光纤通信(第5版)课后习题答案要点

《光纤通信（第五版》习题解答

CHAPTER1

FIBEROPTICCOMMUNICATIONSSYSTEMS

1-1dB=10log10(P2/P1)

LossFractionalPower

(dB)(P2/P1)

01

-10.8

-30.5

-60.25

-100.1

-200.01

-300.001

-400.0001

-500.00001

1-2dB=101oglo(P2/Pi)

dB/10=log1()(P2/Pi)

dB/10

P2/Pt=10

4.cdB/10a，八dB/10

P2=PlX10=0.001x10

1-3Pi=2mW

dB/111/10

P2=P110°=2x10-\10=0.159mW

dB/l9

MP2=Pi10°=10xIO

>P210的0=10x10-9x1。—

-956-9-3

P|=10x10x10=10X10=10W=lmW

《光纤通信（第五版》习题解答

1-5Fromthetext,wcfindthatRG-19/Uweighs1110kg/km.

1mileofcablex1110kg/kmx1.609km/milex2.2Ibs/kg=3929lbs.

1-6Fromthetext,wefindthatRG-19/Uhasanattenuationof22.6dB/kmat100

MHz,

UsingRG-19/U,theallowedlossis:

Loss=10logio*=10logic=-40dB

p2IO2

Maximumcoaxialcablelength=40/22.6=1.8km

Usingafiberwithloss,themaximumlengthoffiberis:

Length=40/5=8km

1-744.7x106bpsx1mcssage/64,000bps=698messages

1-8Withmanuallyoperatedblinkerlights,Iwouldguessabout2or3bps.

1-9ConductingCable

900(pairs/cablc)x24(messages/pair)=21,600messages

Fiber

144(fibers/cable)x672(messages/fiber)=96,768messages

96,768(fibercable)/21,600(coppercable)=4.48

About4.5coppercablesarcneededtocarrythesameamountinformationasthe

singlefibercable.

AttheDS-4rate,eachfibercarries4032messages.Thecomparativemessage

ratesarethen:144x4032/21600=26.88oraboutafactorof27.

12

《光纤通信（第五版》习题解答

1-10

Afiber=冗=兀=126,67mm2

(D等)2=兀(学)2=3,84848

Acopper=兀

Acopper/^tiber=3。

1-11FrequencyWavelength(m)RegionofEM

(Hz)X=c/f=3x10/fSpectrum

103x107Power

605x106Power

1033X1O5Radio

2x104l.5x104Radio

1063x102Radio

1090.3Radio

IO100.03Microwave

10143x10-6Infrared

1-12Visiblewavelengthsrangefrom0.4|xmto0.7gm.

14

WhenX=0.4gm,f==3X-IP®m/s_75x10Hz

入0.4x10-6m

14

When入=0.7gm,f=3&10皿5_43x10Hz

0.7x10-6m

Bandwidth=Af=(7.5-4.3)x1014=3.2x1014Hz

13

《光纤通信（第五版》习题解答

1-13

X入

A.(gm)W(J)

0.63.3xIO*

0.822.4xIO-19

1.31.5x10*19

Avisiblephotonhasmoreenergythananinfraredphoton.

J-14P=W/t=hfN=hNc/X,whereN=numberofphotons/sec

P=6.625^10-34X1010X3J(108/0.8X106=2.5xlO_9IV

1=2.5x10-9w(0.65A/W)=1.6nA

1-15

八耻惠鬻嗑黑Zj=6^109phOt°nS/SeCOnd

1-16

CarrierBitRate(bps)

10kHz102FortheX=1umcarrier

f_3x雇m/s

1MHz104Xlx10-6m

1(X)MHz106=3x1014Hz

10GHz108

1pm3xIO、

14

《光纤通信（第五版》习题解答

1-17

Ji-*lI*—Tp=1A=10'*6*s

h/wwwwMVwww卢漕R

[।MODULATION

v——册

AP=10S-------►MODULATEDCARRIER

WM——WW——卜t

ll—MODULATION

—I-------1------1_I-------------------------------------------------------------------t

TTp=1/FI=0.2x10-6S

AMODULATEDCARRIER

---------------------------------------------------------------------

1-18

8*l412*

f=c=J^J0m/s=2.83x10Hz,BW=O,Olf=2.83x10Hz

X1.06x10-6m

AssumeAf=4000Hzforonevoicechannel.Then

2.83x1012Hzx1channe1/4000Hz=7x108channels

1-19Open-endedsolution.

