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

478 RANDOM PROCESSES AND SPECTRAL ANALYSIS
polar and a periodic component. The polar component is exactly half the polar signal
discussed earlier. Hence, the PSD of this component is one-fourth of the PSD of the
polar signal. The periodic component is of clock frequency R b , and consists of discrete
components of frequency R b and its harmonics.
(b) Bipolar signaling: In this case, ak can take on values 0, 1, and - 1 with probabilities
1/2, 1/4, and 1/4, respectively. Hence,
Also,
ak = (O)Pa, (0) + (l)Pak (l) + (- l)Pak (-1)
1 1 1
= 2 (0) + 4(1) + 4(-1) = 0
Ro = af = (0) 2 Pak (O) + (1) 2 P llk (l) + (-1) 2 Pak (-])
l 2
l
2
l
2
l
= 2 (0) +4(l) +4(-l) = 2
R1 = akak+l = LL akak+1Pakak+1 (akak+1)
k k+I
Because ak and ak+l can take three values each, the sum on the right-hand side has
nine terms, of which only four terms (corresponding to values ±1 for ak and ak+i )
are nonzero. Thus,
R1 = (l)(l)Pakak+l (1, 1) + (-l)(l)Pakak+l (- 1, 1)
Because of the bipolar rule,
+ (1)(- l)Pakak+ 1
(1, -l)+(- 1) (-l)Pakak+l (-1, - 1)
and
Pakak+ 1
(-1, 1) =Pak (-l)Pak+iiak (lJ - 1) =
(i) (1) =
Similarly, we find Pakak+i (1, - 1) = 1/8. Substitution of these values in R1 yields
1
R1 =-- 4
For n 2: 2, the pulse strengths ak and ak+l become independent. Hence,
R n = akak+n = ak ak+n = (0) (0) = 0 n 2: 2
Substitution of these values in Eq. (9.31) and noting that R n is an even function of n,
yields
JP(f) l 2 .
S y (f) = -- sm- (nfT b )
Tb
This result is identical to Eq. (7.2 1b) found earlier by using time averages.