Page 351 - Fiber Optic Communications Fund

332 Fiber Optic Communications

s DCF
L DCF =
2,DCF
1.89 × 10 −21
= = 13.03 km.
145 × 10 −27
(b) Since the amplifiers compensate for the loss due to TF and DCF exactly, we have

G (dB)+ G (dB)= H (dB)+ H DCF (dB),
2
TF
1
H (dB)= 0.18 × 100 = 18 dB,
TF
H (dB)= 0.5 × 13.03 = 6.517 dB,
DCF
G (dB)= 16,
1
G (dB)= 18 + 6.517 − 16 = 8.517 dB.
2
(c) A two-stage amplifier with a DCF in between can be replaced by an equivalent amplifier with gain

G (dB)= G (dB)+ G (dB)− H DCF (dB)= 18 dB,
1
2
eq
G eq = 10 18∕10 = 63.09,
F n,2 1
F n,eq = F n,1 + − ,
G H DCF G 1
1
F (dB)= 5.5dB,
n,1
F = 10 5.5∕10 = 3.548,
n,1
F (dB)= 7.5dB,
n,2
F n,2 = 10 7.5∕10 = 5.62,
H = 10 −H DCF (dB)∕10 = 0.223,
DCF
G = 10 16∕10 = 39.81,
1
5.62 1
F n,eq = 3.548 + − = 4.156.
39.81 × 0.223 39.81
Since we have 70 identical spans, the PSD of the ASE at the receiver is

= 70 × hf(G F − 1)∕2
ASE,eq eq n,eq
12
= 70 × 6.626 × 10 −34 × 193.54 × 10 ×(63.09 × 4.156 − 1)∕2W/Hz
= 1.17 × 10 −15 W/Hz.

For OOK, the peak power is twice the average power:

P = 2 × 10 P in (dBm)∕10 mW
in
= 1.2 × 10 −3 W.