The theoric capacity of fast fading channel is: $$ C_{\rm fading}=E\left\{\log\left(1+\lvert h \rvert^2\textrm{SNR}\right)\right\}, \quad\text{with $E$ is expectance operator.}$$
At high SNR: $$\begin{align} C_{\rm fading}&=E\left\{\log\left(\lvert h \rvert^2\textrm{SNR}\right)\right\}\\ &=\log(\textrm{SNR})&&+E\left\{\log\left(\lvert h \rvert^2\right)\right\}\\ &=C_{\rm AWGN} &&+ E\left\{\log\left(\lvert h \rvert^2\right)\right\} \end{align}$$
The difference regarding AWGN is $\Delta=E\left\{\log\left(\lvert h \rvert^2\right)\right\}$ which is stated as $-0.83\textrm{ bits/s/Hz}$ for standard Rayleigh fading model, or $2.5\textrm{ dB}$ requirement.
- Could anyone explain how they come to this result $-0.83\textrm{ bits}, 2.5\textrm{ dB}$ ?
- Standard Rayleigh fading is $C\mathcal N\left(0,1\right)$ then $E\{\lvert h \rvert^2\} = 1$. Using Jensen's inequality $$\begin{align}\Delta&=E\left\{\log\left(\lvert h \rvert^2\right)\right\}\\ &\ge \log(E\{\lvert h \rvert^2\}) \\&= \log(1) \\&= 0 \\&> -0.83\end{align}$$ and this $\Delta$ should not be $-0.83$? I think I made some mistakes but I don't know what it is.
The (stupid) mistake I have committed is that log is concave, so the inquality is in wrong side. A numerical evaluation of the integral proposed by Marcus confirmed the value 0.83 bits. Thank Marcus.