmax - 0 = Dmax Cheers, Julian PS There is another issue that comes up here - I have assumed that Dynamic range (which until now I would say is the same thing as density range) is Dmax - Dmin where you measure Dmax and Dmin _with_the_same_setup_ - that is, during the one scan. Nikon may argue that their Dmin is measured with the exposure set low, and Dmax with the exposure set as high as possible. This means that they can get up to another 2 to 4 stops(!!!) into their claimed DENSITY RANGE. Which might explain why they use the term Density Range and not Dynamic Range - Dynamic Range certainly means the range that can be covered without changing the setup i.e. the range available at one instant. So I can see it is quite possible that Nikon MAY be able to argue that they cover a Density Range of 3.6 for the LS2000 or 4.2 for the LS4000, although you have to do a couple of separate scans to see it, which is not quite what you would want and certainly not what people are assuming when they read the spec. The mere presence of exposure controls on the Nikon scanner tends to support this idea. So the LS2000 MAY in fact have a density range of 3.6, but it's Dynamic Range could still be 2 (or is it 4) stops less than this - i.e. 3.0 or 2.4. Is it coincidence that most the measurements I have seen are in this range, from memory about 2.6? (I assume people have been measuring Dynamic Range, not Density Range). Julian Robinson in usually sunny, smog free Canberra, Australia
Re: filmscanners: Re: So it's the bits?
At 01:09 13/01/01, Tony wrote: But they aren't AFAIK claiming a DMax figure, nor even an OD range (DMax-DMin), but a wibbly-wobbly bit of slipperiness called 'dynamic range'. Really this is all horribly reminiscent of output power specs for HiFi amps - 'RMS', 'Music Power', 'Peak' and so on, all gibberish without qualifying terms. Caveat emptor! No, they are claiming even more specifically ... and I quote from http://www.klt.co.jp/Nikon/Press_Release/ls-4000.html ... Density range 4.2 ... Contrary to the view put by others (that I am being naive in expecting some vague truth in advertising and that there is no way any action would be successful), there have been some landmark successes in recent times in which advertisers were prevented from lying, some even had to repay money. I am talking Australia here and have no idea what goes on in litigation-central USA or the UK. Since usable density range is one of the single most important characteristics of a scanner, and hence a characteristic which is (or should be) involved in everyone's decision making process when buying, consumers have more than the usual right to know a vaguely defensible (by measurement) figure. I will write to Nikon - whether or not they listen to me I really doubt that this claim will remain for long in these litigatious "truth-in-advertising" times, unless it can be substantiated. I am sure you and others will disagree, but no harm in hoping. Of course there is always the possibility that the useful density range of this scanner _is_ 4.2, in which case I will be very pleased to have Nikon let me know this fact, and be one of the first to line up and buy, even if I have to sell my ... um ... ... house? I remember the Peak Music Power days and used to indulge in a bit of hi-fi salesman baiting on this topic. Often good fun on a hot Saturday afternoon, hi-fi shops being air-conditioned. Anyone want do discuss the crystal clarity of music if you use oxygen-free speaker cables? You know of course that in "ordinary" speaker cables the oxygen molecules get in the way of the electrons, causing them to slow down and rattle around, so the music comes out "muffled". I LOVE hi-fi salesmen. :)!! Julian [This PS is relevant and is copied form another post I just wrote after having a Revelation.] PS There is another issue that comes up here - I have assumed that Dynamic range (which until now I would say is the same thing as density range) is Dmax - Dmin where you measure Dmax and Dmin _with_the_same_setup_ - that is, during the one scan. Nikon may argue that their Dmin is measured with the exposure set low, and Dmax with the exposure set as high as possible. This means that they can get up to another 2 to 4 stops(!!!) into their claimed DENSITY RANGE. Which might explain why they use the term Density Range and not Dynamic Range - Dynamic Range certainly means the range that can be covered without changing the setup i.e. the range available at one instant. So I can see it is quite possible that Nikon MAY be able to argue that they cover a Density Range of 3.6 for the LS2000 or 4.2 for the LS4000, although you have to do a couple of separate scans to see it, which is not quite what you would want and certainly not what people are assuming when they read the spec. The mere presence of exposure controls on the Nikon scanner tends to support this idea. So the LS2000 MAY in fact have a density range of 3.6, but it's Dynamic Range could still be 2 (or is it 4) stops less than this - i.e. 3.0 or 2.4. Is it coincidence that most the measurements I have seen are in this range, from memory about 2.6? (I assume people have been measuring Dynamic Range, not Density Range). Julian Robinson in usually sunny, smog free Canberra, Australia
RE: filmscanners: Re: So it's the bits?
