376632-1-T Three Dimensional Moment Method Simulation of Penetrable Scatterers Consisting of Non-metallic and Circuit Analog Surfaces Erdem Topsakal March 1998 376632-1-T = RL-2520

376632-1-T E. Topsakal M. Carr J. L. Volakis Three Dimensional Moment Method Simulation of Penetrable Scatterers Consisting of Non-metallic and Circuit Analog Surfaces Sikosky Aircraft Corp. 1201 South Ave Bridgeport, CT. 06604 March 1998

PROJECT INFORMATION PROJECT TITLE: REPORT TITLE: Fast integral equation algorithms for penetrable blade and hub scattering in the VHF and UHF bands Three Dimensional Moment Method Simulation of Scatterers Consisting of Non-metallic and Circuit Analog Surfaces U-M REPORT No.: 376632-1-T CONTRACT START DATE: END DATE: May 1998 May 1999 DATE: March 1999 Preliminary Annual Report SPONSOR: SPONSOR CONTRACT No.: U-M PRINCIPAL INVESTIGATOR: Daniel C. Ross Sikorsky Aircraft Corp. Mail Stop BO11A 1201 South Ave. Bridgeport, CT. 06604 Phone: (203) 384-7010 Fax: (203) 384-6701 Email: dross@sikorsky.com John L. Volakis EECS Dept. University of Michigan 1301 Beal Ave Ann Arbor, MI 48109-2122 Phone: (313) 764-0500 FAX: (313) 647-2106 volakis@umich.edu http: //www-personal. engin. umich. edu/ volakis/ PROJECT PEOPLE: E. Topsakal, Michael Carr and John Volakis

THREE DIMENSIONAL MOMENT METHOD SIMULATION OF PENETRABLE SCATTERERS CONSISTING OF NON-METALLIC AND CIRCUIT ANALOG SHEETS Erdem TOPSAKAL, Michael CARR, John VOLAKIS Abstract This report describes the formulation for a mixed integral equation formulation implemented in the code CADRIS. An important aspect of this implementation is the inclusion of Circuit Analog sheet models that is not available in other codes. This capability is combined with modeling capabilities for Resistive sheets, metallic, dielectric and impedance surfaces, and combination of all. The report begins with the introductions of the appropriate integral equations and proceeds to develop their discertization. Several validations are given for PEC and composite scattereres. 1) Surface Integral Equations Consider the general electromagnetic scattering problem depicted in Fig.l.1. Figure 1.1. Geometry showing the various boundaries We are interested in a field point r, located in a closed volume V or on a regular surface Si(i = 1,2,..., n). Beginning with the vector Green's theorem, [(Q.(V' x x T))- (T.(V x V' x Q))dv'] Jv { 1 \ r {L) + / ((T x V' x Q)- (Q x V' x T))ds' = 0 n- St where, Q(r), T(r) E C2;r E V, Si. Here the primes refer to the primed(integration) coordinates. To derive the integral equations for the electric and magnetic currents on the surfaces Si, i = 1, 2,..., n we set 1

T = E(r')(electric field), and Q -= G(r, r') with c-i klr-r'| G(r, r') = 4Ir - k = wv/ (2) Note that when r -r' in V, G, VG and V2G have singularities. To overcome this singularity problem, when integrating G or its derivatives we exclude an infinitesimal sphere of volume V6 - Q and centered at r = r' and shown in Fig. 1.2. We deal with the spherical volume V of radius 6 - 0 separately by invoking the divergence theorem. V s'= zsi+s6 Figure 1.2. Geometry for singularity problem Next we introduce into(1) Maxwell equations V x E + jwpH = -M V x H- jweE = J V.(/(H) = Pm V.(eE) = p After some straightforward vector manipulations, (1) can be written as; f [jwpuJG + M x V'G - (p/e)V'G]dv + / jwp(n' x H)<' - (n' x E) x VG - (n'.E)V'G]ds' = 0 'tl I S,+S (3a) (3b) (3c) (3d) (4) where we have assumed that V in equation (1) is linear, isotropic and homogeneous. When r' is located on one of the Si(i = 1, 2,..) surfaces, we proceed to extract the integral singularity tioted earlier. Refering to Fig.1.3, we rewrite the surface integrals as, 2

Si-S, Figure 1.3. Geometry for singularity extraction / nx Ex V'G = lim [ + +J] (5),+S6 5 —0 s,-S,6 tS6 lim + — E(r)[1 ] (6) <-0 6- Js 27r Here,P is the absolute value of the solid angle subtended by Sib at r in the limit as 6 -b 0. 7r, rE Si [ O, elsewhere ( Based on (6), (4) can now be written as 0(r)E(r) =-[J [jwuJI4 + M x V'< - (p/e)V~']dv' + / ju(n' x H - (n' x E) x V' - (n'.E)V'ds')] ( On invoking duality we also obtain the corresponding integral equation for H as; 0(r(r)r) -[ [/vjwilM - J x V - (m/p)V')]dv' (9) + / -jwp(n' x E( - (n' x H) x V'- (n'.H)V'ds')] Here f denotes the Cauchy Principal Value and 0(r) can be given as follows. 9(r) [2, reSi (10) 0(r) = [, elsewhere (10) In (8) and (9) since aU the sources are contained in the volume V, this volume integral can be refered to as the 'source term'. If there are no sources in V', this integral will be zero. We will assume that the sources are far from the scatterer and represent the source integral by (Ei,H which later be set a plane wave incident in the scatterer. Equation (8) and (9) can then be written as (r(r = E(r)E(r) = Er) + [-jwp(n' x H + (n' x E) x V' + (n'E)V']ds'] 3

