Chapter 5 Field decompositions and the EM potentials 5.1 Spatial symmetry decompositions Spatial symmetry can often be exploited to solve electromagnetics problems. For analytic solutions, symmetry can be used to reduce the number of boundary conditions that must be applied. For computer solutions the storage requirements can be reduced. Typical symmetries include rotation about a point or axis, and reflection through a plane, along an axis, or through a point. We shall consider the common case of reflection through a plane. Reflections through the origin and through an axis will be treated in the exercises Note that spatial symmetry decompositions may be applied even if the ields possess no spatial symmetry. As long as the boundaries and material media are symmetric, the sources and fields may be decomposed into constituents that individually mimic the symmetry of the environment 5.1.1 Planar field symmetry Consider a region of space consisting of linear, isotropic, time-invariant media having laterial parameters E(r), u(r), and o(r). The electromagnetic fields(E, H)within this region are related to their impressed sources (, Jm)and their secondary sources J '=gE through Maxwells curl equations der dEy dHr der dE dey aEr dHz m Hy dEx at +oEr+J ahr dh. dey +oe+ aH aHr ae ②2001
Chapter 5 Field decompositions and the EM potentials 5.1 Spatial symmetry decompositions Spatial symmetry can often be exploited to solve electromagnetics problems. For analytic solutions, symmetry can be used to reduce the number of boundary conditions that must be applied. For computer solutions the storage requirements can be reduced. Typical symmetries include rotation about a point or axis, and reflection through a plane, along an axis, or through a point. We shall consider the common case of reflection through a plane. Reflections through the origin and through an axis will be treated in the exercises. Note that spatial symmetry decompositions may be applied even if the sources and fields possess no spatial symmetry. As long as the boundaries and material media are symmetric, the sources and fields may be decomposed into constituents that individually mimic the symmetry of the environment. 5.1.1 Planar field symmetry Consider a region of space consisting of linear, isotropic, time-invariant media having material parameters (r), µ(r), and σ(r). The electromagnetic fields (E, H) within this region are related to their impressed sources (Ji , Ji m) and their secondary sources Js = σE through Maxwell’s curl equations: ∂Ez ∂y − ∂Ey ∂z = −µ ∂ Hx ∂t − J i mx , (5.1) ∂Ex ∂z − ∂Ez ∂x = −µ ∂ Hy ∂t − J i my , (5.2) ∂Ey ∂x − ∂Ex ∂y = −µ ∂ Hz ∂t − J i mz, (5.3) ∂ Hz ∂y − ∂ Hy ∂z = ∂Ex ∂t + σ Ex + J i x , (5.4) ∂ Hx ∂z − ∂ Hz ∂x = ∂Ey ∂t + σ Ey + J i y , (5.5) ∂ Hy ∂x − ∂ Hx ∂y = ∂Ez ∂t + σ Ez + J i z . (5.6)
We assume the material constants are symmetric about some plane, say z=0. Then (x,y,-3)=∈(x,y,z), H(x,y,-2)=(x,y,z), o(x, y, -z)=o(x,y, z) That is, with respect to z the material constants are even functions. We further assume that the boundaries and boundary conditions, which guarantee uniqueness of solution,are also symmetric about the z=0 plane. Then we define two cases of reflection symmetry. Conditions for even symmetry. We claim that if the sources obey Jm(x, y,2)=-J (x,y,23)=引(x,y,-z),Jm(x,y,z) J(x,y,z)=-/2(x,y,-2),Jm2(x,y,)=Jm2(x,y,-z), then the fields obey E(x, y, z)=E(x, y, -z), Hr(x, y,z)=-H(r, y, -z) Ey(x, y, z)=Ey(, y,-2), Hy(r, y,2)=-H,(, y,-2) E2(x,y,z)=-E2(x,y,-) H2(x,y,x)=H2(x,y,-) The electric field shares the symmetry of the electric source: components parallel to the z=0 plane are even in z, and the component perpendicular is odd. The magnetic field shares the symmetry of the magnetic source: components parallel to the z=0 plane are dd in z, and the component perpendicular is even We can verify our claim by showing that the symmetric fields and sources obey Maxwell's equations. At an arbitrary point z=a>0 equation(5. 1) requires E-21|=-4=2|-m,= By the assumed symmetry condition on source and material constant we get de +Jiz= If our claim holds regarding the field behavior, then dey dHr and we have de- =k=2 +J So this component of Faradays law is satisfied. With similar reasoning we can show that the symmetric sources and fields satisfy (5. 2)-(5.6)as well ②2001
