JohnDKraus《天线第三版》课后习题答案(一)

Preface

ThisInstructors9ManualprovidessolutionstomostoftheproblemsinANTENNAS:

FORALLAPPLICATIONS,THIRDEDITION.Allproblemsaresolvedforwhich

answersappearinAppendixFofthetext,andinaddition,solutionsaregivenforalarge

fractionoftheotherproblems.Includingmultipleparts,thereare600problemsinthe

textandsolutionsarepresentedhereforthemajorityofthem.

Manyoftheproblemtitlesaresupplementedbykeywordsorphrasesalludingtothe

solutionprocedure.Answersareindicated.Manytipsonsolutionsareincludedwhich

canbepassedontostudents.

Althoughanobjectiveofproblemsolvingistoobtainananswer,wehaveendeavored

toalsoprovideinsightsastohowmanyoftheproblemsarerelatedtoengineering

situationsintherealworld.

TheManualincludesanindextoassistinfindingproblemsbytopicorprincipleand

tofacilitatefindingclosely-relatedproblems.

ThisManualwaspreparedwiththeassistanceofDr.ErichPacht.

ProfessorJohnD.Kraus

Dept,ofElectricalEngineering

OhioStateUniversity

2015NeilAve

Columbus,Ohio43210

Dr.RonaldJ.Marhefka

SeniorResearchScientist/AdjunctProfessor

TheOhioStateUniversity

ElectroscienceLaboratory

1320KinnearRoad

Columbus,Ohio43212

TableofContents

Prefaceiii

ProblemSolutions:

Chapter2.AntennaBasics1

Chapter3.TheAntennaFamily17

Chapter4.PointSources19

Chapter5.ArraysofPointSources,Part123

Chapter5.ArraysofPointSources,PartII29

Chapter6.TheElectricDipoleandThinLinearAntennas35

Chapter7.TheLoopAntenna47

Chapter8.End-FireAntennas:TheHelicalBeamAntennaandtheYagi-Uda

Array,Part153

Chapter8.TheHelicalAntenna:AxialandOtherModes,PartII55

Chapter9.Slot,PatchandHomAntennas57

Chapter10.FlatSheet,ComerandParabolicReflectorAntennas65

Chapter11.BroadbandandFrequency-IndependentAntennas75

Chapter12.AntennaTemperature,RemoteSensingandRadarCrossSection81

Chapter13.SelfandMutualImpedances103

Chapter14,TheCylindricalAntennaandtheMomentMethod(MM)105

Chapter15.TheFourierTransformRelationBetweenApertureDistribution

andFar-FieldPattern107

Chapter16.ArraysofDipolesandofAperture109

Chapter17.LensAntennas121

Chapter18.Frequency-SelectiveSurfacesandPeriodicStructures

ByBenA.Munk125

Chapter19.PracticalDesignConsiderationsofLargeApertureAntennas127

Chapter21.AntennasforSpecialApplications135

Chapter23.Baluns,etc.ByBenA.Munk143

Chapter24.AntennaMeasurements.ByArtoLehtoand

PerttiVainikainen147

Index153

iv

Chapter2.AntennaBasics

2-7-1.Directivity.

ShowthatthedirectivityDofanantennamaybewritten

E®，")maxE*®，O)max产

D=__________2_____________

4乃z，"

Solution'.

uav=±^um

u(e,*=s(e,M尸,s(仇。)=网"","0")

Therefore

£®Mma.CeMmax,2

Zj

°=E®梓飞可瓦q£

41JJaZ

)..o.J7

Notethatr~=area/steradian,soU=Sr-or(watts/steradian)=(watts/meter)xmeter

2-7-2.Approximatedirectivities.

Calculatetheapproximatedirectivityfromthehalf-powerbeamwidthsofaunidirectional

2

antennaifthenormalizedpowerpatternisgivenby:(a)Pn=cos"(b)Pn=cos9,(c)Pn

311

=cos9,and(d)Pn=cos0.Inallcasesthesepatternsareunidirectional(+zdirection)

withPnhavingavalueonlyforzenithangles00<0<90°andPn=0for900<0<180°.

Thepatternsareindependentoftheazimuthangle(/).

Solution'.

c40,000/、

(a)®HP=2cos-1(0.5)=2x60。=120°,D=z-=278(a几s.)

