RL 846 RADIATION LAB REPORT PRELIMINARY RESULTS FOR STUDY OF THEORETICAL AND EXPERIMENTAL CHARACTERIZATION OF DISCONTINUITIES IN SHIELDED MICROSTRIP L.P. Dunleavy and P.B. Katehi Electrical Engineering and Computer Science Department University of Michigan Ann Arbor, Michigan August 1987 RL-846 = RL-846

ii TABLE OF CONTENTS LIST OF FIGURES....................................... iii LIST OF APPENDICES.............................. v CHAPTER I. INTRODUCTION:...................................... 1 II. THEORETICAL METHODOLOGY.............................. 7 2.1 SUMMARY OF THEORETICAL APPROACH 2.2 FORMULATION FOR METHOD OF MOMENTS SOLUTION 2.3 DERIVATION OF THE GREEN'S FUNCTION 2.4 IMPEDANCE MATRIX FORMULATION 2.5 EXCITATION VECTOR FORMULATION 2.6 MODIFICATIONS FOR ANALYSIS OF TWO PORT STRUCTURES III. PRELIMINARY RESULTS.............................44 3.1 RESULTS FOR OPEN END DISCONTINUITY 3.2 RESULTS FOR SERIES GAP DISCONTINUITY APPENDICES..................................... 61 BIBLIOGRAPHY.....................................101

iii LIST OF FIGURES Figure 1.1 Typical millimeter-wave integrated circuit structure. Accurate modeling of discontinuities is key to cost effective designs............................... 5 1.2 Two basic classes of microstrip................................ 6 2.1 Basic geometry for the shielded microstrip cavity problem................. 9 2.2 Discontinuities for which thin-strip approximation is valid............... 10 2.3 Flow chart illustrating theoretical approach for characterizing microstrip discontinuities............................................................... 11 2.4 Representation of coaxial feed by a circular aperture with magnetic frill current M........................................... 16 2.5 Total fields Eq, Hq inside cavity produced by magnetic current source MA at aperture and electric current distribution Js on the conducting strip.......... 17 2.6 Test current field Jq on conducting strip and associated fields Eq, Hq........ 18 2.7 Strip geometry for use in basis function expansion of current............... 19 2.8 Geometry used in derivation of the Green's function..................... 24 2.9 Geometry used for numerical integration to compute excitation vector. Note: the relative size of the feed is exaggerated for clarity....................... 36 2.10 Total fields inside cavity Et~t, fjt~t produced by magnetic currents M01, M,2, and electric current Js........................................................... 42 2.11 Strip geometry for basis function expansion with dual excitation........... 43 3.1 Representation for microstrip open end discontinuity..................... 47 3.2 Effective length extension of a microstrip open circuit discontinuity on an alumina substrate (Cr = 9.6), as compared to other numerical results................. 48 3.3 Effective length extension of a microstrip open circuit discontinuity on a quartz substrate (Cr = 3.82), as compared to other numerical results................... 49

iv 3.4 Angle of S11 of an open circuit as compared to measurements and Super Compact results............................................ 50 3.5 Representation for microstrip series gap discontinuity............... 51 3.6 Magnitude of S21 for series gap G1 (G = 5 mil) as compared to measurements, Super Compact, and Touchstone............................... 52 3.7 Angle of S21 for series gap G1 (G = 5 mil) as compared to measurements, Super Compact, and Touchstone..................................... 53 3.8 Angle of Sll for series gap G1 (G = 5 mil) as compared to measurements, Super Compact, and Touchstone...................................... 54 3.9 Magnitude of S21 for series gap G2 (G = 9 mil) as compared to measurements, Super Compact, and Touchstone........................... 55 3.10 Angle of S21 for series gap G2 (G = 9 mil) as compared to measurements, Super Compact, and Touchstone...................................... 56 3.11 Angle of S11 for series gap G2 (G = 9 mil) as compared to measurements, Super Compact, and Touchstone.................................... 57 3.12 Magnitude of S21 for series gap G3 (G = 15 mil) as compared to measurements, Super Compact, and Touchstone........................... 58 3.13 Angle of S21 for series gap G3 (G = 15 mil) as compared to measurements, Super Compact, and Touchstone........................... 59 3.14 Angle of Sll for series gap G3 (G = 15 mil) as compared to measurements, Super Compact, and Touchstone................................. 60 D.1 The current source is raised above the substrate/air interface to apply boundary conditions................................................. 76 F.1 Strip geometry used in evaluation of surface integrals............. 91

