武汉理工大学光纤通信2015年试题

一Consideramultimodesilicafiberthathascorerefractiveindex11.480nandacladdingindex21.478n.Find(a)thecriticalangel,(b)thenumericalaperture,and(c)theacceptanceangel.解：（a）由式21sincnn可以求得临界角（b）由式1222121sin()2ANAnnnn可以计算出数值孔径（c）由式12220,max112sinsinsin()Acnnnnn可以求得出空气中的接受角为二、Consideramultimodestep-indexfiberwitha62.5-mcorediameterandacore-claddingindexdifferenceof1.5percent.Ifthecorerefractiveindexis1.480,estimatethethetotalnumberofmodessupportedinthefiberatawavelengthof850nm.Solution：thenormalizedfrequencyis122aVn,thetotalnumberofmodesis22VM三supposewehaveamultimodestep-indexopticalfiberthathasacoreradiusof25m,acoreindexof1.480andanindexdifference0.01.Whatarethenumberofmodesinthefiberatwavelengths860,1310,and1550nm.解：首先由122aVn计算出V值，再利用212MV可以计算出总模数。四supposewehavethreemultimodestep-indexopticalfiberseachofwhichhasacoreindexof1.48andindexdifference0.01.Assumethethreefibershavecorediametersof50,62.5and100m.Whatarethenumberofmodesinthesefibersatawavelengthof1550nm.五Adouble-heterojunctionInGaAsPLEDemittingatapeakwavelengthof1310nmhasradiativeandnonradiativeinternalqudiativerecombinationtimesof30and100ns,respectively.Thedrivecurrentis40mA.Find(a)thebulkrecombinationtime;(b)theinternalquantumefficiency;and(c)theinternalpowerlevel.解：(a)由111rnr得整体复合寿命rnrrnr（b）由int11rnrr得内量子效率(c)由intintintIhcIphvqq，得内部发光功率六Considera1-kmlongmultimodestep-indexfiberinwhich1n=1.480and=0,01,sothat2n=1.465.whatisthemodaldelayperlengthinthisfiber?Solution:from212LnTcnyields212nTLcn=50ns/km七Considerthefollowingtwomultimodefibers(a)astep-indexfiberwithacoreindex1n=1.458andacore-claddingindexdifference=0.01;(b)aparabolic-profilegraded-indexfiberwiththesamevaluesof1nand.Comparetothermspulsebroadeningperkilometerforthesetwofibers?解：（a）由式211()2343sLnLNAcnc，有114.0/23snnskmLc（b）由式21203sLnc，有2114.0/203snpskmLc八agivenstep-indexfiberhasacorerefractiveindexof1.480,acoreradiusequalto4.5m,andacore-claddingindexdifferenceof0.25percent.Whatisthecutoffwavelengthforthisfiber?解：由式122212122()2caannnVV，当V=2.405时，光纤的截止波长为1221230cannmV。九aparticularLEDhasa5-nsinjectedcarrierlifetime.Whennomodulationcurrentisappliedtothedevice,theopticaloutputpoweris0.250mWforaspecifieddcbias.assumingparasiticcapacitancesarenegligible,whataretheopticaloutputsatmodulationfrequenciesof(a)10MHzand(b)100Mhz?解：（a）由1220()1()ipp可得在10M处输出功率为()239pW在100M处的输出光功率为()76pW由此可见这种器件的输出光功率随着调制速率的增加而增加。十．AgivenGaAlAslaserhasanopticalcavityof300manda100-mwidth.Atanormaloperatingtemperature,thegainfactor332110Acmandthelosscoefficient110cm.Assumethereflectivityisoforeachendface.Find(a)thethresholdcurrentdensityand(b)thethresholdcurrentforthisdevice解：（a）由1211ln()2thtendgLRR和ththgJ可得111ln()thJLR(b)阈值电流thI由下式给出ththIJ腔体横截面积十一consideradouble-heterostructureedge-emittingFabry-PerotAlGaAslaser,whichemitsat900nm.Supposethatthelaserchipis300mlongandtherefractiveindexofthelasermaterialis4.3.(a)Howmanyhalf-wavelengthsspantheregionbetweentheFabry-Perotmirrorsurfaces?(b)WhatIthespacingbetweenthelasingmodes?解：（a）从式22LLnmvnc可得法布里-珀罗腔两镜面间的半波数目m（b）从式22Ln可得十二AGaAsopticalsourcewitharefractiveindexof3.6iscoupledtoasilicafiberthathasarefractiveindexod1.480.whatisthepowerlossbetweenthesourceandthefiber?解：如果光纤端面和光源在屋里上紧密衔接，则由211nnRnn，在光源和光纤头端的分界面上菲涅耳反射可用下式来表示2110.174nnRnn这相当于有17.4%的光功率被反射回光源，与这一R值相应的耦合功率由下式给定(1)coupledemittedPRP用分贝表示的功率损耗L为10lg()10lg(1)0.83coupledemittedPLRdBP这个值有可能因为在光源和光纤端面之间存在折射率匹配物质而减小十三AnInGaAsPopticalsourcethathasarefractiveindexof3.540iscloselycoupledtoastep-indexfiberthathasacorerefractiveindexof1.480.assumethatthesourcesizeissmallerthanthefibercoreandthatthesmallgapbetweenthesourceandthefiberisfilledwithagelthathasarefractiveindexof1.520.(a)whatisthepowerlossindecibelsfromthesourceintothefiber?(b)whatisthepowerlossifnogelisused?解：（a）这里需要考虑两个界面反射率。首先由211nnRnn可以得出凝胶对光源的反射率0.159sgR同样，可得光纤对凝胶的反射率0.040gfR于是，总的反射率为0.0400.0064sggfRRR。功率损耗的分贝值则为10lg(1)0.0028LRdB（b）如果不用凝胶作为折射率匹配材料，切假设光源与光纤之间没有间隙，则由211nnRnn可得此时反射率为R=0.168此时的功率损耗分贝数为10lg(1)0.799LRdB十四6.7agivensiliconavalanchephotodiodehasaquantumefficiencyof65percentanawavelengthof900nm,Suppose0.5Wofopticalpowerproducesamultipliedphotocurrentof10A.WhatisthemultiplicationM?解：初级光电流为pininqIPPhc，倍增因子MpIMI十五AnInGaAspinphotodiodehasthefollowingparametersatawavelengthof1300nm:4,0.90,1000DLInAR,andthesurfaceleakagecurrentisnegligible.Theincidentopticalpoweris30nW(-35dBm),andthereceiverbandwidthis20MHz.findthevariousnoisetermsofthereceiver.解：首先计算初级光电流pI，Pin光电二极管的均方量子噪声电流为22shotpeiqIB均方暗电流为22DBDeiqIB接收机的均方热噪声电流24BTeLkTiBR十六Considerananalogopticalfibersystemoperatingat1550nm,whichhasaneffectivereceivernoisebandwidthof5MHz.assumingthatthereceiverdsignalisquantumnoiselimited,whatistheincidentopticalpowernecessarytohaveasignal-to-noiseratioof50dBatthereceiver?Assumetheresponsivityis0.9A/Wandthatm=0.5.解：注意到50dB的SNR即S/N=105,由式2(/)4eSNqBPm可得入射光功率十七A22biconical