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bits per channel, f-stop range, & shadow detail?

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Michael Long said:
Just what are you expecting out of 35mm format? There was a time
not too long ago I would have been VERY happy with a clean, sharp,
non-grainy 16x20 print using 35mm Velvia.

At 40" you've got maybe 85-90 DPI of original data (after cropping
to 3x4 aspect ratio). There's no way I'd expect NOT to see
posterization issues at that size.
Click to expand...
The sample I posted was from a medium format camera and wasn't interpolated. It has posterization on the shadows. I'd move to medium format if it means deep shadow posterization would be reduced or eliminated. I don't think is has anything to do 35mm format per se. When I blow up images to 40x60, if there's no deep shadows, then I don't get any posterization (I interpolate with S-Spline Pro).

That's why I asked if f-stop range or 16 bit capture would eliminate the problem.

I guess I need to do some more testing with the Leaf back and see.
                            • -- - - - - - - - - - - - SMoody
http://www.pbase.com/smoody
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
 
what your claiming is that A/D converters don't have a dynamic range.

yet this is clearly not the case.

do a google search on A/D and Dynamic Range and you will be flooded with thousands of articles on just this subject.

"The crushing demands of next-generation radios make estimating an A/D converter's dynamic range a daunting task"
http://www.commsdesign.com/design_corner/OEG20021001S0034

"High-Speed CMOS A/D Converters with Wide Dynamic Range Proposal"
http://www.icsl.ucla.edu/aagroup/PDF_files/adc-96.pdf

"Wide Dynamic Range, High-Speed, 18-Bit Sampling A/D Converters"
http://www.analogic.com/Level4/DCPsel/Components/ADC5020.pdf

"Ultrahigh Dynamic Range, High-Speed A/D Converter for Laser Ranging"
http://esto.nasa.gov/programs/act/NRA99-selections/ATI059.html
Gaetan J. said:
--
Gaetan J.
Click to expand...
--
Philip G.
http://www.cgrafx.com
 
Seems both sides are correct. A 14-bit depth may or may not have more dynamic range then a 12-bit depth image. 14 bits gives you more resolving power ie 4 times more levels then 12-bits but its how each level is assigned that really matters. Dynamic range is the number of bit levels times the lumin differential between each bit level OR the dynamic range of the sensor WHICHEVER is smaller. Fuji's new 700 series CCD has two CCD photocites that allow a high and low sensitivity CCD to symultaneously read a brightness and the two values are splinned together and then converted to digital. I would hope the A/D for this camera is more then 8-bits or the extra dynamic range will be wasted.
philip said:
Gaetan J. said:
philip said:
So, theoretically a higher bit-depth converter should have greater
dynamic range, but you really need to know what the noise and
linearity characteristic of the A/D converter subsystem before you
can properly evaluate two systems.

Even though one may be 14-bits of data, it may only really be
12-bits of useful data.
Click to expand...
good explanation but you are mixing up 2 different things. Dynamic
and bits are not related. You can sample a sensor with any number
of bits you want. The only difference will be the number of steps
between 2 values.

So if you have more bits you will eliminate posterization.
Click to expand...
Well.. yes.. but it also is a direct correlation to dynamic range
as well. Think of a digital audio system. The more bits you have
the more subtle the changes in audio levels you have, but also the
more broad the range of levels you can have and still maintain a
smooth enough transistion from descrete step to descrete step. Thus
there is a correlation between number of bits and dynamic range.
philip said:
Sensor dynamic on the other hand will dictate at what intensity you
will loose details. (either way, shadows and highlights).
Click to expand...
On the other hand, here you are absolutely correct. If the sensor
dynamics don't allow for sufficent range of light capture, the rest
of the system is pointless, and thus this is a critical component
any imaging system.
philip said:
Linearity can be handled by software
Click to expand...
Linearity can be handled by software as long as it is sufficiently
algorithmic. If it dances around to randomly from step to step,
just like noise, you will lose low order bits, as the data they
represent is meaningless.
philip said:
--
Gaetan J.
Click to expand...
--
Philip G.
http://www.cgrafx.com
Click to expand...
--

Ken Eis - D100 and S45 Nikon 18-35, 28-105, 70-300, 80-400VR, 500mm and 60mm macro
 
I understand your analogy perfectly, but my engineering background is tugging at me, making me think that this might be an incorrect statement. Don't get upset; bear with me while I explain what I'm thinking, and if I'm wrong, I'll be the first to admit it!