(120)2

-1八40,000.n.,、

(b)ew=2cos(Vo7)=2X45°=90°,D-z-=4.94(ans.)

(90)2

2

-1D=40,0*=73

(c)ew=2cos(Vo7)=2x37.47°=74.93°,(ans.)

(75)2

2-7-2.continued

10,000

(d)^=2cos_|(VCL5),(ans.)

HP(cos-1(V(15))2

*2-7-3.Approximatedirectivities.

Calculatetheapproximatedirectivitiesfromthehalf-powerbeamwidthsofthethree

unidirectionalantennashavingpowerpatternsasfollows:

=PmsinOsin2。

P(Q©=PmsinOsir?(/)

2

P(Q。=PmsinOsir?(/)

P(Q必hasavalueonlyfor0<0<nand0<^<7tandiszeroelsewhere.

Solution'.

TofindDusingapproximaterelations,

wefirstmustfindthehalf^powerbeamwidths.

HPBWHPBW

=90—6or6=90—

22

HPBW)

Forsin0pattern,sin=sin90-2J

2

9。-誓sin]/,-竿sin]；>9。

Forsin20pattern,sin*=sir90-里处

I22

1

sm90一史吧HPBW=90°

I2F

ism-

Forsin30pattern,

I2)2

3

sin[90-HPBW)=_LHPBW=74.9°

I2)V2

*2-7-3.continued

Thus,

〃=41,253sq.deg.41,253—40,000―、

==3.82_=3.70(ans.)

°HP°HP(120)(90)(120)(90).…小

forP(“。)=sinOsin。

・4L253=45”40,000二.

(120)(74.9)(120)(74.9)for=sin,sin-V

41,253,..40,000..、

==6.12==5.93(ans.)

(90)(74.9)(90)(74.9)forpw的=sMdsin)

*2-7-4.Directivityandgain.

(a)Estimatethedirectivityofanantennawith"卬=2°,由p=1°,and(b)findthegainof

thisantennaifefficiencyk=0.5.

Solution'.

40,00040,000"八m/、

(a)nD===2.0x1i0n4or43.0dB(ans.)

%P”HP(2)(1)

(b)G=kD=0.5(2.0x104)=l.Ox104or40.0dB(ans.)

2-9-1.Directivityandapertures.

Showthatthedirectivityofanantennamaybeexpressedas

4万4,双工)‘四力口/*("，丫/办

汇J1,£(x,y)E*(x,y)dxdy

whereE(x,y)istheaperturefielddistribution.

Solution:Ifthefieldovertheapertureisuniform,thedirectivityisamaximum(=Dm)

andthepowerradiatedisP'.Foranactualaperturedistribution,thedirectivityisDand

thepowerradiatedisP.Equatingeffectivepowers

4

£上______」____

DmP'=DP,

Pa/生呼^公办

2-9-1.continued

E-7-f[E(x,y)dxdy

where乙av

A。%

4万U”E(x,)，)dM，££(x,y)dxdy

thereforeq.e.d.

22E(x,y)E*(x,y)dxdy

%稣4号用与二,二&

\\^E[x,y)E\x,y)dxdy2”矶匕日汽Z的痴)‘3、4

A

2-9-2.Effectiveapertureandbeamarea.

Whatisthemaximumeffectiveaperture(approximately)forabeamantennahavinghalf­

powerwidthsof30°and35°inperpendicularplanesintersectinginthebeamaxis?

Minorlobesaresmallandmaybeneglected.

Solution:

22S7§2

Q”“HPOHP=30°X35°,4“=-=一。，。♦=3・1万(ans.)

12人3UXJ)

*2-9-3.Effectiveapertureanddirectivity.

Whatisthemaximumeffectiveapertureofamicrowaveantennawithadirectivityof

900?

2

2驷/564(M)

Solution:0=4乃Aein/A,Aem=——

4兀4万

2-11-1.ReceivedpowerandtheFriisformula.

Whatisthemaximumpowerreceivedatadistanceof0.5kmoverafree-space1GHz

circuitconsistingofatransmittingantennawitha25dBgainandareceivingantenna

witha20dBgain?Thegainiswithrespecttoalosslessisotropicsource.The

transmittingantennainputis150W.

5

Solution'.