1-20Assumetherearc10billion(10*homeseachhavingone4000Hzchannel,then

IO10*(homes)x400()(Hz/home)=4x10^^Hzistherequiredbandwidth.Using

anopticalbeamoffrequency

f=3xIO14Hz

1

M=4x.l0?=0,133

f3x1014

15

《光纤通信（第五版》习题解答

Thebandwidth13.3%ofthecarrierfrequency.Thismightbepossible.

1-21

10】。messages,弋歌=6.4x10"bps

Asingleopticalcarriera(fr3x1014Hzcouldnotbeturnedonandoffthis

fast.

1-22P=100nW=100xItT”W,LetN=numberofphotons

W=Nhf=Pt

N/t=P/hf=(P/hc)X

(a)At800nm

N/t="7(0.310-6)=4x[011photons/second

6.63x10-34。xIQ8)

(b)At1550nm

N/t=4xIO”(1.55/.8)=7.8x1。"photons/second

(c)Thelongerwavelengthrequiresmorephotonsbecausetheenergyper

photonissmalleratthelongerwavelength.

1-23PR=-34dBm

PT=TransmittedpowerindBm

L=-31dB,systemlosses

PR=PT+L

-34=Pr-31

PT=-34+31二・3dBm

PT=0.5mW

16

《光纤通信（第五版》习题解答

1-24T3rateisR=45Mbps

Thenumberoferrorseachsecondis:

-3

Nc=IO,(45x1()6)=45xIOerrors/s

Inoneminute,then

3

Ne=60(45x10)=2.7errors/ininutcs

1-25R=2.3Gbps=2.3xIO9bps

Totalcapacityofthe144fibersis

144(2.3x109)=3.312x1011bps

Allowing64,000bpspervoicemessage,yields

176

3.312x10*-()5175Kio=5.175x10messages=5.175millionmessages

6.4x104

1-26P,=-38dBm,PT=4dBm

L=4-(-38)=42dBm

1-27P,=-60dBm=10logPJmW)

Pi=10"mW=IO-9W

P2=60dBm=10logP式mW)

63

P2=10mW=10W

P2-P)=1000watts(approximately)

1-28L=-5-25-15+10=35dB

1.29f=c/X=3x10V1.55x106=1.93548x10u

Onehundredthofonepercentisadatarateof:

R=104(1.93548x1014)=1.9354x10wbps=19.4Gbps

UseRIID1V=20MbpsforeachJIDTVchannel

WRHDTV=19-4x109/20x106=0.9677x103=967videochannels

1-30Open-endedquestion.

17

《光纤通信（第五版》习题解答

1-31OC-768rateis39,813.12Mb/s

NumberofvoicechannelsisN:

N=(39,813.12x106)/(64x101)=622,080

Theactualnumberislessthanthistoaccommodateoverheadsuchassignaling

andsynchronization.

I-32PhotonenergyWp

Wp=hf=he/入=6.626xIO34x3xIO8/X

Wp=19.878x10々6次加④院

IneV

196

W(,=19,878x10叫(1.6x10A)=1.2423x10/A

Ifthewavelengthisinnm,thephotonenergyineVbecomes:

Wp=1242.3/人

PhotonEnergy

500100015002000

Wavelength

1-33

WavelengthFrequencyEnergy

1.55*m1.935x10'"Hz0.802eV

1.55x105mm1.935x10"kHz1.282x10"J

1.55x10°m1.935x10"MHz

1.55x10"km1.935x105GHz

193.5THz

18

《光纤通信（第五版》习题解答

f=-='Xi。'=1935xlO,4Hz

A1.55x10-6

1242.3_1242.3

=0.802eV

1550

19

《光纤通信（第五版》习题解答

CHAPTER2

OPTICSREVIEW

1=

SinceM=(do/dj)',andtana=aforsmallangles,theny,/y0

Yi-8.32°usingtheexactexpressionor竹=8°usingtheapproximateformula.

