《电磁场》课程教学课件（PPT讲稿，英文）Chapter 10 Electromagnetic Radiation and Principles

Chapter10ElectromagneticRadiationandPrinciplesElectric Current Element, Directivity of AntennasLinear Antennas, Antenna arrayPrinciples ofDuality,Image,ReciprocityHuygens'Principle, Aperture Antennas.1.Radiation by Electric Current Element2. Directivity of Antennas3. Radiation by Symmetrical Antennas4. Radiation by Antenna Arrays5. Radiation by Electric Current Loop6. Principle of Duality7. Principle of Image8.Principle of Reciprocity9.Huygens'Principle10.Radiations byApertureAntennasV

Chapter 10 Electromagnetic Radiation and Principles Electric Current Element, Directivity of Antennas Linear Antennas, Antenna array Principles of Duality, Image, Reciprocity Huygens’ Principle, Aperture Antennas. 1. Radiation by Electric Current Element 2. Directivity of Antennas 3. Radiation by Symmetrical Antennas 4. Radiation by Antenna Arrays 5. Radiation by Electric Current Loop 6. Principle of Duality 7. Principle of Image 8. Principle of Reciprocity 9. Huygens’ Principle 10. Radiations by Aperture Antennas

Linear antennasSurface antennas

Linear antennas Surface antennas

ElectricEcurrentSurfaceelementelectriccurrent1.RadiationbyElectric currentElementA segment of wire carrying a time-varyingelectric current with uniform amplitude andphase is called an electric current element oran electric dipoleandd << l,l<< α _l<<r。Most of radiation properties of an electric current elementarecommon to otherradiators.UV

1. Radiation by Electric current Element A segment of wire carrying a time-varying electric current with uniform amplitude and phase is called an electric current element or an electric dipole. I l d Surface electric current Electric current element and d << l，l <<  ，l << r。 Most of radiation properties of an electric current element are common to other radiators

Assume that the electric currentelement is placed in an unbounddielectric whichis homogeneous,linear,isotropic and losslessSelecttherectangularcoordinateNsystem, and letthe electric currentelementbe placed atthe originandE,uP(x, y, z)aligned with the z-axisWe know that12CVxVxE-oucE=-joJVxVxH-o"u:H=VxJIt is very hard to solve them directly, and using vector magneticpotentialAVV.AH--VxAE=-jo A+ujoμsle-jk/r-rwhere A(r) =dl4元 J1 |r -ru7

Assume that the electric current element is placed in an unbound dielectric which is homogeneous, linear, isotropic and lossless. It is very hard to solve them directly, and using vector magnetic potential A H H J 2  −   =  E  E jJ 2  − = − We know that H =  A  1 r Il z y x  ,  P(x, y, z) o     j j A E A   = − +   −  = − −  l k r r I l r r A r d | | e 4π ( ) j | |  where Select the rectangular coordinate system, and let the electric current element be placed at the origin and aligned with the z-axis

Due to I << a, I <<r, r' </ we can takeThe electric current element has z-component only,dl' = e dl', andHIe-jkrA(r)=e.A.A4元rFor radiation by an antenna,it is more convenient to select thesphericalcoordinatesystem,and we haveA, =0A, = A, cosO A。= -A, sin 0S,uUsing H = - VxA givesuk?I I sin θ11le-jkNH.=4元k2r2krdH。=H, =0U7

z A z A(r) = e kr z r I l A j e 4π − =   r Il z y x   ,  Using H =   A gives  1 kr k r k r k I l H j 2 2 2 e j 1 4π sin −       = +   H = Hr = 0 Due to l  , l  r, r   l we can take ； r 1 1  r − r r   2π j 2π j e e − −  −  r r The electric current element has z-component only, l , and z dl = e d  Ar = A z cos A = −Az sin  A = 0 For radiation by an antenna, it is more convenient to select the spherical coordinate system, and we have Az Ar  -A

VV.AFrom E=-jo A+or VxH=joe Ewefindthejoueelectricfieldsask31 I cos0E. =k2r.2k32元08kI l sin 11e-jkE。=-jkr k?r?+k3r34元08E=0The fields of a z-directed electric current element havethreecomponents: H, E,, and E,only, whileH。= H, =E。=0.The fields area TM wave.U7

