R.L. 3-41 THE GREEN'S FUNCTION FOR A SLOT ON THE GROUND OF A DIELECTRIC SLAB PISTI B. KATEHI RL-841 = RL-841

- 2 - 1. Introduction Recently, a number of papers have been published [1]-[3], on the subject of waveguide slot arrays. These papers describe a procedure one can use to calculate self and mutual admittances between slots. This procedure can consist the basis of a design scheme where the lengths and offsets of individual slots can be computed to produce a specified pattern and input impedance in the presence of mutual coupling. In many practical applications, the slot array is covered by a dielectric slab in order to protect the array as well as modify its characteristics. For this new array, a design scheme can be described only if the theoretical formulas for Yi. are available. Since analytically evaluating Yij is a very complicated boundary value problem, numerical evaluation of Yij may substitute for the needed theoretical formulas. A first step is to obtain the Green's function, that is, the field due to a suitably positioned Hertzian Magnetic Dipole (HMD), since each slot can be viewed as a collection of HMDs. The detailed derivation of the appropriate Green's function will be presented in this report. Sommerfeld first analyzed the cases of both vertical and horizontal Hertzian electric dipoles at a distance h above a semi-infinite lossy dielectric [4]. Since that time, other investigators extended his work to grounded dielectric slabs [5]. A problem very similar to the one presented here is the derivation of the Green's function for a Hertzian electric dipole whichi has

-3 - been solved by R. S. Elliott [6]. Many details of that work will be followed here.

- 4 - 2. Wave Equation Consider a horizontal HMD at the position x'=O, y'=O, z'=-b (Fig. 1). The dielectric slab is assumed infinite in extent with a dielectric constant ~ and a permeability pt. With a eiJ)t time dependance, Maxwell's equations take the form: -- - - - Vx E=-j0olH - M (1) V X H = j(O~E (2) V. B=O (3) V D=O (4) whe re - - B = 1H (5) D = ~E (6) and B = R' (1-jtan6 ) (7) ~ = ~E (1-jtana ). (-) Because of (4), the electric flux density D can be expressed as the curl of a vector potential F as following: D = V x F. (9) When we insert equation (9) into equation (2), the result is that V x (H - jCOF) = 0 (10) or that H = j(-F + V(

- 5 - where em is an unknown scalar function. m When equation (11) is placed in equation (1), then we result in the following equation — a -F -V -— 1 - - 2 — e —, -V2F+ V V ~F= (OqCF - jOCg E (12) V - = ( ~~[ - j3c~Vm Since 'm is not yet defined, we may assume that 1 m- = ---- V -F. jm ( VL (13) This relation is known as the Lorentz transformation. In view of equation (13), equation (12) takes the form -+ 2- - V-2F + k F V2F + k F = $_M (14) i.e., the inhomogeneous wave equation. At points away from the source, the electric and magnetic fields are give by 1 1 E = - V F ~ (15) and H = [k I + V - F. oq)i ( 1. 6)

- 6 - 3. Solutions to Wave Equation The solutions to Wave Equation can be classified to primary and secondary. Primary is the solution to the inhomogeneous equation -Op ^-P V2F + k F = (x-x') 6(y-y') 6(z-z') (1 7) while secondary is the solution to the homogeneous one 2-s 2-s V F + k F = 0. (18) It is known that the primary solution is given by — P - \ e F = x - J (19) In equation (19), the origin of the spherical coordinate system is located at the position of the magnetic dipole. As it has been shown in [4], [6], the function -- can be written in the form -e -ulz+bsI = J (p) e - dX. (20) r o ] 0 2 In equation (20), u = 2. Next, we need to turn our attention to the secondary solution. As Sommerfeld has shown (his proof is reproduced in Appendix A), the secondary solution must have an x and a z component which can be put in the following form: -u Iz+b I F= — 4 A (X) J (Xp) e dX (21) x 47 o

- 7 - and o -u Iz+bJ F = — 4 cos A z( ) J1 e d (22) z 4 J z o in which Ax(L) and Az(X) are functions to be determined from the boundary conditions. In this manner, the solution was transform from the spherical coordinate system to a cylindrical one shown in figure 1. In view of the above, the solutions to the wave equation in regions I, II and III can be put into the form: Region I S A S A ( F = F x + F z (23 I Ix Iz Region II F2= (F x + F2 ) z (24) 2 2x 2x 2z Region III S A A (25) F = + F x + F z (25) 3 33x 3z

