B. P. Lathi, Zhi Ding - Modern Digital and Analog Communication Systems-Oxford University Press (2009)

476 RANDOM PROCESSES AND SPECTRAL ANALYSIS
Figure 9.1 1
Basic pulse for a
random binary
polar signal.
p(t)
- Tb
4
0
Tb
4
t-
In this case, ak can take on values 1 and -1 with probability 1/2 each. Hence,
ak = L akP(ak) = (l)Pak (l) + (- l)Pak (-1)
k=l, - 1
1 l
= ---= 0
2 2
Ro = af = L af P(ak) = (1) 2 Pak (l) + (-1) 2 Pak (- 1)
k=l, - I
l ? 1 2
= 2 (1)- + 2
(- 1) = 1
and because each digit is independent of the remaining digits,
Rn = akak+n = ak ak+n = 0
n I
Hence, from Eq. (9.31),
We already found this result in Eq. (7. 13), where we used time averaging instead of
ensemble averaging. When a process is ergodic of second order (or higher), the ensemble
and time averages yield the same result. Note that Example 9.5 is a special case of this
result, where p(t) is a full-width rectangular pulse TI (t /Tb) with P(f) = Tb sine (nfTb),
and
Example 9. 8
Find the PSD Sy (f) for on-off and bipolar random signals which use a basic pulse for p(t), as
shown in Fig. 9.11. The digits 1 and O are equally likely, and digits are transmitted every Tb
seconds. Each digit is independent of the remaining digits. All these line codes are described
in Sec. 7.2.