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Receive diversity in AWGN

Posted By Krishna Sankar On August 19, 2008 @ 6:39 am In MIMO | 30 Comments

Some among you will be aware that in a wireless link having multiple antenna’s at the receiver (aka receive diversity) improves the bit error rate (BER) performance. In this post, let us try to understand the BER improvement with receive diversity. And, since we are just getting started, let us limit ourselves to additive white Gaussian noise (AWGN) channel (i.e assume that the channel gains are unity).

## Single receive antenna case

Let us begin with the discussion with one transmit antenna, sending signals with energy $E_b$ and one receive antenna. Since we are considering only BPSK modulation, the signals which are sent out are either$+\sqrt{E_b}$ or $-\sqrt{E_b}$. Let there be a single receive antenna having a thermal noise (aka AWGN) with mean $\mu=0$ and variance $\sigma^2 = \frac{N_0}{2}$.

The probability density function of noise is,

$p(n) = \frac{1}{\sqrt{2\pi\sigma^2}}e^{\frac{-(n-\mu)^2}{2\sigma^2}$.

The received signal is of the form,

$y=x+n$, where

$y$is the received symbol,
$x$ is the transmitted symbol (taking values $+\sqrt{E_b}$‘s and$-\sqrt{E_b}$‘s) and
$n$ is the Additive White Gaussian Noise (AWGN).

From our discussion on BER for BPSK modulation in AWGN [1], we know that probability of bit error is,

$P_b=\frac{1}{2}erfc\left({\sqrt{\frac{E_b}{N_0}}}\right)$.

## Receive diversity with two antennas

Now consider the case where we have two receive antennas each having thermal noise [2] (aka AWGN) with mean $\mu=0$ and variance $\sigma^2 = \frac{N_0}{2}$. As the noise on each antenna is independent from each other, in signal processing parlance, we can say that noise on each antennas are i.i.d i.e independent and identically distributed. The transmitter is still sending symbols with energy $E_b$.

The received signal is of the form,

$\left[\begin{eqnarray}y_1 \\ y_2 \end{eqnarray}\right] =x+\left[\begin{eqnarray}n_1\\n_2\end{eqnarray}\right]$, where

$y_1$, $y_2$are the received symbols from receive antenna 1, 2 respectively,
$x$ is the transmitted symbol (taking values $+\sqrt{E_b}$‘s and$-\sqrt{E_b}$‘s) and
$n_1$,$n_2$is the Additive White Gaussian Noise (AWGN) on receive antenna 1, 2 respectively.

For simplicity, let us assume that the signal $+\sqrt{E_b}$ was transmitted. At the receiver, we now have

$y_1=\sqrt{E_b}+n_1$ and

$y_2=\sqrt{E_b}+n_2$

To decode, the simplistic (and the optimal in this scenario) is to take the mean of $y_1$,$y_2$and perform hard decision decoding, i.e

$y_s=\frac{y_1+y_2}{2}$ and if $y_s \ge 0$ implies the transmitted bit is 1 and $y_s \le 0$ implies transmitted bit is 0.

Now let us find out if there is any receive diversity gain by performing this averaging. Splitting $y_s$ into signal term and noise term

$\begin{eqnarray}y_s&=&\frac{\sqrt{E_b} + n_1 +\sqrt{E_b}+n_2}{2}\\&=&\sqrt{E_b}+\left(\frac{n_1+n_2}{2}\right)\end{eqnarray}$.

The first term in the above equation is the signal term – which still transmit symbols with energy $E_b$. The second term is mean of two Gaussian noise terms.

(a) From the discussion on sum of Gaussian random variables [3],

if $X$ is a Gaussian random variable with mean $\mu_1$, variance $\sigma_1^2$ and $Y$ is another independent Gaussian random variable with $\mu_2$, variance $\sigma_2^2$, then $X+Y$ is another Gaussian random variable with $\mu_1 + \mu_2$, variance $\sigma_1^2 +\sigma_2^2$.

(b) Furher, from the discussion on functions of Gaussian random variables [4],

if $X$ is a Gaussian random variable with mean $\mu_1$, variance $\sigma_1^2$, then $aX+b$ is another Gaussian random variable with mean $a\mu_1 +b$, variance $\left(a\sigma_1\right)^2$.

