Why (G-Y) is not transmitted:

• We know that, (G – Y) = – 0.51(R – Y) – 0.186 (B – Y). Since the required amplitudes of both (R – Y) and (B – Y) are less than unity, they may be derived using simple resistor attenuators across the respective signal paths.

  • However, if (G – Y) is to be one of the two transmitted signals then;

  (i) If (R – Y) is the missing signal, its matrix would have to be based on the expression:

The factor 0.59/0.3 (= 1.97) implies gain in the matrix and thus would need an extra amplifier.

(ii) Similarly if (B – Y) is not transmitted, the matrix formula would be:

The factor 0.59/0.11 = 5.4 and 0.3/0.11 = 2.7, both imply gain and two extra amplifiers would be necessary in the matrices.

• This shows that it would be technically less convenient and uneconomical to use (G – Y) as one of the colour difference signals for transmission. In addition, since the proportion of G in Y is relatively large in most cases, the amplitude of (G – Y) is small.
• It is either the smallest of the three colour difference signals, or is atmost equal to the smaller of the other two. The smaller amplitude together with the need for gain in the matrix would make S/N ratio problems more difficult then when (R – Y) and (B – Y) are chosen for transmission.

The G component is having the largest amplitude (0.59V).Therefore (G-Y) will be much smaller than (R-Y) and (B-Y) components.Therefore the (G-Y) component is more vulnerable to interference in its transmission path.

Similar colour objects mean higher bandwidth ,practically it is enough to have a bandwidth of 1.3MHz for chroma signals to produce a picture with satisfactory quality.

This means that the luminance signal will carry all the fine details of the picture and the chroma details will be sent with much less details

This strategy reduces the bandwidth requirements ato a great extent without affecting the quality of reproduced colour picture.This is why it is necessary to weigh down the Chrominance signal