Page 341 - Fiber Optic Communications Fund

322 Fiber Optic Communications

The mean launch power

P =−3dBm.
in
So the peak power is (assuming NRZ rectangular pulses)

P = 2P
in in
= 2 × 10 −(3∕10) mW

= 1mW.
Using Eq. (7.135), we find
√
P in
Q =
4Nn hf(G − 1)f
sp e
√
1 × 10 −3
=
12
4 × 80 × 1.5 × 6.626 × 10 −34 × 193.5 × 10 ×(39.96 − 1)× 7 × 10 9
= 7.71.

7.4.4 Coherent Receiver

Consider the fiber-optic system shown in Fig. 7.13 with balanced coherent detection. Let the received field
envelope be
q(t)= A s(t)+ n (t), (7.137)
r ASE
where A s(t) and n ASE (t) are the signal and ASE noise field envelopes, respectively. In this case, Eq. (7.32) is
r
modified as
{ }
I = 2RA LO Re A s(t)+ n ASE (t) + n shot+ − n shot− . (7.138)
r
Consider a bit ‘1’ of the OOK system. Let us first ignore the shot noise and write Eq. (7.138) as

I = I + I, (7.139)
√
P ,
I = 2R P LO 1r (7.140)
√ { }
I = 2R P Re n . (7.141)
LO ASE
Let
n ASE = |n ASE | exp(i), (7.142)
{ }
Re n = |n | cos(). (7.143)
ASE ASE
Since is a random variable with uniform distribution, it follows that

<I >= 0, (7.144)
2
2
2
2
<I >= 4R P < |n | cos > . (7.145)
LO ASE