Optical Networks: A Practical Perspective. Solutions Manual for Instructors

85 Pages·00.415 MB·English

by  Rajiv Ramaswami, Kumar N. Sivarajan, Galen Sasaki

Optical Networks: A Practical Perspective
Third Edition
Solutions Manual for Instructors
Rajiv Ramaswami, Kumar N. Sivarajan and Galen Sasaki
c 2009 Morgan Kaufmann Publishers. 1
(cid:13)
December 2009
1. This material may not be used or distributed for any commercial purpose without the express written
consentofMorganKaufmannPublishers.

Preface
Inmanycases,theproblemsinthebookrequirefurtherexplorationofthetopicsindetailasopposed
tosimplypluggingnumbersintoequations. Instructorsmaythereforewanttoreviewthesolutions
beforeassigningproblems. Seethebook’swebpagehttp://www.elsevierdirect.com/9780123740922
for the current errata of the book, as well as this solutions manual. If you discover an error that
is not listed there, we would very much appreciate your letting us know about it. You can email
[email protected],[email protected]@hawaii.edu.
Notethatallequationandfigurenumbersusedinthismanualrefertothoseinthethirdedition
ofthebook.
i

2
c h a p t e r
Propagation of Signals in
Optical Fiber
2.1 FromSnell’sLawwehave,
n sinθmax n sinθmax.
0 0 = 1 1
Usingthedefinitionofθmax fromFigure2.3,wehave
0
n sin π/2 θmax n ,
1 − 1 = 2
(cid:0) (cid:1)
or,
n cosθmax n ,
1 1 = 2
or,
n2
sinθmax 1 2.
1 =s − n2
1
Therefore,
1 n2
n sinθmax n − 2 n2 n2
0 0 = 1 s n2 = 1− 2
1 q
whichis(2.2).
2.2 From(2.2),
δT 1n2
11 10ns/km.
L = cn =
2
Therefore,
n cδT
n21 2 .
1 = L
1
2 Propagation of Signals in Optical Fiber
Wehave,
NA n √21 2n cδT/L 2 1.45 3 105 10 8 0.093.
= 1 = 2 = × × × × − =
p p
Themaximumbitrateisgivenby
0.5
2.5Mb/s.
10ns/km 20km =
×
(cid:0) (cid:1)
2.3 Wehave
∂D
H J .
∇× = + ∂t
UsingJ 0andtakingthecurlofbothsides,weget
=
∂( D) ∂ ∂( P)
H ∇× (cid:15)0( E) ∇× .
∇×∇× = ∂t = ∂t ∇× + ∂t
HerewehaveusedtherelationD (cid:15) E P. Using(2.13),thissimplifiesto
0
= +
∂2B ∂( P)
∇×∇×H =−(cid:15)0 ∂t2 + ∇∂×t .
TakingFouriertransforms,wehave
H (cid:15) ω2B iω( P)
∇×∇× ˜ = 0 ˜ − ∇× ˜
(cid:15) ω2µ H iω(cid:15) χ( E)
= 0 0 ˜ − 0˜ ∇× ˜
(cid:15) ω2µ H iω(cid:15) χ(iωµ H)
= 0 0 ˜ − 0˜ 0 ˜
(cid:15) µ ω2(1 χ)H (cid:15) µ ω2n2(ω)H
= 0 0 + ˜ ˜ = 0 0 ˜
ω2n2
H.
= c2 ˜
Using H ( H) 2H,weget
∇×∇× ˜ =∇ ∇· ˜ −∇ ˜
ω2n2
2H H ( H) 0, since B 0.
∇ ˜ + c2 ˜ =∇ ∇· ˜ = ∇· =
2.4 Using 2π a n2 n2 <2.405,
λ 1− 2
q
2πa 2πa
λ n2 n2 n √21.
cutoff = 2.405 1− 1 ≈ 2.405 1
q
Fora 4µmand1 0.003,λ 1.214µm,assumingn 1.5.
cutoff 1
= = = =
2.5 (a) Wehave
2πa
λ n2 n2.
cutoff = 2.405 1− 2
q
3
Usinga 4µm,n 1.45,andλ 1.2µmyields
2 cutoff
= = =
2.405 1.2 2
n1 × 1.452 1.45454.
=s 2 π 4 + =
(cid:18) × × (cid:19)
Therefore1.45<n <1.45454forthefibertobesinglemodedforλ>1.2µm.
1
(b) Wehave
2πa
V n2 n2.
= λ 1− 2
Usinga 4µqm,λ 1.55µm,n 1.45andV 2.0,wehave
2
= = = =
Vλ 2
n n2 1.4552.
1 =s 2πa + 2 =
(cid:18) (cid:19)
Using
0.9960 2
b(V) 1.1428 ,
≈ − V
(cid:18) (cid:19)
weobtainb(2.0) 0.41576. Wealsohave
n2 n2=
b eff− 2.
= n2 n2
2− 2
Therefore,wecancalculaten 1.45218. Thus
eff
2πn =
eff
β 5.887/µm.
= λ =
2.6 Thespecifiednominalvalueofa mustsatisfy
2π(1.05a)
λ < n √2 1.1 0.005
cutoff 1
2.405 × ×
forλ 1.2µmandn 1.5. Thusthelargestvaluethatcanbespecifiedis
