習題答案第5章時變電磁場和平面電磁波解讀

1、第5章時變電磁場和平面電磁波5.15.11已知z2=1j，求復數(shù)z的兩個解。2[解]z=1j=jπjπ2ez1=2e=1.189ej22.5=1.099j0.455j22.5z2=1.189e=1.099j0.4555.25.12已知α是正實數(shù)，試證：(a)若α1[解](a)α1:α2ejtan1α≈e(jααα?α???=cosjsin?≈1j?22?2???jα=α2ejtanα1≈?αe?jπ???ππ??=cojsi?44??=

2、(1j)2=eje，H（t）的復振幅為H=hjh，試證5.35.13設E（t）的復振幅為EiiHejωt，并求E（t）E(t)H(t)≠ReE、H（t）。ejωt=1EejωtEejωt[解]E(t)=ReE[][](2)1jωtejωtHeH21Hej2ωtEHej2ωt得E(t)H(t)=EHEHE41HEHej2ωt≠ReEHejωt=ReE2H(t)=()()[][]E(t)=Re(ejei)ejωt=Re[(ejei)(cos

3、ωtjsinωt)]=ecosωteisinωt1[]H(t)=Re(hjhi)ejωt=hcosωthisinωtE(t)H(t)=ehcos2ωteihisin2ωtehicosωtsinωteihcosωtsinωt[]=1[eheihi(eheihi)cos2ωt(ehieih)sin2ωt]2可見為恒定成分與二倍頻成分的疊加.5.45.14將下列場矢量的瞬時值變換為復矢量，或作相反的變換：?E0sin(ωtkz)y?3E0co

4、s(ωtkz)；(a)(t)=x??E0sinωt3E0cosωt(b)(t)=x??jy?)e(c)=(x?jH0e(d)=y???π???；6???jkz；。jkzsinθjjkz?E0ee2y?3E0ejkz=(jx?y?3)E0ejkz[解](a)=xπππ?j?j??3?31??1???=x???E0e23E0e6?=x?E0?j3?E0(b)=xjj2?2??2????2??????cos(ωtkz)y?cosωtkz(c)

5、(t)=x??π??cos(ωtkz)y?sin(ωtkz)?=x2??H0co(d)(t)=ysωtkzsinθ??π??H0sin(ωtkzsinθ)?=y2?Ib1U1?2π?2?ρdρ=UI[解]P=??ds=Saρb24πl(wèi)na5.95.33在理想導體平面上方的空氣區(qū)域（z＞0）存在時諧電磁場，其電場強度為?E0sinkzcosωt。(t)=x(a)求磁場強度(t)；(b)求在z=0，π4k和π2k處的坡印廷矢量瞬時值及平均值

6、；(c)求導體表面的面電流密度。[解](a)(t)=Rek[]=y?ωμjωt2Eπ???0coskzsinωtη0=E0coskzcosωt?=y2?η0?0ε0E?0sn2kzsin2ωt(b)(t)=(t)?(t)=z4η0z=0(t)=0E?0sin2ωtz=(t)=z4η04kπ2z=π2k(t)=0av1=Tav?T0E1?0sin2kz?(t)?dt=z4η0T2?T0sin4πt=0T或?j??1?2?=Re???=Re

7、?zE0sin2kz?=0?2??4η0???(c)s=n??z=0?j??y=zE0E0η0?j=xE0η0?js(t)=Re?x5.105.34?EEπ???0co?0sinejωt?=xsωt?=xωtη0η2η??00?已知時諧電磁場瞬時值為?2Eecos(ωt30)，Ee(t)=xe和，求坡印廷矢量瞬時值?2Hecos(ωt30)。請寫出其復矢量He(t)=y?EeHe。(t)=e(t)?e(t)，并證明其一周平均值為Sαv=

8、z?2Eee[解]e=xj30=yj30?2Heee2?2EeHecos(t)=Ee(t)?He(t)=zav=?[EH(ωt30)=zeeEeHecos(2ωt60)]1T1T??EeHe得證.()tdt=zEHEHcos2ωt60dt=zeeee??00TT[()]5.115.35設時諧電磁場瞬時值為jωt(t)=Imjωt(t)=Im試求坡印廷矢量瞬時值(t)=(t)?(t)，并求其一周內(nèi)平均值S[解](t)=Imαv][]。[]