helenbach

December 15, 2008

A normal photographic colour meter has no more than three or four (in the new Sekonic) filtered sensors while a spectrophotometer does some kind of spectral analysis (as already mentioned). The Eye-One Pro measures in 3 nm steps, but the readout is usually in 10 nm steps. It will measure ambient light as well as reflected, and can be used as a very fancy colour meter if hooked up to a laptop etc. I use mine with BabelColor and SpectraShop on the laptop. The combination isn't as portable or easy to use as a colour meter (I have the Minolta II, IIIF and Gossen 3F, haven't tested the new Sekonic yet). Let me know if you want a more detailed, more specific comparison.

Best, Helen

December 12, 2008

Haha. Like an old audio optocompressor on the output. You could get that real 60's sound in the highlights.

must go.

December 12, 2008

The maximum dynamic range is set by the noise floor and saturation capacity of the CCD. The ADC should then be designed to handle that. Raw will make the most of it, at the exposure that represents the 'native' ISO. At all other ISOs (or EIs), higher or lower, the dynamic range is likely to be less. In simple, rushed terms.

I didn't think anybody else would be reading this thread all the way down here. I'm amazed.

(I really have to leave for the weekend, much as I would love to continue this pleasant discussion)

Best, Helen

December 12, 2008

Yep, reply edited. In a big rush.

Re EV. The classic use of Ev or EV is that EV represents combinations of shutter speed and aperture that gives the same exposure. Logarithmically

Ev = Tv + Av = Sv + Bv

December 12, 2008

Rishi, (I misread the author's name first time)

"So, then, for the same image taken at the same EV but at ISO 100 vs. ISO 400 -- shouldn't the ISO 400 image, by definition, have less blown-out highlights than the ISO 100 image?"

First, if the EV is not changed between two exposures (and the lighting is not changed) the CCD gets the same exposure, so turning up the ISO (gain) from 100 to 400 will increase the likelihood of blown highlights, not decrease them. But I think you made a typo?

If, however, the exposure is reduced and the gain is increased you can work out what happens from the CCD information I gave earlier.

Here is the information again. It is genuine stuff.

600 mV/lux s linear to 2100 mV, 3200 mV at full well, 0.2 mV RMS noise.

The analogue out from the CCD goes to the analogue in of the PGA (programmable gain amplifier), which is typically packaged with the ADC. The PGA could be 0-36 dB, with the gain set by a 10-bit controller. The analogue out from the PGA goes to the ADC as you might imagine.

That should be enough information for you to figure out that as you turn up the gain, the dynamic range (the exposure range over which detail is recorded, ie the meaning used in this thread) falls. You can also work out what the dynamic range is likely to be...

The noise floor of the CCD is amplified, but the upper limit of the ADC input stays at 2000 mV or just under. So the ADC ouput maxes out from the amplified signal even though the sensor may not be saturated.

Sorry, I'm rushing off for the weekend - hope this isn't too brief, no doubt someone else will help!

Best, Helen

December 12, 2008

Vijay,

"I see. So you agree that Tech Pan in Dektol has zero dynamic range. And so if I increase contrast, I decrease dynamic range?

And how do you define the dynamic range of slide film then? Or perhaps a print?"

You sure took me for a ride. I thought that you were being genuine for a while there.

This has to be one of the most surreal threads I have ever seen. So proud to have been a part of it.

Best,

Helen

December 12, 2008

Vijay,

No, not what a DAC would do. Given the CCD and ADC characteristics, what can you tell for certain about the exposure a photosite received that is represented by 000 in the ADC output; what can you tell for certain about the exposure a photosite that is represented by 001 in the ADC output etc... That kind of thing.

Best,

Helen

December 12, 2008

Vijay,

"Umm, I meant that when I see 0000 (or 0001 or 0010 etc) as a code output from the ADC what should I do with it? You want me play detective, right; and I need this piece of information to continue."

You should work out the maximum amount of data that you can deduce for certain about the exposure that the CCD received, as I suggested originally. No need to go into every step of the ADC output, just the output steps at the ends of the range and those adjacent (eg 000, 001, 110, 111 for the 3-bit ADC), and why not use the ADC data you already quoted. Just my suggestion.

Best,

Helen

December 12, 2008

Vijay,

What do you think everyone else has been talking about? Have you realised that everyone else is referring to the exposure range, not the density range? That is the common meaning of 'dynamic range' in photography.

What film that you use has a density range of 4.0 (or 80 db)?

This is rather funny, don't you think?

Best,

Helen

December 12, 2008

Vijay,

"1. How is the exposure that the CCD received related to the dynamic range of that CCD?

2. How do I make sense of the ADC data unless I know what each code means? What code to voltage (or luminance) conversion table should I use?"

