I found the probabilities of error (symbol error rate) for 16-QAM and 16-PSK.
Pe_16-QAM = 1 - (1 - 1.75 * Q(sqrt(0.2 * Es/No)))^2
Pe_16-PSK = 2 * Q(sin(pi/16) * sqrt(2 * Es/No))
Then I graphed the SER vs. Es/No. How do I find which modulation scheme is more power efficient, in decibels? To reach the same symbol error rate, one has to use more power than the other. From the graph, for Pe = 0.4, 16-PSK is using 9 dB while 16-QAM is using 8 dB. Does that mean 16-PSK is the more power efficient modulation?
Pe_16-QAM = 1 - (1 - 1.75 * Q(sqrt(0.2 * Es/No)))^2
Pe_16-PSK = 2 * Q(sin(pi/16) * sqrt(2 * Es/No))
Then I graphed the SER vs. Es/No. How do I find which modulation scheme is more power efficient, in decibels? To reach the same symbol error rate, one has to use more power than the other. From the graph, for Pe = 0.4, 16-PSK is using 9 dB while 16-QAM is using 8 dB. Does that mean 16-PSK is the more power efficient modulation?