v LIST OF APPENDICES Appendix A. REVIEW OF THE METHOD OF MOMENTS.......................... 62 B. DERIVATION OF INTEGRAL EQUATION FOR ELECTRIC FIELD.... 64 C. EIGENFUNCTION SOLUTION FOR GREEN'S FUNCTION............... 67 D. BOUNDARY CONDITIONS AT SUBSTRATE/AIR INTERFACE........... 75 E. EVALUATION OF THE MODIFIED DYADIC GREENS FUNCTION. 85 F. EVALUATION OF CLOSED FORM INTEGRALS OVER SUBSECTIONAL SURFACES....................................................... 89 G. EVALUATION OF THE MAGNETIC FIELD COMPONENTS.............. 96

CHAPTER I INTRODUCTION: Millimeter-wave integrated circuits are becoming increasingly important in a variety of scientific and military applications, and a wide range of solid state circuitry has been demonstrated in both hybrid and monolithic form. However, the inability to accurately predict the electrical characteristics of various circuit components is a serious barrier to the widespread and cost effective application of these technologies. Accurate microstrip discontinuity modeling is key to improving the cost effectiveness of microwave and millimeter-wave circuit designs. Typical millimeter-wave IC's contain various active and passive elements interconnected by microstrip transmission lines as illustrated in Figure 1.1. In the vicinity of transmission line junctions and other discontinuities, evanescent fields are excited which cause unwanted parasitic effects, and generate space and surface waves that can significantly affect circuit operation. These discontinuity effects can be modeled by the use of lumped equivalent circuits; however, there are a number of different approaches that can be used to approximate the values of the equivalent circuit elements. The majority of existing approaches are based on either quasi-static solutions or use a planar waveguide model. Despite their inherent accuracy limitations, equivalent circuits derived from such solutions are adequate for many applications since an approximate design, at the least, provides a starting point after which the circuit may be tuned to achieve 1

2 the desired performance. This is generally true of hybrid integrated circuits operating in the lower microwave range (at or below X-band). Quasi-static techniques are well established and described in standard texts [1]-[3]. With these techniques, equivalent circuits are derived in terms of static (i.e. frequency independent) capacitances and low frequency inductances. Convenient analytical formulas for discontinuity parasitics are possible, yet their accuracy is questionable for frequencies above a few GHz. Planar waveguide models provide a frequency dependent solution. In this approach, an equivalent planar waveguide geometry is proposed for the microstrip problem. This transformed problem is then solved using an appropriate analytical technique such as mode matching [1]-[4]. Models derived from this technique are generally considered accurate to higher frequencies than quasi-static models. However, although they can include dispersion effects in the solution, planar waveguide models cannot take surface wave effects into account. Even if the model were strictly valid, the accuracy is limited by the method used to approximate the effective width, and dielectric constant of the equivalent planar waveguide. Further, this approach does not provide a means to account for radiation effects when present. In many cases, the limitations of the above two techniques cannot be tolerated. Often, circuit tuning is difficult or impossible; as a result, inaccurate circuit models lead to long design cycles with many costly circuit iterations. One example where this is true is in Monolithic Microwave Integrated Circuit(MMIC) design, where the small size and the fragility of MMICs make tuning virtually impossible. Also, for both hybrid and monolithic circuits, the need for greater modeling accuracy increases with frequency. Parasitics have a greater effect on circuit behavior, and it becomes ever more important to accurately model surface wave effects, mutual interactions between circuit elements and radiation losses (if present). Hence for MMICs and for

