EINDHOVEN UNIVERSITY OF TECHNOLOGY Department of Mathematics and Computer Science. CASA-Report February 2011

Transcription

1 EINDHOVEN UNIVERSITY OF TECHNOLOGY Department of Mathematics and Computer Science CASA-Report February 211 Notes on solving Maxwell equations Part 2: Green s function for stratified media by R. Rook Centre for Analysis, Scientific computing and Applications Department of Mathematics and Computer Science Eindhoven University of Technology P.O. Box MB Eindhoven, The Netherlands ISSN:

2

3 Notes on solving Maxwell equations Part 2: Green's function for stratied media R.Rook February 3, Introduction In the previous report (part 1), the problem and its governing equations are described and is discarded in this report. The nite element method in part 1, or any other method for that matter, determines the elds in and close to the scatterer (near-eld) that is used to construct the elds in the far-eld. The goal of part 2 is to nd far-eld expressions formulated as total elds or the Radar Cross Section (RCS) of the scattered elds. The far-eld is calculated from the scatterer problem in the contrast formulation. The scatterer then acts as a radiating object with a known source J. Using Green's function theory, the far-eld solution is just the convolution of that source with the fundamental solution G to the Maxwell equation. Without loss of generality, the expressions are formulated in total elds E and H. Again, the time convention for the time-harmonic term exp( iωt) is used, but in contrary to part 1, the quantities are in full dimensions, following closely the notation used by Chew, Balanis and others. 2 Green's function The electric eld dyadic Green's function ḠE in a homogeneous medium is the starting point. It consists of the fundamental solutions to Helmholtz equation, which can be written in a Fourier expansion of plane waves. This expansion allows embedding in a multilayer medium. Finally, the vector potential approach is used to derive the potential Green's function ḠA. The latter is a two-step derivation, where the electric and magnetic elds are functions of the vector potentials and involves less evaluation in case of far-eld calculations. 1

4 2.1 Homogeneous medium Recall the Maxwell equation in homogeneous medium (part 1) E k 2 E = iωµj. This is written as three Helmholtz equations (Cartesian coordinates) which solution is the convolution [ 2 E(r) + k 2 E(r) = iωµ Ī + ] k 2 J(r), (1) E(r) = iωµ Ω g(r r) [Ī + k 2 ] J(r )dr or (Chew, page 27) E(r) = iωµ Ω J(r ) [Ī + k 2 ] g(r r)dr. with g(r) the fundamental solution to the Helmholtz equation (see next subsection). Alternatively, this can be written as E(r) = iωµ Ω J(r )ḠE (r, r)dr, with the dyadic Green's function ḠE (r, r ) (second rank tensor) Ḡ E (r, r) = [Ī + ] g(r r), that satises the following equation (Chew, page 31) k 2 ḠE (r, r ) k 2 Ḡ E (r, r ) = Īδ(r r ). It can be shown that (Chew, page 28) (ḠE (r, r)) T = ḠE (r, r ) = [ Ī + ] k 2 g(r r ), So, the convolution of this dyadic Green's function and the source nally is 2

5 E(r) = iωµ Ω Ḡ E (r, r ) J(r )dr. This integral is dened properly provided that r / Ω as is of order 1/ r r 3 when r r and should de redened in that case (Chew, page 28). 2.2 Helmholtz equation The fundamental solution to the scalar wave equation or Helmholtz equation is ( 2 + k 2 )g(r, r ) = δ(r r ) g(r, r ) = g(r, r) = g(r r ) = exp(ik r r ) 4π r r. Suppose that the Fourier transform of the solution exists g(r, r ) = 1 (2π) 3 g(k x, k y, k z ) exp(ik x (x x )+ik y (y y )+ik z (z z ))dk x dk y dk z. The Fourier transform of the Helmholtz equation (k 2 k 2 x k 2 y k 2 z) g(k x, k y, k z ) exp(ik x (x x )+ik y (y y )+ik z (z z ))dk x dk y dk z = exp(ik x (x x ) + ik y (y y ) + ik z (z z ))dk x dk y dk z implies the Fourier components 1 g(k x, k y, k z ) = k 2 kx 2 ky 2 kz 2. Eliminating k z, with use of the poles k z = ± k 2 kx 2 ky 2 and Jordan's Lemma (allowing a small loss) results in the Weyl identity (Chew, page 65) 3

6 exp(ik r r ) r r = i 2π 1 k z exp(ik x (x x ) + ik y (y y ) + ik z z z ) dk x dk y, (2) where k 2 z = k 2 k 2 x k 2 y, Im(k z ) > and Re(k z ) >, for all k x and k y in the integration. The commonly used spectral decomposition of point source solutions can be formulated as Fourier integrals. 2.3 Fourier integrals For short notation we introduce the (shifted) 2D Fourier transforms (cf. Michalski 199) f = F (f(x x, y y )) = f(x x, y y ) exp( ik x (x x ) ik y (y y ))dxdy. ( ) f = F 1 f(kx, k y ) = 1 4π 2 f(k x, k y ) exp(ik x (x x ) + ik y (y y ))dk x dk y, which can be used to switch easily from dierent representations. By introducing cylindrical coordinates, x x = ρ cos(φ), y y = ρ sin(φ), k x = k ρ cos(α), k y = k ρ sin(α), ρ = ( ) y y (x x ) 2 + (y y ) 2, φ = arctan x x, and k 2 ρ = k 2 x + k 2 y, we can express various inverse Fourier integrals that arise in Sommerfeld-type integrals [ ] S n G(kρ ) (k ρ ρ) = 1 G(k ρ )J n (k ρ ρ)kρ n+1 dk ρ, (3) 2π 4

7 where J n is the Bessel's function of order n. Explicitly, the inverse Fourier transforms are F 1 ( G) [ ] = G = S G, F 1 [ (ik x G) = x G = cos(φ)s 1 G], F 1 [ (ik y G) = y G = sin(φ)s 1 G], F 1 (kx 2 G) ( [ ] [ = 2 x 2 G = 1 cos(2φ)s 2 G S kρ 2 2 G ]), F 1 (ky 2 G) = 2 y 2 G = 1 ( [ ] [ cos(2φ)s 2 G + S kρ 2 2 G ]), F 1 (k x k y G) = 2 x y G = 1 2 sin(2φ)s 2 [ G]. The Fourier component of the Sommerfeld identity are easily obtained ( exp(ik r r ) ) F 4π r r where the Sommerfeld identity is given by = i exp(ik z z z ), (4) 2 k z 4πg(r, r ) = exp(ik r r ) r r = i = i 2 k ρ k z J (k ρ ρ) exp(ik z z z ) dk ρ k ρ k z H (1) (k ρρ) exp(ik z z z ) dk ρ. (5) The physical interpretation is that a spherical wave is expanded as an integral summation of conical (or cylindrical) waves multiplied by a plane wave in z- direction (Chew, page 66). For postprocessing purposes, one may use the standard (non-shifted) Fourier transform H of f is ˆf (hat). It is given by a well-known of the Fourier transform 5