I was going to leave the choice of those values to you. You have already given some ADC data - why not compare the 3-bit ADC with a higher-bit ADC?. I suggest 600 mV/lux s linear to 2100 mV, 3200 mV at full well, 0.2 mV RMS noise for the CCD if you have nothing else that you would prefer to use. Zero dB for the PGA seems sensible.

Best,

Helen

December 12, 2008

Vijay,

Quotes from you, in this thread:

"Umm, I don't get the whole log vs linear thing - it is dynamic range that matters, so lets say a dynamic range of 80 db (10,000:1) is a dynamic range of 80 dB = 13 plus stops = 10,000:1 = 4.0 and so on. Shapes of curves don't really matter, you can always convert between curve shapes. If it is log, apply an antilog function. If it is linear and you want log, just take the logarithm etc.

Film may well have 80 dB dynamic range, but digital sensors will eventually get there if they aren't already there."

"I stick to strict definitions (like you haven't figured that out by now) - and the strict defintion of dynamic range of film is its opacity range or its density range (if you use bels as the unit, dynamic range is exactly equal to (Dmax - Dmin) * 2 )."

Please reconcile these.

Thanks,

Helen

December 12, 2008

This is funny.

What do you think the title of this thread should have been? Honestly.

Best,

Helen

December 12, 2008

Vijay,

I'm still finding that there is a complete logical disconnect between what Bernie and I have been trying to explain and the unrelated argument that you are locked in to. I have no idea how to break through that right now, other than to ask you to think about my previous suggestion:

Try to look at it as if you were a detective. All you have are the CCD-PGA-ADC characteristics and the output data. What is the maximum amount of data that you can deduce for certain about the exposure that the CCD received?

Let's take this one step at a time...

Best,

Helen

December 12, 2008

Vijay,

"Helen, let's discuss your question to me for a bit first, please. You asked me about dynamic range, Dmax and Dmin; evidently you thought that somehow your and my definitions differed."

Happy to do so.

Here's the first answer you gave to my request for your definition of dynamic range:

"Dynamic range is just Dmax:Dmin; unless Dmin is 0, the dynamic range is finite, regardless of the slope of the characteristic curve. The dynamic range is the ratio of the height of curve at the top to the height of the curve at the bottom, regardless of the slope."

Now I understand that to be density range, not dynamic range. Generally when people refer to dynamic range of, say, film vs digital (people talk about that occasionally) they are referring to the ability of the medium to capture scene brightness range - ie the exposure range (x-axis on a standard characteristic curve) between the toe and the shoulder, not the density range (y-axis on a standard characteristic curve, or 'height' as you call it).

You threw me another curve ball by referring to it as a ratio of DMax to DMin, because DMax and DMin are conventionally expressed as log values, so the density range is conventionally expressed as the difference between DMax and DMin. It is finite even when DMin (being a log) is zero. If the minimum density isn't expressed as a log it cannot be zero. That means that no matter whether the density range is calculated from arithmetic densities as a ratio or from logarithmic densities as a difference it is always finite (as long as the material is not opaque) - that makes sense. It would not make sense to have a method of calculating density range that could give an infinite result - unless the maximum density is infinite (total opacity), when infinite density range makes sense.

You later gave this similar meaning for dynamic range:

"for film, it is opacity range; which is identical to density range."

Again, that is what I would call the density range of the film, not the dynamic range. If you call the density range of the film the dynamic range, what do you call what I call the dynamic range?

Best,

Helen

December 12, 2008

Vijay,

Somehow we need to find the key that lets you look at this in a different way from the one that you (and I) have been taught to look at it as designers. Try to look at it as if you were a detective. All you have are the CCD-PGA-ADC characteristics and the output data. What is the maximum amount of data that you can deduce for certain about the exposure that the CCD received? Please forget the DAC or the printer. These are irrelevant, as Bernie so rightly says. It is a very simple, logical process.

Best,

Helen

December 11, 2008

Vijay,

1. I'm not assigning anything - I used your assignments. If you want to argue with them, argue with yourself, not me.

2. Please read my purpose statement. I'm trying to explain why the limits I'm suggesting have a valid use for the purposes of this discussion, even though they are not the ones in your 'book'. It is a matter of logic and reason, not rote.

3. I agree, and have always referred to the CCD-PGA-ADC system as having to prevent 'out-of-range' signals. No argument there. I'm just using the voltages so I don't have to go through the conversion from lux s to mV. As we are assuming a linear analogue conversion there should be no problem. If you want me to, I'll start with the lux s values and go through the whole exercise again. The reason and logic remain the same, and the conclusions will be the same.

"4. You keep ignoring the extensibility of ranges requirement: I could build a 4-bit ADC with an input dynamic range of 2048 mV. Its table could look like this: 1 uV - 127 mV : 0 128 mV - 255 mV : 1 256 mV - 383 mV : 2 384 mV - 511 mV : 3 512 mV - 639 mV : 4 640 mV - 767 mV : 5 768 mV - 895 mV : 6 896 mV - 1023 mV : 7 1024 mV - 1151 mV : 8 1152 mV - 1279 mV : 9 and so on.