A-D'VAr"2

2=C//=3X108/109=0.3m,

“4万"4乃

2-11-1.continued

B；12a2

zi316x032,100

p=0.0108W=10.8mW(ans.)

Pr=P,r=,(4万)2r222

■71(4TT)*25*002

*2-11-2.Spacecraftlinkover100Mm.

Twospacecraftareseparatedby100Mm.EachhasanantennawithD=1000operating

at2.5GHz.IfcraftA'sreceiverrequires20dBover1pW,whattransmitterpoweris

requiredoncraftBtoachievethissignallevel?

Solution'.

Q22

89

2=C//=3X10/2.5X10=0.12m,A„=A,r=

4乃

^.(required)=100x10-12=1O-10W

r222_(4乃)2/%2

mTOTO10”(47)2

=10966Ws11kW(ans.)

，0.4£>U21060.122

2-11-3.Spacecraftlinkover3Mm.

Twospacecraftareseparatedby3Mm.EachhasanantennawithD=200operatingat2

GHz.IfcraftA'sreceiverrequires20dBover1pW,whattransmitterpowerisrequired

oncraftBtoachievethissignallevel?

Solution'.

DA2

2=c//=3xl08/2xl09=0.15mA"

4zr

/；.=100x10-12=10-l°W

r2A2(44)2r222=叱黑黑”w

'La2

2-11-4.MarsandJupiterlinks.

(a)Designatwo-wayradiolinktooperateoverearth-Marsdistancesfordataandpicture

transmissionwithaMarsprobeat2.5GHzwitha5MHzbandwidth.ApowerofIO_19

6

WHz-1istobedeliveredtotheearthreceiverand10'1!WHz/totheMarsreceiver.The

Marsantennamustbenolargerthan3mindiameter.SpecifyeffectiveapertureofMars

andearthantennasandtransmitterpower(totaloverentirebandwidth)ateachend.Take

earth-Marsdistanceas6light-minutes,(b)Repeat(a)fbranearth-Jupiterlink.Takethe

earth-Jupiterdistanceas40light-minutes.

2-11-4.continued

Solution:

(a)A=c//=3xl08/2.5xl09=0.12m

196-13

Pr(earth)=1O-x5x10=5x10W

^(Mars)=10-17x5xl06=5xl0-nW

2

Take.(Mars)=(1/2)乃L5?=3.5m(fap=0.5)

TakeE(Mars)=lkW

2

Take儿(earth)=(1/2)7IS?=350m(£ap=0.5)

P,(earth)=Pr(Mars)

Ael(earth)A(1,(Mars)

D/田'<-H(360X3X108)20.122..

P(earth)=5x1i0n=6.9OMW

t3.5x350

Toreducetherequiredearthstationpower,taketheearthstationantenna

A«=(1/2)乃502=3927m2(ans.)

so

^(earth)=6.9xl06(15/50)2=620kW(ans.)

E(earth)=£(Mars)(Mar?A(earth)=43.5x3%——=8x10-14w

r'r2A2(360X3X108)20.122

whichisabout16%oftherequired5x10-13W.Therequired5xIO-13Wcouldbe

obtainedbyincreasingtheMarstransmitterpowerbyafactorof6.3.Otheralternatives

wouldbe(1)toreducethebandwidth(anddatarate)reducingtherequiredvalueofP,or

(2)toemployamoresensitivereceiver.

7

AsdiscussedinSec.12-1,thenoisepowerofareceivingsystemisafunctionofits

systemtemperatureTandbandwidthBasgivenbyP=kTB,wherek=Boltzmann's

constant=1.38X10-23JK-1.

ForB=5x106Hz(asgiveninthisproblem)andT=50K(anattainablevalue),

P(noise)=1.38xIO-23x50x5x106=3.5x10-15W

2-11-4.continued

Tliereceivedpower(8XIO-'4W)isabout20timesthisnoisepower,whichisprobably

sufficientforsatisfactorycommunication.Accordingly,witha50Kreceivingsystem

temperatureattheearthstation,aMarstransmitterpowerof1kWisadequate.

(b)ThegivenJupiterdistanceis40/6=6.7timesthattoMars,whichmakesthe

requiredtransmitterpowers6.7=45timesasmuchortherequiredreceiverpowers1/45

asmuch.

Neitherappearsfeasible.Butapracticalsolutionwouldbetoreducethebandwidthfor

theJupiterlinkbyafactorofabout50,makingB=(5/50)X106=100kHz.

