I am reading the Poynton's book, where he says: "For a contrast ratio 100:1 about 463 code values would be required".

Well, indeed the equation will look like this to ensure smooth 1% step of brightness:

lg(100)/lg(1.01) ~ 463

Umm, this requires 9 bits per color for the display unit. I know that my Panasonic has 10 bits per color internally, so its max theoretic contrast ratio (ensuring that gradients are smooth) is 1.01^1024 ~ 26000, whoa! 9 bits is barely enough: 1.01^512 == 612. What about 8 bits which is used in computers and in DVDs (24 bits per RGB, that is 8 bits per color component): 1.01 ^ 256 == 12.7 What? Do I miss something?

How is it possible to use 8 bits per color yet achieving good contrast ratio and smooth gradients? Is it implied that EVERY digital display device must have a wider dynamic range and must "expand" 8-bit picture to 10, 12 or 16 bits of the display panel, filling gaps in between? So basically 8 bits is the transport format only and cannot be displayed directly? What should I search for to read something about this? I could not find relevant info in Poynton's book so far.

Well, for one thing, you forgot to factor in gamma. Displays are not linear in their light output versus drive signal.

-Steve

As steve said, you're not accounting for gamma. Digital displays are linear, which is why they need to de-gamma and use a much higher bitdepth. The content is nonlinear, which allows a much lower bitdepth to acheive the same performance. That's what Poynton is illustrating in this section, assuming you're reading there at the beginning of chapter 1.

If I understand correctly, "10-bit" is not 'per color' bit-depth as in a computer - it's 10-bit sampling of an analog waveform. Analogous to 16-bit sampling in CDs.

those are the same thing...

As steve said, you're not accounting for gamma. Digital displays are linear, which is why they need to de-gamma and use a much higher bitdepth.
Ok I got that, digital displays MUST have higher bitdepth because of de-gamma.
The content is nonlinear, which allows a much lower bitdepth to acheive the same performance. That's what Poynton is illustrating in this section, assuming you're reading there at the beginning of chapter 1.
I am reading it from section 1. The signal is non-linear, this is exactly what Poynton writes (slightly different wording from his Gamma FAQ):
If you use nonlinear coding, then the 1.01 “delta” required at the black end of the scale applies as a ratio, not an absolute increment, and progresses like compound interest up to white. This results in about 460 codes, or about nine bits per component.
I totally understand where 460 codes came from, the formula is above. At the same time he writes:
Eight bits, nonlinearly coded according to Rec. 709, is sufficient for broadcast-quality digital television at a contrast ratio of about 50:1.
In this case I don't see where he got 50 from, as lg(50)/lg(1.01) == 393, which requres the same 9 bits, not 8.

On another note, his Gamma FAQ has somewhat enhanced info on contrast ratio compared to the book:

Contrast ratio is the ratio of intensity between the brightest white and the darkest black of a particular device or a particular environment. Projected cinema film, or a photographic reflection print, has a contrast ratio of about 80:1. Television assumes a contrast ratio, in your living room, of about 30:1. Typical office viewing conditions restrict the contrast ratio of a
CRT display to about 5:1.
So average TV contrast ratio is only 30 ! The how the checkerboard contrast of 200, 300 or even 500 is achieved? Does it mean that this contrast is not smooth all the way from black to white if we render a gradient? What is the point of advertised contrast ratios of 10000? I see that 10 bits allow to achieve a very high contrast ratio -- on paper -- but what is the point if DVD provides information for only 50 as Poynton writes? And why is it 50 and not 12 as one can see using the same formula? I am in the dark.

Eight bits, nonlinearly coded according to Rec. 709, is sufficient for broadcast-quality digital television at a contrast ratio of about 50:1.In this case I don't see where he got 50 from, as lg(50)/lg(1.01) == 393, which requres the same 9 bits, not 8.
Rec709 is a power function, not a logarithmic function. So the lg/lg formula does not apply to Rec709.

Poynton discusses different kinds of functions such as:
- linear
- logarithmic, which is non-linear
- power, which is non-linear

It is important to keep track of which kind he is referring to in each context.

So average TV contrast ratio is only 30 ! The how the checkerboard contrast of 200, 300 or even 500 is achieved? Does it mean that this contrast is not smooth all the way from black to white if we render a gradient? What is the point of advertised contrast ratios of 10000? I see that 10 bits allow to achieve a very high contrast ratio -- on paper -- but what is the point if DVD provides information for only 50 as Poynton writes? And why is it 50 and not 12 as one can see using the same formula? I am in the dark.

You often can and do have a higher contrast ratio, but without adequate DAC resolution (bit depth per color channel) you will see banding in gradients; the exact amount will vary based on the mapping between code values and brightness ("gamma").

I found his whole discussion rather confused since contrast ratio has nothing to do with bit depth at all. It is the ratio between the darkest and brightest elements that can be displayed. It has nothing to do with how many steps there are in between.

Certainly you could have a 1 bit display with a 10,000 to 1 contrast ratio. It just means that 1 is 10,000 times brighter than 0.

Larry

That is correct. Though if you add in the caveat that you want to avoid banding, then the two can be related. Obviously a light bulb is a 1-bit system with infinite CR.

