I. INTRODUCTION Solution of complicated boundary value problems in the spectral domain is a very well known technique. Several years ago Itoh and Mittra used spectral analysis to solve planar geometry problems occurring in microwave integrated circuits, [1]. There have been numerous papers subsequently published utilizing this method, [2]. In this paper, Fourier analysis is used to formulate the solution for reflected and transmitted surface current densities in back to back shielded microstrip structures coupled by a transverse slot in their common wall. By applying Galerkin's procedure in the spectral domain on complementary field quantities, a solvable system of equations containing the desired reflection and transmission coefficient results. II. THEORY Application of the Galerkin method on the spectral domain electromagnetic quantities supported by the structure illustrated in Figure 1 requires a matrix equation which expresses electric and magnetic fields in terms of complement source densities. The appropriate expression is given in equation (1). EZ (a, h) G11 (a,) G1 (a4,) G13(a ) K (a,h,a) la() (a,04A) = G21(a,4) G22(at ) G23(at ) |M (at0 )] EZ (a, -hb, f3) G31 (a, G (a,1) 33 (a, ) K Zb h aZ~~ 1 1 1 2 G13 t ~~~~~~1za a (1)

Explicit expressions for the Giij(a), along with an outline of their derivation are given in Appendix A. In this derivation, we have assumed narrow microstrip lines and a narrow coupling slot. These acceptable restrictions allow us to assume uni-directional surface current densities which have a Maxwellian distribution in the direction of their narrow dimension, [3]. Recent theoretical and experimental research by Dunleavy has shown that the narrow microstrip line current variation mentioned above is very accurate, [4]. Unfortunately, no such verification exists for a magnetic current on a narrow aperture. However, since a narrow microstrip line supporting an electric current and a narrow aperture supporting a magnetic current are eseentially dual quantities, [5], it seems that an assumed Maxwellian distribution for the aperture problem would be accurate. III. SPECTRAL DOMAIN REPRESENTATION OF CURRENT DENSITIES The detailed geometry of this problem is given in Figure 2. In the spatial domain the two electric surface current densities away from the discontinuities are written as: KZ (x',h,z') = (e Re ) z (2) Zb b Kzb(X',-hbz') -Te 0g(x') Z<-Z where 2

j 2 { 2 (x'-xi) 2-1/2 W W,1 [ rW. Wi 2 (i = a,b) 0 Otherwise (3) Ms ms It should be mentioned that Y and Y are known constants a Zb for their respective guiding structures. Their values may be obtained from the literature, [2]. In the vicinity of the discontinuities, all surface current densities in equation (1) must be expressed in terms of appropriate basis functions. KZ (x',h,z') and KZ (x',-hb,z') are written in the spatial domain as N KZ (x',hi,z) = Ii g. (x') f (Z') i = a,b (4) Where the I are unknown amplitudes and ni sin[K(zI-zn-1)] z < Z < sin (KlZ) n-l n (z') sin K(z -z) ni nni n+1 sin(Kl) n Z n+l 0 Otherwise For n * 1

sin [K (~zTl -z'] ) ] Iz + 1 <z'_<+z sin (K1) o z f (z) ii ) 7 (5b) (+) i f = () Otherwise where 1 =Z -Zn Zn -Zn- The magnetic current on the coupling aperture is represented as M Mx (x',0',z ) = Vn fn(x') go(z') (6) n=1 where fn(x') and go(z') are given as equations (7) and (8), respectively, and the Vm are unknown amplitude coefficients. sin [ K(x'-xn_) ] sin (Kl) n-1 n fn(x') = sin [K(x n+-x')] sin(Kl) n n+l 0 Otherwise -1/2 -t 2 2 g0(z') = (8) 0 Otherwise %~~~~~

