Page 313 - Fiber Optic Communications Fund

294 Fiber Optic Communications

Using Eqs. (6.253) and (6.267), we obtain

2 2
2
2
B .
sp−sp = R ASE oe (6.271)
Next, let us consider the case in which the optical filter is an ideal band-pass filter with full bandwidth f and
o
the electrical filter is an ideal low-pass filter with cutoff frequency f .
e
( )
f
̃
H (f)= rect , (6.272)
opt
f
o
( )
f
̃
H (f)= rect , (6.273)
e
2f e
̃
2
 (f)= |H (f)| =[rect(f∕f )] 2 (6.274)
opt opt o
= rect(f∕f ), (6.275)
o
 (t)=  −1 [rect(f∕f )] (6.276)
opt o
= f sinc(f t), (6.277)
o o
where
sin (x)
sinc(x)= , (6.278)
x
2
2
2
 (t)= f sinc (f t). (6.279)
opt o o
Eq. (6.268) may be rewritten as
′′
′′
′′
B 2 oe = ∫ u(t )H (t )dt , (6.280)
e
′′ 2 ′′ ′′
u(t )=  (t )∗ H (t ). (6.281)
opt e
Here ∗ denotes the convolution. Since the convolution in the time domain becomes a multiplication in the
frequency domain, the Fourier transform of Eq. (6.281) is
̃
̃ u(f)=  (f)H (f). (6.282)
opt2 e
′′
2
 ̃ opt2 (f)= [ (t )] (6.283)
opt
2
2
= [f sinc (f t)] (6.284)
o o
= triang (f∕f )f , (6.285)
o o
where
triang (f∕f )= 1 − |f|∕f o if |f| < f , (6.286)
o
o
= 0 otherwise. (6.287)