天线 第三版 英文版 (约翰﹒克劳斯 JohnD.Kraus 著) 电子工业出版社 课后答案 157页

'PrefaceThisInstructors’ManualprovidessolutionstomostoftheproblemsinANTENNAS:FORALLAPPLICATIONS,THIRDEDITION.AllproblemsaresolvedforwhichanswersappearinAppendixFofthetext,andinaddition,solutionsaregivenforalargefractionoftheotherproblems.Includingmultipleparts,thereare600problemsinthetextandsolutionsarepresentedhereforthemajorityofthem.Manyoftheproblemtitlesaresupplementedbykeywordsorphrasesalludingtothesolutionprocedure.Answersareindicated.Manytipsonsolutionsareincludedwhichcanbepassedontostudents.Althoughanobjectiveofproblemsolvingistoobtainananswer,wehaveendeavoredtoalsoprovideinsightsastohowmanyoftheproblemsarerelatedtoengineeringsituationsintherealworld.TheManualincludesanindextoassistinfindingproblemsbytopicorprincipleandtofacilitatefindingclosely-relatedproblems.ThisManualwaspreparedwiththeassistanceofDr.ErichPacht.ProfessorJohnD.KrausDept.ofElectricalEngineeringOhioStateUniversity2015NeilAve课后答案网Columbus,Ohio43210Dr.RonaldJ.MarhefkaSeniorResearchScientist/AdjunctProfessorTheOhioStateUniversitywww.hackshp.cnElectroscienceLaboratory1320KinnearRoadColumbus,Ohio43212iii TableofContentsPrefaceiiiProblemSolutions:Chapter2.AntennaBasics............................................................................................1Chapter3.TheAntennaFamily..................................................................................17Chapter4.PointSources.............................................................................................19Chapter5.ArraysofPointSources,PartI..................................................................23Chapter5.ArraysofPointSources,PartII.................................................................29Chapter6.TheElectricDipoleandThinLinearAntennas.........................................35Chapter7.TheLoopAntenna.....................................................................................47Chapter8.End-FireAntennas:TheHelicalBeamAntennaandtheYagi-UdaArray,PartI...............................................................................................53Chapter8.TheHelicalAntenna:AxialandOtherModes,PartII.............................55Chapter9.Slot,PatchandHornAntennas..................................................................57Chapter10.FlatSheet,CornerandParabolicReflectorAntennas...............................65Chapter11.BroadbandandFrequency-IndependentAntennas....................................75Chapter12.AntennaTemperature,RemoteSensingandRadarCrossSection............81Chapter13.SelfandMutualImpedances....................................................................103Chapter14.TheCylindricalAntennaandtheMomentMethod(MM)......................105Chapter15.TheFourierTransformRelationBetweenApertureDistributionandFar-FieldPattern................................................................................107Chapter16.ArraysofDipolesandofAperture...........................................................109Chapter17.LensAntennas..........................................................................................121Chapter18.Frequency-SelectiveSurfacesandPeriodicStructuresByBenA.Munk......................................................................................125Chapter19.PracticalDesignConsiderations课后答案网ofLargeApertureAntennas................127Chapter21.AntennasforSpecialApplications..........................................................135Chapter23.Baluns,etc.ByBenA.Munk..................................................................143Chapter24.AntennaMeasurements.ByArtoLehtoandPerttiVainikainen....................................................................................147www.hackshp.cnIndex153iv 1Chapter2.AntennaBasics2-7-1.Directivity.ShowthatthedirectivityDofanantennamaybewritten()∗()Eθ,φmaxEθ,φmax2rZD=()()∗1Eθ,φEθ,φ2∫∫rdΩ4π4πZSolution:U(θ,φ)maxD=Uav21,U(θ,φ)=S(θ,φ)r,U=U(θ,φ)dΩmaxmaxav∫∫4π4π()()∗2Eθ,φEθ,φU(θ,φ)=S(θ,φ)r,S(θ,φ)=ZTherefore()()∗Eθ,φmaxEθ,φmax2rZD=q.e.d.()()∗1Eθ,φEθ,φ2∫∫rdΩ4π4πZ2222Notethatr=area/steradian,so课后答案网U=Sror(watts/steradian)=(watts/meter)×meter2-7-2.Approximatedirectivities.Calculatetheapproximatedirectivityfromthehalf-powerbeamwidthsofaunidirectionalantennaifthenormalizwww.hackshp.cnedpowerpatternisgivenby:(a)Pn=cosθ,(b)Pn23n=cosθ,(c)Pn=cosθ,and(d)Pn=cosθ.Inallcasesthesepatternsareunidirectional(+zdirection)withPnhavingavalueonlyforzenithangles0°≤θ≤90°andPn=0for90°≤θ≤180°.Thepatternsareindependentoftheazimuthangleφ.Solution:−1oo40,000(a)θ=2cos)5.0(=2×60=120,D==278(ans.)HP2(120)−1oo40,000(b)θ=2cos()5.0=×245=90,D==.494(ans.)HP2(90) 2−13oo40,000(c)θ=2cos()5.0=×237.47=74.93,D==3.7(ans.)HP2(75)2-7-2.continued−1n10,000(d)θ=2cos()5.0,D=(ans.)HP−1n2(cos(5.0))*2-7-3.Approximatedirectivities.Calculatetheapproximatedirectivitiesfromthehalf-powerbeamwidthsofthethreeunidirectionalantennashavingpowerpatternsasfollows:2P(θ,φ)=Pmsinθsinφ3P(θ,φ)=Pmsinθsinφ23P(θ,φ)=PmsinθsinφP(θ,φ)hasavalueonlyfor0≤θ≤πand0≤φ≤πandiszeroelsewhere.Solution:TofindDusingapproximaterelations,wefirstmustfindthehalf-powerbeamwidths.HPBWHPBW=90−θorθ=90−2课后答案网2⎛HPBW⎞1Forsinθpattern,sinθ=sin⎜90−⎟=,⎝2⎠2HPBW−1⎛1⎞HPBW−1⎛1⎞o90−sin⎜⎟,−sin⎜⎟−90,∴HPBW120=2⎝2⎠www.hackshp.cn2⎝2⎠222⎛HPBW⎞1Forsinθpattern,sinθ=sin⎜90−⎟=,⎝2⎠2⎛HPBW⎞1osin⎜90−⎟=,∴HPBW90=⎝2⎠2333⎛HPBW⎞1Forsinθpattern,sinθ=sin⎜90−⎟=,⎝2⎠2 3⎛HPBW⎞1osin⎜90−⎟=,∴HPBW74.9=23⎝⎠2*2-7-3.continuedThus,41,253sq.deg.41,25340,000D===.382≅=.370(ans.)θHPφHP(120)(90)(120)(90)2forP(θ,φ)=sinθsinφ41,25340,000==.459≅=.445(ans.)(120)(74)9.(120)(74)9.3forP(θ,φ)=sinθsinφ41,25340,000==.612≅=.593(ans.)(90)(74)9.(90)(74)9.forP(θ,φ)=sin2θsin3φ*2-7-4.Directivityandgain.(a)EstimatethedirectivityofanantennawithθHP=2°,φHP=1°,and(b)findthegainofthisantennaifefficiencyk=0.5.Solution:40,00040,0004(a)D===0.2×10or43.0dB(ans.)θφ2()()1HPHP44(b)G=kD=0.2(5.0×10课后答案网)=0.1×10or40.0dB(ans.)2-9-1.Directivityandapertures.Showthatthedirectivityofanantennamaybeexpressedaswww.hackshp.cn()∗()4π∫∫ApEx,ydxdy∫∫ApEx,ydxdyD=2∗λ∫∫E()()x,yEx,ydxdyApwhereE(x,y)istheaperturefielddistribution.Solution:Ifthefieldovertheapertureisuniform,thedirectivityisamaximum(=Dm)andthepowerradiatedisP′.Foranactualaperturedistribution,thedirectivityisDandthepowerradiatedisP.Equatingeffectivepowers 4*EavEavP′4πApDP′=DP,D=D=AZmm2p∗PλE()()x,yEx,y∫∫dxdyApZ2-9-1.continued1whereEav=∫∫E(x,y)dxdyApAp∗4π∫∫ApExydxdy(),,∫∫ApE()xydxdythereforeD=q.e.d.2∗λ∫∫ExyExydxdy()(),,Ap∗∗∗EEAEEEAavavpavavavewhere===ε=,,∗1()2ap∫∫ExyExydxdy()()∫∫ExyE()(),,∗xydxdyEAavpApAp2-9-2.Effectiveapertureandbeamarea.Whatisthemaximumeffectiveaperture(approximately)forabeamantennahavinghalf-powerwidthsof30°and35°inperpendicularplanesintersectinginthebeamaxis?Minorlobesaresmallandmaybeneglected.Solution:22ooλ573.22Ω≅AθφHPHP=30×课后答案网35,Aem=≅ooλ=1.3λ(ans.)ΩA30×35*2-9-3.Effectiveapertureanddirectivity.Whatisthemaximumeffectiveapertureofwww.hackshp.cnamicrowaveantennawithadirectivityof900?22Dλ90022Solution:DA=4/πλem,Aem==λ=716.λ(ans.)4π4π2-11-1.ReceivedpowerandtheFriisformula.Whatisthemaximumpowerreceivedatadistanceof0.5kmoverafree-space1GHzcircuitconsistingofatransmittingantennawitha25dBgainandareceivingantennawitha20dBgain?Thegainiswithrespecttoalosslessisotropicsource.Thetransmittingantennainputis150W. 5Solution:2289DtλDrλλ==cf/310/10×=0.3m,A=,A=eter44ππ2-11-1.continued222AetAerDtλDrλ316×3.0×100Pr=Pt22=Pt222=15022=.00108W=10.8mW(ans.)rλ4(π)rλ4(π)500*2-11-2.Spacecraftlinkover100Mm.Twospacecraftareseparatedby100Mm.EachhasanantennawithD=1000operatingat2.5GHz.IfcraftA"sreceiverrequires20dBover1pW,whattransmitterpowerisrequiredoncraftBtoachievethissignallevel?Solution:289Dλλ==cf/310/2.510××=0.12m,AA==eter4π−−1210P(required)10010=×=10Wr2222222162rrλπ(4)λπr(4)−1010(4)πP==PP=P=10=≅10966W11kW(ans.)tr22r42r262ADDλλ100.12et2-11-3.Spacecraftlinkover3Mm.课后答案网Twospacecraftareseparatedby3Mm.EachhasanantennawithD=200operatingat2GHz.IfcraftA"sreceiverrequires20dBover1pW,whattransmitterpowerisrequiredoncraftBtoachievethissignallevel?Solution:www.hackshp.cnDλ289λ==cf/310/210××=0.15mAA==eter4π−−1210P=×=1001010Wr22222212rrλπ(4)λ−10(4)910π×P==PP=10=158W(ans.)trr22242AADλλ4100.15××eter2-11-4.MarsandJupiterlinks.(a)Designatwo-wayradiolinktooperateoverearth-Marsdistancesfordataandpicture-19transmissionwithaMarsprobeat2.5GHzwitha5MHzbandwidth.Apowerof10-1-17-1WHzistobedeliveredtotheearthreceiverand10WHztotheMarsreceiver.The 6Marsantennamustbenolargerthan3mindiameter.SpecifyeffectiveapertureofMarsandearthantennasandtransmitterpower(totaloverentirebandwidth)ateachend.Takeearth-Marsdistanceas6light-minutes.(b)Repeat(a)foranearth-Jupiterlink.Taketheearth-Jupiterdistanceas40light-minutes.2-11-4.continuedSolution:89(a)λ==cf/310/2.510××=0.12m−−19613P(earth)10=×510×=510×Wr−−17611P(Mars)10=×510×=510×Wr22TakeAe(Mars)=(1/2)π1.5=5.3m(εap=5.0)TakePt(Mars)=1kW22TakeAe(earth)=(1/2)π15=350m(εap=5.0)22rλPt(earth)=Pr(Mars)Aet(earth)Aet(Mars)822−11(360×3×10).012Pt(earth)=5×10=9.6MW5.3×350Toreducetherequiredearthstationpower,taketheearthstationantenna课后答案网22Ae=)2/1(π50=3927m(ans.)so62P(earth)=×6.910(15/50)=620kW(ans.)twww.hackshp.cnAet(Mars)Aer(earth)35.3×3930−14Pr(earth)=Pt(Mars)22=10822=8×10Wrλ(360×3×10).012−13−13whichisabout16%oftherequired5x10W.Therequired5x10WcouldbeobtainedbyincreasingtheMarstransmitterpowerbyafactorof6.3.Otheralternativeswouldbe(1)toreducethebandwidth(anddatarate)reducingtherequiredvalueofPror(2)toemployamoresensitivereceiver.AsdiscussedinSec.12-1,thenoisepowerofareceivingsystemisafunctionofitssystemtemperatureTandbandwidthBasgivenbyP=kTB,wherek=Boltzmann’s−23−1constant=1.38x10JK. 76ForB=5x10Hz(asgiveninthisproblem)andT=50K(anattainablevalue),−236−15P(noise)=1.38×10×50×5×10=5.3×10W2-11-4.continued−14Thereceivedpower(8x10W)isabout20timesthisnoisepower,whichisprobablysufficientforsatisfactorycommunication.Accordingly,witha50Kreceivingsystemtemperatureattheearthstation,aMarstransmitterpowerof1kWisadequate.(b)ThegivenJupiterdistanceis40/6=6.7timesthattoMars,whichmakesthe2requiredtransmitterpowers6.7=45timesasmuchortherequiredreceiverpowers1/45asmuch.Neitherappearsfeasible.Butapracticalsolutionwouldbetoreducethebandwidthfor6theJupiterlinkbyafactorofabout50,makingB=(5/50)x10=100kHz.*2-11-5.Moonlink.Aradiolinkfromthemoontotheearthhasamoon-based5λlongright-handedmono-filaraxial-modehelicalantenna(seeEq.(8-3-7))anda2Wtransmitteroperatingat1.5GHz.Whatshouldthepolarizationstateandeffectiveaperturebefortheearth-based-14antennainordertodeliver10Wtothereceiver?Taketheearth-moondistanceas1.27light-seconds.Solution:课后答案网89λ==cf/310/1.510××=0.2m,From(8-3-7)thedirectivityofthemoonhelixisgivenby2DλD=12www.hackshp.cn×5=60andAet(moon)=4πFromFriisformula2222−1482PrrλPr4(π)rλ10(3×10×.127)4π2Aer==2==152mRCPorPtAetPtDλ2×60about14mdiameter(ans.)2-16-1.Spaceshipnearmoon.Aspaceshipatlunardistancefromtheearthtransmits2GHzwaves.Ifapowerof10Wisradiatedisotropically,find(a)theaveragePoyntingvectorattheearth,(b)thermselectricfieldEattheearthand(c)thetimeittakesfortheradiowavestotravelfromthe 8spaceshiptotheearth.(Taketheearth-moondistanceas380Mm.)(d)Howmanyphotonsperunitareapersecondfallontheearthfromthespaceshiptransmitter?2-16-1.continuedSolution:Pt10−18−2−2(a)PV(atearth)===5.5×10Wm=5.5aWm(ans.)2624πr4π(380×10)22/1(b)PV=S=E/ZorE=(SZ)−182/1−9−1orE=5.5(×10×377)=45×10=45nVm(ans.)68(c)t=r/c=380×103/×10=.127s(ans.)−349−24−34(d)Photon=hf=.663×10×2×10=3.1×10J,whereh=.663×10Js−18−1−2Thisistheenergyofa2.5MHzphoton.From(a),PV=5.5×10Jsm−185.5×106−2−1Therefore,numberofphotons==2.4×10ms(ans.)−243.1×102-16-2.MorepowerwithCP.ShowthattheaveragePoyntingvectorofacircularlypolarizedwaveistwicethatofalinearlypolarizedwaveifthemaximumelectricfieldEisthesameforbothwaves.Thismeansthatamediumcanhandletwiceasmu课后答案网chpowerbeforebreakdownwithcircularpolarization(CP)thanwithlinearpolarization(LP).Solution:22E1+E2From(2-16-3)wehaveforrmsfieldsthatwww.hackshp.cnPV=Sav=Zo2E1ForLP,EE(or)==0,soS21avZo22E1ForCP,EE==,soS12avZoThereforeSCP=2SLP(ans.)2-16-3.PVconstantforCP.ShowthattheinstantaneousPoyntingvector(PV)ofaplanecircularlypolarizedtravelingwaveisaconstant. 9Solution:ECP=Excosωt+EysinωtwhereEx=Ey=Eo2-16-3.continued22221/2221/2EEt=+=+=(cosωωEtEsin)(cosωtsinωtE)(aconstant)CPoooo2EoThereforePVorS(instantaneous)=(aconstant)(ans.)Z*2-16-4.EPwavepowerAnellipticallypolarizedwaveinamediumwithconstantsσ=0,µr=2,εr=5hasH-fieldcomponents(normaltothedirectionofpropagationandnormaltoeachother)of-12amplitudes3and4Am.Findtheaveragepowerconveyedthroughanareaof5mnormaltothedirectionofpropagation.Solution:12212/12212/122−2Sav=Z(H1+H2)=377(µr/εr)(H1+H2)=377)5/2(3(+4)=2980Wm222P=ASav=5×2980=14902W=14.9kW(ans.)2-17-1.CrosseddipolesforCPandotherstates.课后答案网Twoλ/2dipolesarecrossedat90°.Ifthetwodipolesarefedwithequalcurrents,whatisthepolarizationoftheradiationperpendiculartotheplaneofthedipolesifthecurrentsare(a)inphase,(b)phasequadrature(90°differenceinphase)and(c)phaseoctature(45°differenceinphase)?www.hackshp.cnSolution:(a)LP(ans.)(b)CP(ans.)(c)From(2-17-3)sin2ε=sin2γsinδ−1Dwhereγ==tan(EE/)4521Dδ=45ε=221D2AR===cotεε1/tan2.41(EP)...(ans.) 10*2-17-2.PolarizationoftwoLPwaves.Awavetravelingnormallyoutofthepage(towardthereader)hastwolinearlypolarizedcomponentsE=2cosωtx()DE=3cosωt+90y(a)Whatistheaxialratiooftheresultantwave?(b)Whatisthetiltangleτofthemajoraxisofthepolarizationellipse?(c)DoesErotateclockwiseorcounterclockwise?Solution:(a)From(2-15-8),AR=2/3=5.1(ans.)o(b)τ=90(ans.)(c)At0,tE==E;attT==/4,EE−,thereforerotationisCW(ans.)xy2-17-3.SuperpositionoftwoEPwaves.Awavetravelingnormallyoutwardfromthepage(towardthereader)istheresultantoftwoellipticallypolarizedwaves,onewithcomponentsofEgivenbyE′=2cosωtandE′=6cos()ωt+πy课后答案网x2andtheotherwithcomponentsgivenbyE′′=1cosωtandE′′=3cos()ωt−πyx2(a)Whatistheaxialratiooftheresultantwave?(b)DoesErotateclockwiseorcounterclockwise?www.hackshp.cnSolution:EEE=+=′′′2cosωωωt+cost=3costyyyEEE=+=′′′6cos(ωπt+/2)3cos(+ωπt−/2)=−6sinωt+3sinωt=−3sinωtxxx(a)ExandEyareinphasequadratureandAR==3/31(CP)(ans.)(b)Att==0,3Eyˆ,attT==/4,3Ex−ˆ,thereforerotationisCCW(ans.) 11*2-17-4.TwoLPcomponents.Anellipticallypolarizedplanewavetravelingnormallyoutofthepage(towardthe-1reader)haslinearlypolarizedcomponentsExandEy.GiventhatEx=Ey=1VmandthatEyleadsExby72°,(a)Calculateandsketchthepolarizationellipse.(b)Whatistheaxialratio?(c)Whatistheangleτbetweenthemajoraxisandthex-axis?Solution:−1oo(b)γ=tan(E2/E1)=45,δ=72oFrom(2-17-3),ε=36,thereforeAR=/1tanε=.138(ans.)o(c)From(2-17-3),sin2τ=tan2ε/tanδorτ=45(ans.)2-17-5.TwoLPcomponentsandPoincarésphere.AnswerthesamequestionsasinProb.2-17-4forthecasewhereEyleadsExby72°as-1-1beforebutEx=2VmandEy=1Vm.Solution:−1o(b)γ==tan263.4oδ=72oε=248.andAR=.217(ans.)o(c)τ=112.(ans.)课后答案网*2-17-6.TwoCPwaves.Twocircularlypolarizedwavesintersectattheorigin.One(y-wave)istravelinginthepositiveydirectionwithErotatingclockwiseasobservedwww.hackshp.cnfromapointonthepositivey-axis.Theother(x-wave)istravelinginthepositivexdirectionwithErotatingclockwiseasobservedfromapointonthepositivex-axis.Attheorigin,Eforthey-waveisinthepositivezdirectionatthesameinstantthatEforthex-waveisinthenegativezdirection.WhatisthelocusoftheresultantEvectorattheorigin?Solution:Resolve2wavesintocomponentsormakesketchasshown.Itisassumedthatthewaveshaveequalmagnitude. 12*2-17-6.continuedoLocusofEisastraightlineinxyplaneatanangleof45withrespecttox(ory)axis.*2-17-7.CPwaves.AwavetravelingnormallyoutofthepageistheresultantoftwocircularlypolarizedD-1jωtj()ωt+90componentsE=5eandE=2e(Vm).Find(a)theaxialratioAR,(b)rightleftthetiltangleτand(c)thehandofrotation(leftorright).Solution:课后答案网www.hackshp.cn2+5(a)AR==−3/7=−.233(ans.)[NoteminussignforRH(right-handed2−5polarization)]o(b)Fromdiagram,τ=−45(ans.)(c)SinceErotatescounterclockwiseasafunctionoftime,RH.(ans.) 132-17-8.EPwave.Awavetravelingnormallyoutofthepage(towardthereader)istheresultantoftwoDlinearlypolarizedcomponentsE=3cosωtandE=2cos()ωt+90.Fortheresultantxywavefind(a)theaxialratioAR,(b)thetiltangleτand(c)thehandofrotation(leftorright).Solution:(a)AR=3/2=1.5(ans.)o(b)τ=0(ans.)(c)CW,LEP(ans.)*2-17-9.CPwaves.Twocircularlypolarizedwavestravelingnormallyoutofthepagehavefieldsgivenby−jωtjωt-1E=2eandE=3e(Vm)(rms).Fortheresultantwavefind(a)AR,(b)theleftrighthandofrotationand(c)thePoyntingvector.Solution:2+3(a)AR==−5(ans.)23-(b)REP(ans.)22EL+ER4+9课后答案网−2−2(c)PV===.0034Wm=34mWm(ans.)Z3772-17-10.EPwaves.www.hackshp.cnAwavetravelingnormallyoutofthepageistheresultantoftwoellipticallypolarized(EP)waves,onewithcomponentsE=5cosωtandE=3sinωtandanotherwithxyjωt−jωtcomponentsE=3eandE=4e.Fortheresultantwave,find(a)AR,(b)τandrl(c)thehandofrotation.Solution:(a)Ex=5cosωt+3cosωt+4cosωt=12cosωtEy=3sinωt+3sinωt−4sinωt=2sinωt 142-17-10.continuedAR=122/=6(ans.)o(b)SinceExandEyareintime-phasequadraturewithEx(max)>Ey(max),τ=0.