My training suggests to me that bit depth IS dynamic range. Since dynamic range is defined as the ratio of the smallest measurable unit to the largest, then with an 8-bit signal the d.r. would be 2 to the 8th power (256) expressed in decibels, or db. With 12 bits, 2 to the 12th power, which is 4096. Tells me that a sensor with 8-bit resolution has only 255 steps of resolving capability, while a 12-bit sensor has 4096 steps of resolving capability.

By using JPEGs, which are 8-bit units generally, if you want to lighten a channel, you basically "shift" the 8 bits up within the 12 bits of information. By converting to 16 bit TIFF, you actually cannot create data where none exists, so PS effectively interpolates 12 bits to get 16. I'm not certain if it just truncates and loses the other 4 bits or if it interpolates ... I'll have to look into that. But the point is, it does not throw away 4 bits going to 16/TIFF like you would going to 8/JPEG.

Thus, the camera has a fixed dynamic range, which is defined by the A/D (Analog-to-Digital) converter for each luminosity measurement, by color. If you have a 12-bit sensor (a la EOS-10D), you have a dynamic range of 4096, which in binary (the language of computers) is expressed from 000 to FFF, where each character is 4 hexadecimal digits.

So, after all this background, what I'm not clear on what you were describing was how the number of sensor bits is NOT the dynamic range. Assuming a well-balanced histogram, where it's not shifted too bright, cutting off some range in the bottom (shadows), you should have greater dynamic range and resolving capability to recover shadows. What I haven't checked also is to see if PS only uses 8 bits for curves and levels when it's a TIFF. I don't have a TIFF pic on this machine to test. I'll try that later as an experiment.

So I humbly request clarification. I am only trying to understand, and I hope I have presented it that way. No offense intended whatsoever. I seek enlightenment, sayeth the seeker!

Thanks for any insight,
Wynn
Andrew Rodney said:
smoody said:
Andrew Rodney said:
My question is...do the higher bits-per-channel of digital backs
Click to expand...
help in capturing shadow detail more accurately? Or maybe the 12
stops of range?
Click to expand...
More bit depth does NOT provide more dynamic range. They are
completely different. Higher bit depth will allow you to edit a
file with the range it has and greatly reduce posterization. But
you'll gain no more shadow detail.

Dynamic range is the range of tones between black and white. So
more dynamic range allows you to alter shadows without blowing
highlights. But bit depth doesn't do anything in this regard.

Think of a stair case that's 50 feet tall. The length is the
dynamic range. The number of actual steps in the staircase is the
bit depth. If you have a staircase that's 75 feet and has 75 steps,
you have more tones than the 50 foot staircase with 75 steps.
Having more steps doesn't make the staircase any larger, it only
breaks up the steps finer.
--
Andrew Rodney
http://www.digitaldog.net
Click to expand...
 
Wynn Aker wrote:
...
Wynn Aker said:
My training suggests to me that bit depth IS dynamic range. Since
dynamic range is defined as the ratio of the smallest measurable
unit to the largest,
Click to expand...
...

exact,

but then a bit is not a unit of mesure. It as no value in itself. So, 0 (zero) can represent total darkness and 1 (one) total light (white). or even the reverse. You just wont have details between the too.

Or if you will you can mesure something using inch or millimeter (no fractions allowed :-) with a ruler (somebody allready used this analogy) the dynamic range is the lenght of the ruler, the bit depth is how much unit available to "describe" what you are mesuring.

--
Gaetan J.
 
Wynn Aker said:
My training suggests to me that bit depth IS dynamic range.
Click to expand...
Consider an 8-bit number that holds a distance. What's the range of distances that it can describe? It can describe any range of distances, the only limitation is that it can only describe them with limited precision, it can only describe 256 different distances. But depending on how we map these 256 values to real-world distances, we can describe any range of small to large distances.