8 ˆf(k x, k y ; x, y ) = H (f(x x, y y )) = f(x x, y y ) exp( ik x x ik y y)dxdy = f(x x, y y ) exp( ik x (x x ) ik y (y y )) exp( ik x x ik y y )dxdy = f(k x, k y ; x, y ) exp( ik x x ik y y ) The spectral representation is used to derive the Green's function in terms of Herzian dipoles where we derive the elds given point sources. 2.4 Hertzian dipoles In 3D, a general source is composed of the three (linear independent) vector sources: the Hertzian dipoles. Suppose a current source is dened as J(r) = ˆαIlδ(r r ), radiating from r = r, Il is constant and ˆα is the direction of the dipole). The electric and magnetic elds result from the convolution with the Green's function (note that J(r) = ˆαIl in r = r and J(r) = elsewhere) ( E(r) = iωµ Ī + ) k 2 ˆαIl exp(ik r r ) 4π r r H(r) = ˆαIl exp(ik r r ) 4π r r. If there is a spectral representation Ẽz(k ρ, r) of E z, the other components can be derived using the Maxwell equations (see Chew, page 76) E(r) = Ẽ(k ρ, r)dk ρ, H(r) = H(k ρ, r)dk ρ, where the x- and y-components for Ẽ and H are given by ( Ẽx Ẽ y ) = 1 kρ 2 [ ( ) ( ) ] kx ky i k y z Ẽz + ωµ k H z x (6) 6

9 ( ) Hx = 1 [ ( kx H y kρ 2 i k y ) ( ) ] z H ky z ωɛẽz, (7) k x where Ẽz and H z are the spectral components of E z and H z, respectively. The next sections list the explicit relations for the eld due to the electric dipoles (Chew, page 71); similar results are obtained by duality (Chew, page 74). Recall that the z-components determines the TE and TM polarization Vertical electric dipole (VED) The VED ˆα = ẑ gives the components (Chew, page 71), with E z = iil ) (k exp(ik r r ) 4πωɛ z 2 4π r r, H z =. (8) Assuming that the dipole is of unit strength (Il = 1) and using the Sommerfeld identity (5), the Fourier components of the z-components of the electric and magnetic elds are (a function of k ρ ) k 2 ρ ωɛẽz = 1 exp(ik z z z ), Hz =. 2 k z By using (7), the other components are ωɛẽx = ± 1 2 k x exp(ik z z z ), ωɛẽy = ± 1 2 k y exp(ik z z z ) H x = 1 k y exp(ik z z z ), Hy = 1 k x exp(ik z z z ). 2 k z 2 k z Horizontal electric dipole (HED) The HED ˆα = ˆx is given by the components E z = iil ωɛ 2 z x exp(ik r r ) 4π r r, H z = Il y exp(ik r r ) 4π r r. (9) Assuming that Il = 1 and using the Sommerfeld identity (5), the Fourier components are ωɛẽx = 1 2 (k2 k 2 x) 1 k z exp(ik z z z ), ωɛẽy = 1 2 k xk y 1 k z exp(ik z z z ) 7

10 ωɛẽz = ± 1 2 k x exp(ik z z z ) H x =, Hy = 1 2 exp(ik z z z ) H z = 1 2 k y k z exp(ik z z z ). For the HED in y-direction, the z-components are E z = iil ωɛ 2 z y with the Fourier components exp(ik r r ) 4π r r, H z = Il exp(ik r r ) x 4π r r ωɛẽx = 1 2 k xk y 1 k z exp(ik z z z ), ωɛẽy = 1 2 (k2 k 2 y) 1 k z exp(ik z z z ) ωɛẽz = ± 1 2 k y exp(ik z z z ), H x = 1 2 exp(ik z z z ), Hy = H z = 1 2 k x k z exp(ik z z z ). The Ẽ components form the elements of the electric Green's function Ḡ E. The H is here the magnetic eld due to the electric dipoles Summary The spectral components of the dyadic Green's function ḠE for the electric eld in an homogeneous medium is simply the 2D Fourier transform of the dyad Ḡ E (r, r ) = [ Ī + /k 2] g(r r ), explicitly Ḡ E (k x, k y ; z, z ) = 1 k 2 k 2 kx 2 k x k y ik x z k x k y k 2 ky 2 ik y ik x z ik y z z kρ 2 g(k x, k y ; z, z ). Given the relation iωµ H = Ẽ, it can readily be seen that the columns of 8

11 z region 1... region m... region N z=z source z= d_1 z= d_m z= d_{n 1} Figure 1: stratied medium Ḡ E = Ī g = z ik y z ik x ik y ik x g(k x, k y ; z, z ) are the magnetic elds due to the two HEDs and the VED, respectively. When dipoles are placed in a stratied medium, reection from the interfaces should be incorporated. The general solution is a linear sum of planar waves, which propagation through stratied media can be calculated explicitly. 2.5 Stratied Media The construction of this (dyadic) Green's function is done via the homogeneous Green's function for homogeneous media and its convolution. The Sommerfeld (5) or Weyl identities (2) expand the solution in planar waves travelling in z- direction. The z-variation of the solution in free space in z is given by F (z, z ) = exp(ik z z z ). Suppose the stratied medium consists of N layers and a (dipole) source is located in medium m (see Figure 1), the solution is constructed inside this medium and the solution outside this medium is constructed recursively Planar waves The simplest case of a stratied medium is the half-plane, which consists of two layers or N = 2. An incoming TE planar wave in region 1 is reected by the layer in region 2 and is written as E m 1y(z) = e [exp( ik 1z z) + R 12 exp(2ik 1z d 1 + ik z z)]. 9

12 At z = d 1 the eld matches to the transmitted wave in region 2 E m 2y(z) = e T 12 exp( ik 2z z). Using the boundary conditions for the electric and magnetic elds, the Fresnel coecients are calculated R 12 = µ 2k 1z µ 1 k 2z µ 2 k 1z + µ 1 k 2z, T 12 = 2µ 2 k 1z µ 2 k 1z + µ 1 k 2z. This procedure can be generalized for multiple layers Generalized reection coecient The R ij are the generalized reection coecients in the stratied media. For both TE and TM these are derived from a recursive relation (Chew, pages 52+53) and R i,i+1 = R i,i+1 + R i+1,i+2 exp(2ik i+1,z (d i+1 d i )) 1 + R i,i+1 Ri+1,i+2 exp(2ik i+1,z (d i+1 d i )), RN,N+1 =, R i,i 1 = R i,i 1 + R i 1,i 2 exp(2ik i 1,z (d i d i 1 )) 1 + R i,i 1 Ri 1,i 2 exp(2ik i 1,z (d i d i 1 )), R1 =, with the Fresnel reection and transmission coecients for layers i and i + 1 as if they would be in half-space (Chew, page 49) R T E i,i+1 = µ i+1k iz µ i k i+1,z µ i+1 k iz + µ i k i+1,z, R T M i,i+1 = ɛ i+1k iz ɛ i k i+1,z ɛ i+1 k iz + ɛ i k i+1,z. Furthermore, R ji = R ij and the transmission coecient is T ij = 1 + R ij and k iz /µ i (1 R ij ) = k jz /µ j T ij. 1

13 2.5.3 Inside the source region If the source is embedded in same medium m, this wave is accompanied by the reection from the layer beneath and transmission wave from above. For TE or TM modes, we replace F by F (z, z ) = exp(ik mz z z ) + A m exp( ik mz z) + C m exp(ik mz z), (1) where k mz is the z-component of the wave vector in region m. The two extra terms are reections at the boundaries z = d m 1 and z = d m and constraints can be found for A m and C m. For z > z at z = d m 1, A m exp(ik mz d m 1 ) = R m,m 1 [exp(ik mz d m 1 + z ) + C m exp( ik mz d m 1 )] and for z < z at z = d m, C m exp( ik mz d m ) = R m,m+1 [exp(ik mz d m + z ) + A m exp(ik mz d m )], where R ij are the generalized reection coecients. Solving A m and D m from the two relations gives A m = exp( ik mz d m 1 ) R m,m 1 [ exp( ik mz (d m 1 + z )) + R m,m+1 exp(ik mz (2d m d m 1 + z ))] Mm C m = exp(ik mz d m ) R m,m+1 [ exp(ik mz (d m + z )) + R m,m 1 exp( ik mz (2d m 1 d m + z ))] Mm, with M m = [ 1 R m,m+1 Rm,m 1 exp(2ik mz (d m d m 1 )] 1. Using this F, the contributions by the reection and transmission waves to the dipoles can be derived. 11