snip PS There is another issue that comes up here - I have assumed that Dynamic range (which until now I would say is the same thing as density range) is Dmax - Dmin where you measure Dmax and Dmin _with_the_same_setup_ - that is, during the one scan. I brought up this point a while ago, I believe...and that is the Leafscan uses a 12 bit converter, but does the conversion at three different ranges, from which it gets basically a 16 bit result. If a design uses this multiple range technique, you can't just go by the number of bits in the converter to figure out what the dynamic range of the system is without knowing what the multiple ranges add to the picture (no pun intended).
Re: filmscanners: Re: So it's the bits?
Julian wrote: Because it is an 8-bit D/A, the lowest level we can read is 2^8 lower than 1024 = 1024/256 = 4mV. This is the value of one least significant bit (LSB). Also, let's assume that this is an optimally engineered 8-bit system. Because it is optimally engineered, let's say that the CCD/amplifier noise is a quarter of the LSB level i.e. 1mV RMS. edited here Now, let's unplug the 8 bit D/A, and plug in a 12 bit D/A instead, to the same circuit. My point is - NOTHING CHANGES. Won't the 12bit a/d converter allow the information between 4mv and the 1mv noise level to be resolved? Colin Maddock
Re: filmscanners: Re: So it's the bits?
"Colin Maddock" [EMAIL PROTECTED] wrote: Won't the 12bit a/d converter allow the information between 4mv and the 1mv noise level to be resolved? It may, but I think Julian's point is valid which is that for a given sensitivity from the analog circuitry, changing the A/D won't make any difference to the density ranges that the analog circuitry resolves. It only increases the accuracy with which we read the range of analog values that the CCD *does* resolve. Rob
RE: filmscanners: Re: So it's the bits?
Because it is an 8-bit D/A, the lowest level we can read is 2^8 lower than 1024 = 1024/256 = 4mV. The number of bits has NOTHING to do with what voltage it can read. Different converters have different voltage ranges, AND the input voltage range can be changed via an analog front end to the converter. Typical converters have a voltage range from +3 to -3 volts, or a 6V swing. An 8 bit converter would therefore have 6/256 or .023V or 23mV resolution per bit.
RE: filmscanners: Re: So it's the bits?
for a given sensitivity from the analog circuitry, changing the A/D won't make any difference to the density ranges that the analog circuitry resolves. It only increases the accuracy with which we read the range of analog values that the CCD *does* resolve. May be I'm slow today...but that paragraph is really unclear to me, and I know this stuff quite well. What exactly do you mean by 'for a given sensitivity from the analog circuitry'? Sensitivity can describe one of many characteristics, so this seems ambiguous. Also, what do you mean by 'to the density ranges that the analog circuitry resolves'. The above paragraph seems to intermix (confuse) different concepts/terms, and really comes across, at least to me, as not very comprehdable. I don't think 'resolves' is the right word there. Also, 'analog values that the CCD *does* resolve'? Again, resolve doesn't really sound right here... The only thing in the described system that is 'resolved' is the A/D.
Re: filmscanners: Re: So it's the bits?
