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I am to use the following equation as part of a model for a wireless communications network:

Channel Model

I would like to know the specific parts of the eqautions:

"k" - is this the Boltzmann constant?

"v" - I have read that the minimum value of this can be 2, but how do I calculate it or what is the formula for it?

"xi" (zero-mean log-normal shadowing) - how do I calculate this/what is the formula for this? (could it be the Friis equation?)

"F" the equation just says that this corresponds to Rayleigh fading but it isn't clear what this means.

This part of a model is taken from the following paper: http://thesai.org/Downloads/Volume9No1/Paper_55-An_Energy_Efficient_User_Centric_Approach.pdf

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A full and complete answer to your question would be be too long to write as your question touches on many different losses. If you want a complete treatment you can read Chapter 2 and 3 of the free wireless book by Goldsmith linked HERE.

The "channel gain" is the sum of all losses that the transmitted signal experiences before it finally reaches the receiver. The three losses are the i) path loss, ii) slow fading loss and the iii) fast fading loss. The slow fading loss follows a log-normal distribution and the fast fading loss follows either a riciean distribution or a rayleigh distribution. Both these distributions take a parameter \$\sigma\$ which depends on the environment so you have to search and find the empirical value of \$\sigma\$ which corresponds with your particular environment. If you calculating an instantaneous loss you can just pick a value from these distributions for each time instance but if you are performing a link budget then you need to integrate each of these distributions and find the worst case loss.

In the model you have linked to, the authors have the path loss represented as \$\left(\kappa - \upsilon\log _{10} d_{i,j} \right)\$ (Which is odd as this is usually represented as something like \$\left(\kappa - 10 \upsilon\log _{10} d_{i,j} \right)\$, even in the reference the authors cite). The path loss is the loss you get from Friis equation.

In the paper, the slow fading loss has been represented in the form \$(\xi _{i,s,j})\$ and the fast fading loss \$\left(10 \log_{10} F_{i,s,j}\right)\$.

So in general when perfoming a link budget you would have \$P_r = P_t - G_c\$ where \$P_r\$ is power received, \$P_t\$ is power transmitted and \$G_c\$ is channel link gain. The channel link gain is evaluated as

$$ G_c = L_{\text{PATH}} + L_{\text{SLOW}} + L_{\text{FAST}} $$

where

$$ L_{\text{PATH}} \approx \text{(FRIIS LOSS)} = G_t + G_r + 10\log_{10} \left(\frac{\lambda}{4\pi d}\right)^2 $$

For the slow fading loss with a margin of \$p\$ in an environment with a value of \$\sigma\$ of \$\sigma _s\$. The value of \$p\$ is the probability (i.e between 0 and 1) that an instantaneous slow fading loss will be more than \$L_{\text{SLOW}}\$. You can take \$p\$ to be a high value such as 0.95. \$L_{\text{SLOW}}\$ = \$\xi\$ will be found by solving for \$\xi\$ in the formula

$$ p = \text{CDF of Normal Distribution} = \frac{1}{2} \left[ 1 + \text{erfc}\left( \frac{\xi}{\sigma _s \sqrt{2}} \right) \right] $$

Similarly for fast fading you have \$L_{\text{FAST}} = 10\log _{10} F\$ where \$F\$ is found by solving

$$ p = \text{CDF of Rayleigh Distribution} = 1 - e^{\frac{-F}{2 \sigma _f ^2}} $$

You can get typical values of \$\sigma _s\$ for various environments and get information on evaluating the instantaneous loss from the answer to this QUESTION.

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