Values of luminance (y) and colour difference signals on colours
When televising colour scenes even when voltages R, G and B are not equal, the ‘Y ’ signal still represents monochrome equivalent of the colour because the proportions 0.3, 0.59 and 0.11 taken of R, G and B respectively still represent the contribution which red, green and blue lights make to the luminance. The following section calculates the Desaturated Purple luminance and colour difference signals in details.
DE saturated Purple:
- Considering a desaturated purple colour, which is a shade of magenta.
- Since the hue is magenta (purple) it implies that it is a mixture of red and blue.
- The white light content will develop all the three i.e., R, G and B voltages, the magnitudes of which will depend on the intensity of desaturation of the colour.
- Thus R and B voltages will dominate and both must be of greater amplitude than G.
- As an illustration let R = 0.7, G = 0.2 and B = 0.6 volts.
- The white content is represented by equal quantities of the three primaries and the actual amount must be indicated by the smallest voltage of the three, that is, by the magnitude of G.
- Thus white is due to 0.2 R, 0.2 G and 0.2 B.
- The remaining, 0.5 R and 0.4 B together represent the magenta hue.
- The luminance signal Y = 0.3 R 0.59 G 0.11 B.
Substituting the values of R, G, and B we get Y = 0.3 (0.7) 0.59 (0.2) 0.11(0.6) = 0.394 (volts).
- The colour difference signals are:
(R – Y) = 0.7 – 0.394 = 0.306 (volts)
(B – Y) = 0.6 – 0.394 = 0.206 (volts)
- Reception at the colour receiver—at the receiver after demodulation, the signals, Y, (B – Y) and (R – Y), become
Then by a process of matrixing the voltages B and R are obtained as:
R = (R – Y) Y = 0.306 0.394 = 0.7 V
B = (B – Y) Y = 0.206 0.394 = 0.6 V
- (G – Y) matrix—The missing signal (G – Y) that is not transmitted can be recovered by using a suitable matrix based on
the explanation given below:
Y = 0.3 R 0.59G 0.11B
also (0.3 0.59 0.11)Y = 0.3R 0.59G 0.11B
Rearranging the above expression we get:
0.59(G – Y) = – 0.3 (R – Y) – 0.11 (B – Y)
Substituting the values of (R – Y) and (B – Y) (G – Y) = – (0.51 × 0.306) – 0.186(0.206) = – 0.15606 – 0.038216 = – 0.194
∴ G = (G – Y) Y = – 0.194 0.394 = 0.2, and this checks with the given value.
- Reception on a monochrome receiver—Since the value of luminance signal Y = 0.394V, and the peak white corresponds to 1 volt (100%) the magenta will show up as a fairly dull grey in a black and white picture.
This is as would be expected for this colour.