We also see that lx = xn+1 - Xn = Xn - Xn-1 and xn = (n-1) lx. Under these conditions the requirement that the magnetic current is non-zero only on the coupling aperture is satisfied. The next step in the analysis is to obtain the Fourier transforms of (4) and (6). The transforms of interest are written as: jax' jZ z' K (a,h,a) = K (x',h,z') e e dx'dz' (9) 00 K (a-h = KZ (,-hbz ) e e dx'dz' (10) Zb J Zb rxM,0Oz' jax' jz MXa (a,0o) = x',O,z) e e dx'dz' (11) -00 For KZ (a,fh,4) we obtain a oo Zn- 1 C jax' CjAz Kz (a,h P) = ga(x') e dx' fn(Z') e dz' n=2 a zn+ 1 + I f1(zI) e dz' + e - Re Z) ej dz' ] zO ZZ Kz(aha) = I 2 + I3 + I4 (12) Wa +Xa -1/2 a I ~2 (x'-x) 2 a 2 Xaa

Zn Zn +1 Ia = sin K(z'-z n-1) eZ dz sin K(z 1-z') e d e dz' 2 na sin(K sin(K1)sin(Kl n=2 z1 z (14a) zo a jrz 1I3 IJ f1(z') e dz' (1 4b) zo -1 0 Z a (r + JP)z' (-Y + jp)z' I4 [e -Re dz' (15) zo a a a After some algebra I1, I2, and I3 are found to be: a Jaxa. wa I= e J~a 2J (16) a j23 e JPzn I I e (17a) I K2= 2 na (17a) J (Zo - )n ) a Ila J1lz I = (KI2a e [K1K-e (jPsinKl - kcosKl)] (17b) (K- ) sin(Kl ) To evaluate I4, some special considerations are necessary. 14 is written alternatively as A~~~~~~~~~

a r a (7z7+ jP) z' (-TZs + jP) z'J 1 u(z'-z) - Re z (18) 4 JRe ms and we assume a lossless system so that M is a purely imaginary a rs ms quantity, Yza = j k. The Fourier transform of the product of two ZaZa functions is equal to the convolution of the individual Fourier transforms of the two functions, [6]. Using this idea to evaluate (18) shows jk, z jk,, a * f jk0zae -Rea (19) U (jZ~d'-o ej1 dgz' U(z-z () dz' + e p (20a) I aZIejPdzI = 2K + k)(O a From equatios' 1-29R) a (20c) From equations (19) and (20) we express Ia as 4

I4 2 i e' O~ A)( + [ + k Zk - R 8-Z - k) d Loo jjXZ = 2i L ie o (X) 8 (P-A + k) + (- + k Za jx Za - ( ) Re Ms (,- - ks ) | Za ja Za a 2 2 (kmS)ok a4 2 2 ( + kms) - 22 RS ( - k) 4 Za Za j (P+kz) o (-) zo + 27 +k R (21) za za Substituting (16), (17), and (21) into (12) shows: jaxa -i( +kz) Zo { wal 21c e e a Kz (a, ha,f) = - Jo(a Za jaxa -j (zka) Wo j2pe JO (a 2 ) N 2K e e a 2Jo 2 JZn j (j-km) (K2-p2) n=2 na a + Ia ( 2 K 2) ) (a ) [K - e (jIsinK1 - kcosK1 ) + K a (K +) sin (kl e + 22 8 (+k) - RS (-k) e Ja ) (22) ~2.[6B i;)-i~(& 2mS,1:~~~ \8

In a completely analogous manner Kz (a,-hb,P) is found to be 7reb i ( P+k ) wbZ 2n e e Kzb(a(,-hb,) -T Joa jaXb -jP(zolZ) J( 1Wb Ilb e e ( b1 + ei e ) ( [K + e (jPsinKl + kcosKl)] (K2-2) sin(kl ) t pn2jaxb r ( w ms) (K 2n=) (23) The discretized microstrip currents near the discontinuity region are divided in'N' partitions in both guides'a' and'b'. More severe disruption of the current occurs in guide'a' due to the open end upon which KZ is incident. If'N' partitions are Za required to accurately characterize KZ then the same number will Za represent K in its respective discontinuity region. Zb The Fourier transform of Mx (x',O,z') is expressed from (6), a (7), (8) and (11) as M l( ~ o00 M a (Oc ) = Vn J ffn(x') go(z') ei ei dx' dz = I4 5 n=l (24)

t/2 -1/2 14 _ j 2 - 1-[2z 1 " e dz' -t/2 and clearly t/2 1/2 I4 _= | 2 (_2z [cosz' + jsinjz'] dz' t/2 I4 | xt i @ cot/sli2 -1/2 =t2c r1- C cosiz' dz' -t/2 t/2 2 1 2 S s4i= J- os )r osm ea sin(7 do 2t L e n: I~t J~[B t I = V f(xC) j-xSdx' (25) j edx (2) - result yields

