Brief Summarizing Remarks on Rayleigh Fading

A Rayleigh-faded channel $h$ is often written as

$$h \sim \mathcal{N}_{\mathbb{C}}(0,1)$$

which is short-hand for saying that 

$$\Re\{h\} \sim \mathcal{N}(0,1/2) \\ \Im\{h\} \sim \mathcal{N}(0,1/2)$$

where $\mathcal{N}(\mu,\sigma^2)$ is a normal distribution with mean $\mu$ and variance $\sigma^2$.

We have $\sigma^2 = 1/2$ in the case of $h \sim \mathcal{N}_{\mathbb{C}}(0,1)$.

The magnitude of the channel $\lvert h \rvert$ follows a Rayleigh distribution where $\sigma = 1/\sqrt{2}$ whose PDF is

$$f(x) = \frac{x}{\sigma^2} \exp \left( \frac{-x^2}{2\sigma^2} \right)$$

which, taking $\Omega = 2 \sigma^2$, is equivalent to

$$f(x) = \frac{2x}{\Omega} \exp \left( \frac{-x^2}{\Omega} \right).$$

Note that when $\sigma^2 = 1/2$, $\Omega = 1$.

Sometimes the Rayleigh distribution is parameterized by $\sigma$. Other times it is parameterized by $\Omega = 2 \sigma^2$.

The distribution of $\lvert h \rvert^2$ follows an exponential distribution with PDF

$$f(x) = \lambda \cdot \exp \left( -\lambda x \right)$$

where $\lambda = \frac{1}{2\sigma^2} = 1$ (when $\sigma^2 = 1/2$) and whose mean is $\lambda$, implying that $\mathbb{E}\left[ \lvert h \rvert^2 \right] = 1$.

In other words, the average channel power is 1. This is convenient for abstracting the large-scale path loss (e.g., Friis) and large-scale fading (e.g., log-normal shadowing) from the small-scale fading distribution (e.g., Rayleigh fading).

One could also use $\Omega = 2 \sigma^2 \implies \lambda = \frac{1}{\Omega}$ to write the PDF as

$$f(x) = \frac{1}{\Omega} \cdot \exp \left( \frac{-x}{\Omega} \right).$$

Note that $h$ can also be written as having distribution

$$h \sim \mathcal{N}_{\mathbb{C}}(0,\Omega)$$

where now it’s clear that $\Omega$ is the average power of $h$ (i.e., its variance is its average power since $h$ is zero mean).