14 2.5.4 Vertical electric dipole (VED) If the reected waves from the top and bottom layers are included and the components of the electric elds are ωɛẽx = 1 2 k x [± exp(ik z z z ) + A e v exp(ik z z) B e v exp( ik z z)] ωɛẽy = 1 2 k y [± exp(ik z z z ) + A e v exp(ik z z) B e v exp( ik z z)] k 2 ρ ωɛẽz = 1 [exp(ik z z z ) + A e v exp(ik z z) + Bv e exp( ik z z)], 2 k z and the magnetic eld components are H x = 1 2 k y k z [exp(ik z z z ) + A e v exp(ik z z) + B e v exp( ik z z)] H y = 1 2 k x k z [exp(ik z z z ) + A e v exp(ik z z) + B e v exp( ik z z)] (11) H z = Horizontal electric dipole (HED) In the stratied medium, the reected waves from the top and bottom layers are included and the electric eld components are ωɛẽx = 1 2 [ (k 2 k 2 x) 1 k z exp(ik z z z )+ 1 k 2 ρ ( k 2 x k 2 zb e h + k 2 yk 2 A e h) 1 k z exp( ik z z)+ 1 k 2 ρ ( k 2 x k 2 zd e h + k 2 yk 2 C e h) 1 k z exp(ik z z) ] ωɛẽy = 1 [ 1 2 k xk y exp(ik z z z )+ k z 1 ( k 2 kρ 2 z Bh e + k 2 A e 1 h) exp( ik z z) + 1 ] ( k 2 k z kρ 2 z Dh e + k 2 Ch) e 1 exp(ik z z) k z 12

15 ωɛẽz = 1 2 k x [± exp(ik z z z ) + B e h exp( ik z z) + D e h exp(ik z z)] and the magnetic eld components H x = 1 2 k x k y k 2 ρ [(A e h + B e h) exp( ik z z) + (D e h C e h) exp(ik z z)] H y = 1 2 exp(ik z z z ) [ (k 2 y A e h kxb 2 h) e exp( ik z z)+ k 2 ρ ( k 2 yd e h k 2 xc e h) exp(ik z z) ] (12) H z = 1 2 k y k z [exp(ik z z z ) + A e h exp( ik z z) + C e h exp(ik z z)] Note that the reected up-going wave D e h exp(ik zz) will have a minus sign due to the negative sign of the primary down going eld exp(ik z z z ) in ωɛẽz. For the HED in y-direction, change the role of x and y gives the z-components ωɛẽz = 1 2 k y [± exp(ik z z z ) + B e h exp( ik z z) + D e h exp(ik z z)] H z = 1 2 k x k z [exp(ik z z z ) + A e h exp( ik z z) + C e h exp(ik z z)]. Note the sign change in the magnetic eld component H z = Il x g(r, r ) Coecients The coecients A, B, C and D represent TM and TE waves as given in the table A e h A m h A e v A m v B e h B m h B e v B m v TE TM TM TE TM TE TM TE C e h C m h C e v C m v D e h D m h D e v D m v TE TM TE TM TM TE TM TE and are functions of the source location z. In the above notation we have (the superscripts e and m are omitted, d m < z < d m 1 ) 13

16 B v = A h = exp( ik mz d m 1 ) R m,m 1 [exp( ik mz (d m 1 + z )) + R ] m,m+1 exp(ik mz (2d m d m 1 + z (13) )) Mm A v = C h = exp(ik mz d m ) R m,m+1 [exp(ik mz (d m + z )) + R ] m,m 1 exp( ik mz (2d m 1 d m + z (14) )) Mm D v = B h = exp( ik mz d m 1 ) R m,m 1 [exp( ik mz (d m 1 + z )) R ] m,m+1 exp(ik mz (2d m d m 1 + z (15) )) Mm C v = D h = exp(ik mz d m ) R m,m+1 [ exp(ik mz (d m + z )) + R m,m 1 exp( ik mz (2d m 1 d m + z ))] Mm. (16) Note the minus signs appearing in the square brackets of coecients D v = B h and C v = D h. This is due to the minus sign of the down going wave from the HED (and HMD) solutions. Also useful are the relations between the coecients For m = 1, we may also write z A h = ik mz B h z C h = ik mz D h z B h = ik mz A h z D h = ik mz C h, A e v = C m h, B e v = A m h, D e v = B m h, C e v = D m h. and for m = N, D e h C e h = D m h C m h, A e h + B e h = A m h + B m h. 14

17 2.5.7 Summary These components build up the electric eld dyadic Green's function Ḡ E for problem dened on a stratied medium. The next subsection uses the Fourier modes of dipole solutions derived previously in the vector potential approach. 2.6 Vector potential approach In a source free medium, the magnetic ux B and electric ux D are soleniodal, i.e. B =, D =. Then the arbitrary vectors A and F, the potential vectors, exist such that B A = µh A = A, D F = ɛe F = F. The denition of the curl above together with the divergence of A and F, uniquely dene the vector potentials. We choose the Lorentz condition or gauge to dene the divergence (Balanis, page 257). Substituting this into the Maxwell equations, the total electric E = E A + E F and magnetic H = H A + H F elds yield (Balanis, page 26) [ E = iω Ī + ] k 2 A iω ɛ F (17) [ H = iω Ī + ] k 2 F + iω A. (18) µ So, a eld is a result of both vector potentials; the scatterer problem gives the vector potentials in terms of Green's functions and sources and is described in the next sections Green's function The Green's function is not unique and one of the commonly used is the traditional form for the vector potentials ḠA and ḠF (Michalski 199) and are given by 15

18 Ḡ A,F = G xx G xx. (19) G zx G zy G zz For the x- and y-horizontal dipoles only two components per column (direction) need to be specied (Sommerfeld, page 257). Using this form, the following integrals dene vector potentials for volume sources, with J the electric eld density and M the (non-physical) magnetic eld density, are dened (Balanis, page 276) A = Ḡ A (r, r ) J(r )dv (2) V V F = Ḡ F (r, r ) M(r )dv. (21) For a radiating surface S (in free space) with linear densities J S and M S A = Ḡ A (r, r ) J S (r )ds (22) S S F = Ḡ F (r, r ) M S (r )ds. (23) For magnetic and electric currents I e and I m these reduce to line integrals over C A = Ḡ A (r, r ) I e (r )dl C C F = Ḡ F (r, r ) I m (r )dl. The direct relation between the electric/magnetic Green's function from the previous section and vector potential Green's functions are µ ḠE (r, r ) = ɛ ḠH (r, r ) = [ Ī + ] Ḡ A (r, r ). (24) k 2 [ Ī + ] k 2 Ḡ F (r, r ). (25) In the next section, the elements in the vector potential Green's function are derived for an observer at z and source at z situated in the same layer. The solution outside the source layer is determined in the Appendix. 16

19 2.6.2 Observer inside source layer By using VED, HED, VMD and HMD dipoles the contributing elements to dyadic ḠA,F can be constructed. Use the TE z - and TM z -components of the elds generated by the dipoles and derive the components as explained in the previous section. Then look for a vector potential A in spectral domain such that or in spectral form A = µh A ik y à z ( z Ãy ) ik x à z z Ãx = ik x à y ik y à x µ H x µ H y µ H z The vector potential A only needs two components to be specied. The result will be the rst column in the dyadic ḠA. For a HED in x-direction, the spectral components of the magnetic eld is given by (12). The vector potential components for a HED become, with the choice à y =,. or ik y à z = µ H x, ik y à x = µ H z à x = µ 2ik z [exp(ik z z z ) + A e h exp( ik z z) + C e h exp(ik z z)], à y =, à z = µ k x 2i kρ 2 [(A e h + Bh) e exp( ik z z) + (Dh e Ch) e exp(ik z z)]. The HED in y-direction one chooses Ãx =, which implies or ik x à y = µ H z, ik x à z = µ H y à x =, 17

