1
$\begingroup$

I am trying to understand about correlation between random variables in context of wireless communication.

In research papers related to 6G, I came across the following statements:

Suppose there is a base station (BS) having an array of $N$ multiple antennas and a single antenna user. Then the uplink channel vector from user to BS is given as $\textbf{h}\in \mathbb{C}^{N\times 1}$ and authors assume the communication channel as the Rayleigh fading model.

Assuming that the receive antennas at the BS are widely separated, hence the correlation between receive antennas can be omitted, so we have

$\textbf{h}\sim \mathcal{C}\mathcal{N}(\textbf{0},\textbf{I})$---(1)

My query is I am not getting how absence of correlation between receive antennas results in equation (1).

Any help in this regard will be highly appreciated.

$\endgroup$
2
  • 2
    $\begingroup$ Do you know what $\mathcal{CN}(\mathbf {\boldsymbol{\mu}}, \mathbf \Gamma)$ means? It's the multidimensional complex normal distribution; the covariance matrix $\mathbf \Gamma$ being $\mathbf I$, a diagonal matrix, tells us that any two different elements in $\mathbf h$ are uncorrelated. That's really just the definition of that notation! You probably want to look up what a covariance matrix is. $\endgroup$ Commented Apr 21 at 10:13
  • $\begingroup$ Thanks a lot sir...Understood the concept.... $\endgroup$ Commented Apr 21 at 10:43

2 Answers 2

1
$\begingroup$

The notion that TX antennas are uncorrelated, for example, is shorthand for saying that the channel responses from TX antennas to any RX antenna are uncorrelated ($\mbox{E}[h_{i,0,n}^\ h_{j,1,n}^*]=0$ for TX antennas 0, 1 to RX antenna $n$, where $h_{k,m,n}$ is tap $k$ of the channel from TX antenna $m$ to RX antenna $n$). Of course, this presumes a type of channel and doesn't hold for all channels that those TX antennas might be placed in.

For example, we may talk about uncorrelated channels for a fading channel from TX to some non-line-of-sight theoretical RX placement, but if the RX channel were placed in sight of the TX antennas, the channels would become dependent and correlated, since the response from TX to RX has a deterministic component.

$\endgroup$
1
$\begingroup$

One common model for a wireless channel is the Rayleigh model where the channel coefficient is modeled as a circularly-symmetric complex Gaussian random variable with mean zero and variance $\sigma^{2}$ denoted by ${\cal{CN}}(0,\sigma^{2})$. Typically, it's assumed that the channel has a unit variance as its impact can be lumped into the transmitted power or to the additive white Gaussian noise (AGWN) that define the overall system SNR.

For the SIMO communication system at your hand, the received signal at the $n$th base station antenna can be written as $y_{n} = h_{n}d + z_{n}$ where $d$ is the transmitted symbol by the user, $h_{n}\sim{\cal{CN}}(0,1)$ is the channel between the user and the $n$th receiver antenna at BS, and $z_{n}$ is the AWGN noise at the $n$th received antenna. In a vector format, the received signal at base station, stacking all the received signals at $N$ base station antenna, you can write

$$ \boldsymbol{y} = \begin{bmatrix}y_{1} \\ \vdots \\ y_{N}\end{bmatrix} = d\underbrace{\begin{bmatrix}h_{1} \\ \vdots \\ h_{N}\end{bmatrix}}_{\boldsymbol{h}} + \boldsymbol{z} $$

From physics point of view, the channel between the transmitter and the $i$th and $j$th receiver antenna are somewhat correlated if these antenna are relatively close to each other (relative to the wavelength of transmitting wave). If the antennas are co-located and are placed at a proper distance, on can assume that the channels seen by different receiving antenna are statistcally independent which means that the joint probability density $f_{h_{i},h_{j}}(x,y) = f_{h_{i}}(x)f_{h_{j}}(y), \forall i,j = 1,\ldots,N$. Since each channel coefficient is distributed as a Gaussian random variable, we have $\mathbb{E}\{h_{i}h_{j}^{*}\} = \delta_{ij}\sigma_{i}^{2}$ where $\delta_{ij}$ is the Kronecker delta

$$ \delta_{ij} = \left\{\begin{array}{ll} 1 & i = j\\ 0 & i\neq j\end{array}\right . $$

Hence,

$$ \mathbb{E}\{\boldsymbol{h}\boldsymbol{h}^{H}\} = \begin{bmatrix} \sigma_{1}^{2} & 0 & \cdots & 0\\ 0 & \sigma_{2}^{2} & \cdots & 0\\ \vdots & \vdots & \ddots & \cdots \\ 0 & 0 & \cdots & \sigma_{N}^{2} \end{bmatrix} $$

If all channels have unit power, $\sigma_{i} = 1,\forall i = 1,\ldots, N$, then $\mathbb{E}\{\boldsymbol{h}\boldsymbol{h}^{H}\} = \boldsymbol{I}$ where $\boldsymbol{I}$ is the $N\times N$ identity matrix.

Note that Gaussian random vectors are fully defined by their mean and covariance matrix, hence, $\boldsymbol{h}\sim{\cal{CN}}(\boldsymbol{\mu} = \boldsymbol{0},\boldsymbol{\Sigma} = \mathbb{E}\{(\boldsymbol{h}-\boldsymbol{\mu})(\boldsymbol{h}-\boldsymbol{\mu})^{H}\} = \boldsymbol{I})$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.