(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.
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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