Finally then, M (a, 0,) is written as: a2jan M = Vn e (28) Xa 2 (K -a2 ) n=l Equations (22), (23), and (28) are very important quantities. IV. FORMULATION OF THE MATRIX EQUATION The results obtained in the previous section will allow us to solve equation (1). In the following discussion, spatial and/or spectral dependence of functions will be implied and not explicitly written. If there is any ambiguity, it will be clearly explained from equation (1): - (la) E = 11 Z + G1 M (29) -' (Ia) H K M (30) Hz G21Za + G22 Mxa (30) ~ (lb).. Ez G32M + G33 K (31) Since, from Appendix A we know that G13 = G23 = G31 = 0. Also, note that the surface current densities are expressed in terms of summations of appropriate basis functions as derived earlier. Before going any further with the analysis it will be convenient to write K, Kz, and M as: z Za1 xa 11

~a ~a n K = - + R k1 + In K (32) n=l N Kz = T k1 + (33) b n b b n=l1 M ~n M = V M (34) Xa n xa n=l 1 a a a n ~b ~n ~n where k, kl, and KZ are easily deduced from (22). kl, K and M are similarly obtained, respectively, from (23) and (28). We next multiply equation (29) by Kq and integrate from -oo to oo with a respect to a and f for different values of q. This shows: @0 — q -'(la) |f| Kzq E Kda d K K d d -00 0 + KZa G12 Mx da do q = 1,2,... N (35) but Kq E(la) JKz Ez da do = O -0 Because, from Parseval's theorem 00 00 jjKq E da d= 2 Kza Ez dxdz= -(2Ki) - 12

~q -- (la) Since it is clear that Kz and Ez are complementary in the za z x-direction. Using (32) and (34), equation (35) becomes: -k + R + I'fla Kz dadp _J Za co M ~q ~n ||f K GGk dV M ddad = 0 q = 1,2,...N Za G 12 l dad Performing similar operations on equations (30) and (31) yields: N 00 -||M G,, ko d + R|x G-1 kq da + Ina Za G1 KZa dado (36) n=1 00 13 /,v n Kz G12 M dado 0 q,1 1,...,I — _0 Performing similar operations on equations (30) and (31) yields: From (30) ~ q a M ~ a M G2 k dado + R M G k d(Xdp ~~ M~q ~Kn + I —a xa G21KZ dado (37) n=l M 00 n~l xa 21 x

From (31) @00 M 00 -T J Kqb G33 k1 dadp + Vn b G32a n=l (38) N 0 + Inb G33 Kzb dad = ~ q = 1f2f. N n=l -00 Equations (36), (37), and (38) represent an inhomogeneous system of 2N+M equations containing 2N+M+2 unknowns. We get two additional relations by the requirement of electric surface current continuity at (x=ha, z=zo) and (x=-hb, z=-zo). Look at Figure 2. From equations (2), (3), (4), and (5) we see ms ms jkz, z -Jkz zo Ii = e - Re (39a) -jks z I = T e (39b) Equations (39) represent the necessary relations for solving for R and T. We shall now express (36), (37), and (38) more compactly: (36) becomes N M S =RP+ In Dqn + Vn G q = 1,2,...,N n=l n=l where 14

r q ~a Sq = J G11 k d(ld3 (00 K Gll k1 dadf (40) 00 DGn = Kz Gl KZ d(ldp -00 qn | Z a G12 dad (37) becomes N M $q R Q + In Lq + V F q=1,2,...,M nil n=l where S = G2 k dado q xa 21 o co Qq= JJ M G21 kl dadl (41) qn MG2 1 Kza dd3 A II — 00 (38) becomes 15