2-2NA=n]sin0=sin0

0NA

0°0.0

10°0.174

20"0.342

30°0.50()

4500.707

20

《光纤通信（第五版》习题解答

2.3

do/fM

1infinity

1.25

1.42.5

1.61.67

2.01.0

3.00.5

4.00.33

。0%1

d<)=ftii/(dif)=20d(/(di-20)

4(mm)do(mm)

20infinity

25100

3060

6030

8026.7

20

21

《光纤通信（第五版》习题解答

2.5f=20mm,D=10mm,X=0.8pm,d=2.44/D=3.9gm

2.6w=1mm,A.=0.8gm,f=20mm

Focusedspotsize-wo=Xf/TIW=5.09pm

2.7I/Io=e'2r2/w2,w=Imm

r(mm)1/1。

01.0

0.20.923

0.40.726

0.60.486

0.80.278

1.00.135

1.50.011

8

2.8X=0.81am,w=1mm,zm00n=3.8x10m

DivergenceAngle=0=2X/TTW=5.09x10'4r

Spotsizeonthemoon=womoon=0zmoon!2=96.7km

Inmiles,womoon=96.7kmx(1mi./1,609km)=60miles

IfR=1km,

w。=峥=0,QO05号Xa=0255m

IfR=10km,

wo=fiR=SQ005Q9XIO4=2.55m

22

《光纤通信（第五版》习题解答

2.9Forglassn-1.5

(a)Thevelocityis:

v=C=3XIQ®=2x108m/s

n],5

Thetraveltimeis:

t=6x106/2X108=3x10'2seconds

i=30ms

(b)Thesatellitepathisapproximately

d=2x22000=44,000miles

d=44000xl609meters/milc

d=7.05x107m

Thetraveltimeis:

t=7103，=0.236seconds=236ms

3x108

(c)Thetwo-waysatellitedelayisabouthalfasecond.Thisisnoticeable.The

fiberdelayisnotnoticeable.

2-10

sin6t_ni

Since6t>仇，thensin0t>sin0tandthenwemusthaveni>n2.

23

《光纤通信（第五版》习题解答

INCIDENTANGLE

2-12sin0t=(m/m)sin仇=(1.48/1.46)sin&=1.01369sin0,

24

《光纤通信（第五版》习题解答

2-134=%

Fusedsilica:n=1.46

Silicon:n=3.5

Allwavelengthsinnminthefollowingtable:

入oZ(Silica)人(Silicon)

800548229

1300890371

----

Thewavelengthshortenswhenanopticalbeamentersamaterialfromfree

space.Thefrequencyremainsthesame.Thephotonenergy(hf)remainsthe

same.

2-14Usearefractiveindexof1.5.Thewavevelocityis;

8

C3x10r1八87

p=-=---------=2x10mis

n1.5

25

《光纤通信（第五版》习题解答

Thedistancetraveledinonesecond(and,thus,thefiberlength)is：

c/=v/=2x10xx1=2x10B/n

2-15，=e-2//w：

I。

0A.c5-ec-2产/M

SolveforD=2r,withtheresult:

D=2r=-I.177w

26

《光纤通信(第五版》习题解答

CHAPTER3

LIGHTWAVEFUNDAMENTALS

3-1A(T/L)=(T/L)2-(T/L)J=-M△/.,AX=-入1，入2>入卜

ForMpositive(X<1.3pm),thelongerwavelengthtravelstaster.A(T/L)<0

ForMnegative(X>1.3pm),theshorterwavelengthtravelsfaster.A(T/L)>0

3-2Z=0.85pm,AA=30nm,M=90ps/nmxkm

A(T/L)=-MAX=-30x90=-2700ps/km

ForA入=2nm,

△(T/L)=-2x90=-180ps/km

3-3X=1.55jim,M=-20ps/nmxkm,A(r/L)=-MAX

For△/.=30nm,

A(T/L)=-MAk=-(-20)x30=600ps/km

For△<=2nni,

A(T/L)--20)x2=40ps/km

3-4f3(JBxL=0.5/A(T/L),Optical.R&xL=0.35/A(T/L)

f3dBxL=0.35/A(T/L),Electrical.RNRZxL=0.7/A(T/L)

入=0.85pm:(1)△兀=30nm,A(T/L)=2.7ns/km

(2)AX=2nm,A(T/L)=0.18ns/km

X=1.55[im:(1)AX.=30nm,A(r/L)=0.6ns/km

(2)△九=2run,A(T/L)=0.04ns/km

27

《光纤通信（第五版》习题解答

,3dB(opt)f3dB(elect)(MHZ)RNRZ(Mb/s)

XAX(MHz)RRZ(Mb/s)

