第六章平面电磁波6-1.在εr=2,μr=1的理想介质中,频率为f=150MHZ的均匀平面波沿y方向传播,y=0处,E=z?10V/m,求E,E(y,t),H,H(y,t),Sc,υp.解:smccvrrp/2,mfcfvp22222kyjjkyezeEE2010?Z=120π/2ZezyZEkHyj/10??/?2=x?(2/12π)yje2E(y,t)=z?102cos(2π*150*106t-2πy)H(y,t)=x?/6πcos(2π*150*106t-2πy)Sc=*HE=y?52/6π6-2.在真空中H=x?xH=x?0Hzje2,求E,E(z,t),λ,f,Z,Sc。解:Z=120πE=kHZ?=zjeHzx20120)?(?=y?120π0Hzje2E=y?120π02H)21032cos(8ztk=2πλ=k2=1m,Hzcvfp8103Sc=*HE=-z?120π0H26-3.在理想介质中E(x,t)=y?80π2cos(10*107πt+2πx)H(x,t)=-z?2cos(10*107πt+2πx)求:f,εr,μr,λ.解:71010,f=2=5*107Hz2k,λ=k2=1m,mfc60由k=2π=(εrμr)2/1和Z=80π=120π(μr/εr)2/1得:εr=9,μr=46-4.均匀平面电磁波在真空中沿k?=1/2(y?+z?)方向传播,0E=10x?,求E,E(y,z,t),H,H(y,z,t),Sc解:设λ已知,则k=2π/λ,E=0Erkje?=x?10))(/2(zyjeH=1/Zk?E=2/24π(y?-z?))(/2zyjeE(y,z,t)=x?102cos(2πc/λt-(2π/λ)(y+z))H(y,z,t)=1/12π(y?-z?)cos(2πc/λt-(2π/λ)(y+z))Sc=*HE=(5/62π)(y?+z?)5、在均匀理想介质中)sin(2?)cos(2?)(00kztEykztExtE.求)(tH及平均坡印亭矢量。解:)sin(2?)cos(2?)(00kztEykztExtEjkzeyjxEE)??(0jkzexjyZEZEzH)??(/?0))sin(2?)cos(2?()(0kztxkztyZEtHZEzHES20*2?]Re[平6、证明电磁波E=5(x?+3y?))3(yxjeH=(5/120π)z?)3(yxje为均匀平面波.解:因为0HE,0?kE,0?kH因此是TEM波;因为Cyx3是平面方程,因此是平面波;)?3?(50yxE、zH?)120/5(0是常矢量。所以此波是均匀平面波。7、由(6.2-5)和(6.2-6)式推导(6.2-7)式。解:由于)(jkcc'''jkkkc因此'''jkk)(j上式两边取平方,得jkkjkk222"'2"'由此得222"'kk"'2kk解此二元方程得]1)(1[2'2k]1)(1[2''2k8、求f=100kHz,1MHz,100MHz,10GHz时电磁波在铝(σ=3.6*107/欧米,εr=1,μr=1)中的集肤深度.解:δ=1/f=77106.31041ff=100kHz,δ=2.6526*104mf=1MHz,δ=8.3882*105mf=100MHz,δ=8.3882*106mf=10GHz,δ=8.3882*107m9、银的σ=6.1*107(1/欧米),在什么频率上,δ=1mm?解:由δ=1/f得:f=21,当δ=1mm,0,f=21=4.156KHz10、电磁波的频率为100MHz,媒质参数为εr=8,μr=1,σ=0.5*103(1/欧米),求υ'p,λ',ck.解:因为1011.0854.882510854.88102105.01283所以该媒质为弱损耗媒质,ck/8102'8=5.9212082105.02"3k2100608.1mck06.1108'2'8υ'p='k=1/=1.0607*108m/s11、设地球的εr=8,μr=1,σ=5*103(1/欧米),在什么频率范围可将地球近似看作低损耗媒质?求该频率上的"k.解:当σ<<ωε时,可看作介质,工程中大于10倍即可,故:10*σ<2πf*ε即f>210=8.988*108Hz"k=2=0.333212、50MHz的均匀平面波透入到湿土(mSrr/02.0,1,16)中。求相位常数、衰减常数、相速、波长、波阻抗、集肤厚度。解:45.010854.816105202.0127]1)(1[2]1)(1[2'22rrck68.02]145.01[81031052287]1)(1[2]1)(1[2"22rrck15.02]145.01[81031052287mNp/smkvp/1074.068.01050]1)(1[21''862mk47.1'2')1(1)1(jjZcc45.01130j135.04.27jemk06.115.021"113、设2)(0k,其中σ=0.5*103(1/欧米),求在MHz1020的群速和相速。解:smkvp/10447.0105.01041022222377000smddkdkdvg/10894.02222111200014、设2)(1)(cck,其中GHzc52,求在GHz820的群速和相速。解:smckvcp/1084.310328.1)85(1103)(1882802smcddkdkdvccg/10667.1)(1)(1180215、分析下面波的极化类型1)xjxjezeyE2210?10?2)xjjxjjeezeeyE24/24/10?10?3)xjjxjjeezeeyE24/24/20?10?解:1)xjxjxjezyezeyE)??(1010?10?22相位差,是线极化波;2)xjjjxjjxjjeezyeeezeeyE22/4/24/24/)??(1010?10?xjjezjye24/)??(10是左旋圆极化波;3)xjjxjjeezeeyE24/24/20?10?xjjezjye24/)?2?(10是右旋椭圆极化波。16、在真空中,均匀平面波E=[x?(-1+j2)+y?(-2-j)]jze是什么极化波?求λ,H。解:E=[x?(-1+j2)+y?(-2-j)]jzejzjzeyjxjeyjjxj]??)[21(]?212?)[21(是右旋圆极化波;1k;22kjzejxyjZEzH]?[12021/?17、均匀平面波E(y,t)=x?20Esin(ωt+4πy)+z?20Esin(ωt+4πy-π/3)是什么极化波?求H。解:E(y,t)=x?20Esin(ωt+...