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Sec. 6–2 Power Spectral Density 429
a = ;1
Example 6–4 EVALUATION OF THE PSD FOR A POLAR BASEBAND SIGNAL
Let x(t) be a polar signal with random binary data. A sample function of this signal is illustrated
in Fig. 6–5a. Assume that the data are independent from bit to bit and that the probability of
1
obtaining a binary 1 during any bit interval is
2. Find the PSD of x(t).
The polar signal may be modeled by
x(t) =
q
a a n f(t - nT b )
(6–56)
n = -q
where f(t) is the signaling pulse shape, shown in Fig. 6–5b, and T b is the duration of one bit.
{a n } is a set of random variables that represent the binary data. It is given that the random
variables are independent. Clearly, each one is discretely distributed at and
P(a n = 1) = P(a n = -1) = 1 2 , as described in the statement of the problem.
n
The PSD for x(t) will be evaluated first by using method 1, which requires that X T ( f ) be
obtained. We can obtain x T (t) by truncating Eq. (6–56); that is,
x T (t) =
where T>2 = AN + 1 2 B T b . Then
n = N
a
n =-N
a n f(t - nT b )
or
X T (f) = [x T (t)] =
N
N
a a n [f(t - nT b )] = a a n F(f)e -jvnT b
n =-N
n =-N
N
X T (f) = F(f) a a n e -jvnT b
n =-N
(6–57)
where F( f ) = [f(t)]. When we substitute Eq. (6–57) into Eq. (6–42), we find that the PSD is
x (f) =
lim
T: q
a 1 T ƒ F(f) ƒ 2 `
2
= ƒ F(f) ƒ lim a 1
T: q T
N
a
n =-N
a
N
n =-N
2
a n e -jvnT b ` b
N
a
m =-N
a n a m e j(m-n)vTb b
(6–58)
The average a n a m
now needs to be evaluated for the case of polar signaling (a n =;1). We have
a n a m = e a n 2 , n = m
a n a m, n Z m
where a n a m = a a n m for n m, since a n and a m are independent. Using the discrete distribution
for a n , we get
a n = (+1) 1 2 + (-1) 1 2 = 0
Similarly, a m = 0. Further,
a n 2 = (+1) 2 A 1 2 B + (-1) 2 A 1 2 B = 1