《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