Page 397 - Fiber Optic Communications Fund

378 Fiber Optic Communications

Channel
+
s(t)
∑ H (ω) I
1
PD1 1
+ n(t) |∙| 2 Select bit ‘1’ if I > I
T b * Comparator 1 0
+ PD2 I 0 Select bit ‘0’ if I < I
∑ 1 0
|∙| 2
Figure 8.27 DPSK receiver.

the FSK with direct detection or asynchronous detection is applicable in this case except that the energy E 1
appearing in Eq. (8.297) should be replaced by 2E (= 2E ), where E is the average energy of s(t):
1 av av
( )
1 E av 1 DD
P = exp − = exp (− ). (8.307)
b
2 ASE 2
Comparing Eqs. (8.307) and (8.298), we see that the direct detection orthogonal FSK requires 3 dB more
power than DPSK to reach the same BER.

Example 8.3
50% duty cycle rectangular RZ pulses are used in a direct detection long-haul 40-Gb/s DPSK system operating
at 1550 nm. The peak transmitter power is 3 dBm. The fiber-optic link consists of N spans of 80-km standard
SMF with a loss of 0.2 dB/km followed by an optical amplifier with n = 1 and gain G being equal to the
sp
−5
fiber loss. Find the maximum transmission distance so that the error probability is <= 10 . Ignore receiver
noise.
Solution:
For 50% duty RZ pulses, we have
peak power
average power = , (8.308)
2
average power (dBm) = peak power (dBm) + 10 log 10 (1∕2)

= 3dBm − 3dB
= 0 dBm, (8.309)

average power = 10 0∕10 mW = 1 mW, (8.310)
average energy of a pulse = E = average power × bit interval. (8.311)
av
Bit interval,
1
T = s = 25 ps, (8.312)
b 9
40 × 10
E = 1mW × 25 ps = 2.5 × 10 −14 J. (8.313)
av
Operating frequency, 8
3 × 10
f = = 193.54 THz, (8.314)
1550 × 10 −9