So let me ask you: for this ADC, what is the dynamic range of the input based on the first 8 codes and the range based on the next 8 codes?"

The answer is as before for the first 8 codes: anything below 128 mV gives an output of 0, anything above 896 mV gives an output of 7. You didn't disagree with that before, and there's no reason to disagree with it this time.

If you take the second 8 codes (8 to 15) anything below 1151 mV produces an output of 8 and anything 1920 mV or above produces an output of 15.

By separating the two ranges you lose the switch between 7 and 8 in the output. Continuity of input is maintained, but not continuity of output. There is no logical flaw here - it is the consequence of splitting the range. Extensibility is maintained. You can set out the full table, and everything fits. There are no discontinuities - that is why I have ignored them: they are in your imagination. If you wish to argue with this it might be an idea to take the reasoning step-by-step and make sure that you are applying the principles correctly. You will find that there is no logical flaw. It is all solid. Any 'principle' that says otherwise is either wrong or, more likely, misapplied.

You have 16 pieces of string. You want to join as many steel rings together as you can in a line (rather than a circle) and each ring has to have two pieces of string attached. You begin by taking two bunches of string, each with 8 pieces. With each bunch you can connect 7 rings. String-ring-string-ring-string etc. If you tie the two sets together you can connect one more ring. You are measuring the ends of the string for your purposes, I'm measuring the locations of the rings for my purposes. I'm trying to explain why there are reasons to do that.

If you use the whole range, the limits for difference in output become 128 mV and 1920 mV.

"Are you confusing dynamic range with tonal range or gamma? Dynamic range is just Dmax:Dmin; unless Dmin is 0, the dynamic range is finite, regardless of the slope of the characteristic curve. The dynamic range is the ratio of the height of curve at the top to the height of the curve at the bottom, regardless of the slope. This has nothing to do with infinite dynamic range. Infinite gamma is not infinite dynamic range. Gamma is mathematically a derivative (dy/dx) and dynamic range is a ratio (y/x). A derivative can be infinity for a curve which has finite endpoints and finite dynamic range."

No, I'm not confusing dynamic range with tonal range or gamma. You are now confusing dynamic range with density range, and you also seem confused about what DMax and DMin mean in conventional film terminology. I think that the best thing for you to do is to make a clear statement of what you call the dynamic range of film and of a digital camera, then we can revisit this part of the logic.

Best, Helen

December 11, 2008

Vijay,

"You need more bits to represent more tones, but not the two endpoints. One bit carries enough information to distinguish two states that define dynamic range."

If we accept your definition, then every digital camera and every film, including Tech Pan developed in Dektol, has infinite dynamic range, and the whole thing becomes a non-issue. There's no need to discuss it any more, anywhere. Thank goodness.

Best, Helen

December 11, 2008

Vijay,

OK, I'll have one more attempt.

Purpose: To demonstrate that there is an important practical distinction between the input range that an ADC can tolerate and the input dynamic range that an ADC responds to.

As the example, we will take Vijay's 3-bit ADC:

"Now if we move to a 3-bit ADC with the same input range, the table looks like: 1 uV - 127 mV : 0 128 mV - 255 mV : 1 256 mV - 383 mV : 2 384 mV - 511 mV : 3 512 mV - 639 mV : 4 640 mV - 767 mV : 5 768 mV - 895 mV : 6 896 mV - 1023 mV : 7 Once again, the minimum is 1 uV and the maximum is 1023 mV, with the exact same DR."

I have already explained that an 'out of range' signal will not work with an ADC (or CCD-PGA-ADC chain) used in a camera, and Vijay has not challenged that, so Vijay's table means that the ADC outputs as follows:

Input under 128 mV; output = 0

Input 896 mV or over; output = 7

Vijay says that the dynamic range is 1023 mV/1 µV = 1 023 000:1

An input of 1 µV will produce an output of 0, an input of 1023 mV will produce an output of 7.

That is the 'standard' way that the input range of an ADC is defined.

But think about it. Does this have any bearing on the discussion? I've suggested that the Integer linear output from an ADC sets an intrinsic relationship between the bit depth and the maximum dynamic range (of the incoming signal) that can possibly be represented by those bits.

