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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.

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  • $\begingroup$ you might want to specify what about Rayleigh fading $\sim C\mathcal N(0,1)$; I think that might make it much easier to spot and explain possible errors/problems. $\endgroup$ – Marcus Müller Jan 22 '17 at 12:22
  • $\begingroup$ I mean fading ~$C\mathcal N\left(0,1\right)$, I expect that $\Delta=E\left\{\log\left(\lvert h \rvert^2\right)\right\} > \log(E\left\{\left(\lvert h \rvert^2\right)\right\}) = \log(1) = 0 > -0.83$. I think I made some mistakes but dont know what it is. Thanks $\endgroup$ – AlexTP Jan 22 '17 at 16:19
  • $\begingroup$ The $-0.83$ bits or $2.5$ dB is a typical result you can find in most of books. $\endgroup$ – AlexTP Jan 22 '17 at 16:45
  • $\begingroup$ "fading $\sim C\mathcal N(0,1)$" really doesn't make any sense to me, sorry, that's why I ask. In a Rayleigh scenario, $|h|\sim \text{Rayleigh}(\sigma^2)$, and you'd use the unconscious statistician's lemma to find the expectation of $\log(|h|^2)$; it's a bit of an ugly integral that you'd have to solve, there. And if you could edit your question and include your formula from your comment, and explain where the $>$ comes from, that would be really interesting! $\endgroup$ – Marcus Müller Jan 22 '17 at 17:23
  • $\begingroup$ thank @MarcusMüller, I have modified my question. The > comes from Jensen inequality. $\endgroup$ – AlexTP Jan 22 '17 at 18:14
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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 $$\begin{align} E\left\{\log_2\left(\lvert h\rvert^2\right)\right\} &= \int_{-\infty}^\infty \log_2\left(\lvert x\rvert^2\right)\frac {x}{\sigma^2}e^{\frac{x^2}{2\sigma^2}}\,dx\\ &= \int_0^\infty 2\log_2\left(\lvert x\rvert\right)\frac {x}{\sigma^2}e^{\frac{x^2}{2\sigma^2}}\,dx\\ &= \frac2{\sigma^2}\int_0^\infty x\log_2\left( x\right) e^{\frac{x^2}{2\sigma^2}}\,dx \end{align}$$

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  • $\begingroup$ sorry I should add a comment instead of modifying your answer. I am not used to dealing with stackexchange GUI. Could you please update the old answer with my comment if you agree with it ? Thanks a lot. $\endgroup$ – AlexTP Jan 22 '17 at 22:39
  • $\begingroup$ @portabili you tried to reply to me within my answer; that's not how things are supposed to work. I think if you spotted your own mistake, it's perfectly OK if you post an answer of your own! If it highlights and explains what mistake you made (and your edit was pretty good with respect to that), I don't see a reason not to upvote your answer! $\endgroup$ – Marcus Müller Jan 22 '17 at 22:42

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