3 both hybrid and monolithic millimeter-wave IC's, a more rigorous solution for microstrip discontinuities is necessary. A technique that meets this requirement was recently developed by Katehi and Alexopoulos to treat discontinuites in open microstrip geometries [5]. In this technique, the currents on the microstrip conductors are first computed by a Galerkin's implementation of the method of moments. Based on this current, a transmission line model is used to evaluate frequency dependent equivalent circuits and scattering parameters [6]. This technique has so far been applied to solve for various discontinuities in open microstrip. Here, the top of the substrate is left open to the air as shown in Figure 1.2a. One application where open microstrip is used is in a monolithic antenna array where radiating elements are integrated along with passive and active components on the same or adjacent substrates. In open microstrip, radiation from circuit elements is unavoidable and requires accurate modeling at high frequencies. Radiation is often avoided in practical circuit designs by enclosing parts of the circuitry in shielding boxes (or housings); hence, it is important to also establish an accurate method of solution for microwave and millimeter-wave circuits operating in a shielded microstrip environment. To this end, the new analytical methodology presented here is an extension of the approach of Katehi and Alexopoulos to shielded microstrip configurations of the type shown in Figure 1.2b. This report presents theoretical methodology and preliminary results for this new method that promises to provide more accurate circuit models for discontinuities in shielded microstrip. While rigorous solutions to shielded discontinuities have been advanced by others, their accuracy is unclear and there are several structures which have not been adequately analyzed. One approach has been developed by Jansen et. al., who use what is referred to as a spectral domain iterative technique [7]-[10]. Although reasonable results have been demonstrated for several microstrip structures, the accuracy of their methods is

4 unclear since there has been little or no accompanying experimental verification, and only limited comparisons to other rigorous numerical solutions. Furthermore, there are important discontinuity structures that have not been addressed by these authors, leaving considerable room for contribution. Another solution has been presented more recently by Rautio and Harrington [11], [11] who use a method of moments technique. However, the only results from this work we are aware of is for the input impedance and current distribution for an open circuited stub. Their solution is, in some respects, more general since they assume a two directional current distribution, while the present work uses one directional currents. Our approach also differs in the choice of basis functions and the method of circuit excitation employed. The new analytical methods developed here have been applied to obtain results for open end and series gap discontinuities in shielded microstrip. Work is currently in progress to obtain results for coupled line filter structures. To test these analytical methods, an experimental study is being conducted in cooperation with the Microwave Products Division of Hughes Aircraft Company. This work includes a study of "de-embedding" techniques, from which it has been concluded that; the technique most suitable for this work is the thru-short-delay (TSD) method. With this method, the major factor influencing accuracy is microstrip connection repeatability. To explore this, a connection repeatability study was conducted [13], and the results were extended to evaluate the associated uncertainty in de-embedding accuracy. S-parameter measurements were then obtained for selected discontinuity structures. The numerical and experimental results are compared wherever possible, and demonstrate the accuracy and usefulness of the new theoretical methods. The combination of theoretical and experimental work presented here represents an important contribution in the area of millimeter-wave IC design by lending more insight into the high frequency behavior of microstrip discontinuities than previously possible.

5 \ tran microstrip Figure 1.1 rate modeling of dscon of iscntiuites s ey to cost effective de

6 k\\\^\^^\\\\\X\\\^\\\ a) Open microstrip I I \\\\\\\\\\\\\\ b) Shielded microstrip Figure 1.2: Two basic classes of microstrip.

CHAPTER II THEORETICAL METHODOLOGY 2.1 SUMMARY OF THEORETICAL APPROACH 2.1.1 Problem geometry As a first step in the development of the theoretical technique for analyzing shielded microstrip discontinuities, consider the shielded microstrip geometry shown in Figure II. The shielding box forms a waveguide cavity, which -for most practical uses- is cut-off for the highest frequency of operation. That is, the cavity dimensions are usually such that non-evanescent modes are suppressed. However, the solution presented here, as far as the computation of the current distribution is concerned, is applicable whether the cavity is cut-off or not. as far as the computation of the current distribution is concerned 2.1.2 Theoretical assumptions In this solution, a few simplifying assumptions are made to reduce unnecessary complexity, and excessive computer time. Throughout the analysis, it is assumed that the width of the conducting strips is small compared to the effective (or guided) wavelength. In 7