For instance, we could map it linearly, and use it to represent 0 to 255 millimeters. Or 0 to 255 miles.

Or we could map it logarithmically. 0 could represent 0 miles, and 1 could be 10 miles, and 2 could be 100 miles. Then we could represent the range of 0 to 10^255 miles.

The number of bits we have just determines our precision. If we read out 12 bits from our sensor, and convert that to 8 bits when building a jpeg, we map our range of values into the smaller space, and lose 4 bits of "precision", but we don't change our ability to represent a broad range of values.
-harry
 
Gaetan J. said:
Gaetan J. said:
My training suggests to me that bit depth IS dynamic range. Since
dynamic range is defined as the ratio of the smallest measurable
unit to the largest,
Click to expand...
...

exact,

but then a bit is not a unit of mesure. It as no value in itself.
So, 0 (zero) can represent total darkness and 1 (one) total light
(white). or even the reverse. You just wont have details between
the too.
Gaetan J. said:
Or if you will you can mesure something using inch or millimeter
Click to expand...
(no fractions allowed :-) with a ruler (somebody allready used this
analogy) the dynamic range is the lenght of the ruler, the bit
depth is how much unit available to "describe" what you are
mesuring.
Click to expand...
your correct with respect to bit depth by itself, but not in respect to bit depth as it relates to an A/D system. A/Ds have a fixed scale. The reference system (voltage) for the A/D is almost always a fixed scale value (In some cases there may be a selectable scale for a few ranges).

Thus the A/D converter does have a dynamic range.

A true 16-bit A/D converter does have more dynamic range than an 8-bit converter, within the scope of a specific hardware application. When you are design a system, you are specifically assigning a conversion range for the A/D system and as such assigning a speicific quantity to each bit. So if you replace an 8-bit A/D with a 12-bit A/D you are either resolving smaller quantities or increasing the range that can be caputured. Or better still you can do both. Increase the dynamic range and resolve more detail.

--
Philip G.
http://www.cgrafx.com
 
Gaetan J. said:
but then a bit is not a unit of mesure. It as no value in itself.
So, 0 (zero) can represent total darkness and 1 (one) total light
(white). or even the reverse. You just wont have details between
the too.

Or if you will you can mesure something using inch or millimeter
(no fractions allowed :-) with a ruler (somebody allready used this
analogy) the dynamic range is the lenght of the ruler, the bit
depth is how much unit available to "describe" what you are
mesuring.
Click to expand...
would differ with that, if I recall. The dynamic range is not the one foot length of the ruler, rather the ability to measure the ruler. If you have a ruler which only indicates inches, then you only have a total of 12 discrete steps of measurement, which is a very small dynamic range. However, if you use thousandths of an inch, you have 12 thousand total discrete measurements, which is a much bigger dynamic range (bigger, in fact, than the 4096 discrete steps of a 12-bit digital measurement).

The CMOS sensor measure in analog measurements, in very small micro- or milli-volt levels. Each site is then converted via an Analog-to-Digital conversion process, only because computers and microprocessors (such as the one in a digital camera) can read digital, and not analog.

That digital audio (CD) system at home? 16 bit digital, with 2 to 16th power or 96db dynamic range. Lots! Without getting off-topic into digital audio and its relative strengths or weaknesses (which is surely debated in many other, audiophile-related forums!!!!), it merely illustrates that more bits in the A/D conversion is betterer. Presumably, the A/D conversion process will one day reach a greater number of bit depth, possibly 16 instead of 12, when the speeds increase to the point that you don't slow up the system by having to push the added data through whatever the smallest portion of the data pipeline currently is.

Anyway, the dynamic range is not the length of the ruler. It's a representation of the degree to which you can measure USING the ruler, within the boundaries of the length of the ruler. If all-black is 000 and all-white is FFF, you can measure almost-all-black as 001, and almost-almost-all-black as 002 with a 12-bit system. With a system using fewer bits, or a compression scheme that uses fewer bits, you lose the granularity between 000, 001 and 002 that you had with 12 bits. You might instead jump from what was 000 directly to 003, because you can no longer resolve any measurements in-between.