20 à y = µ 2ik z [exp(ik z z z ) + A e h exp( ik z z) + C e h exp(ik z z)], à z = µ k y 2i kρ 2 [(A e h + Bh) e exp( ik z z) + (Dh e Ch) e exp(ik z z)], which is the same as switching role x and y in the derivation for the HED in x-direction. For a VED the magnetic components are given by (11). Thus, the choice Ãy = in the potential vector results in or à x =, ik z à z = µ H y, ik y à z = µ H x à x =, à y =, à z = µ 2ik z [exp(ik z z z ) + A e v exp(ik z z) + B e v exp( ik z z)]. All components of the vector potential Green's functions G A,F can be obtained (cf. Dural & Aksun 1995) G A xx = µ 2ik z [exp(ik z z z ) + A e h exp( ik z z) + C e h exp(ik z z)] G A zx = µ [ kx k z 2ik z kρ 2 (A e h + Bh) e exp( ik z z) + k ] xk z kρ 2 (Dh e Ch) e exp(ik z z) G A zz = µ 2ik z [exp(ik z z z ) + A e v exp(ik z z) + B e v exp( ik z z)] G F xx = ɛ [exp(ik z z z ) + A m h exp( ik z z) + Ch m 2ik z exp(ik z z)] G F zx = ɛ [ kx k z 2ik z kρ 2 (A m h + Bh m ) exp( ik z z) + k ] xk z kρ 2 (Dh m Ch m ) exp(ik z z) 18

21 G F zz = ɛ [exp(ik z z z ) + A m v exp(ik z z) + Bv m exp( ik z z)]. 2ik z The elements from the y-oriented HED and HMD follow directly by rotation in the x, y-plane G A,F yy = A,F G xx, GA,F zy /k y = G A,F zx /k x. The spatial Green's function are obtained using the inverse Fourier transforms G A xx = F 1 ( G) = S [ G] = S [ G A xx] = 1 G A 2π xxj (k ρ ρ)k ρ dk ρ = µ 4πi [exp(ik z z z ) + A e h exp( ik z z) + C e h exp(ik z z)] k ρ k z J (k ρ ρ)dk ρ = µ exp(ik r r ) 4π r r µ 4πi [A e h exp( ik z z) + C e h exp(ik z z)] k ρ k z J (k ρ ρ)dk ρ G A zx = F 1 (ik x G) = cos(φ)s1 [ G] = cos(φ)s 1 [ GA zx ik x ] = cos(φ) 1 G A zx J 1 (k ρ ρ)k 2π ik ρdk 2 ρ x = µ 4π cos(φ) [(A e h + B e h) exp( ik z z) + (D e h C e h) exp(ik z z)] J 1 (k ρ ρ)dk ρ, G A zz = F 1 ( G) = S [ G] = S [ G A zz] = 1 G A 2π zzj (k ρ ρ)k ρ dk ρ = µ 4πi [exp(ik z z z ) + A e v exp(ik z z) + B e v exp( ik z z)] k ρ k z J (k ρ ρ)dk ρ = µ exp(ik r r ) 4π r r µ 4πi [A e v exp(ik z z) + B e v exp( ik z z)] k ρ k z J (k ρ ρ)dk ρ. 19

22 1 2 3 Figure 2: Waves reection Example In the example problem N = 3, where sources are in region 1 and 2. interfaces are located at z = d 1 and z = d 2. First, a source at z = z in region 1 (m = 1 and n = 1). reection coecient is with k 2z = R 12 = R 12 + R 23 exp(2ik 2z (d 2 d 1 )) 1 + R 12 R 23 exp(2ik 2z (d 2 d 1 )) The The generalized k 2 2 k2 ρ (and M 1 = 1). The spectral Green's functions become G A xx = µ 1 2ik 1z [exp(ik 1z z z ) + C e h exp(ik 1z z)] G A zx = µ [ ] 1 kx k 1z 2ik 1z kρ 2 (Dh e Ch) e exp(ik 1z z) G A zz = µ 1 2ik 1z [exp(ik 1z z z ) + A e v exp(ik 1z z)] G F xx = ɛ 1 2ik 1z [exp(ik 1z z z ) + C m h exp(ik 1z z)] G F zx = ɛ [ ] 1 kx k 1z 2ik 1z kρ 2 (Dh m Ch m ) exp(ik 1z z) G F zz = ɛ 1 2ik 1z [exp(ik 1z z z ) + A m v exp(ik 1z z)], with coecients (either TE or TM) A v = C h = R 12 exp(ik 1z (2d 1 + z )) 2

23 D h = R 12 exp(ik 1z (2d 1 + z )). To obtain the closed-form spatial Green's functions by doing the Fourier inverses (3). The construction of the solution for observers outside the source layer is described in the Appendix. Now, we have relations for the elds as function of the sources applied in the whole domain. We use these to calculate the elds in the desired subdomain Summary The electric Green's function is obtained using the dipole solution directly (sections and 2.5.5) or by applying (24) to the vector potential Green's function µ Ḡ E = 1 k 2 k 2 kx 2 k x k y ik x z k x k y k 2 kx 2 ik y ik x z ik y z z k 2 k 2 z Ḡ A and similarly ɛ Ḡ H = 1/k 2 [ ] Ḡ F by duality. 3 Far-eld reconstruction The ingredient for the calculation of RCS for a scatterer is a far-eld description of the electric eld E. The far-eld is constructed using Green's functions for a given electric and magnetic currents on the scatterer. The idea is to evaluate or approximate the convolutions to obtain the far-eld quantities as function of the currents in the scatterer. 3.1 Radar Cross Section The bi-static RCS is dened in terms of the time-averaged Poynting vector [ RCS(θ, φ) lim 4πr 2 P ] scat, (26) r P inc where P i = 1 2 Re(E H ) is the power density. A property of solutions for point sources in homogeneous media is the exp(ikr)/r behavior in the far eld. Furthermore, it can be shown that P i = Eθ i 2 + Eφ i 2 for the far-eld, where the radial components E r and H r vanish (Balanis, page 28). 21

24 For 2D problems, the bistatic scattering width is dened as SW (θ) = lim r [2πr Es 2 E i 2 The observation points far from the scatterer, r 2D 2 /λ, denes the far-eld (Fraunhofer) region, where D is the maximal dimension of the scatterer and λ is the wavelength. In this region the maximum phase error is π/ (Balanis, page 286). The limit is the minimum distance for the RCS to be valid. Another limit denes the near-eld r.62 D 3 /λ 3 (Fresnel). For r in a homogeneous medium, the elds are orthogonal to the radial direction (plane waves and 1/r 2 ); we may write in terms of vector potentials A and F (Balanis, page 285). ]. 4 Numerical evaluation All representations involve (multiple) integrals over innite (frequency) domains of rapid oscillating functions containing singularities. Also the sheer number of integrals, for each location requires a new evaluation, makes the evaluation of the solution costly. Straightforward numerical integration requires many sample points or do not even converge due to singularities. The next subsections describe techniques to deal with the problems. The path in the Sommerfeld intergrals can be extended to the complex plane to avoid singularities. Some singularities can be subtracted analytically. Finally, the integrand can be approximated by exponentials, which integral is derived analytical. 4.1 Sommerfeld integration path (SIP) Sommerfeld integrals can be evaluated using Cauchy theorem. The integral I = f(x)dx is an integration of complex function f along the real x-axis. Suppose the function f has a singularity at x = x. A new contour C is parametrized by s(t), with s() = and s( ) = can be dened and I = C f(z)dz = f(s(t)) ds dt dt. 22