8 this case, unidirectional currents may be assumed with negligible loss in accuracy. Also, while substrate losses are accounted for, it is assumed that the strip conductors, and the walls of the shielding box are lossless. These assumptions are valid for the high frequency analysis of the discontinuity structures of Figure 2.2, provided good conductors are used in the metalized areas. 2.1.3 Description of theoretical approach The theoretical technique is based on a Galerkin's method formulation of the method of moments. The required integral equation is derived by first representing the coaxial feed by an equivalent magnetic current source. Reciprocity theorem is then applied to relate this magnetic current source and the electric current on the conducting strips to the electromagnetic fields inside the cavity. By expanding the electric current into a series of sinusoidal subsectional basis functions, the integral equation is transformed into a matrix equation. The matrix equation is then solved to compute the current distribution. Finally, based on the current, either an equivalent circuit, or scattering parameters, or both are derived to characterize the discontinuity being considered. A flow chart illustrating this approach is shown in Figure 2.3. 2.2 FORMULATION FOR METHOD OF MOMENTS SOLUTION The method of moments is a well established numerical technique for solving electromagnetic problems [14],[15]. A review of the basic approach is given in Appendix A. This section makes use of the method of moments to set up a matrix equation that provides for a computer solution to shielded microstrip discontinuity problems. 2.2.1 Formulation of integral equation using reciprocity theorem

9 shielded cavity microstrip (or housing) coaxial input coaxial output dielectric substrate Figure 2.1: Basic geometry for the shielded microstrip cavity problem.

10 T \ w OPEN END - T SERIES GAP SERIES GAP 1\\\ \\\\\. k\\ \\\\^i I k"",,', W < ' 4 COUPLED LINES W Figure 2.2: Discontinuities for which thin-strip approximation is valid

11 METHOD OF MOMENTS I ii CURRENT DISTRIBUTION ON CONDUCTING STRIPS I TRANSMISSION LINE MODEL 1 - EQUIVALENT CIRCUIT MODEL v SCATTERING PARAMETERS I,, Figure 2.3: Flow chart illustrating theoretical approach for characterizing microstrip discontinuities

12 Consider the problem geometry of Figure 2.1. In most practical applications, the coaxial feed (or launcher) is designed to allow only transverse electromagnetic (TEM) propagation, and the feed's cnter conductor is small compared to a wavelength (i.e. kra <: 1). In these cases, the radial electric field will be dominant on the aperture and we can replace the feed by an equivalent magnetic surface current whose only component is in the ~ direction (i.e.Ms = Moo) where 4 refers to the cylindrical coordinate referenced to the center of the feeding aperture (see Figure 2.4). This method of modeling the feed with a magnetic current source will be discussed further in Section 2.5. The magnetic current source is coupled with the current distribution Jr on the conducting strip to produce the total electric field Etot and the total magnetic field Htot inside the cavity as indicated in Figure 2.5. We now propose an independent test current source Jq existing only on a small subsection of the conducting strip as shown in Figure 2.6. Using reciprocity theorem, the two sets of current sources are related according to J J (J *Eq - Hq. Ms)dv= J J Jq. Ettdv (2.1) where the volume V is the interior of the cavity. Since Jq is zero everywhere except over one subsection of the conducting strip, and x-directed (the thin strip approximation), the right hand side of ( 2.1) reduces to J JJJq.Etotdv= If oJqjEt(z= h)ds =0 (2.2) where Sq is the surface of an arbitrary subsection and Etot(z = h) is the a-component of the total electric field which must vanish on the surface of the conductors (z = h) since they are assumed to be perfectly conducting.

13 Reducing the remaining volume integrals in ( 2.1) to surface integrals results in jt Eq(z = h) * Jds = / Hq(x = 0) Mds. (2.3),Sstrip Sfeed Note that this equation is not explicitly in the form of the operator equation of (A.1). This is because in using the reciprocity theorem formulation we have inherently placed it in the inner product form of (A.3). 2.2.2 Expansion of current with sinusoidal subsectional basis functions In order to solve the integral equation (2.3), the current distribution Js is expanded into a series of orthogonal functions as follows. Consider the strip geometry shown in Figure 2.7, let NSECT Js = (y) E Ipap(X)z (2.4) p=l where Ip are unknown current coefficients. The function / (y) describes the variation of the current in the transverse direction and is given by 2 w Yo - W/2< y < Yo + W/2 ()= — [ 2(Y-Y]2 (2.5) 0 else This variation was chosen to agree with that derived by Maxwell for the charge density distribution on an isolated conducting strip [16], and it has been used successfully by others to describe the transverse variation of microstrip currents [5], [17],[18]. As we will see later, with this choice, the y part of the surface integral on the left hand side of (2.3) can be solved in closed form. The basis functions ap(x) comprise an orthonormal set and are given by sin[K(xp+;-x)] < < cl()= <sinK( -l)] (2.6) p I sin(Kle ) Xp1 (2 < ) 0 else k

