Example 6.3
A transmitter for a digital radio system is peak
power limited to 150 W. Determine the average power that can be supported for both 16-PSK
and square 16-QAM transmission.
Solution

With reference to one quadrant of the 16-QAM
constellation, the average power developed by each of the vectors A, B, C, D is as follows:

The maximum vector power is given as 150 W, therefore:

Therefore the average power for all symbol states is:
Average power (QAM) = 10a2 / R = 83.33 W
The average power for 16-PSK is the same for
all symbol states and is equal to the peak symbol power since unfiltered PSK is a constant
envelope modulation format. Thus:
Average Power (PSK) = s2 / R = 150 W