- 8 (23)-(25) in integral form may be written as: Equations F I = A~ (j AP,?) 0 -u z Jo (p) edX + + 2 0-u 0 coso (k) 1(xp) e 0k 0 (26) F 2 Af = x K ~,j) j (X) Leu (z+bS) 2x+ 0 (?X) e + -U Z ~ B2 (X) eu dX,+ + 2 -- cos4 J j1?p A2z 0 ~ B 2 z(?) e ( 2 7) F I CI k- + A (. U 3Y + + B (X) e] dX, + 3x + 2 ( C -~j cos'o j (xp) ~A3z (X) eZ + B 3 (X) eUZcX 0 (2 8)

- 9 - In equations (26)-(28) u0 = 2 - k, u = - k with k2 = (O2~ and k2 = o. 2 2 and k = cO0~l..

- 10 - 4. Boundary Conditions In this boundary value problem, there are three interfaces to be considered. The first is the ground plane (z=-h), the second is an imaginary horizontal plane going through the center of the magnetic dipole and the third one coincides with the air-dielectric interface. Continuity of the tangential electromagnetic field through each of these interfaces give the following relations: E3x E = 3y 2x s E = 2y S H 2x S H 2y E = 2x E = y H = 2x 2y 0 0 z=-h S S E3x Elx ly Hl3x ly ly z=-bs } =0 (29) (30) (31) In order to translate these equations into boundary conditions for the electric vector potential F, equations (15) and (16) have to be considered. derived From equation (15), the following relations can be

- 11 - 1 aFZ E = - x - ay E = - - - y ~ 3x az (32) (33) (34) 1 ('x E - - ~ dy Also from equation (16), the components of the magnetic field are given by j H = g Ox (~ k2F + VF x ax i35) H y (EgL ay '(36) H = k + - V F z Ou~ z az (37) In view of equations (32)-(37), equations (29) -(31) give the following relations aF 3 -OF -=0 3p aF3 = 0 oz (38) z=-h z=-b s (39) (40) aF aFs 3z ap

- 12 - s s S S 2x 3x F s aF 2x 3x az az a -- —.S a - -*s - V- F = - V F3 ap a p Fx = F 1 aFlz 1 aF;z e ap ~ ap s (41) z=-b s (42) (43) z=-b s z=0 (44) (45) (46) 1 a.~ — -- 1a -- —. E 3p ( ly e ap 2 1 aF x 1 aF E az E az z=0 z=0 (47) Equations (38)-(47) provide ten necessary conditions for the ten unknown functions appearing in the expressions for the electric vector potential F which is also called the Green's function for the problem. Specifically we have: uh + B ( A (X)e + B3 (X) e 3z 3z -uh 0 (48)

- 13 - A (k) eu 3x -e + B 3x,i e =O (4 9) ubs -ubs A z(X) e + B 2z(X) e ubs - A (k) e 3 z -ubs + B 3 (X) e (5 0) ub A 2x(X) e + B 2 (X) e = A (?)L)e 3x + B 3 (?.) e (51) ubs -ubs A 2 (?~) e - B 2 (A)~)e ubs = A (?)4e 3x -B3x (&) e (52) [ub -ub5]- ub A. [A2 (200 e + B 2 x() e J + uA 2 z(.)e s - B2z (X.) e - uB3z (?~) e [ u b + B 3 (?~) e -ub] ub~ J+ uA 3 (?-)es ( 53) C0Ax (,)= Ee~e - + A 2 (A) + B2 (?L) (54 ) (55) A 1z(X) = Az (%k) + Bz k XA lx(X) + uOAI (X) = X e -ubs? A X + X + 2x ()+B2x k 1i ~ uA 2 (24 - uB 2 (24 (5 6)

- 14 -u1Alx (X) = u eub u + A 2x(k) - uB 2x(?) (57)

- 15 - 5. Green's Function The simultaneous satisfaction of equations (48) to (57) gives cosh [u(b -h) ] A1 = 2 X — lx fl ( r, h) (58) cosh[u(b -h) ] A = 2(1-~ )X2 z f (X h) sinh (uh) f2 (X,~Ih) (59) k -h ~6 u sinh(ub ) + ucosh(ub ) A = -- e 2x U f (X,- h) (6() cosh [u (b -h) ] -uh 2 s A = ( -1)e -- 2z r f (X,~ h) I r 1 (61) k cosh [u(b -h) ] B = (u-~ u ) -- x r o u fk,e h) uh 2 cosh [u(b.-h)] B2 = (1-e )e ek z r fl (,, h) A 3() = A2 (k) B x () = B2x () (62) 1 f2 (,~, h) (6o3) (61 ) (65) A3z () = A2z () ) (66)

