In literature, I am seeing that MRC almost always works better than SC. Also, in my simulations for Rayleigh and Rician fading channel, I am getting the same results. Could anyone please explain what are the reasons behind this? I tried to read some papers but did not get a clear idea.

  • $\begingroup$ removed the unrelated matlab tag. Just because you're using matlab doesn't mean a question is related to that. Try to keep your tags as precise and descriptive as possible. $\endgroup$ Commented Jul 16, 2018 at 11:13

1 Answer 1


Very quick explanation: SC just throws away $N-1$ of $N$ observations, whereas MRC combines all $N$.

MRC does to a signal in spatial domain what a matched filter does in frequency domain: it maximizes the inner product of signal vector and weights. The exact same derivation as for the matched filter's optimal performance under AWGN apply to the MRC under the same assumption.

You can also do the same consideration on a basic stochastic level.

Assume the combiner has correct knowledge of each observation's SNR.

Then, SC picks the best one. Derivation is simple: if channel realizations are Rayleigh distributed, then the best of these channels has $\sum_{n=1}^N \frac1N$ increased SNR in expectation, $N$ being the number of concurrent observations.

For MRC, you literally get the weighted average of the received signals, so your overall SNR literally is $\sum_{n=1}^N \frac {S_k}{N_k}$, which yields a diversity gain of $N$ (the maximum possible).

Regarding your "I've read papers": the 1959 Brennan paper is the only one you need to read for this :) But, honestly, this is knowledge that's 60 years old, so you don't have to read the original paper, and should probably rather read a good modern textbook or lecture slides explaining combining methods, as these tend to didactically process the paper's content to make it more accessible. A really quick google for "diversity gain MRC" instantly yielded a link to these lecture notes, and you could have found them, too! They're easy to read, I find after quickly scanning them.

  • $\begingroup$ Is there a case that SC outperforms MRC? And is there a case that MRC is not available that one must use SC? $\endgroup$
    – AlexTP
    Commented Jul 16, 2018 at 15:00
  • $\begingroup$ @AlexTP define "outperform"! If you mean "SC being constantly better than MRC": yes, but only if your channel isn't Rayleigh, or your noise isn't uncorrelated. If both is given, then: see the formula, $\sum_{n=1}^N\frac 1n < \sum_{n=1}^N 1 = N \,\forall N > 1$, hence MRC better than SC for any number of antennas greater than one. $\endgroup$ Commented Jul 16, 2018 at 15:13
  • $\begingroup$ I do agree with your answer and yeah, the definition of "outperform" is exactly what I am asking. I mean I just wonder if there is any criteria that SC is "better" (in normal sense) than MRC. In other words, do people still use SC if two assumptions above are satisfied? $\endgroup$
    – AlexTP
    Commented Jul 16, 2018 at 23:39
  • 1
    $\begingroup$ well, it's certainly simpler to only have to deal with one signal. But no, as you effectively always increase the sum SNR in MRC, if you can do MRC, you'll always do MRC. Reality is that especially for bad SNRs, your CSI estimator might be off, so to make sure you're not mis-estimating a bad signal for a good one, you might be cutting of your MRC sum below some SNR threshold. But, really, the cases where you need to do that are relatively hard to construct – how many 16-antenna receivers do you have of which some are borderline noise, so you're not sure if your CSI estimate is good enough? $\endgroup$ Commented Jul 16, 2018 at 23:52
  • $\begingroup$ So, under perfect CSI, you'll never prefer SC, but in reality, CSI doesn't come for free, and so you might be willing to make a trade-off: being unable to estimate the CSI for worse channels without spending too much time on channel estimation, you might want to simply pick one (or a few) channel and roll with that – but that might indicated that you're actually working with a channel that's more Rician than Rayleigh! $\endgroup$ Commented Jul 16, 2018 at 23:56

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