From or ，we find the electric fields as   j j A E A   = − +  H = j E e j 1 2π cos j j 2 2 3 3 3 kr r k r k r k I l E −       = − +   kr k r k r k r k I l E j 2 2 3 3 3 e 1 j 1 4π sin j −       = − − + +    = 0 E The fields of a z-directed electric current element have three components: , , and only, while . H Er E H = Hr = E = 0 The fields are a TM wave

k3I lcoseIn summary, we have1E,=k2r2元081ZE,k' I I sin 0le-jkhEeS,uk24元0krHAEek?I I sin eJH。=I4元kry@E。=H。= H, =0r > a is called the far-fieldregion, and where the fields arecalledthefar-zonefields.The absolutelength is not of main concern. The dimensionona scale with the wavelength is as the unit determining the antennacharacteristics.U7

kr k r k r k I l H j 2 2 2 e j 1 4π sin −       = +   e j 1 2π cos j j 2 2 3 3 3 kr r k r k r k I l E −       = − +   kr k r k r k r k I l E j 2 2 3 3 3 e 1 j 1 4π sin j −       = − − + +    E = H = Hr = 0  r Il z y x   ,  E Er H In summary, we have r >  is called the far-field region, and where the fields are called the far-zone fields

2元Near-zone field: Since r << a and kr<< l, the lower order terms入can be omitted, and e-jkr = 1, we haveofI I sin eI lcoseI I sin 0HE =E。=4元r24元0832元06r3Comparing the above equation to those for static fields, we seethat they are just the magnetic field produced by the steady electriccurrent element ll and the electric field by the electric dipole qlThe fields and the sources arein phase, and have no time delay.The near-zone fields are called quasi-staticfieldsUEV

Near-zone field: Since and , the lower order terms of can be omitted, and , we have r   1 2π kr = r   ) 1 ( kr e 1 j  − kr 4π sin 2 r I l H   = 3 2π cos j r I l Er   = − 3 4π sin j r I l E    = − Comparing the above equation to those for static fields, we see that they are just the magnetic field produced by the steady electric current element Il and the electric field by the electric dipole ql . The near-zone fields are called quasi-static fields. The fields and the sources are in phase, and have no time delay

The electric field and the magnetic field have a phase difference元ofso that the realpart of the complex energy flow densityvector2is zero.No energy flow, onlyan exchange ofenergy between the sourceand thefield.The energy is bound around the source, and accordingly thenear-zonefields are alsocalled bound fields2元Far-zone field: Since r >> and kr:>> l, the higher order2terms of ()can be neglected, we only haveH, and E。 askrI I sin ZIl sin ee-jkre~kHE。=22r2Arμ is the intrinsic impedance of around medium.Where Z-u上7

The electric field and the magnetic field have a phase difference of , so that the real part of the complex energy flow density vector is zero. 2 π The energy is bound around the source, and accordingly the near-zone fields are also called bound fields. No energy flow, only an exchange of energy between the source and the field. kr r I l H j e 2 sin j − =    kr r ZI l E j e 2 sin j − =    Where is the intrinsic impedance of around medium.   Z = Far-zone field: Since and , the higher order terms of can be neglected, we only have and as r   1 2π kr = r   H E ) 1 ( kr

I I sin 0ZI I sin 0e-jke-jkoH =E。2r2rThe far-zone field has thefollowing characteristics:(a) The far-zone field is the electromagnetic wave traveling alongEthe radial directionr. It is a TEM wave andH(b) The electric and the magnetic fields are in phase, and thecomplex energy flow densityvector has only thereal part.It meansthat energy is being transmitted outwardly, and the field is calledradiationfield(c) The amplitudes of the far-zone fields are inversely propor-tionalto thedistancer.This attenuationisnot resultedfromdissipationin the media, but due to an expansion of the area of thewavefront.UV

kr r I l H j e 2 sin j − =    kr r ZI l E j e 2 sin j − =    The far-zone field has the following characteristics: (a) The far-zone field is the electromagnetic wave traveling along the radial direction r . It is a TEM wave and Z . H E =   (b) The electric and the magnetic fields are in phase, and the complex energy flow density vector has only the real part. It means that energy is being transmitted outwardly, and the field is called radiation field. (c) The amplitudes of the far-zone fields are inverselypropor￾tional to the distance r. This attenuation is not resulted from dissipation in the media, but due to an expansion of the area of the wave front