−1oOrfrom(2-17-3),sin2τ=tan2ε/tanδ,ε=tan/1(AR)=.946obutδ=90sotanδ=∞Eatt=T/4oThereforeτ=0(ans.)(c)Att=,0Ex=12,Ey=0CCWEatt=0oAtt=T4/(ωt=90),Ex=,0Ey=2ThereforerotationisCCW,sopolarizationisrightelliptical,REP(ans.)*2-17-11.CPwaves.Awavetravelingnormallyoutofthepageistheresultantoftwocircularlypolarized()Djωt−jωt+45componentsE=2eandE=4e.Fortheresultantwave,find(a)AR,(b)τrland(c)thehandofrotation.Solution:E1+Er4+26(a)AR====3(ans.)E1−Er4−22oo(b)WhenωtE==0,2∠__0andE=4∠__−45r1课后答案网WhenωtE=−2211oo,=2∠∠__−22andE=4__−221o22r12sothatE+E=6E=∠__−221oorτ=−221o(ans.)1rmax22Notethattherotationdirectionsareoppositeforwww.hackshp.cnErandE1sothatfor−ωt,Et=−2b∠∠__ωωutEt=+__r1Also,τcanbedeterminedanalyticallybycombiningthewavesintoanExandEycomponentwithvaluesofooEE=−5.60∠∠__30.4and=2.9516.3__xyofromwhichδ=−467. 15*2-17-11.continuedSincefrom(a)AR=3,εcanbedeterminedandfrom(2-17-3),thetiltangleoτ=−22.5(ans.)(c)E1>ErsorotationisCW(LEP)(ans.)2-17-12.Circular-depolarizationratio.IftheaxialratioofawaveisAR,showthatthecircular-depolarizationratioofthewaveisgivenby.AR1−R=AR1+Thus,forpurecircularpolarizationAR=1andR=0(nodepolarization)butforlinearpolarizationAR=∞andR=1.Solution:Anywavemayberesolvedinto2circularly-polarizedcomponentsofoppositehand,ErandE1foranaxialratioEmaxEr+E1AR==EminEr−E1课后答案网E1AR−1fromwhichthecirculardepolarizationratioR==ErAR+1Thusforpurecircularpolarization,AR=1andthereiszerodepolarization(R=0),whileforpurelinearpolarizationAR=www.hackshp.cn∞andthedepolarizationratioisunity(R=1).WhenAR=3,R=½. 16课后答案网www.hackshp.cn 17Chapter3.TheAntennaFamily3-4-1.Alpine-hornantenna.ReferringtoFig.3-4a,thelowfrequencylimitoccurswhentheopen-endspacing>λ/2andthehighfrequencylimitwhenthetransmissionlinespacingd≈λ/4.Ifd=2mmandtheopen-endspacing=1000d,whatisthebandwidth?Solution:D=openedendspacing,d=transmissionlinespacingλmax2DBandwidth===1000(ans.)λmind2*3-4-2.Alpine-hornantenna.Ifd=transmissionlinespacing,whatopen-endspacingisrequiredfora200-to-1bandwidth?Solution:Ifd=transmissionlinespacing=λmin/2andD=open-endspacing=λmax2/,λ课后答案网maxD2for200-to-1bandwidth,wemusthave==200,orDd=200(ans.)dλmin2www.hackshp.cn*3-5-2.Rectangularhornantenna.Whatistherequiredapertureareaforanoptimumrectangularhornantennaoperatingat2GHzwith16dBigain?Solution:FromFig.3-5forf=2GHz(λ=0.15m),27.5wh63.1λ2Dw===∴18dBi63.1,h==0.19m(ans.)2λ7.5 18*3-5-3.Conicalhornantenna.Whatistherequireddiameterofaconicalhornantennaoperatingat3GHzwith14dBigain?Solution:FromFig.3-5forf=3GHz(λ=0.1m),226.5πλr15.82Dr===∴12dBi15.8,==0.09m,d=2r=0.18m(ans.)2λπ6.53-7-2.BeamwidthanddirectivityFormostantennas,thehalf-powerbeamwidth(HPBW)maybeestimatedasHPBW=κλ/D,whereλistheoperatingwavelength,Distheantennadimensionintheplaneofinterest,andκisafactorwhichvariesfrom0.9to1.4,dependingonthefiledamplitudetaperacrosstheantenna.Usingthisapproximation,findthedirectivityandgainforthefollowingantennas:(a)circularparabolicdishwith2mradiusoperatingat6GHz,(b)ellipticalparabolicdishwithdimensionsof1m×10moperatedat1GHz.Assumeκ=1and50percentefficiencyineachcase.Solution:FromFig.3-9forf==1600MHz(λ0.1875m),GD==17dBi50=(for100%efficiency)课后答案网15L50(a)DL==50,so==λλ3.33λ15LIfspacing=λ/π,numberofwww.hackshp.cnturns=n==105.≈10(ans.)λ/π(b)Turndiameter=λπ/=≈0.05966cm(ans.)2n+121(c)AxialratioAR===.105(ans.)2n20 19Chapter4.PointSources*4-3-1.Solarpower-1-2Theearthreceivesfromthesun2.2gcalmincm.(a)WhatisthecorrespondingPoyntingvectorinwattspersquaremeter?(b)Whatisthepoweroutputofthesun,assumingthatitisanisotropicsource?(c)Whatisthermsfieldintensityattheearthduetothesun’sradiation,assumingallthesun’senergyisatasinglefrequency?-1Note:1watt=14.3gcalmin,distanceearthtosun=149Gm.Solution:−−122.2gcalmincm−−22(a)S==0.1539Wcm=1539Wm(ans.)−114.3gcalmin222226(b)P(sun)=×Sr4ππ=15394××1.49×10W=4.2910W×(ans.)21212−1(c)SEZES==/,(Z)=(1539377)×=762Vm(ans.)oo4-5-1.Approximatedirectivities.(a)ShowthatthedirectivityforasourcewithatunidirectionalpowerpatterngivenbynU=UmcosθcanbeexpressedasD=2(n+1).Uhasavalueonlyfor0°≤θ≤90°.Thepatternsareindependentoftheazimuthangleφ.(b)Comparetheexactvaluescalculatefrom(a)withtheapproximatevaluesforthedirectivitiesoftheantennasfoundinProb.2-7-2andfindthedBdifferencefromtheexactvalues.Solution:课后答案网n42π(a)IfUU==cosθ,D==2(n+1)(ans.)mπ2π2nn+12sπθθ∫incosdθcosθ0−n+10(b)www.hackshp.cnForn=1,Forn=2,Forn=3,D≈2.78⇒4.4dBiD≈4.94⇒6.9dBiD≈7.3⇒8.6dBiapprox.approx.approx.D=46⇒.0dBiD=6⇒7.8dBiD=8⇒9.0dBiexactexactexactDD−=1.6dBDD−=0.9dBDD−=0.4dBexactapprox.exactapprox.exactapprox. 20*4-5-2.Exactversusapproximatedirectivities.(a)Calculatetheexactdirectivitiesofthethreeunidirectionalantennashavingpowerpatternsasfollows:2P(θ,φ)=Pmsinθsinφ3P(θ,φ)=Pmsinθsinφ23P(θ,φ)=PmsinθsinφP(θ,φ)hasavalueonlyfor0≤θ≤πand0≤φ≤πandiszeroelsewhere.(b)Comparetheexactvaluesin(a)withtheapproximatevaluesfoundinProb.2-7-3.Solution:44ππ(a)Dd==,Ω=sinθθddφΩA∫∫Pdn(,)θφΩ4π244ππForP(θ,φ)=Pmsinθsinφ,D==2ππππPsinsinθφsin22θsinφddθφm∫∫sinθθφdd∫∫0000Pmπ2⎛⎞θππ14π16∫sinθθdD=−⎜⎟sin2θ=,∴===5.09(ans.)0⎝⎠2402⎛⎞⎛⎞πππ⎜⎟⎜⎟⎝⎠⎝⎠22Usingthesameapproach,wefind,344ππforP(θ,φ)=Pmsinθsinφ,D===6.0(ans.)ππ课后答案网sin23θφsinddθφ⎛⎞⎛⎞π4∫∫00⎜⎟⎜⎟⎝⎠⎝⎠232344ππforP(θ,φ)=Pmsinθsinφ,D===7.1(ans.)ππsin33θφsinddθφ⎛⎞⎛⎞44∫∫00⎜⎟⎜⎟www.hackshp.cn⎝⎠⎝⎠33(b)Tabulating,wehave5.1vs.3.8,6.0vs.4.6,and7.1vs.6.1(ans.)4-5-3.Directivityandminorlobes.Provethefollowingtheorem:iftheminorlobesofaradiationpatternremainconstantasthebeamwidthofthemainlobeapproacheszero,thenthedirectivityoftheantennaapproachesaconstantvalueasthebeamwidthofthemainlobesapproacheszero. 214-5-3.continuedSolution:44ππD==ΩΩ+ΩAMmwhereΩ=totalbeamareaAΩ=mainlobebeamareaMΩ=minorlobebeamareamasΩ→Ω→Ω0,,soD=Ω4π(aconstant)(ans.)MAmm4-5-4.Directivitybyintegration.(a)CalculatebygraphicalintegrationornumericalmethodsthedirectivityofasourcewithaunidirectionalpowerpatterngivenbyU=cosθ.ComparethisdirectivityvaluewiththeexactvaluefromProb.4-5-1.Uhasavalueonlyfor0°≤θ≤90°and0°≤φ≤360°andiszeroelsewhere.2(b)RepeatforaunidirectionalpowerpatterngivenbyU=cosθ.3(c)RepeatforaunidirectionalpowerpatterngivenbyU=cosθ.Solution:Exactvaluesfor(a),(b),and(c)are:4,6,and8.(ans.)4-5-5.Directivity.课后答案网CalculatethedirectivityforasourcewithrelativefieldpatternE=cos2θcosθ.Solution:πAssumingaunidirectionalpattern,(0≤≤θ),D=24(ans.)www.hackshp.cn2 22课后答案网www.hackshp.cn 23Chapter5.ArraysofPointSources,PartI5-2-4.Two-sourceend-firearray.(a)Calculatethedirectivityofanend-firearrayoftwoidenticalisotropicpointsourcesinphaseopposition,spacedλ/2apartalongthepolaraxis,therelativefieldpatternbeinggivenby⎛π⎞E=sin⎜cosθ⎟⎝2⎠whereθisthepolarangle.(b)Showthatthedirectivityforanordinaryend-firearrayoftwoidenticalisotropicpointsourcesspacedadistancedisgivenby2D=.1+()λ4πdsin()4πdλSolution:(a)D=2(ans.)5-2-8.Foursourcesinsquarearray.(a)DeriveanexpressionforE(φ)foranarrayof4identicalisotropicpointsourcesarrangedasinFig.P5-2-8.Thespacingdbetweeneachsourceandthecenterpointofthearrayis3λ/8.Sources1and2arein-phase,andsources3and4inoppositephasewithrespectto1and2.(b)Plot,approximately,thenormalizedpattern.课后答案网www.hackshp.cnFigureP5-2-8.Foursourcesinsquarearray.Solution:(a)Edd()φβ=−cos(cos)cos(φβsin)φ(ans.)n5-5-1.Fieldandphasepatterns.Calculateandplotthefieldandphasepatternsofanarrayof2nonisotropicdissimilarsourcesforwhichthetotalfieldisgivenby 24E=cosφ+sinφ∠ψπwhereψ=dcosφ+δ=()cosφ+12Takesource1asthereferenceforphase.SeeFig.P5-5-1.FigureP5-5-1.Fieldandphasepatterns.Solution:SeeFigures5-16and5-17.5-6-5.Twelve-sourceend-firearray.(a)Calculateandplotthefieldpatternofalinearend-firearrayof12isotropicpointsourcesofequalamplitudespacedλ/4apartfortheordinaryend-firecondition.(b)Calculatethedirectivitybygraphicalornumericalintegrationoftheentirepattern.Notethatitisthepowerpattern(squareoffieldpattern)whichistobeintegrated.Itismostconvenienttomakethearrayaxiscoincidewiththepolarorz-axisofFig.2-5sothatthepatternisafunctionof课后答案网θ.(c)Calculatethedirectivitybytheapproximatehalf-powerbeamwidthmethodandcomparewiththatobtainedin(b).Solution:www.hackshp.cn(b)Da=17(ns.)(c)D=10(ans.) 255-6-7.Twelve-sourceend-firewithincreaseddirectivity.(a)Calculateandplotthepatternofalinearend-firearrayof12isotropicpointsourcesofequalamplitudespacedλ/4apartandphasedtofulfilltheHansenandWoodyardincreased-directivitycondition.(b)CalculatethedirectivitybygraphicalornumericalintegrationoftheentirepatternandcomparewiththedirectivityobtainedinProb.5-6-5and5-6-6.(c)Calculatethedirectivitybytheapproximatehalf-powerbeamwidthmethodandcomparewiththatobtainedin(b).Solution:(b)Da=26(ns.)(c)Da=35(ns.)5-6-9.Directivityofordinaryend-firearray.Showthatthedirectivityofanordinaryend-firearraymaybeexpressedasnD=n−11+()λ2πnd∑[](n−k)ksin()4πkdλk=1Notethat2n−1⎡sin()nψ2⎤ψ⎢⎥=n+∑2()n−kcos2k⎣()ψ2⎦k=12课后答案网Solution:Changeofvariable.Itisassumedthatthearrayhasauniformspacingdbetweentheisotropicsources.Thebeamareawww.hackshp.cn212ππ⎡⎤sin()nψ2Ω=⎢⎥sinθθddφ(1)An2∫∫00sin()ψ2⎣⎦whereθ=anglefromarrayaxis.Thepatternisnotafunctionofφso(1)reducesto22ππ⎡⎤sin()nψ2Ω=⎢⎥sinθθd(2)An2∫0sin()ψ2⎣⎦where/2ψ=−πθd(cos1)(2.1)λ 265-6-9.continuedψDifferentiatingddd=πθsinθ(3)λ21ψorsinθθd=(4)πd2λandintroducing(4)in(2)222πdλ⎡⎤sin()nψ2ψΩ=⎢⎥d(5)And2∫0sin()ψ22λ⎣⎦Notenewlimitswithchangeofvariablefromθto/2.ψWhenθ==0,ψ/20andwhenθπ==,ψ/22πd.λ2⎡⎤sin()nψ2n−1Since⎢⎥=+nn∑2(−kk)cos(2ψ/2)(6)⎣⎦()ψ2k=1n−122πdλψ(5)canbeexpressedΩ=A2∫[2nn+∑()−kkdcos(2ψ/2)](7)nd0k=12λ22⎡ψn−1(nk−)⎤2πdλIntegrating(7)Ω=A2⎢nk+∑sin(2ψ/2)⎥(8)ndλ⎣22k=1k⎦0n−12⎡nk−⎤orΩ=A2⎢2sππndλλ+∑in(4kd)⎥(9)课后答案网ndλ⎣k=1k⎦24π2πndλandD==n−1(10)Ωnk−A2sππnd+∑in(4kd)λλk=1kwww.hackshp.cnnThereforeD=q.e.d.(11)n−1λnk−1s++∑in(4πλkd/)2πndk=1kWenotethatwhendD==λ/4,oramultiplethereof,thesummationtermiszeroandnexactly.Thisproblemandthenextoneareexcellentexamplesofintegrationwithchangeofvariableandchangeoflimits.ThefinalformforDin(11)aboveiswelladaptedforacomputerprogram. 275-6-10.Directivityofbroadsidearray.ShowthatthedirectivityofabroadsidearraymaybeexpressedasnD=n−11+()λπnd∑[]()n−kksin()2πkdλk=1Solution:ψThesolutionissimilartothatforProb.5-6-9with=πθdcosλ2whereθ==0,ψ/2πθddandwhen=,/2=πψ−πsothat(8)ofProb.5-6-9becomesλλ2(⎡⎤ψn−1nk−)−πdλΩ=A2⎢⎥nk+∑sin(2ψ/2)ndλ⎣⎦2k=1k+πdλn−1−−2⎡nk⎤Ω=A2⎢22sππndλλ+∑in(2)kd⎥ndλ⎣k=1k⎦n−14⎡nk−⎤Ω=A2⎢ππndλλ+∑sin(2kd)⎥ndλ⎣k=1k⎦24ππndnλD===q.e.d.n−1n−1Ωnk−⎛⎞λnk−Aππnd+∑sin(2kd)1s+⎜⎟∑in(2πλkd/)λλk=1k课后答案网⎝⎠πndk=1kNotethatwhend=λ/2,oramultiplethereof,thesummationtermiszeroandDn=exactly.Seeapplicationoftheaboverelationstotheevaluationofwww.hackshp.cnDandofthemainbeamareaΩofanarrayof16pointsourcesinProb.16-6-7(c)and(d).A 28课后答案网www.hackshp.cn 29Chapter5.ArraysofPointSources,PartII5-8-1.Threeunequalsources.Threeisotropicin-linesourceshaveλ/4spacing.Themiddlesourcehas3timesthecurrentoftheendsources.Ifthephaseofthemiddlesourceis0°,thephaseofoneendsource+90°andphaseoftheotherendsource-90°,makeagraphofthenormalizedfieldpattern.Solution:PhasoradditionEn0.6North0.6South0.2East1.0West0.24North-East课后答案网0.96North-West5-8-7.Strayfactoranddirectivegain.Theratioofthemainbeamsolidanglewww.hackshp.cnΩMto(total)beamsolidangleΩAiscalledthemainbeamefficiency.Theratiooftheminor-lobesolidangleΩmtothe(total)beamsolidangleΩAiscalledthestrayfactor.ItfollowsthatΩM/ΩA+Ωm/ΩA=1.Showthattheaveragedirectivitygainovertheminorlobesofahighlydirectiveantennaisnearlyequaltothestrayfactor.Thedirectivegainisequaltothedirectivitymultipliedbythenormalizedpowerpattern[=DPn(θ,φ)],makingitafunctionofanglewiththemaximumvalueequaltoD.Solution:ΩmStrayfactor=ΩA 305-8-7.continuedwhereΩ=totalbeamareaAΩ=mainlobebeamareaMΩ=minorlobebeamareamP(,)θφdΩ∫∫nΩm4π−ΩM=ΩA∫∫Pdn(,)θφΩ4π1Averagedirectivegainoverminorlobes=DG(minor)=DP(,)θφdΩav∫∫n4π−ΩM4π−ΩMwhereD=Ω4/πA4(πθφPd,)Ω∫∫n144π−ΩMπΩmTherefore(minor)DG==av44ππ−ΩΩ−ΩΩMAMAIfΩ<<4π(antennahighlydirective),MΩmDG(minor)≅(strayfactor)q.e.d.avΩA*5-9-2.Three-sourcearray.课后答案网Thecentersourceofa3-sourcearrayhasa(current)amplitudeofunity.Forasidelobelevel0.1ofthemainlobemaximumfield,findtheDolph-Tchebyscheffvalueoftheamplitudeoftheendsources.Thesourcespacingd=λ/2.Solution:Lettheamplitudes(currents)ofthe3sourcesbeasinthesketchwww.hackshp.cnA2AA1O1dR==λ/2,10λ/2λ/2Letamplitudeofcentersource==12Ao2nT−=12,()x=2x−=1R2222xx−=1102=11oo2xx==5.5±2.345oo 315-9-2.continuedψψ22EAA=+22cos2=+2AA2(2cos−=+1)2AAw2(2−1)3o1o1o122Letwxx=/soo2xA122EAA=+22(21−=+)24AxAx−=−2213o122o1xxoo22EA=+0.728x(A−2)A=2x−131o1Therefore,0.728AA==2and2.75112AA−=2−1and2A=−5.514.5=o1oThus,normalizing2AA==1and2.754.5=0.61(ans.)o1Amplitudedistributionis0.611.000.61Patternhas4minorlobes.Forcentersource,amplitude=1ThesidesourceamplitudesfordifferentRvaluesare:R8101215A10.640.610.590.575-9-4.EightsourceD-Tdistribution.(a)FindtheDolph-Tchebyscheffcurrentdistributionfortheminimumbeamwidthofalinearin-phasebroadsidearrayofeightis课后答案网otropicsources.Thespacingbetweentheelementsisλ/4andthesidelobelevelistobe40dBdown.Takeφ=0inthebroadsidedirection.(b)Locatethenullsandthemaximaoftheminorlobes.(c)Plot,approximately,thenormalizedfieldpattern(0°≤φ≤360°).(d)Whatisthehalf-powerbeamwidth?www.hackshp.cnSolution:(a)0.14,0.42,0.75,1.00,1.00,0.75,0.42,0.14(b)Max.at:oooooooooooo±21,±27,±36,±48,±61,±84,±96,±119,±132,±144,±153,±159Nullsat:oooooooooooo±18,±23,±32,±42,±54,±71,±109,±126,±138,±148,±157,±162o(d)HPBW=12(ans.) 32*5-18-1.Twosourcesinphase.Twoisotropicpointsourcesofequalamplitudeandsamephasearespaced2λapart.(a)Plotagraphofthefieldpattern.(b)Tabulatetheanglesformaximaandnulls.Solution:2(a)PowerpatternP=EnnIn课后答案网www.hackshp.cnInstructionalcommenttopassontostudents:oThelobeswithnarrowestbeamwidthsarebroadside(±90),whilethewidestbeamwidthoolobesareend-fire(0and180).Thefourlobesbetweenbroadsideandend-fireareintermediateinbeamwidth.Inthreedimensionsthepatternisafigure-of-revolutionooaroundthearrayaxis(0and180axis)sothatthebroadsidebeamisaflatdisk,theend-firelobesarethickcigars,whiletheintermediatelobesarecones.Theaccompanyingfigureissimplyacrosssectionofthethree-dimensionalspacefigure. 335-18-2.Twosourcesinoppositephase.Twoisotropicsourcesofequalamplitudeandoppositephasehave1.5λspacing.Findtheanglesforallmaximaandnulls.Solution:oooooooMaximumat:0,180,±70.5,±109.5,Nullsat:±48.2,±90,±131.8课后答案网www.hackshp.cn 34课后答案网www.hackshp.cn 35Chapter6.TheElectricDipoleandThinLinearAntennas*6-2-1.Electricdipole.(a)TwoequalstaticelectricchargesofoppositesignseparatedbyadistanceLconstituteastaticelectricdipole.ShowthattheelectricpotentialatadistancerfromsuchadipoleisgivenbyQLcosθV=24πεrwhereQisthemagnitudeofeachchargeandθistheanglebetweentheradiusrandthelinejoiningthecharges(axisofdipole).ItisassumedthatrisverylargecomparedtoL.(b)FindthevectorvalueoftheelectricfieldEatalargedistancefromastaticelectricdipolebytakingthegradientofthepotentialexpressioninpart(a).Solution:QQrLθ1(a)(at)Vr=−,cosθ44πεrr12πε2+QrrL=−(/2)cos,θθrrL=+(/2)cosθr12L/2Q⎛⎞11V=−⎜⎟4(πε⎝⎠rL−+/2)cos(θrL/2)cosθLr2QrL⎛⎞+−(/2)cosθθrL+(/2)cos=⎜⎟L/24(πε⎝⎠rL22+/2)cos2θLQLcosθ−QcosθForrLV>>,=q.e.d.2课后答案网24πεr(b)Er=−∇=V??∂∂VV+θφ?11+∂V=−QL⎛⎞⎜⎟−r2cosθθ−θφ1sin+032∂∂rrθθrsin∂φ4πε⎝⎠rrr=+rˆQLcosθθθˆQLsinwww.hackshp.cn3324πεrrπεQLcosθθQLsinorEEE===,,0(ans.)r33θφ24πεrrπε*6-2-2.Shortdipolefields.Adipoleantennaoflength5cmisoperatedatafrequencyof100MHzwithterminalcurrentIo=120mA.Attimet=1s,angleθ=45°,anddistancer=3m,find(a)Er,(b)Eθ,and(c)Hφ. 36*6-2-2.continuedSolution:(a)From(6-2-12)jtr()ωβ−Ilecosθ⎛⎞11oE=+⎜⎟r232πεo⎝⎠crjrω⎡⎤⎛⎞(2)10010π×66j⎢⎥()2π10010(1)×−⎜⎟(3)⎢⎥⎣⎦⎜⎟⎝⎠310×8−3oe⎛⎞11=×(12010)(0.05)cos45⎜⎟+−1282632(8.8510ππ×××)⎝⎠310(3)j(2)10010(3)−−−232o=×−×=×−2.8310j(4.510)2.8610∠__9V/m(ans.)(b)From(6-2-13)jtr()ωβ−Ileosinθ⎛⎞jω11−−22Ej=+⎜⎟+=1.4110×+(8.6510)×θ2234πεo⎝⎠crcrjrω−2o=×8.7710∠__81V/m(ans.)(c)From(6-2-15)jtr()ωβ−Ileosinθ⎛⎞jω1−−54Hj=+⎜⎟=3.7510×+(2.3610)×φ24π⎝⎠crr−4o=×2.3910∠__81A/m(ans.)*6-2-4.Shortdipolequasi-stationaryfields.课后答案网ForthedipoleantennaofProb.6-2-2,atadistancer=1m,usethegeneralexpressionsofTable6-1tofind(a)Er,(b)Eθ,and(c)Hφ.Comparetheseresultstothoseobtainedusingthequasi-stationaryexpressionsofTable6-1.www.hackshp.cnSolution:UsingthesameapproachforEE,,andHasinsolutiontoProb.6-2-2,wefindforrθφr=1m,E=282mV/mrE=242mV/mθH=784mA/mφ*6-2-4.continuedUsingquasi-stationaryequations, 37−3oqloocosθθIlcos12010(0.05cos45)×−3Ej====0(−1211×0)r3326−1232(πεrjrjωπε2)(2π)(100××10)(8.8510)1oo=121mVm(ans.)qloosinθθIlsin−3E===0−j(6110)×=61mV/m(ans.)θ334(πεrjrωπε4)ooIlosinθ−4H==3.381×0=338µA/m(ans.)φ24πr*6-3-1.Isotropicantenna.Radiationresistance.-1Anomnidirectional(isotropic)antennahasafieldpatterngivenbyE=10I/r(Vm),whereI=terminalcurrent(A)andr=distance(m).Findtheradiationresistance.Solution:2210IE100IES==so=2rZrZ22LetP==poweroversphere4πrS,whichmustequalpowerIRtotheantenna22terminals.ThereforeIR=4aπrSnd212100I400Rr==43π=.33Ω(ans.)22课后答案网Ir120π120*6-3-2.Shortdipolepower.www.hackshp.cn(a)Findthepowerradiatedbya10cmdipoleantennaoperatedat50MHzwithanaveragecurrentof5mA.