*2-11-5.Moonlink.

Aradiolinkfromthemoontotheearthhasamoon-based5入longright-handedmono-

filaraxial-modehelicalantenna(seeEq.(8-3-7))anda2Wtransmitteroperatingat1.5

GHz.Whatshouldthepolarizationstateandeffectiveaperturebefortheearth-based

antennainordertodeliver1014Wtothereceiver?Taketheearth-moondistanceas1.27

light-seconds.

Solution'.

2=C//=3X108/1.5X109=0.2m,

From(8-3-7)thedirectivityofthemoonhelixisgivenby

DA2

D=12x5=60andAet(moon)=

44

FromFriisformula

生以J(旬-=10*(3x108x]27/452m2RCPor

P,AelW2x60

about14mdiameter(ans.)

8

2-16-1.Spaceshipnearmoon.

Aspaceshipatlunardistancefromtheearthtransmits2GHzwaves.Ifapowerof10W

isradiatedisotropically,find(a)theaveragePoyntingvectorattheearth,(b)therms

electricfieldEattheearthand(c)thetimeittakesfortheradiowavestotravelfromthe

spaceshiptotheearth.(Taketheearth-moondistanceas380Mm.)(d)Howmany

photonsperunitareapersecondfallontheearthfromthespaceshiptransmitter?

2-16-1.continued

Solution:

(a)PV(atearth)=—=—~==5.5xlO-18Wm-2=5.5aWm-2(ans.)

4万/4^(380xlO6)2

(b)PV=S=E2/ZorE=(SZ)”2

orE=(5.5xIO*x377)"2=45xio-9=45nVmT(ans.)

(c)t-r/c-380x106/3x108=1.27s(ans.)

(d)Photon="=6.63xIO'x2x1()9=1.3x10-24j,where/?=6.63xlO3^

Thisistheenergyofa2.5MHzphoton.From(a),PV=5.5xlO-18Js-Im-2

Therefore,numberofphotons=5.5.1O*-42x106m-2s_1(ans.)

1.3x10

2-16-2.MorepowerwithCP.

ShowthattheaveragePoyntingvectorofacircularlypolarizedwaveistwicethatofa

linearlypolarizedwaveifthemaximumelectricfieldEisthesameforbothwaves.This

meansthatamediumcanhandletwiceasmuchpowerbeforebreakdownwithcircular

polarization(CP)thanwithlinearpolarization(LP).

Solution'.

From(2-16-3)wehaveforrmsfieldsthatPV=Sav=

E2

ForLP,E(or&)=0,soS=—

2avZQ

ForCP,=E2,SOSav=—L-

ThereforeS(、p=2sLp(ans.)

9

2-16-3.PVconstantforCP.

ShowthattheinstantaneousPoyntingvector(PV)ofaplanecircularlypolarizedtraveling

waveisaconstant.

Solution'.

ECP=Excosm+EysincotwhereEx=Ey=Eo

2-16-3.continued

222222

|£CP|=(E(^coscot+E;sincoty=Eo(cos69/4-sincoty=EQ(aconstant)

E2

ThereforePVorS(instantaneous)=-9-(aconstant)(ans.)

*2-16-4.EPwavepower

Anellipticallypolarizedwaveinamediumwithconstantscr=0,〃=2,&=5hasH-

fieldcomponents(normaltothedirectionofpropagationandnormaltoeachother)of

amplitudes3and4Am'1.Findtheaveragepowerconveyedthroughanareaof5m2

normaltothedirectionofpropagation.

Solution'.

1/2222

Sav=gz("[2+M)=；377(〃r/J)“2(”12+用)=1377(2/5)(3+4)=2980Wm-

P=ASav=5x2980=14902W=14.9kW(ans.)

2-17-1.CrosseddipolesforCPandotherstates.

TwoA/2dipolesarecrossedat90°.Ifthetwodipolesarefedwithequalcurrents,whatis

thepolarizationoftheradiationperpendiculartotheplaneofthedipolesifthecurrents

are(a)inphase,(b)phasequadrature(90°differenceinphase)and(c)phaseoctature(45°

differenceinphase)?

Solution:

(a)LP(arts.)

(b)CP(arts.)