We assume the material constants are symmetric about some plane, say z = 0. Then (x, y, −z) = (x, y,z), µ(x, y, −z) = µ(x, y,z), σ(x, y, −z) = σ(x, y,z). That is, with respect to z the material constants are even functions. We further assume that the boundaries and boundary conditions, which guarantee uniqueness of solution, are also symmetric about the z = 0 plane. Then we define two cases of reflection symmetry. Conditions for even symmetry. We claim that 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) = Ex (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). The electric field shares the symmetry of the electric source: components parallel to the z = 0 plane are even in z, and the component perpendicular is odd. The magnetic field shares the symmetry of the magnetic source: components parallel to the z = 0 plane are odd in z, and the component perpendicular is even. We can verify our claim by showing that the symmetric fields and sources obey Maxwell’s equations. At an arbitrary point z = a > 0 equation (5.1) requires ∂Ez ∂y z=a − ∂Ey ∂z z=a = −µ|z=a ∂ Hx ∂t z=a − J i mx |z=a. By the assumed symmetry condition on source and material constant we get ∂Ez ∂y z=a − ∂Ey ∂z z=a = −µ|z=−a ∂ Hx ∂t z=a + J i mx |z=−a. If our claim holds regarding the field behavior, then ∂Ez ∂y z=−a = −∂Ez ∂y z=a , ∂Ey ∂z z=−a = −∂Ey ∂z z=a , ∂ Hx ∂t z=−a = −∂ Hx ∂t z=a , and we have −∂Ez ∂y z=−a + ∂Ey ∂z z=−a = µ|z=−a ∂ Hx ∂t z=−a + J i mx |z=−a. So this component of Faraday’s lawis satisfied. With similar reasoning we can showthat the symmetric sources and fields satisfy (5.2)–(5.6) as well.
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 ②2001
Conditions 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
etric field deco ition. Field sym source distributions through a symmetry decomposition of the sources and fields. Con sider the general impressed source distributions (J, Jm). The source set #(xy,z)=5[H(x,y,2)+f(x,y,-) J(,y,z)=5[(x,,2)+(x,y,-) J,(x, y, z) 2 [(x,y,z) Jm(x, y, z) 2 Jm.(,y,z) I(x, y, z) (x,y,z)-my(x,y,-2) learly of even symmetric type while the source set (a,y,)=1 [(x,y,x)-J(x,y,-2 1 J(x [(x,y.z)-(x,y,-2) (x,y,x)=2(x,y.3)+(,y-], Jm(,y, z)=5Jm(x, y, z)+Jmx(x, y, -2) m(,y,x)=5[m(x,y,x)+ J is of the odd symmetric type. SinceJ'=J+J and Jm= Jin +Jm, we can decompose any source into constituents having, respectively, even and odd symmetry with respect to a plane. The source with even symmetry produces an even field set, while the source with odd symmetry produces an odd field set. The total field is the sum of the fields Planar symmetry for frequency-domain fields. The symmetry conditions intro- duced above for the time-domain fields also hold for the frequency-domain fields. Because both the conductivity and permittivity must be even functions, we combine their effects and require the complex permittivity to be even. Otherwise the field symmetries and urce decompositions are identical Example of symmetry decomposition: line source between conducting planes. Consider a z-directed electric line source lo located at y =h, x=0 between conducting planes at y=td, d>h. The material between the plates has permeability i(o) and complex permittivity E(o). We decompose the source into one of even symmetric type ith line sources lo/2 located at y= th, and one of odd symmetric type with a line ②2001
Symmetric field decomposition. Field symmetries may be applied to arbitrary source distributions through a symmetry decomposition of the sources and fields. Consider the general impressed source distributions (Ji , Ji m). The source set J ie x (x, y,z) = 1 2 J i x (x, y,z) + J i x (x, y, −z) , J ie y (x, y,z) = 1 2 J i y (x, y,z) + J i y (x, y, −z) , J ie z (x, y,z) = 1 2 J i z (x, y,z) − J i z (x, y, −z) , J ie mx (x, y,z) = 1 2 J i mx (x, y,z) − J i mx (x, y, −z) , J ie my (x, y,z) = 1 2 J i my (x, y,z) − J i my (x, y, −z) , J ie mz(x, y,z) = 1 2 J i mz(x, y,z) + J i mz(x, y, −z) , is clearly of even symmetric type while the source set J io x (x, y,z) = 1 2 J i x (x, y,z) − J i x (x, y, −z) , J io y (x, y,z) = 1 2 J i y (x, y,z) − J i y (x, y, −z) , J io z (x, y,z) = 1 2 J i z (x, y,z) + J i z (x, y, −z) , J io mx (x, y,z) = 1 2 J i mx (x, y,z) + J i mx (x, y, −z) , J io my (x, y,z) = 1 2 J i my (x, y,z) + J i my (x, y, −z) , J io mz(x, y,z) = 1 2 J i mz(x, y,z) − J i mz(x, y, −z) , is of the odd symmetric type. Since Ji = Jie + Jio and Ji m = Jie m + Jio m , we can decompose any source into constituents having, respectively, even and odd symmetry with respect to a plane. The source with even symmetry produces an even field set, while the source with odd symmetry produces an odd field set. The total field is the sum of the fields from each field set. Planar symmetry for frequency-domain fields. The symmetry conditions introduced above for the time-domain fields also hold for the frequency-domain fields. Because both the conductivity and permittivity must be even functions, we combine their effects and require the complex permittivity to be even. Otherwise the field symmetries and source decompositions are identical. Example of symmetry decomposition: line source between conducting planes. Consider a z-directed electric line source ˜I0 located at y = h, x = 0 between conducting planes at y = ±d, d > h. The material between the plates has permeability µ(ω) ˜ and complex permittivity ˜ c(ω). We decompose the source into one of even symmetric type with line sources ˜I0/2 located at y = ±h, and one of odd symmetric type with a line
source Io/2 located at y= h and a line source -lo/2 located at y= -h. We solve each of these problems by exploiting the appropriate symmetry, and superpose the results to find the solution to the original problem For the even-symmetric case, we begin by using(4.407) to represent the field EI (,)=-x+A For y> h this becomes E(x, y, o= 2 cos k The secondary(scattered) field consists of waves propagating in both the ty-directions +jΔ [A*(r, o)- /k, '+A(kx,oeky ]e ik dkr.(5.7) ∞+j△ The impressed field is even about y =0. Since the total field E,= E;+Es must be even in y(E is parallel to the plane y=0), the scattered field must also be even. Thus, A+=a and the total field is for y > h E 2A(kr, a)cos ky,'so lo(o)2 cos ky h -jk, e -jk, dks Now the electric field must obey the boundary condition E,=0 at y=+d. How since E is even the satisfaction of this condition at y= d automatically implie atisfaction at y=-d. So we set 2A*(, o)cos kyd-a lo(o)2 cos k,h and invoke the Fourier integral theorem to get AT(Kx, o)=op lo(o) cos kyh The total field for this case is E ∞+ For the odd-symmetric case the impressed field is El (x, y, o) 一 hksi dkx ②2001
source ˜I0/2 located at y = h and a line source −˜I0/2 located at y = −h. We solve each of these problems by exploiting the appropriate symmetry, and superpose the results to find the solution to the original problem. For the even-symmetric case, we begin by using (4.407) to represent the impressed field: E˜ i z(x, y,ω) = −ωµ˜ ˜I0(ω) 2 2π ∞+ j −∞+ j e− jky |y−h| + e− jky |y+h| 2ky e− jkx x dkx . For y > h this becomes E˜ i z(x, y,ω) = −ωµ˜ ˜I0(ω) 2 2π ∞+ j −∞+ j 2 cos kyh 2ky e− jky y e− jkx x dkx . The secondary (scattered) field consists of waves propagating in both the ±y-directions: E˜ s z(x, y,ω) = 1 2π ∞+ j −∞+ j A+(kx ,ω)e− jky y + A−(kx ,ω)e jky y e− jkx x dkx . (5.7) The impressed field is even about y = 0. Since the total field Ez = Ei z + Es z must be even in y (Ez is parallel to the plane y = 0), the scattered field must also be even. Thus, A+ = A− and the total field is for y > h E˜ z(x, y,ω) = 1 2π ∞+ j −∞+ j 2A+(kx ,ω) cos ky y − ωµ˜ ˜I0(ω) 2 2 cos kyh 2ky e− jky y e− jkx x dkx . Nowthe electric field must obey the boundary condition E˜ z = 0 at y = ±d. However, since E˜ z is even the satisfaction of this condition at y = d automatically implies its satisfaction at y = −d. So we set 1 2π ∞+ j −∞+ j 2A+(kx ,ω) cos kyd − ωµ˜ ˜I0(ω) 2 2 cos kyh 2ky e− jkyd e− jkx x dkx = 0 and invoke the Fourier integral theorem to get A+(kx ,ω) = ωµ˜ ˜I0(ω) 2 cos kyh 2ky e− jkyd cos kyd . The total field for this case is E˜ z(x, y,ω) = −ωµ˜ ˜I0(ω) 2 2π ∞+ j −∞+ j e− jky |y−h| + e− jky |y+h| 2ky − − 2 cos kyh 2ky e− jkyd cos kyd cos ky y e− jkx x dkx . For the odd-symmetric case the impressed field is E˜ i z(x, y,ω) = −ωµ˜ ˜I0(ω) 2 2π ∞+ j −∞+ j e− jky |y−h| − e− jky |y+h| 2ky e− jkx x dkx ,