I am calculating the capacity of $K$ channels between a transmitter and $N$ receivers in a cellular scenario. I use the following equation:

$$C_n^k = B\log_2\left(1 + \frac{Ph_n^k}{N_0B + I_n^k}\right)$$


  • $C_n^k$ is the capacity between the transmitter and receiver $n$ over channel $k$
  • $B$ is the bandwidth
  • $P$ is the power
  • $N_0$ is a noise term
  • $I_n^k$ is the power signal from the other transmitters (interference)
  • $h_n^k$ is the channel gain

In my scenario I need to simulate the $h_n^k$ using Rayleigh fading. I understand that it relies on the distance between the sender and receiver and as my scenario is in urban area, so I expect to have $(d^{-4})$ in the $h_n^k$ equation as this is how fading is affecting when the signal is traversing buildings. However, when I searched for the Rayleigh fading equation, I could not find similar one.


In the literature, the path loss is commonly omitted. The main reason is that all users experience a similar amount of path loss effect. Therefore, the comparison without path loss is a fair technique. On the other hand, if you desire to obtain realistic results by adding path loss terms, you should also take the equivalent bandwidth of the system, noise temperature, and so on. Secondly, in your equations, you did a typo. I think the channel gains should be written as $|h_n^k|^2$, not $h_n^k$ in logarithm.

  • $\begingroup$ Thanks for your reply. I am interested to model the Preceived in an urban environment. That is the Power_transfer * channel gain. That is why I need to know the channel gain. I do not believe I have a typo in my equation, here is a reference paper: Check equation 5 at page 9 in here: arxiv.org/abs/2012.03414 The term $h_n^k$ is explained in the last paragraph in page 8. $\endgroup$
    – Kasparov92
    Sep 5 '21 at 9:00

Okay I ended up with 2 solutions:

(1) Based on [1] equation (17-19)

$h_n^k = 20 \;log_{10}{(\frac{\lambda}{4\;\pi \; d_0})} * (\frac{d_0}{d})^\gamma$


  • $\lambda$ is the wavelength,
  • $d_0$ is a reference distance for the antenna far field and usually set to 10m
  • $d$ is the distance between the sender and receiver
  • $\gamma$ is the path loss exponent that depends on the propagation environment. For urban environment it ranges 3.7-6.5, which in my case I would set to 4. Table 2 in 1 further details more value for different environments.

In this case, the $\lambda$ will be different for each channel, and will is calculated as follow $\lambda = \frac{c}{f}$ where c is speed of light and f is the channel frequency.

So this approach is more accurate but very dependent on the application. The other approach is more statistical approach.

(2) Based on [2] (section 2.3, page 3)

The channel gain between Tx and Rx for a channel k is $h_n^k = (e^1 / r^\gamma)$


  • e is a number drawn from exponential distribution
  • $\gamma$ is the path loss exponent that depends on the propagation environment

Both solutions are used in the literature and are valid for research.

  • $\begingroup$ Hello, the mathematical expression for $h_n^k$ is the path-loss itself. It has a multiplicative effect on the received signal. On the other hand, channel gain is a complex term that represents the small-scale behaviors of the channel. In your $h_n^k$ expressions, there is no random variable, thus, the channel gain is not varying. For this reason, the channel model becomes AWGN not Rayleigh. $\endgroup$ Oct 17 '21 at 0:00
  • $\begingroup$ $h_n^k$ is random as e is not a constant, but rather a random number drawn from exponential distribution as I explained in my answer $\endgroup$
    – Kasparov92
    Oct 18 '21 at 3:14

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