We have a Shannon formula for Capacity as \begin{align} C=B\log_2(1+\frac{h*P}{B*N_0}) \end{align} where h is a random variable which denotes the channel gain with channel mean power \begin{align} \Omega=E[h] \end{align}

When I do the simulation such as for Rayleigh fading. I have to set a value for $\Omega$. But I don't know if $\Omega > 1$ or it must be smaller than 1. I have read many papers from IEEE. Some of papers set value of $\Omega > 1$, some of them is smaller than 1.

Could you please explain for me?

1- When it is greater than 1 and when it is smaller than 1.

2- In practice, how the range of $\Omega$ is ?

• Please explain how the channel capacity $C$ affected by $\Omega$ since $\Omega$ does not appear anywhere in the expression for $C$, and indeed $C$ seems to be a random variable since it is a function of $h$ (and $h$ is a random variable since you seem to be talking of its expected value $E[h]$. Also, please edit your question to eliminate the boldface large letters. Commented Sep 20, 2012 at 12:15
• In the above equation, h is a random variable which expresses the fading environment. Accordingly, C is also a random variable. Commented Sep 20, 2012 at 16:28
• Could you post references to the papers that you're talking about? Commented Sep 20, 2012 at 22:48
• I am sorry to answer late. Commented Oct 11, 2012 at 6:48
• I am sorry to answer late. Jianghong Luo, Rick S. Blum,"Power Allocation in a Transmit Diversity System with Mean Channel Gain Information", IEEE COMMUNICATIONS LETTERS, VOL. 9, NO. 7, JULY 2005. the channel mean power is less than or equal 1. However, in the paper, Vahid Asghari and Sonia Aıssa, "Adaptive Time-Sharing and Power Allocation for Cognitive Radio Fading Broadcast Channels", IEEE ICC 2010. The channel mean power is greater than 1. Commented Oct 11, 2012 at 6:55

Consider two random variables $x \sim N(0, \sigma^2)$ and $y \sim N(0, \sigma^2)$. Then $h = \sqrt{x^2 + y^2}$ is Raylegh-distributed with parameter $\sigma$, whose p.d.f is

$$f(h) = \frac{h}{\sigma^2} \exp(-h^2/2\sigma^2)$$

According to Wikipedia,

$$\Omega = E[h] = \sigma \sqrt{\frac{\pi}{2}} \implies \Omega \propto \sigma$$

Hence, the range of $\Omega$ depends on the variance of your input signals $x$ and $y$. If you think of $x$ and $y$ as "signal + AWGN", then $\sigma$ is essentially the signal-to-noise ratio. So, whether $\Omega < 1$ or $\Omega > 1$ depends on whether you are modelling low SNR conditions or high SNR conditions, respectively.

• How can $x$ and $y$ be "signal + AWGN?" There is no signal component, as the Gaussian random variables have zero mean. I wouldn't expect $\Omega > 1$ to straightforwardly correspond to high-SNR conditions, as it would imply that, in accordance with the OP's equation, fading would cause an enhancement in channel capacity in high-SNR conditions, which doesn't square with intuition. Commented Sep 25, 2012 at 12:52
• @JasonR In a Rayleigh fading environment, the received signal has random phase (uniformly distributed on $[0,2\pi)$) and so the signal components are treated as zero-mean Gaussian random variables. There is energy in the signal but the energy is variable and leads to high error rates during deep fades. That is why the BER, which is the error rate averaged over deep fades as well as strong signals, decreases only inversely with increasing SNR in a Rayleigh fading channel whereas in an AWGN channel, BER decreases exponentially as SNR increases. Commented Sep 25, 2012 at 14:22
• @JasonR: It's certainly possible, say, if the signal is a stream of BPSK $\{+1, -1\}$ symbols. then "signal + AWGN" is still zero-mean Gaussian-distributed. And OP's equation is not correct either. Channel capacity is defined as the expected value of what OP has written. For ex, for SISO channel, it is $C = E\{\log_2(1 + \rho |h|^2) \}$ where $|h|$ is rayleigh-distributed channel coefficient. Commented Sep 25, 2012 at 15:50
• @bdsatish The statement "if the signal is a stream of BPSK {+1,−1} symbols. then "signal + AWGN" is still zero-mean Gaussian-distributed" is not quite correct. Yes, the sum is zero-mean, but its distribution is what is called a mixed Gaussian distribution, something like $(\phi(x+a) + \phi(x-a))/2$ where $\phi(x)$ is the (Gaussian) density function of the noise. Commented Sep 25, 2012 at 18:12