Or so it seems to me.

Wynn
 
hm said:
hm said:
My training suggests to me that bit depth IS dynamic range.
Click to expand...
Consider an 8-bit number that holds a distance. What's the range of
distances that it can describe? It can describe any range of
distances, the only limitation is that it can only describe them
with limited precision, it can only describe 256 different
distances. But depending on how we map these 256 values to
real-world distances, we can describe any range of small to large
distances.

For instance, we could map it linearly, and use it to represent 0
to 255 millimeters. Or 0 to 255 miles.

Or we could map it logarithmically. 0 could represent 0 miles, and
1 could be 10 miles, and 2 could be 100 miles. Then we could
represent the range of 0 to 10^255 miles.

The number of bits we have just determines our precision. If we
read out 12 bits from our sensor, and convert that to 8 bits when
building a jpeg, we map our range of values into the smaller space,
and lose 4 bits of "precision", but we don't change our ability to
represent a broad range of values.
-harry
Click to expand...
But your sensor output does not produce a broad range of values. It produces a fixed range of values. It is not a variable that can be adjusted on a whim.

The sensors range will be an output from noise floor (dark) to max output (lights), and it is a very specific maximum output (either a voltage or current depending on the specifics of the electronics).

--
Philip G.
http://www.cgrafx.com
 
hm said:
The number of bits we have just determines our precision.
Click to expand...
Agree, but not "just"

If we
hm said:
read out 12 bits from our sensor, and convert that to 8 bits when
building a jpeg, we map our range of values into the smaller space,
and lose 4 bits of "precision", but we don't change our ability to
represent a broad range of values.
Click to expand...
Disagree. If you throw away 4 bits, losing "precision," you lose the ability to represent the "broad range of value," since we had predetermined that 0 to 4095 is "full range." Otherwise, there's no point in 12 bits.

A corollary is to envision a well-balanced histogram: if you have a good spread, you have used a fairly full-range of values. If you're grossly overexposed, you won't even use the bottom X percent of the range, so there won't BE any data at 000, 001, 002, etc. FWIW, the histogram is a Fast Fourier Transform of the luminosity data. The presence and magnitude of data in a "bucket," each bucket being data position 000, 001, 002, etc.

Whew! This is cool! And my dad thought I drank the whole time in college!
 
...the information fed to it.

In the film world, it's customary to think of dynamic range in terms of stops, based on the lightest and darkest values it can register. Slide film typically has about 5 stops of range, film around 7.

DSLR sensors are much the same. Feed them too much light and they clip. Too little and you can't recover the information from the noise floor.

So if a sensor has a 5 stop range, and can deliver the precision, then a better a/d (12 bit vs 10) can deliver more accurate values and subtleties, but the dynamic range of the camera system is still only 5 stops.
 
Michael Long said:
...the information fed to it.

In the film world, it's customary to think of dynamic range in
terms of stops, based on the lightest and darkest values it can
register. Slide film typically has about 5 stops of range, film
around 7.

DSLR sensors are much the same. Feed them too much light and they
clip. Too little and you can't recover the information from the
noise floor.

So if a sensor has a 5 stop range, and can deliver the precision,
then a better a/d (12 bit vs 10) can deliver more accurate values
and subtleties, but the dynamic range of the camera system is still
only 5 stops.
Click to expand...
I believe that a "stop" is typicall a doubling or halving, correct? in which case, 2 to the 5th or 2 to the 7th would be the dynamic range, right?

I'm really surprised at the level of knowledge base in this thread. There are obviously a lot of knowledgeable people! My only point in the whole thread was that the dynamic range was the end-to-end resolving capability, and that by using 8 bits instead of the 12 available, data is lost in the process, and you can't pull the same degree of shadow detail out that you could by using a 16-bit TIFF converted from 12 bit RAW. That's all.
 
Wynn Aker said:
Disagree. If you throw away 4 bits, losing "precision," you lose
the ability to represent the "broad range of value," since we had
predetermined that 0 to 4095 is "full range." Otherwise, there's
no point in 12 bits.
Click to expand...
No, because if you map to 8 bits then 4095 is mapped to 255, and is still "white". 0 is 0, and is still black, and so on.

