Page 203 - Fiber Optic Communications Fund

184 Fiber Optic Communications

where f = 1∕2T . In Eq. (4.138), we have used the fact that a phase shift in time domain leads to a frequency
0
b
shift in frequency domain. Using Eqs. (4.137) and (4.138), we find
[ ( ) ( )] [ ( )] [ ( )]
t t 1 1
 cos rect = T sinc 2T b f − + sinc 2T b f +
b
T b 2T b 2T b 2T b
[ ]
1 1
=−T sin (2T f) +
b b
(2T f − 1) (2T f + 1)
b b
−T sin (2T f)4T f
b
b
b
= , (4.139)
4T f − 1
2 2
b
where we have used [ ( )]
1 sin (2T f ± )
b
sinc 2T f ± = . (4.140)
b
2T (2T f ± 1)
b b
The Fourier transform of p(t) is
[ ]
1 2T sin (2fT ) T sin (2T f)4T f
b
b
b
b
b
̃ p(f)= −
2 2
2 2fT b (4T f − 1)
b
T sinc (2fT )
b
b
= . (4.141)
2 2
(1 − 4T f )
b
The PSD is given by Eq. (4.13) as
2 2 2 2
0 b
b
A |̃p(f)| A T sinc (2fT )
0
(f)= = . (4.142)
m
2 2 2
T b (1 − 4T f )
b
Example 4.7
Show that the Fourier transform of the duobinary pulse
sin (Bt)
p(t)= (4.143)
Bt(1 − Bt)
is ( ) ( )
2 f f
̃ p(f)= cos rect exp (if∕B). (4.144)
B B B
Solution:
Note that
1 1 1
= − .
t(1 − Bt) t t − T b