O(r)H(r) = 0 [H(r) + J [jE(n' x E)F lt S( (12) + (n' x H) x V'( + (n'.H)V'(]ds' Next we rewrite these in terms of surface current densities (J, M) where J = n x H6(S) (13a) M=E x n6(S). (13b) and it follows that -t -t n.E = V.(n x H) = V.J (14a) Jae JWE and n.H = V.(n x E)= -V.M. (14b) J3'l J.wp Sustituting (13a, b) and (14a, b) into (11) and (12) gives the integral representation 0(r)E(r) = E'(r) - AJ + QM (15a) 0(r)H(r) = Hi(r) - J - -AM (156) 7II where J and M are unknown surface current densities. Here, A and Q are the integro-differantial operators given bye, Abfr(r) = j[jirF(r') + V(V' r(r'))]G(r - r')ds (16a) and Qr(r) = r(r) x VG(r - r')ds', (16b) and 71 = /l/ is the characteristic impedance of the medium. 2) Formulation for Different Type of Boundary Conditions 2.1) PEC Boundary(Metallic Surfaces) Different types of surface integral formulations have been developed for these kind of surfaces. We give here the very well-known EFIE(Electric Field Integral Equation), MFIE(Magnetic Field Integral Equation) and CFIE(Combined Field Integral Equation) formulations. a)EFIE Formulation Consider the PEC surface depicted in Fig.2.1. 4

Ei,Hi Region Bj(sj,) / / E2,H-= 0 aBI 2 RRegion B(2 Figure 2.1.Geometry for imposing the PEC boundary condition Using (15a) and (15b) for the fields outside the PEC surface, we can write, 0(r)E(r) = E(r) - AJ + M, r E B1 UB12 (17a) 0(r)H(r) = H'(r) - QJ - AM, rE B1 U B12 (17b) 712 E2(r) = H2(r)- 0, r E2 (17c) To construct the integral equation, we note that on the metallic surface, n x [E1 - E2] = n x E = 0 (18a) n x [H1 - H2] = n x H1 = J (18b) These imply, M = 0 and thus (17a) become, El(r) = E(r)-AJ(r), r E 1 (19a) and nx E/(r)= n x AJ(r), r E B12 (19b) The above are the so called EFIE whose solution gives the unknown current on the surface. b)MFIE Formulation From (17b) and (18b) with M = 0, the magnetic field on the boundary B12 is given by ) = n x H(r)-n x QJ(r), r E B12 (20) 2 c)CFIE Formulation 5

The solution of EFIE and MFIE formulation can return non physical results at internal resonant frequencies. In this case we resort to the C'FIE to overcome this problem(ref Peterson-Wilton). CFIE combines the EFIE and MFIE equations in a linear fashion as a(EFIE) + (1 - ac)r(MFIE) = CFIE (21a) or 1 c(Ei(r)) + (1 - c)r/(n x Hi(r)) = (aA + (1 - a)r7n x Q + J(r)) (21b) The coefficient a is arbitrary and possibly complex. Typical a is set to 1/2 but other choices can be made. Basically the CFIE shifts the resonances of the MFIE and EFIE outside the range of interest. 2.2) Resistive Boundary Consider the Resistive Boundary surface depicted in (Fig.2.2). / Ei, EIHIH ___ / n,7 Region B(sE1,) ) 3 EH2 a Bl2 Re egion B2(s-2,j2) M 1 Figure 2.2. Geometry for application of the resistive boundary condition For a resistive boundary, the boundary condition on 3B12 are n x [E1 - E2] = 0 (22a) n x [Ei + E2] = 2Re7U1n x n x [H1 - H2] (22b) n x E1 = -M1; n x H1 = J1 (22c) n x E =M2; nx H2 =-J2. (22d) where Re refers to the normalized surface resistivity in ohms per square. Fields in the regions 61 and 62 can be given as 01(r)E1 = Ei - AiJ(r) + Q1M(r) (23a) 1 02(r)HE = H- A1Jl(r) + -2A1M(r) (236) 02(r)E2 = -A2J2(r) + Q2M2(r) (23c) 6