Using the above two equations, the noise term $\frac{n_1+n_2}{2}$ is another Gaussian random variable with mean $\mu =0$ and variance $\sigma^2 = \frac{N_0}{4}$.

When compared with the single antenna case, we can see the variance of the noise term is scaled by a factor of 2. This implies that the effective bit energy to noise ratio in a two receive antenna case is twice the bit energy to noise ratio for single antenna case.

$\left[\frac{E_b}{N_0}\right]_{eff,2}=\frac{2E_b}{N_0}$.

So the bit error probability for two receive antenna case is,

$P_b=\frac{1}{2}erfc\left({\sqrt{\frac{2E_b}{N_0}}}\right)$.

Expressing in decibels, with two receive antennas, we need only $10\log_{10}(2) = 3dB$ lower bit energy $E_b$! No wonder people like receive diversity.

## Receive diversity with N receive antenna

With a general N receive antenna case, the received symbol is,

$\left[\begin{eqnarray}y_1 \\ y_2 \\ \vdots\\y_N\end{eqnarray}\right] =x+\left[\begin{eqnarray}n_1\\n_2\\ \vdots\\n_N\end{eqnarray}\right]$, where

$y_1$, $y_2$,… $y_N$are the received symbols from receive antenna 1, 2 respectively,
$x$ is the transmitted symbol (taking values $+\sqrt{E_b}$‘s and$-\sqrt{E_b}$‘s) and
$n_1$,$n_2$,… $n_N$is the Additive White Gaussian Noise (AWGN) on receive antenna 1, 2,… N respectively.

For demodulation, we compute$y_s$ which is the mean of all the N received symbols, and if $y_s \ge 0$ implies the transmitted bit is 1 and $y_s \le 0$ implies transmitted bit is 0.

The variance of the noise term  $\frac{\left(n_1 + n_2 + \cdots + n_N\right)}{N}$ is $\frac{\sigma^2}{2N}$.

Effective bit energy to noise ratio in a N receive antenna case is N times the bit energy to noise ratio for single antenna case.

$\left[\frac{E_b}{N_0}\right]_{eff,N}=\frac{NE_b}{N_0}$.

So the bit error probability for N receive antenna case is,

$P_b=\frac{1}{2}erfc\left({\sqrt{\frac{NE_b}{N_0}}}\right)$.

## Simulation Model

Simple Matlab/Octave example simulating a BPSK transmission and reception in AWGN only scenario with multiple receive antennas. The script performs the following

(a) Generate random binary sequence of +1’s and -1’s.

(b) Add white Gaussian noise on the received symbol on all the receive chains

(c) For each symbol, compute the mean of the received symbol over all the receive chains

(d) Perform hard decision decoding and count the bit errors

(e) Repeat for receive chains = 1, 2, 3, 4 for multiple values of $\frac{E_b}{N_0}$

(f) Plot the simulation and theoretical results.

Figure: BER for BPSK in AWGN with receive diversity (number of receive antennas = 1, 2, 3, 4)

## Observations

1. As desired, the simulation results show good agreement with theory.

2. With an N receive antenna system, the gain compared to single receive antenna case is around $10\log_{10}(N)$. For example in the above figure, for bit error rate of $10^{-4}$,

• with two receive antennas the required $\frac{E_b}{N_0}$ is 3dB lower,
• with three receive antennas, the required $\frac{E_b}{N_0}$ is 4.7dB lower and
• with four three receive antennas, the required $\frac{E_b}{N_0}$ is 6dB lower.

Hope this helped you get started with receive diversity. If not, please do drop in a comment. Happy learning.

URL to article: http://www.dsplog.com/2008/08/19/receive-diversity-in-awgn/

URLs in this post:

[1] BER for BPSK modulation in AWGN: http://www.dsplog.com/2007/08/05/bit-error-probability-for-bpsk-modulation/

[2] thermal noise: http://en.wikipedia.org/wiki/Thermal_noise

[3] sum of Gaussian random variables: http://en.wikipedia.org/wiki/Sum_of_normally_distributed_random_variables

[4] functions of Gaussian random variables: http://en.wikipedia.org/wiki/Normal_distribution#Properties

[5] Matlab/Octave script for BER of BPSK in AWGN with receive diversity: http://www.dsplog.com/db-install/wp-content/uploads/2008/08/script_ber_bpsk_awgn_receive_diversity.m

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