25 Figure 3: Real part of ln of a typical integrand. Im(k ) ρ k =k ρ Re(k ) ρ Figure 4: FSIP, dot represents a pole. For semi-innite Sommerfeld integrals, a pole is at k ρ = k. In this context, the contour s(t) is the so-called Sommerfeld integration path (SIP) or folded SIP (FSIP). 4.2 Singularity subtraction For sources and observation point are close, the integrand in the Sommerfeld integrals can be highly oscillatory for large frequencies and singular in the limit. However, we have (Sommerfeld) exp(ik z α) k z J (k ρ ρ)dk ρ = exp(ikr), r = ρ r 2 + α 2 exp(k ρ α)j 1 (k ρ ρ)dk ρ = 1 ρ ( 1 + ) α. ρ2 + α 2 Take the limit of the integrand for k ρ and use the above identities to add and subtract the singularity from the integrand. 23

26 Im k Ρ Re k Ρ Figure 5: Deformed SIP for GPOF method 4.3 GPOF The general pencil-of-function method (GPOF) can be used to approximate the integrand in the Sommerfeld integrals. The idea is to approximate the Fourier components, for given z and z, as function of k ρ in a sum of exponentials, i.e G(k ρ (t)) z,z = µ 2ik z M b i exp(s i t) where t is the running variable in a linear parametrization of k z (t)/k = α + βt and by using k ρ (t) = k 2 kz(t). 2 Then we can write as a function of k z (k ρ ) = k 2 kρ 2 i=1 G(k ρ ) z,z = µ 2ik z M i=1 ( ) k z (k ρ )/k α b i exp s i = β µ 2ik z M i=1 ( ) ( α b i exp s i exp ik z (k ρ ) s ) i. β iβ Use the Sommerfeld identity here to get a closed-form 3D Greens' function G(r, r ) z,z = F 1 ( G) = µ M i=1 ( ) α exp(ikri ) b i exp s i β 4πr i with r i = (ρ ρ ) 2 (s i /β) 2 (which are complex functions). The residues b i and the poles s i are obtained using GPOF method (Cheng & Yang, 25). The approximation can be improved by taking two piecewise linear parametrizations of the SIP in the complex k z plane. One starting from k going to the positive imaginary axis and the other is on the positive imaginary axis. This is the two-level DCIM (Aksun 1996, Aksun 25). The G A zx and G A zy elements are approximated in the same way. We have F 1 ( G A zx/ik x ) = G A zxdx, when using G prim (r) = exp(ikr)/(4πr), we get 24

27 ( ( )) G A zx = Gzx F 1 x ik x = µ M i=1 ( ) α dg prim r i b i exp s i β dr x, with dg prim dr = d dr ( ) exp(ikr) 4πr = (ikr 1) exp(ikr) 4πr 2. For second-order derivatives (e.g. in G E xx and G E yy) 2 G prim x 2 = d2 G prim dr 2 ( ) 2 r + dgprim 2 r x dr x 2 and mixed derivatives (e.g. in G E xy) 2 G prim x y = d2 G prim dr 2 r r x y + dgprim dr 2 r x y both using d 2 G prim dr 2 = ( (ikr 1) ) exp(ikr) 4πr 3. The partial derivatives of the radius r = r(x, y) are r x = x x r 2 r x = 1 2 r (x x ) 2 r 3 2 r x y = (x x )(y y ) 1 The derivatives / x(f 1 (...)) can be done in a straightforward fashion (either symbolically or numerically). r Summary Assuming that the current density J(r ) in a domain Ω is given, the solution to the Maxwell equations can be constructed by the following steps in the vector potential approach (no need for the equivalence principle). The main ingredient is the Green's function ḠA. 25

28 4.4.1 Green's function 1. Each column in the (dyadic) Green's function (19) is derived from the solutions for 2 HEDs and a VED, using equations (9) and (8). 2. These solutions are expanded in planar waves in z-direction using Fourier transformations (3), resulting in Sommerfeld type integrals. 3. The stratied medium is included by adding the reected waves from the adjacent layers as in (1), which are either TE or TM waves. 4. The vector potential A for the electric eld is the integral (2). 5. The eld E outside that domain is evaluated by relations (17) and (18). The resulting expression for the total eld can be evaluated numerically Numerical evaluation 1. Dene observer points in which the Ī + /k2 operator can be approximated by nite dierence (central dierence and bi-linear interpolation). 2. Do volume integration of the potential vector (2) involving the product Ḡ A (r, r ) J(r ) for the components (ḠA J) x, (ḠA J) y and (ḠA J) z for each observer point (adaptive rules like Romberg or Filon integration; Gauss-Chebychev or Clenshaw-Curtis integration rules). 3. Evaluate the Sommerfeld integrals contained in ḠA using a SIP as depicted in Figure 4 (integration rules for slowly decaying, highly oscillating integrands) or create closed-form approximations using GPOF method. The evaluation of -operator can be avoided by using the Green's function for the electric eld ḠE. This involves more Sommerfeld integrals to be evaluated. 5 2D problem The relations for the full 3D Maxwell equations reduce when the solution is (semi)constant in one direction (say y-direction). The derivations in the next section follow the same recipe as in the 3D case. 5.1 Homogeneous medium The Green's function is a dyadic function and has the same functionality as in 3D. However, the fundamental solution g(r, r ) is the solution of the 2D Helmholtz equation. 26

29 5.2 Helmholtz equation The fundamental solution to the 2D Helmholtz equation, is ( 2 + k 2 )g(r, r ) = δ(x x )δ(z z ) g(r, r ) = i 4 H(1) (k r r ). Suppose that the Fourier transform g of the solution exists g(r, r ) = 1 2π g(k x, k z ) exp(ik x (x x ) + ik z (z z ))dk x dk z. The Fourier transform of the 2D Helmholtz equation (k 2 kx k 2 z) g(k 2 x, k z ) exp(ik x (x x )+ik z (z z ))dk x dk z = exp(ik x (x x )+ik z (z z ))dk x dk z implies the Fourier components 1 g(k x, k z ) = k 2 kx 2 kz 2 eliminating k z with poles k z = ± k 2 k 2 x and using Jordan's Lemma (allowing a small loss) gives g(r, r ) = i 4π exp(ik x (x x ) + ik z z z ) k z dk x = i 2π exp(ik z z z ) k z cos(k x (x x ))dk x with k z = k 2 k 2 x. Here we recognize the F (z, z ) = exp(ik z z z ) term, which allows the use of equation (1) and the other following derivations. 27

30 5.3 Fourier integrals The inverse Fourier transform F 1 ( G) = 1 π G(k x ) cos(k x (x x ))dk x, F 1 (ik x G) = x G = 1 G(k x ) sin(k x (x x ))k x dk x, π F 1 (kx 2 G) = 2 x 2 G = 1 G(k x ) cos(k x (x x ))k π xdk 2 x. The Fourier transform of the fundamental solution is (cf. Weyl identity) F ( ) i 4 H(1) (k r r ) = i exp(ik z z z ). (27) 2 k z The next subsections discusses the oblique incidence case (TE/TM) and the conical case (general). 5.4 Hertzian dipoles The 2D solution satises the 3D Maxwell equations. So, a dipole at r = r (is actually a line source at x = x and z = z ) gives ( E(r) = iωµ Ī + ) k 2 ˆαIl i 4 H(1) (k r r ) H(r) = ˆαIl i 4 H(1) (k r r ). We use the dipoles to derive the columns of the dyadic function. Suppose Il = 1, then the HED ˆα = ˆx (e.g. J(r) = ˆxIlδ(r r )) gives the TM and TE components E z = i i ωɛ z x 2 4 H(1) (k r r ), H z =, the HED ˆα = ŷ gives the TM and TE components 28