02(r)H2 -QJ2(r)2 -JA2M2(r) (23d) In (23a - d), indices 1 and 2 are corresponding to the fields and currents in regions BI(E - E1, /1 i i), and 2(e - e2, [I - /12), respectively. Using (22a - d) in (23a - d) we obtain 1A + Re - Re J2 Ei Qi 1 A1+M+ -A2+ -Q M1= Hi (24) Re Q2 A2 + Re, J2 O If (e1 = E2, P1 = /2), (24) reduses to, [Ai + 2Re]J = Ei, J = J1 + J2 (25) which refers to the case of a resistive sheet boundary in free space. 2.3) Dielectric Boundary A homogenous penetrable body is depicted in Fig.2.3. E,,H, Region B1(s,L) / I2,H2 B12 M, Figure 2.3. Geometry for constructing integral equations for dielectric boundary Here we will give four different integral equation formulations for dielectric boundaries. For this situation, the expression for E1,2 and H1,2 are given in (23a- d). A) EFIE formulation: The problem given in Fig 2.3 can be seperated into two sub-problems each gives the field in one of the regions. EH, E,,Ht o Region (E^) Region ( ) / E2,Ht o 0 E,Ht 3,B,2 asE,2 Region B~(oC) Rga, -B \QB 7

iFigre 2.4. ( eorrietry for the a)l)lication of the EFFIE an(d MFIE for (dielectric,oiIlri(ldary a) External Problem 1)) Internal Problem [lounrl(lary conditions for the E field for the external and the internal prol)lerrs are; n x E1 =0 (26a) and n x E2 = 0 (26b) respectively. From (23a - d) and (26a, b), we obtain the EFIE equations, n x Ei = n x A1JI - n x Q iM1 (27a) 0 = n x A2J1 - n x 2M1 (27b) B) MFIE formulation: Boundary conditions for the Magnetic field for the external and internal problems are; n x H1 = 0 (28a) and n x H2 = 0 (28b) respectively, on the surface of the dielectric body. Using (28) in (23a - d), we arrive to MFIE. I n x Hi = n x Q1J1 + — n x AIMI (29a) 0 =n x Q2J + -r1n x A2Mi (29b) C) CFIE formulation: Combining the EFIE and MFIE formulations as outlined in (21), yields [aEi + (1 - a)rqn x H'] = (aAi + (1 - a)n x Qi)JI -(fli + (1 -a) n x AI)M1 (30a) 0 = (a(\A2 + (1 - ct)n x Q2)Ji - (aQ2 + (1 - a)-I i x A2)Mi (306) 71; 8

These are the most general integral equations to be solved for J1 and M1. They can in general be combined with similar integral equations from other dielectric boundaries for simulating rather complex geometries. D) PMCHW formulation: The PMCHW formulation is a special case of the CFIE and is also robust at interior resonant frequencies. The method relies on implying the continuity condition on the surface of the dielectric body. That is, n x [E - E2] =0 (31a) and n x [H1 -H2] = 0. (31b) Employing these conditions to the fields given in (23a-d), we obtain the PMCHW equations. f2Ql + Q2 [ Al+] (32) 2.4) Impedance Boundary The impedance boundary condition is of the form, n x E = Zn x (n x Hi). (33) / Ei, E,,H f Region (Ei)..)/ eRegio n B (, M1 M, Figure 2.5 Geometry for applying the impedance boundary condition Substituting (33) into (23a, b), we get the surface integral equation. I Ql Ai + [M M1 [H (34) In (34), t = / represents the characteristic impedance of the surrounding medium. 2.4) CA (Circuit-Analog Boundary) 9

Consider a thin(penetrable or impenetrable) multilayered sheet(see fig.2.6). Relations between the tangential components on the two sides of the sheet(consistent with duality and reciproity) can be written as EiHi EH, 4 / Region B,(s,,-) 2,H2 aB,I. Region B2(E2) / J/ M, M, Figure 2.6 Geometry for applying the CA boundary condition n x [E+(r) + E-(r)] = Rn x n[H+(r)- H-(r)] - Ren x n[E+(r) - E-(r)] (35a) n x [H+(r) + H-(r)] = Rmn x n[E+(r) - E-(r)] + Rcn x n[H+(r) - H-r)] (356) Here E~(r) and H+(r) represent the fields on the upper and the lower surfaces of the sheet, respectively; Re and Rm are the electric and the magnetic resistivities and Re is a cross coupling term. Rewriting (35a) and (35b) in matrix form, we obtain [ E- X1 X12 E+ (36) n x H-j - X21 X22 7n x H+ (36) where ij (i, j = 1, 2) are given by Xl = [1 + (1/2 - Rc)2/ReRm]/[1- {(1/4 - R)/ReRm}] (37a) X12 = -/[Rm[l - {(1/4 - R)/ReRm}] (37b) X21 /[Re[1 - {(1/4- R2)/ReRm}] (37b) X22 = [1 + (1/2 + Rc)2 {(1/4-/ReRm}] (37d) in which Rn(Re = 27Re, Rm = -(2/7i)Rm, Rc = 2Rc), (n = e, m, c) stand for the normalized resistivities. Assuming that the thin multilayered sheet can be characterized by its reflection and transmission properties, R,(n = e, m, c) can be determined by relating them to the reflection and transmission coefficients of the sheet. For general layered structure we need two reflection (rf) and one transmission coefficient (T). The corresponding reflection and transmission coefficients from (35a, b) are F+(O) = (2ReCosO - 2Rm/cosO ~ 4Rc)/(4(ReRm + R2) + 1 + 2 * RecosO + 2Rm/cosO) (38a) 10