10

(c)From(2-17-3)sin=sin2/sin3

-1

wherey-tan(E2/E1)=45°

5=45°

c=22%°

AR=cot6,=1/tan61=2.41(EP)...(Q〃S.)

*2-17-2.PolarizationoftwoLPwaves.

Awavetravelingnormallyoutofthepage(towardthereader)hastwolinearlypolarized

components

Ex=2cos&

Ey=3cos依+90，)

(a)Whatistheaxialratiooftheresultantwave?

(b)WhatisthetiltangleTofthemajoraxisofthepolarizationellipse?

(c)DoesErotateclockwiseorcounterclockwise?

Solution'.

(a)From(2-15-8),AR=3/2=1.5(ans.)

(b)T=90°(ans.)

(c)Atr=0,E=Ex\atz=T/4,E=-Ey,thereforerotationisCW(a〃s.)

2-17-3.SuperpositionoftwoEPwaves.

Awavetravelingnormallyoutwardfromthepage(towardthereader)istheresultantof

twoellipticallypolarizedwaves,onewithcomponentsofEgivenby

E'y=2cos创andE；-6cos[x+f)

andtheotherwithcomponentsgivenbyE\=1coscotandE"=3cos(cot一f)

(a)Whatistheaxialratiooftheresultantwave?

(b)DoesErotateclockwiseorcounterclockwise?

Solution'.

Ey-E'y+E；-2COS69/+COS69/=3coscot

Ex-E'x+=6cos(69/+/2)+3cos(69/-TT/2)--6sin69/+3sin69/=-3sin69/

11

(a)ExandEyareinphasequadratureandAR=3/3=1(CP)(cms.)

(b)Atf=0,E=y3,att=T/4,E=-x3,thereforerotationisCCW(Q〃S.)

*2-17-4.TwoLPcomponents.

Anellipticallypolarizedplanewavetravelingnormallyoutofthepage(towardthe

reader)haslinearlypolarizedcomponentsExandEy.GiventhatEx=4=1Vm"and

thatEyleadsExby72°,

(a)Calculateandsketchthepolarizationellipse.

(b)Whatistheaxialratio?

(c)WhatistheangleTbetweenthemajoraxisandthex-axis?

Solution'.

-I

(b)y=tan(£2/用)=45°,b=72°

From(2-17-3),£=360,thereforeAR=1/tan£,=1.38

(c)From(2-17-3),sin2x=tan26：/tanorT=45°(ans.)

2-17-5.TwoLPcomponentsandPoincaresphere.

AnswerthesamequestionsasinProb.2-17-4forthecasewhereEyleadsExby72°as

1

beforebutEA-=2Vm/andEy=1Vm.

Solution:

(b)/=tan-12=63.4°

5=72°

s-24.8°andAR=2.17(ans.)

(c)T=11.2°(ans.)

*2-17-6.TwoCPwaves.

Twocircularlypolarizedwavesintersectattheorigin.One(y-wave)istravelinginthe

positiveydirectionwithErotatingclockwiseasobservedfromapointonthepositivey-

axis.Theother(x-wave)istravelinginthepositivexdirectionwithErotatingclockwise

asobservedfromapointonthepositivex-axis.Attheorigin,Eforthey-waveisinthe

positivezdirectionatthesameinstantthatEforthex-waveisinthenegativezdirection.

WhatisthelocusoftheresultantEvectorattheorigin?

Solution:

12

Resolve2wavesintocomponentsormakesketchasshown.Itisassumedthatthewaves

haveequalmagnitude.

*2-17-6.continued

T/4

Att-0

andt=T/2

,totalEs0

LocusofEisastraightlineinxyplaneatanangleof450withrespecttox(ory)axis.

*2-17-7.CPwaves.

Awavetravelingnormallyoutofthepageistheresultantoftwocircularlypolarized

JC0ty+901

components加=5eandEleft=2e^)(Vm').Find(a)theaxialratioAR,(b)

thetiltangleTand(c)thehandofrotation(leftorright).

Solution:

Im

(a)AR=__7/3_一2.33(ans.)[NoteminussignforRH(right-handed

2-5polarization)]

13

(b)Fromdiagram,T=-45°(Q〃S.)

(c)SinceErotatescounterclockwiseasafunctionoftime,RH.(ans.)

2-17-8.EPwave.

