# What is the relation between noise (AWGN) variance and number of antennas?

I'm working in two Matlab scripts trying to simulate a MIMO system using OFDM (downlink) and SC-FDE (uplink).

My question is: for different numbers of transmit and receive antennas, what is the relationship between the AWGN variance and the number of antennas? Does it depend if it is uplink or downlink?

I'm using matricial equations like: Y = HX + N , with matrix dimensions: Y(NR x length) , H(NR x NT) , X(NT x length) , N(NR x length)

## 2 Answers

Usually, the model is that every receiver antenna picks up noise of the same PSD, $N_0/2$. This means that the variance of noise samples in each quadrature branch is $N_0/2$. This noise is local to each antenna and is independent of the rest of the system.

To calculate the total amount of noise in the receiver, you multiply by the number of receiver antennas. This is usually not needed, though, since the figure of merit is the ratio of the energy per information bit (or symbol) and the variance of the noise affecting that bit (or symbol).

• But if i don't multiply my noise variance by a factor of NR or NT , the results will be too good. In OFDM/downlink, i've seen multiplying Eb (energy per bit) by NT, but in SC-FDE/uplink, they multiply Eb by NR. Which got me confused – Guilherme Gaspar Sep 11 '16 at 14:14
• @MBaz - Isn't there a coherent gain based on the number of antennas (didn't see that mentioned in your answer): The signal if we assume received equally at each antenna and properly delay aligned (which is essentially what we do in beam steering) will increase by 20 Log N while the noise will increase by 10 Log N where N is the number of antennas. – Dan Boschen Apr 9 '17 at 22:14
• @DanBoschen Isn't that a kind of "processing" gain, though? The SNR at the output of each matched filter (here we go again :) does not depend on the number of antennas. Of course, you can later combine the signals from each antenna to improve the overall SNR (e.g. MRC in spatial diversity or Alamouti in space-time diversity). In any case, I understood the OP's question as regarding the channel model, where I believe my answer is correct. – MBaz Apr 9 '17 at 23:03
• @MBaz - Yes it is, our friend the "processing gain" again! It's everywhere! – Dan Boschen Apr 10 '17 at 1:19

There is no relationship if the noise if white. For the system $$\mathbf{y}=\mathbf{H}\mathbf{x}+\mathbf{n}$$ where $$\mathbf{n}$$ is $$K$$-dimensional, the covariance matrix is $$\sigma_n^2\mathbf{I}_{K\times K}$$, where $$\sigma_n^2$$ is the noise variance, and $$\mathbf{I}_{K\times K}$$ is the $$K\times K$$ identity matrix.