I am trying to understand how to calculate the power of a received signal under Raleigh fading.

What I gather is that the Rayleigh fading is a complex Gaussian random variable and that the multipath components are summed.

I also understand that the fading is applied at the symbol level.

Would it be possible to calculate the received power of a signal without having to deal with the number of symbols transmitted in that signal?

So, I have made a mini working example in MATLAB.

R = 500; % Cell radius in meters
eta=4; % path loss exponent
P_BS = 251.189; % BS transmission power in mWatts
D_BS_to_UE = 450; % Distance between UEand BS in meters

N_multipath = 3; % Number of multipaths
h = (randn(1,N_multipath))+1i*(randn(1,N_multipath)); % Rayleigh Fading

P_received = sum(abs(h./sqrt(2)).^2) * P_BS * D_BS_to_UE.^(-eta)

I guess what I am trying to understand are the following:

  1. Whether the sum(abs(h./sqrt(2)).^2) is correct as it concerns the signal in its entirely and not per symbol.

  2. What is the value of N_multipath that can be considered practical? Is there a way to calculate that or is it arbitrarily chosen as I have done?

  3. I have also seen a discussion on taps. What I gather is that it entirely depends on frequency-selective or flat fading. If flat then one tap, if frequency-selective then more than one tap. How do I add the taps if I choose frequency-selective?

  • $\begingroup$ You're asking three different questions, at least some of which already have answers on this site. I'd suggest doing a little bit more research, and then, if things are still unclear, breaking up this question into two or three shorter questions. $\endgroup$
    – MBaz
    Sep 7 '20 at 13:53
  • $\begingroup$ @MBaz I did that already but I felt that they still do not answer my questions. Or at least not in a way that I can grasp. $\endgroup$
    – nashynash
    Sep 7 '20 at 13:56
  • $\begingroup$ I still recommend breaking this up into two or three questions. Most importantly, include in your question(s) what you have researched and what it is that you still don't understand. The way your question is phrased, you're asking for a re-hash of answers that are already out there. $\endgroup$
    – MBaz
    Sep 7 '20 at 14:56
  • $\begingroup$ @MBaz Okay. I will revisit those posts again and refine my question. $\endgroup$
    – nashynash
    Sep 7 '20 at 15:03

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