Page 412 - Fiber Optic Communications Fund

Channel Multiplexing Techniques 393

Solution:
From Eq. (9.13), we have
B
= . (9.14)
Δf
For QAM-64, from Section 4.9, we find

B = B log 64 = 6 × 10 Gb/s. (9.15)
s 2
(a) The channel spacing is given by
B 60
Δf = = GHz
6
= 10 GHz. (9.16)
(b) For a Nyquist pulse, the one-sided bandwidth f ∕2is B ∕2. Therefore, the signal bandwidth in a channel
s s
f = B = 10 GHz.
s s
Total bandwidth of the WDM system =(N − 1)Δf + 2f ∕2
s
=(11 × 10 + 10) GHz

= 120 GHz. (9.17)
(c) Total data rate = NB = 12 × 60 Gb/s

= 720 Gb/s. (9.18)

Example 9.2
A WDM system consists of 11 channels with a channel spacing of 100 GHz. The signal in each channel is
band-limited to 50 GHz. The average power per channel is 0 dBm. The WDM signal is transmitted over a
fiber of length 50 km. Fiber loss = 0.2 dB/km. Find the total power at the fiber output.

Solution:
Let the signal in channel k at the fiber input be
∑ i2kΔft
q (t)= a f(t − nT )e , k =−5, −4, … , 5
k
s
n,k
n
= g (t)e i2kΔft , (9.19)
k
where Δf = 100 GHz. The total signal field at the fiber input is
5
∑
q (t)= q (t). (9.20)
in
k
k=−5
Taking the Fourier transform of Eq. (9.20), we find
5 5
∑ ∑
̃q (f)= ̃q (f)= ̃g (f − kΔf), (9.21)
k
in
k
k=−5 k=−5