Page 433 - Fiber Optic Communications Fund

414 Fiber Optic Communications

Therefore, the total bit rate is
NB = 2NB = 10 Gb/s, (9.100)
s
1
where N is the number of subcarriers. Since the frequency separation between the subcarriers Δf = = B ,
s
T s
Eq. (9.86) may be rewritten as
T g
L = , (9.101)
max
| |(2NB )
2 s
where L max is the maximum reach up to which carrier orthogonality is preserved. From Eq. (9.101), we find
3 −27 9
T = 5000 × 10 × 22 × 10 × 10 × 10 × s
g
= 3.4558 ns. (9.102)
Since T = 0.07T , T = 49.368 ns. Using Eq. (9.100), with T = 1∕B , we find
g s s s s
{ 9 −9 }
10 × 10 × 49.368 × 10
N = floor
2
= 246. (9.103)

Example 9.9

In a polarization-multiplexed OFDM system, there are 64 subcarriers and each carrier is modulated by
QAM-64. OFDM symbol period = 12.8 ns, launch power to the fiber = 2 dBm, fiber loss = 0.19 dB/km,
fiber length = 70 km. Calculate (a) the signal power/subcarrier/polarization at the fiber output, (b) the data
rate and (c) the spectral efficiency.

Solution:
(a)
Total loss = 0.19 × 70 = 13.3 dB. (9.104)
Input power = 2 dBm. (9.105)

Output power P (dBm) = 2 − 13.3dBm
out
=−11.3 dBm. (9.106)

P (mW) = 10 −11.3∕10 mW
out
= 0.0741 mW. (9.107)

No. of subcarriers = 64. (9.108)
No. of polarizations = 2. (9.109)

0.0741
Signal power/subcarrier/polarization = mW
2 × 64
= 5.789 × 10 −4 mW. (9.110)