31 E z =, H z = i x 4 H(1) (k r r ), and the VED ˆα = ẑ gives the TM and TE components E z = i ) (k i ωɛ z 2 4 H(1) (k r r ), H z =. From these relations, the spectral z-components are easily extracted. The other components are derived from (6) and (7), with k ρ = k x, or Ẽ x = i k x z Ẽz, Ẽ y = 1 ωµ k H z x H x = i k x z H z, Hy = 1 ωɛẽz k x For the ˆα = ˆx, the components are (TM) ωɛẽx = 1 2 k z exp(ik z z z ), Ẽy =, ωɛẽz = ± 1 2 k x exp(ik z z z ) for ˆα = ŷ, (TE) H x =, Hy = 1 2 exp(ik z z z ), Hz = Ẽ x =, ωɛẽy = 1 k 2 exp(ik z z z ), 2 k Ẽz =, z H x = ± 1 2 exp(ik z z z ), Hy =, and for ˆα = ẑ, (TM) H z = 1 2 k x k z exp(ik z z z ) ωɛẽx = ± 1 2 k x exp(ik z z z ), Ẽy =, 29

32 ωɛẽz = 1 kx 2 exp(ik z z z ) 2 k z H x =, Hy = 1 2 k x k z exp(ik z z z ), H z =. These are columns of the electric and magnetic dyadic Green's function. 5.5 Vector potential The 2D Green's function ḠA for stratied media will be (call it the traditional form) Ḡ A = G A xx G A yy G A zz. For the construction of the vector potential we have to satisfy or the spectral representation µh A = A z Ãy ( ) ik x à z z Ãx ik x à y = µ H x µ H y µ H z For HED ˆα = ˆx, from H x = H z = directly follows Ãy = and with the choice à z = we have. à x = µ 2ik z exp(ik z z z ), à y = Ãz =. HED ˆα = ŷ, from H y = with the choice Ãz = follows à x =, à y = µ 2ik z exp(ik z z z ), à z =. VED ˆα = ẑ, from H x = (and H z = ) that Ãy = with the choice Ãx = à x = Ãy =, à z = µ 2ik z exp(ik z z z ). 3

33 5.6 Stratied media Adding the stratied media to the expressions do not add extra elements to the dyad ḠA. For HED ˆα = ˆx the previous results are replaced by (TM) ωɛẽx = 1 2 k z [exp(ik z z z ) B e h exp( ik z z) + D e h exp(ik z z)] Ẽ y = ωɛẽz = 1 2 k x [± exp(ik z z z ) + B e h exp( ik z z) + D e h exp(ik z z)], H x = H y = 1 2 [± exp(ik z z z ) + B e h exp( ik z z) + D e h exp(ik z z)] thus with z Ãx = µ 1 k x ωɛẽz we get H z = Ã x = µ 2ik z [exp(ik z z z ) B e h exp( ik z z) + D e h exp(ik z z)]. The Ãx is the rst element of the rst column in the dyad (TM) G A xx = µ 2ik z [exp(ik z z z ) + B e h exp( ik z z) + D e h exp(ik z z)]. The other components are derived in a similar way (TE and TM, respectively) G A yy = µ 2ik z [exp(ik z z z ) + A e h exp( ik z z) + C e h exp(ik z z)], G A zz = µ 2ik z [exp(ik z z z ) + A e v exp(ik z z) + B e v exp( ik z z)]. Applying the inverse Fourier transform nally gives 31

34 G A xx = F 1 ( G A xx) = 1 π µ 2πi G A xx(k x ) cos(k x (x x ))dk x = [exp(ik z z z ) + B e h exp( ik z z) + D e h exp(ik z z)] 1 k z cos(k x (x x ))dk x = µ i 4 H(1) (k r r ) µ 2πi [B e h exp( ik z z) + D e h exp(ik z z)] 1 k z cos(k x (x x ))dk x, G A yy = F 1 ( G A yy) = 1 π µ 2πi G A yy(k x ) cos(k x (x x ))dk x = [exp(ik z z z ) + A e h exp( ik z z) + C e h exp(ik z z)] 1 k z cos(k x (x x ))dk x = µ i 4 H(1) (k r r ) µ 2πi [A e h exp( ik z z) + C e h exp(ik z z)] 1 k z cos(k x (x x ))dk x, G A zz = F 1 ( G A zz) = 1 π µ 2πi G A zz(k x ) cos(k x (x x ))dk x = [exp(ik z z z ) + A e v exp(ik z z) + B e v exp( ik z z)] 1 k z cos(k x (x x ))dk x = µ i 4 H(1) (k r r ) µ 2πi [A e v exp(ik z z) + B e v exp( ik z z)] 1 k z cos(k x (x x ))dk x. The coecients B e h, De h, Ae h, Ce h, Ae v and C e v are solved at the boundaries of the source region and are given by relations (13) to (16). In a similar way the total elds for a multi-layer medium are derived. For ˆα = ŷ, we get (TE) Ẽ x = ωɛẽy = 1 k 2 [exp(ik z z z ) + A e h exp( ik z z) + Ch e exp(ik z z)] 2 k z 32

35 Ẽ z =, H x = 1 2 [± exp(ik z z z ) A e h exp( ik z z) + C e h exp(ik z z)] H y = For ˆα = ẑ, (TM) H z = 1 2 k x k z [exp(ik z z z ) + A e h exp( ik z z) + C e h exp(ik z z)] ωɛẽx = 1 2 k x [± exp(ik z z z ) + A e v exp(ik z z) B e v exp( ik z z)] Ẽ y = ωɛẽz = 1 kx 2 [exp(ik z z z ) + A e v exp(ik z z) + Bv e exp( ik z z)], 2 k z H x = H y = 1 2 k x k z [exp(ik z z z ) + A e v exp(ik z z) + B e v exp( ik z z)] H z =. The columns in dyadic Green's function Ḡ E are the corresponding dipole solutions Ẽ. The non-zeros in this dyadic for the electric eld are G E xx ( = i ) k x z G E zx = 1 1 k 2 2i k z [exp(ik z z z ) Bh e exp( ik z z) + Dh e exp(ik z z)] G E zx = 1 k 2 1 2i k x [± exp(ik z z z ) + B e h exp( ik z z) + D e h exp(ik z z)] G E yy = 1 1 [exp(ik z z z ) + A e h exp( ik z z) + Ch e exp(ik z z)] 2i k z 33

36 G E xz ( = i ) k x z G E zz = 1 1 k 2 2i k x [± exp(ik z z z ) + A e v exp(ik z z) Bv e exp( ik z z)] G E zz = 1 1 kx 2 k 2 [exp(ik z z z ) + A e v exp(ik z z) + Bv e exp( ik z z)]. 2i k z Note again that G A µg E E/(iωµ) or µg E ωɛe/ik 2. This is also obtained by applying (24) to the vector potential Green's function, giving the expression µ Ḡ E = 1 k 2 k 2 k 2 x ik x z k 2 ik x z k 2 k 2 z Ḡ A. Here you can see that for the TM z case, only the G yy component is needed. 5.7 Conical case In the conical case, the solution has the same y-dependency as the incoming eld, namely g(y) = exp(ik y y). We need a reformulation of the Maxwell equation to get the 2D equation for this case. Start with (1) and solve this with the separation of variables E(x, y, z) = Ê(x, z)g(y) and J(x, y, z) = Ĵ(x, z)g(y). The -operator reduces to ˆ = [ x + ik y + z ] T. So becomes [ 2 E(r) + k 2 E(r) = iωµ Ī + ] k 2 J(r) ( ) [ 2 x z 2 Ê(x, z) + ˆk 2 Ê(x, z) = iωµ Ī + with ˆk 2 = k 2 k 2 y and ˆ ˆ = 2 x ik y x 2 xz ik y x k 2 y ik y z 2 zx ik y z 2 z ] ˆ ˆ Ĵ(x, z), This again leads to the solution using 2D Green's function (free space). k 2 34