(b)Howmuch(average)currentwouldbeneededtoradiatepowerof1W?Solution:(a)26⎛⎞(2)5010π×−3⎜⎟(510)0.1×28µβo()Ilav⎝⎠310×−6P==377=2.7410W×=2.74µW(ans.)επ1212πo 38*6-3-2.continued12−3⎛⎞1(b)ForPI==1W,510×⎜⎟=3.0A(ans.)av−6⎝⎠2.710×6-3-4.Shortdipole.Forathincenter-feddipoleλ/15longfind(a)directivityD,(b)gainG,(c)effectiveapertureAe,(d)beamsolidΩAand(e)radiationresistanceRr.Theantennacurrenttaperslinearlyfromitsvalueattheterminalstozeroatitsends.Thelossresistanceis1Ω.Solution:(a)E()sinθθ=n44ππ4π4πD====22πππΩA∫∫sinθdΩ∫∫sin23θθθsinddΩ2π∫sinθθd0004π43π===1.5or1.76dBi(ans.)422π3(d)From(a),Ω=8/38.38srπ=(ans.)A222⎛⎞Iav2⎛⎞⎛⎞11(e)From(6-3-14),RL===790⎜⎟790⎜⎟⎜⎟0.878Ω(ans.)rλ⎝⎠Io⎝⎠⎝⎠2150.878(b)Gk==D×=1.50.70or1.54dBi−(ans.)0.8781+2课后答案网λ32(c)Ak==AwhereA=λeememΩ8πA0.87832ThereforeA=×=0.058λ(ans.)e0.87818+πwww.hackshp.cn*6-3-5.Conicalpattern.Anantennahasaconicalfieldpatternwithuniformfieldforzenithangles(θ)from0to60°andzerofieldfrom60to180°.Findexactly(a)thebeamsolidangleand(b)directivity.Thepatternisindependentoftheazimuthangle(φ).Solution:oooo360606060(a)Ω=ddΩ=2sπθinθπ=−2cosθπ=sr(ans.)A∫∫00∫00 39*6-3-5.continued44ππ(b)D===4(ans.)ΩπA6-3-6.Conicalpattern.Anantennahasaconicalfieldpatternwithuniformfiledforzenithangles(θ)from0to45°andzerofieldfrom45to180°.Findexactly(a)thebeamsolidangle,(b)directivity-1and(c)effectiveaperture.(d)FindtheradiationresistanceiftheE=5Vmatadistanceof50mforaterminalcurrentI=2A(rms).Thepatternisindependentoftheazimuthangle(φ).Solution:o45(a)Ω=2πθsindθ=1.84sr(ans.)A∫044ππ(b)D===6.83(ans.)Ω1.84A22λλ2(c)AA====0.543λ(ans.)eemΩ1.84A2222E152(d)IR=Ωr,R=1.8450×=76.3Ω(ans.)rAr2Z2377*6-3-7.Directionalpatternin课后答案网θandφ.Anantennahasauniformfieldpatternforzenithangles(θ)between45and90°andforazimuth(φ)anglesbetween0and120°.IfE=3Vm-1atadistanceof500mfromtheantennaandtheterminalcurrentis5A,findtheradiationresistanceoftheantenna.E=0exceptwithintheanglesgivenabove.www.hackshp.cnSolution:ooo120902π90(a)Ω=sinθθφdd=−cosθ=1.48sr(ans.)A∫∫045o345o2211322E(c)Rr=Ω=1.48500×=Ω354(ans.)rA22IZ5377 40*6-3-8.Directionalpatterninθandφ.-1AnantennahasauniformfieldE=2Vm(rms)atadistanceof100mforzenithanglesbetween30and60°andazimuthanglesφbetween0and90°withE=0elsewhere.Theantennaterminalcurrentis3A(rms).Find(a)directivity,(b)effectiveapertureand(c)radiationresistance.Solution:ooo9060π60(a)Ω=sinθθφdd=−cosθ=0.575sr(ans.)A∫∫030o230o4πDa==21.9(ns.)0.57522λλ2(b)AA====1.74λ(ans.)eemΩ0.575A2211222E(c)Rr=Ω=0.575100×=Ω6.78(ans.)rA22IZ3377*6-3-9.Directionalpatternwithbacklobe.Thefieldpatternofanantennavarieswithzenithangle(θ)asfollows:En(=Enormalized)=1betweenθ=0°andθ=30°(mainlobe),En=0betweenθ=30°andθ=90°andEn=1/3betweenθ=90°andθ=180°(backlobe).Thepatternisindependentofazimuth-1angle(φ).(a)Findtheexactdirectivity.(b)Ifthefieldequals8Vm(rms)forθ=0°atadistanceof200mwithaterminalcurrentI=4A(rms),findtheradiationresistance.课后答案网Solution:oo302π180(a)Ω=2πθsinddθ+sinθθ=2(0.1340.111)π+=2(0.245)πA∫∫09320o4πwww.hackshp.cnD==8.16(ans.)2(0.245)π221122E8(b)Rr=Ω=2(0.245)200π=Ω653(ans.)rA22IZ4120π6-3-10.Shortdipole.Theradiatedfieldofashort-dipoleantennawithuniformcurrentisgivenbyEl=30β()Irsinθ,wherel=length,I=current,r=distanceandθ=patternangle.Findtheradiationresistance. 416-3-10.continuedSolution:122ThecurrentIgivenintheproblemisapeakvalue,soweputIR=ΩSrdr∫∫2Power4πPowerinputradiated2EwhereSE=andisasgivenZ2222230βlI23π222soRr==2πθsindθ80π(/)lλ=790(/)lλΩ(ans.)rIr22120π∫06-3-11.Relationofradiationresistancetobeamarea.ShowthattheradiationresistanceofanantennaisafunctionofitsbeamareaΩAasgivenby2SrR=Ωr2AIwhereS=PoyntingvectoratdistancerindirectionofpatternmaximumI=terminalcurrent.Solution:222SrTakingIasthermsvaluewesetIRSr=Ω,thereforeR=Ωq.e.d.rArA2PowerPowerIinputradiated*6-3-12.Radiationresistance.课后答案网Anantennameasuredatadistanceof500misfoundtohaveafar-fieldpatternof|E|=1.5withnoφdependence.IfEEo(sinθ)o=1V/mandIo=650mA,findtheradiationresistanceofthisantenna.www.hackshp.cnSolution:From(6-3-5)2120ππE1202ππ2Rd==ssinθθrsindθdφrIZ22∫∫(0.65)(377)2200∫oosπ−34=×(6.2810)(500)2πθ∫sindθ0π⎡⎤3xxxsin(2)sin(4)⎛3π⎞=−+==19.719.7⎜⎟23.2(Ωans.)⎢⎥⎣⎦84320⎝8⎠ 42*6-5-1.λ/2antenna.Assumethatthecurrentisofuniformmagnitudeandin-phasealongtheentirelengthofaλ/2thinlinearelement.(a)Calculateandplotthepatternofthefarfield.(b)Whatistheradiationresistance?(c)Tabulateforcomparison:1.Radiationresistanceofpart(b)above2.Radiationresistanceatthecurrentloopofaλ/2thinlinearelementwithsinusoidalin-phasecurrentdistribution3.Radiationresistanceofaλ/2dipolecalculatedbymeansoftheshortdipoleformula(d)Discussthethreeresultstabulatedinpart(c)andgivereasonsforthedifferences.Solution:(a)E()tansin[(/2)cos]θθπθ=(ans.)n(b)R=Ω168(ans.)(c)R[from(b)]=168(Ωans.)RIa(sinusoidal)=Ω73(ns.)Ra(shortdipole)=197(Ωns.)(d)168Ωisappropriateforuniformcurrent.73Ωisappropriateforsinusoidalcurrent.197Ωassumesuniformcurrent,buttheshortdipoleformuladoesnottakeintoaccountthedifferenceindistancetodifferentpartsofthedipole(assumes课后答案网λ>>L)whichisnotappropriateandleadstoalargerresistance(197Ω)ascomparedtothecorrectvalueof168Ω.6-6-1.2λantenna.www.hackshp.cnTheinstantaneouscurrentdistributionofathinlinearcenter-fedantenna2λlongissinusoidalasshowninFig.P6-6-1.(a)Calculateandplotthepatternofthefarfield.(b)Whatistheradiationresistancereferredtoacurrentloop?(c)Whatistheradiationresistanceatthetransmission-lineterminalsasshown?(d)Whatistheradiationresistanceλ/8fromacurrentloop? 436-6-1.continuedFigureP6-6-1.2λantenna.Solution:cos(2cos)1πθ−(a)E()θ=(ans.)nsinθ(b)RI(at)=Ω259(ans.)max(c)R(atterminals)=∞Ω(ans.)(d)RI(at/8from)λ=Ω518(ans.)max6-7-1.λ/2antennasinechelon.Calculateandplottheradiation-fieldpatternintheplaneoftwothinlinearλ/2antennaswithequalin-phasecurrentsandthespacingrelationshipshowninFig.P6-7-1.Assumesinusoidalcurrentdistributions.课后答案网www.hackshp.cnFigureP6-7-1.λ/2antennasinechelon.Solution:cos[(/2)cos]πθπ⎛⎞2E()θπ=+cos⎜⎟cos[(/4)]θnsinθ⎜⎟4⎝⎠ 44*6-8-1.1λand10λantennaswithtravelingwaves.(a)Calculateandplotthefar-fieldpatternintheplaneofathinlinearelement1λlong,carryingasingleuniformtravelingwavefor2casesoftherelativephasevelocityp=1and0.5.(b)Repeatforthesinglecaseofanelement10λlongandp=1.Solution:sinφ1(a)From(6-8-5),E()φ=−[sin(πcos)]φ,patternshave4lobes.n1c−pposφ(b)Patternhas40lobes.6-8-2.Equivalenceofpatternfactors.Showthatthefieldpatternofanordinaryend-firearrayofalargenumberofcollinearshortdipolesasgivenbyEq.(5-6-8),multipliedbythedipolepatternsinφ,isequivalenttoEq.(6-8-5)foralonglinearconductorwithtravelingwaveforp=1.Solution:nψsin2(1)Fieldpattern=(5-6-8)ψsin2whereψβ=+dcosφδωbsin[(1−pcos)]φ2pc(2)Fieldpattern=sinφ课后答案网(6-8-5)1c−posφForordinaryend-fire,ψβ=−d(cosφ1)⎛⎞βndwww.hackshp.cnsin⎜⎟(1cos)−φ⎝⎠2Alsoifdissmall(1)becomesβd(1cos)−φ2Forlargernndb,.≅Alsomultiplyingbythesourcefactorsinφandtakingthecon-stantβd/21=inthedenominator,(1)becomes⎛⎞βdsin⎜⎟(1cos)−φ⎝⎠2sinφ1cos−φ 456-8-2.continuedwhichisthesameas(2)forp=1ωπbf2bbβsince==q.e.d.222pcfλNotethatforagivenlengthb,thenumbernisassumedtobesufficientlylargethatdcanbesmallenoughtoallowsinψ/2in(1)tobereplacedbyψ/2.课后答案网www.hackshp.cn 46课后答案网www.hackshp.cn 47Chapter7.TheLoopAntenna7-2-1.Loopanddipoleforcircularpolarization.Ifashortelectricdipoleantennaismountedinsideasmallloopantenna(onpolaraxis,Fig.7-3)andbothdipoleandlooparefedinphasewithequalpower,showthattheradiationiseverywherecircularlypolarizedwithapatternasinFig.7-7forthe0.1λdiameterloop.Solution:Uniformcurrentsareassumed.2120πθIAsinE()(loop)=θ(1)φ2rλj60πθILsinE()(dipole)=θ(2)θrλ24⎛⎞AR(loop)=320π⎜⎟Ω(3)r2⎝⎠λ22RL(dipole)=80π(4)rλForequalpowerinputs,22IR(loop)=IR(dipole)looprrdipole课后答案网2222IloopRr(dipole)80πLLλλ===(5)2422222IR(loop)320πλπλ(/A)4(/A)dipolerILloopλ=(6)2IA2(/)πλwww.hackshp.cndipoleTherefore2120πθLIAsin60πILsinθλdipoledipoleE()(loop)=θ=(7)φ22rAλπλ2(/)rλwhichisequalinmagnitudetoE()θ(dipole)butintime-phasequadrature(noj).θSincethe2linearlypolarizedfields(EoftheloopandEofthedipole)areatrightφθangles,areequalinmagnitudeandareintime-phasequadrature,thetotalfieldoftheloop-dipolecombinationiseverywherecircularlypolarizedwithasinθpattern.q.e.d. 487-2-1.continuedEquatingthemagnitudeof(1)and(2)(fieldsequalandcurrentsequal)weobtainLA=2π(8)2λλwhichsatisfies(6)forequalloopanddipolecurrents.Thus(8)isaconditionforcircularpolarization.2SubstitutingAd=π(/4),whered=loopdiameterin(8)andputtingCd=π22LπdC1==2π(9)22λλλ4212weobtainCL=(2)(10)λλasanotherexpressionoftheconditionforcircularpolarization.Thus,forashortdipoleλ/10long,theloopcircumferencemustbe12C=×(20.1)=0.45(11)λ0.45λandtheloopdiameterd==0.14λπor1.4timesthedipolelength.Ifthedipolecurrenttaperstozeroattheendsofthedipole,theconditionforCPis课后答案网LA=4π(12)2λλand12CL=()(13)λλwww.hackshp.cn12Foraλ/10dipolethecircumferencemustnowbeC==(0.1)0.316andtheloopλ0.316λdiameterd=≅0.1λorapproximatelythesameasthedipolelength.πTheconditionof(10)isappliedintheWheeler-typehelicalantenna.SeeSection8-22,equation(8-22-4)andProb.8-11-1. 497-4-1.The3λ/4diameterloop.Calculateandplotthefar-fieldpatternnormaltotheplaneofacircularloop3λ/4indiameterwithauniformin-phasecurrentdistribution.Solution:3C==π2.36λ4From(7-3-8)orTable7-2,theEpatternisgivenbyφJC(sin)θ1λSeeFigure7-6.*7-6-1.Radiationresistanceofloop.WhatistheradiationresistanceoftheloopofProb.7-4-1?Solution:From(7-6-13)forloopofanysize2Cλ2RCJ=60π()ydyrλ∫20whereCC==π342.36,2=4.71λλ22CCλλFrom(7-6-16),J()ydy=−Jydy()2(2)JC∫∫2o1λ00课后答案网2CλByintegrationoftheJ()ycurvefrom0to2C(4.71)=,Jydy()=0.792oλ∫o0Fromtables(JahnkeandEmde),www.hackshp.cnJCJ(2)==(4.71)−0.281611λ2CλandJydy()==0.792020.28161.355+×=∫202ThereforeR=×60π2.361.3551894(Roundoffto1890)=ΩΩ(ans.)r 507-6-2.Small-loopresistance.(a)UsingaPoyntingvectorintegration,showthattheradiationresistanceofasmallloop4222isequalto320π()AλΩwhereA=areaofloop(m).(b)Showthattheeffective2apertureofanisotropicantennaequalsλ/4π.Solution:222SrΩErΩAmaxA(a)R==r22IZIFrom(7-5-2)andTable7-2,2120πIAEE==sinθθsinφ2maxrλπ482Ω=2sπθinsinθdθ=2ππ=A∫033242222120ππIAr844⎛⎞ATherefore,RC==320π⎜⎟197Ω=Ωq.e.d.r2422λrIλπ1203⎝⎠λ22λλ(b)Ω=4,πA==q.e.d.AeΩ4πA7-7-1.Theλ/10diameterloop.课后答案网Whatisthemaximumeffectiveapertureofathinloopantenna0.1λindiameterwithauniformin-phasecurrentdistribution?Solution:www.hackshp.cnΩisthesameasforashortdipole(8/3sr).=πSeeProb.6-3-4a.A2λ⎛⎞322Therefore,A==⎜⎟λλ=0.119(ans.)emΩ⎝⎠8πA7-8-1.Pattern,radiationresistanceanddirectivityofloops.Acircularloopantennawithuniformin-phasecurrenthasadiameterd.Whatis(a)thefar-fieldpattern(calculateandplot),(b)theradiationresistanceand(c)thedirectivityforeachofthreecaseswhere(1)d=λ/4,(2)d=1.5λand(3)d=8λ? 517-8-1.continuedSolution:SincealltheloopshaveC>1/3,thegeneralexpressionforEinTable7-2mustbeλφused.FromTable7-2andFigures7-10and7-11,theradiationresistanceanddirectivityvaluesare:DiameterCRDirectivityλrλ/40.78576Ω1.51.5λ4.712340Ω3.828λ25.114800Ω17.1*7-8-2.Circularloop.Acircularloopantennawithuniformin-phasecurrenthasadiameterd.Find(a)thefar-fieldpattern(calculateandplot),(b)theradiationresistanceand(c)thedirectivityforthefollowingthreecases:(1)d=λ/3,(2)d=0.75λand(3)d=2λ.Solution:SeeProbs.7-4-1and7-8-1.Radiationresistanceanddirectivityvaluesare:DiameterCRDirectivity课后答案网λrλ/31.05180Ω1.50.75λ2.361550Ω1.22λwww.hackshp.cn6.284100Ω3.6*7-9-1.The1λsquareloop.Calculateandplotthefar-fieldpatterninaplanenormaltotheplaneofasquareloopandparalleltooneside.Theloopis1λonaside.Assumeuniformin-phasecurrents. 52*7-9-1.continuedSolution:Patternisthatof2pointsourcesinoppositephase.ReferringtoCase2ofSection5-2,wehaveford/2==2(/2)πλπ,rE()sin(cos)φ=πφnooresultingina4-lobedpatternwithmaximaatφ=±60and120±andnullsatooo0,90and180.±7-9-2.Smallsquareloop.Resolvingthesmallsquareloopwithuniformcurrentintofourshortdipoles,showthatthefar-fieldpatternintheplaneoftheloopisacircle.Solution:ThefieldpatternE(1,2)ofsides1and2ofthesmallsquareloopistheproductofthepatternof2pointsourcesinoppositephaseseparatedbydasgivenby课后答案网sin[(d/2)cos]φrandthepatternofshortdipoleasgivenbycosφwww.hackshp.cnorEd(1,2)=cossin[(φφ/2)cos]r2ForsmalldthisreducestoE(1,2)=cosφnoThepatternofsides3and4isthesamerotatedthrough90orintermsofφisgivenby2E(3,4)=sinφnThetotalpatternintheplaneofthesquareloopisthen22EEE()φφ=+=+=(1,2)(3,4)cossinφ1nnnThereforeE()φisaconstantasafunctionofφandthepatternisacircle.q.e.d. 53Chapter8.End-FireAntennas:TheHelicalBeamAntennaandtheYagi-UdaArray,PartI8-3-1.A10-turnhelix.Aright-handedmonofilarhelicalantennahas10turns,100mmdiameterand70mmturnspacing.Thefrequencyis1GHz.(a)WhatistheHPBW?(b)Whatisthegain?(c)Whatisthepolarizationstate?(d)Repeattheproblemforafrequencyof300MHz.Solution:8310×(a)λπ==0.3mC==(0.1)0.3149100.3140.07CS==1.047==0.233λλ0.30.3From(8-3-4)OO5252oHPBW===32.5(ans.)1212CnS()1.047(100.233)×λλ2(b)From(8-3-7),DC≅=12nS30.7or14.9dBi(ans.)λλIflossesarenegligiblethegain=D.(c)PolarizationisRCP.(ans.)86(d)At300MHz,λ=×310/300101m×=,0.314/10.314.C==课后答案网λThisistoosmallfortheaxialmodewhichrequiresthat0.7<andnotequalasinthisproblem.HEλλ 61*9-9-1.continuedInanoptimumhorn,thelength(whichisnotspecifiedinthisproblem)isreducedbyrelaxingtheallowablephasevariationattheedgeofthemouthbyarbitraryamountsoo(90=×2ππ0.25radinthe-planeand144EH=×20.4radinthe-plane).Thisresultsinlessgainthancalculatedabove,whereuniformphaseisassumedovertheaperture.From(9-9-2),whichassumes60%apertureefficiency,thedirectivityofthe10λsquarehornis22DA=×7.5/λ=×=7.510750or29dBipTosummarize:whenuniformphaseisassumed(0ε=.81)asintheinitialsolutionapabove,D=1018or30dBibutforanoptimum(shorter)horn(ε=0.6),D=750orap29dBi.9-9-2.Hornpattern.(a)CalculateandplottheE-planepatternofthehornofProb.9-9-1,assuminguniformilluminationovertheaperture.(b)Whatisthehalf-powerbeamwidthandtheanglebetweenfirstnulls?Solution:(a)From(5-12-18)thepatternofauniformapertureoflengthaisψ′课后答案网sin2sin(πθasin)λE==(1)n′ψπθasinλ2wherea=aperturelength=10www.hackshp.cnλθ=anglefrombroadsideo(b)FromTable5-8,HPBW==50.8/105.08(ans.)ooIntroducing5.08/2=2.54into(1)yieldsE=0.707whichconfirmsthat5.08isthen22trueHPBWsincePE==0.707=0.5nnUsing(5-7-7)andsettingnd=aforacontinuousaperture,λλ−−11oBWFN===2sin(1/a)2sin(1/10)11.48(ans.)λ 629-9-2.continuedSettingnd=aassumesnverylargeanddverysmall,butwehavenotassumedthatλλλtheirproductndisnecessarilyverylarge.Ifwehad,wecouldwriteλBWFN=2/aradλandobtainoBWFN=2/10rad=11.46oforadifferenceof0.02.9-9-3.Rectangularhornantenna.Whatistherequiredapertureareaforanoptimumrectangularhornantennaoperatingat2GHzwith12dBigain?Solution:7.5ADp2From(9-9-2)orFig.3-5b,DA≅=,λ2pλ7.51.2D==1015.85,λ=0.15m15.8522A=×=(0.15)0.0475mp7.5课后答案网9-9-4.Conicalhornantenna.Whatistherequireddiameterofaconicalhornantennaoperatingat2GHzwitha12dBigain?www.hackshp.cnSolution:26.5πrFromFig.3-5b,D≅2λD1.2Thediameterdr=2isd=2λ,D==1015.85,λ=0.15m6.5π15.85d=×20.15=26.4cm6.5π 639-9-5.Pyramidalhorn.(a)DeterminethelengthL,apertureaHandhalf-anglesinEandHplanesforapyramidalelectromagnetichornforwhichtheapertureaE=8λ.ThehornisfedwitharectangularwaveguidewithTE10mode.Takeδ=λ/10intheEplaneandδ=λ/4intheHplane.(b)WhataretheHPBWsinbothEandHplanes?(c)Whatisthedirectivity?(d)Whatistheapertureefficiency?Solution:(a)Fora0.1λtoleranceintheE-plane,therelationwithdimensionsinwavelengthsisshowninthesketch.222FromwhichLa+=/4L++0.2L0.01λλEλλL+0.1λa/2Eλθ/2E2witha=8(given),La==/.880(ans.)LEλλλEλIntheH-planewehavefromthesketchthat80.25a/2Hλθ/2Haa/2==6.33and12.780HHλλ−1oθ/2==tan4/802.9(ans.)E−1oθ/2==tan6.33/804.5(ans.)H(c)Ifthephaseovertheapertureisuniformε=0.81(seesolutiontoProbs.19-1-7apand9-9-1),D=×××=课后答案网4π812.70.811034or30.1dBioHowever,thephasehasbeenrelaxedto36=×2π0.1radintheE-planeandtoo90=×2π0.25radintheH-plane,resultinginreducedapertureefficiency,soεmustapoobelessthan0.8.IftheE-planephaseisrelaxedtowww.hackshp.cn90andtheH-planephaseto144,ε∼0.6,whichisappropriateforanoptimumhorn.Thus,fortheconditionsofthisapproblemwhicharebetweenanoptimumhornanduniformphase,0.6<<ε0.8.Takingapε≅0.7,apD=×××=4π812.70.7894or29.5dBi(ans.)(b)AssuminguniformphaseintheE-plane,oo50.850.8oo(HPBW)≅==≅6.356.4(ans.)Ea8Eλandfromtheapproximation 649-9-5.continued4100041000D===894(HPBW)(HPBW)6.4(HPBW)EHHoso(HPBW)≅7.2HFromTable9-1foranoptimumhorn,o56o(HPBW)≅=7E8(ans.)o67o(HPBW)==5.3H12.7oThetrue(HPBW)forthisproblemisprobablycloseto6.4.Whilethetrue(HPBW)EHoisprobablycloseto5.3.(d)ε=0.7frompart(c).