14 for p $ 1, and sin[K(Ix- x)] 0 < X < a() sKlx) (2.7) 0 else for p = 1, where K = wv/,toer0o is the real part of the wave number in the dielectric region W is the width of the microstrip line Yo is the y-coordinate of the center of the strip with respect to the origin of Figure 2.1 Xp is the x-coordinate of the pth subsection (= (p - 1)lx) Il is the subsection length (lx = p+l - Xp) The x part of (2.3 can also be solved in closed form due to our choice of sinusoidal subsectional basis functions ap (x). 2.2.3 Transformation of integral equation into a matrix equation The integral equation (2.3) can now be transformed into a matrix equation by substituting the expansion given by (2.4) for the current J8. This results in the following: [ j Eq( = h) (y) ap (x) Xds Ip Hq, Mds (2.8) p=l 'f eed which can be expressed as [Zqp] [Ip] = [Vq] (2.9) where Sp is the surface area of the pth subsection [Zqp] is the impedance matrix, which has the dimensions of NSECT x NSECT [Ip] is the unknown current vector, which has the dimensions of NSECT x 1

15 [Vq] is the excitation vector, which has the dimensions of NSECT x 1 The individual elements of the impedance matrix are given by Zqp = E,(z= h) h l, (y)ap(x) ds. (2.10) The elements of the excitation vector are given by Vq= ee Hq Mds. (2.11) We can now solve for the current vector by matrix inversion and multiplication according to [Ip] = [Zqp]- [Vq]. (2.12) 2.3 DERIVATION OF THE GREEN'S FUNCTION To compute the elements of the impedance matrix, we must derive the Green's function associated with the electric and magnetic fields Eq, Hq. We will first define the problem geometry and outline the electromagnetic theory to be used; then, the boundary value problem will be solved for the Green's function. 2.3.1 Geometry and electromagnetic theory The geometry used in the Green's function derivation is shown in Figure 2.8. The cavity is divided into two regions. Region 1 consists of the volume contained within the substrate (z < h), while region 2 is the volume above the substrate surface (z > h). The Green's function will be defined as the electric field due to an infinitesimal current source located on the substrate surface of Figure 2.8. After deriving the Green's function, the fields associated with the test source Jq will be evaluated by integrating over the surface of the qth subsection.

16 COAXIAL FEED MAGNETIC FRILL CURRENT M 0 ANNULAR. APERTURE Figure 2.4: Representation of coaxial feed by a circular aperture with magnetic frill current Mo.

17 y STRIP CONDUCTOR Figure 2.5: Total fields Eq, IIq inside cavity produced by magnetic current source MO at aperture and electric current distribution J, on the conducting strip

18 Figure 2.6: Test current field Jq on conducting strip and associated fields Eq, Hq.

19 y b y 0 x x = (p-l)l p x Figure 2.7: Strip geometry for use in basis function expansion of current.

20 The test source Jq and the associated fields within the cavity (Eq,Hq) are related through Maxwell's equations, which may be put in the following form: V x Ei = -jwuiHi (2.13) V x Hi = jwiE' +J (2.14) V J = -jwp (2.15) V(E)= P (2.16) V(iH) = 0 (2.17) where pi is the permeability of medium i, Ei is the complex permittivity of medium i, and p is the charge density associated with J. It is assumed that both regions are non-magnetic and that region 2 is air, hence l = u2= -o = 47r x 10-7H/m (2.18) E s*o for i = 1 i = (2.19) eo for i = 2 where -j. (2.20) In (2.20) a denotes the conductivity of the substrate material, and the quantity w. is referred to as the loss tangent. In (2.13) - (2.17), i = 1,2 indicates that these equations hold in each of the regions respectively. In addition, and the assumed time dependence is ejt, and it is suppressed throughout the dissertation. To simplify the notation of this subsection, the subscript q is suppressed with the understanding that all the field quantities discussed here are associated with the test source Jq (i.e. Ei = E~ etc.). 2.3.2 Solution to boundary value problem for Green's function We now introduce the vector potentials A' such that

