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I start with BPSK achieving 1 bits/sec/Hz over passband AWGN. Factoring in 1/3 rate coding this becomes 0.333 bits/sec/Hz. This is not correct. The Shannon noisy channel coding theorem states that the reliable discrete-time rate $r$ (whose unit is bits per symbol, or bits per channel-use, or bpcu) is upper-bounded $$r \lt \frac{1}{2}\log_2\left(1 + \frac{S}{...


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You've computed the bound on the capacity of an AWGN channel. But with the additional constraint that the signal must use BPSK modulation the capacity will be lower (I don't know offhand what it is but there should be a reference in the text). Edit: There's a good description of the difference between continuous- and discrete-input AWGN channels here: ...


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