(ans.)ap课后答案网www.hackshp.cn 65Chapter10.FlatSheet,CornerandParabolicReflectorAntennas10-2-1.Flatsheetreflector.Calculateandplottheradiationpatternofaλ/2dipoleantennaspaced0.15λfromaninfiniteflatsheetforassumedantennalossresistanceRL=0and5Ω.Expressthepatternsingainoveraλ/2dipoleantennainfreespacewiththesamepowerinput(andzerolossresistance).Solution:From(10-2-1)thegainoveraλ/2referencedipoleisgivenby12⎛⎞R11GS()2φφ=⎜⎟sin(cos)(1)fr⎝⎠RRR+−11L12where,S=spacingofdipolefromreflectorφ=anglefromperpendiculartoreflector(SeeFig.10-2.)Notethat(1)differsfrom(10-2-1)inthatR=0inthenumeratorunderthesquarerootLsignsincetheproblemrequeststhegaintobeexpressedwithrespecttoalosslessreferenceantenna.Maximumradiationisatφ=0,so(1)becomes,课后答案网12⎛⎞73.1G()2φπ=×⎜⎟sin(20.15)f⎝⎠73.1+−R29.4LandforR=0LG()φ=2.09or6.41dB(=8.56dBi)(ans.)fwww.hackshp.cnNotethatRisforaspacingof0.3(20.15)λλ=×.SeeTable13-1.12NotethatRG=Ω10,()1.89or5.52dB(=7.67dBi)φ=(ans.)LfNotethatG()φisthegainwithrespecttoareferenceλ/2dipoleandmoreexplicitlyfcanbewrittenGA()[/φHW].fThelossresistanceR=Ω10resultsinabout0.9dBreductioningainwithrespecttoaLlosslessreferencedipole.Ifthereferencedipolealsohas10Ωlossresistance,thegainreductionisabout0.3dB. 6610-2-1.continuedTheabovegainsagreewiththoseshownforR=0andextrapolatedforR=Ω10LLat0.15S=λinFig.10-4.NotethatinFig.10-4anequallossresistanceisassumedinthereferenceantenna.ThepatternforR=0shouldbeintermediatetothoseinFig.10-3forspacingsofL0.125(=/8)and0.25(=/4).λλλλThepatternforR=Ω10issmallerthantheoneforLR=0butofthesameshape(radiusvectordifferingbyaconstantfactor).L10-3-1.Square-cornerreflector.Asquare-cornerreflectorhasadrivenλ/2dipoleantennaspaceλ/2fromthecorner.Assumeperfectlyconductingsheetreflectorsofinfiniteextent(idealreflector).Calculateandplottheradiationpatterninaplaneatrightanglestothedrivenelement.Solution:From(10-3-6)thegainofalosslesscornerreflectoroverareferenceλ/2dipoleisgivenby12⎛⎞R11GS()φφ=−2⎜⎟[cos(cos)cos(Ssin)]φfrr⎝⎠RRR+−2111412oForS=λ/2andmaximumradiationdirection(0φ=)thisbecomes12⎛⎞课后答案网73.1G()φ==4⎜⎟3.06or9.7dB(=11.9dBi)f⎝⎠73.13.8224++×SeeTable13-1andFig.13-13forthemutualresistancevaluesforRat1λseparation14andRat0.707λseparation.Theabovecalculatedgainagreeswiththevalueshownby12www.hackshp.cnthecurveinFig.10-11.ThepatternshouldbeidenticaltotheoneinFig.10-12a.10-3-2.Square-cornerreflector.(a)Showthattherelativefieldpatternintheplaneofthedrivenλ/2elementofasquare-cornerreflectorisgivenby()Dcos90cosθE=[]1−cos()Ssinθrsinθwhereθistheanglewithrespecttotheelementaxis.Assumethatthecorner-reflectorsheetsareperfectlyconductingandofinfiniteextent. 6710-3-2.continued(b)Calculateandplotthefieldpatternintheplaneofthedrivenelementforaspacingofλ/2tothecorner.Comparewiththepatternatrightangles(Prob.10-3-1).Solution:(a)Thepatternintheplaneofthedipole(Eplane)isthatofanarrayofthreeλ/2elementsarrangedasinthesketchwithamplitudes1:2:1andphasingasindicated.I=−1PhasorsketchS−−1s∠__SinθrθI=+2o−2cos(sin)Srθ20∠__S−1∠__SsinθrI=−1Bypatternmultiplicationthepatternistheproductofthepatternofanarrayof3isotropicsourceswithamplitudesandphasing−+−1:2:1andthepatternofλ/2dipole(6-4-4).Thus,cos(90cos)oθES=−(21∠∠__sinθθ−−1__Ssin)rrsinθor,seephasorsketch,课后答案网ocos(90cos)θES=−2[1cos(sin)]θrsinθDroppingthescalefactor2yieldstheresultssought,q.e.d.www.hackshp.cn*10-3-4.Square-cornerreflector.(a)Calculateandplotthepatternofa90°cornerreflectorwithathincenter-fedλ/2drivenantennaspaced0.35λfromthecorner.Assumethatthecornerreflectorisofinfiniteextent.(b)Calculatetheradiationresistanceofthedrivenantenna.(c)Calculatethegainoftheantennaandcornerreflectorovertheantennaalone.Assumethatlossesarenegligible. 68*10-3-4.continuedSolution:(a)From(10-3-6)thenormalizedfieldpatternforS=0.35λisoo[cos(126cos)cos(126sin)]φφ−E()φ=n1.588(b)RRRR=+−=−+=Ω273.124.82573.3(ans.)r111412(c)From(10-3-6)forφλ==0andS0.3512G()φ=×2(73.1/73.3)1.588=3.17or10.0dB(=12.1dBi)(ans.)f10-3-5.Square-cornerreflectorversusarrayofitsimageelements.AssumethatthecornerreflectorofProb.10-3-4isremovedandthatinitsplacethethreeimagesusedintheanalysisarepresentphysically,resultingin4-elementdrivenarray.(a)Calculateandplotthepatternofthisarray.(b)Calculatetheradiationresistanceatthecenterofoneoftheantennas.(c)Calculatethegainofthearrayoveroneoftheantennasalone.Solution:(a)4-lobedpatternasinFig.10-9withshapeofpatternofProb.10-3-4a.(b)R=Ω73.3(ans.)r课后答案网(c)G()1.59or4.0dB(=6.1dBi)φ=(ans.)fsincepowerisfedtoall4elementsinsteadoftoonlyone(Powergaindownbyafactorof4orby6dB).www.hackshp.cn*10-3-6.Square-cornerreflectorarray.Four90°corner-reflectorantennasarearrangedinlineasabroadsidearray.Thecorneredgesareparallelandside-by-sideasinFig.P10-3-6.Thespacingbetweencornersis1λ.Thedrivenantennaineachcornerisaλ/2elementspaced0.4λfromthecorner.Allantennasareenergizedinphaseandhaveequalcurrentamplitude.Assumingthatthepropertiesofeachcornerarethesameasifitssideswereofinfiniteextent,whatis(a)thegainofthearrayoverasingleλ/2antennaand(b)thehalf-powerbeamwidthintheHplane? 69*10-3-6.continuedFigureP10-3-6.Square-cornerreflectorarray.Solution:(a)From(10-3-6)thegainofonecornerreflectorwithS=0.4λisgivenby12⎛⎞73.1ooG()φ=−2⎜⎟(cos144cos0)f⎝⎠73.118.642−+=×20.8701.813.15or10dB(=12.1dBi)×=≅Underlosslessconditions,2DG=×()f()φ1.6416.3=Thus,themaximumeffectiveapertureofonecorneris22Dλλ16.32A=≅=1.3λem44ππTheeffectiveapertureofasinglecorner课后答案网maythenberepresentedbyarectangle11λ×.3λasinthesketchbelow.www.hackshp.cn1.3λAemterminals1λλ/2dipoleλo90cornerreflector0.4λ 70*10-3-6.continuedInanarrayof4reflectorsasinFig.P10-3-6theedgesoftheaperturesoverlap0.3λsothatthereflectorsaretooclose.However,atthe1λspacingthetotalapertureis2414λλλ×=andthetotalgainofthearrayunderlosslessconditionsis4πAemGD===×≅4π450or17dBi(ans.)2λNointeractionbetweencornerreflectorshasbeenassumed.Withwiderspacing(1.3)=λtheexpectedgain=×16.34=65or18dBi.(b)Assumingauniformaperturedistribution,theHPBWisgivenapproximatelyfromTable5-8byoooHPBW=50.8/L==50.8/412.7λTodeterminetheHPBWmoreaccurately,letususethetotalantennapattern.Bypatternmultiplicationitisequaltotheproductofanarrayof4in-phaseisotropicpointsourceswith1λspacingandthepatternofasinglecornerreflectorasgivenby1sin(4sin)1πφE()φ=−[cos(0.8cos)cos(0.8sin)]πφπφn4sin(sin)1.809πφThe¼isthenormalizingfactorforthearrayand1/1.809forthecornerreflector.Thus,owhenφ=0,E()1.φ=Notethatφmustapproachzerointhelimitinthearrayfactorntoavoidanindeterminateresult.ooHalfoftheaboveapproximateHPBWis12.7/2=6.35.Introducingitintotheabove课后答案网oequationyieldsEE()φφφ===0.703.For6.30,()0.707astabulatedbelow.nnφE()φno6.350.703owww.hackshp.cn6.300.707oThus,HPBW=×26.3012.6=(ans.)The4-sourcearrayfactorismuchsharperthanthecornerreflectorpatternandlargelydeterminestheHPBW.Returningtopart(a)forthedirectivity,letuscalculateitsvaluewiththeapproximateorelationof(2-7-9)usingtheHPBWofpart(b)fortheH-planeandtheHPBWof78fortheE-planefromExample6-4.1. 71*10-3-6.continuedThus,40000D≅=40.7(=16dBi)12.67.8×ascomparedtoD≅50(=17dBi)ascalculatedinpart(a).Althoughthedirectivityof16.3forasinglecornerreflectorshouldbeaccurate,sinceitisdeterminedfromthepatternviatheimpedances*,thedirectivityof50forthearrayof4cornerreflectorsinvolvessomeuncertainty(aperturesoverlapping).Nevertheless,thetwomethodsagreewithin1dB.____________________________*Assuminginfinitesides10-3-7.Cornerreflector.λ/4tothedrivenelement.Asquare-cornerreflectorhasaspacingofλ/4betweenthedrivenλ/2elementandthecorner.ShowthatthedirectivityD=12.8dBi.Solution:Forthecaseofnolosses,2DG=×()f()φλ1.64,andforS=/4andφ=0,(10-3-6)becomes12⎛⎞73.1G()2φ==⎜⎟3.39课后答案网f⎝⎠73.112.735−−2Therefore,DG=×()()φ1.6418.9or12.8dBi=(ans.)fwww.hackshp.cn10-3-8.Cornerreflector.λ/2tothedrivenelement.Asquare-cornerreflectorhasadrivenλ/2elementλ/2fromthecorner.(a)Calculateandplotthefar-fieldpatterninbothprincipalplanes.(b)WhataretheHPBWsinthetwoprincipalplanes?(c)Whatistheterminalimpedanceofthedrivenelement?(d)Calculatethedirectivityintwoways:(1)fromimpedancesofdrivenandimagedipolesand(2)fromHPBWs,andcompare.Assumeperfectlyconductingsheetreflectorsofinfiniteextent. 7210-3-8.continuedSolution:(a)FromProb.10-3-2thepatternintheE-planeisgivenbyo1cos(90cos)θE()θπ=−[1cos(sin)]θ(1)n2sinθFrom(10-3-6)thepatternintheH-planeisgivenby1E()θπ=−[cos(cos)cos(sin)]φπφ(2)n2ooNotethatE()θ=maximumforθ=90whileE()φ=maximumforφ=0.nn(b)AssuminginitiallythatHPBW()θ≅HPBW()φandnotingfromFig.10-11thatforS=λ/2thedirectivityisabout12dBi,wehavefrom(2-7-9)thato40000oHPBW()θ50oD≅≅16orHPBW()θ≅50and==252HPBW()θ22oooIntroducingθ=−=902565in(1)yieldsE()θwhichistoohigh.Bytrialanderror,noweobtainE()θθ≅0.707when=34.5.nooTherefore,HPBW()θ≅×234.5=69(ans.)课后答案网oIntroducingφ=25in(2)yieldsE()φ=0.60whichistoolow.Bytrialanderror,wenoobtain()Eφ≅0.707whenφ=21.nooTherefore,HPBW()www.hackshp.cnφ≅×=22142(ans.)(c)Theterminalimpedanceofthedrivenelementis(seeProb.10-3-1solution),R=++73.13.8224125×=Ω(ans.)TFromProb.10-3-1solution,2DG=×()f()φ1.6415.4(=11.9dBi)b=yimpedance(ans.)40000D≅=13.8(=11.4dBi)bybeamwidths(ans.)oo69×42 7310-3-8.continuedTheD=15.4valueis,ofcourse,moreaccuratesinceitisbasedonthepatternviatheimpedances.Thetwomethodsdiffer,however,byonly0.4dB.*10-7-2.Parabolicreflectorwithmissingsector.2Acircularparabolicdishantennahasaneffectiveapertureof100m.Ifone30°sectoroftheparabolaisremoved,findtheneweffectiveaperture.Therestoftheantenna,includingthefeed,isunchanged.SectorSolution:removedo45φφ2ThefulldishhasaneffectiveapertureA=100m.Assumingthatthedishcharacter-eoisticsareindependentofangle(),φremovingone45sectorreducestheeffectiveaper-tureto7/8ofitsoriginalvalueprovidedthe课后答案网feedismodifiedandsoasnottoilluminatetheareaofthemissingsector.However,thefeedisnotmodifiedand,therefore,its2efficiencyisdownto7/8.Therefore,thenetapertureefficiencyis(7/8)andtheneteffectiveapertureis22(7/8)×=10076.6m(ans.)www.hackshp.cn 74课后答案网www.hackshp.cn 75Chapter11.BroadbandandFrequency-IndependentAntennas*11-2-2.The2°cone.Calculatetheterminalimpedanceofaconicalantennaof2°totalangleoperatingagainstaverylargegroundplane.Thelengthloftheconeis3λ/8.Solution:oFrom(11-2-2)forZ,andnotingthatwhenθissmall(2θ<0)k⎛⎞θ4Thencot⎜⎟≅⎝⎠4θ4orZ=120lnkθandwith(11-2-4)and(11-2-5)forZZand(11-2-3)formi,soZj=+Ω270350(ans.)i11-5-1.Logspiral.Designaplanarlog-spiralantennaofthetypeshowninFig.11-11tooperateatfrequenciesfrom1to10GHz.Makea课后答案网drawingwithdimensionsinmillimeters.Solution:89Highfrequencylimit=10GHz,λ=×310/1010×=30mmwww.hackshp.cnLowfrequencylimit=1GHz,λ=300mmoTakeβ=77.6(seeFig.11-10)From(11-5-5),r=antiln(/tan)=antiln(/4.55)θβθ 7611-5-1.continuedθrR0rad11.5mmπ/21.412.12π2.003.003/2π2.824.232π4.006.005/2π5.668.503π8.0012.07/2π11.317.04π16.024.09/2π22.634.05π32.048.011/2π45.268.06π64.096.0SpiralislikeoneinFig.11-10.Ifgapdatcenterisequaltoλ/10athighfrequencylimit,thengapshouldbe30/10=3mmandradiusRofactualspiral=3/2=1.5mm.课后答案网Ifdiameterofspiralisλ/2atlowfrequencylimit,thentheactualspiralradiusshouldbe300/(22)×=75mm.Thisrequiresthatθπ==4.55ln(75/1.5)17.8=5.7www.hackshp.cnForgoodmeasurewemakeθπ=6.Thus,thespiralhas3turns(θπ=6).ThetablegivesdatafortheactualspiralradiusRinmmversustheangleθinrad.Theoveralldiameterofthespiralis96×=2192mmwhichat1GHzis192/300=0.64λ.Callingtheabovespiralnumber1,drawanidenticalspiralrotatedthroughπ/2rad,athirdrotatedthroughπradandafourthrotatedthrough3/2πrad.Metalizetheareasbetweenspirals1and4andbetweenspirals2and3,leavingtheremainingareasopen.ConnectthefeedacrossthegapattheinnermostendsofthespiralsasinFig.11-11. 7711-7-1.Logperiodic.Designan“optimum”log-periodicantennaofthetypeshowninFig.11-17tooperateatfrequenciesfrom100to500MHzwith11elements.Give(a)lengthoflongestelement,(b)lengthofshortestelement,and(c)gain.Solution:oFromFig.11-19letusselectthepointwhereα=15intersectstheoptimumdesignlinewhichshouldresultinanantennawithslightlymorethan7dBigain.Fromthefigure,k=1.195.Thedesiredfrequencyratiois5=F=250/50.Thus,from(7)therequirednumberofelementsmustsbeatleastequaltologFlog5n===9.0logklog1.195at250MHz,λλ==1.2m,/20.6m(ans.)at50MHz,λλ==6m,/23m(ans.)9Ifelement1is0.6inlong,thenelement10(=n+1)is0.6×1.195=2.98morapproximately3masrequired.Addingadirectorinfrontofelement1andareflectorinbackofelement10bringsthetotalnumberofelementsto12.(ans.)ThelengthAofanyelementwithrespecttothelengthA,ofthenextshorterelementis2givenby课后答案网AA/1==k.195(ans.)21From(11-7-5)(notegeometryofFig.11-18),thespacingsbetweenanytwoelementsisrelatedtothelengthAoftheadjacentshorterelementbywww.hackshp.cnAA(1k−)0.195s===0.364A(ans.)o2tanα2tan15[Notethat(11-7-6)givesswithrespecttoadjacentlongerelements.]Finally,connecttheelementsasinFig.11-17. 7811-7-2.StackedLPs.TwoLParrayslikeintheworkedexampleofSec.11-7arestackedasinFig.11-21a.(a)Calculateandplottheverticalplanefieldpattern.Notethatpatternmultiplicationcannotbeapplied.(b)Whatisthegain?Solution:FromtheworkedexampleofSec.11-7,oα====15,kFn1.2,4,7.6(8)andn+19=Considerthatλλ==1mand4m.minmaxn8Therefore,AA==0.5mandAk=0.51.2×=A=2m11n+19From(11-7-6)thedistancebetweenelements1and9isn−2n(1kk−)S==∑AA11where0.5mn=02tanαand7n0.21.2×S=∑o=0.187+0.224+0.268+0.322+0.387+0.464+0.557+0.669=3.08mn=04tan15ThestackedLPsareshowninsideviewinth课后答案网esketchbelow.Elements1through9areincludedinthecalculation.Element1isλ/2resonantat1mwavelengthandelement9isλ/2resonantat4mwavelength.Adirectorelementisaddedaheadofelement1andareflectorelementisaddedafterelement9makingatotalof11elements.DInthe60anglestackingarrangementthestwww.hackshp.cnackingdistanceis0.5morλ/2at1mwavelengthand3.65mor0.91λat4mwavelength.Atthegeometricmeanwavelength(2m)thestackingdistanceis1.25mor0.625λ. 7911-7-2.continued课后答案网Letuscalculatetheverticalplanepatternatwww.hackshp.cn2mwavelengthwhereelements3,4and5oftheupperandlowerLPsareactive.Asanapproximation,letusconsiderthatthe3activeelementsareauniformordinaryend-firearraywithspacingequaltotheaverageofthespacingbetweenelements3and4andbetween4and5.Forelement4wetakeA=λ/2.Thenfrom(11-7-6)4k−−11.21S===0.187λ45o4tanα4tan15 8011-7-2.continuedandS==0.187/1.20.155λ340.155.187+S==0.171λav21sinnψ/2Theend-firearrayfieldpatternisgivenbyE=3sin/2ψwhere2π×0.171oψφ/2=−(cos1)=30.8(cosφ−1),n=32EachLP(end-firearray)hasabroadcardiod-shapedpatternliketheonesshowninFig.oo11-21awithonepatterndirectedup30andtheotherdown30.Thetotalfieldpatternintheresultantofthesepatternsandabroadsidearrayof2in-phaseisotropicsourcesstackedverticallyandspaced0.625λwithpatterngivenbyoE=×cos[(2π0.625/2)sin]cos(112.5sin)φφ=ThispatternisshowninFig.16-11.NumericaladditionoftheLPpatternsandmulti-plicationoftheresultantbythebroadsidepatternyieldsthetotalfieldpatternforλ=2mshowninthesketch.Atλ=1mtheup-and-downminorlobesdisappearbutthemainbeamisaboutthesame.Atλ=4mthemainbeamisnarrowerbuttheup-and-downminorlobesarelarger.课后答案网(b)EachLPhasagain≅7dBi.FromtheequationofProb.5-2-3,thedirectivityof2in-phaseisotropicsourceswith0.625λspacingis2.44or3.9dBi.Soforλ=2m,thegainmaybeasmuchas73.910.9dBi+≅(ans.)Atbothλλ==1mand4mwww.hackshp.cnthegainmaybelessthanthis.ooTheHPBWintheverticalplaneisabout47andinthehorizontalplaneabout76.Fromtheapproximatedirectivityrelation,wehave,neglectingminorlobes,41000D==11.5or10.6dBi(ans.)4776×Actualdirectivityisprobably∼10dBi. 81Chapter12.AntennaTemperature,RemoteSensingandRadarCrossSection*12-2-1.Antennatemperature.Anend-firearrayisdirectedatthezenith.Thearrayislocatedoverflatnonreflectingground.If0.9ΩAiswithin45°ofthezenithand0.08ΩAbetween45°andthehorizoncalculatetheantennatemperature.Theskybrightnesstemperatureis5Kbetweenthezenithand45°fromthezenith,50Kbetween45°fromthezenithandthehorizonand300Kfortheground(belowthehorizon).Theantennais99percentefficientandisataphysicaltemperatureof300K.Solution:0.9Ωisat5KA0.08Ωisat50KATherefore0.02Ωisat300KAFrom(12-1-8),1T=×[50.9Ω+500.08×Ω+×Ω3000.02]=4.54614.5K++=(ans.)AAAAΩA*12-2-2.Earth-stationantennatemperature.课后答案网2Anearth-stationdishof100meffectiveapertureisdirectedatthezenith.Calculatetheantennatemperatureassumingthattheskytemperatureisuniformandequalto6K.Takethegroundtemperatureequalto300Kandassumethat1/3oftheminor-lobebeamareaisinthebackdirection.Thewavelengthis75mmandthebeamefficiencyis0.8.www.hackshp.cnSolution:From(12-1-8),121T=[60.8×Ω+××Ω+60.2300××Ω=++=0.2]4.80.82025.6K(ans.)AAAAΩ33A 82*12-3-4.SatelliteTVdownlink.Atransmitter(transponder)onaClarkeorbitsatelliteproducesaneffectiveradiatedpower(ERP)atanearthstationof35dBover1Wisotropic.(a)DeterminetheS/Nratio(dB)iftheearthstationantennadiameteris3m,theantennatemperature25K,thereceivertemperature75Kandthebandwidth30MHz.Takethesatellitedistanceas36,000km.Assumetheantennasaparabolicreflector(dish-type)of50percentefficiency.(SeeExample12-3.1).(b)Ifa10-dBS/Nratioisacceptable,whatistherequireddiameteroftheearthstationantenna?Solution:(a)SatelliteERP=35dB(over1Wisotropic)22ERP=PD==P4/,πλAPAλERP/4πtttettetFrom(12-3-3),2SPAAλERPAteterer==,ERP=35dBor31622222NkTrBλπλ4kTrBsyssys11222Ar===ππ1.53.53m,2575100T=+=Kersys22S31623.53×==16.6or12.2dB(ans.)−232147N4π××××××1.3810课后答案网1003.610310×(b)Ifonly10dBS/Nratioisacceptable,thedishaperturecouldbe2.2dB(12.210dB)=−lessor60%oftheπ1.52=7.07m2areaspecifiedinpart(a).22Thereforeacceptablearea=×=7.070.6www.hackshp.cn4.24m=πr12Foradiameter==2r2(4.24/)π=2.3m(ans.)