Page 102 - Fiber Optic Communications Fund

Optical Fiber Transmission 83

Example 2.14
Consider a fiber-optic system as shown in Fig. 2.38. Fiber loss = 0.2 dB/km, length = 80 km, loss in optical
filter = 0.5 dB, and amplifier gain = 15 dB. If the minimum power required at the receiver to have a good
signal-to-noise ratio is −3 dBm, calculate the lower limit on the transmitter power in dBm and mW units.

Solution:
Fiber loss, F (dB)=−0.2 × 80 =−16 dB.
1
Filter loss, F (dB)=−0.5dB.
2
Amplifier gain, G(dB)= 15 dB.
The minimum power required at the receiver is

P (dBm)=−3dBm,
out
P (dBm)= P (dBm)+ F (dB)+ F (dB)+ G(dB).
out in 1 2
Therefore, the lower limit on the transmitter power is
P (dBm)=−3 + 16 + 0.5 − 15 dBm =−1.5dBm.
in
Using Eq. (2.125), the transmitter power in mW units is

P = 10 0.1P in (dBm) = 0.7079 mW.
in

Figure 2.38 A fiber-optic system with loss and gain.

Example 2.15
The electric field envelope at the fiber input is

s (t)= A cos (2f t). (2.246)
in m
Show that the electric field envelope at the fiber output is
[ 2 ]
s (t)= A cos (2f t) exp i(2f ) L∕2 . (2.247)
out m m 2
Ignore fiber loss and .
1