Awavetravelingnormallyoutofthepage(towardthereader)istheresultantoftwo

linearlypolarizedcomponentsEx=3coscotandEy=2cos(cot+900).Fortheresultant

wavefind(a)theaxialratioAR,(b)thetiltangleTand(c)thehandofrotation(leftor

right).

Solution:

(a)AR=3/2=1.5(ans.)

(b)T=0°(ans.)

(c)CW,LEP(ans.)

*2-17-9.CPwaves.

Twocircularlypolarizedwavestravelingnormallyoutofthepagehavefieldsgivenby

1

Elefi=andEright=(Vm)(rms).Fortheresultantwavefind(a)AR,(b)the

handofrotationand(c)thePoyntingvector.

Solution'.

2+3

(a)AR=、3=-5(ans.)

(b)REP(cms.)

(c)PV=琢,+琉==0.034Wm-2=34mWm_2(ans.)

Z377

2-17-10.EPwaves.

Awavetravelingnormallyoutofthepageistheresultantoftwoellipticallypolarized

(EP)waves,onewithcomponentsEx=5coscotandEy=3sin67/andanotherwith

jM

componentsEr=3*andEf=4e~.Fortheresultantwave,find(a)AR,(b)Tand

(c)thehandofrotation.

Solution'.

14

(a)

Ex=5COSM+3COSO/+4COSGI=12COSM

Ey=3sin①，+3sinGf—4sinGf=2smcot

2-17-10.continued

AR=12/2=6(ans.)

(b)SinceExandEvareintime-phasequadraturewithEv(max)>Ev(max),T=0°.

Orfrom(2-17-3),sin2i=tan2e/tan5,s-tan-1(1/AR)=9.46°

but8-90°sotan=oo

Eat/=774

ThereforeT=0°(ans.)j

(c)Atr=0,E=*12,E=0CCW

I__JEatr=0

Atr=T/4(3t=90°),Ex=0,Ev=2

ThereforerotationisCCW,sopolarizationisrightelliptical,REP(a〃s.)

*2-17-11.CPwaves.

Awavetravelingnormallyoutofthepageistheresultantoftwocircularlypolarized

jM

componentsEr=2eandE1=公-/3+右).Fortheresultantwave,find(a)AR,(b)T

and(c)thehandofrotation.

Solution:

ADE]+E4+26.、

(a)AR=--==—=3((ans.)

E1-Er4-22

(b)Whencot=0,Er=2/0°andE,=4Z=L45°

WhenG1=—22%°,纥=2222%°andg=4z22%°

sothatEj+=Emax=6/-22ort=-22%°(ans.)

NotethattherotationdirectionsareoppositeforErandE\

sothatfor-cot,Er=2/-cotbutEx-/+cot

Also,TcanbedeterminedanalyticallybycombiningthewavesintoanExandEy

componentwithvaluesof

15

Ex=5.60ZZ30.4°andEx=2.95/16.3°

fromwhich8--46.7°

*2-17-11.continued

Sincefrom(a)AR=3,6*canbedeterminedandfrom(2-17-3),thetiltangle

T=-22.5°

(c)Ei>ErsorotationisCW(LEP)(ans.)

2-17-12.Circular-depolarizationratio.

IftheaxialratioofawaveisAR,showthatthecircular-depolarizationratioofthewave

isgivenby.

心切

AR+1

Thus,forpurecircularpolarizationAR=1andR=0(nodepolarization)butforlinear

polarizationAR=ooand/?=1.

Solution'.

Anywavemayberesolvedinto2circularly-polarizedcomponentsofoppositehand,Er

andE\foranaxialratio

F"+当

AR=max

%in

EAR-I

fromwhichthecirculardepolarizationratio/?=L=

ErAR+l

Thusforpurecircularpolarization,AR=1andthereiszerodepolarization(R=0),while

forpurelinearpolarizationAR=ooandthedepolarizationratioisunity(R=1).When

AR=3,R=%.

16

17

Chapter3.TheAntennaFamily

3-4-1.Alpine-hornantenna.

ReferringtoFig.3-4a,thelowfrequencylimitoccurswhentheopen-endspacing>A/2

andthehighfrequencylimitwhenthetransmissionlinespacingd«%4.Ifd=2mmand

theopen-endspacing=1000d,whatisthebandwidth?

Solution*.

D=openedendspacing,d=transmissionlinespacing