*12-3-5.Systemtemperature.Thedigitaloutputofa1.4GHzradiotelescopegivesthefollowingvalues(arbitraryunits)asafunctionofthesiderealtimewhilescanningauniformbrightnessregion.Theintegrationtimeis14s,with1sidletimeforprintout.Theoutputunitsareproportionaltopower. 83*12-3-5.continuedTimeOutputTimeOutputmsms3130234324522931452353300236320022433152333215226333023032302393345226Ifthetemperaturecalibrationgives170unitsfor2.9Kapplied,find(a)thermsnoiseatthereceiver,(b)theminimumdetectabletemperature,(c)thesystemtemperatureand(d)theminimumdetectablefluxdensity.Thecalibrationsignalisintroducedatthereceiver.Thetransmissionlinefromtheantennatothereceiverhas0.5dBattenuation.The2antennaeffectiveapertureis500m.Thereceiverbandwidthis7MHz.Thereceiverconstantk′=2.Solution:(a)Noiseoutputreadings234,235,etc.withrespecttoaveragevalue(231),aresquared,averagedandthensquarerootedforroot-mean-square(rms)noisevalue4.71.Thermsnoiseatreceiveristhen4.71×=2.90.08K(ans.)170(b)Transmissionlineattenuation=0.5dBforefficiencyof0.89.Therefore,∆=T0.08/0.89=0.09K(ans.)min课后答案网(c)From(12-2-3),12612∆∆Tf()t0.09(710××14)minT===445K(ans.)syskwww.hackshp.cn′2(d)From(12-1-7),−232kT∆21.3810×××0.09min∆=S==497mJy≅500mJy(roundedoff)(ans.)minA500e 8412-3-6.Systemtemperature.Findthesystemtemperatureofareceivingsystemwith15Kantennatemperature,0.95transmission-lineefficiency,300Ktransmission-linetemperature,75Kreceiverfirst-stagetemperature,100Kreceiversecond-stagetemperatureand200Kreceiverthird-stagetemperature.Eachreceiverstagehas16dBgain.Solution:From(12-2-1)and(12-2-2),⎛⎞⎛11100200⎞T=+15300⎜⎟⎜−+175++⎟sys⎝⎠⎝.95.954040⎠=++++=1515.878.92.65.3117.6K(ans.)*12-3-7.Solarinterferencetoearthstation.TwiceayearthesunpassesthroughtheapparentdeclinationofthegeostationaryClarke-orbitsatellites,causingsolar-noiseinterferencetoearthstations.AtypicalforecastnoticeappearingonU.S.satelliteTVscreensreads:ATTENTIONCHANNELUSERS:WEWILLBEEXPERIENCINGSOLAROUTAGESFROMOCTOBER15TO26FROM12:00TO15:00HOURS课后答案网(a)Iftheequivalenttemperatureofthesunat4GHzis50,000K,findthesun’ssignal-to-noiseration(indecibels)foranearthstationwitha3-mparabolicdishantennaat4GHz.Takethesunsdiameteras0.5°andtheearth-stationsystemtemperatureas100K.(b)Comparethisresultwiththatforthecarrier-to-noseratiocalculatedinExample12-3.1foratypicalClarke-orbitTVtransponder.www.hackshp.cn2(c)Howlongdoestheinterferencelast?NotethattherelationΩ=λAgivestheAesolidbeamangleinsteradiansandnotinsquaredegrees.(d)WhydotheoutagesoccurbetweenOctober15and26andnotattheautumnalequinoxaroundSeptember20whenthesuniscrossingtheequator?(e)Howcansatelliteservicesworkaroundasolaroutage? 85*12-3-7.continuedSolution:2kTsys(a)EarthstationS=minAe22kT∆∆kTΩ2kTΩAAAssS===sun2AAΩλeeAAssuming50%apertureefficiencyasinProb.12-3-4,22λ0.075−3Ω===1.5910sr×=5.22sq.deg.AA12eπ1.524o2SSTΩ510××π(.25)sunssTherefore,====18.8or12.7dB(ans.)NSTΩ×1005.22minsysAsq.deg.(b)FromProb.12-3-4,12.712.2−=0.5dBmorethansatellitecarrier,resultingindegradationofTVpicturequality.Thephenomenonmaybedescribedasnoisejammingbythesun.(c)Athalf-power,solarnoisewillbereducedtoonly30.52.5−=dBbelowcarrier.Assuminglowside-lobes,thesolarinterferenceshouldnotlastmorethanthetimeittakesthesuntodriftbetweenfirstnulls.Usingthiscriterionwehave课后答案网Time=×4(min/deg)BWFN(deg)SolardriftrateFromTable15-1,oooHPBW==66/D66/(30.075)1.65andTime=≅421.6513.2min××=(ans.)λwww.hackshp.cnoAllowingfortheangularextentofthesun(approx.12)increasesthetimebyabout2minutes. 86*12-3-9.Criticalfrequency.MUF.Layersmaybesaidtoexistintheearth’sionospherewheretheionizationgradientissufficienttorefractradiowavesbacktoearth.[Althoughthewaveactuallymaybebentgraduallyalongacurvedpathinanionizedregionofconsiderablethickness,ausefulsimplificationforsomesituationsistoassumethatthewaveisreflectedasthoughfromahorizontalperfectlyconductingsurfacesituatedata(virtual)heighth.]Thehighestfrequencyatwhichthislayerreflectsaverticallyincidentwavebacktotheearthiscalledthecriticalfrequencyfo.Higherfrequenciesattheverticalincidencepassthrough.Forwavesatobliqueincidence(φ>0inFig.P12-3-9)themaximumusablefrequency(MUF)forpoint-to-pointcommunicationontheearthisgivenbyMUF=fo/cosφ,whereφ=angleofincidence.Thecriticalfrequencyf=9N,whereN=electrondensityo-3(numberm).Nisafunctionofsolarirradiationandotherfactors.Bothfoandhvarywithtimeofday,season,latitudeandphaseofthe11-yearsunspotcycle.FindtheMUFfor(a)adistanced=1.3MmbyF2-layer(h=325km)reflectionwithF2-layerelectron11-3densityN=6x10m;(b)adistanced=1.5MmbyF2-layer(h=275km)reflection12-3withN=10m;and(c)adistanced=1MmbysporadicE-layer(h=100km)11-3reflectionN=8x10m.Neglectearthcurvature.课后答案网FigureP12-3-9.Communicationpathviareflectionfromionosphericlayer.www.hackshp.cnSolution:MUF=MaximumUsableFrequencyforcommunicationviaionosphericreflection.(a)51112−−11⎛⎞6.510×ofN==×=99(610)6.97MHz,φ=tan[(/2)/]dh=tan⎜⎟=63.43o5⎝⎠3.2510×cosφφ==0.477,MUFf/cos=6.97/0.44715.6MHz=(ans.)o 87*12-3-9.continued51212-1⎛⎞7.510×o(b)f==9(10)9MHz,φφ=tan⎜⎟=69.86,cos=0.344o5⎝⎠2.7510×MUF==9/0.34426.1MHz(ans.)51112-1⎛⎞510×o(c)f=×9(810)=8.05MHz,φφ=tan⎜⎟=78.69,cos=0.196o5⎝⎠106MUF8.0510/0.196=×=41.0MHz12-3-10.mUFforClarke-orbitsatellites.Stationarycommunication(relay)satellitesareplacedintheClarkeorbitatheightsofabout36Mm.Thisisfarabovetheionosphere,sothatthetransmissionpathpassescompletelythroughtheionospheretwice,asinFig.12-3-10.Sincefrequenciesof2GHzandaboveareusuallyusedtheionospherehaslittleeffect.Thehighfrequencyalsopermitswidebandwidths.Iftheionosphereconsistsofalayer200mthickbetween12-1heightsof200and400kmwithauniformelectrondensityN=10m,findthelowestfrequency(orminimumusablefrequency,mUF)whichcanbeusedwithacommunicationsatellite(a)forverticalincidenceand(b)forpaths30°fromthezenith.(c)Foranearthstationontheequator,whatisthemUFforasatellite15°abovetheeasternorwesternhorizon?课后答案网www.hackshp.cnFigureP12-3-10.CommunicationpathviageostationaryClarke-orbitrelaysatellite.Solution:mUF=minimumusablefrequencyfortransmissionthroughionosphere121212(a)mUF===9()N9(10)9MHz(ans.) 8812-3-10.continuedAlthoughMUF=mUF,whetheritisoneortheother,dependsonthepointofview.ItisMUFforreflectionandmUFfortransmission.Belowcriticalfrequency,waveisreflected;abovecriticalfrequency,waveistransmitted.o(b)mUF==9MHz/cos3010.4MHz(ans.)o(c)mUF==9MHz/cos7534.8MHz(ans.)96AtypicalClarkeorbitsatellitefrequencyis4GHz,whichis115(410/34.810)=××timeshigherinfrequencythanthemUF,sothatat4GHz,transmissionthroughtheoionospherecanoccuratmuchlowerelevationanglesthan15.*12-3-11.Minimumdetectabletemperature.Aradiotelescopehasthefollowingcharacteristics:antennanoisetemperature50K,receivernoisetemperature50K,transmission-linebetweenantennaandreceiver1dBlossand270Kphysicaltemperature,receiverbandwidth5MHz,receiverintegration2time5s,receiver(system)constantk′=π/2andantennaeffectiveaperture500m.Iftworecordsareaveraged,find(a)theminimumdetectabletemperatureand(b)theminimumdetectablefluxdensity.Solution:(a)For1dBloss,ε==1/1.260.792课后答案网From(12-2-1),⎛⎞150T=+50270⎜⎟−+=++=15071.863.3185Ksys⎝⎠.79.79From(12-2-3),www.hackshp.cnkT′sysπ185∆=T==0.058K≅0.06K(roundedoff)(ans.)min12612()∆×ftn2(510×5×2)wheren=numberofrecordsaveraged−232kT21.3810×××0.058min(b)From(12-1-7),∆=S==320mJy(ans.)minA500e 8912-3-12.Minimumdetectabletemperature.Aradiotelescopeoperatesat2650MHzwiththefollowingparameters:systemtemperature150K,predetectionbandwidth100MHz,postdetectiontimeconstant5s,2systemconstantk’=2.2andeffectiveapertureofantenna800m.Find(a)theminimumdetectabletemperatureand(b)theminimumdetectablefluxdensity.(c)Iffourrecordsareaveraged,whatchangeresultsin(a)and(b)?Solution:2.250×(a)∆=T=0.015K(ans.)min812(10×5)−2321.3810×××0.015(b)∆=S=52mJy(ans.)min8000.015(c)∆=T=0.008Kmin4(ans.)52∆==S26mJymin4*12-3-13.Interstellarwirelesslink.6Ifanextraterrestrialcivilization(ETC)transmits10W,10spulsesofright-handcirc-ularlypolarized5GHzradiationwitha100mdiameterdish,whatisthemaximumdistanceatwhichtheETCcanbereceivedwithanSNR=3.Assumethereceivingantennaontheearthalsohasa100mdiameterantennaresponsivetorightcircularpolarization,thatbothantennas(theirsandour课后答案网s)have50percentapertureefficiency,andthattheearthstationhasasystemtemperatureof10Kandbandwidthof0.1Hz.Solution:(Note:ProblemstatementshouldspecifyS/N=3.)8ScPAA310×teretFrom(12-3-3),==www.hackshp.cn,0λ==.06m229NrBλkTf510×sys6222PAAteret10[(50)0.5][(50)0.5]ππ382r===1.0310m×22−23λBkT(/SN)(0.6)(0.1)(1.3810×)(10)(3)sys19Therefore,r=×1.0210m1000lightyears≅(ans.) 90*12-3-14.Backpackingpenguin.Thispenguin(Fig.P12-3-14)participatedinastudyofAntarcticpenguinmigrationhabits.Itsbackpackradiowithλ/4antennatransmitteddataonitsbodytemperatureanditsheartandrespirationrates.Italsoprovidedinformationonitslocationasitmovedwithitsflockacrosstheicecap.Thebackpackoperatedat100MHzwithapeakpowerof1Wandabandwidthof1-kHzoftone-modulateddatasignals.IfTsys=1000KandSNR=30dB,whatisthemaximumrange?Thetransmittingandreceivingantennasareλ/4stubs.FigureP12-3-14.Antarcticbackpackingpenguin.Solution:8ScPAA310×teretFrom(12-3-3),==,3λ==m226NrBλkTf10010×sys22⎛⎞1.5λλ⎛⎞1.5Pt⎜⎟⎜⎟2PAAteret⎝⎠44ππ⎝⎠r==22λλBkT(/)SNBkT(/)SNsyssys2课后答案网⎛⎞1.5(3)1⎜⎟⎝⎠4π112==9.310m×23−233(3)(1010)(1.3810××)(1000)(10)5Therefore,r=×9.610m≅960kmwww.hackshp.cn(ans.)*12-3-17.Lowearthorbitcommunicationssatellite.Acommunicationssatelliteinlowearthorbit(LEO)hasrup=1500kmandrdown=1000km,withanuplinkfrequencyof14.25GHzandadownlinkfrequencyof12GHz.Findthefull-circuitC/NifthetransmittingearthstationERPis60dBWandthesatelliteERPis25dBW.AssumethesatellitereceiverG/Tis5dB/Kandtheearth-stationG/Tis30dB/K. 91*12-3-17.continuedSolution:4πA2FromProb.12-3-15,CPAAteterG=e,wehaveCPGGttrλ=andusing=22222NrkλTλNr(4)πkTsyssys222Wenowdefineeffectiveradiatedpower=ERP=PG,Powerloss==Lr(4)πλ/ttC⎛⎞⎛⎞11⎛⎞GrThisgives=(ERP)⎜⎟⎜⎟⎜⎟Nk⎝⎠⎝⎠L⎜⎟⎝⎠TsysFortheuplink,cC310×8⎛⎞⎛⎞⎛1⎛⎞1G(sat)⎞rλ===0.021m,and⎜⎟=(ERP)⎜⎟⎜⎜⎟⎟upfN14.2510×9⎝⎠up⎝⎠⎝⎜⎟⎜L⎝⎠kT(sat)⎠⎟upupupsys26⎛⎞4(1.510)π×WorkingindB,ERPup=60dBW,L==10log⎜⎟179.1dBup⎝⎠0.021−23Gr(sat.)-1k=×10log(1.3810)=−228.6dB,=5dBKT(sat.)sys⎛⎞CSo⎜⎟=−60179.1228.65114.5dB++=⎝⎠Nup⎛⎞CSimilarly,⎜⎟课后答案网=−+25174228.630109.6+=⎝⎠Ndown−−−111⎛⎞⎛⎞⎛⎞CCC11−11FromProb.12-3-1,⎜⎟⎜⎟⎜⎟=+=+=1.4510×11.4510.96⎝⎠⎝⎠⎝⎠NNNupdown1010www.hackshp.cnC110==6.8910×=108.4dB(ans.)−11N1.4510×*12-3-18.Directbroadcastsatellite(DBS).DirectbroadcastsatelliteservicesprovideCDqualityaudiotoconsumersviasatellitesingeosynchronousorbit.TheWorldAdministrativeRadioConference(WARC)hasestablishedtheserequirementsforsuchservices. 92*12-3-18.continuedFrequencyband11.7to12.2GHz(Kuband)Channelbandwidth27MHz2Minimumpowerfluxdensity-103dBW/mReceiverfigureofmerit(G/T)6dB/KMinimumcarrier-to-noiseratio14dB(a)Findtheeffectiveradiatedpower(ERP)over1WisotropicneededtoproducethespecifiedfluxdensityattheEarth’ssurfacefromaDBSsatelliteina36,000kmorbit.(b)Ifthesatellitehasa100Wtransmitterandisoperatedat12GHz,whatsizecircularparabolicdishantennamustbeusedtoachievetherequiredERP?Assume50percentefficiency.(c)Doesaconsumerreceiverwithcircular1mdishantennawith50percentefficiencyandsystemnoisetemperatureof1000KmeetthespecifiedG/T?(d)Byhowmuchdoesthesystemspecifiedinparts(a)through(c)exceedtherequiredcarrier-to-noiseratio?Solution:PGtt(a)SinceERP=PGandthepowerfluxdensityψ()r=,tt24πr−2ERPwehave−=103dBWm724(3.610)π×72−10.35∴=×ERP4(3.610)10π=×8.1610W(ans.)24EππAdRP4(π)(/2)(b)Gk===k22λλP课后答案网t252ERP(λ)48.1610(0.025)4×1.0d===m(ans.)22Pkt4π(100)(4)(3.14)(0.5)(c)Fora1-mreceiveantennaand1000Knoisetemperature,www.hackshp.cn2G4πAe4()(1/2)ππ−−11==k0.5==5.48K7.4dBK(yes)(ans.)22TTλ(0.03)(1000)(d)TheC/NforachannelbandwidthBis2CG⎛⎞⎛⎞⎛⎞⎛⎞1115⎛⎞0.03⎛1⎞⎛1⎞==ERP⎜⎟⎜⎟⎜⎟⎜⎟8.1610×⎜⎟⎜⎟(5.5)⎜⎟NL⎝⎠⎝⎠⎝⎠⎝⎠kTB⎝⎠4(3.610)π××72⎝1.3810−37⎠⎝2.710×⎠==52.917.2dB 93*12-3-18.continuedThereforeC/Nexceedsrequirement(14dB)by17.214−=3.2dB(ans.)12-3-19.SimplifiedexpressionforC/N.TheexpressionforC/NprovidedinP12-3-15maybesimplifiedbymakingthefollowingsubstitutions:Effectiveisotropicradiatedpower=ERP=PtGt(W)22Linkpathloss=Llink=4πr/λThecarrier-tonoiseratiomaythenbewrittenasC11Gr=ERPNLkTlinksyswhereGr/Tsysisthereceiveantennagaindividedbythesystemnoisetemperature.Thisratio,referredtoas“GoverT,”iscommonlyusedasafigureofmeritforsatelliteandearthstationreceivers.FindtheC/NratiofortheuplinktoasatelliteattheClarkeorbit(r=36,000km)equippedwitha1mparabolicdishantennawithefficiencyof50percentandareceiverwithnoisetemperatureof1500K.Assumethatthetransmittingearthstationutilizesa1kWtransmitteranda50percentefficient10mdishantennaandoperatesatafrequencyof6GHz.Solution:课后答案网22Cr⎛⎞⎛⎞11⎛⎞G⎛4ππ⎞⎛⎞4(3.61×07)r19==ERP⎜⎟⎜⎟⎜⎟,λ0.05m,L=⎜⎟=⎜⎟=8.1810×NL⎝⎠⎝⎠K⎜⎟⎝⎠Tsys⎝λ⎠⎝⎠0.0524()(5)ππwww.hackshp.cn20.52⎛⎞4()(5)ππ81Gr(0.05)−ERP==PG10000.5⎜⎟=×1.9710W,==1.3Ktt2⎝⎠(0.05)Tsys1500C8⎛⎞11⎛⎞Therefore,=×(1.9710)⎜⎟⎜⎟(1.3)19−23N⎝⎠8.1810××⎝⎠1.381011=×=2.310113.5dB(ans.)12-3-22.Galileo’suncooperativeantenna.WhentheGalileospacecraftarrivedatJupiterin1995,groundcontrollershadbeenstrugglingfor3yearstoopenthespacecraft’s5mhigh-gaincommunicationsdish,which 94wastooperateat10GHz(Xband).Unabletodeploythisantennabecauseofprelaunchlossoflubricant,alow-directivity(G=10dB)S-bandantennaoperatingat2GHzhadtobeusedtorelayallpicturesanddatafromthespacecrafttotheEarth.Foraspacecraft11transmitpowerof20W,distancetoEarthof7.6x10m,and70mdishwith50percentefficiencyatthereceivingstation,find(a)themaximumachievabledatarateifthe5mX-bandantennahaddeployedand(b)themaximumdatarateusingthe1mS-bandantenna.Solution:C⎛⎞⎛⎞11⎛⎞Gr=ERP⎜⎟⎜⎟⎜⎟NL⎝⎠⎝⎠K⎜⎟⎝⎠Tsys2⎛⎞4()(2.5)ππ6(a)Forthe5-mX-bandantenna,ERP==PG200.5⎜⎟=×2.7410Wtt2⎝⎠(0.03)24()(35)ππ220.5112⎛⎞44ππr⎛⎞(7.6×10)29Gr(0.03)41−L==⎜⎟⎜⎟=1.0110,×==5.3710K×⎝⎠λ⎝⎠0.03Tsys500C64⎛⎞11⎛⎞5Therefore,=×(2.7410)⎜⎟⎜⎟(5.3710)1.0510×=×29−23N⎝⎠1.0110××⎝⎠1.3810C5ThemaximumdatarateM==×1.441.44(1.0510)152kBits/s≅(ans.)N*12-4-1.Antennatemperaturewithabsorbingcloud.课后答案网Aradiosourceisoccultedbyaninterveningemittingandabsorbingcloudofunityopticaldepthandbrightnesstemperature100K.Thesourcehasauniformbrightnessdis-tributionof200Kandasolidangleof1squaredegree.Theradiotelescopehasan2effectiveapertureof50m.Ifthewavelengthis50cm,findtheantennatemperaturewww.hackshp.cnwhentheradiotelescopeisdirectedatthesource.Thecloudisofuniformthicknessandhasanangularextentof5squaredegrees.Assumethattheantennahasuniformresponseoverthesourceandcloud. 95*12-4-1.continuedSolution:ΩΩ−−ττFrom(12-4-1)and(12-1-6),∆=TTcs(1−+eTcc)eAcsΩΩAA2Ω=5/57.3=0.00152src2Ω=1/57.3=0.00030srs22Ω=λ/A=0.5/50=0.005srAem.00152−−11.0003Therefore,∆=Te100(1−+)200e=23.6K(ans.)A.005.00512-4-3.Forestabsorption.Anearth-resourcesatellitepassiveremote-sensingantennadirectedattheAmazonRiverBasinmeasuresanight-timetemperatureTA=21°C.IftheearthtemperatureTe=27°CandtheAmazonforesttemperatureTf=15°C,findtheforestabsorptioncoefficientτf.Solution:−τf∆−TTAf294288−From(12-4-2),e===0.5andτ=0.693(ans.)fTT−−300288fe课后答案网*12-4-4.Jupitersignals.-20-2-1Fluxdensitiesof10WmHzarecommonlyreceivedfromJupiterat20MHz.Whatisthepowerperunitbandwidthradiatedatthesource?Taketheearth-Jupiterdistanceas40light-minutesandassumethatthesourceradiatesisotropically.www.hackshp.cnSolution:−−18111r=×40lightmin60smin×310ms×=7.2010m×PP1rt=×2∆∆ff4πrPtPr22−−02221or=×=×××=4ππr1047.21065.1kWHz(ans.)∆∆ff 9612-5-1.Radardetection-12Aradarreceiverhasasensitivityof10W.Iftheradarantennaeffectiveapertureis12mandthewavelengthis10cm,findthetransmitterpowerrequiredtodetectanobject2with5mradarcrosssectionatadistanceof1km.Solution:42−−12342Pr4πλ10××4π(10)(0.1)rP===25mW(ans.)t22Aσ(1)×5*12-5-3.RCSofelectron.Thealternatingelectricfieldofapassingelectromagneticwavecausesanelectron(initiallyatrest)tooscillate(Fig.P12-5-3).ThisoscillationoftheelectronmakesitequivalenttoashortdipoleantennawithD=1.5.Showthattheratioofthepower22scatteredpersteradiantotheincidentPoyntingvectorisgivenby(sµθemin/4)π,owhereeandmarethechargeandmassoftheelectronandθistheangleofthescatteredradiationwithrespecttothedirectionoftheelectricfieldEoftheincidentwave.Thisratiotimes4πistheradarcrosssectionoftheelectron.SuchreradiationiscalledThompsonscatter.课后答案网FigureP12-5-3.Solution:ωθIlsinoFrom(6-2-17),themagnitudeoftheelectricdipolefar-fieldisE=θ24πεcrwww.hackshp.cno2TheforceontheelectronduetotheincidentfieldEisF===eEmamωlooeEoSo,l=2mω⎛⎞eEoωω()e⎜⎟2sinθ⎝⎠mωSinceIe==ω,Eoθ24πεcro22224πrE⎛⎞eEsinθ1θoThus,thescatteredpowerdensityS==4π⎜⎟scat2Z4πεcmZ⎝⎠o 97*12-5-3.continued2EoandsincetheincidentpowerdensityS=,theratioofscatteredtoincidentpowerisincZ2⎛⎞eEsinθ1o4π⎜⎟2222222Se⎝⎠4πεcmZ1s⎛⎞eeinθθ1⎛⎞sin1⎛⎞µsinθscatooσ===⎜⎟=⎜⎟=⎜⎟ScE244πε2mπεµε⎜⎟()1/m4πminco⎝⎠oo⎝⎠oo⎝⎠Z−282Forsinθσ=≅1,110×m(ans.)