What you lost was was the subtle difference between 2046 and 2047, both of which are now mapped to 127.
 
Wynn Aker said:
Michael Long said:
...the information fed to it.

In the film world, it's customary to think of dynamic range in
terms of stops, based on the lightest and darkest values it can
register. Slide film typically has about 5 stops of range, film
around 7.

DSLR sensors are much the same. Feed them too much light and they
clip. Too little and you can't recover the information from the
noise floor.

So if a sensor has a 5 stop range, and can deliver the precision,
then a better a/d (12 bit vs 10) can deliver more accurate values
and subtleties, but the dynamic range of the camera system is still
only 5 stops.
Click to expand...
I believe that a "stop" is typicall a doubling or halving, correct?
in which case, 2 to the 5th or 2 to the 7th would be the dynamic
range, right?

I'm really surprised at the level of knowledge base in this thread.
There are obviously a lot of knowledgeable people! My only point
in the whole thread was that the dynamic range was the end-to-end
resolving capability, and that by using 8 bits instead of the 12
available, data is lost in the process, and you can't pull the same
degree of shadow detail out that you could by using a 16-bit TIFF
converted from 12 bit RAW. That's all.
Click to expand...
 
Michael Long said:
Michael Long said:
Disagree. If you throw away 4 bits, losing "precision," you lose
the ability to represent the "broad range of value," since we had
predetermined that 0 to 4095 is "full range." Otherwise, there's
no point in 12 bits.
Click to expand...
No, because if you map to 8 bits then 4095 is mapped to 255, and is
still "white". 0 is 0, and is still black, and so on.

What you lost was was the subtle difference between 2046 and 2047,
both of which are now mapped to 127.
Click to expand...
 
Wynn Aker said:
My training suggests to me that bit depth IS dynamic range. Since
dynamic range is defined as the ratio of the smallest measurable
unit to the largest, then with an 8-bit signal the d.r. would be 2
to the 8th power (256) expressed in decibels, or db.
Click to expand...
There are all kinds of devices that can provide you with 8 bits per color and all will have vastly different dynamic ranges.

There are all kinds of scanners on the market that produce 8bits per color and have dynamic range specs* that range for 3.0 to over 4.0

there is no universally accepted way of measuring dynamic range so the figures are, like milage-yours may vary.

If you take a device that can capture a dynamic range of 4.0 and shoot with the lens cap on, what range do you really end up with in your 8 bit file .

If you have a device with a 4.0 dynamic range, that's the total tonal range from highlight to shadow you can capture. It's like chrome film has about a 5 stop range where a neg might have a 8 stop range. Anyway, the range has, well a range. The bit depth is only the number of finite steps from one end to the other.

You can have a device that produces a 3.8 dynamic range in 12 bits and you can have a device that has a dynamic range of 3.0 in 16 bits. One has nothing to do with the other.
Wynn Aker said:
With 12 bits,
2 to the 12th power, which is 4096.
Click to expand...
Bits, but not range!
Wynn Aker said:
Tells me that a sensor with
8-bit resolution has only 255 steps of resolving capability, while
a 12-bit sensor has 4096 steps of resolving capability.
Click to expand...
No because we're talking steps, not range.

If you have a camera with an Fstop range from 2.8 to 22, that's the range. That you can do half stops verses quarter stops by moving the appature ring in now way changes the range you have. If someone builds a lens that has 1/10 stops between Fstops, that doesn't make your 2.8 lens a 1.2 does it?

