As I understand it luminance is the amplitude of the signal, and color is the phase. You need 2 color signals (the third one can be derived from the other 2, like B = 1 - R - G). How can 1 phase signal encode 2 colors?

  • \$\begingroup\$ There are actually 3 separate carriers in a PAL signal. The luminance carrier, which is compatible with BW television, the Chroma carrier, that carries the color signals in quadrature and the audio carrier. More information: en.wikipedia.org/wiki/PAL \$\endgroup\$ – drxzcl Nov 6 '12 at 12:04

I don't know the details of PAL, so I'll answer for NTSC although I think PAL works pretty much the same way.

It takes a total of three degrees of freedom to specify how to light a color pixel. These three degrees of freedom can be expressed various different ways as long as they are sufficiently linearly independent. Eventually for CRT TVs, whatever the three signals are need to be converted to drive levels for the red, green, and blue electron guns.

Black and white TV is easy to understand. The video signal (after demodulation from the RF carrier) during a horizontal sweep is simply the intensity. This was used to directly drive the single electron gun and all was fine.

When color came along, it was made backwards compatible so that old B+W TVs could pick up the new color broadcasts, just show only the black and white part of the picture. The amplitude of the signal was already the intensity, so that was left alone. That's one degree of freedom.

Somehow two more degrees of freedom needed to be added to the signal. This was done by adding a high frequency signal on top of the basic intensity signal. This new signal was called the color carrier. It's frequency was just a bit higher than what B+W TVs could resolve, so it was unnoticable on old sets. The additional two degrees of freedom are encoded on this color carrier by its phase and its amplitude. A color TV has a fairly tight filter for the color carrier to isolate it, then it separately demodulates its amplitude and phase to make the two additional signals needed beyond the basic intensity information already there without the color carrier.

Demodulating the color carrier amplitude is conceptually straight forward, but how to demodulate the phase? Compare the phase angle to what? Where is the reference? This is where the color burst comes in. During the vertical blanking interval after the sync tip and before the intensity signal for the new scan line (this area of the video signal is called the "front porch"), the color burst was added to provide the 0 phase reference. A color TV needs a good low-drift oscillator that locks onto the color burst signal and stays in sync with very little phase drift until the next color burst. This phase-referenced internal color oscillator in the TV is then phase-compared to the immediate phase of the color carrier, and the result becomes the third degree of freedom.

The three values decoded in this way are not directly R, G, and B, but they are linearly independent enough to map out a color space of their own. A linear transformation is required to convert these three signals into the separate R, G, and B that ultimately drive the electron guns, but that's just a bunch of resistors to perform the 3x3 matrix multiply by fixed coefficients.

The meaning of the three directly decoded signals is roughly intensity from the baseband signal, color saturation from the color carrier amplitude, and hue from the color carrier phase.

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  • \$\begingroup\$ My understanding: in NTSC the phase of the colour subcarrier differs by half a cycle between one scan line and the next. That helps any part of the colour signal that bleeds into luminance visually to cancel itself out by doing the opposite from one line to the next. But phase shift resulting from transmission path isn't addressed — hence tint controls. PAL has phase align across four lines rather than two, and switches the phase of one subcarrier every line (thinking QAM, not FM/AM), which cancels out transmission phase errors — hence no tint controls. PAL is NTSC, modified with hindsight. \$\endgroup\$ – Tommy Aug 5 '15 at 17:52

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