Page 434 - Fiber Optic Communications Fund

Channel Multiplexing Techniques 415

(b) −9
OFDM symbol period T = 12.8 × 10 s. (9.111)
s
1
Symbol rate/subcarrier B = = 78.125 MBaud. (9.112)
s
T
s
For QAM-64, we have
bit rate B = log 64B = 468.75 Mb/s. (9.113)
2 s
Since there are 64 subcarriers and two polarizations, the total data rate is
6
B = 468.75 × 10 × 2 × 64
tot
= 60 Gb/s. (9.114)
1
(c) Separation between subcarriers Δf = = 78.125 MHz. (9.115)
T
s
Total bandwidth = no. of subcarriers ×Δf
= 5 GHz. (9.116)

total data rate
Spectral efficiency =
total bandwidth
= 12 b/s/Hz. (9.117)

Exercises

9.1 In a polarization-multiplexed WDM system, number of channels = 20, total data rate = 2 Tb/s, and
spectral efficiency = 4 b/s/Hz. Calculate the channel spacing.
(Ans: 25 GHz.)

9.2 In a polarization-multiplexed WDM system based on NRZ-OOK, the first null of the NRZ spectrum
occurs at f = 40 GHz. WDM signal bandwidth = 34.3 nm. The channel spacing = 2.5f and spectral
0
0
efficiency = 0.2 b/s/Hz. Calculate (a) the number of channels and (b) the total data rate.
(Ans: (a) 43, (b) 0.86 Tb/s.)
9.3 A polarization-multiplexed WDM signal is transmitted over a 60-km-long fiber. Number of channels
= 20, fiber loss = 0.18 dB/km, channel spacing = 100 GHz, signal in each channel band-limited to
40 GHz. If the total power at the fiber output is −12.8 dBm, find the signal power/channel/polarization
at the fiber output.
(Ans: 0.0158 mW.)

9.4 Explain the operating principles of an AWG multiplexer/demultiplexer.
9.5 In a WDM system, an AWG is used to demultiplex two channels. Find the length difference Δl of the
adjacent single-mode waveguides such that the corresponding phase-shift difference is 10. Assume
6
−1
= 5.8 × 10 m .
0
(Ans: 5.416 μm.)