Conditions for odd symmetry. We can also show that if the sources obey Ji(x,y, z)=-(r, y,-2), Jm(x, y, z)=Jm(x, y, -z) Jy(x,y,2)=-x, y,-z), my(x, y, z)=Jmy (x, J(x,y,3)=J(x,y,-z),m:(x,y,x)=-/m2(x,y,-z) then the fields obey E(x, y, 2)=-ECx, y, -z), H,(r, y, z)=H(x, y, -z) Ey(r, H,(x,y,z)=H,(x,y,-x) E2(x,y,x)=E2(x,y,-) H2(x,y,z)=-H2(x,y,-x) Again the electric field has the same symmetry as the electric source. However, in this case components parallel to the z=0 plane are odd in z and the component perpendicular is even. Similarly, the magnetic field has the same symmetry as the magnetic source. Here components parallel to the z=0 plane are even in z and the component perpendicular is odd Field symmetries and the concept of source images. In the case of odd symmetr the electric field parallel to the z=0 plane is an odd function of z. If we assume that the field is also continuous across this plane then the electric field tangential to z=0 must vanish: the condition required at the surface of a perfect electric conductor(PEC) We may regard the problem of sources above a perfect conductor in the z=0 plane equivalent to the problem of sources odd about this plane, as long as the sources in both cases are identical for z>0. We refer to the source in the region z <0 as the image of the source in the region z>0. Thus the image source (', Jm)obeys I(x,, -z)=-J(, y, z), JA(x, y, -2)=Jm (x, y, z) J!(x,y,-)=(x,y,z),Jm2(x That is, parallel components of electric current image in the opposite direction, and the perpendicular component images in the same direction; parallel components of the magnetic current image in the same direction, while the perpendicular component images in the opposite direction. In the case of even symmetry, the magnetic field parallel to the z=0 plane is odd, d thus the magnetic field tangential to the z=0 plane must be zero. We therefore have an equivalence between the problem of a source above a plane of perfect magnetic conductor(PMC) and the problem of sources even about that plane. In this case we identify image sources that obey Jl(x,y, -z)=Ji(x, y, z), Jm (x, y, -z)=-Jmi(, y, z) J(x, y, -z)=J(x, y, z), Jmy J2(x,y,-2)=-12(x,y,z),m2 Jm(x, y, z) Parallel components of electric current image in the same direction, and the perpendicular component images in the opposite direction; parallel components of magnetic current image in the opposite direction, and the perpendicular component images in the same direction In the case of odd symmetry, we sometimes say that an"electric wall"exists at z=0 The term"magnetic wall"can be used in the case of even symmetry. These terms are particularly common in the description of waveguide fields ②2001Conditions for odd symmetry. We can also showthat if the sources obey J i x (x, y,z) = −J i x (x, y, −z), J i mx (x, y,z) = J i mx (x, y, −z), J i y (x, y,z) = −J i y (x, y, −z), J i my (x, y,z) = J i my (x, y, −z), J i z (x, y,z) = J i z (x, y, −z), J i mz(x, y,z) = −J i mz(x, y, −z), then the fields obey Ex (x, y,z) = −Ei x (x, y, −z), Hx (x, y,z) = Hx (x, y, −z), Ey (x, y,z) = −Ey (x, y, −z), Hy (x, y,z) = Hy (x, y, −z), Ez(x, y,z) = Ez(x, y, −z), Hz(x, y,z) = −Hz(x, y, −z). Again the electric field has the same symmetry as the electric source. However, in this case components parallel to the z = 0 plane are odd in z and the component perpendicular is even. Similarly, the magnetic field has the same symmetry as the magnetic source. Here components parallel to the z = 0 plane are even in z and the component perpendicular is odd. Field symmetries and the concept of source images. In the case of odd symmetry the electric field parallel to the z = 0 plane is an odd function of z. If we assume that the field is also continuous across this plane, then the electric field tangential to z = 0 must vanish: the condition required at the surface of a perfect electric conductor (PEC). We may regard the problem of sources above a perfect conductor in the z = 0 plane as equivalent to the problem of sources odd about this plane, as long as the sources in both cases are identical for z > 0. We refer to the source in the region z < 0 as the image of the source in the region z > 0. Thus the image source (JI , JI m) obeys J I x (x, y, −z) = −J i x (x, y,z), J I mx (x, y, −z) = J i mx (x, y,z), J I y (x, y, −z) = −J i y (x, y,z), J I my (x, y, −z) = J i my (x, y,z), J I z (x, y, −z) = J i z (x, y,z), J I mz(x, y, −z) = −J i mz(x, y,z). That is, parallel components of electric current image in the opposite direction, and the perpendicular component images in the same direction; parallel components of the magnetic current image in the same direction, while the perpendicular component images in the opposite direction. In the case of even symmetry, the magnetic field parallel to the z = 0 plane is odd, and thus the magnetic field tangential to the z = 0 plane must be zero. We therefore have an equivalence between the problem of a source above a plane of perfect magnetic conductor (PMC) and the problem of sources even about that plane. In this case we identify image sources that obey J I x (x, y, −z) = J i x (x, y,z), J I mx (x, y, −z) = −J i mx (x, y,z), J I y (x, y, −z) = J i y (x, y,z), J I my (x, y, −z) = −J i my (x, y,z), J I z (x, y, −z) = −J i z (x, y,z), J I mz(x, y, −z) = J i mz(x, y,z). Parallel components of electric current image in the same direction, and the perpendicular component images in the opposite direction; parallel components of magnetic current image in the opposite direction, and the perpendicular component images in the same direction. In the case of odd symmetry, we sometimes say that an “electric wall” exists at z = 0. The term “magnetic wall” can be used in the case of even symmetry. These terms are particularly common in the description of waveguide fields