VALUES OF LUMINANCE (Y) AND COLOR DIFFERENCE SIGNALS
When televising color scenes even when voltages R, G and B are not equal, the ‘Y’ signal still represents monochrome equivalent of the color 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. This aspect can be illustrated by considering some specific colors.
Consider a de-saturated purple color, which is a shade of magenta. Since the hue is magenta (purple) it implies that it is a mixture of red and blue. Two word de-saturated indicates that some white light is also there. 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 color.
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 color difference signals are:
(R – Y) = 0.7 – 0.394 = + 0.306 (volts) (B – Y) = 0.6 – 0.394 = + 0.206 (volts)
Reception at the color receiver—At the receiver after demodulation, the signals, Y, (B – Y) and (R – Y), become available. 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
(iv) (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)
(G – Y) =− 03059.(R – Y) –0 11059.
(B – Y) = – 0.51(R – Y) – 0.186 (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.
(v) 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 color.
A de-saturated orange having the same degree of desaturation as in the previous example is considered now. Taking R = 0.7, G = 0.6, and B = 0.2, it is obvious that the output voltages due to white are R = 0.2, G = 0.2 and B = 0.2.
Then red and green colors which dominate and represent the actual color content with, R = 0.5 and G = 0.4 give the orange hue. Proceeding as in the previous example we get:
(i) Luminance signal Y = 0.3 R + 0.59G + 0.11 B.
Substituting the values of R, G and B we get Y = 0.586 volt.
(ii) Similarly, the color difference signal magnitudes are:
(R – Y) = (0.7 – 0.586) = + 0.114
(B – Y) = (0.2 – 0.586) = – 0.386 and
(iii) At the receiver by matrixing we get
R = (R – Y) + Y = 0.7
G = (G – Y) + Y = 0.6
B = (B – Y) + Y = 0.2
This checks with the voltages developed by the three camera tubes at the transmitting end.
(v) Reception on a monochrome receiver Only the luminance signal is received and, as expected, with Y = 0.586, the orange hue will appear as bright grey.