*12-5-5.Detectingoneelectronat10km.IftheAreciboionospheric300mdiameterantennaoperatesat100MHz,howmuchpowerisrequiredtodetectasingleelectronataheight(straightup)of10kmwithanSNR–0dB?SeeFig.P12-5-5.Thebandwidthis1Hz,Tsys=100Kandtheapertureefficiency=50percent.FigureP12-5-5.Solution:课后答案网8SPGAtterσλ==310×3m,A==0.5(150)π243.5310m×2=24,6erNr(4)πkTB10010×44πAe4(3.5310)π×44.9310G===×t22www.hackshp.cnλ3Thus,24244−23(4)ππrkTBSN(/)(4)(10)(1.3810×)(100)(1)(1)16P===1.2510W×(ans.)t44−28GAσ(4.9310)(3.5310)(110×××)ter 9812-5-6.Effectofresonanceonradarcrosssectionofshortdipoles.(a)Calculatetheradarcrosssectionofalosslessresonantdipole(ZL=-jXA)withlength=λ/10anddiameter=λ/100.(SeeSecs.2-9,14-12).634L(b)Calculatetheradarcrosssectionofthesamedipolefrom,whereL4[]()2λlnL2b−1isthedipolelengthandbisitsradius.(c)ComparebothvalueswiththemaximumradarcrosssectionshowninFig.12-9.Commentonresults.*A(matcheddipole,ZZ=)emLASolution:Directivity22(a)0.714λλ=×40.1191.5×(ans.)−52A==σ(resonantdipole,Z−jX)(b)210×λ(ans.)stLA2(c)0.83λ(ans.)Summaryandcomparison:CaseσCondition(a)20.714λResonantshortdipole,length0.1(λZjLA=−X)(b)−52210×λNon-resonant0.1λdipole(ZL=0)(c)20.83λResonant0.47dipole(λZL=0)In(a)resonanceisobtainedbymakingZ=−jX.In(c)resonanceisobtained课后答案网LA(0X=)byincreasingthelengthto0.47λ(withZ=0,terminalsshort-circuited).ALThecrosssectionin(c)islargerthanin(a)becausethedipoleisphysicallylonger.In(b)thedipoleisnon-resonantbecausewww.hackshp.cnZL=0(terminalsshort-circuited)andthelength(0.1)λismuchlessthantheresonantlength(0.47),λresultinginaverysmallradarcrosssection.Itappearsthatifashortdipole(length≤0.1)λisresonatedbymakingZj=−X,itsLA2radarcrosssection(0.714λ)approachesthecrosssectionofaresonantλ/2dipole(length=0.47λ),regardlessofhowshortitis,provideditislossless.Ashortdipolemayberesonated(Zj=−X)byconnectingastubofappropriatelengthLAacrossitsterminals, 9912-5-6.continuedThus,DipoleStuborbyconnectinglumpedinductance,thus*12-5-12Fastballvelocity.A20GHzradarmeasuresaDopplershiftof6kHzonabaseballpitcher’sfastball.Whatisthefastball’svelocity?Solution:83∆∆fv2(cf3×10)(6×10)==,v==45m/s162km/h101mi/h==(ans.)9fc22f(20×10)*12-5-13.Radarpowerforfastballmeasurement.课后答案网TomeasurethevelocityofthefastballofProb.12-5-12withthe20GHzradaratadistanceof100m,whatpowerisrequiredforanSNR=30dB?Theradarusesaconicalhornwithdiameter=8cmandapertureefficiencyεap=0.5.Theballdiameter=7cmandithasaradarcrosssection(RCS)halfthatofaperfectlyconductingsphereofthesamediameter.www.hackshp.cnSolution:8SPGGσ310×ttrλ==0.015m,=,349Nr(4)πkTB2010×226.5ππr6.5(0.04)23−20.5GGk===Dk==72.6,σπ==0.5(0.035)1.910m×tr22λ(0.015) 100*12-5-13.continued3434−2363(4)ππrkTBSN(/)(4)(100)(1.3810×)(600)(10)(10)P==≈160mW(ans.)t23−GGσ(72.6)(1.910)×tr*12-5-14.Anticollisionradar.Toprovideanticollisionwarnings,forward-lookingradarsonautomobiles,trucksandothervehicles(seeFig.P12-5-14)canalertthedriverofvehiclesaheadthataredeceleratingtoofastorhavestopped.Thebrakelightonthevehicleaheadmaynotbeworkingoritmaybeobscuredbypoorvisibility.Towarnofclear-distancedecreaseratesof9m/sormore,whatdopplershiftmusta20GHzradarbeabletodetect?FigureP12-5-14.Solution:8∆×fv23课后答案网c10==,λ===0.015m,v9m/s,9fcf2010×222vfv(9)∆=fa===1.2kHz(ns.)cλ0.015www.hackshp.cn*12-5-18.Policeradar.Apulsedspeedmeasuringradarmustbeabletoresolvethereturnsformtwocarsseparatedby30m.Findthemaximumpulsewidththatcanbeusedtopreventoverlappingofthereturnsfromthetwovehicles.Notethatittakest=2R/csecondsforasignaltotravelfromthetransmittertoatargetandreturn,wheretisthetime,Ristherangeandcisthevelocityinthemedia.Ifτisthepulsewidth,thenitcanbeshownthattherangeresolution(theminimumrangedifferencebetweenobjectsforwhichthereturnsdonotoverlapintime)isgivenby 101*12-5-18.continuedcτ∆R=R−R=212wherethesubscriptsdenotethedifferentobjects.Solution:cRτ2∆∆=Rorτ=2c2(10)−8Thepulsewidthmustbelessthanτ==6.610s×=66ns(ans.)8310×*12-5-20.Seaclutter.Search-and-rescueaircraftusingradartolocatelostvesselsmustcontendwithbackscatterfromthesurfaceoftheocean.Theamplitudeofthesereturns(knowasseaclutter)dependsonthefrequencyandpolarizationoftheradarwaveform,thesizeoftheilluminatedpatchonthesurface,theangleofincidence,andtheseastate.ThescatteringgeometryisshowninFig.P12-5-20.Tocharacterizeseaclusterindependentlyontheradarfootprintonthesurface,thescatteringcrosssectionoftheoceanmaybespecifiedperunitarea.Thisparameter,designatedσo,hasdimensionsofsquaremetersofdBaboveasquaremeter(dBsm).ThetotalRCSofapatchofoceansurfaceisfoundbymultiplyingσobytheareaofthepatch.Forareascattering,theradarequationiswritten课后答案网22AλePP=σA.rt4opatch4πrForapulsedradarwithpulsewidthτand3dBantennabeamwidthofθrad,theilluminatedareaforlowgrazingangleisapproximately(www.hackshp.cncτ/2)(rθ).Theradarequationmaythereforebewrittenas2222Aeeλλ⎛⎞cτAcσoτθPPrt==43σθo⎜⎟()rPr42ππrr⎝⎠8(a)Determinethereceivedpowerfromseaclutteratarangeof10kmforamonostaticpulsedradartransmittinga1µspulsewith1kWpeakpoweratafrequencyof6GHz.Assumea1.5mcirculardishantennawith50percentefficiencyandseastate4(σo=-302dBsm/matCband).(b)Forareceiverbandwidthof10kHzandnoisefigureof3.5dB,findthereceivernoisepower.(c)IfthisradarusedtosearchforashipwithRCS=33 102*12-5-20.continueddBsmundertheseconditions,whatsignal-to-noiseandsignal-to-clutterratioscanbeexpected?FigureP12-5-20.Seasearch-and-rescuegeometry.Solution:228Acλστθ310×eoPP==,0λ=.05mtr3986πr×10λ0.05(a)θ≅==0.033radians3dBD1.5222−−301086P==(1000)[(0.75)](0.05)(10π课后答案网)(310)(110)(0.033)××3.110×−15W(ans.)r438(10)π(b)Thereceivernoisepoweris3.510−−23417PNk==TB(10www.hackshp.cn)(1.3810×)(290)10=8.9610×W(ans.)nf2(c)Forashipwithσ==33dBmatr10km.222223310Aeλσ[(0.75)](0.05)(10π)−3PP==1000=1.2410W×rt44444ππr(10)Thus,thesignal-to-noise(S/N)andsignal-to-clutter(S/C)ratiosare:−−33SS1.2410××1.210==1384=31.4dB(ans.),==4016dB(=ans.)−−1715NC8.9610××3.110 103Chapter13.SelfandMutualImpedances*13-4-1.A5λ/2antenna.Calculatetheself-resistanceandself-reactanceofathin,symmetricalcenter-fedlinearantenna5λ/2long.Solution:From(13-5-2),Zn=+−+30[0.577ln(2πππ)Ci(2n)jSi(2n)],where5n=11sin(10)πSince22πππn==>5101>,wehavefrom(13-3-18),Ci(10)π==010πAndfrom(13-3-22),ππcos(10)π1Si(10)π=−=−=1.5392102ππ10andR=+−30[0.577ln(10)0]120.7π=Ω(ans.)11X=×301.539=Ω46.2(ans.)1113-6-1.Parallelside-by-sideλ/2antennas.Calculatethemutualresistanceandmutualreactancefortwoparallelside-by-sidethinlinearλ/2antennaswithaseparationof0.15课后答案网λ.Solution:From(13-7-6),www.hackshp.cn2222Rd=−+30{2Ci(ββ)Ci[(dL+Ld)]Ci[(−+βL−L)]}21wheredL==0.15,λ0.5λandR=−30[2Ci(0.942)Ci(6.42)Ci(0.138)]−21From(13-3-16)orCitableorfromFig.13-5andfrom(13-3-18),wehaveR=−30(0.6000.577ln0.138)−−=60.1Ω(ans.)(comparewithFig.13-13)21 10413-6-1.continuedFrom(13-7-7),2222X=−30{2Si(ββdd)Si[(−+L+Ld)]Si[(−β+L−L)]}21andX=−30[2Si(0.942)Si(6.42)Si(0.138)]−−21From(13-3-20)orSitableorfromFig.13-5andfrom(13-3-21),wehaveX=−30(1.81.420.138)−−=−7.3Ω(ans.)(comparewithFig.13-13)11*13-6-3.Threeside-by-sideantennas.ThreeantennasarearrangedasshowninFig.P13-6-3.Thecurrentsareofthesamemagnitudeinallantennas.Thecurrentsarein-phasein(a)and(c),butthecurrentin(b)isinanti-phase.Theself-resistanceofeachantennais100Ω,whilethemutualresistancesare:Rab=Rbc=40ΩandRac=-10Ω.Whatistheradiationresistanceofeachoftheantennas?Theresistancesarereferredtotheterminals,whichareinthesamelocationinallantennas.FigureP13-6-3.Threeside-by-sideantennas.课后答案网Solution:RRRR=−+=−−=Ω100401050asabacRRR=−=−=Ω21008020(ans.)bsawww.hackshp.cnbRR==Ω50ca13-8-1.Twoλ/2antennasinechelon.Calculatethemutualresistanceandreactanceoftwoparallelthinlinearλ/2antennasinechelonforthecasewhered=0.25λandh=1.25λ(seeFig.13-16).Solution:Use(13-9-1),(13-9-2),(13-3-16)and(13-3-20)wherehd==1.25andλ0.25λ 105Chapter14.TheCylindricalAntennaandtheMomentMethod(MM)14-10-1.Chargedistribution.Determinetheelectrostaticchargedistributiononacylindricalconductingrodwithalength-diameterratioof6.Solution:QQQQQQ123321Dividerodinto6equalsegments.Bysymmetrychargesareasshown.Neglectingendfaces,thepotentialatpointP12is1⎛⎞QQQQQQ123321V(P)=+⎜⎟++++124πε⎝⎠2aaaaaa210265082WritingsimilarexpressionforVV(P)and(P),equatingthem(sincethepotentialis2334constantalongtherod)andsolvingforthechargesyields:课后答案网QQQ::=1.582:1.062:1.000(ans.)12314-12-2.λ/10dipoleimpedance.www.hackshp.cnShowthattheconvergenceortruevalueoftheselfimpedanceZsofthedipoleofTable14-4is1.852-j1895Ω.Solution:UsingZvaluesofTable14-4,plotRNvs.tosuitablelargescaleandsuppressedzerossandnotethatRapproachesaconstant(convergence)valueasNbecomeslargewhichsshouldagreewithRichmond’svaluegivenfollowing(14-12-34).CalculateRforlargersvaluesofNthan7,ifdesired.DosameforX.s 106课后答案网www.hackshp.cn 107Chapter15.TheFourierTransformRelationBetweenApertureDistributionandFar-FieldPattern*15-3-1.Patternsmoothing.Anidealizedantennapattern-brightnessdistributionisillustratedbythe1-dimensionaldiagraminFig.P15-3-1.ThebrightnessdistributionconsistsofapointsourceoffluxdensitySandauniformsource2°wide,alsooffluxdensityS.Thepointsourceis2°fromthecenterofthe2°source.Theantennapatternistriangular(symmetrical)witha2°beamwidthbetweenzeropointsandwithzeroresponsebeyond.(a)Drawanaccurategraphoftheobservedfluxdensityasafunctionofanglefromthecenterofthe2°source.(b)Whatisthemaximumratiooftheobservedtotheactualtotalfluxdensity(2S)?FigureP15-3-1.Patternsmoothing.Solution:Ratio=1/2(ans.)课后答案网15-6-1.Numberofelements.InFig.15-15howmanyelementsnhavebeenassumed?Solution:www.hackshp.cn211FromFig.15-15,=×nd4dλλorn=8 108课后答案网www.hackshp.cn 109Chapter16.ArraysofDipolesandofApertures*16-2-1.Twoλ/2-elementbroadsidearray.(a)Calculateandplotthegainofabroadsidearrayof2side-by-sideλ/2elementsinfreespaceasafunctionofthespacingdforvaluesofdfrom0to2λ.Expressthegainwithrespecttoasingleλ/2element.Assumeallelementsare100percentefficient.(b)Whatspacingresultsinthelargestgain?(c)Calculateandplottheradiationfieldpatternsforλ/2spacing.Showalsothepatternsoftheλ/2referenceantennatotheproperrelativescale.Solution:(a)From(16-2-27),12GA()(/φφHWRRR)[2/(=+)]cos[(cos)/2]dfr001112Inbroadsidedirectionφ=π/212soGA()(/φHWRRR)[2/(=+)]f001112whereRR==Ω73.10011andRisafunctionofthespacingasgiveninTable13-1(p.453).Afewvaluesofthe12gainforspacingsfrom0to1λarelistedbelow:SpacingλGainoverλ/2reference课后答案网0.01.000.11.020.21.080.31.190.41.36www.hackshp.cn0.51.560.61.720.71.740.81.640.91.491.01.38etc.Notethat2DA(/2)λπ===4(/2)/λλ4(30/73.1)1.64or2.15dBiππem 110*16-2-1.continuedsoDof2in-phaseλ/2elementsat0.67λspacingisequalto4.92.15+=7.1dBiasabove.(b)Byinterpolation,thehighestgainoccursforaspacingofabout0.67λforwhichthegainisabout1.76or4.9dB(=7.1dBi).Atspacingsover1λnogainsexceedthis.16-3-1.Twoλ/2-elementend-firearray.A2-elementend-firearrayinfreespaceconsistsof2verticalside-by-sideλ/2elementswithequalout-of-phasecurrents.Atwhatanglesinthehorizontalplaneisthegainequaltounity:(a)Whenthespacingisλ/2?(b)Whenthespacingisλ/4?Solution:(a)From(16-3-18),12GA()(/φφHWRRR)[2/(=−)]sin[(cos)/2]dfr001112Whend=λ/2,12GA()(/φHWRRR)[2/(=−)]sin[(/2)cos)]πφf001112whereRR==Ω73.10011课后答案网andfromTable13-1,R=−12.7Ω12soGA()(/φHW)1.31sin[(/2)cos)]=πφwww.hackshp.cnfForunitgain,sin[(/2)cos)]1/1.310.763πφ==ooooorcosφ==49.8/900.553andφ=±56,124±(ans.)(b)Whenλ/4,12GA()(/φHWRRR)[2/(=−)]sin[(/4)cos]πφf001112whereRR==Ω73.10011andfromTable13-1,R=Ω40.912 11116-3-1.continuedsoGA()(/φHW)=2.13sin[(/4)cos]πφfForunitgain,sin[(/4)cos]1/2.13πφ==0.47oooocosφφ===28/450.62and±52,128±(ans.)*16-3-2.Impedanceandgainof2-elementarray.Twothincenter-fedλ/2antennasaredriveninphaseopposition.Assumethatthecurrentdistributionsaresinusoidal.Iftheantennasareparallelandspaced0.2λ,(a)Calculatethemutualimpedanceoftheantennas.(b)Calculatethegainofthearrayinfreespaceoveroneoftheantennasalone.Solution:(a)Thisisasingle-electronW8JKarray.FromSec.13-7,Zj=−Ω5221(ans.)12(b)From(16-5-8)andassuminglosses,12⎛⎞273×oGA()(max)(/φHW)=⎜⎟sin36f课后答案网⎝⎠7352−=×=2.640.5881.55or3.8dB(=6.0dBi)(ans.)16-4-3.Two-elementarraywithunequalcurrents.www.hackshp.cn(a)Considertwoλ/2side-by-sideverticalelementsspacedadistancedwithcurrentsrelatedbyI2=aI1/δ.Developthegainexpressioninaplaneparalleltotheelementsandthegainnormaltotheelements,takingaverticalλ/2elementwiththesamepowerinputasreference(0≤a≤1).CheckthatthesereducetoEq.(16-4-15)andEq.(16-4-13)whena=1.(b)Plotthefieldpatternsinbothplanesandalsoshowthefieldpatternofthereferenceantennainproperrelativeproportionforthecasewhered=λ/4,a=½andδ=120°.Solution:212212(a)GR()θδ=+{/[R(1aa)2+Rcos]}×(1+a+2cos)aψf111112 11216-4-3.continuedwhereψ=+dsinθδrGG()φ==()butwithθψφdcos+δffr*16-6-1.ImpedanceofD-Tarray.(a)Calculatethedriving-pointimpedanceatthecenterofeachelementofanin-phasebroadsidearrayof6side-by-sideλ/2elementsspacedλ/2apart.ThecurrentshaveaDolph-Tchebyscheffdistributionsuchthattheminorlobeshave1/5thefieldintensityofthemajorlobe.(b)Designafeedsystemforthearray.Solution:(a)FromProb.5-9-5andthe6sourceshavethedistribution:1234560.930.841.001.000.840.93Normalizingthecurrentforelement1,thedistributionis1.000.901.081.080.901.00UsingimpedancedatafromChap.13andassumingthinelements,thedrivingpointimpedanceofelement1is课后答案网Rj=++−−+73430.9(12j29)1.08(3++−−++−−j18)1.08(2j12)0.9(1jj10)131=−+−+−+−+−+−=−Ω7310.83.220.91jj(432619.41393)6329=R(ans.)6Inlikemanner,RR==−Ω46j2,(ans.)R==−ΩR53j10,(ans.)25www.hackshp.cn3416-6-3.Squarearray.Fourisotropicpointsourcesofequalamplitudearearrangedatthecornersofasquare,asinFig.P16-6-3.Ifthephasesareasindicatedbythearrows,determineandplotthefar-fieldpatterns. 11316-6-3.continuedFigure16-6-3.Squarearray.Solution:Patternisaroundedsquare.oooE==1.00atφ0,90,180±nooE==0.895atφ±45,135±n*16-6-4.Sevenshortdipoles.4-dBangle.Alinearbroadside(in-phase)arrayof7shortdipoleshasaseparationof0.35λbetweendipoles.Findtheanglefromthemaximumfieldforwhichthefieldis4dB(tonearest0.1°).Solution:课后答案网Thedipolesareassumedtobealignedcollinearlysothatthepatternofasingledipoleisproportionaltosinφwhereφistheanglefromthearray.Thuswww.hackshp.cnφ0.35λooSincethedipolesarein-phase,themaximumfieldisatφφ==90or(E)90.maxThenormalizedpatternisgivenby1sinnψ/2E=sinφ(1)nnsinψ/2wheren=7ψ=×(2/)0.35cos,πλλφ4=20logxx,1=.585 114*16-6-4.continuedTherefore,E(4dB)1/1.585−==0.631noSetting(1)equalto0.631,n=7andsolving(seenotebelow)yieldsφ(4dB)−=78.3oooAnglefromφ()Eis90−=78.311.7(ans.)maxNote:Usetrialanderrortosolve(1)forφ(4dB)−orcalculatepatternwithsmallincrementsinφ.16-6-5.Squarearray.Fouridenticalshortdipoles(perpendiculartopage)arearrangedatthecornersofasquareλ/2onaside.Theupperleftandlowerrightdipolesareinthesamephasewhilethe2dipolesattheothercornersareintheoppositephase.Ifthedirectiontotheright(xdirection)correspondstoφ=0°,findtheanglesφforallmaximaandminimaofthefieldpatternintheplaneofthepage.Solution:oooooPatternmaximaatφ=±45,135±Patternminimaatφ=±0,90,180*16-6-7.Sixteen-sourcebroadsidearray.Auniformlineararrayhas16isotropicin课后答案网-phasepointsourceswithatspacingλ/2.Calculateexactly(a)thehalf-powerbeamwidth,(b)thelevelofthefirstsidelobe,(c)thebeamsolidangle,(d)thebeamefficiency,(e)thedirectivityand(f)theeffectiveaperture.Solution:www.hackshp.cn(a)From(5-6-9),o1sin(1440cos)φE(HP)==0.707o16sin(90cos)φoooooBytrialanderror,φ==86.82andHPBW2(90−86.82)=6.36=622(′ans.)(b)From(5-18-10),K=1(firstminorlobe)1E≅=0.215or-13.3dBML16sin[(21)/32]+π 115*16-6-7.continuedThisisonlyapproximate(becomesexactonlyforverylargen).Todeterminethelevelmoreaccurately,wefindtheapproximateangleforthemaximumofthefirstminorlobefrom(5-18-5).−−11o±+(2K1)±3φ≅==coscos79.2m2nd1λ216××2ooThenfrom(5-6-9)wecalculateEatanglescloseto79.2andfindthatEpeaksat79.7withEa=0.22012or-13.15dB(ns.)Although(5-18-5)locatestheanglewherethenumeratorof(5-18-5)isamaximum(=1),thedenominatorisnotconstant.SeediscussionofSec.5-18(p.159)andalsoFig.5-47(p.100).(c)FromequationforDinProb.5-6-10,thesummationtermiszeroford=λ/2sothatD=16exactly.SinceDD=ΩΩ4/,ππ=4/=4/16π=π/4sr(ans.)AA(d)HPBW1/≅=×=nd1/(160.5)1/8radindirectionφ,λBWinθdirection=2radπTherefore,Ω=×2ππε(1/8)=/4srand=ΩΩ=/1or100%MMMAThisresultistoolargesincewithanyminorlobesεmustbelessthanunity(or课后答案网MΩ<Ω).MAForanexactevaluation,wehavefromProb.5-6-10that2www.hackshp.cn⎡⎤n−1nk−θ2Ω=M2⎢⎥ndπθλλcos+∑sin(2πθkdcos)ndλ⎣⎦k=1kθ1owhereθγ=−901ooθγ=+902oγ=angletofirstnullo−−11−1oFrom(5-7-7),γ==×sin(1/nd)sin[1/(160.5)]sin(1/8)==7.18oλooTherefore,θθ==82.82and97.1812 116*16-6-7.continuedThuso97.18n−14⎡⎤nk−Ω=M2⎢⎥ndπθλλcos+∑sin(2πθkdcos)ndλ⎣⎦k=1k90on−120⎡⎤nk−.5π4⎡15Ω=M2⎢⎥0.125ndππλλ+∑sin(0.25kd)=+2⎢sin(0.125)πndλ⎣⎦k=1k1616×0.5⎣11413121110++++sin(0.25)πππsin(0.375)sin(0.5)sin(0.625)ππ+sin(0.75)2345698765+++++sin(0.875)πππππsin(1.00)sin(1.25)sin(1.375)sin(1.5)78910114321⎤++++sin(1.625)ππππsin(1.75)sin(1.875)sin(2.00)⎥12131415⎦Ω=0.0982+0.03125[5.740+4.950+4.003+3.000+2.033+1.179+0.492M+0–0.550–0.554–0.455–0.308–0.163–0.055+0]=0.0982+0.0312519.312×=0.702sr=ΩMεπ=Ω/Ω=0.702/(/4)=0.894(ans.)MMABygraphicalintegration(seeExample4-5.6andFig.4-8b)εwasfoundtobeMapproximately0.90,ingoodagreementwiththeaboveresult.Thegraphicalintegrationtookafractionofthetimeoftheaboveanalyticalintegrationandalthoughlessaccurate,providedconfidenceintheresultbecauseitismuchlesssusceptibletogrosserrors.