T(O) = (4(R' + R Rm) - 1)/(4(ReRm+R, + R I 1 + 2ReCosO + 2Rm/cosO) (38b) where 0 stands for the incident angle. At normal incidence (0 = 0) the above can be inverted to yield Re [T2(0) -(1 + r+(o))(1 + r-(O))]/[r+(O)-() - (1 - T2(0))2]/2 (39a) R = [T2(0)- (1 - F+(0))( - r-())]/[r+(O)r-() - (1 - T2(0))2]/2 (39b) R, = [r-(o) - F+(o)]/[+()r-(O) - (1 - T2(0))2]/2 (39c) Thus, upon having the reflecyion/transmission coefficients we can extract the corresponding Rn(n = e, 7n, c) values. 2.4.1) Reduction to simple sheet condition The transmission line model can be used to relate the above R,(n = e, m, c) parameters to the resistivity (Zp) and conductivity (Zs) values for simple sheets. Equating the reflection and transmission coefficients for the above circuit with those from (4) yields series: — Re 00, Rm = -Z/Zo, Rc = 0 (40a) parallel: Re -oo, Rm =-Z,/Zo, R, = 0 (40b) and the ij(i,j = 1, 2) matrices reduce to [1 Z,/Zo Xser = [ 1 (41a) and Xpar. -ZOZp 1] (1ib) When (41a, b) is used in (36), we conclude that a single parallel impedance circuit represents a resistive boundarycondition, whereas a single series imprdance circuit represent a magnetically conductive boundary. 2.4.2) Surface Integral Equation To construct a surface integral equation let us refer to Fig.2.3.. In this case interior and the exterior fields can be expressed with (23a - d). The relation in between the tangential field components and the surface currents can also be given with (22c, d). Substituting (23a - d) into (36) with the identification that E1 = E+, E2 = E-, H1 = H+, H2 = H- we obtain the integral equation; A 1- 2 -Q -t 0 J1 -Ei nl1_ 1 __ Xl1 QA 2lX12 +~ -02 M1 Hi X7X1212 (42) 1(X12- XlX22 0 A2- -Q2 J2 0X21 2X22 X11X22 X22 22 2l(AX21 - Q2 A2 + 72 nx2. [ 0 XL 2 i7 l2 112 11

3) Composite Structures Consider the geometry depicted in Fig. 2.8. We are going to give the surface integral equations for different problems implemented in the code CADRIS. 3.1) Problem 1 In our first problem S1 and S2 are dielectric and 53 and S4 are the perfectly conducting surfaces. Electric and the magnetic fields can be written as follows in three different region(see Fig. 2.8.). 0(r)Ei(r) = Ei - Ai(Ji + J4) + QlM1 (43a) 1 O(r)H1(r) = Hi - 1(J1 + J4) + — rA1M1 (43b) 0(r)E2(r) = -A2(J2 - J1) + Q2(M2 - Mi) (43c) 0(r)H2(r) = -Q2(J2 - J) +- — A2(M2 - Ml) (43d) W2 0(r)E3(r) = -A3(J3 - J2) + -Q3M2 (43e) 0(r)H3(r) = - (J3 - J2)- - A3M2 (43f) boundary conditions on the surfaces S1 - 54 can be given as OnS:n x [E - E2] =; n x [H1-H2] = 0 (44a) OnS2:n x [E2-E3] = O; n x [H2-H3] = O (44b) OnS3: n x E3 = 0 (EFIE);n x H3 = J3 (MFIE) (44c) OnS4: n x E1 = 0 (EFIE); n x H1 = J4 (MFIE) (44d) Using (44a - d) in (43a - f), and after some straight forward manipulations we find the surface integral equation as ZI = V (45a) Al + A2 -Q1-Q2 -A2 Q2 0 A1 Q1 + Q2 2A1 + i A2 -Q2 — A2 0 Q1 -A2 Q2 A2 + A3 -2 - Q3 -A3 0 Z = (45b) -Q2 — A2 Q2 +3 n A2 + 1A3 -Q3 0 0 0 -A3 Q3 A3 0 A1 -Q2 0 0 0 A1 12

