Page 376 - Fiber Optic Communications Fund

Performance Analysis 357

I(t) T b
ʃ (∙)dt
0
+ If r(T ) > 0
b
b
Acos + ∆ω t ∑ r(T ) Decision select ‘1’
ω IF
2 device Otherwise
* select ‘0’
T b
ʃ (∙)dt
0

∆ω
Acos * t
ω IF
2
Figure 8.15 Matched filter for FSK signal in a fiber-optic system with heterodyne receiver.

where
sin (x)
sinc(x)= . (8.171)
x
The superscript e is introduced to indicate that E represents the energy of a bit in the electrical domain. When
≫ 1∕T , the second term on the right-hand side of Eq. (8.170) can be ignored. Similarly,
IF
b
T b [( Δ ) ]
e
e
E = E = A 2 cos 2 + t dt
0 1 ∫ IF 2
0
A 2 T b { [ ( Δ ) ]}
= 1 + cos 2 + t dt
2 ∫ 0 2
IF
2
A T b
= . (8.172)
2
The contribution from the second term on the right-hand side of Eq. (8.172) is negligible since ≫ 1∕T .
IF b
e
Here, E denotes the normalized energy of the bit ‘j’, j = 0, 1 in the electrical domain. The corresponding
j
energy in the optical domain is
T b
opt 2 2
E = |s (t)| dt = A T = E . (8.173)
1 ∫ 1 b av
0
Substituting Eqs. (8.172) and (8.170) in Eq. (8.38), we obtain
2E av
max = [1 − sinc(ΔT ∕)], (8.174)
b
N het
0
( )
√
1 max
P = erfc . (8.175)
b
2 8
The above equation is valid for arbitrary frequency difference Δ. To find the minimum achievable BER,
we need to minimize P with respect to Δ or equivalently maximize . In other words, the sinc function
b max
should be minimum. From Fig. 8.16, we see that the minimum value is −0.217 at ΔT = 1.43. Choosing
b