课后答案网(e)Asnotedin(c),D=16(ans.)2222(f)FromDA=4/πλ,AD===λπλπ/416/41.27λ(ans.)emwww.hackshp.cnem*16-8-6.Four-towerbroadcastarray.Abroadcastarrayhas4identicalverticaltowersarrangedinaneast-westlinewithaspacingdandprogressivephaseshiftδ.Find(a)dand(b)δsothatthereisamaximumfieldatφ=45°(northeast)andanullatφ=90°(north).Therecanbeothernullsandmaxima,butnomaximumcanexceedtheoneat45°.Thedistancedmustbelessthanλ/2.Solution:φWEd 117*16-8-6.continued(a)oooNullatφ=90requiresthatδ=±90or180±Formaximumfield(fieldsofalltowersinphase)setoψβ=+dcosφδ==0andδ−90=−π/2radmaxδπ/2sod=−==0.354λ(ans.)oβφcos(2/)cos45πλmaxoπIfδπ=−180=−rad,d==0.707,λbutthisexceeds0.5λo(2/)cos45πλo(b)Therefore,δ=−90(ans.)16-10-1.Eight-sourcescanningarray.Alinearbroadsidearrayhas8sourcesofequalamplitudeandλ/2spacing.Findtheprogressivephaseshiftrequiredtoswingthebeam(a)5°,(b)10°and(c)15°fromthebroadsidedirection.(d)FindBWFNwhenallsourcesareinphase.Solution:o课后答案网φ=90oγ=0γφddwww.hackshp.cnoBroadsideissetatφ=90.Setψβ=+dcosφδ=0max22ππλooTherefore,δφ=−dcos=−cos95=+15.7(ans.)maxλλ2oooδ=−180cos85=−15.7oooThus,dependingonwhetherδis+15.7or15.7,−thebeamis5leftorrightofbroadside.Inthesameway,wehave 11816-10-1.continuedoo(b)δ=±31.3(ans.)forbeam10leftorrightofbroadsideoo(c)δ=±46.6(ans.)forbeam15leftorrightofbroadside(d)From(5-7-7)theangleofthefirstnullfrombroadside,whenthesourcesarein-phase(0δ=),isgivenbythecomplementaryangle−−11oγφλ=−=90sin(/nd)=sin(1/4)14.48=oooTherefore,BWFN=×214.48=28.96≅29(ans.)Fromthelongbroadsidearrayequation(5-7-10),ooBWFN≅===2/λπnd1/2rad180/228.65TheHPBWisabitlessthanBWFN/2.Forlongbroadsidearrays,wehavefromTable5-8(p.155)thatooHPBW===50.8/L50.8/3.514.5λ*16-16-1.TerminatedV.Travelingwave.(a)Calculateandplotthefar-fieldpatternofaterminated-Vantennawith5λlegsand45°includedangle.(b)WhatistheHPBW?课后答案网Solution:(a)ThefieldpatternforeachlegoftheVisshownattheleftandthecombinedfieldpatternattheright.MinorlobesareneglectedexceptfortheprincipalsidelobeoftheV.www.hackshp.cno(b)HPBW≅17(ans.) 119*16-16-2.E-typerhombic.DesignamaximumE-typerhombicantennaforanelevationangleα=17.5°.Solution:FromTable16-1(p.590)foramaximumErhombic,oH==1/(4sin)1/(4sin17.5)0.83(α=ans.)λooφα=−=9072.5(ans.)2La==0.5/sinα5.5(ns.)λ16-16-3.Alignmentrhombic.Designanalignment-typerhombicantennaforanelevationangleα=17.5°.Solution:FromTable16-1(p.590)foranalignmentrhombic,oH==1/(4sin17.5)0.83(ans.)λooφα=−=9072.5(ans.)2L==0.371/sinα4.1(ans.)λ课后答案网*16-16-4.Compromiserhombic.Designacompromise-typerhombicantennaforanelevationangleα=17.5°ataheightabovegroundofλ/2.www.hackshp.cnSolution:FromTable16-1(p.590)foracompromiserhombic,H=0.5(ans.)λoooφ=−=9017.572.5(ans.)2otan[(πL)sin17.5]⎛⎞10.5λL=−⎜⎟λooosin17.5⎝⎠2sin17.5ππtan(sin17.5) 120*16-16-4.continuedorLλ=0.56otan(16.3L)λBytrialanderror,L=5.14(ans.)λ16-16-5.Compromiserhombic.Designacompromise-typerhombicantennaforanelevationangleα=17.5°withleglengthof3λ.Solution:FromTable16-1(p.590),oH==1/(4sin17.5)0.83(ans.)λ−1o⎛⎞30.371−φ==sin⎜⎟67(ans.)o⎝⎠3cos17.5*16-16-6.Compromiserhombic.Designacompromise-typerhombicantennaforanelevationangleα=17.5°ataheightabovegroundofλ/2andaleglengthof3λ.Solution:课后答案网FromTable16-1(p.590,bottomentry),HL1λλ=−sintanφαπtan(2Hsin)απ4ψtan(2ψπL)www.hackshp.cnλλwhereψφ=−(1sincos)/2αoBytrialanderror,φ≅60(ans.) 121Chapter17.LensAntennas17-2-1.Dielectriclens.(a)Designaplano-convexdielectriclensfor5GHzwithadiameterof10λ.ThelensmaterialistobeparaffinandtheFnumberistobeunity.Drawthelenscrosssection.(b)Whattypeofprimaryantennapatternisrequiredtoproduceauniformaperturedistribution?Solution:89()aλ=×310/(510)0.06m60mm×==FL==1sod==10λ600mm(d=diameter)n=1.4(seeTable17-1)Thereforefrom(17-2-7),(1.41)600−R=1.4cosθ−1θRRsinθo0600mm0mmo10634110o20761260o22805∼300=d/2课后答案网Rd/2300mm=θ=22owww.hackshp.cnFocusL=600mmLens(lower-halfmirrorimage)(b)From(17-2-14),powerdensityatedgeoflensiso3S(1.4cos22−1)θ==0.35or4.6dBdown2oS(1.41)(1.4cos22)−−o 12217-2-1.continuedToreducesidelobes,thismuchorevenmoretapermaybedesirable.Toobtainauniformaperturedistribution,asrequestedintheproblem,requiresafeedantennaattheofocuswithmoreradiation(upabout4.6dB)at22offaxisthanonaxis.Thisisdifficulttoachievewithoutunacceptablespilloverunlessthelensisenclosedinaconicalhorn,exceptthatatedgelocationswhereEisparalleltotheedge,Emustbezero.Toreducethiseffectacorrugatedhorncouldbeused.17-3-1.Artificialdielectric.Designanartificialdielectricwithrelativepermittivityof1.4foruseat3GHzwhentheartificialdielectricconsistsof(a)copperspheres,(b)copperdisks,(c)copperstrips.Solution:FromTable17-2,3(a)επ(sphere)14=+Na=1.4r81−9At3GHz,λ=×310ms/310Hz×=0.1m100mm=Fora<<λ,takea(radius)=5mmfromwhich1.41−0.4−3N===255,000m(ans.)33−344ππa(5×10)课后答案网−63Thedielectricvolumepersphere=1/255,000=×410mWhilethevolumeofeachsphereisgivenby33−−373(4/3)ππa=×(4/3)(510)=5.210m×www.hackshp.cn−6volumeofdielectric410×Therefore,==7.7−7volumeofsphere5.210×−−6132(410)×=1.5910×=15.9mm=sideofcubeversusspherediameter=2510mm×=Thus,thereis15.9−=105.9mmbetweenadjacentspheresinacubicallatticesothereisroomforthesphereswithouttouching,providedthelatticeuniform. 12317-3-1.continued3(b)ε(discs)=1+5.33Na,andtakinga(radius)=5mm(diameter=10mm),r1.41−−3N==600,000m(ans.)−335.33(510)×−63Thedielectricvolumeperdisc==1/600,0001.710m×foracubesidelengthof−613(1.710)×≅12mm,sothatthereis1210−=2mmminimumspacingbetweenadjactentdiscsinauniformlattice.2(c)ε(strips)=1+7.85NwrTakingw(width)=10mm,1.41−−2N==51,000m(ans.)−237.85(10)asviewedincrosssection(seeFig.17-8a).Thesquareareaperstripisthen1/51,000−52−512=×210mforacross-sectionalareasidelength(210)×=4.5mm.Thisislessthanthestripwidth.However,iftheCrosssectionsquareischangedtoarectangleofthesameareawithsidelengthratioof9asinthesketch,theedgesofthestripsareseparatedby3.5mmandtheflatsidesby1.5mm.10mm课后答案网TheaboveanswersarenotuniqueandarenotRectangularnecessarilythebestsolutions.areaStripwww.hackshp.cn*17-4-1.Unzonedmetal-platelens.Designanunzonedplano-concaveE-planetypeofmetalplatelensoftheunconstrainedtypewithanaperture10λsquareforusewitha3GHzlinesource10λlong.Thesourceistobe20λfromthelens(F=2).Maketheindexofrefraction0.6.(a)Whatshouldthespacingbetweentheplatesbe?(b)Drawtheshapeofthelensandgivedimensions.(c)Whatisthebandwidthofthelensifthemaximumtolerablepathdifferenceisλ/4? 124*17-4-1.continuedSolution:89λ=×310/(310)×=0.1m100mm=nFA===0.6,2soL/2(Fig.17-13)(b)Expressingdimensionsinλ,wehavefrom(17-4-4)(1−nL)(10.6)20−8λR===λ1−−−ncosθθθ10.6cos10.6cosθRsinRθλλo0200o1019.63.4o15.2519.05.0212212(a)From(17-4-2),nb=−[1(λλtoo/2)]orbn=/2(1−)For(ans.)inesourcewithR⊥toline(c)From(17-4-12),5λLens2nδBandwidthλ(lower-half2Eθ=15.25oL=20λ(1−nt)λmirrorimage)otLR=−cosθ=−2019cos15.25=1.67λλλ课后答案网212212(a)From(17-4-2),nb=−[1(λλ/2)]orbn=/2(1−)ooFornb===0.6,0.625λ62.5mm(ans.)owww.hackshp.cn2(b)From(17-4-12),Bandwidth=−2/nnδ(1)tλλotLR=−cosθ=−2019cos15.25=1.67λλλ20.60.25××Therefore,Bandwidth==0.28or28%(ans.)2(10.6)1.67− 125Chapter18.Frequency-SelectiveSurfacesandPeriodicStructures.ByBenA.Munk18-9-1.Unloadedtripole.Determinetheapproximatelengthofthelegsofanunloadedtrislotoperatingatf=15GHzwith(a)Nodielectricsubstrate.(b)Dielectricsubstrateεr=2.2andthickness0.50mmlocatedonbothsidesoftheFSS(usearithmeticaverageofεrandεoforεeff).(c)DetermineDxjustshortenoughthatnogratinglobesarepresentwhenscanninginthexy–planeforanyangleofincidence.Solution:(a)Foronelegofthetripole,i.e.,themonopolelengthλ2.0==0.5cm(ans.)44(b)ε=2.2sinceitisthesameonbothsideseff22λ===1.35cmeff2.21.48λ1.35eff==0.337cm(ans.)44课后答案网(c)Forthescatteringcase,(15-6-1)canbewrittenasmsinηη+=sinisD/λxwww.hackshp.cnoSincegratinglobesstartintheplaneofthearray,ηη==90andm=1isSo2/,==λλDD/22=/21=cm(ans.)xx18-9-2.Four-Leggedloadedelement.Determinetheapproximatedimensionsforafourleggedloadedelementoperatingatf=15GHzwith(a)Nodielectricsubstrate.(b)Dielectricsubstrateεr=2.2andthickness0.50mmlocatedononlyonesidetheFSS(estimateεeff). 12618-9-2.continued(c)Leaveaseparationof1mmbetweenadjacentelements(rectangulargrid);determinethelowestonsetfrequencyforgratinglobesforanyangleofincidence.Solution:(a)Foralooptypeelement,thesizeshouldbeλ/4across.effSoλ/42.0/40.50cm==(ans.)2.21.0+3.2(b)ε===1.6,λ===2/1.62/1.2651.58cmeffeff22λ/40.4cm=(ans.)eff(c)AsinProb.18-9-1,theconditionwewanttomeetis2/o==λλDDr2xxWithnodielectric,D=+=0.50.10.6cmx8310×soλ==1.2cm,f=25GHz(ans.)−21.210×Withdielectric,D=+=0.40.10.5cm课后答案网x8310×soλ==1cm,f−2=30GHz(ans.)110×www.hackshp.cn 127Chapter19.PracticalDesignConsiderationsofLargeApertureAntennas*19-1-3.Efficiencyofrectangularaperturewithpartialtaper.Calculatetheapertureefficiencyanddirectivityofanantennawithrectangularaperturex1y1withauniformfielddistributionintheydirectionandacosinefielddistributioninthexdirection(zeroatedges,maximumatcenter)ifx1=20λandy1=10λ.Solution:FromProb.19-1-6solution,(a)ε=0.81or81%(ans.)ap(b)D=×××=4π10200.812036or33dBi(ans.)*19-1-4.Efficiencyofrectangularaperturewithfulltaper.RepeatProb.19-1-3forthecasewheretheaperturefieldhasacosinedistributioninboththexandydirections.Solution:FromProb.19-1-7solution,(a)0.657ε=≅66%(ans.)ap课后答案网(b)D=×××4π10200.6571651or32dBi=(ans.)19-1-5.Efficiencyofaperturewithphaseripple.www.hackshp.cnAsquareunidirectionalaperture(x1y1)is10λonasideandhasadesigndistributionfortheelectricfieldwhichisuniforminthexdirectionbuttriangularintheydirectionwithmaximumatthecenterandzeroattheedges.Designphaseisconstantacrosstheaperture.However,intheactualaperturedistributionthereisaplus-and-minus-30°sinusoidalphasevariationinthexdirectionwithaphasecycleperwavelength.Calculate(a)thedesigndirectivity,(b)theutilizationfactor,(c)theactualdirectivity,(d)theachievementfactor,(e)theeffectiveapertureand(f)theapertureefficiency.Solution:ReferringtoSec.19-1, 12819-1-5.continuedLetExy′(,)==Exy(,)1maxmaxDesignfieldActualfield1Design:E′′=Exydxdy(,)(1)av∫∫Ap2110λλ5yE==∫∫dxdyA0052λp10λyNote:Thisresultcanbededuceddirectlyfromthefigurebynotingthataverageheightoftriangleis½max.10λx(b)Utilizationfactor,k:u1k=(2)u*1(⎛⎞Exy′′,)(⎛⎞Exy,)∫∫⎜⎟⎜⎟dxdyAEE⎝⎠′′⎝⎠pavav13==(ans.)(3)21210λλ5⎛⎞y42∫∫⎜⎟dxdyA(1/2)00⎝⎠5λp课后答案网Notethatforin-phasefields(19-1-50)isasimplifiedformof(2)giving22E(1/2)av==3/4=kasin(3)(4)2u()E1/3www.hackshp.cnav(a)Designdirectivity,D(design):44ππ2DA(design)==k(100λ)(3/4)=940(ans.)(5)22puλλTurningattentionnowtotheeffectofthephasevariation:2210λλ⎛⎞ππxy51Ed==cos⎜⎟sinxdy(0.933)(6)av∫∫A00⎝⎠65λλ2p 12919-1-5.continuedExEExy(,)av10.866+Notethatfromfiguresabove,20E≅=.933av2(d)Achievementfactor,:ka∗1(⎛⎞Exy′′,)(⎛⎞Exy,)∫∫⎜⎟⎜⎟dxdyAEE⎝⎠′′⎝⎠pavavk=a∗1(⎛⎞Exy,)(⎛⎞Exy,)∫∫⎜⎟⎜⎟dxdyAEEpa⎝⎠va⎝⎠v4/3ka==0.87(ns.)(7)a11课后答案网22(/5)ydλxdy2∫∫A[(1/2)0.933]p*whereExEx()()1=www.hackshp.cnNotethatgainlossduetototalphasevariationacrossaperture(notsurfacedeviation)isfrom(19-2-3)oo2o⎛⎞δ′δ′30×0.70721.2k=cos⎜⎟360,where==goo⎝⎠λλ3603602oorkk==cos21.20.87=asin(7)ga19-1-5.continued 130(c)Directivity:4πA3pDk==k4π×100××0.87=818(ans.)2uaλ4(e)Effectiveaperture,:Ae2λ2A===DAkk65.2λ(ans.)epua4π(f)Apertureefficiency,:εapε==kk0.65(ans.)apauNote:Althoughphaseerrorswithsmallcorrelationdistance()≅λasinProb.19-1-5reducethedirectivityand,hence,increaseΩ,theHPBWisnotaffectedappreciable.AHowever,forlargercorrelationdistances()>>λthescatteredradiationbecomesmoredirective,causingthenearsidelobestoincreaseandultimatelythemainbeamandtheHPBWmaybeaffected.*19-1-6.Rectangularaperture.Cosinetaper.Anantennawithrectangularaperturex1y1hasauniformfieldintheydirectionandacosinefielddistributioninthexdirection(zeroatedges,maximumatcenter).Ifx1=16λandy1=8λ,calculate(a)theapertureefficiencyand(b)thedirectivity.Solution:课后答案网yy1y1=8λx=www.hackshp.cn16λ1EExxo1EEoπxExE()=sinox1*19-1-6.continued 131Althoughthetaperinthex-directionisdescribedasacosinetaper,letusrepresentitbyasinefunctionasfollows:2()Ex()av(a)From(19-1-50),ε=ap2()Ex()avx11xx11EEππxx⎛⎞−x2Eoo1owhereEx()==Exdx()sindx=⎜⎟cos=av∫∫xx00xxx⎝⎠ππ11111022221xx11EEoo2πx[()Ex]==Exdx()sindx=av∫∫xx00x21112⎛⎞2⎜⎟Eo⎝⎠π8Therefore,ε===≅0.811or81%ap122πEo22(b)AA==×ελ0.818×16λ=103.7λeapem4πAe22D==×(4πλ103.7)/λ=1304or31.2dBi2λ19-1-7.Rectangularaperture.Cosinetapers.课后答案网RepeatProb.19-1-6forthecasewheretheaperturefieldhasacosinedistributioninboththexandydirections.Solution:www.hackshp.cnLetthedistributionberepresentedbyππxxExyE(,)=sinsinoxy11 13219-1-7.continued1xy11Eox1ππxy1y(a)(,)Exy==Ex(,)ydxdysindxsindyavxy∫∫00xy∫0x∫0y111111xy11EE⎡⎤xyππxy⎡⎤4oo11=−⎢⎥cos⎢⎥−cos=2xyπππxy11⎣⎦100⎣⎦1222221xy11EEoox1ππxyy12[(,)]Exy==Exydxdy(,)sindxsindy=avxy∫∫00xy∫0x∫0y4111111Therefore,2⎛⎞4⎜⎟2Eo⎝⎠π164×ε===≅0.657or66%ap142πEo42(b)AA==×ελ0.6578×16λ=84.1λeapem24πA48πλ×4.1eD===1057or30.2dBi22λλ*19-1-8.A20λlinesource.Cosine-squaredtaper.(a)Calculateandplotthefar-fieldpatternofacontinuousin-phaselinesource20λlongwithcosine-squaredfielddistribution.课后答案网(b)WhatistheHPBW?Solution:www.hackshp.cnE(x)Thefieldalongthelinemaybe−x+x1representedby12πxEx()cos=20λ2x1(a)ThefieldpatternE()θistheFouriertransformofthedistributionEx()alongtheline.Thus,++x1jx(2/)cosπλθ10λ2jx(2/)cosπλθE()θπ==∫∫Exe()dxcos[(/2)(/10)]xλedx−−x110λ 133*19-1-8.continuedLetsx=/λfromwhichdx=λdsThen++102js2πθcos101cos(+πs/10)js2πθcosEs()θλ==∫∫cos(/20)πedsλeds−−10102λλ++1010js2cπθosjs2cπθos=+∫∫edscos(πsed/10)s22−−1010andsin(20cos)πθ1sin(20cos⎡θπ+−1)sin(20cosθπ1)⎤E()θ=+⎢+⎥n2cosπθ2[2cos⎣θ+−(1/10)]π[2cosθ(1/10)]π⎦2sin(20cos)πθ⎛⎞4cosθ=−⎜⎟1(ans.)(1)22cosπθ⎝⎠4cosθ−0.01(b)Fromgraphorbytrialanderrorfrom(1),oooHPBW=−2(9087.9)=4.2(ans.)FromTable4-3fora20λuniformaperture,oHPBW=50.8/L==50.8/202.5λThus,thecosine-squaredaperturedistributionhasnearlytwicetheHPBWoftheuniformaperture,butitssidelobesaremuchlowerw课后答案网ithfirstsidelobedown31dBascomparedtoonly13dBdownfora20λuniformaperturedistribution.www.hackshp.cn 134课后答案网www.hackshp.cn 135Chapter21.AntennasforSpecialApplications21-4-2.Horizontaldipoleaboveimperfectground.Calculatetheverticalplanefieldpatternbroadsidetoahorizontalλ/2dipoleantennaλ/4-3-1-1aboveactualhomogeneousgroundwithconstantsεr’=12andσ=2x10Ωmat(a)100kHzand(b)100MHz.Solution:−−−311µµ===,1εσ′2,2×10mΩ,h=λ/4or212sinαε−−(cosα)rρ=(1)⊥212sinαε+−(cosα)rE=+1ρ__cos(2βαhjhsin)+sin(2βαsin)(2)⊥⊥∠−3σ210×(b)ε′′===0.36at100MHzr81−2ωε2108.8510π×oεεε=−=−≅′′jj′120.412rrrIntroducingεinto(1),(1)into(2)andevaluating(2)asafunctionofαresultsintherpatternshown.Thepatternforperfectlyconductingground()σ=∞isalsoshownforcomparison(sameaspatternof2isotropicsourcesinphaseoppositionandspacedλ/2).oForperfectlyconductinggroundthefielddoubles(课后答案网E=2)atthezenith(α=90),butwiththeactualgroundoftheproblem,itisreducedtoabout1.55(down2.2dB)becauseofpartialabsorptionofthewavereflectedfromtheground.www.hackshp.cn(a)At100kHz,ε′′=≅360andρ−1,sothepatternisapproximatelythesameasforr⊥σ=∞inthesketch. 13621-9-1.Squareloop.Calculateandplotthefar-fieldpatternintheplaneofaloopantennaconsistingoffourλ/2center-feddipoleswithsinusoidalcurrentdistributionarrangedtoformasquareλ/2onaside.Thedipolesareallinphasearoundthesquare.Solution:Squarishpatternwithroundededges.Maximum-to-minimumfieldratio=1.14*21-9-3.DFandmonopulse.Manydirection-finding(DF)antennasconsistofsmall(intermsofλ)loopsgivingafigure-of-eightpatternasinFig.P21-9-3a.Althoughthenullissharpthebearing(directionoftransmittersignal)mayhaveconsiderableuncertaintyunlesstheS/Nratioislarge.Toresolvethe180°ambiguityofthelooppattern,anauxiliaryantennamaybeusedwiththelooptogiveacardiodpatternwithbroadmaximuminthesignaldirectionandnullintheoppositedirection.Themaximumofabeamantennapattern,asinFig.P21-9-3b,canbeemployedtoobtainabearingwiththeadvantageofahigherS/Nratiobutwithreducedpatternchangeperunitangle.However,if2receiversand2displacebeamsareused,asinFig.P21-9-3c,alargepower-patternchangecanbecombinedwithahighS/Nratio.Anarrangementofthiskindforreceivingradarechosignalscangivebearinginformationonasinglepulse(monopulseradar).Ifthepowerreceivedonbeam1isP1andonbeam2isP2,thenifP2>P1thebearingistotheright.IfP1>P2thebearingistotheleftandifP1=P2thebearingisonaxis(boresight).(With4antennas,bearinginformationleft-rightandup-downcanbeobtained.)课后答案网4(a)Ifthepowerpatternisproportionaltocosθ,asinFig.P21-9-3c,determineP2/P1iftheinterbeam(squint)angleα=40°for∆θ=5and10°.(b)Repeatforα=50°.www.hackshp.cn(c)DeterminetheP0/P1ofthesinglepowerpatternofFig.P21-9-3bfor∆θ=5and10°if4thepowerpatternisalsoproportionaltocosθ.(d)Tabulatetheresultsforcomparisonandindicateanyimprovementofthedoubleoverthesinglebeam. 137*21-9-3.continuedFigureP21-9-3.Directionfinding:(a)withloopmull,(b)withbeammaximumand(c)withdoublebeam(monopulse).Solution:o(a)α=404oooP2cos(20−5)∆=θ5,==1.290or1.1dB4ooPcos(20+5)14oooP2cos(20−10)∆=θ10,==1.672or2.2dB4ooPcos(20+10)1o(b)α=50课后答案网4oooP2cos(25−5)∆=θ5,==1.386or1.4dB4ooPcos(25+5)14oooP2cos(25−10)∆=θ10,==1.933or2.9dB4ooPcos(25+10)www.hackshp.cn14ooP0cos0(c)α==5,=1.015or0.06dB4oPcos514ooP0cos0α==10,=1.063or0.26dB4oPcos101oo(d)Over1dBmoreat5andabout2dBmoreat10. 