4 ms·
Can’t you calculate the CRLB for any given distribution if you wanted? That’s what my lab did for microscopy anyway. Saying you’re beating the Shannon limit is
by ramraj07 2y ago
Can’t you calculate the CRLB for any given distribution if you wanted? That’s what my lab did for microscopy anyway. Saying you’re beating the Shannon limit is like saying you’re beating the second law of thermodynamics to me.. but I could be wrong.
- deleted 2y ago[deleted]
- Sanzig 2y agoYou are correct. People often say "Shannon limit" (the general case) when they are really referring to the "Shannon-Hartley Limit" (the simplified case of an additive white Gaussian noise channel). For example, MIMO appears to "break" the Shannon-Hartley limit because it does exceed the theoretical AWGN capacity for a simple channel. However, when you apply Shannon's theory to reformulate the problem for the case of a multipath channel with defined mutual coupling, you find that there is a higher limit you are still bounded by.
- s1dev 2y agoI often wondered why MIMO was such an investigated topic. It would make sense if the Shannon limit is higher for this channel. Is there a foundational paper or review that shows this?
- cycomanic 2y agoShannon theory assumes Gaussian noise, however in the very low power regime that's just not true. I agree it's unintuitive. Have a look at the Gordon paper I posted earlier.