Page 249 - Fiber Optic Communications Fund

230 Fiber Optic Communications

Solution:
The power at the transmitter,
P (dBm)= 12 dBm.
T
Fiber loss,
loss(dB) = 0.2dB/km × 100 km = 20 dB.

The power at the receiver,

P (dBm)= P (dBm)− loss(dB) =(12 − 20) dBm =−8dBm.
T
r
(a)
P LO (dBm)= 10 dBm,
P = 10 0.1P LO (dBm) mW = 10 mW,
LO
P = 10 0.1P r (dBm) mW = 0.1585 mW.
r
For a BPSK signal assuming rectangular NRZ pulses, s(t) takes values ±1. From Eq. (5.95), we have
√ √
Peak current = |I | = R P P = 0.9 × 10 × 0.1585 mA
d
r LO
= 1.1331 mA.
If we use Eq. (5.88) after ignoring the d.c. term, we find the peak current as

√
|I| = RP ∕2 + R P P = 1.204 mA. (5.97)
r LO
r
Note that |I| ≈ |I |. The difference |I − I | is known as intermodulation cross-talk.
d
d
(b)
P LO (dBm)=−10 dBm,
P = 10 0.1×−10 mW = 0.1mW,
LO
√
|I | = 0.9 0.1 × 0.1585 = 0.1133 mA.
d
The peak current |I| after ignoring the d.c. term is
√
|I| = RP ∕2 + R P P = 0.1846 mA. (5.98)
r r LO
In this case, the intermodulation cross-talk is comparable to I .
d

5.6.1.2 Heterodyne Single-Branch Receiver
When the frequency offset between the transmitter laser and LO is in the microwave range, the signal I (t)
d
given by Eq. (5.91) may be interpreted as the message s(t) modulating the microwave carrier of frequency .
IF
Let us assume that s(t) is real and Eq. (5.91) can be written as
I (t)= I s(t) cos ( t +Δ), (5.99)
d 0 IF