21 1 - H =-Vx A. (2.21.) IPo In view of (2.21), the electric field may be written as (Appendix B,(B.13) ) E= - (1 + VV-) A (2.22) where At satisfies the inhomogeneous wave equation V2 A' + kcAi = -iJ. (2.23) The integral form of the electric field is given by (B.18) or in integral form (B.18) = -jwo ff + 2-iVV-) (G T] * Jdv' (2.24) where k? = P2poci and G is a dyadic Green's function [ref. Tai] satisfying the following equation (22.5 2 Gi + kGi = i (r - r). (2.25) In (2.24) and (2.25), the superscript T denotes the dyadic transpose operation, and I is the unit dyadic given by xx + yy + zi. Because of the existence of an air dielectric interface, a two component vector potential is necessary to satisfy the boundary conditions [19]. Accordingly, let Ai = Axi + A.z (2.26) From (B.17) Ai is related to G by the following volume integral: A' = P J. G'dv'. (2.27) G may be expressed in most general form as follows: GXXXX + Gxxy + Gxzz G= + Gyx + Gyy + G yzy (2.28) + Gozx + Gzy + Gzz

22 Assuming an infinitesimal x-directed current source given by J = 6 (r- ') x (2.29) in (2.27) allows for reducing Gi to = Gx:x + GXZi: (2.30) The functional forms for the dyadic components of (2.30) are found by applying appropriate boundary conditions at the walls: x = 0, and a; y = 0, and b; and z = 0, and c. As detailed in Appendix C, these components may be expressed as 00 00 G() = E E A() cos kx sin kyy sink(1)z (2.31) m=l n=O G = B (1) sin k.x sin ky ycosk(1)z (2.32) m=l n=O G(2 = y E A(2) cos kxx sin kyy sin k(2)(z - c) (2.33) m=l n=O G) = - E B2 sin kxx sin ky cos k2)(z-c) (2.34) m=1 n=O where kx = nr/a (2.35) ky = mnr/b (2.36) 1) -= kl2 - k2 - 2 (2.37) k(2) = /k- k2 - k2 (2.38) k = wv//eT (2.39) ko = wx/7o o. (2.40) The coefficients A(', Amn2 B(1) and B2) are found by applying boundary conditions at the substrate/air interface (z = h). The details of this analysis can be found in Appendix D. The results are: -A() C -p. cos kx' sin kyy' tan k(2)(h - c) dn cos 241) abdlmn cos kz(1) h

23 A(2) -Sn cos kIx' sin kyy' tan k(l)h A(m2) -2n COS mn(2) -(2.42) abdlmn cos k 2(h - c) B(1) — yn l- k2 - Ocs kx2' sin kyy' tan k!')h tan k2)(h -c) 243 abdlmnd2mn cos kO)h B)n -(- er)kx cos kxx'sin kyy'tan k(1)htan k(2)(h - c) (2.44) abdlmnd2mn cos k - )(h - c) where (2 for n =- 0 4 n~ O dlmn = k(2)tan k()h- k1) tan k(2)(h- c) (2.46) d2mn = k(2)tan k(2)(h - c)- k()tan k(1)h (2.47) Having derived the Green's function, we are now ready to proceed to the formulation for the elements of the impedance matrix and excitation vector. 2.4 IMPEDANCE MATRIX FORMULATION The elements of the impedance matrix are given by (2.10) Zqp = / / Eq(z = h). b (y') ap (x') xds' (2.48) which reduces to Zqp= j Eqx(z = h),/ (y') ap (x') ds' (2.49) after performing the inner product in (2.48). Hence, to evaluate the impedance matrix elements we need Eqx(z = h); that is, the x-component of of the electric field due to the test currents Jq at the substrate/air interface (z = h). Once Eqx(z = h) has been found the above surface integration will be carried out -which, as we will see, can be completed in closed form. 2.4.1 Evaluation of the electric field due to the test currents Since Jq is a surface current distribution, the volume integral in (2.24) is reduced to a

y z=c -- a) Cutaway view -y=b x=a z z=c - z=h x I x=a b) Cross section In x-z plane Figure 2.8: Geometry used in derivation of the Green's function

