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It is well known that spatial diversity can combat multipath small scale Rayleigh fading in a SIMO system by offering the receiver multiple options of independent receptions using more than one antenna ( this is called selection combining, which is one of several spatial diversity schemes). The basic idea is that the probability of several antennas all simultaneously seeing destructive fading is much lower than the same probability in case of only one antenna

Since by adding more and more antennas we get closer and closer to an AWGN channel, this means that the channel becomes more and more independent from multipath fading, and this is also called channel hardening (channel converges more and more to a deterministic value)

The question is the following: Is there a mathematical derivation that shows how can a random Rayleigh PDF (probability distribution function) in case of one antenna converge more and more to a deterministic value in case of a large number of receiver antennas

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  • $\begingroup$ Can't you just take the equations for MRC and find the limit as the number of receivers goes to infinity? $\endgroup$ – MBaz Dec 21 '20 at 15:06
  • $\begingroup$ Thank you very much for the editing (I upvoted)! $\endgroup$ – Marcus Müller Dec 22 '20 at 17:30
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multiple options

So, we're talking about Selection Combining (SC); there's other combining methods, too!

independent receptions

Good that you're stating that, this will be handy in a minute.

Is there a mathematical derivation that tackles this subject from this point of view?

The very definition of a CDF tackles that, but only if the channel realizations have a distribution that have a finite magnitude maximum, which is usually not the case.

Let $|h_i|$ be the magnitude of the $i$th of the $N$ individual channels. Then, the cumulative density function of the "best of all $N$ channels" is simply defined as $\eqref{def}$

\begin{align} F_{\text{SC},N}(x) &= P\left(\max_{i=1,\ldots,N}|h_i| < x\right) \label{def}\tag{1}\\ &=P\left(|h_1|<x \,\wedge\, |h_2|<x\,\wedge\,\ldots\,\wedge\,|h_N|<x\right)\tag2{}\\ &=\prod_{i=1}^N P\left(|h_i|<x\right)\tag3&\text{due to independence}\\ &=\prod_{i=1}^N F_{|h_i|}\left(x\right)\tag3 & \text{definition of "CDF"}\\ &=\prod_{i=1}^N F_{|h_1|}(x)\tag4 &\text{all identically distr.}\\ &= \left(F_{|h_1|}(x)\right)^N\tag5\label{ord} \end{align}

I'm almost certain you've seen $\eqref{ord}$ already – you're looking for the concept of diversity order.

Now, let's see about your infinite number of antennas. Two ways to say the same thing:

  1. You know that this doesn't make sense. If I'm sampling any continuous distribution, such as an absolute normal distribution, an infinite amount of times, I will surely hit any non-zero-mass interval. So, even from basic definition of probability, the answer is already clear: you can assume that you get an arbitrarily large channel coefficient! Congratulations!
  2. $\lim_{N\to\infty} k^N = 0\quad\forall |k|<1$, no matter what $k$ actually is. But if $k=F(x)$ is a cumulative probability function (CDF), it's always $<1$ (unless you hit the maximum value), by definition of CDF. In other words: you surely never get anything but 0; that means, the probability of getting a value out of your diversity combining that is smaller than any value that's not the maximum of your possible channel realizations is 0, which means you always get the maximum possible channel.

This makes no sense physically, because it especially means that with finite transmit energy, one of your infinitely many receive antennas gets infinitely much receive energy, if you assume any fat-tailed distribution of channel coefficient magnitudes (e.g. normal). That's (one of the many reasons) why your considerations don't matter for communication engineering.

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  • $\begingroup$ I would like to reformulate my question: How can a random Rayleigh fading probability distribution function in case of one antenna converge to a deterministic value (which is the definition of channel hardening) in case of a large number of receive antennas? $\endgroup$ – ali khalil Dec 21 '20 at 21:59
  • $\begingroup$ Can you then actually state your exact question in the question itself (instead of as a comment here), please? You basically asked the opposite – a fixed value is not infinity, you didn't mention Rayleigh fading, and you basically stated a selection combining question, which I don't think you want to ask here. Also, bad news: for finite numbers of receive antenna, it doesn't ever reach a deterministic value (it can't, see my answer). $\endgroup$ – Marcus Müller Dec 22 '20 at 7:33
  • $\begingroup$ I am afraid you didn't understand the question... so I do not think you can answer. Never mind! Thank you $\endgroup$ – ali khalil Dec 22 '20 at 14:06
  • $\begingroup$ I really think I do. I teach this kind of stuff. But: no matter whether I understand it or not – edit your question so that others can read your clarification; nobody but me will read the comments under my answer. $\endgroup$ – Marcus Müller Dec 22 '20 at 14:14
  • $\begingroup$ If you do then you would have understood what channel hardening is all about: The channel itself tends to be deterministic with a very large number of antennas: when you say a fixed value is not infinity I have the impression that you are confused between what the channel converges to and the number of antennas, I hope I am wrong! $\endgroup$ – ali khalil Dec 22 '20 at 14:18

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