0=- T U+ I V +n n X q 1,2,..,N q nb qn n q = n=l n=l where Uq Z= I G33 k1 dadj r00 V KG K dadD (42) qn Zb 33 Zb q ~n X K= G3 M dadj3 qn JZb 32 Xa V. DISCUSSION OF NUMERICAL SOLUTION It is clear from the results of the last section that the problem under consideration is very difficult. Since no numerical results will be obtained, a discussion of how to do this is in order. Equations (40), (41), and (42) contain integrals which must be evaluated from -oo to oo with respect to two variables: a and f. A brief glance at Appendix A indicates that numerical integration of these terms will surely be required. Although it 1 1 1 1 is not obvious, these integrals vary at worst as 3/2 2 I 3/2 2 a a in one part or another. The squared terms will converge rapidly I 1 and the same can be said for the 3 and -- terms. Ultimately, one 3/22 wpould need to program a computer to do these integrations to find 16

how the convergence progresses. However, in [7], Mittra and Itoh discuss so called'numerically efficient techniques' for solving boundary value problems. The approach used in this report closely resembles the algorithms discussed there. As a result, it would seem that satisfactory convergence is obtainable for the equations derived here. Besides having to evaluate the integrals mentioned, we must also solve a linear system of equations. There are many viable alternatives for this part of the problem and no further discussion is necessary. VI. CONCLUSION The solution for reflection and transmission coefficients in slot-coupled microstrip lines has been formulated in the spectral domain. Although no numerical results have been obtained, pretinent numerical considerations have been discussed. 17

REFERENCES [1] T. Itoh and R. Mittra, "Spectral-Domain Approach for Calculating the Dispersion Characteristics of Microstrip Lines," IEEE Trans. Microwave Theory Tech., vol. MTT-21, No. 7, pp. 496-499, July 1973. [2] T. Itoh, ed., "Planar Transmission Line Structures," IEEE Press, 1987. [3] P. Katehi and N. Alexopoulos, "Frequency-Dependent Characteristics of Microstrip Discontinuities in Millimeter-Wave Integrated Circuits," IEEE Trans. Microwave Theory Tech., vol. MTT-33, No. 10, pp. 1029-1035, October 1985. [4] L. Dunleavy, Ph.D. thesis, University of Michigan, 1988. [5] R. Harrington, "Time-Harmonic Electromagnetic Fields," New York: McGraw-Hill, 1961. [6] A. Papoulis, "The Fourier Integral and Its Applications," New York: McGraw-Hill, 1962. [7] R. Mittra, ed., "Computer Techniques in Electromagnetics," pp. 305-342, New York: Pergamon Press, 1973. 18

Appendix A Derivation of the Green's matrix Hybrid mode analysis with E(x,y,z) = -jOA + VV ~ A + V x F (A-l) H(x,y,z) = jiF - VV * F + -V x A A A Choose A = zA and F = zF where z z (i) W1 1 _(i)_ -jax -jgz (Xy, ) z) (a,y,J) e e dado z z (2ic) (A-2) =(- ) (a,y,) = lalb,2a,2b) _____i_ -jax -jgz (x,y,z) 2 (a,y3 ) e e doadl z z (2ic) (A-3) F = ( ) (i = la,lb,2a, 2b) (i) (i) A (x,y,z) and F (x,y,z) must satisfy z z (A-4) (V + ki0 F (x,y,1z) 19

Substituting (A-2) and (A-3) into (A-4) implies: " (i) [ (a2 )] i (a( Xy) -o Lti 2 12+32 2 Letting yi = + k. shows (A-5) 4y2 = 0 i) (, y, Appropriate solutions to (A-5) in each region of the structure shown in Figure 1 are: -la) ~(la) -(la) A = A sinh (TaY) + B cosh (alay) VA = D sinh[ b)]2a - (lb) -(lb) (lb) (A-6);jA = A sinh (YbY) + B cosh (alby) =2 P s inh [ 2b(y+b)] (la) "(la) -(la)'F = C sinh (lY) + D cosh (aiaY) "F == N cosh[Y72a(y-b (2a) "(lb) [ (lb) (A-7) "'F = C sinh (lby) + D cosh (albY) (2b) "h2b) Y2 =F =N cosh T2b\ ) Note: The - notation means Fourier transform domain. 20

