# New RGBY TV.

by Archosaur
Tags: rgby
 P: 328 I just saw a commercial for this. I understand how additive color mixing works, and how the addition of a yellow channel would allow for the creation of more saturated yellows, but if we only have RGB cone cells, then we should be able to detect no more colors than can be created by the mixing of red, green, and blue light, so an RGBY tv should not be able to create any detectable colors that an RGB tv can't. So, what's the deal? Is Sharp just full of it, or am I missing something?
 P: 15,319 Think about the inks on a high quality poster or label for a food product. They could print using only C, M, Y and K (in fact, in principle, even the K is redundant), but to get better colours, they usually print with about 8 colours. A 2ndary purple made from a primary red & blue is a pretty murky purple.
 P: 328 But, if we only have RGB cone cells, then we interpret every color by breaking it down into its R, G, and B components. We therefore could not detect a 100% saturated yellow, because it cannot be expressed completely in RGB components. The color we detect then is less saturated. I don't deny that an RGBY tv can create more saturated yellows. I'm saying that we wouldn't be able to detect them. I've made three drawings below. The first is what we can detect, the second is what an RGB tv can create, and the third is what an RGBY tv can create. I'm guessing that I don't understand cone cells well enough. I'm going to do some reading on them.
 Sci Advisor P: 2,470 New RGBY TV. The RGB cones in your eye don't just pick out a specific frequency. There is an entire spectrum of frequencies that each cone will respond to. Some stronger than others. Similarly, RGB in the TV aren't pure spectral lines either. They are also distributed. Now, imagine some spectrum p(f). What your eye will see is a set of 3 intensities: $$I_R = \int df C_R(f)*p(f)$$ $$I_G = \int df C_G(f)*p(f)$$ $$I_B = \int df C_B(f)*p(f)$$ Where CR(f), CG(f), and CB(f) are response spectra of the cones. The screen has three emissions with spectra LR(f), LG(f), and LB(f), and that gives you an overall spectrum: $$p'(f) = R*L_R(f) + G*L_G(f) + B*L_B(f)$$ Where R, G, and B are the three RGB components you send to the screen. So in general, you want to find a set of R, G, and B such that: $$\int df (R*L_R(f) + G*L_G(f) + B*L_B(f)) C_R(f) = \int df p(f)*C_R(f)$$ $$\int df (R*L_R(f) + G*L_G(f) + B*L_B(f)) C_G(f) = \int df p(f)*C_G(f)$$ $$\int df (R*L_R(f) + G*L_G(f) + B*L_B(f)) C_B(f) = \int df p(f)*C_B(f)$$ For any fixed choice of LR(f), LG(f), and LB(f), this problem cannot be solved in general for arbitrary p(f). This means that no matter what you choose for your RGB elements, there will be spectra that will be perceived differently by human eye from anything you can produce on the screen. Classical example of that is color violet. It cannot be reproduced on RGB screen. The best these can come up with is some shade of purple that resembles it. But if you have a prism, which lets you produce a pure spectral violet, you can convince yourself that it's not something you can recreate on the screen. Adding another channel helps significantly. With RGBY, the complete set of colors that can be reproduce increases. However, here is where I have a problem with it. In order for it to be of any use, you need a signal that encodes 4 components. Standard TV broadcast is YCrCb, which has 1-to-1 with RGB (give or take quantization errors). I'm not sure if HD uses the same or actual RGB. Either way, while the screen itself will be capable of reproducing more colors, I doubt you'd get much use out of it, unless they change broadcast standards.
P: 15,319
 Quote by Archosaur I've made three drawings below. The first is what we can detect, the second is what an RGB tv can create, and the third is what an RGBY tv can create.
The triangles are definitely not equilateral. Our red and green receptors are very close together, leaving blue (yellow's complement) way out on the edge.
P: 5,523
 Quote by Archosaur I've made three drawings below. The first is what we can detect, the second is what an RGB tv can create, and the third is what an RGBY tv can create.
That's not really the best way to think about color perception. The CIE diagram is better:

http://en.wikipedia.org/wiki/CIE_1931_color_space

Halfway down, there's the region of color vision that can be replicated by an RGB system. It's not clear to me how adding Y enlarges the region- I would have guessed that the 4th color should be 510 nm. I wonder, since color representation also involves *negative* values, if the real benefit is 'adding' -Y rather than adding 'Y'.
P: 15,319
 Quote by K^2 However, here is where I have a problem with it. In order for it to be of any use, you need a signal that encodes 4 components. Standard TV broadcast is YCrCb, which has 1-to-1 with RGB (give or take quantization errors). I'm not sure if HD uses the same or actual RGB. Either way, while the screen itself will be capable of reproducing more colors, I doubt you'd get much use out of it, unless they change broadcast standards.
It's possible they simply use a gain algorithm to boost saturation of any yellows. I'm not sure but it's possible that the signal (source) is an ideal spectrum and it's simply the output (diodes,LCD) that's weak in the yellow area.

So, adding a yellow output would allow yellows to be boosted without a corresponding unwanted red/green boost.

But I'm just speculating.
 P: 328 Ok, it was my understanding of cone cells that was flawed. Like K^2 said, they are triggered by a range of wavelengths, and indeed your green and red cones are highly sensitive to yellow light. And, yes, my drawings were crude, but my intention was to show that you can't see a color on the edge of the circle (a 100% saturated color) by mixing any two other colors.
 P: 328 Also, K^2, I like the way you write integrals. Too many people see the dx term as just an end parenthesis.
P: 2,470
 Quote by DaveC426913 It's possible they simply use a gain algorithm to boost saturation of any yellows. I'm not sure but it's possible that the signal (source) is an ideal spectrum and it's simply the output (diodes,LCD) that's weak in the yellow area.
You couldn't send the actual spectrum across, but I think I see what you mean.

If you built a camera that perfectly replicates cones' spectra in its RGB elements, I suppose, you could make it work. Send across the IR, IG, and IB that correspond to excitation level in neurons actually going back to visual cortex, and have the circuitry figure out the rest depending on how many parameters it can play with.

That would work, except that the modern cameras are actually designed to work as well as possible with modern screens. So their intensities are are already mixed up somewhat. You'd still need an entirely new infrastructure, not to mention remaking of all the movies you plan to show, for it to be truly useful.

But I may be missing something as well.
 Also, K^2, I like the way you write integrals. Too many people see the dx term as just an end parenthesis.
Habit, I guess. I have to admit, I don't always use strictly mathematically correct treatment of dx. It's just too convenient to shift it around as an actual infinitesimal element very often. I guess that's the sort of shortcut that results in mathematicians making fun of physicists. But I haven't thought of it as a "parenthesis" at least since differential equations class way back when.