Capacity loss of fast fading regarding AWGN

Question

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.

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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.

Answer 1

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.

Rayleigh fading is when the amplitudes of the real and imaginary part of $h$ are normal distributed; so $|h|$ follows a Rayleigh distribution, which has a probability density function of

$$ p(x=|h|) = \frac {x}{\sigma^2}e^{\frac{x^2}{2\sigma^2}}\text. $$

Hence, $E\left\{\lvert h\rvert^2\right\}$ is not the variance of the Normal distribution with variance $\sigma^2$.

Instead, the value you're looking for is $$\begin{align} E\left\{\log_2\left(\lvert h\rvert^2\right)\right\} &= \int_{-\infty}^\infty \log_2\left(\lvert x\rvert^2\right)\frac {x}{\sigma^2}e^{\frac{x^2}{2\sigma^2}}\,dx\\ &= \int_0^\infty 2\log_2\left(\lvert x\rvert\right)\frac {x}{\sigma^2}e^{\frac{x^2}{2\sigma^2}}\,dx\\ &= \frac2{\sigma^2}\int_0^\infty x\log_2\left( x\right) e^{\frac{x^2}{2\sigma^2}}\,dx \end{align}$$