Page 235 - Schaum's Outlines - Probability, Random Variables And Random Processes

ANALYSIS AND PROCESSING OF RANDOM PROCESSES [CHAP 6

Hence py(t) = E[Y(t)] = (1)P[Y(t) = 11 + (- I)P[Y(t) = - 11
= e-*'(cash It - sinh It) = ePzAt
(b) Similarly, since Y(t)Y(t + z) = 1 if there are an even number of events in (t, t + z) for z > 0 and
Y(t)Y(t + z) = - 1 if there are an odd number of events, then for t > 0 and t + z > 0,

(Adn
= (1) e-" W + (- 1) e-" -
n even n ! nodd n!

which indicates that R,(t, t + z) = RY(z), and by Eq. (6.13),

Note that since E[Y(t)] is not a constant, Y(t) is not WSS.

6.18. Consider the random process

where Y(t) is the semirandom telegraph signal of Prob. 6.17 and A is a r.v. independent of Y(t)
and takes on the values + 1 with equal probability. The process Z(t) is known as the random
telegraph signal.
(a) Show that Z(t) is WSS.
(b) Find the power spectral density of Z(t).
(a) Since E(A) = 0 and E(A2) = 1, the mean of Z(t) is

and the autocorrelation of Z(t) is

Thus, using Eq. (6.130), we obtain

Thus, we see that Z(t) is WSS.
(b) Taking the Fourier transform of Eq. (6.132) (see Appendix B), we see that the power spectrum of Z(t) is
given by

6.19. Let X(t) and Y(t) be both zero-mean and WSS random processes. Consider the random process
Z(t) defined by

(a) Determine the autocorrelation function and the power spectral density of Z(t), (i) if X(t) and
Y(t) are jointly WSS; (ii) if X(t) and Y(t) are orthogonal.
(b) Show that if X(t) and Y(t) are orthogonal, then the mean square of Z(t) is equal to the sum
of the mean squares of X(t) and Y(t).