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

9.3 Power Spectral Density 475
Recall that a is uniformly distributed over the interval O to T b . Hence, p(a) = I/T b over
the interval (0, T b ), and is zero otherwise. Therefore,
oo
oo
1 ( Tb
R y
(r) = L Rn L T Jo
n=-oo k=-oo
b O
oo oo
1
/ t- kTb
T b
n=-oo k=-oo t-(k+l)Tb
p(t - kT b - a)p(t + r - [k + n] T b - a) da
= - L Rn L p(f3)p(f3 + r - nT b ) d/3
00
1
1 00
= - L Rn p(f3)p(f3 + r - nT b ) d f3
T b
n=-oo
-oo
The integral on the right-hand side is the time autocorrelation function of the pulse p(t)
with the argument r - nT b . Thus,
(9.28)
where
(9.29)
and
i/r p
(r) = 1_: p(t)p(t + r) dt (9.30)
As seen in Eq. (3.74), if p(t) P(j), then i/r p
(r) IP(f)J 2 . Therefore, the PSD of
y(t), which is the Fourier transform of R y
(r), is given by
2
00
_ JP(J) I
- "
,.,-,
,ne-jn2TCfJ'b
T b
n= -oo
(9.31)
This result is similar to that found in Eq. (7.llb). The only difference is the use of the
ensemble average in defining R n in this chapter, whereas R n in Chapter 7 is the time
average.
Example 9.7
Find the PSD S y
(J) for a polar binary random signal where 1 is transmitted by a pulse p(t)
(Fig. 9.11) whose Fourier transform is P(f), and O is transmitted by -p(t). The digits 1 and 0
are equally likely, and one digit is transmitted every T b seconds. Each digit is independent of
the other digits.