Page 201 - Fiber Optic Communications Fund

182 Fiber Optic Communications

where
√
m (t)= P (2x + 1) x ∈[−X∕2, X∕2 − 1], (4.122)
I
0
√
m (t)= P (2y + 1) y ∈[−Y∕2, Y∕2 − 1]. (4.123)
0
Q
Here, x and y are integers. In complex notation, Eq. (4.121) can be written as
√
2
A out = m + m 2 Q e i(2f c t+) , (4.124)
I
where [ ]
m Q
= tan −1 , (4.125)
m I
2 2 2
P = |A | = m + m , (4.126)
out out I Q
X∕2−1 Y∕2−1
1 ∑ ∑ 2 2
̄
P = m + m
out I Q
XY
x=−X∕2 y=−Y∕2
[ X∕2−1 Y∕2−1 ]
P 0 ∑ ∑
2
= Y (2x + 1) + X (2y + 1) 2
XY
x=−X∕2 y=−Y∕2
2
2
P [(X − 1)+(Y − 1)]
0
= . (4.127)
3
4.10 Additional Examples

Example 4.5
Find the power spectral density of the unipolar signals.
Solution:
From Eq. (4.9), we have
L L
∑ ∑
2
2
(f)= A |̃p(f)| lim 1 < a a > e i2f(l−k)T b . (4.128)
m 0 L→∞ (2L + 1)T l k
b l=−L k=−L
For unipolar signals, a takes values 1 or 0 with equal probability. Let us write
k
1
a = b + , (4.129)
k k
2
where b is a random variable that takes values ±1∕2 with equal probability similar to the random variable
k
associated with the polar signal. Using Eq. (4.129), Eq. (4.128) can be expanded as
L L
1 ∑ ∑ [ 1 1 1 ]
2
2
(f)= A |̃p(f)| lim ⟨b b ⟩ + ⟨b ⟩ + ⟨b ⟩ + exp (i2f(l − k)T ). (4.130)
l
k
m
k l
b
0
L→∞ (2L + 1)T
b l=−L k=−L 2 2 4