cutoff 1
= =
1.2 2.405
a × 2.78µm.
= 2π 1.05 1.5 √2 1.1 0.005 =
× × × × ×
Notethatwehaveusedthepropertythatλ increaseswithincreaseinaor1sothatthelargest
cutoff
possiblevaluesofa and1areusedincalculatingthecutoffwavelength.
2.7 Wehave
∂A i ∂2A
β 0.
∂z + 2 2 ∂t2 =
TakingFouriertransforms,weget
∂A i
˜ β ( iω)2A 0, or,
∂z + 2 2 − ˜ =
∂A iβ ω2
˜ 2 A 0.
∂z − 2 ˜ =
SolvingthisforA(z,ω),weget
˜
iβ ω2
2
A(z,ω) A(0,ω) exp z .
˜ = ˜ 2
" #
4 Propagation of Signals in Optical Fiber
Notethat
∞ ∞
A˜(0,ω) = A(0,t)eiωt dt = A0e−t2/2T02 eiωtdt
Z Z
−∞ −∞
A0 ∞ e−12 (cid:18)Tt022−2iωt+(iωT0)2(cid:19) e21(iωT0)2 dt
=
Z
−∞
A0e−ω2T02/2 ∞ e−21 Tt0−iωT0 2 dt
= (cid:16) (cid:17)
Z
−∞
= A0e−ω2T02/2 2πT02.
q
Therefore,
iβ ω2
A˜(z,ω)=A0T0√2πe−ω2T02/2 exp i 22 z .
!
Usingthis,weobtain
A(z,t) = 21π ∞ A˜(z,ω)e−iωt dω= A√02Tπ0 ∞ e−ω22T02 eiβ22ω2z e−iωtdω
Z Z
−∞ −∞
A0T0 ∞ e−21(cid:20)ω2 T02−iβ2z +2iωt+T02(−it)iβ22z(cid:21)e−t2/2(T02−iβ2z) dω
= √2π (cid:0) (cid:1)
Z
−∞
2
= A√02Tπ0 ∞ e−21 qω2(cid:0)T02−iβ0z(cid:1)+√T0i2t−iβz! e−t2/2(T02−iβ2z) dω
Z
−∞
2
A0T0 e−2(T02t−2iβ2z) ∞ e−21(T02−iβ2z)(cid:18)ω+T02−itiβ2z(cid:19) dω
= √2π
Z−∞
1
A0T0e−t2/2(T02−iβ2z) .
= T2 iβ z
0 − 2
q
Notethatinthelaststepweusedtheformulagivenintheproblemwith
1 T iβ z
0 2
α + ,
= T0−iβ2z = T02+β22z2
with
T
0
Re(α) >0.
= T2 β2z2
0 + 2
5
2.8 From(E.8),wederive(E.9)and(E.10)asdiscussedinAppendixE.(2.13)nowfollowsfrom(E.10).
2.9 From(2.28),withκ 0,
=
A T 1 (t β z)2
0 0 2
A(z,t) exp − ,
= T02−iβ2z −2 (T02−iβ2z)!
q
which is the envelope of a Gaussian pulse for all z. Letting t t β z (so that we choose a
0 1
= =
referenceframemovingwiththepulse),wehave
A T 1 t2
0 0 0
A(z,t ) exp
0 = T02−iβ2z −2 (T02−iβ2z)!
q
A0T0 exp 1 t02(T02+iβ2z) .
= T02−iβ2z −2 T04+(β2z)2 !
q
Thephaseofthispulseis
β zt2
2 0
ω t .
0 0+ 2(T4 (β z)2)
0 + 2
Hencethechirpfactoris,comparingwith(2.26),
κ β2zT02 β2z/T02 sgn(β2)z/LD.
= T04+(β2z)2 = 1+ βT22z 2 = 1+ LzD 2
(cid:18) 0 (cid:19) (cid:16) (cid:17)
2.10 AGaussianpulseisdescribedby
A(t) A0e−21t2/T02.
=
Itsrms widthisgivenby
∞ t2 A(t)2 dt
Trms −∞ | | .
=vuR ∞ A(t)2dt
u −∞ | |
t
R
Wehave
T A2
∞ A(t)2dt ∞ A2e t2/T2 dt √2π 0 0 T √πA2,
| | = 0 − = √2 = 0 0
Z−∞ Z−∞
and
T T2
∞ t2|A(t)|2dt =A20 ∞ t2e−t2/T02dt =A20√02√2π 20 .
Z−∞ Z−∞
6 Propagation of Signals in Optical Fiber
Therefore,
T2 T
Trms 0 0 .
=s 2 = √2
2.11 From(2.13),
2 2
T κβ z β z
z 2 2
| | 1 .
T0 =vu + T02 ! + T02 !
u
t
Forpositiveκ andnegativeβ ,
2
T κz 2 z 2
z
| | 1 .
T0 =s(cid:18) − LD(cid:19) +(cid:18)LD(cid:19)
(a) Differentiating the equation above, the minimum pulse width occurs for z z which
min
=
solves
κz z
κ 1 0.
− − LD + LD =
(cid:18) (cid:19)
Thisyields
κ
z L .
min = 1 κ2 D
Forκ 5, +
= 5
z L 0.192L .
min D D
= 26 =
(b) Thepulsewidthequalsthatofanunchirpedpulseif
κz 2 z 2 z 2
1 1 ,
− LD + LD = + LD
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
thatis,if
2L
D
z .
= κ
Forκ 5,wegetz 0.4L .
D
= =
2.12 Weleavethistothereadertogothroughthealgebraandverify.
2.13 Forafirstordersoliton,
γP
N2 0 1.
= β /T2 =
| 2| 0
Usingγ 1/W-km,β 2ps2/km,andP 50mW,
2 0
= = =
β
2
T0 | | 6.32ps.
=sγP0 =
Recallthatasolitonpulseisdescribedby
t 2
sech
(cid:18)T0(cid:19)= e−t/T0 +et/T0