Page 331 - Fiber Optic Communications Fund

312 Fiber Optic Communications

P (dBm)= P (dBm)− loss(dB) (7.69)
in
1r
= 4 − 0.2 × 240 =−44 dBm
= 10 P 1r (dBm)∕10 mW = 3.98 × 10 −5 mW,

P LO = 10 P LO (dBm)∕10 mW = 10 mW. (7.70)

The mean of bit ‘1’ is
√
I = 2R P P (7.71)
1r LO
1
√
= 2 × 0.874 3.98 × 10 −5 × 10 mA
= 3.48 × 10 −2 mA.
The total noise variance of ‘1’ is given by Eq. (7.43),

2
= 2qB R(P + P LO )+ 4k TB ∕R L (7.72)
B
e
e
1r
1
−2
9
= 2 × 1.602 × 10 −19 × 7.5 × 10 × 0.874 ×(3.98 × 10 −8 + 10 )
9
+ 4 × 1.38 × 10 −23 × 290 × 7.5 × 10 ∕100 A 2
2
= 2.22 × 10 −11 A .
The mean and variance of ‘0’ are calculated as follows:
I = 0. (7.73)
0
Since P ≪ P , comparing Eqs. (7.43) and (7.46), we find
1r LO
2
2
2
= = 2.22 × 10 −11 A , (7.74)
1 0
I − I 0
1
Q = (7.75)
+ 0
1
3.48 × 10 −5
= √ = 3.7.
2 × 2.22 × 10 −11
The approximate Q-factor is given by Eq. (7.53),
√
P 1r
Q = = 3.8. (7.76)
2hfB e
(b) For PSK, the peak power is the same as the average power,
P = P . (7.77)
in in
P (dBm)= 1 dBm. (7.78)
in
The peak power at the receiver is
P (dBm)= P (dBm)− loss(dB) (7.79)
1r in
= 1 − 0.2 × 240

=−47 dBm, (7.80)