Page 78 - Fiber Optic Communications Fund

Optical Fiber Transmission 59

second term on the right-hand side. To simplify Eq. (2.98), let us choose the variable Ω= − .Using
0
Eq. (2.102) in Eq. (2.98), we obtain
+∞
1 2
̃
F(t, z)= B(Ω) exp [−z∕2 − i( t − z)+ i Ωz + i Ω z∕2] exp (−iΩt) dΩ
2 ∫ −∞
2
0
0
1
exp [−z∕2 − i( t − z)] +∞ 2
0
0
̃
= B(Ω) exp (i Ωz + i Ω z∕2 − iΩt) dΩ
2
1
2 ∫ −∞
exp [−i( t − z)] +∞
0
0
̃
= B(Ω)H (Ω, z) exp (−iΩt) dΩ, (2.106)
f
2 ∫ −∞
where
2
H (Ω, z)= exp (−z∕2 + i Ωz + i Ω z∕2) (2.107)
2
1
f
is called the fiber transfer function and
̃
̃
B(Ω) ≡ A( +Ω). (2.108)
0
The linear phase shift Ωz corresponds to a delay in time domain. To see that, set = 0 in Eq. (2.107) and
1 2
the fiber output at z = L,
exp [−z∕2 − i( t − L)] +∞
0
0
̃
F(t, L)= B(Ω) exp [−iΩ(t − L)] dΩ
1
2 ∫ −∞
= exp [−z∕2 − i( t − L)]B(t − L). (2.109)
0 0 1
In a dispersion-free fiber ( = 0), the pulse is simply delayed by L at the fiber output without any change
1
2
in pulse shape, as in the free-space propagation. Using Eqs. (2.98) and (2.106), the optical field distribution
can be written as
(x, y, z, t)= Φ(x, y) exp [−i( t − z)] s(t, z) , (2.110)
0
0
⏟⏟⏟ ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟ ⏟⏟⏟
transverse field carrier field envelope
where +∞
1
̃
s(t, z)= B(Ω)H (Ω, z) exp (−iΩt) dΩ (2.111)
f
2 ∫ −∞
and
+∞
̃
B(Ω) = s(t, 0) exp (iΩt)dt. (2.112)
∫
−∞
Eqs. (2.111) and (2.112) can be rewritten as
[ ]
s(t, z)=  −1 ̃ B(Ω)H (Ω, z) , (2.113)
f
̃
B(Ω) = [s(t, 0)], (2.114)
̃
B(Ω) ⇌ s(t, 0), (2.115)
where  and  −1 denote Fourier and inverse Fourier transforms, respectively, and ⇌ indicates that they are
Fourier transform pairs. In this section, we focus mainly on the field envelope s(t, z). Let us assume that the
transverse field distribution of the laser output is the same as that of the fiber, and therefore there is no change
in the transverse field distribution along the fiber. Let the field envelope of the laser output be s (t),
i
s (t)= s(t, 0) (2.116)
i