Obviously a light bulb is a 1-bit system with infinite CR.
There's more here than one initially reads into it.... what it means is that no matter how bright or dim it is when on, if it truly goes "off", i.e. outputs zero light, the contrast ratio is infinite.... black is beautiful :D

:cool:

assuming the light bulb works...

assuming the light bulb works...
twice we agree :D

:cool:

In this case I don't see where he got 50 from, as lg(50)/lg(1.01) = 393, which requres the same 9 bits, not 8.235 - 16 = 219

(100 / x ) = (463 / 219)

x = 47.3 or ~ 50:1 :)

1.01 ^ 256 == 12.7 What? Do I miss something?FWIW if you can comprehend the following quote from Dr. Soneira, you may find a missing piece regarding this puzzle, BTW Mr. Poynton fails to mention the following detail for some reason?

http://www.displaymate.com/ShootOut_Part_3.htm
:rolleyes:
The next question is how to distribute the brightness levels among the allowed digital values.
….each step would then be 1.01 times the brightness of the previous step. Each step would also be a bit wider than the previous step, so the spacing between the steps would vary and be non-linear. (The brightness for step n would be proportional to 1.01^n)

..but it’s not the method that’s used because a linear spacing is more convenient and makes signal processing a lot easier. So, in reality, the intensity steps are all separated by equal differences rather than equal ratios. That means that the brightness ratio between adjacent steps will then increase as the brightness decreases, so the granularity artifacts will show up first at the dim-end of the intensity scale.

If I understand correctly, "10-bit" is not 'per color' bit-depth as in a computer - it's 10-bit sampling of an analog waveform. Analogous to 16-bit sampling in CDs.219 / 100 = 2.19 digital steps per IRE, and 1 IRE is ~ the threshold of (HVS) human perception.

While each discrete level of quanta i.e. in a audio waveform, the calibrated voltage level per step has linear weighting with respect to the scoped waveform...video undergoes non-linear voltage (analog) gamma correction.

This gamma corrected (Y') waveform has pre-emphasis (a non-linear voltage/distortion) component or "power function", this shifts the tonal values, which in effect robs contrast potential (DR that would have been perceptually wasted) from the region above the gamma cross point and re-distributes it toward the darker (1-50IRE) tones.

**Remember: Y' voltage is NOT proportional to intensity**

1)Gamma linearizes (CRT) display response

2)Gamma corrects for for a lossy 8-bit encoding

3)Gamma accounts for a change in perception to maintain rendering intent.

219 / 100 = 2.19 digital steps per IRE, and 1 IRE is ~ the threshold of (HVS) human perception.

No it most certainly is not.

As for anyone reading, please feel free to ignore thomas, he is a wonderful example of how a little bit of knowledge and a great deal of stupidity can be a dangerous thing.

Seriously Chris if you can't read simple math...please forget about comprehending the complex SM of gamma!
http://www.avsforum.com/avs-vb/showthread.php?t=767179
(235 - 16) / 100 = 2.19 digital levels per IRE (with black at 0 IRE)
BTW I hope your not challenging the ~1% delta required for human perception :)

I am saying that 1 IRE is not the same as 1% luminance.

I am saying that 1 IRE is not the same as 1% luminance.Pssst,

SMPTE EG 28 recommends the symbol Y' to denote luma and the symbol Y to denote luminance.

Please read my postings more carefully, I used the prime symbol to indicate non-linear gamma correction, and you incorrectly substituted the wrong adjective "luminance" ;)

From Tektronix Standards overview:
“Eventually the IRE (later to be the IEEE) established a unit of measure for video signals. This "IRE unit'' was defined as 1% of the video range from blanking to peak white, without reference to the actual signal voltage. Although defined as a ratio"

Have a nice day!

FYI SMPTE seems to agree with Poynton's 50:1 CR, Weber Fechner, and my postings regarding the HVS perception threshold?http://voyager.ee.fau.edu/Document/smpte95.htm
:rolleyes: Figure 2: Number of bits required for gray scale rendition at various spatial frequencies to keep each gray scale step below the perception threshold. The upper curve is the maximum camera signal level limited by the modulation transfer function (MTF). The lower curve is the lowest signal that can be detected by the visual system, limited by the (CSF) threshold contrast sensitivity function.

The Weber-Fechner law would indicate that the minimum number of bits required to depict gray scale for a reasonable contrast ratio is achieved if gamma correction is used in the camera with a linear transfer curve near black and a logarithmic curve for high light levels. The 1/2.2 gamma correction currently used in cameras is a close approximation to this transfer curve. With a 50:1 contrast ratio for the display, 8 bits are required to have each gray scale step below the threshold of perception.References:
William Glenn, John Marcinka, and Robert Dhein, "Subband Coding Compression System for Program Production, " presented at the 136th Technical Conference and World Media Expo, SMPTE, Los Angeles, October, 1994.

Are you related to Lyndon LaRouche?

No, but if you click on the following link, you can explain all you know about "image science" to:

Imaging Technology Space Center
(ITSC) one of 16 NASA Research Partnership Centers

Dr. William Glenn / Director
Tel: (561) 297-3411
glenn@fau.edu

Thanks in advance :)