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7 votes

Why don't we transmit at rates higher than the Shannon capacity if we are going to get a nonzero probability of error anyways ?

Look at the strong converse to Shannon's theorem: for rates above the channel capacity, if $n$ bits are to be transmitted, the probability of error is exponentially close to 1, so $1-e^{c n}$ for ...
Peter Shor 's user avatar

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