(N/ SS3 S4 FIg*.2 General Problem- Geom-etryr

0 0 0 (It

J1 -Ez(S') M Hi (S'1) J2 0 I = V (45c) M2 0 J3 0.J4 EiE(S4) Problem 2 In the second problem S is a dielectric, S2 is a resistive and S3 and S4 are the perfectly conducting surfaces. In this case fields can be as O(r)E1 (r) = Ei - Al(J1 + J4) + QlMi (46a) 1 0(r)Hl(r) = Hi - Ql(J1 + J4) +- - AiM1 (46b) o(r)E2(r) = -A2(J2 - J1) + Q2(M2 - Mi) (46c) I O(r)H2(r) = -Q2(J2 - J) + 2A2(M2 - M1) (46d) 1 2 0(r)E3(r) = -A3(J3 + J22) + -Q3M2 (46e) 0(r)H3(r) = -Q3(J3 + J22) - - A3M2 (46f) 1r3 boundary conditions on the surfaces S1 - S4 can be given as OnS1:n x [E - E2] = 0 n x [H - H2] = 0 (47a) OnrS2:n x[E-E]= nx[E2 +E3] = 2Rn x n x [H2-H3] (47b) OnS3: n x E3 = O(EFIE);n x H3 = J3(MFIE) (47c) On54:n x E1 = O(EFIE);n x HI = J4(MFIE) (47d) Using (47a - d) in (46a - f), and after some straight forward manipulations we find the surface integral equation as 13

9 CD m ' - F4 -14 mog ~Tl ohmm * uq (To 2 D 0 0 -C) CD 0 CD ID O!

A2 + A2 - A2~ -Q9 0 -Al -Q1 - Q, 77 -1, 71'kI+ 1T A j 1 ~21A Q7 ') rl 0 0 - Q I1 -A,) -Q2 A2 + I% Q,) 0 0 J2 J22 J3 L 4 - 2A2 77 -Q2 212 + 1 R rl r?2A3 2771 2 3 Q33 Q3 0 Et(S1) HZ(S) 0 v= 0 0 0 0 0 - 77 -K3 A3 + gIR A3 0 0 0 0 -Q33 A3 A3 0 Al - Q, 0 0 0 A1 - (48a) (48b) Problem 3 In the third problem SI is a dielectric, S2 is a CA-boundary and S3 and S4 are the perfectly conducting surfaces. In this case fields can be as 0(r)El(r) = E" - Al1(J1 + J4) + Q1M1 0(r)H1(r) = H' - Q1(J1 + J4) ~ - AIM1 l i 0(r)E2(r) = -A2(J2 - Jl) + Q2(M2 - M1) 0(r)H2(r) = -Q2(J2 - J1)+ I-A2(M2 - Ml) 2/ O(r)E3(r) = -A3(J3 + J22) + +Q3M22 I O(r)H3(r) = -2Q3(J3 + J22) - boundary conditions on the surfaces S, - S4 can be given as OnS1: n x [E1 - E2] = O n x [H1 - H2] = O 14 (49a) (49b) (49c) (49d) (49e) (49f) (-Oa)

t OEM * CD 0 M, m "I 0=90 m Omni 0 ulq 0 tl-j O-A CD 0 CD r-0 0 ----i CD I (..*i 0 9

OnS2 InxH = K iln~xH~] (50b) OnS3: n x E3 =O(EFIE) ii x H3 J3(MFIE) (t'c) 0C OniS4: n x El = O(EF IE) ni x HI J4(MF IE) (50d) Using (50a - d) in (49a - f), and after some straight forward manipulations we find the surface integral ecluation as Al + A2 Ql + Q2 -A2 0Q 0 0 Al 'rAl+ -'A2 0 0 -A2 A2 7X22 2X21 Q1 7 X2- XllX22 ) 0 0 4rA2+ Xi o A2 rl 7X 12 W(1 0 0 0 E-(S4) - 0 0 0 X22 ~23 A3 0 0 0 0 2712 -27 -~23 0 0 0 0 0 A3 ~23 A3 0 A1 -~21 0 0 0 0 0 A1 (5 I a) I = - Ji - J3 - J4 - (51lb) Problem 4 In the fourth problem S1 and 52 are dielectric, S3 is a PEC and 54 is an impedance surface. In this case fields can be written as 0(r)El (r) = E'- Al (J1 ~ J4) + Q 1 (Ml + M4) 15 (52a)

Dielectric PEC Impedance Fig.2.12 Geometry for Problem-4

6?(r)H1(r) =H' - Q1(JI + 34) + - -IA1(M1 ~ M4) Ifi O(r)E2(r) = -A9)(J3 - J1) + Q2-('M2 - MI) 0(r)H9)(r) =-Q9(32 - JO) -~A2(M2 - Ml) 0(r)E3(r) =-A3(J3 - 32) + - Q3 M2 O(r)H3(r) A3M23 -32 boundary conditions on the surfaces S, -.4can be given as ( 52b) (52c) (52 d) (52e) (52f) 0nS,: n x[El -E2]=0 nx[H1-H2] =0 (53a) OnIS2: n x [E2- E3] =0,nx [H2- H3] = 0 ( 5 3b ) OnS53:n x E3 = 0 (EFIE) -nx~H3 = 33 (MFIE) ('54c) OnS4:nxE=Zn xn xH1 (54d) Using (54a - d) in (53a - f), and after some straight forward manipulations we find the surface integral equation as -Al -HA2 Q1 +Q22 -Ak2 0 Al -Ql - Q 'Ai + -'A2 0 771 I = -A2 A2 + A3 Q2 + Q3 -A3 0 0 J2 M2 33 34 1 A2 ~121 rj3A 0 0 El(Si) HW(S1) 0 V = 0 0 Et(S4 ) Hz(S4 ) -~ 0 0 -A3 -Q3 A3 0 0 Al 0 0 0 A1 + z ~1 0 0 '~Al+ 1 7 2Z, (55a) (5 5b) Problem 5 16