On Thu, 11 Jan 2001 13:06:00 +1100 Julian Robinson ([EMAIL PROTECTED]) wrote: My conclusion from all this is that the manufacturers cheat by saying that the Dmax is defined by the D/A resolution as a shorthand, which is true if the IMPLICATION which follows is that the rest of the system is engineered so that the whole of the D/A range is useful - that is, that the noise level is significantly less than one LSB. I don't think that this implied design constraint is true, at least in the case of consumer/semi-pro scanners. Yup. It's self-evidently not, else we'd not see any noise, whereas with most scanners we see rather more than we'd like, at least until cancellation by multiscanning is applied. Hopefully, these are very good scanners, and there's no reason to think they aren't. But Nikon's figures, unqualified as they are, tell us absolutely nothing useful at all, except that someone in marketing thinks we're a bit gullible. Of course if they read lists like this, they'd know better :) Regards Tony Sleep http://www.halftone.co.uk - Online portfolio exhibit; + film scanner info comparisons
RE: filmscanners: Re: So it's the bits?
Paragraph is clear enough for me to understand. And is perfectly correct to my judgement. Slava --- Austin Franklin [EMAIL PROTECTED] wrote: for a given sensitivity from the analog circuitry, changing the A/D won't make any difference to the density ranges that the analog circuitry resolves. It only increases the accuracy with which we read the range of analog values that the CCD *does* resolve. May be I'm slow today...but that paragraph is really unclear to me, and I know this stuff quite well. What exactly do you mean by 'for a given sensitivity from the analog circuitry'? Sensitivity can describe one of many characteristics, so this seems ambiguous. Also, what do you mean by 'to the density ranges that the analog circuitry resolves'. The above paragraph seems to intermix (confuse) different concepts/terms, and really comes across, at least to me, as not very comprehdable. I don't think 'resolves' is the right word there. Also, 'analog values that the CCD *does* resolve'? Again, resolve doesn't really sound right here... The only thing in the described system that is 'resolved' is the A/D. = --- NOTE: EMAIL HAS CHANGED !!! - Slava Zilberfayn| Home +1(416)7838430 | Work +1(416)5931122x2486 EMAIL: [EMAIL PROTECTED]OR [EMAIL PROTECTED] ADDRESS: appt 1219, 377 Ridelle ave, M6B1K2, Toronto, ON, CANADA ___ Do You Yahoo!? Get your free @yahoo.ca address at http://mail.yahoo.ca
RE: filmscanners: Re: So it's the bits?
It appears to me the word 'sensitivity' was meant as 'range'. Sensitivity of an analog system is the rate of change. A higher bit A/D could give higher sensitivity, but would not give a better range (which I believe is what the paragraph was trying to say), since the range is fixed for a given system. Also, the analog front end filter for the A/D does not 'resolve' anything. The CCD doesn't 'resolve' either. The only thing doing any 'resolving' in the described system is the A/D. Also, it is not a given that a higher bit A/D, in a particular system, will improve accuracy, though it MAY, depending on the system. Where noise in the system is greater than the resolution of the A/D, it will not. -Original Message- Paragraph is clear enough for me to understand. And is perfectly correct to my judgement. Slava --- Austin Franklin [EMAIL PROTECTED] wrote: for a given sensitivity from the analog circuitry, changing the A/D won't make any difference to the density ranges that the analog circuitry resolves. It only increases the accuracy with which we read the range of analog values that the CCD *does* resolve. May be I'm slow today...but that paragraph is really unclear to me, and I know this stuff quite well. What exactly do you mean by 'for a given sensitivity from the analog circuitry'? Sensitivity can describe one of many characteristics, so this seems ambiguous. Also, what do you mean by 'to the density ranges that the analog circuitry resolves'. The above paragraph seems to intermix (confuse) different concepts/terms, and really comes across, at least to me, as not very comprehdable. I don't think 'resolves' is the right word there. Also, 'analog values that the CCD *does* resolve'? Again, resolve doesn't really sound right here... The only thing in the described system that is 'resolved' is the A/D.
