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1、CHAPTER1111.1.ShowthatE=Aejk0z+φisasolutiontothevectorHelmholtzequation,Sec.11.1,Eq.(30),xs√fork0=ωµ0≤0andanyφandA:Wetaked2Aejk0z+φ=(jk)2Aejk0z+φ=−k2Edz200xs11.2.A100-MHzuniformplanewavepropagatesinalosslessmediumforwhich≤r=5andµr=1.Find:√√a)v:v=c/≤=3×108/5=1.34×108m/s.pprb)β:β=ω/v=(2π×108
2、)/(1.34×108)=4.69m−1.pc)λ:λ=2π/β=1.34m.d)Es:AssumerealamplitudeE0,forwardztravel,andxpolarization,andwriteEs=E0exp(−jβz)ax=E0exp(−j4.69z)axV/m.√√e)Hs:First,theintrinsicimpedanceofthemediumisη=η0/≤r=377/5=169Ω.ThenHs=(E0/η)exp(−jβz)ay=(E0/169)exp(−j4.69z)ayA/m.f)=(1/2)Re{E×H∗}=(E2/337)aW
3、/m2ss0z12.3.AnHfieldinfreespaceisgivenasH(x,t)=10cos(108t−βx)aA/m.Findya)β:Sincewehaveauniformplanewave,β=ω/c,whereweidentifyω=108sec−1.Thusβ=108/(3×108)=0.33rad/m.b)λ:Weknowλ=2π/β=18.9m.c)E(x,t)atP(0.1,0.2,0.3)att=1ns:UseE(x,t)=−ηH(x,t)=−(377)(10)cos(108t−0βx)=−3.77×103cos(108t−βx).Thevect
4、ordirectionofEwillbe−a,sincewerequirezthatS=E×H,whereSisx-directed.Atthegivenpoint,therelevantcoordinateisx=0.1.Usingthis,alongwitht=10−9sec,wefinallyobtainE(x,t)=−3.77×103cos[(108)(10−9)−(0.33)(0.1)]a=−3.77×103cos(6.7×10−2)azz=−3.76×103aV/mz11.4.Smallantennashavelowefficiencies(aswillbeseeni
5、nChapter14)andtheefficiencyincreaseswithsizeuptothepointatwhichacriticaldimensionoftheantennaisanappreciablefractionofawavelength,sayλ/8.a)Anantennaisthatis12cmlongisoperatedinairat1MHz.Whatfractionofawavelengthlongisit?Thefreespacewavelengthwillbec3.0×108m/s1.2λ===300m,sothatthefraction==4.
6、0×10−3airf106s−1300b)Thesameantennaisembeddedinaferritematerialforwhich≤r=20andµr=2,000.Whatfractionofawavelengthisitnow?λair3001.2λferrite=√=p=1.5m⇒fraction==0.8µr≤r(20)(2000)1.522111.5.A150-MHzuniformplanewaveinfreespaceisdescribedbyH=(4+j10)(2a+ja)e−jβzsxyA/m.a)Findnumericalvaluesforω,λ
7、,andβ:First,ω=2π×150×106=3π×108sec−1.Second,forauniformplanewaveinfreespace,λ=2πc/ω=c/f=(3×108)/(1.5×108)=2m.Third,β=2π/λ=πrad/m.b)FindH(z,t)att=1.5ns,z=20cm:UseH(z,t)=Re{Hejωt}=Re{(4+j10)(2a+ja)(cos(ωt−βz)+jsin(ωt−βz)}sxy=[8cos(ωt−βz)−20sin(ωt−βz)]ax−[10cos(ω