) l 5i I - - I 2 Q0 0; *00 0 (N;_4 T0o * - ^ M fn.-.4 *MO * o L; r2 Q f P4 *~Il

In the fifth probleni S1 is a dielectric, S2 is a resistive 53 is a PEC and S4 is an impedance surface. In this case fields can be as O(r)Ej(r) - A1(J1 + J4) + Q1(M1 + M4) (56a) O(r) H 1(r) =Hz.-Q I(Ji 1 4 I lM (56b) 71 0(r)E2(r) = -A2(J2 - J1) + Q2(M2 - M1) (56c) I O(r)H2(r)= -Q2(J2 - J1) + - I A2(M2 - M1) (56d) 2/ 172 O(r)E3(r) = -A3(J3 + J22) + -Q3M2 (56e) t O(r)H3(r) = -Q3(J3 + J22) - - ~A3M2 (56f) 1)3 boundary conditions on the surfaces Si - S4 can be given as OnS1: n x [E1 - E2] = O n x [H1 - H2] = O (57a) OnS2:nx [E2 - E3] = O n x [E2 + E31 = 217iRn x n x [H2 - H31 (57b) OnS3 n x E3 = O(EFIE) n x H3 = J3(MFIE) (57c) OnS4 n x El = Zn x n x H1 (57d) Using (57a - d) in (56a - f), and after some straight forward manipulations we find the surface integral equation as Z = A- + A2 -Q1 - Q2 -A2 Q2 0 0 A1 2 1 A 0 0 Q, 1 0l 771 '77 1 -A2 2 A2 + ~L1 -Q2 - lR 0 0 0 2 2 -4A2 1 A Q2 4A2 + + R -423 -Q3 0 0;72 2 733 271 0 0 2, A3 A3 + ~%R 03 0 0 0 0 0 A3 A3 A3 0 0 Al -Q1 0 0 0 0 A+Sl + -Q, Al0 0 0 0 A (5)8 a) 17

E 1 M1 HHz(S) J2 0 M2 0 1= V (58b) J22 0 J3 0 J4 EP(S4) M4- Hi(S4) Problem 6 In the sixth problem S1 is a dielectric, S2 is a CABC S3 is a PEC and S4 is an Impedance surface. In this case fields can be as 0(r)Ei(r) = Ei - A1(Ji + J4) + iQ(M1 + M4) (59a) 1 0(r)Hi(r) = H' - Q1(J1 + J4) +- — A(M1 + M4) (59b) 1/i O(r)E2(r) = -A2(J2 - J) + Q2(M2 - M1) (59c) O(r)H2(r) = -Q(J2 - J) + -1A2(M2 - M1) (59d) ~2 O(r)E3(r) = -A3(J3 + J22) + Q3M22 (59e) 1 0(r)H3(r) = -Q3(J3 + J22) - rA3M22 (59f) 1/3 boundary conditions on the surfaces S1 - S4 can be given as OnSl:n x [E 1-E2] = 0 n x [H1-H2] = 0 (60a) E- [Xii X12 E+ 77n x H [X21 X22 [ 77n x H+ (60b) OnS3: n x E3 = O(EFIE) n x H3 = J3(MFIE) (60c) OnS4:n x E1 = Zn x n x H1 (60d) Using (60a - d) in (59a - f), and after some straight forward manipulations we find the surface integral equation as 18

C-0 O" 4 0-4 0 r-4 0-t r) 0 cr 0 CL w "I t-o

z = A, + A+2 Q21 +4 2 -A2 -S 22 0 0 0 A1 LQ -~1 - 22 -Al + 1A2 ) -A2 0 0 0 -f1 7 -A2 A2 - 2x22 Ql (XI2- X1X22) 0 0 0 0 ~22 ~ -1 -~A2 -1A2+ x. 0 (x121 - XlX22 ) 0 0 0 0 0 2X21 0 A2 - T7XII 2X21 Q3 A3 0 0 0 0 0 2X 1 -Q2;A+ 2x12 -Q3 0 0 0 0 A3 Q3 A3 0 0 A i -2, ~2, 1, A 0 0 0 0 0 0 0 0 A1 + z -fl1 Q(61a) L + 2 (6 1a) 1= -Ji - Ml J2 M2 J22 M22 J3 J4 LM4 - -Ei (51) ' H'(Si) 0 0 0 0 0 E'(S4) -HO5) (61b) Problem 7 In the second problem S1 is a dielectric, S3 and In this case fields can be as S4 are PEC surfaces. 0(r)Ei(r) = E' - Ai(JI + J4) + QilM 0(r)Hi (r) = H' - Q2 (JI + J4) + - 2 AIM, 0(r)E,2(r) = -A2(J3 - Ji) - Ql2M - )'/2 (62a) (62b) (62c) (62d) 19