The transformed fields must satisfy -(lq) (2q) Ez (aihq E) = E (aIhq ) - (lq) (2q) EX (a hqI?) = E (ahqi) Hz (ah,) = H (a, hq,) A (lq) - (2q) A-8 (ahqf) H= (a,hqI,) K (a hq) (lq) Ex (aO, ) = 0 (lq) EZ (a,0,o) = M (at,0o) (q = a,b) Kz (a,hq, ) is the Fourier transform of the electric surface current on the qth microstrip. M (a(,O,f) is the Fourier transform of magnetic surface current on the aperture which is common to the two microstrip structures. At this point in the derivation it should be noted that symmetry about the line x=O, y=O exists between guides (a) and (b). In order to derive Gll((a,3) and G12 (aC,), we apply the ~(la) boundary conditions (A-8) and express Ea (ah a, ) appropriately. G32(aP) and G33(ap) are obtained from, respectively, Gl2(a4,) and Gll(a4,) as G32(a ) = G12 (a, ) ha = -hb; kla klb (A-9) G33 (aL, ) = Gl (at) I ha = hb; kla klb 21

Consequently, solving for G12((a,) and Gil(a,) yields G32(aP) and G33(aP) in a direct manner. Imposing the boundary conditions (A-8) and appropriately manipulating terms yields the components of the Green's matrix as: Gil(a.)= (jal)(kla- a) (k2a ) a/ {a (kla k2) _'2k2 (ka 2 32) - Y2 k1 k2 2) k k 0 k kk ~+ yrya(k2 2) (k2a 2) {2 k2 tanh[Y2ahb)] coth(Yla ha) 2 h(h -bl I + ka tanh(laha) coth[Y2 h-b } -jah( ka kla) cosh(laha) G (((a ) G21 (a,, )= 2a Y~ala a2a G[l(O (kla~ a2) tanh[y2a(h b)] - YIa(k2a P2) tanh( laha)] =G 0 31 2 22 2 2 2 (X Ih G1 (a, ) = cosh(l h) - 2f(kla 82) (k2a 02) (ka k2a (a,1,h) (X P, h) ira(aliha) = sin(Yiaha) { tanh[Y2a(ha b)] [f32y (k2a- 52) (k~a+ 72a) -a25 y 2a(kla -k2a) + a y 2a(kla+'Y 2a) (kla~ 52) - kiaY 2a(kla~ 52) ] +tn(Ylaha) [ kl>a1a( k2- 132) - 7132i(k2a+ 82 ) (k2a _132) 222 Fa a sa) = in Yia a tanh YY 2a Y',,a la 8 0 2aa+'2a 2 2k2 2 k 2 2k 2 2 k2 2 ap Yy 2a l-2a )+'X ^ 2 a la+' 2a kla a)- 0la'Y2 a

+ CSh(Ylakha) [a ia(k2a k2)- aY laY2a k2a )] aa(aP,ha) = a13 (kla- k2a) + k 2 ( 132)2 + 22(21 ay l + rYl k2 (k2~ 32 ) (k-la- 1) tanh('Ylaha) cOth[f2a(ha b) + Y la')'-2a ( 2 - ka 2 32 2 2 COth 2 2 tanh2a(hab) =1(a5) = ka k2a k2( a 2(k - (k + k )(a2+ 2) + k ( k2a- 132 [ 2 COSh(laha) 2 2 tanh(ha) th[ 2 a(h a-b] 2 2 -z23 laY2akla k- a 2) cothI71haI tanh[r2a(ha-b 2a'Y](a k,la coth ) = - j.... + jap p) -j ( 2s(2 [2 k22, la(2- 2) la 2a a

z~~ (a, k-> 3) { i [y2klaka 132) tanh[72a(h-a b)]'- a(ka 2) tanh(Ylaha) - ( - k2a) / [ 2aY2a(ka 13) (0 coth['Y2a ha -b - eliaya(k~a 12) coth(tlaha)] } G32 (a,P) is obtained through equation A-9. G13(a,J) = 0 G23 (a, ) = 0 G33 (a,s) is obtained through equation A-9. 24