Page 344 - Fiber Optic Communications Fund

Transmission System Design 325

To have Q = 6, the signal photon-to-noise photon ratio N ∕N should be 18. When N is very small, the
n
s
n
amplifier noise variance becomes comparable with the shot noise and Eq. (7.166) becomes less accurate.
When the shot noise is included, Eq. (7.166) is modified as (see Example 7.8)
2N s
2
Q = . (7.167)
PSK N + 1∕2
n
Example 7.5

Ina1.55-μm coherent long-haul fiber-optic system based on PSK as shown in Fig. 7.13, fiber loss =
0.2 dB/km, amplifier spacing L = 100 km. Fiber loss is fully compensated by the amplifier placed periodi-
a
cally along the transmission line. The mean fiber launch power =−2dBm, n = 1.4, and B = 5 GHz. Find
sp e
−9
the transmission distance at which the BER becomes equal to 10 . Ignore shot noise, thermal noise, and
spontaneous–spontaneous beat noise.
Solution:

loss(dB)= 0.2 × 100 = 20 dB.
When the loss is fully compensated by the gain, we have

G(dB)= loss(dB)
= 20 dB,

G = 10 20∕10 = 100.
The ASE power spectral density
eq
= Nn hf(G − 1)
ASE sp
12
= N × 1.4 × 6.626 × 10 −34 × 193.54 × 10 ×(100 − 1)
= N × 1.77 × 10 −17 W/Hz.

From Eq. (7.163), we have
√ √
√ √
P in √ P in
√
Q PSK = √ eq = √ ,
B Nn hf(G − 1)B
ASE e sp e
P = 10 −2∕10 mW = 6.309 × 10 −4 mW.
in
−9
To have a BER of 10 , Q should be 6,
√
6.309 × 10 −4
6 = ,
N × 1.77 × 10 −17 × 5 × 10 9
or
N = floor(197.24)= 197.

Total transmission distance = 197 × 100 km
= 19, 700 km.