What sets the dynamic range that is represented by the output from this ADC? What does an output of 0 mean? It means nothing other than the incoming signal was below 128 mV. An input analogue value of 1 µV produces an output of 0. An input of 24.896 mV produces an output of 0. The lowest input signal that causes a change in the output is a change that occurs at 128 mV. That is the lowest input that can be considered as being represented (however coarsely) by the output. Nothing and nobody can analyse the output pixel value map and say 'The input value here was 1 µV', or 100 mV, or any other value below 128 mV. However the presence of two adjacent pixels, one of which has a value of 0 and one of which has a value of 1 could lead to the deduction that here the input value changed from one side of 128 mV to the other side of 128 mV and the output switched accordingly. 128 mV is represented in some manner. This cannot be said of any lower input voltage.

In the absence of any other value, 128 mV looks like the lowest input value that we can say is represented in some way by the output. If we choose any other lower input value, we can equally well choose any other value that is lower than 128 mV. If 1 µV, why not 0.5 µV? It would produce the same output of 0.

I will not bore you with the same argument at the other end, but using Vijay's numbers it comes out at 896 mV.

We have, therefore, two values: the lowest and the highest values that can be said to be represented by something in the output. The dynamic range that produces a response in the output is a mere 896/128 = 6.5.

Now it doesn't matter what the exact numbers are. What I am trying to get across is that in a camera, in terms of the output, it does not matter what the input range of the ADC is, but it does matter what the input range that causes a change in the output is.

Best,

Helen

December 11, 2008

Vijay,

When you are in a hole, stop digging. At least for a while.

Best,

Helen

December 11, 2008

That confirms it. You are confused about why it is very important to realise the difference between the input range that an ADC can handle and the input range that produces a change in the ADC's output, and I have no idea how to explain it another way right now. Maybe someone else could try. This might be the crux of the problem.

Best, Helen

PS

"I know it tells you nothing more but this is consistent with information theory."

That is priceless. I couldn't agree more.

December 11, 2008

Vijay,

"I don't see the distinction, because the ADC is useless without the DAC; you can't see binary values, so they have to be converted back to voltages or tones. DACs follow the same conventions as ADCs; they use the range endpoints as I described. So if DACs don't see the distinction, neither do I."

In the post that you are responding to I did not mention a DAC, I simply referred to an ADC. You added the DAC. You keep doing this - you add things in to my posts that aren't there. Why why why?

I'll try again. I believe that you are confusing the input range that an ADC can tolerate with the input range that produces a response - ie a change in the output. You gave an example of a 3-bit ADC that changed the output from 0 to 1 when the input changed from 127 mV to 128 mV; and from 6 to 7 when the input changed from 895 mV to 896 mV. The ADC could accept a signal higher or lower than that, but the output would not change. I don't mind if you want to change those values - it is the principle I'm still trying to get across. There is no point in quoting a wider input range than the input range that will cause a change in the ADC output. That is a very simple, but important concept. DACs have nothing to do with it, and I did not mention them.

Best,

Helen

December 11, 2008

Vijay,

"So if you take a 1 bit ADC that takes 10uV - 512 mV and converts it to 0; and 512 mV to 1024 mV and converts it to 1 and you feed this to a one bit DAC whose output dynamic range is 0 to 100 V, then it will convert the binary 0 to 1 uV (or its noise floor), and the binary 1 to 100 V.

If you take the DAC's output and feed it to a CRT or to a printer, you will get black and white; the full dynamic range that the monitor or paper can handle."

That tells you absolutely nothing about the signal that was input to the original ADC other than some of it was below 512 mV and some of it was above 512 mV. That is all it tells you.

Best, Helen

December 11, 2008

A one-bit system can't be said to be linear or logarithmic or anything, because there is no proportional response. There is only one change in input that produces a change in output. Therefore it can't really be used as an example, I think - it will only mislead...

I think that Vijay is confusing the dynamic range that a device can handle with the dynamic range that a device will respond to. They may not be the same thing, as demonstrated by the examples he gave. His example ADC can handle a higher dynamic range than it responds to. That is not unusual.

Best,

Helen

December 11, 2008

Vijay,

I chose the endpoints for exactly the reasons I explained. (referring to your 3-bit ADC example) Every input below 127 mV gets output as the lowest ADC output value and every input above 896 mV gets output as the highest output value. That was given in your data. Therefore you cannot set the lower limit any lower than 127 mV because you cannot know where it is from the information the ADC gives you. You cannot set the upper limit any higher than 896 mV because you cannot know where it is from the information the ADC gives you.

If you disagree with this, please explain how you can set any other upper or lower value as a valid limit, and what happens if that limit is exceeded (ie if you accept any wider limit you may as well call it infinity - which is a conclusion that has no practical use). Remember that 'out of range' just won't work for a camera (unless you want to challenge that, which you appear not to have done).

Best,

Helen

December 11, 2008

Helen,

"The ADC may only know the top of the lowest riser and the bottom of the top riser."

I think that you meant 'the output from the ADC can only tell you the top of the lowest riser and the bottom of the top riser'. Maybe you were in a rush, or maybe you are just clueless.

Best,

Helen

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Posts posted by helenbach