563489578934

418
Random Processes and Spectral Analysis Chap. 6
CASE 1: A STATIONARY RESULT. First, assume that A and v 0 are deterministic
constants and u 0 is a random variable. t is the time parameter. In addition, assume that u 0 is uniformly
distributed over -p
to p. Then c ! u 0 + w 0 t is a random variable uniformly distributed
over the interval w 0 t - p 6 c 6 w 0 t p. The first-order PDF for x(t) can be obtained by the
transformation technique developed in Sec. B–8 of Appendix B. This is essentially the same
problem as the one worked out in Example B–9. From Eq. (B–71), the first-order PDF for x(t) is
1
,
f(x) = c
p2A 2 - x 2 |x| … A
0, x elsewhere
(6–5a)
Because this PDF is not a function of t, x(t) is a first-order stationary process for the assumptions
of Case 1, where u 0 is a random variable. This result would be applicable to problems in which u 0
is the random start-up phase of an unsynchronized oscillator.
CASE 2: A NONSTATIONARY RESULT. Second, assume that A, v 0 , and u 0 are
deterministic constants. Then, at any time, the value of x(t) is known with a probability of
unity. Thus, the first-order PDF of x(t) is
f(x) = d(x - A sin(v 0 t + u 0 ))
(6–5b)
This PDF is a function of t; consequently, x(t) is not first-order stationary for the assumption of
Case 2, where u 0 is a deterministic constant. This result will be applicable to problems in which
the oscillator is synchronized to some external source so that the oscillator start-up phase will
have the known value u 0 .
DEFINITION. A random process is said to be ergodic if all time averages of any sample
function are equal to the corresponding ensemble averages (expectations).
Two important averages in electrical engineering are DC and RMS values. These values
are defined in terms of time averages, but if the process is ergodic, they may be evaluated by
the use of ensemble averages. The DC value of x(t) is x dc ! 8x(t)9.
When x(t) is ergodic, the
time average is equal to the ensemble average, so we obtain
where the time average is
the ensemble average is
x dc ! 8x(t)9 K [x(t)] = m x
8[x(t)]9 = lim
T: q
q
[x(t)] = [x]f x (x)dx = m x
L
- q
1
T L
T/2
-T/2
[x(t)] dt
(6–6a)
(6–6b)
(6–6c)