Page 104 - Fiber Optic Communications Fund

Optical Fiber Transmission 85

Solution:
The effective transfer function is given by Eq. (2.232) as
{ [ 2 ]}
( TF TF DCF DCF ) (2f)
H (f)= exp i L + 2 L 2
eff
2
eff eff
2
= exp [i L (2f) ∕2], (2.254)
2
where
eff eff
TF TF
L
L = L + DCF DCF . (2.255)
2 2 2
√
2
In Example 2.6, we found that when | L| = 3T , the output pulse width is twice the input pulse width.
2
0
eff eff
Replacing L by L , we find
2
2
√
2
TF TF
| L + DCF DCF | = 3T , (2.256)
L
2 2 0
TF 2 TF
=−21 ps ∕km, L = 80 km,
2
2
DCF = 130 ps ∕km, T = 12.5ps,
2 FWHM
T = T FWHM ∕1.665 = 7.507 ps.
0
Eq. (2.256) may be written as
√
2
−21 × 80 + 130L DCF =± 3 × 7.507 .
Therefore, L DCF = 12.17 km or 13.67 km. As the pulse propagates in the DCF, the pulse undergoes compres-
TF TF
L
sion. At L DCF = 12.17 km, the pulse width is twice the initial pulse width. If L =− DCF DCF , the pulse
2 2
width becomes equal to the initial pulse width. This corresponds to a propagation distance of 12.92 km in the
DCF. After this, pulse broadening takes place and when L DCF = 13.67 km, the output pulse width is twice the
initial pulse width again.
Example 2.17
The zero dispersion wavelength of a transmission fiber (TF) is chosen as 1490 nm, so that the local dispersion
in the desired wavelength range 1530–1560 nm is not zero (so as to avoid the enhancement of nonlinear
effects). Find the accumulated dispersion of the DCF so that the net accumulated dispersion does not exceed
2
2
1100 ps/nm. Assume that the dispersion slopes of the TF and DCF are 0.08 ps/nm /km and 0 ps/nm /km,
respectively. Total transmission distance = 800 km. Other parameters are the same as in Example 2.10.
Solution:
In the absence of DCF, the dispersion at 1560 nm is
D (1560 nm)= 0.08(1560 − 1490) ps/nm/km = 5.6 ps/nm/km. (2.257)
TF
The accumulated dispersion at 1560 nm is
D L = 4480 ps/nm. (2.258)
TF TF