Explanation of the Shannon Limit

I dont understand shannon limit.
For example I see that "single-link code is approximately 1.3dB away from the shannon limit at an error rate of 10^-5.

so how this comparison is made .

if you could help please ...

Thankyou very much for this..

shannon limit 1.6

Shannon limit: Eb/N0 = -1.6 dB, where Eb - a bit enegy, N0 - one-sided power spectral density of noise, which is usually considered to be Gaussian.

This expression means, that if Eb/N0<-1.6 dB there are definitely no any noise combating codes at all, which can provide arbitrarily small error rate. Otherwise, such codes must exist.

Your quotation: "For example I see that "single-link code is approximately 1.3dB away from the shannon limit at an error rate of 10^-5". It means, that at an aerror rate 10^-5 the signal-to-noise ratio is 1.3 dB much than the Shannon limit, that is, you have the reserve of 1.3 dB against the Shannon limit .

With respect,

Dmitrij

1.6 db shannon limit

It means, that at an aerror rate 10^-5 the signal-to-noise ratio is 1.3 dB much than the Shannon limit, that is, you have the reserve of 1.3 dB against the Shannon limit .

Does it mean that to gain error rate of 10^-5 you need 1.3 dB more power than required for Shannon capacity ..does it mean like that ??
If so that what power in dB is needed according to Shannon to achieve error rate of 10^-5 and how to find it.

With Respect..

shannon-limit

Just expanding on what Dimtrij said..

>>Does it mean that to gain error rate of 10^-5 you need 1.3 dB more power than required for Shannon capacity ..does it mean like that ??

Yes. In other words, if you were to find out that ultimate code, you would be able to reduce your power by 1.3 db and still maintain your error rate to be 10^-5 (or lower). But you were to reduce power beyond 1.3 db from current, there is no scheme that will allow you maintain 10^-5 or lower BER

>>If so that what power in dB is needed according to Shannon to achieve error rate of 10^-5 and how to find it.

-1.6db. The closer your scheme is to that, closer you are to the limits of the channel. Actually, you can achieve any lesser BER also, if you have sucha scheme.

The results applies for gaussian channels.

-b