- 1 6 - B 3z(X) = Bz (A) ( 6 7) In the above equations f1 (X,c -, h) = F-ruocosh (uh) + usinh (uh) f2(X,, -h) = ucosh(uh) + u 0sinh (uh) (6'8 ) (69) When these results are placed in equations (2 6) -(28), one has the Green's function for an HMD a distance b 5 below the airL — deleIectlr IC interface. F, F I z F 2x F2 z F- I?cosh[u (b -h) ] - - m 1(~,,h J(&p) e dX (7 0) 6"0 (1 -F6) Ccosh [ u(bs-h)I cosoh 2it f, (A., Ih) 0 - r 00 s inh (uh) f2 (XI ceI h) -0z (-71) (uz) -C-uosinh (uz) f 1X Fr' h) J_ (;-P) dX (72) (73) - - (l )csFcosh(u (b s h)] sinh[u(h+z)] 2 0 f(;k 2 h

- 17 -Fx - J2 cosh[u(h+z)] Cr u sn ub )+ucosh(ub S)J(X dX 74 ox2T f( Ferh) cosubh) - (1osh __u________ sinhllu(h+z)]2 3 C r f f1( - h) f (X- h)

- 18 -Conclusions The Green's function for a horizontal Hertzian magnetic dipole has been derived by solving the appropriate boundary value problem. This Green's function can be used in formulating solutions to the problems of slots on the ground of a dielectric substrate, cavity-backed slots and dielectric covered waveguide slot arrays.

- 19 - References 1. R. S. Elliott, "The Scattering Properties of Slots Cut in the Broad Wall of a Rectangular Waveguide, Part II: Longitudinal Slots," Hughes Technical Internal Correspondence No. 5G13.00/17, December 1984. 2. R. S. Elliott and W. R. O'Loughlin, "The Design of Slot Arrays Including Internal Mutual Coupling," IEEE Trans. on Antennas and Propagat., Vol. 34, No. 9, September 1986. 3. R. S. Robertson and R. S. Elliott, "The Design of Transverse Slot Arrays Fed by the Meandering Strip of Boxed Stripline," IEEE Trans. on Antennas and Propagat., Vol. 35, No. 3, March 1987. 4. A. Sommerfeld, "Partial Differential Equations in Physics," Chapter VI, Academic Press; 1964. 5. J. R. Wait, "Characteristics of Antennas Over Lossy Earth," Chapter 23 in volume II of "Antenna Theory," ed. Collin and Zucker, McGraw-Hill; 1969. 6. R. S. Elloitt, "The Green's Function for Electric Dipcles Parallel To and Above or Within a Grounded Dielectric Slab," Hughes Technical Internal Correspondence No. 5752.00/072, February 1978.

Appendix A S Proof of the Need for a F Component S Assume that there is only a F component in all three regions. Then from equations (35)-(37), we have H = i k2F + 2 F x it x x2 X j a2 H = F Y 0c)L ayax x (A. 1) (A. 2) (A. 3) j __ a2Fx H = ~O)~ azax Continuity of Hy along the air-dielectric interface results in the following equation 1 aF x 1 aF2x ~ ayax ~ ayax o (A.4) Integration of equation (A.4) with respect to y and then differentiation with respect to x gives

-A.2 - 1 ia2Fx 2x (A.5) o ax2 E x2 From continuity of Hx along the same interface, we get F = F Fx 2x (A. 6) However, differentiation of equation (A.6) twice with respect to x will give a2F1x a2F2x (A. 7) ax2 ax2 Equations (A.5) and (A.7) agree only if E= o which is not possible. Therefore Fs should have one more component along the z direction.

1 L Z 0 0 -7 = (i - - - - - - - - - - - - - - - - 9 9 II z - / cjI t/ z=liib / (I) 00 X. (I) (III) I.-I-I-.-.-.-.-I A WA V, 2FA W, -, -, V, -4 -, W, W, W, W, W, WI...WA -1 WI 20 -1 W, WA.ff Al r ff - - - - - Figure:i