Page 85 - Fiber Optic Communications Fund

66 Fiber Optic Communications

Using Eqs. (2.149) and (2.150), the inverse Fourier transform of ̃s (f) is
o
Ab [ 2 ]
s (t)= exp −(tb) . (2.155)
o
a
Using Eqs. (2.151) and (2.154), we have
2
[T − i L] 1∕2
a 0 2
= , (2.156)
b T
0
t 2
2 2
b t = . (2.157)
2
2(T − i L)
0 2
Therefore, the output field envelope is
[ ]
AT 0 t 2
s (t)= exp − ( ) . (2.158)
o
2
2
(T − i L) 1∕2 2 T − i L
0 2 0 2
To find the pulse width at the output, let us first calculate the output power
[ ] 2
2 2
A T 0 | | t 2 | |
2
P (t)= |s (t)| = | exp − ( ) | . (2.159)
o
o
( ) 2 | 2 |
| 2 1∕2| | 2 T − i L |
| T − i L | | 0 |
2
2
| 0 |
Since 2 2
t 2 t (T + i L)
2
0
= , (2.160)
2
4
2 2
2(T − i L) 2(T + L )
2
0 0 2
we obtain ( )
T A 2 t 2
0
P (t)= exp − (2.161)
o 2
T
1 T
1
2
2 2
2
4
where T =(T + L )∕T . The FWHM at the output is given by
1 0 2 0
4
2 2 1∕2
(T + L )
2
0
T out = 2(ln 2) 1∕2 T = 2(ln 2) 1∕2 . (2.162)
FWHM 1
T 0
To determine the amount of pulse broadening, the ratio of output to input pulse widths is calculated as
2 2 1∕2
4
T out (T + L )
2
0
FWHM
= in = 2 . (2.163)
T T
FWHM 0
When
√
| L| = 3T 2 0 (2.164)
2
we find = 2, which means that the output pulse width is twice the input pulse width. Note that the amount
of pulse broadening is independent of the sign of the dispersion coefficient .
2
The frequency chirp or instantaneous frequency deviation is defined as
d
(t)=− , (2.165)
dt