Question

A modulation order M=2 corresponding to the 2 BPSK symbols ' +1 ' and ' -1 '. The channel was FIR order N=1, i.e., the delay-spread channel model includes the present and one delayed symbol. It was shown that the number of states needed to be tracked by the Viterbi algorithm per symbol time was 4 , and the best 2 of these 'survivors' were used to generate the states for the next symbol time. Suppose that the channel is FIR order 1 but M-QAM modulation is used instead of BPSK. Determine the number of states needed by the Viterbi algorithm per symbol time.

Answer 1

The number of states needed by the Viterbi algorithm per symbol time for M-QAM modulation is 256.

Step-by-step explanation:

The number of states needed by the Viterbi algorithm per symbol time can be determined by considering the modulation order M and the channel order N. In this case, the channel is FIR order 1 and M-QAM modulation is used instead of BPSK. For M-QAM modulation, the number of states needed by the Viterbi algorithm per symbol time is given by 2^((M^2)*(N+1)).

Substituting M=2 and N=1 into the formula, we get 2^((2^2)*(1+1)) = 2^8 = 256. Therefore, the number of states needed by the Viterbi algorithm per symbol time for M-QAM modulation is 256.