(um)(nm)100m1km10km100mIkm10km100m1km10km

0.8530185018518.5130013013260026026

0.852277802778277.8194441944194388803888389

1.5530833083383.3583058358.3116601166IJ7

L55212500012500012500875008750875175000175001750

3-5

k2r=%=缸n=一""-二7,66xIO6rad/m

6

Zo.82xio-

6

kg]ass=k0n-(7.66x10)(1.5)-1.15x10,rad/m

入glass='glass、=0.82/1.5=0.55pm

3-6

¥H,f=a,△f=(川AAbX=0.82pm

WhenA九=20run,Af/f=20/820=0.024,or2.4%

△f=8.78xIO1？Hz

WhenAX=1nm,Af/f-1/820-0.0012,or0.12%

△f=4.39xIO11Hz

S'RYU”AlGaAsAir

nj=3,6n2=1

10log100.68=-1.67dB

32%68%

Transmissionloss=1.67dB

28

《光纤通信（第五版》习题解答

3-8臼=1.48,n2=146

ps=,Rs=|pj2,perpendiculai

Q|cos0j+Vi)5-sin20j

2

2・o

n2oo+n

sl

pR=|pJ2,parallel

C22p

n0

—n

n2SI

0B=tan」(n2/n])=44.61°,%=sin'㈣/卬)=80.57°

5

o.

0.0

01020304050&)7080\90

IncidentAnglo,5,徒g)

3-9Provetan0B=叱/可

Setpp=0inEq.(3-29).Solvefor9)=0B

少

2

n5-n,sin0j=0

成cosOj=niVn^-nysin2优

ngcos20j=n1(n[-nfsin2仇)

n；cos20j+njsin2仇=nfn!

29

《光纤通信（第五版》习题解答

n^(l-sin20,)+nfsin2缶=n那

sin20,(nj-ng)=n；n与-城

(n1-n$)(nf+n?)sin2仇=nj仔彳-n,)

sin26i=T1T

码

nf+n5

田2

Jn彳+吗

Thus,0jisasshowninthetriangle,sothat

tan9j=皿/叼

or

tanGo=n7/ni

ArA■o-

9sin-1^=sin-1第=80.572

cC

nt1.48

ko=i2L=__=7.66xIO6

X.82x104

0,(deg)a(m-1)e-az

00.5124z/X

%011111

5

829.8x0.04

6

841.4x0.01

6

861.7x6.004

6

881.8x5.003

6--

901.9x101,47.22.05.002

30

《光纤通信（第五版》习题解答

3-11Lf=-10log[exp（-0.693尸）]

LetLf=1dB,solvefbr叫四：

fyfj-dB=0,58

+P+P

3-12PT=Pi+P2=PoiO211COS（9mt+M）+P22cos（3mt+『2）

=4+2cos（cumt+<|>]+侧>/2）cos/2）

where%一如andweusedtheidentity

cosA+cosB=2cos

⑴利.加:。

PT=4+2cos（cnmt+（j）|）

P]=P2=2+cos（comt+如）

31

《光纤通信（第五版》习题解答

P

11

6-.

(2)P|=2+cos(o)mt+(|)i)

P2=2+cos(<omt+机+n/2)

PT=4+cos(a>mt+<j>i)

3焉

+cos((nmt+0|+n/2)uti+icr11+2x10t

⑶P]=2+cos(tOmt+^i)

P2=2+COS(G)mt+M+71)

PT=4

(4)PT=4+2cos(<omt+如+△巾/2)cos(A(|)/2)

A。2cosg。/2)Pac*P'P

024

n/41.853.7

n/21.4142.8

3n/20.771.58

n00

3-13nj=1.48

n2-1.465

X=1,3x10'6

(a)FromEq.3-29

.一”4652cos85+1.48V1.4652-L482sin285

p♦j—h

1.4652cos85+1.48V1.4652-1.482sin285

32

《光纤通信（第五版》习题解答

=-0.187+j0.2456=0,309日皿=

P—0.187+j0.2456—0.309ei52.7

(b)TheattenuationfactorisgivenbyEq.(3-35):

a=ko7n?sin2Gi-n^=―V1.482sin285-1.4652

1.3xIO-6

a------2a-----0.1659=0.802radian/gm

1.3xIO-6

Thedecayisgivenby:

屋8=0.1

产=10

az=In10

z=2.3/0.802=2.87RITI

3-14TheproofisoutlinedusingEq.(3-8)inthetext.