37 ˆḠ E (r, r ) = [ Ī + ] ˆ ˆ ĝ(r, r ), k 2 with ĝ(r, r ) is the fundamental solution to the 2D Helmholtz equation ĝ(r, r ) = i 4 H(1) (ˆk2 r r ) = i 2π exp(ik z z z ) k z cos(k x (x x ))dk x and k z = ˆk2 k 2 x. The stratied media is added as done before and the relation between the eld and vector potential Green's function is given by µ ˆḠ E = 1 k 2 k 2 kx 2 k y k x ik x z k y k x k 2 ky 2 ik y ik x z ik y z z k 2 k 2 z ˆḠ A. (28) Note that k y is a constant xed by the incident wave. Following the denition in part 1, set k y = k sin(θ) sin(φ). 5.8 GPOF The 2D Green's functions are easily obtained by substituting the term G prim (r) = (i/4)h (1) (kr) in which the radius reduces to ρ ρ = x x. For the derivatives use dg prim dr = d dr ( ) i 4 H(1) (kr) = ik 4 H(1) 1 (kr) and d 2 G prim dr 2 = ik2 8 ( H (1) ) (kr) H(1) 2 (kr). For the conical case we apply (28) to obtain the eld Green's function. In this Ĝ A xx, Ĝ A yy, case 7 function need to be approximated, i.e. z Ĝ A zz and 2 z ĜA 2 zz, or if the primary term is subtracted Ĝ A zz, z Ĝ A xx Ĝ prim = µ 1 [Bh e exp( ik z z) + Dh e exp(ik z z)] 2i k z Ĝ A xx, z Ĝ A yy, 35

38 Ĝ A yy Ĝ prim = µ 1 [A e h exp( ik z z) + Bh e exp(ik z z)] 2i k z Ĝ A zz Ĝ prim = µ 1 [A e v exp(ik z z) + Bv e exp( ik z z)] 2i k z ( ĜA z xx Ĝ prim) = µ 2 [ Be h exp( ik z z) + Dh e exp(ik z z)] ( ĜA z yy Ĝ prim) = µ 2 [ Ae h exp( ik z z) + Bh e exp(ik z z)] ( ĜA z zz Ĝ prim) = µ 2 [Ae v exp(ik z z) Bv e exp( ik z z)] 2 z 2 ( ĜA zz Ĝ prim) = µ 2i k z [A e v exp(ik z z) + B e v exp( ik z z)]. The GPOF sum can directly be transformed to the closed-form Green's functions using identity (27). The x-derivatives are applied analytically to these functions using the derivatives of the primary Green's functions as described above, with k = ˆk. 5.9 Far-eld For the 2D case you can nd the expressions in the book of Allen Taove (Taove 2), where the Hankel functions are approximated. In order to obtain the leading order far-eld spectral eld, we can use a stationaryphase approximation. This gives a relation of the spectral eld at innity in a xed direction from the origin. Suppose the electric eld at z component can be written as E(x, z) = + Ê(k x ; ) exp(ik x x + ik z z)dk x. We are interested in the far-eld value of this eld at r in the direction s = ( x r, z r ), where r = x 2 + z 2 is the distance from r to the origin. So, we may write 36

39 E (s x, s z ) = k x k ( ( kx Ê(k x ; ) exp ikr k s x + k )) z k s z dk x. if the evanescent waves decay at innity (homogeneous waves). Changing variable p = k x /k we then have E (s x, s z ) = p 1 kê(kp; ) exp (ikr (ps x + m(p)s z )) dp, with m(p) = 1 p 2. Now, we have the far-eld E written in the desired form. Given a direction s = (s x, s z ) (remember that s z = 1 s 2 x), approximate the function F (κ; s x ) = 1 1 a(p) exp (iκg(p; s x )) dp, with F (κ; s x ) = E (s x, s z ), κ = kr, a(p) = kê(kp; ) and g(p; s x) = ps x + ms z, using the stationary-phase method as κ. It can be shown that the critical points of the second kind (end points of the integration path) are of order 1/κ. Let we continue with the critical points of the rst kind, thus the points p 1 where g (p 1 ) =. We have g (p) = s x + m s z = s x p m s z ( ) m pm g (p) = s z m 2 = s ) z (1 + p2 m m 2. The (rst and only) critical point is at p 1 = s x and concequently m 1 = s z. So, [ ( ) ] 2 g sx (p 1 ) = 1 + = 1 s z s 2 z < and we have the approximation, as κ F (1) (κ; s x ) The electric eld is then given by ( 2π κg (p 1 ) a(p 1) exp(iκg(p 1 )) exp iπ ). 4 37

40 E (s x, s z ) s 2 z ( 2π kr kê(ks x; ) exp(ikr) exp iπ ) 4 where the far-eld approximation of the Hankel function H (1) (kr) can be recognized, or E (s x, s z ) ks z πê(ks x; )H (1) (kr). For a given direction s, the mode k x = ks x (k z = ks z ) is the only contribution for r and is given by Ê(k x ; ) = E (k x /k, k z /k) πk z H (1) (kr). Plugging this back into the Fourier transformation we started with gives E(x, z) = πh (1) 1 (kr) k x k E (k x /k, k z /k) 1 k z exp(ik x x + ik z z)dk x. For a analytical solution to a problem, the far-eld E should be derived. For instance, the general solution to a conducting or dielectric cylinder E(x, z) = n= n= A n H n (1) (kr) exp(inφ) using the far-eld aproximation for the Hankel functions, we arrive at Ê(k x ; ) = This simplies to An 2 πkr Ê(k x ; ) = 1 πk z exp(i(kr π/4)) exp( inπ/2) exp(inφ) 2 πk z exp(i(kr π/4)) n= πkr n= A n exp(in(π/2 φ)). Higher-order terms can be obtained using the Method of Steepest Descent (Chew, page 82). 38

41 6 1D problem For the sake of competeness, the one-dimensional Helmholtz equation has the trivial solution g(z, z ) = i 2 exp(k z z ). k Using this in the derivation of the vector potential, the (free space) Green's function reduces to µ ḠE (z, z ) = ḠA (z, z ) = µ 1 1 which convolution gives TEM z waves (H z = E z = ). g(z, z ), Inclusion of multiple layers follows the same way as in 2D and 3D. Note that k z k is the only wave number present in a 1D problem. Basically, g(z, z ) is a single Fourier mode. Also for the 1D problem the oblique and conical case the Green's functions can be derived in the same way. 7 Integral equation The Green's function convolutions give rise to integral equations (integro-dierential equations) for the electric elds (EFIE and MPIE). The equations can be solved using for instance the Method of Moments (MoM). 7.1 Equivalence principle Huygens' principle or surface equivalence theorem uses the Gauss' divergence theorem to replace the volume integral with a integral over surface S and S inf. Assuming that the solution and Green's functions vanish at S inf, we may write (Chew, page 32) E(r ) = S n E(r) ḠE (r, r ) iωµ(n H(r)) ḠE (r, r )ds. It can be shown (Chew, page 32) that this is equivalent to 39

