Re: Color Reference of NTSC Formula?

[Mark McDougall <markm@vl.com.au>]

Bryan Parkoff wrote:

> Think of color in a square.  It is truly RGB.  Darkness color is in 
> the bottom of this square and brightness color is in the top of this square. 
> You fill 0-255 value on R, G, and B signal.  It allows the line of RGB to be 
> moving from the bottom to the top of this square until you get the correct 
> color.

An RGB colour space incorporates both chrominance and luminence which is
why you have black and white included in the space. In, YIQ colour space
(which is the same as CIE XYZ) the 'wheel' mapped out by I&Q describes
only the chrominance - it does *not* include the luminance, Luminance
(Y) represents the "power" or intensity of the light and is not encoded
in the phase shift but rather the amplitude of the grey-scale signal.
That's why there's no black or white on the "VECTOR SCOPE".

>     Think of color in a circle what I am referring Deep Red, Dark Blue, Dark 
> Green, and Brown.  Apple IIgs attempts to emulate a color in the circle to 
> display HGR and DHGR using phase shift by the following 0 to 360 degree. 
> You place one bit on position 0 and three zero bits on position 1 through 3 
> of DHGR.  The phase shift is on 0 degree.  It displays Deep Red pixel.  It 
> is where an arrow in the color circle is pointed to the 0 degree.

I think you've sort of got the idea, but it's twisted a little. A group
of 4 dots does not represent a "phase shift". Phase shift is relative to
the colour burst, and is a property of a single pixel, not a group of
pixels. I *think* what may be confusing is that the Apple can only
adjust the phase of the signal by +/- 90 degrees for each bit, so it
would take 4 shifts (4 pixels) to move all the way around the colour
wheel. Of course, you can go backwards as well, so at most you need 2
shifts to get to any desired pixel value.

> It does the same process to clearing and setting bit.  It is like a bit is 
> moving from phase shift to phase shift.  1000, 0100, 0010, 0001, 1000.

Look at the *shape* of the waveform produced a bitstream.
eg. Say we have a colour burst encoded digitally at 4X the burst
frequency. It may look like this...
	0011001100110011
So it takes 4 bits for once cycle, which looks more like a sine wave
after it's passed through a low-pass filter.

Now we choose a colour exactly *in phase* with the burst (which is a bit
confusing, because the I axis is defined as being 57 deg from the burst,
so our colour would be whatever is represented at -57 deg on the vector
scope)... the encoding would be:
burst:	00110011001100110011001100110011
IQ:	00110011001100110011001100110011
which is of course exactly the same as the burst.

Now say we wish to move 90 degrees around the colour circle. So we need
to shift the chrominance signal by 90 degrees. We then have...
burst:	00110011001100110011001100110011
IQ:	01100110011001100110011001100110
The above signals show the encoding over *several* pixels.

When you start changing the phase *every* pixel, it gets messy and
difficult to actually show the phase shift.

>     Do you agree what I provided 16 colors in a table with 4 bits and 
> degree?  Is it wrong?

I haven't checked all your numbers but yes, I think some of them are wrong.

>     How do you get the value of Q and I with your knowledge?  Do you guess 
> the values by trying to pick up the correct degree for the color?  

No I didn't guess at all. It's all derived from the formula that
converts RGB to YQI...
	Y = 0.3R + 0.59G + 0.11B
which is the luminance or grey-scale component
	Q = 0.21R - 0.52G + 0.31B
	I = 0.6R - 0.28G - 0.32B
which give the two quadrature components of chrominance.

So for RGB red (255,0,0) which we scale to (1,0,0) for simplicity...
	Y = 0.3*1 + 0.59*0 + 0.11*0 = 0.3
	Q = 0.21*1 - 0.52*0 + 0.31*0 = 0.21
	I = 0.6*1 - 0.28*8 - 0.32*0 = 0.6
I&Q are two vectors at right-angle (quadrature). The representative
phase is the angle subtended when you join the vectors (this is the
angle on the colour wheel, which gives the hue) and the magnitude is how
far from the centre of the wheel the colour lies, which gives the
saturation. The *angle* is given by the arctan of the length of Q
divided by the length of I.
So for our example of red, with Q=0.21 and I=0.6, the phase is
	arctan(0.21/0.6) = 19.3 deg.
Since the I axis is defined as being 57deg from the burst, looking at
the vector scope burst is at 180, so I is at 123deg. Subtracting 19.3
from 123 gives 103.7 which corresponds with the Red square on the vector
scope.

For yellow, which is RED+GREEN in RGB space, I used (255,255,0) to get
167 deg.

Regards,

-- 
Mark McDougall, Engineer
Virtual Logic Pty Ltd, <http://www.vl.com.au>
21-25 King St, Rockdale, 2216
Ph: +612-9599-3255 Fax: +612-9599-3266