Skip to main content
7 votes
Accepted

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

Only top scored, non community-wiki answers of a minimum length are eligible