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Sec. 6–1 Some Basic Definitions 419
and m x denotes the mean value. Similarly, the RMS value is obtained as
! 28x 2 (t)9 K 3x 2 = 2s 2 2
x + m x (6–7)
where is the variance of x(t).
In summary, if a process is ergodic, all time and ensemble averages are interchangeable.
Then the time average cannot be a function of time, since the time parameter has been averaged
out. Furthermore, the ergodic process must be stationary, because otherwise the ensemble
averages (such as moments) would be a function of time. However, not all stationary
processes are ergodic.
s x
2
X rms
Example 6–3 ERGODIC RANDOM PROCESS
Let a random process be given by
x(t) = A cos(v 0 t + u)
(6–8)
where A and v 0 are constants and u is a random variable that is uniformly distributed over
(0, 2p).
First, we evaluate some ensemble averages. The mean and the second moment are
and
q
2p
x = [x(u)]f u (u) du = [A cos(v 0 t + u)] 1
L L 2p du = 0
-q
0
(6–9)
x 2 (t) =
L
2p
0
[A cos(v 0 t + u)] 2 1 A2
du =
2p 2
(6–10)
In this example, the time parameter t disappeared when the ensemble averages were evaluated,
which would not be the case unless x(t) was stationary.
Second, we evaluate the corresponding time averages by using a typical sample function
of the random process. One sample function is x(t, E 1 ) = A cos v 0 t, which occurs when u = 0,
corresponding to one of the events (outcomes). The time average for any of the sample functions
can be evaluated by letting u be the appropriate value between 0 and 2p. The time averages for
the first and second moments are
8x(t)9 = 1 T 0
A cos(v
T 0 t + u) dt = 0
0 L
0
(6–11)
and
8x 2 (t)9 = 1 T 0
[A cos (v
T 0 t + u)] 2 dt = A2
0 L 2
0
(6–12)
where T 0 = 1f 0 and the time-averaging operator for a periodic function, Eq. (2–4), has been used.
In this example, u disappears when the time average is evaluated. This is a consequence of x(t)
being an ergodic process. Conversely, for a nonergodic example, the time average would be a
random variable.