Random Signals for Engineers using MATLAB and Mathcad
Copyright  1999 Springer Verlag NY
Example 5.8 Correlation Functions Sinusoidal Processes
In this example we will compute the correlation function for a Random Processes that contains
sinusoidal functions of time. The first function contains two random parameters, a and b. We will first
2
2
form the R(t,) = E[x(t) x(t + )] and then find the conditions upon E[a] E[b], E[a ] , E[b ] and the E[a b]
to make R() wide sense stationary. We begin with the definition of the x(t) function
syms a b om t tau
x=a*cos(om*t)-b*sin(om*t);
The correlation function is formed by finding x(t) x(t + ) and then taking the expected value of the
product. Taking the product
pr=x*subs(x,t,t+tau);
pretty(pr)
(a cos(om t) - b sin(om t)) (a cos(om (t + tau)) - b sin(om (t +
tau)))
Multiplying out using Matlab we obtain
pr=simplify(pr);
pretty(pr)
2
a
cos(om t) cos(om (t + tau)) - a cos(om t) b sin(om (t + tau))
2
- b sin(om t) a cos(om (t + tau)) + b
2
sin(om t) sin(om (t + tau))
2
The term containing a , b and ab are collected. Substitution of the trigonometric identities into the a 2 term
we have for
maple('combine','cos(om*t)*cos(om(t+tau)),trig')
ans =
1/2*cos(om*t-om(t+tau))+1/2*cos(om*t+om(t+tau))
Multiplying out
simplify(1/2*cos(om*t-om*(t+tau))+1/2*cos(om*t+om*(t+tau)))
ans =
1/2*cos(om*tau)+1/2*cos(2*om*t+om*tau)
The b2 term can be similarly found to be
maple('combine','sin(om*t)*sin(om*(t+tau)),trig')
ans =
1/2*cos(om*tau)-1/2*cos(2*om*t+om*tau)
When we take the expected value of the product and have E[a 2] = E[b2] = 2 the term containing  t is
cancelled and we have
R    2  cos   
In addition, to make the correlation function independent of t, we require that E[a b ] = 0 . In order to have
the Random Function x(t) be wide sense stationary in the first and second order we require that E[a] = E[b]
= 0 in addition to the requirement that E[a b] = 0.
The next function we consider is
syms rho phi
x=cos(rho*t+phi);
where  is a constant and is a random variable. We would like to determine the functional form of
density function of  that will make x(t) wide sense stationary. Let us begin by expanding
'E[x(t)]=E['
expand(cos(rho*t+phi))
ans =
E[x(t)]=E[
ans =
cos(rho*t)*cos(phi)-sin(rho*t)*sin(phi)
moving the Expectation operator into the brackets we have
E[x(t)] = cos( t) E[cos()] - sin( t) E[sin()]
The characteristic function of  is defined by
() = E[ej] = E[cos()] + j E[sin()]
The right hand side was obtained by an euler expansion of the exponent and applying the expectation
operation. The condition that makes E[x(t)] = 0 is E[cos()] = E[sin()] = 0
This condition can be rewritten in terms of the characteristic function as ( 1 ) = 0. The correlation
function can be expressed as
cr=E[x(t+) x(t)]
cr=cos(rho*(t+tau)+phi)*cos(rho*t+phi);
Using the trigonometric identities we have
maple('combine','cos(rho*(t+tau)+phi)*cos(rho*t+phi),trig')
ans =
1/2*cos(rho*tau)+1/2*cos(2*rho*t+rho*tau+2*phi)
Since only the second term contains the random variable when we take the Expected value of the product
above, we obtain E[cos()] = E[sin()] = 0. This calculation follows the same procedure that we
used for the expectation operator computation.
This condition can be expressed in terms of the characteristic function as ( 2 ) = 0. When this condition is
true the correlation function is only a function of and thus the process x(t) is wide sense stationary in the
second order.
The first and second order results that were determined are the conditions for a signal x(t) to be wide sense
stationary expressed in terms of the characteristic function. From example 5.5 we have shown that when
is uniformly distributed in {-  }, then x(t) is wide sense stationary. We may verify that these condition
on that are given in terms of the characteristic function of are satisfied when the characteristic function is
computed for a uniform distribution. The Characteristic function for a uniform distribution is
syms u
phiom=int(exp(i*om*u),u,-pi,pi)
phiom =
-i/om*exp(i*pi*om)+i/om*exp(-i*pi*om)
Replacing the exponents by a sin term
pretty(sin(pi*om)/pi/om)
sin(pi om)
---------pi om
We can easily verify that, as required,
1  2  0