J4 J3 Dielectric PEC Fig.2.14 Geometry for Problem-7

boundary conditions on the surfaces S, - '4 can be given as OnS'1:n x [E1 - E2] = 0; x [H - H2] = 0 (63a) OnS3 n x E = O(EFIE);n x H3 = J3(MFIE) (63b) OnS4:n x E1 = O(EFIE) n x H1 = J4(MFIE) (63c) Using (63a - c) in (62a - d), and after some straight forward manipulations we find the surface integral equation as A + A2 -Q1 - Q2 -A2 A1 Ql +~Q2 12Al + A2 -Q2 Q1 Z= 2 (64a) -A2 Q2 A2 0 - A1 -21 0 A1 1 E(5'1) M1 Hi (S1) I V= (646) J3 0 - J4 E(S4) Problem 8 In the eight problem S1 is a dielectric, S3 is a PEC and S4 is an impedance surface. In this case fields can be as 0(r)El(r) = E' - Ai(Ji + J4) + Q1(M1 + M4) (65a) 0(r)Hl(r) = H1 - Q1(J1 + J4)- + — AlM (656) 71i 0(r)E2(r) = -A2(J3 - J1) - Q2M1 (65c) 0(r)H2(r) = -Q2(J3 - J1) + - A2M1 (65d) T2 boundary conditions on the surfaces S1 - S4 can be given as OnS1: n x [E - E2] =; n x [H1-H2] = 0 (66a) OnS3: n x E3 = O(EFIE) n x H3 = J3(MFIE) (66b) OnS4:n x E1 = Zn x n x H1 (66c) 20

J1, M1 J3 Dielectric PEC Impedance Fig.2.14 Geometry for Problem-8

V(sing (66a - c) in (65a - d), and after some straight forward manipulations we find tih-e surface integral equation as A1 + A2 -Q1-Q2 -A2 A1 -Qi Q1+Q2 -A1 +-2A9 -Q2 Q1, Z = -A2 Q2 A2 0 (67a) A1 — i 0 A1 + - Q1 - A1 0 Qi1 A1i 2+ - J1 J1 - -Ei(S1) M1 H(S1) I= J3 V= 0 (67b) J4 Et (S4) M4- Hi (S4) Problem 9 In the eight problem S2 is a resistive, S3 and S4 are PEC surfaces. In this case fields can be as 0(r)Ei(r) = Ei - Al(J2 + J4) + Q1M2 (68a) 0(r)H (r) = Hi - 1 (J2 + J4) + -2-A1M2 (68b) 0(r)E2(r) = -A2(J3 + J22) - Q2M2 (68c) 0(r)H2(r) = -Q2(J3 + J22) + -22M2 (68d) 72 boundary conditions on the surfaces S1 - S4 can be given as OnS2:n x [E2 - E3] = 0; x [E2 + E3] = 271iRn x n x [H2 - Ha] (69a) On5S3:n x E3 = O(EFIE); n x H3 = J3(MFIE) (69b) OnS4: n x E1 = 0 (EFIE); n x H = J4 (MFIE) (69c) Using (69a - c) in (68a - d), and after some straight forward manipulations we find the surface integral equation as 21

J4 J22 J3 Resistive Card v PEC Fig.2.14 Geometry for Problem-9

-A_ + -i o A A1 + -1 0 Al - 2 Z= r Q2 a A2 + Ao 0 (70Oa) 0 A2 A2 A2 0 A1 -Q 0 0 A1 J - Ei(S1) M2 H'(S1) I J22 V= 0 (7Ob) J3 0 J4- Ei(S4) Problem 10 In the eight problem S2 is a resistive, S3 is a PEC and S4 is an impedance surface. In this case fields can be as O(r)Ei(r) = Ei - A1(J2 + J4) + Q1(M2 + M4) (71a) 0(r)Hl(r) = Hi - Ql(J2 + J4) + -2 A(M2 + M4) (71b) r/1 0(r)E2(r) = -A2(J3 + J22) - Q2M2 (71c) 1 0(r)H2(r) = -Q2(J3 + J22) + - 2M2 (72d) '72 boundary conditions on the surfaces S1 - 54 can be given as OnS2:n x [E- E3] = 0; n x [E2+ E3] = 2771Rn x n x [H2- H3] (73a) OnS3:n x E3 = O(EFIE); n x H3 = J3(MFIE) (73b) OnS4 n x E1 = Zn x n x H1 (73c) Using (73a - c) in (72a - d), and after some straight forward manipulations we find the surface integral equation as 22

