## 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