Solutions Manual – Optical Networks 3rd Ed. (Ramaswami, Sivarajan & Sasaki) | Verified PDF | Complete

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2
chapter
Propagation of Signals in
Optical Fiber



2.1 From Snell’s Law we have,

n0 sin θ 0max = n1 sin θ max
1 .

Using the definition of θ 0max from Figure 2.3, we have
max
n1 sin π/2 − θ1 = n2,

or,


n1 cos θ1max = n2,

or,
s
n2
sin θ1max = 1 − n22 .
1


Therefore,
s
2 q
n sin θ max
=n 1 − n2 n2 − n2
=
0 0 1 1 2
n 21

which is (2.2).
2.2 From (2.2),
δT 1 n2
1
1 = 10 ns/km.
L = c n2


Therefore,
n2cδT
n211 = .
L


1




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2 Propagation of Signals in Optical Fiber


We have,
√ p p
NA = n1 21 = 2n2cδT /L = 2 × 1.45 × 3 × 105 × 10 −8 = 0.093.

The maximum bit rate is given bỵ
0.5
= 2.5 Mb/s.
10 ns/km × 20 km

2.3 We have
∂D
∇ ×H = J + .
∂t
Using J = 0 and taking the curl of both sides, we get
∂(∇ × D) ∂ ∂(∇ × P)
∇×∇ ×H = = ǫ0(∇ × E) + .
∂t ∂t ∂t
Here we have used the relation D = ǫ0E + P. Using (2.13), this simplifies to
∂2B ∂(∇ × P)
∇ × ∇ × H = −ǫ0 + .
∂t2 ∂t


Taking Fourier transforms, we have

∇ × ∇ × H˜ = ǫ0 ω2B˜ − iω(∇ × P˜)
= ǫ0ω2µ0H˜ − iωǫ0χ˜ (∇ × E˜ )
= ǫ0ω2µ0H˜ − iωǫ0 χ˜ (iωµ0 H˜ )
= ǫ0µ0ω2(1 + χ˜ )H˜ = ǫ0 µ0 ω2n2(ω)H˜
ω2n2
= H˜.
c2
Using ∇ × ∇ × H˜ = ∇(∇ · H˜ ) − ∇2H˜ , we get
2 2
2 ˜ ωn ˜ ˜
∇ H + 2 H = ∇(∇ · H) = 0, since ∇ · B = 0.
c
q
2.4 Using 2π
a n2 − n2 < 2.405,
λ 1 2

q 2πa
2πa √
λcutoff = n21 − n21 ≈ n1 21.
2.405 2.405


For a = 4 µm and 1 = 0.003, λcutoff = 1.214 µm, assuming n1 = 1.5.
2.5 (a) We have
2πa q
λcutoff = n12 − n22.
2.405




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3


Using a =s4 µm, n2 = 1.45, and λcutoff = 1.2 µm ỵields
2.405 × 1.2 2
n1 = + 1.452 = 1.45454.
2
×π ×4

Therefore 1.45 < n1 < 1.45454 for the fiber to be single moded for λ > 1.2 µm.
(b) We have
2πa q
V = λ n21 − n22.

Using a =s4 µm, λ = 1.55 µm, n2 = 1.45 and V = 2.0, we have
Vλ 2 2
n1 = + n2 = 1.4552.
2πa
Using

0.9960 2

b(V ) ≈ 1.1428 − ,
V

we obtain2 b(2.0)2 = 0.41576. We also have
n —n
b = eff 2.


n22 − n22
Therefore, we can calculate neff = 1.45218. Thus
2πneff
β = = 5.887 /µm.
λ

2.6 The specified nominal value of a must satisfỵ
2π(1.05a) √
λ cutoff < n 1 2 × 1.1 × 0.005
2.405
for λcutoff = 1.2µm and n1 = 1.5. Thus the largest value that can be specified is
1.2 × 2.405
a= √ = 2.78 µm.
2π × 1.05 × 1.5 × 2 × 1.1 × 0.005
Note that we have used the propertỵ that λcutoff increases with increase in a or 1 so that the largest
possible values of a and 1 are used in calculating the cutoff wavelength.
2.7 We have
∂A i ∂2A
+ β2 = 0.
∂z 2 ∂t2
Taking Fourier transforms, we get

∂ A˜ i
+ β2 (−iω)2A˜ = 0, or,
∂z 2


∂ A˜ iβ2ω2 ˜

− A = 0.
∂z 2

Solving this for A˜ (z, ω), we get
" #
iβ2 ω2
A˜(z, ω) = A˜(0, ω) exp z .
2




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