Page 103 - Fiber Optic Communications Fund

84 Fiber Optic Communications

Solution:
Taking the Fourier transform of Eq. (2.246), we find
A
̃ s (f)= [s (t)] = [(f − f )+ (f + f )], (2.248)
m
in
in
m
2
where is the Dirac delta function. From Eq. (2.107), we have the fiber transfer function (after ignoring
1
and loss)
2
H(f)= exp [i (2f) L∕2]. (2.249)
2
The output spectrum is
̃ s (f)= ̃s (f)H(f)
out
in
A [ ] [ 2 ]
= (f − f )+ (f + f ) exp i (2f) L∕2 . (2.250)
m
2
m
2
Taking the inverse Fourier transform of Eq. (2.250), we obtain
∞ [ ]
L
A 2 2
s (t)= 2 ∫ −∞ [(f − f )+ (f + f )] exp −i2ft + i(2f) 2 df. (2.251)
o
m
m
Using the following relation:
∞
(f − f )X(f)df = X(f ), (2.252)
∫ m m
−∞
Eq. (2.251) is simplified as
A [ ] [ 2 ]
s (t)= exp (−i2f t)+ exp (i2f t) exp i(2f ) L∕2
m
o
m
2
m
2
2
= A cos (2f t) exp [i(2f ) L∕2]. (2.253)
m
2
m
Comparing Eqs. (2.246) and (2.253), we find that if the field envelope is a sinusoid, it acquires only a
phase shift.
Example 2.16
Consider a fiber-optic system as shown in Fig. 2.39. A Gaussian pulse is launched into the transmission fiber.
Find the length of DCF so that the pulse width (FWHM) at the output of the DCF is twice the pulse width
2
2
at the input of the TF. Assume TF =−21 ps /km, DCF = 130 ps /km, L TF = 80 km. FWHM at the input of
2 2
TF = 12.5 ps. Ignore loss and .
1
Tx Rx
TF DCF
Figure 2.39 Fiber-optic system consisting of TF and DCF. TF = transmission fiber, DCF = dispersion-compensating
fiber, Tx = transmitter, and Rx = receiver.