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For a channel of bandwidth B Hz and signal-to-noise ratio of S/N, i have used Shannon-formula to calculate the maximum channel capacity C. For transmitting at the nyquist baud rate (2 symbol/sec), when i substituded in nyquist formula to find the number of levels, and used the number of levels to find the number of bits per symbol, i found that the number of bits/symbol equals 0.5.
What does this mean? How can this be achieved?

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  • $\begingroup$ this mean in average, one symbol convoys 0.5 bits. For example 2 symbols are needed to send out 1 bit. $\endgroup$ – AlexTP Feb 24 '18 at 19:21
  • $\begingroup$ @ Alex TP The number of bits per symbol is log-2(M), and M equals 2,4,8,16,... None of these values yield 0.5. $\endgroup$ – user24907 Feb 24 '18 at 20:44
  • $\begingroup$ $\log_2(M)$ is the number of bits sent out per symbol, not the number of bits per symbol that guarantees arbitrary low codeword error probabilities assuming codeword length approachs infinity. $\endgroup$ – AlexTP Feb 25 '18 at 2:24

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