One Problem on Communication Systems Engineering with Answer | ELEG 403, Assignments of Electrical and Electronics Engineering

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Problem 6.16 The probability of symbol error for 16-QAM is given by 3E, Pi = (1 Zone soni Setting P, = 107, we get 3 1 BE ay 10° = -= = {1 5) ext 30N, Solving this equation for E,,/No, av = 58 Ny = 17.6dB The probability of symbol error for 16-PSK is given by E. Pi,= eat | sirens) Setting P, = 107, we get 10° = erfel [Fanceo) No Solving this equation for E/No, we get =~ = 142 = 21.5aB No Hence, on the average, the 16-PSK demands 21.5 - 17.6 = 3.9 dB more symbol energy than the 16-QAM for P, = 10°. Thus the 16-QAM requires about 4 dB less in signal energy than the 16-PSK for a fixed Ng and P, = 103, However, for this advantage of the 16-QAM over the 16-PSK to be realized, the channel must be linear. 320