42 S inf r S Figure 6: Equivalence principle. E(r ) = S Ḡ H (r, r) n E(r)dS + iωµ S Ḡ E (r, r) n H(r)dS, which is similar to the potentional approach (22) and (23), with J S = n H and M S = n E (use /k 2 =, relations (24) and (25)). From the uniqueness theorem (Chew, page 32), the solution can be derived either from given J S or M S, provided that the Green's functions satisfy the boundary conditions ḠH = or ḠE = on the surface S, respectively. For an unbounded homogeneous medium ḠH = ḠE. 7.2 EFIE The Green's function can be dened via the contrast eld E c = E E m, where E m is the known background solution. This eld saties the Maxwell equation or E c k 2 E c = k 2 E m E m E c k 2 me c = (k 2 k 2 m)(e m + E c ), where k m is the wave number in the stratied media (without the scatterer). Using the Green's function ḠE for multi-layer media we obtain the following electric eld integral equation (EFIE) for E c (use iωµj = (k 2 k 2 m)(e m + E c )) E c (r) = V Ḡ E (r, r )(k 2 res k 2 )(E m (r ) + E c (r ))dr, 4

43 with V is the scattering object area, k res and k constant material properties in the scatterer (k 2 k 2 m vanishes outside this area). Rearranging terms gives E c (r) k 2 res k 2 V Ḡ E (r, r )E c (r )dr = V Ḡ E (r, r )E m (r )dr. The 2D TE z half plane problem, with E m = ŷe m y, simplies the EFIE considerably. In this case, Ḡ E = ḠA /µ resulting in E c y(r) k 2 res k 2 1 µ V G A yy(r, r )E c y(r )dr = 1 µ V G A yy(r, r )E m y (r )dr, Take E c y(r) = N n=1 a ng n (r) and N test functions w m (r) = δ(r r m ), with r m the locations in the center of segment S m. Note that this method of moments is now similar to the point-matching method. The basis functions are piecewise constant g n (r) = 1 for r in segment S n and zero outside. E m y (r) is the background plane wave solution for the problem without scatterer (normal incident) E m y (r) = exp( ik 1z z) + R 12 exp(2ik 1z d 1 + ik 1z z) which is evaluated in all r m and approximated by E m y (r) = N m=1 Em y (r m )g m (r). The EFIE can be written as a linear system [Z mn ] [I n ] = [V m ] with A mn = 1 µ 7.3 MPIE 1 Z mn = kres 2 k 2 A mn, V m = A Ey m (r m ), I n = a n, S n G A yy(r m, r )dr. The Mixed Potential Integral Equation (MPIE) is a reformulation of the EFIE and is given by E(r) = iω V Ḡ A (r, r ) J(r )dr + 1 k 2 φa (r), with φ A (r) = V [ Ḡ A (r, r ) ] J(r )dr 41

44 The equation of continuity states that J = iωq. By assuming that there exist a scalar function K A φ (r, r ) and a vector function P A (r, r ), such that ḠA (r, r ) = K A φ (r, r ) + P A (r, r ), then (provided that the scatterer does not intersect with the interfaces, Michalski 199) φ A (r) = V Kφ A (r, r ) J(r )dr + = iω V V P A (r, r ) J(r )dr Kφ A (r, r ) q(r )dr + P A (r, r ) J(r )dr. Add P A to ḠA to obtain an alternative dyadic Green's function K A = ḠA + P A. Together with the scalar potential kernel Kφ A, the solution can be written as the mixed eld integral equation E(r) = iω V K A (r, r ) J(r )dr + iω k 2 V V Kφ A (r, r ) q(r )dr. The choice P A x = P A y = result in the alternative Green's function and has all the continuity properties in z and z at the interfaces (Michalski 199). 8 Appendix In the appendix, various problems are presented which have known analytical solutions. Furthermore, the Green's function for observers that are not in the same layer as the sources is given. At the end, some useful identities are given. 8.1 Problems The following problems can be solved analytically and are used to validate the methods Conducting cylinder The perfectly conducting cylinder is oriented in the direction of the z-axis (Balanis, page 63 converted to exp( iωt) convention). Dene the total electric eld as E t = E i + E s, with 42

45 E i = ẑe i z = ẑe exp(ikx) = ẑe n= i n J n (kρ) exp(inφ). The boundary condition for PEC, n E t = implies E t z =, and the outgoing waves are combination of Hankel functions H (1) n E s z = n= Then, for all n at the boundary ρ = a, c n H n (1) (kρ). Solving coecients c n The total eld is E i n J n (ka) exp(inφ) + c n H (1) n (ka) =. E s z(φ, ρ) = E c n = i n J n (ka) E H n (1) (ka) exp(inφ) n= i n J n (ka) H (1) n The induced current is (Balanis, page 65) J S (φ) = ẑ 2E πaωµ (ka) H(1) n n= i n exp(inφ) (kρ) exp(inφ) H n (1) (ka) The vector potential A for this problem is dened as A = ẑµ 2π i 4 H(1) (k r r )J S (φ )a dφ. Note that from the vector potential approach E s = iω(i + /k 2 )A = iωa as in this problem the vector A only has a z-component. 43

46 8.1.2 Dielectric cylinder The material inside the cylinder with radius a has refraction index n 2 and outside the cylinder n 1, with the corresponding wave numbers k i = 2πn i /λ. The the scattered eld is given by (exp(iωt)-convention) E s y(ρ, φ) = B ext n n= A n H n (2) (k 1 ρ) exp(inφ) A n = Bn ext /C n [ ( π )] = i n exp in 2 + θ C n = H(2) n (k 1 a)j n(k 2 a) H n (2) (k 1 a)j n (k 2 a) J n (k 1 a)j n(k 2 a) J n(k. 1 a)j n (k 2 a) 8.2 Green's function for the stratied medium The case that the observer is outside the source region is presented here Outside source region The solution has amplitudes that are related to the amplitudes on the boundary of the source region (Chew, page 79). If z R m and z R n, n < m (above the source), the up going eld can be written as F + (z, z ) = A + n [ exp(ik nz z) + R ] n,n 1 exp( 2ik nz d n 1 ik nz z) Factor A + n is related to the amplitude at the upper boundary of the source region z = d m 1 A + m = exp( ik mz (z + d m 1 )) + C m exp( ik mz d m 1 ) + A m exp(ik mz d m 1 ) or A + m = [exp( ik mz z ) + exp(ik mz (z + 2d m )) R m,m+1 ] exp( ik mz d m 1 ) M m 44

47 in a recursive way as A + i exp(ik izd i ) = A + i+1 exp(ik i+1,zd i )S + i+1,i (29) S + i+1,i = T i+1,i 1 R i,i+1 Ri,i 1 exp(2ik iz (d i d i 1 )). Similarly, if z R m and z R n, n > m (beneath source), we have the down going wave F (z, z ) = A n [ exp( ik nz z) + R ] n,n+1 exp(2ik nz d n + ik nz z). Factor A n z = d m is related to the amplitude at the lower source region boundary A m = exp(ik mz (z + d m )) + A m exp(ik mz d m ) + C m exp(2ik mz d m ) or A m = [exp(ik mz z ) + exp( ik mz (z + 2d m 1 )) R m,m 1 ] exp(ik mz d m ) M m in a recursive way as A i exp(ik izd i ) = A i 1 exp(ik i 1,zd i 1 )S i 1,i (3) S i 1,i = T i 1,i 1 R i,i 1 Ri,i+1 exp(2ik iz (d i d i 1 )) Observer outside source layer The elds outside the source region m are related to the amplitude A + m and A m of outgoing elds at the boundary of the source region. At z = d m 1 and z = d m, the amplitudes of the x-component of the waves from a HED/HMD dipole A + m = exp( ik mz (z + d m 1 )) + C h exp( ik mz d m 1 ) + A h exp(ik mz d m 1 ) A m = exp(ik mz (z + d m )) + C h exp( ik mz d m ) + A h exp(ik mz d m ), 45