'J22 J3 Resistive Card Impedance PECme e Fig.2.15 Geometry for Problem-10

A+r -21 0 A1 1 12 1+ -229 -12 1 - rlR 2, A2+i R A 0 0O Z = (74a) 0 22 A, A 0 0 Al -Q, 0 0 Ai + - QI 1A1 0 0 Qi rl - +. - J2 -E(S1) M2 Hi(S1) J22 0 I= V= (74b) J3 0 J4 E(S4) M4 J - H(S4) Problem 11 In the eight problem S2 is a CA boundary, S3 and S4 are PEC surfaces. In this case fields can be as 0(r)Ei(r) = Ei - Ai(J2 + J4) + QiM2 (75a) 1 0(r)Hl(r) = H/ - Q2(J2 + J4) + 2 —AiM2 (75b) 0(r)E2(r) = -A2(J3 + J22) - Q2M2 (75c) O(r)H2(r) = -Q2(J3 + J22) - -22M22 (75d) 7/2 boundary conditions on the surfaces S1 - S4 can be given as E- 1 X1 X12\ E 1 OHS2' tn x H q- x T- (76a) [rnxXHJ [21 X22 [7rn x H+ (7a) OnS3 n x E = O(EFIE); n x H3 = J3(MFIE) (76b) OnS4:n x E1 = O(EFIE); n x H1 = J4(MFIE) (76c) Using (76a - c) in (75a - d), and after some straight forward manipulations we find the surface integral equation as 23

J4 J22 J3 CA boundary v PEC Fig.2.16 Geometry for Problem-11

A1- 2 1 Qi 9(Y12 - Y22) 0 0 A1 -/1 A] + \" 2 77, 12 0 - 11 )(22 ) M2 J22 I = M,2 J3 J4 77 -2\21 0 - (21 A2 0 -El(-) H'(Si) 0 0 0 E(S4) - 0 1 -Q2 02A + 2X2 - l2 0 0 0 A2 Q2 A2 0 A 1 0 0 0 A1i - (77a) (77b) Problem 12 In the eight problem S2 is a CA boundary, 53 is a PEC and S4 is an Impedance surface. In this case fields can be as 0(r)Ei(r) = Ei - A1(J2 + J4) + Q1(M2 + M4) (78a) 0(r)Hi(r) = H' - Q1(J2 + J4) + -- Ai(M2 + M4 (786) 0(r)E2(r) = -A2(J3 + J22) - Q2M22 (78c) 0(r)H2(r) = -Q2(J3 + J22)- - Q2M22 (78d) boundary conditions on the surfaces SI - S4 can be given as OnS2: H - [Xi 2 n x HJ (79a) [_7nxH ] [_X21 X22 7 r7nxH+ v OnS3: n x E3 = O(EFIE); n x H3 = J3(MFIE) (79b) OnS4: n x El = Zn x n x H1 (79c) Using (79a - c) in (78a - d). and after some straight forward manipulations we find the surface integral equation as 24

r) 010. ar 0 CL w "I omi0 0 0

Q 1 rl ( k 1 2 k 11 k22 - k.1k2 0 0 Al L 12Al + Y,'< 0 T~l(-(21- 'X 11k22 )(1 0 -Q1 Li' 'II 2, I 1?? Ti 3-'(~21 0 A2- 2k__ Q2i2 A2 0 0 - J9 M2 M22 J3 J4 0?T'(7~12 1~1 71 2 +77X 1 -Q22 0 0 Hz (Si) 0 ( = 0 0 E'(S4) WH(S4) - 0 0 A2) A2 0 0 Al 0 0 0 A1 + 2 Q I "r I1 Ti, 0 0 0 — Q I A,+ 71 2Z I (8 Oa) (80b) Problem 13 In the thirteenth problem 33 is a PEC and 34 is an Impedance surface. In this case fields can be as 0(r)EI(r) = E' - Al(J3 +J4) + Q1M4 1 0(r)Hi(r) = W' - 1(J3 + J4) + - A1M4;71 boundary conditions on the surfaces Si - S4 can be given as OnS3 n x E3 = 0(EF I) n x H3 = J3(MFIE) (81a) (ilb) (82a) OnS4 n xE1 = Zn x n x H1 (82b) Using (82a, b) in (Sla, b), and after some straight forward manipulations we find the surface integral equation as 25

J3 J4 9 A14 PEC lt"Pedance 0 0 i etr.V. Or am

A! A 1 - -1 Z = Ai Ai + z -Qi (y3a) -83a) Al + 2 -J3 -E'(,S3)I = J4 V= E(S'4) (83b) LM4 H'(,5(S4) Problem 14 In the thirteenth problem S3 is an impedance and S4 is a PEC surface. In this case fields can be as 0(r)E1 (r) - Ei - Ai(J3 + J4) + Q1M3 (84a) 0(r)Hl(r) = H - Q (J3 + J4) + - R- AiM3 (84b) boundary conditions on the surfaces S1 - S4 can be given as OnS3:n x E1 = Zn x n x H1 (85a) OnS4: n x E1 = O(EFIE); n x H1 = J4(MFIE) (85b) Using (85a, b) in (84a, b), and after some straight forward manipulations we find the surface integral equation as -Ai+f -Qi A1 -Z = -1 -A1 + &-Q1 (86a) - A1 -Ql A1i - J3 - E(S3) 1= M3 V H'(S3) (86b) J4 E- (S4) 26

143 J3 J4 iftlPedance 0 0 jeo -0 I or Oro ern -

TEST STRUCTURES (SINGLE SURFACES)

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