RE: filmscanners: Re: So it's the bits?
Austin wrote: May be I'm slow today...but that paragraph is really unclear to me, and I know this stuff quite well. What exactly do you mean by 'for a given sensitivity from the analog circuitry'? OK, let me put it another way and try to avoid some of the ambiguous terms. You have an analogue circuit which starts with a CCD and produces meaningful variations in voltage for a certain minimum and a certain maximum light intensity (I'm ignoring colour for simplicity's sake). This voltage is fed into an analog to digital converter (ADC). The real minimum and maximum light intensities which the circuit can resolve into a digital number is determined by the analog CCD circuit NOT by the number of bits used in the ADC. The number of bits in the ADC *only* determines the number of steps between the minimum voltage it can resolve into a number and the maximum voltage it can resolve into a number. The only reason I can see that a greater number of bits would help is that when you are at the extremities of the CCD's range, more bits should help resolve meaningful data from noise, or by reducing the size of the steps, reduce the loss of image information which lies between the steps at a lower bit depth. Ultimately, it's the CCD circuitry which determines the minimum and maximum light intensities that the scanner can (in theory) resolve, not the number of bits used to convert it to a digital image which just determines the smoothness of the conversion. What Ed's demonstration of a relationship between bits and dynamic range demonstrated to me was that the numbers for some scanners simply *seem* to be the theoretical maximum determined by a mathematical relationship from a given bit depth *not* a real measurement of the DR as determined by the kind of tests Dave Hemingway described. In fact, referring back to my argument above, there's no reason why an 8 bit per channel scanner couldn't have a dynamic range of (say) 4 if the analog circuitry is capable of measuring that range of light intensities. Rob Rob Geraghty [EMAIL PROTECTED] http://wordweb.com
RE: filmscanners: Re: So it's the bits?
Rob, I agree with what you wrote, except that having read some of Tony's old posts I think this last point quoted below is not true - rather, there is a definite limit to dynamic range prescribed by the number of bits. An 8 bit scanner can never do better than a "Dmax" or ~dynamic range of log10 2^8 = 2.4. This is because the lowest usable level "step" has to be around one LSB to be meaningful. Even introducing offsets to set the LSB decision point to the top of the noise level doesn't help you because the next usable level is still one bit away, and it is this step height which effectively sets the minimum usable level, not the threshold you set the LSB decison point to. This is a change of view for me, I started out the other way round! Julian At 10:53 12/01/01, you wrote: In fact, referring back to my argument above, there's no reason why an 8 bit per channel scanner couldn't have a dynamic range of (say) 4 if the analog circuitry is capable of measuring that range of light intensities. Julian Robinson in usually sunny, smog free Canberra, Australia
RE: filmscanners: Re: So it's the bits?
Austin this was an ILLUSTRATION, not based on an actual D/A - I was using an illustrative range of 0 -1024mV just to make a point which is valid whatever range you choose. I could have talked about -3 to +3 V but the point would have been even more obscure than it already is. As you point out and as I mentioned, there is no problem introducing an offset and/or amplification to give any range to a D/A, or to modify linearity through log pre-amps etc, but it is irrelevant and confusing when related to my main point. The detail of voltage ranges is not important to the principle that determines measurable density ranges, and in fact I don't think linearity is relevant either. Julian R At 01:50 12/01/01, you wrote: Because it is an 8-bit D/A, the lowest level we can read is 2^8 lower than 1024 = 1024/256 = 4mV. The number of bits has NOTHING to do with what voltage it can read. Different converters have different voltage ranges, AND the input voltage range can be changed via an analog front end to the converter. Typical converters have a voltage range from +3 to -3 volts, or a 6V swing. An 8 bit converter would therefore have 6/256 or .023V or 23mV resolution per bit. Julian Robinson in usually sunny, smog free Canberra, Australia
RE: filmscanners: Re: So it's the bits?
