Generate coefficients of fading channel (lognormal + Rayleigh)

Question

This question concerns fading channels in wireless communication, but its essence is still statistics and distributions, so I pose it here.

I want to generate (sample) complex-valued coefficients that model the wireless channel by taking into account a constant, a lognormal distribution (slow fading) and a Rayleigh distribution (fast fading).

I begin with a small example considering only the constant and Rayleigh fading. Let $P$ be the average received power representing the constant. Then, channel coefficients distribute as

$$A=\sqrt{\frac{P}{2}}(V+iW)$$

where $V,W\sim \mathcal{CN}(0,1)$. This means that the channel envelope $|A|$ is Rayleigh distributed since $\sqrt{V^2+W^2}$ is essentially a Rayleigh random variable. Note that the average received power is still $P$ since

$$\mathbb{E}\left[|A|^2\right]=\frac{P}{2}\mathbb{E}\left[(V^2+W^2)\right]=P$$

Now to the real question. I wish to include also the slow fading part into the expression above. In dB scale the mean and standard deviation of the slow fading is given to be $\mu_{dB}=0$ dB and $\sigma_{dB}=5$ dB, respectively. I believe the new expression can be written as

$$A=\sqrt{\frac{P\cdot X}{2}}(V+iW)$$

where $X=e^{Y}=e^{(\mu_{dB}+\sigma_{dB}Z)/10}=e^{\sigma_{dB}Z/10}$ and $Z\sim \mathcal{CN}(0,1)$.

Is this correct? The average receiver power should still be $P$.

Remark 1: Is it paramount that the base for the lognormally distributed random variables is $e$ as in $X=e^{Y}$? Can't the base be $10$ and $X$ still be lognormally distributed?

Remark 2: Assume that $P'=10\log(P)+\sigma_{dB}Z$, where $Z$ is as before. Is it then equivalent to the proposed expression above to generate the coeffcients as the following?

$$A=\sqrt{\frac{10^{P'/10}}{2}}(V+iW)$$

Answer 1

In regard to your first remark: it depends on how you want $X$ be related to the underlying normal distribution $Y$. If $Y = (\mu_{dB} + \sigma_{dB} Z)/10$ where $Z$ is standard normal, then the distributions of $10^Y$ and $e^Y$ do not have the same lognormal parameters, since $$10^Y = e^{Y \log 10},$$ thereby introducing an additional scale factor. But both $10^Y$ and $e^Y$ belong to the lognormal family.

In regard to the second remark: no. If $P' = P + \sigma_{dB} Z$, and $P$ is a constant, $\sigma_{dB}$ is the same as in $X = e^Y = \exp((\mu_{dB} + \sigma_{dB} Z)/10)$, then $PX \ne P'$.

In regard to your initial question about the average power, if $Z$ is independent of $V$ and $W$, then the expectation of $X$ is not $1$: it is $e^{\sigma_{dB}^2/200}$ if $\mu_{dB} = 0$; and the expectation of $\sqrt{X}$ is $e^{\sigma_{dB}^2/800}$.

Note that for any lognormal distribution, if the parameter $\mu$ is nonnegative, the expectation is strictly greater than $1$, since $\sigma > 0$ implies $e^{\mu + \sigma^2/2} > 1$. Indeed, we must have $\mu = -\sigma^2/2$ for the expectation to equal $1$.

Answer 2

Thanks for the discussion so far. Please keep replying, this topic is real urgent for me. Based on the discussion so far I believe I am ready to state a few things.

Let $X=e^{\sigma Z}$, where $Z$ is standard normal variable. The expectation of $X$ is known to be $\mathbb{E}[X]=e^{\sigma^2/2}$. Based on earlier discussion we have that

$$A=\sqrt{\frac{P\cdot X}{2\cdot \mathbb{E}[X]}}(V+iW)$$

in order for $\mathbb{E}[|A|^2]=P$ to be true.

In my case, $\sigma_{dB}$ is known but $\sigma$ is not. At first I thought maybe $\sigma_{dB}=10\log_{10}(\sigma)$, but this may not be true based on the following exposition. Since $\sigma_{dB}$ is known, the received power can be written in dB scale as

$$P'_{dB}=10\log(P)+\sigma_{dB}Z$$

and in linear scale:

$$P'=10^{P'_{dB}/10}=10^{(10\log(P)+\sigma_{dB}Z)/10} = P\cdot 10^{\sigma_{dB}Z/10}$$

The expectation of $P'$ is $\mathbb{E}[P']=\mathbb{E}[P\cdot 10^{\sigma_{dB}Z/10}]=P\cdot e^{(\sigma_{dB}/10)^2/2}$. It means that

$$A=\sqrt{\frac{P'}{2\cdot e^{(\sigma_{dB}/10)^2/2}}}(V+iW)$$

in order for $\mathbb{E}[|A|^2]=P$ to be true. Here $P',V,W$ and $\sigma_{dB}$ are all known, thus enabling us to generate the channel coefficients accordingly.

Is this correct?