Page 377 - Fiber Optic Communications Fund

358 Fiber Optic Communications

1

0.8

0.6
Sinc(∆ωT b /π) 0.4

0.2

0

*0.217

*0.4
0 0.5 1 1.5 2 2.5 3 3.5 4
∆ωT b /π

Figure 8.16 Sinc function.

this value of Δ in Eq. (8.174), we obtain

(√ )
1 1.217 het
P = erfc . (8.176)
b
2 4
Comparing Eqs. (8.144) and (8.176), we find that a system based on OOK requires an average energy 1.217
times that of a system based on FSK to achieve the given BER.

8.4.3.1 Orthogonal FSK
The signals I (t) and I (t) are said to be orthogonal if
1 0
T b
E e = I (t)I (t) dt = 0. (8.177)
10 ∫ 1 0
0
From Eq. (8.170), we find that E e is zero if
10
ΔT = n, n = 1, 2, 3, … (8.178)
b
or
n
Δf = . (8.179)
2T
b
As Δf increases, the signal bandwidth increases too. The minimum frequency separation occurs when n = 1:
1
Δf = . (8.180)
min
2T
b