Ditto. Frank Paris [EMAIL PROTECTED] http://albums.photopoint.com/j/AlbumList?u=62684 -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]]On Behalf Of Viacheslav Zilberfayn Sent: Thursday, January 11, 2001 1:12 PM To: [EMAIL PROTECTED] Subject: RE: filmscanners: Re: So it's the bits? Paragraph is clear enough for me to understand. And is perfectly correct to my judgement. Slava --- Austin Franklin [EMAIL PROTECTED] wrote: for a given sensitivity from the analog circuitry, changing the A/D won't make any difference to the density ranges that the analog circuitry resolves. It only increases the accuracy with which we read the range of analog values that the CCD *does* resolve. May be I'm slow today...but that paragraph is really unclear to me, and I know this stuff quite well. What exactly do you mean by 'for a given sensitivity from the analog circuitry'? Sensitivity can describe one of many characteristics, so this seems ambiguous. Also, what do you mean by 'to the density ranges that the analog circuitry resolves'. The above paragraph seems to intermix (confuse) different concepts/terms, and really comes across, at least to me, as not very comprehdable. I don't think 'resolves' is the right word there. Also, 'analog values that the CCD *does* resolve'? Again, resolve doesn't really sound right here... The only thing in the described system that is 'resolved' is the A/D. = --- NOTE: EMAIL HAS CHANGED !!! - Slava Zilberfayn| Home +1(416)7838430 | Work +1(416)5931122x2486 EMAIL: [EMAIL PROTECTED]OR [EMAIL PROTECTED] ADDRESS: appt 1219, 377 Ridelle ave, M6B1K2, Toronto, ON, CANADA ___ Do You Yahoo!? Get your free @yahoo.ca address at http://mail.yahoo.ca
RE: filmscanners: Re: So it's the bits?
(I'm ignoring colour for simplicity's sake). Color isn't relevant. The sensor doesn't have any color information, only intensity information. The color is deterministic...ie, a particular sensor has a particular color filter over it. The real minimum and maximum light intensities which the circuit can resolve into a digital number is determined by the analog CCD circuit NOT by the number of bits used in the ADC. True. And therefore this 'range', as it is typically called, can be represented by one bit or 1000 bits. Which, I believe, was the point that someone else was making... The number of bits in the ADC *only* determines the number of steps between the minimum voltage it can resolve into a number and the maximum voltage it can resolve into a number. Basically, true. Typically, that 'number of steps' is called the resolution, ie, if you have an 8 volt range, and you have three bits (therefore can represent 8 different values), your resolution is 1V. The only reason I can see that a greater number of bits would help is that when you are at the extremities of the CCD's range, more bits should help resolve meaningful data from noise, or by reducing the size of the steps, reduce the loss of image information which lies between the steps at a lower bit depth. I don't follow you here. Ultimately, it's the CCD circuitry which determines the minimum and maximum light intensities that the scanner can (in theory) resolve, not the number of bits used to convert it to a digital image Agreed... which just determines the smoothness of the conversion. I know what you mean, but that's the first time I've heard it called that... What Ed's demonstration of a relationship between bits and dynamic range demonstrated to me was that the numbers for some scanners simply *seem* to be the theoretical maximum determined by a mathematical relationship from a given bit depth *not* a real measurement of the DR as determined by the kind of tests Dave Hemingway described. I was saying somewhat the same thing, but apparently for different reasons? In fact, referring back to my argument above, there's no reason why an 8 bit per channel scanner couldn't have a dynamic range of (say) 4 if the analog circuitry is capable of measuring that range of light intensities. No, that's wrong (and you were doing so well ;-). Dynamic range IS resolution over ANY range, and 8 bits won't give you a DMax of 4. In fact, 8 bits is 48db, or a DMax of log10(2^8) or 2.4. Perhaps you are confusing the meaning of DMax? The 'capture' range (range from min voltage to max voltage), as you have aptly stated, is completely separate from resolution (within that capture range). Just a note on 'resolution'. Most people consider resolution to be, say, the number of pixels wide the scan is times the number of pixels long the scan is. That is true (though only by common use, not by correct use ;-), but there are other resolutions. Consider that X and Y resolution. There is also Z resolution, which is the bit depth of the pixel, more bits is higher resolution. That is the resolution I am talking about in this discussion, not the X by Y resolution of the image.