--
Andrew Rodney
http://www.digitaldog.net
 
Andrew Rodney said:
If you have a device with a 4.0 dynamic range, that's the total
tonal range from highlight to shadow you can capture. It's like
chrome film has about a 5 stop range where a neg might have a 8
stop range
Click to expand...
I have no knowledge in this area
Andrew Rodney said:
Anyway, the range has, well a range. The bit depth is
only the number of finite steps from one end to the other.
Click to expand...
Totally agree
Andrew Rodney said:
You can have a device that produces a 3.8 dynamic range in 12 bits
and you can have a device that has a dynamic range of 3.0 in 16
bits. One has nothing to do with the other.
Click to expand...
Disagree. The "dynamic range" is a measurement of the resolving ability of a given "range." Remember: dynamic range is defined as the ratio, in decibels, of the largest measureable value to the smallest. Not the same this as absolute "range," which in this case I take to be the length of the ruler itself. When we go to measure that perfect one-foot ruler, did we measure it with a device having a resolving ability (i.e. dynamic range) of 4 bits, or 12 bits? Big difference in the result.
Andrew Rodney said:
Andrew Rodney said:
With 12 bits,
2 to the 12th power, which is 4096.
Click to expand...
Bits, but not range!
Click to expand...
Yes, bits, no, not range, but IS dynamic range, if expressed in decibels
Andrew Rodney said:
Andrew Rodney said:
Tells me that a sensor with
8-bit resolution has only 255 steps of resolving capability, while
a 12-bit sensor has 4096 steps of resolving capability.
Click to expand...
No because we're talking steps, not range.
Click to expand...
Agree. I was talking steps, which is analogous to dynamic range.
Andrew Rodney said:
If you have a camera with an Fstop range from 2.8 to 22, that's the
range.
Click to expand...
Agree
Andrew Rodney said:
That you can do half stops verses quarter stops by moving
the appature ring in now way changes the range you have.
Click to expand...
Agree
Andrew Rodney said:
If someone
builds a lens that has 1/10 stops between Fstops, that doesn't make
your 2.8 lens a 1.2 does it?
Click to expand...
No. But that was never the point. I don't understand that particular analogy, but maybe I'll catch it later when re-reading.

Again: range does not equal dynamic range. That lens from 2.8 to 22 has that given range. If you use a device to resolve/measure what the f-stop is, how many bits did it use to indicate it's position? The number of bits implies the dynamic range, due to more bits equals more dynamic range.

I'm thinking that I shouldn't mess with "the digital dog!"

Wynn
Andrew Rodney said:
--
Andrew Rodney
http://www.digitaldog.net
Click to expand...
 
Wynn Aker said:
I understand your analogy perfectly, but my engineering background
is tugging at me, making me think that this might be an incorrect
statement. Don't get upset; bear with me while I explain what I'm
thinking, and if I'm wrong, I'll be the first to admit it!

My training suggests to me that bit depth IS dynamic range. Since
dynamic range is defined as the ratio of the smallest measurable
unit to the largest, then with an 8-bit signal the d.r. would be 2
to the 8th power (256) expressed in decibels, or db. With 12 bits,
2 to the 12th power, which is 4096. Tells me that a sensor with
8-bit resolution has only 255 steps of resolving capability, while
a 12-bit sensor has 4096 steps of resolving capability.
Click to expand...
You're training isn't wrong. It's just that you're not talking about what you think you're talking about.

Specifically, an image format with 8 bits per pixel has a theoretical maximum dynamic range of 8 stops. Likewise a 12-bit format has a theoretical maximum dynamic range of 12 stops.

BUT... that's only refering to the theoretical capacity of the image format itself. How many steps it can represent without clipping or compression. It doesn't say anything about the image being captured or the actual real-world capabilities of the capture device.

The dynamic range of the actual image is determined by the capabilities of the capture device, i.e. the image sensor. If the image sensor is going to saturate at an input level that is 6 stops brighter than the point where it begins to register in the first place, then you've only a maximum of 6 stops of dynamic range.

As long as the image sensor outputs at least 6-bits per pixel, you should be able to save the image without clipping off highlights or shadows, or without compressing the range and dropping out bits and pieces in between.

If the image sensor outputs 8-bits or 12-bits per pixel, then it can do a better job of maintaining fine detail within that 6-stop range, but it doesn't change the fact that only 6 stops of information is being captured.

Look at it a different way. You can take an old cassette recording and transfer it to Compact Disc, and then you can take a new state of the art digital recorder and make a new recording and disc. The dynamic range of the recording format is fixed, but the dynamic range of the actual recordings themselves will be drastically different. You don't expect that transfering the cassette recording to a CD is going to magically improve the dynamic range.

Mike
 
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