Page 277 - Fiber Optic Communications Fund

258 Fiber Optic Communications

(b) From Eq. (6.65), we have

2 2
2
sp−sp = R ASE (2f − f )f
e e
o
2
= 0.8 ×(6.48 × 10 −16 2 9 9 2
) ×(2 × 16 − 9)× 10 × 9 × 10 A
2
= 5.56 × 10 −11 A .
(c) The total variance is
2
2
tot = 2 s−sp + sp−sp
= 2.657 × 10 −8 + 5.56 × 10 −11 A 2
2
≅ 2.663 × 10 −8 A ≅ 2 .
s−sp
When P is −60 dBm,
in
P (dBm)= P (dBm)+ G(dB)
out in
=−60 + 30 dBm =−30 dBm,
P out = 10 P out (dBm)∕10 mW = 10 −6 W.

(a)
2
9
2 s−sp = 4 × 0.8 × 6.48 × 10 −16 × 10 −6 × 8 × 10 A 2
2
= 1.328 × 10 −11 A .
(b) 2 sp−sp is unaffected by the change in signal power, i.e.,

2 −11 2
= 5.56 × 10 A .
sp−sp
2
(c) In this case, 2 s−sp is smaller than sp−sp . The total variance is
2
2 tot = 6.89 × 10 −11 A .

6.5.4 Polarization Effects
So far, we have considered the signal and noise in a single polarization. Since a single-mode fiber has two
polarization modes, the electric field envelopes in x- and y-polarization components after the optical filter
may be written as

E = + n , (6.72)
x out, x F,x
E = + n , (6.73)
y out,y F,y
where out and n represent the signal and noise outputs of the optical filter, respectively. The total power
F
incident on the photodetector is
2
2
2
2
P = |E | + |E | = | out, x + n F,x | + | out,y + n F,y | . (6.74)
x
y