-az

E=Eoesin(a)t-kz)

Theintensityis:

I=EQe-2az$in2(cot-kz)

Dividingtheaverageintensityatz=Lbytheaverageintensityatz=0yields

thefractionalloss:

33

《光纤通信(第五版》习题解答

Loss=e-2aL/e0=e*2aL

ConvertingtodB:

Loss(dB)-10loge*2aL=-20aLlogc=-8.685aL

Ifaisinunitsofkm'1,thenLmustbeinkm,and

Loss(dB)/L(km)=-8.685a

or

dB/km=-8.685a

3-]5Thepowerisreducedbythefactor

q-2az

Leta=2x10'5cm”,L=1km=103m=105cm

55

©•2(2x10*)(10)-e-4o.O183fractionalloss

dB=10log0.0183=-17.4dB

AlternativelyusingtheconversionbetweendB/kmanda

dB/km=-8,658a

a=2x104cm/=2..Xx1°「・皿=2km"1

cmkm

dB=-8.085x2=-17.4dB

3-16Using

dB/km=・8.685a

-0.2=-8.685ct

a=0.023km"

3-17UsingEq.(3-17),

6=-10log{e'693(2/f3dB)}

34

《光纤通信（第五版》习题解答

-0.6=loge-2772“3-dB

2

e3

4=2.772/fj.dB

2

3

2.772/dB=In4=1.386

f3_dB=1.414GHzOptical3-dBbandwidth

FromEq.(3-18)

fj.dB(electrical)=0.7!fj.dB(optical)

^-dB(electrical)=0.71x1.414=1GHz

3-18

M阁闾

M。=-0.095ps/(nm2xkni)

,=1300nm

X=1290nni

ps

0.967

(nmxkm)

3-19

ps1s(1nm12ikm

Mo=-0.0951012psUO'9nJ103m

nm2xkm

33

(a)Mo=-0.095x10s/m

32

(b)Mo=-0.095xWns/(nmxkm)

3-20A(T/L)=IMAk|=3ps/km

35

《光纤通信（第五版》习题解答

△?v=2nm

Thus,M=1.5ps/nmxkm

Assume

Xo=1300nm

3呼卜喏

63=x-130.04=-73004

X4-13004-63入3=0

Solvebyiteration

AX4-13004-63入3

13203.00x1012-2.856x1012-0.144x1012=0.03x1012

13152.99x1012-2.856x1012-0.143x1012=0.009x1012

人0=1315-1300=15nm

PlotMvs.入from1275to1325nmtographicallysolvethisproblem.

Extraforthisproblem:

At:1310nm

[1310.L3Q0；)=0.93ps/nmxkm

At:1315nm

M=。,乎](1315=1.4ps/nmxkni

36

《光纤通信(第五版》习题解答

3-21T=20ps,solitonpulsewidth

(a)Themaximumrateis:

R=-L=------1---------0.05x1012=50Gbps

T20X10/2

(b)Thesystemlosseswilllimitthelengthoffiberthatcanbeused.

3-22(MJ4)[入-入；/入3]

TheslopeisthevalueofdM/dXevaluatedat入。.

dM/d入=(Mo/4)[1-屋”(-3/〃)]

EvaluatingalA.=Xoyields

(dlVVdX)=M。

aswastobeshown.

3-23P°/Pm=10必°

IndB,thelossis:

Loss=10logPom/Pm=10log10TU10=10(yL/10)log10

Loss=yL,YindB/kmandLinkm.

Comparethiswiththealternativelosscalculation

dB=101oge2oL

=-8.685a

Loss(dB)=10logr卬-8685=JO[ogeO.23yL=晔23>L)loge

Loss=2.3yL(0.434)=yL

Thus,thetworesultsareidentical.

3-24Pout/Pin=10yL/l°

Y=-0,5dB/km

37

《光纤通信（第五版》习题解答

L(km)Pout/PinEfficiency(%)

10.8989

100.31631.6

10010-50.001

3-25L=0.2dB/km,X=1550nni

UseBeefsLaw

PM/PM=10YS0

Y=-0.2dB/km

L(km)Pou/PmEfficiency(%)

10.95595.5

100.63163.1

1000.011

3-26X=1.55|im,AX=2nm

M=-3ps/(nm.kin)

△(T/L)=・M△入=3x2=6ps/km

Thisismuchlessthancalculatedinproblem3-2,asthedispersionMismuch

lessatthelongerwavelength.

3-27ForRZcoding;

0350.35

RxL==58.3X109ATMXZ?/5

A(r/Z)6xl0-12

RxL=5&3Gbls

ForNRZcoding:

38

《光纤通信（第五版》习题解答

0.70.7

=116.7xlOyZr/MX5/5

RxL=A(r/Z)-6xlOZli

RxL=\\6.7Gb/s

,,八0.50.5

源（。即必=而力=菽谈=83.3GHzxkm

0.35

f3dB(electrical)==58.3GHzxkm

A(r/Z)