138*21-10-1.OverlandTVforHP,VPandCP.(a)AtypicaloverlandmicrowavecommunicationscircuitforAM,FMorTVbetweenatransmitteronatallbuildingandadistantreceiverinvolves2pathsoftransmission,onedirectpath(lengthro)andoneanindirectpathwithgroundreflection(lengthr1+r2),assuggestedinFig.P21-10-1.Leth1=300mandd=5km.Forafrequencyof100MHzcalculatetheratioofthepowerreceivedperunitareatothetransmittedpowerasafunctionoftheheighth2ofthereceivingantenna.Plottheseresultsindecibelsasabscissaversush2asordinatefor3caseswithtransmittingandreceivingantennasboth(1)verticallypolarized,(2)horizontallypolarizedand(3)right-circularlypolarizedforh2valuesfrom0to100m.Assumethatthetransmittingantennaisisotropicandthatthereceivingantennasarealsoisotropic(allhavethesameeffectiveaperture).Considerthatthegroundisflatandperfectlyconducting.(b)Comparetheresultsforthe3typesofpolarization,andshowthatcircularpolarizationisbestfromthestandpointofboththenoncriticalnessoftheheighth2andtheabsenceofechoorghostsignals.Thus,forhorizontalorverticalpolarizationthedirectandground-reflectedwavesmaycancelatcertainheightswhileatotherheights,wheretheyreinforce,theimagesontheTVscreenmaybeobjectionablebecausethetimedifferenceviathe2pathsproducesadoubleimage(adirectimageanditsghost).(c)Extendthecomparisonof(b)toconsidertheeffectofotherbuildingsorstructuresthatmayproduceadditionalpathsoftransmission.Notethatdirectsatellite-to-earthTVdownlinksaresubstantiallyfreeofthesereflectionandghostimageeffects.课后答案网www.hackshp.cnFigureP21-10-1.Overlandmicrowavecommunicationcircuit.Solution:(a)and(b)answersinAppendixF,pg.919-920.(c)Theeffectofreflectionfromotherbuildingsorstructures(orfromaircraft)canbeminimizedbytheuseofCPtransmitandreceiveantennasofthesamehand,particularlywhenthesestructuresaremanywavelengthsinsizeandreflectionisspecular.Trouble-somereflectionscanbereducedbyplacingnon-reflectingabsorbersonthestructure. 139*21-12-1.Signalingtosubmergedsubmarines.-1Calculatethedepthsatwhicha1µVmfieldwillbeobtainedwithEatthesurface-1equalto1Vmatfrequenciesof1,10,100and1000kHz.Whatcombinationoffrequencyandantennasismostsuitable?Solution:FromTableA-6,takeεσ′==80and4forseawater.Atthehighestfrequency(1000rkHz),σω>>ε,sothatαω=µσ/2canbeusedatallfourfrequencies.At1kHz,37−21ππ××××04104−1α==0.13Npm2SinceE−−6αy613.8==10eye,=log=Eααoandat1kHZ,depthy=106mat10kHz,y=35m(ans.)at100kHz,y=11mat1000kHzy=3.5mFromthestandpointoffrequency,1kHzgive课后答案网sgreatestdepth.However,from(21-2-3)theradiationresistanceofamonopoleantennaasafunctionofitsheight()hisp2⎛⎞hpR=Ω400⎜⎟rwww.hackshp.cn⎝⎠λForh=300mat1kHzp2⎛⎞300−4R==400⎜⎟410(or400µ)×ΩΩr5⎝⎠310×Withsuchasmallradiationresistance,radiationefficiencywillbepoor.At10kHztheradiationresistanceisahundredtimesgreater.Apracticalchoiceinvolvesacompromiseofseawaterloss,land(transmitting)antennaeffectiveheight,andsubmarineantennaefficiencyasafunctionofthefrequency. 140*21-13-1.Surface-wavepowers.-3A100-MHzwaveistravelingparalleltoacoppersheet(|Zc|=3.7x10Ω)withE(=-1100Vmrms)perpendiculartothesheet.Find(a)thePoyntingvector(wattspersquaremeter)paralleltosheetand(b)thePoyntingvectorintothesheet.Solution:22Ey100−2(a)S===26.5Wm(ans.)&tosheetZ377o232−2E⎛⎞1003.710×−2(b)SH===RZRZ⎜⎟=182µWm(ans.)intosheetec2ecZo⎝⎠377221-13-2.Surface-wavepowers.A100-MHzwaveistravelingparalleltoaconductingsheetforwhich|Zc|=0.02Ω.IfE-1isperpendiculartothesheetandequalto150Vm(rms),find(a)wattspersquaremetertravelingparalleltothesheetand(b)wattspersquaremeterintothesheet.Solution:22Ey150−2(a)S===59.7Wm(ans.)&tosheetZ377o222−2E⎛⎞150210×−2(b)S===HRZRZ⎜⎟=2.24mWm(ans.)intosheetec课后答案网2ecZo⎝⎠3772*21-13-3.Surface-wavepower.www.hackshp.cnAplane3-GHzwaveinairistravelingparalleltotheboundaryofaconductingmedium7withHparalleltotheboundary.Theconstantsfortheconductingmediumareσ=10-1-1-1Ωmandµr=εr=1.Ifthetraveling-wavermselectricfieldE=75mVm,findtheaveragepowerperunitarealostintheconductingmedium.Solution:22ES==HRZRZintosheetec2ecZo−79µω41ππ××0231×0oRZ===Ω0.034ec722σ×10 141*21-13-3.continued2⎛⎞0.75−2Therefore,S==⎜⎟0.0341.35nWm(ans.)intosheet⎝⎠37721-13-4.Surface-wavecurrentsheet.ATEMwaveistravelinginairparalleltotheplaneboundaryofaconductingmedium.ShowthatifK=ρsv,whereKisthesheet-currentdensityinamperespermeter,ρsisthesurfacechargedensityincoulombspersquaremeterandvthevelocityofthewaveinmeterspersecond,itfollowsthatK=H,whereHisthemagnitudeoftheHfieldofthewave.Solution:QmQ1−−11Kv==ρ==Am=Ims2mssmByAmperes’slaw,integralofHaroundstripofwidthwequalscurrentenclosedor课后答案网v∫Hids=Iw=KwH==wKandHK(notethatH⊥K)(ans.)*21-13-6.Coated-surfacewavecutoff.www.hackshp.cnAperfectlyconductingflatsheetoflargeextenthasadielectriccoating(εr=3)ofthicknessd=5mm.FindthecutofffrequencyfortheTMo(dominant)modeanditsattenuationperunitdistance.Solution:28π.89−1α=−31=Npm(ans.)f=0(ans.)cλλoo 142课后答案网www.hackshp.cn 143Chapter23.Baluns,etc.ByBenA.Munk23-3-1.Balun200Ω,antenna70Ω.ATypeIIIbalunhasthecharacteristicimpedanceequaltoZcp=200Ωandtheelectricallengthisequaltolp=7.5cm.ItisconnectedtoanantennawithimpedanceZA=70Ω.(a)FindthebalunimpedancejXpatf=500,1000and1500MHz.(b)CalculatetheparallelimpedancesZA||jXpat5001000and1500MHzandplottheminaSmithChartnormalizedtoZo=50Ω.Checkthatalltheseimpedanceslieonacirclewithadiameterspanningover(0,0)andZA=70Ω.Alternatively,youmaydetermineZA||jXpgraphicallyinaSmithChart.(c)ExplainwhateffectitwouldhaveonthebandwidthifwechangedZcpto150Ωor250Ω.Solution:8A310×p7.5(a)λ==60cm,==0.125L8510×λ60L8A310×p7.5λ==30cm,==0.250M9110×λ30M8A310×p7.5λ==20cm,==0.375H91.510×λ20HFromtheSmithChart,bymoving课后答案网thenumberofwavelengthsaroundfromtheshort(zero)position,itisfoundthatforfj==500MHz,Xj200Ω(ans.)pforfj==1000MHz,Xj∞Ω(ans.)www.hackshp.cnpforfj==1500MHz,X−j200Ω(ans.)pAlternatively,thetransmissionlineequationcanbeused.22ZjXZX+jZXApApAp(b)Zj&X==Ap22Zj++XZXApApFormidfrequency,X=∞,ZXZ&=pApA 14423-3-1.continued2270200×+×j70200Forlowfrequency,XZ==200,&jX=62.36+j21.83pAp2270+200Zj&XApNormalizedtoZ=50,=+1.25j0.44oZoSimilarly,forhighfrequency,XZ=−200,&jXj=62.36−21.83pApZj&XAp=−1.25j0.44ZoSeeaccompanyingfigureofSmithChart课后答案网www.hackshp.cn 14523-3-1.continuedTofindthesevaluesbytheSmithChart,itisamatterofaddingthevaluesasadmit-tances.Thisisaccomplishedbyfindingtheirpositionasimpedances,projectingthevaluesthroughtheoriginanequaldistance,addingthem,thenprojectingtheaddedvaluesanequaldistancetotheothersideoftheorigin.(c)ForZZ=Ω150,itisfoundthat&±=±jX57.48j26.82cpApornormalizedas=±1.15j0.54ForZZ=Ω250,itisfoundthat&±=±jX64.91j18.18cpApornormalizedas=±1.30j0.36Itiseasilyseenthatthe150Ωvaluedecreasesthebandwidthandthe250Ωvalueincreasesthebandwidth.Note:Thecloserthevaluesaretotheorigin,thebettertheVSWR.23-3-5Stubimpedance.(a)Whatistheterminalimpedanceofaground-planemountedstubantennafedwitha50-Ωair-filledcoaxiallineiftheVSWRonthelineis2.5andthefirstvoltageminimumis0.17λfromtheterminals?(b)DesignatransformersothattheVSWR=1.Solution:课后答案网ZTZ+jZtanβxTo.17λVSWR=2.5ZZm=o(1)Z+jZtanβxwww.hackshp.cnoTVminZo=50ΩwhereZV==impedanceonlineatR+j0mmminZj==lineimpedance50+0ΩoZ=+stubantennaterminalimpedance=RjXTTTRearranging(1)intermsofrealandimaginaryparts: 14623-3-5continued⎛⎞XRTmRR−=⎜⎟tanβxbyequatingreals,(2)mTR⎝⎠oandRRTmtanββxXR=+tanxbyequatingimaginaries(3)ToRooRR===50/2.520,50,tanβx=×tan(360.17)1.82=mo20From(2),20−=RX×=1.820.728XTTT5020From(3),RX×=+×1.82501.82,0.728R=+X91TTTT50Fromwhich,ZRj=+=−ΩX56j50(ans.)TTT课后答案网www.hackshp.cn 147Chapter24.AntennaMeasurements.ByArtoLehtoandPerttiVainikainen24-3-1.Uncertaintyofpatternmeasurementduetoreflectedwave.Thelevelofawavereflectedfromthegroundis45dBbelowthelevelofthedirectwave.Howlargeoferrors(indB)arepossibleinthemeasurementof:(a)mainlobepeak;(b)-13dBsidelobe;(c)-35dBsidelobe?Solution:−(45/20)FromSec.24-3bandsincethereflectedwaveis−=45dBor100.0056,(a)10.0056−=0.9944or0.049dB−(ans.)10.00561.0056or+0.049dB+=(ans.)−(13/20)(b)−=13dBsidelobesprovides100.22380.22380.0056−so=−0.9749or0.22dB(ans.)0.22380.22380.0056+=1.0251or+0.22dB(ans.)0.2238课后答案网−(35/20)(c)−=35dBsidelobesprovides100.01780.01780.0056−so=−0.6838or3.30dB(ans.)0.0178www.hackshp.cn0.01780.0056+=1.3162or+2.38dB(ans.)0.017824-3-2.Rangelengthrequirementduetoallowedphasecurvature.Themaximumallowedphasecurvatureinthemeasurementofaverylow-sidelobeantennais5°.Thewidthoftheantennais8manditoperatesat5.3GHz.FindtherequiredseparationbetweenthesourceandAUT. 14824-3-2.continuedSolution:dRD/2RSimilartoFig.24-5,letdbethedistancecausingthephaseerror.222⎛⎞DThen()Rd+=+R⎜⎟⎝⎠222222DDRd++=+2,RdRR≅48doFora5phaseerror,25πkd==dπ(rad)λ180d51so,==λ360722219DDTherefore,R==728λλ课后答案网8310××964Since,λ==0.0566m,R≥=10,176m95.310×0.0566www.hackshp.cn24-4-1.Designofelevatedrange.Designanelevatedrange(rangelength,antennaheights,sourceantennadiameter)forthemeasurementofa1.2mreflectorantennaoperatingat23GHz.Solution:8310×λ==0.013m102.310×2222D×(1.2)so,R≥==221m(ans.)λ0.013 14924-4-1.continuedFromcombiningrequirementsin(24-4-1)and(24-4-2)HD≅×=×=551.26m(ans.)RandsimilarlyforH=HTR1.5λR1.50.013221××From(24-4-1),D≥==0.72m(ans.)TH6R24-4-2.Timerequiredfornear-fieldscanning.Estimatethetimeneededforaplanarnear-fieldmeasurementofa2mantennaat300GHz.Thesamplingspeedis10samplespersecond.Solution:8310×2mλ==0.001m,D==2000λ11310×0.001mSampleat2perwavelength,sosamples=4000perlineperside326Totalsamples=2(410)3210××=×63210×6t==3.210sec×=≅888hrs54min37days(ans.)10samples/sec课后答案网24-5-1.Powerrequirementforcertaindynamicrange.www.hackshp.cnTheAUThasagainof40dBiat10GHz.Thegainofthesourceantennais20dBi.Theseparationbetweentheantennasis200m.Thereceiversensitivity(signallevelthatissufficientformeasurement)is–105dBm.Findtheminimumtransmittedpowerthatisneededforadynamicrangeof60dB.Solution:22⎛⎞⎛λ0.03⎞−10From(24-5-2)andsince⎜⎟⎜==⎟1.4210×=−98dB⎝⎠⎝44ππR×200⎠ 15024-5-1.continued⎛⎞PR⎜⎟=−40dBi+20dBi98dB=−38dBP⎝⎠TdBWith−105dBmneededataminimumforthereceptionanda60dBdynamicrange,thenP=−105dBm38dB60dB++=−7dBmtPa=0.2mW(ns.)t24-5-2.Gainmeasurementusingthreeunknownantennas.Threehornantennas,A,B,andCaremeasuredinpairsat12GHz.Theseparationofantennasis8m.Thetransmittedpoweris+3dBm.Thereceivedpowersare-31dBm,36dBm,and-28dBmforantennaspairsAB,AC,andBC,respectively.Findthegainsoftheantennas.Solution:−28P⎛⎞λ310×TFrom(24-5-2),GG==⎜⎟,λ=0.025mTR10PR⎝⎠41π.2×10R22⎛⎞⎛⎞λ0.025−8⎜⎟⎜⎟==6.1810or72dB×−⎝⎠⎝⎠44ππR8then课后答案网GG==C−31dBm3dBm72dB−+=38dBABABGG==C−36dBm3dBm72dB−+=33dBACACGG==C−28dBm3dBm72dB−+=41dBBCBCwww.hackshp.cnGCBABCAB⎛⎞CAB2==,,GG⎜⎟G=CBCCBCGCCCCACAC⎝⎠ACSoCC1BCACG==(41dB+33dB38dB)18dBi−=(ans.)CC2ABCACG==−=33dB18dB15dBi(ans.)AGC 15124-5-2.continuedCBCG==−=41dB18dB23dBi(ans.)BGC24-5-3.Gainmeasurementusingcelestialradiosource.At2.7GHztheantennatemperatureincreases50Kasa20mreflectorispointedtoCygnusA.Findtheantennagainandapertureefficiency.Solution:−238πkT∆A81π×××.3810505From(24-5-7),G===1.7910×=52.5dBi22−62Sλ78510××(0.111)252Gλ1.7910××(0.111)2A===175.5me44ππFora20mcircularreflector,A175.5eε===0.56or56%ap2Aπ(10)p24-5-4.Impedanceinlaboratory.Youtrytomeasuretheimpedanceofahorn课后答案网antennawith15dBigainat10GHzinanormallaboratoryroombypointingthemainlobeoftheantennaperpendicularlytowardsawall2maway.Thepowerreflectioncoefficientofthewallis0.3anditcanbeassumedtocoverpracticallythewholebeamoftheAUT.EstimatetheuncertaintyofthemeasurementofthereflectioncoefficientoftheAUTduetothereflectionofthewall.Solution:www.hackshp.cnThenormalizedreceivedpowerfromthehorntothewallandbackintothehorn2PR⎛⎞λ=ρGG⎜⎟TRPT⎝⎠4πR22⎛⎞⎛λ0.03⎞−7ρλ==0.3−5dB,=0.03m,⎜⎟⎜=⎟=×=3.610−64dB⎝⎠⎝4Rππ422××⎠ 15224-5-4.continuedPR=−5dB15dBi15dBi64dB++−=−39dBPT=0.000126inpower=0.01122involtageSotheuncertaintyisabout1%.课后答案网www.hackshp.cn 153INDEXIndexreadsasfollows:Entry(Problemnumber)pageADApertureDepolarizationratio(2-17-2)15withphaseripple(19-1-5)127Detectingoneelectron(12-5-5)97withtapereddistribution(19-1-3)127,Dielectric,artificial(17-3-1)122(19-1-4)127,(19-1-6)130,(19-1-7)131,Dipole(19-1-8)132electric(6-2-1)35Arrayshort(6-2-2)35,(6-2-4)36,(6-3-2)37,broadside(5-6-10)27,(16-2-1)109(6-3-4)38,(6-3-10)40sevenshortdipoles(16-6-4)113Directionfinding(21-9-3)136sixteensource(16-6-7)114Directionalpattern,withbacklobe(6-3-9)40end-fire(16-3-1)110Directionalpatterninθandφ(6−3−7)39,foursourcesinsquare(5-2-8)23(6−3−8),40ordinaryend-fire(5-6-9)25Directivegain(5-8-7)29square-cornerreflector(10-3-6)68Directivity(2-7-1)1,(2-7-2)1,(2-7-3)2,(2-7-4)three-source(5-9-2)303,(2-9-1)3,(2-9-3)4,(3-7-2)18,(4-5-1)19,threeunequalsources(5-8-1)29(4-5-2)20,(4-5-3)20,(4-5-4)21,(4-5-5)21,twelve-sourceend-fire(5-6-5)24,(5-6-7)25(5-2-4)23,(5-6-5)24,(5-6-9)25,(5-6-10)27two-element,unequalcurrents(16-4-3)111Dolph-Tchebyscheffarray(5-9-4)31,(16-6-1)two-sourceend-fire(5-2-4)23112two-sourcesinoppositephase(5-18-2)33Dynamicrange,powerrequirement(24-5-1)149Artificialdielectric(17-3-1)122EBEchelonarray(6-7-1)43Backpackingpenguin(12-3-4)90Eelevatedrangedesign(24-4-1)148Balun(23-3-1)143Effectiveaperture(2-9-2)4,(2-9-3)4Beamwidth(3-7-2)18Ellipticallypolarizedwave(2-16-4)9,(2-17-3)Broadcastarray,four-tower(16-8-6)116课后答案网10,(2-17-4)11,(2-17-8)13,(2-17-10)13BroadsidearrayEnd-firearray,twoelement(16-3-1)110twoelement(16-2-1)109sevenshortdipoles(16-6-4)113Fsixteensource(16-6-7)114Fieldpattern(5-5-1)24,(5-6-5)24two-sourcesinphase(5-18-1)32Forestabsorption(12-4-3)95Cwww.hackshp.cnFriisformula(2-11-1)4Carrier-to-noiseratio(C/N)(12-3-19)93Circularlypolarizedwave(2-16-2)8,(2-16-3)8,G(2-17-6)11,(2-17-7)12,(2-17-9)13,(2-17-11)14,Gain(2-7-4)3,(9-9-1)60Coated-surfacewavecutoff(21-13-6)141GainmeasurementConicalantenna(11-2-2)75usingcelestialradiosource(24-5-3)151Conicalpattern(6-3-5)38,(6-3-6)39usingthreeunknownantennas(24-5-2)150Criticalfrequency(12-3-9)86Galileospacecraft(12-3-22)94 154HMHelicalantennasMaximumuseablefrequency(MUF)(12-3-9)86axialmode(8-3-1)53,(8-3-2)53,(8-3-3)54,Microstripline(9-7-3)59(8-8-1)55,(8-15-1)56Minimumusablefrequency(mUF)(12-3-10)87normalmode(8-11-1)55Momentmethod,chargedrod(14-10-1)105Hornantenna(3-4-1)17,(3-4-2)17,(3-5-2)17,Monopulse(21-9-3)136(3-5-3)18Horns(9-9-1)60,(9-9-2)61,(9-9-3)62,(9-9-4)N62,(9-9-5)63Near-fieldscanning,timerequired(24-4-2)149IOImpedanceOverlandTV(21-10-1)1382-elementarray(16-3-2)1115λ/2antenna(13-4-1)103Pantennasinechelon(13-8-1)104dipole(14-12-2)105Patches,50and100Ω(9-7-1)59ImpedancePatternDolph-Tchebyscheffarray(16-6-1)112directionalwithbacklobe(6-3-9)40inlaboratory(24-5-4)151directionalinθandφ.(6-3-7)39,(6-3-8)40open-slot(9-5-3)58elements(15-6-1)107side-by-sideantennas(13-6-1)103,(13-6-3)factors(6-8-2)44104horn(9-9-2)61slot(9-5-1)57measurement,uncertainty(24-3-1)147stubantenna(23-3-5)145smoothing(15-3-1)107Isotropicantenna(6-3-1)37Patternsoverimperfectground(21-4-2)135Phasepattterns(5-5-1)24JPoincarésphere(2-17-5)11Polarization(2-16-4)9,(2-17-1)9,(2-17-2)10,Jupitersignals(12-4-4)95(2-17-3)10,(2-17-4)11,(2-17-5)11,(2-17-6)11,(2-17-7)12,(2-17-8)13,(2-17-9)13,L(2-17-10)13,(2-17-11)14,Lenses(17-2-1)121,(17-4-1)123课后答案网Poyntingvector(2-16-2)8,(2-16-3)8Linearlypolarizedwaves(2-17-2)10LinkQinterstellarwireless(12-3-13)89Quad-helixantenna(8-15-1)56MarsandJupiter(2-11-4)5Moon(2-11-5)7Rsatellite(12-3-4)82www.hackshp.cnspacecraft(2-11-2)5,(2-11-3)5Radarcrosssection(12-5-6)98spaceshipnearmoon(2-16-1)7Radardetection(12-5-1)96,(12-5-12)99,Loadedelement(18-9-2)125(12-5-13)99,(12-5-14)100,(12-5-18)100,Log-periodicantenna(11-7-1)77,(11-7-2)78(12-5-20)101Log-spiralantenna(11-5-1)75Radiationresistance(6-3-1)37,(6-3-11)41,Loop(6-3-12)41,(6-5-1)42,(6-6-1)42,(7-6-1)49λ/10(7-7-1)50Rangelength(24-3-2)1473λ/4(7-4-1)49RCSofelectron(12-5-3)96circular(7-8-2)51Reflectordirectivity(7-8-1)50flatsheet(10-2-1)65radiationresistance(7-6-1)49,(7-8-1)50parabolicwithmissingsector(10-7-2)73square(7-9-1)51,(7-9-2)52square-corner(10-3-1)66,(10-3-2)66,Loop-dipoleforCP(7-2-1)47(10-3-4)67,(10-3-5)68,(10-3-6)68,Lowearthorbitsatellite(12-3-17)90(10-3-7)71,(10-3-8)71Rhombic 155alignment(16-16-3)119Surface-wavecompromise(16-16-4)119,(16-16-5)120,currentsheet(21-13-4)141(16-16-6)120cutoff(21-13-6)141E-type(16-16-2)119powers(21-13-1)140,(21-13-2)140,(21-13-3)140STSatellitedownlink(12-3-4)82,SeealsoLinkTemperaturedirect-broadcast(DBS)(12-3-18)91withabsorbingcloud(12-4-1)94lowearthorbit(LEO)(12-3-17)90antenna(12-2-1)81,(12-2-2)81Scanningarrayminimumdetectable(12-3-11)88,(12-3-12)eight-source(16-10-1)11789Slots(9-2-1)57,(9-5-1)57,(9-5-2)58,(9-5-3)system(12-3-5)82,(12-3-6)8458Solarinterference(12-3-7)84Travelingwaveantennas(6-8-1)44Solarpower(4-3-1)19Squarearray(16-6-3)112,(16-6-5)114USquareloop(21-9-1)136Unloadedtripole(18-9-1)125Strayfactor(5-8-7)29Stubimpedance(23-3-5)145VSubmarines,communicationwith(21-12-1)139Vantenna(16-16-1)118课后答案网www.hackshp.cn'