# 5.5 Crosscorrelation of random processes

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(Blank Abstract)

Before diving into a more complex statistical analysis of random signals and processes , let us quickly review the idea of correlation . Recall that the correlation of two signals or variables is the expectedvalue of the product of those two variables. Since our main focus is to discover more about random processes, a collectionof random signals, we will deal with two random processes in this discussion, where in this case we will deal with samplesfrom two different random processes. We will analyze the expected value of the product of these two variables and how they correlate to one another, where theargument to this correlation function will be the time difference. For the correlation of signals from the same randomprocess, look at the autocorrelation function .

## Crosscorrelation function

When dealing with multiple random processes, it is also important to be able to describe the relationship, if any,between the processes. For example, this may occur if more than one random signal is applied to a system. In order to dothis, we use the crosscorrelation function , where the variables are instances from two different wide sensestationary random processes.

Crosscorrelation
if two processes are wide sense stationary, the expected value of the product of a random variable from one randomprocess with a time-shifted, random variable from a different random process
Looking at the generalized formula for the crosscorrelation, we will represent our two random processes by allowing $U=U(t)$ and $V=V(t-)$ . We will define the crosscorrelation function as
${R}_{uv}(t, t-)=(UV)=\int \,d u$ v u v f u v
Just as the case with the autocorrelation function, if ourinput and output, denoted as $U(t)$ and $V(t)$ , are at least jointly wide sense stationary, then the crosscorrelation does not depend on absolute time; it is justa function of the time difference. This means we can simplify our writing of the above function as
${R}_{uv}()=(UV)$
or if we deal with two real signal sequences, $x(n)$ and $y(n)$ , then we arrive at a more commonly seen formula for the discrete crosscorrelation function. See the formula belowand notice the similarities between it and the convolution of two signals:
${R}_{xy}(n, n-m)={R}_{xy}(m)=\sum$ x n y n m

## Properties of crosscorrelation

Below we will look at several properties of the crosscorrelation function that hold for two wide sense stationary (WSS) random processes.

• Crosscorrelation is not an even function; however, it does have a unique symmetryproperty:
${R}_{xy}(-)={R}_{yx}()$
• The maximum value of the crosscorrelation is not always when the shift equals zero; however, we can prove thefollowing property revealing to us what value the maximum cannot exceed.
$\left|{R}_{xy}()\right|\le \sqrt{{R}_{xx}(0){R}_{yy}(0)}$
• When two random processes are statistically independent then we have
${R}_{xy}()={R}_{yx}()$

## Examples

Let us begin by looking at a simple example showing the relationship between two sequences. Using , find the crosscorrelation of the sequences $x(n)=\{, 0, 0, 2, -3, 6, 1, 3, 0, 0, \}()$ $y(n)=\{, 0, 0, 1, -2, 4, 1, -3, 0, 0, \}()$ for each of the following possible time shifts: $m=\{0, 3, -1\}()$ .

• For $m=0$ , we should begin by finding the product sequence $s(n)=x(n)y(n)$ . Doing this we get the following sequence: $s(n)=\{, 0, 0, 2, 6, 24, 1, -9, 0, 0, \}()$ and so from the sum in our crosscorrelation function we arrive at the answer of ${R}_{xy}(0)=22$
• For $m=3$ , we will approach it the same was we did above; however, we will now shift $y(n)$ to the right. Then we can find the product sequence $s(n)=x(n)y(n-3)$ , which yields $s(n)=\{, 0, 0, 0, 0, 0, 1, -6, 0, 0, \}()$ and from the crosscorrelation function we arrive at the answer of ${R}_{xy}(3)=-6$
• For $m=-1$ , we will again take the same approach; however, we will now shift $y(n)$ to the left. Then we can find the product sequence $s(n)=x(n)y(n+1)$ , which yields $s(n)=\{, 0, 0, -4, -12, 6, -3, 0, 0, 0, \}()$ and from the crosscorrelation function we arrive at the answer of ${R}_{xy}(-1)=-13$

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