Re: filmscanners: Re: So it's the bits?
At 05:58 12/01/01, Tony wrote: . But Nikon's figures, unqualified as they are, tell us absolutely nothing useful at all, except that someone in marketing thinks we're a bit gullible. Of course if they read lists like this, they'd know better :) Actually in thinking about it, it is worse than that - they should be open to consumer legal action. If they state an unqualified figure for Dmax, then when measured by some "reasonable" process it should meet that figure. With the likelihood that it will not, this would mean that they are just plain lying, and therefore should be open to action through consumer protection laws. I agree that the LS2000 does not appear to meet it's stated figure just by observing significant amounts of noise in my scans, as you state in another post. So why do they get away with making what seems to be a plain untrue statement which is designed to affect a would-be purchaser's decision-making? I assume it is not just Nikon who make outlandish claims, but from what we have heard from Polaroid, they may be a more conservative and actually have some more-or-less justifiable measurement basis for their lower claims to date. Julian Julian Robinson in usually sunny, smog free Canberra, Australia
RE: filmscanners: Re: So it's the bits?
At 09:53 AM 1/12/01 +1000, Rob wrote: The only reason I can see that a greater number of bits would help is that when you are at the extremities of the CCD's range, more bits should help resolve meaningful data from noise, or by reducing the size of the steps, reduce the loss of image information which lies between the steps at a lower bit depth. Not quite. There's no point going for extra bits, without a corresponding decrease in overall system noise. If the noise is equal to one LSB at 8 bits, then it's 2 LSBs at 9 bits, 4 LSBs at 10 bits, etc. If you do the math, you'll find that using a 14-bit A/D on most CCD scanners is kind of silly; in such cases, one LSB generally equates to about 10-50 microvolts of signal. A highly dynamic, time- variant signal, at that. One way to reduce the noise is averaging -- ie., scanning with several passes, and averaging the results. Alternatively, just scan more slowly, and accumulate a larger charge in each CCD cell. Of course, the CCD and the analog front end must allow for this without saturation. It's true that at the shoulder and toe of the film's response curve, those extra bits -- if they were real -- would be most useful and welcome. rafe b.
Re: filmscanners: Re: So it's the bits?
Colin Maddock wrote: Julian wrote: Because it is an 8-bit D/A, the lowest level we can read is 2^8 lower than 1024 = 1024/256 = 4mV. This is the value of one least significant bit (LSB). Also, let's assume that this is an optimally engineered 8-bit system. Because it is optimally engineered, let's say that the CCD/amplifier noise is a quarter of the LSB level i.e. 1mV RMS. edited here Now, let's unplug the 8 bit D/A, and plug in a 12 bit D/A instead, to the same circuit. My point is - NOTHING CHANGES. Won't the 12bit a/d converter allow the information between 4mv and the 1mv noise level to be resolved? Colin Maddock Is there anyone out there other than the participants who has any idea what they are saying? Ray Amos
filmscanners: Re: So it's the bits?
Frank Paris asks: Where along the path from sensor to scanned image is the mapping performed that actually corresponds to the psychological way we perceive brightness levels? When the gamma correction is applied, I think. Colin Maddock
filmscanners: Re: So it's the bits?
