Page 188 - Probability, Random Variables and Random Processes

RANDOM PROCESSES [CHAP 5

and by Eq. (5.91),

which depends only on k. Thus, (X,} is a WSS process.

5.17. Show that if a random process X(t) is WSS, then it must also be covariance stationary.
If X(t) is WSS, then
E[X(t)] = p (constant) for all t
Rx(t, t + r)] = Rx(7) for all t
NOW Kx(t, t + T) = Cov[X(t)X(t + T)] = Rx(t, t + z) - E[X(t)]E[X(t + z)]
= R,(z) - p2
which indicates that Kx(t, t + z) depends only on z; thus, X(t) is covariance stationary.

5.18. Consider a random process X(t) defined by
X(t)= U cos cot + V sin cot -a < t < KI
where ~r> is constant and U and V are r.v.'s.
(a) Show that the condition
E(U) = E(V) = 0

is necessary for X(t) to be stationary.
(b) Show that X(t) is WSS if and only if U and V are uncorrelated with equal variance; that is,
E(UV) = o E(u~) E(v~) c2 (5.95)
=
=
(a) Now
px(t) = E[X(t)] = E(U) cos wt + E(V) sin cot
must be independent of t for X(t) to be stationary. This is possible only if px(t) = 0, that is,
E(U) = E(V) = 0.
(6) If X(t) is WSS, then

But X(0) = U and X(n/2w) = V; thus
E(U2) = E(V2) = ax2 = a2
Using the above result, we obtain
Rx(t, t + 7) = E[X(t)X(t + T)]
= E((U cos wt + V sin ot)[U cos o(t + z) + V sin o(t + z)]}
= o2 cos oz + E(UV) sin(2wt + wz) (5.96)

which will be a function of z only if E(UV) = 0. Conversely, if E(UV) = 0 and E(U2) = E(V2) = 02,
then from the result of part (a) and Eq. (5.96), we have

Hence, X(t) is WSS.