Tony thanks for pointing me to the archives, which I have now read. (If anyone is interested, the topic was Bit Depth OD, in late March, early April 2000). I think as usual I did not express myself clearly in my last post, judging by responses, but I take the point that it has been discussed at length before so interested folk would do well to read the above mentioned archived posts. To avoid my example and ramblings, the guts of this post is in the para below the *, below. I accept that the number of bits defines the MAXIMUM density range which can be recorded if other components are good enough, what I was trying to get at was that it has nothing to do with ACTUAL or USABLE density range. I admit I mixed in the problem of smaller number of bits still covering the same density range which is not true, but a larger number of bits still tells you nothing about the actual density range capability of the system. For example, let us say I have a CCD and amplifier/electronics and a light source which can read a max brightness (corresponds to Dmin) of 1024 units (lumens, candelas /m2 or volts or anything), let's say this gives us 1024mV output. (to say that this is a maximum implies that something saturates or becomes unusably non-linear with higher light levels, so this is the mimimum density (highest light level) that can be read by this scanner in a calibrated fashion). Because it is an 8-bit D/A, the lowest level we can read is 2^8 lower than 1024 = 1024/256 = 4mV. This is the value of one least significant bit (LSB). Also, let's assume that this is an optimally engineered 8-bit system. Because it is optimally engineered, let's say that the CCD/amplifier noise is a quarter of the LSB level i.e. 1mV RMS. This is feasible and optimal in the sense that the engineered noise level is only just good enough to suit the D/A, with the result that a reading of one LSB is fairly meaningful if somewhat noisy. (This is the sort of noise that could be reduced greatly by multi-scanning). Now because a reading of one LSB is more-or-less meaningful, then we can fairly say that the dynamic range or range of densities that can be usefully measured is 1024/4 = 256 = 2^8 = 2.4 on the log scale. So this scanner could be accurately sold as a Dmax=2.4 scanner. Now, let's unplug the 8 bit D/A, and plug in a 12 bit D/A instead, to the same circuit. My point is - NOTHING CHANGES. Your maximum usable signal is still 1024mV, your useful minimum signal is still around 4mV so your density range is still 2.4. The fact that it is a 12 bit D/A has no effect whatsoever on the performance of the scanner, except that the bottom 4 bits are wasted because they only measure noise. The manufacturer might trumpet a Dmax of 2^12 = 3.6, but it is NOT TRUE. My intended point yesterday was that the manufacturer might even choose to establish an offset so that the whole D/A range is used i.e. so that the LSB measures around 4mV and the MSB still measures 1024 mV, but this does not change the density range, only the resolution within that range. My conclusion from all this is that the manufacturers cheat by saying that the Dmax is defined by the D/A resolution as a shorthand, which is true if the IMPLICATION which follows is that the rest of the system is engineered so that the whole of the D/A range is useful - that is, that the noise level is significantly less than one LSB. I don't think that this implied design constraint is true, at least in the case of consumer/semi-pro scanners. As Tony said... "but anyone can stick a 150mph speedometer on a 120mph car". In going from 12 bits to 14 bits and thus from an implied Dmax of 3.6 to an implied Dmax of 4.2, Nikon also needed to make sure that the CCD/amp noise level is reduced by a factor of 4 - assuming the noise level was well below 1 LSB in the old design (which I don't think was true anyway). I have read the published specs (UK site) of the new LS-4000, and I didn't see any claim for Dmax, although they do of course mention 14 bit data. My bet is that not all of this 14bits is useful, or that it only achieves a different objective - better resolution over a density range smaller than 4.2. Having ranted and rampaged around this old topic for an hour now, I feel comfortable with it, so if you think I am on a loony path could you please tell me before I do myself some damage! Cheers, Julian R (I am amazed at the number of Julians who frequent photography related lists, now 4 by my count) At 00:36 11/01/01, you wrote: On Wed, 10 Jan 2001 19:41:35 +1100 Julian Robinson ([EMAIL PROTECTED]) wrote: I am having a fundamental problem comprehending why the number of bits is even vaguely related to any supposed density range. I understand the maths quoted here and in many other posts, but fail to understand why the fact that the ratio of smallest bit size to largest number