25 surface integral as follows: - jo [(/ + [-VV.) (G )T] Jqdx'dy' (2.50) where Sq is the surface of the qth subsection. For better accuracy, the test currents Jq are expressed in terms of functions which are identical to the basis functions (Galerkin's method) Jq = b(y')aq(x')& (2.51) where b(y') is given by (2.5) with y' replacing y, and eq(x') is given by (2.6) and (2.7) with p replaced by q and x replaced by x'. We now substitute (2.51) for Jq in (2.50) to yield E= -j S [( + AVV'-) (G i)].'(y')aq(X')&dx'dy' (2.52) Let us define a modified dyadic Green's function rF by F = -jwulo [( -2 VV Gi)1 (2.53) Then, Eq can be expressed as E= / I y)a(x'dx'dy') (') (2.54) The dyadic transpose of (2.30) yields (Gi)T Gxx^ + Gi (2.55) When this expression is substituted in (2.53) and the divergence and gradient operations are performed we can express r as ( see Appendix E, (E.4) ) rF = ri xx + riyx + riF z (2.56) where + k da +9 z 9G (2 r.. = G T'9X 9xd2 dz

26 = 1 X ( +Gx aGxz) (2.58) r = + (aG' dG + Z ( &X + Z (2.59) Substituting this expression into (2.54) gives Eq= J F xx (')q(x')dx'dy'x /rSq + IfJs Ftx(y')aq(x')dx'dy'y + j s rFi(Y')aq(x')dx'dy'z. (2.60) Recall that we only need the x-component of this field which is given by Et = f IS rF (y')aq(x')dx'dy'. (2.61) Furthermore, at z = h, boundary conditions require that E(x)(z = h) and E(x)(z = h) be identical. From the above equation it is obvious that this implies r(l)(z = h) = F(2)(z = h) = rFx(z = h). (2.62) This equality is verified in Appendix E, and the result of (E.18) may be put in the equivalent form rX(z =: h) = 00 00 ~ JWO E E mn [cos kzx sin kyy cos kxx sin kyy] (2.63) m=l n=O abdlmnd2mn m ---1 n0 — where fn = n tank(1)htank(2)(h-c) [k (2) (1-k) ta k2) (h ) r 1n tOn tZ 2 zr k2- tan 2( ko (2.64) — k(1) (- tan k(1)h] (2.64) If we place (2.63) into (2.61) we obtain EM)(z=h) = E2)(Z=h)= Eq(z = h) = J I rx,(y')q(z')dx'dy' 00 00 fmn -- joWo abd l 2m- cos k,x sin kyyqmn. (265) m=l n=O

27 where Iiqmn = / j cos kxx' sin kyy'/(y')aq(x')dxldy'. (2.66) This surface integration is evaluated in closed form in Appendix F; the result is qm,K cos kxxq sin Rln sin R2n sin(kYo)Jo ( (267) d3n sin K lx 2 where 2 for q = 1 Cg = (2.68) 4 else K = wv//oeor (2.69 ) d3 = K2 - k2 (2.70) 1 Rln = (k + kx)lx (2.711) R2n = (k - k)lx (2.72) xq = (q- 1)l. (2.73) Hence, the formulation for the x-component of the electric field Eq evaluated at the substrate surface is given by (2.65) where Iqmn is specified in (2.67), fmn is given by (2.64), and dlmn and d2mn are defined by (2.46) and (2.47) respectively. We are now ready to evaluate the impedance elements. 2.4.2 Evaluation of the impedance matrix elements From (2.65) and (2.49) we have 00 00 fm m Zqp = j/ Z I abdlmnd2m pmn (2.74) mnO abdimnd2mjn where ipmn = / / cos kxx sin kyyO(y)ap(x)dxdy. (2.75) Since (2.75) is the same integral as (2.66), we again use the results of Appendix F (by

28 substituting p for q) to produce (pK cos kYxp sin Rl sin R2n ik ) ( (2.76) mn d3n sino) ( (276) where 2 forp=l e = (2.77) 4 else and P = (P - 1)X. (2.78) Finally, from (2.74), we can write out the entire expression for Zqp by replacing the abbreviated notation for dimn, d2mn, fnm, 1qmn, and Ipmn with their defined expressions to yield: Zqp = 0 00 9n tan k() h tan k2)(h - c) m=l n=0 k( tan kh - k) tan k(2h - c)] [k2)ec (1 - ) tan k(2)(h - c) - k(1) (1-.) tan k()h] [kC(2) tan k2)(h - c) - k1) tan k) hi (q(p cos kxxq cos kzXp [K sin R1n sin R2n]2 ( (2.79') ab sin kl [K2 - k2]2 We now turn to the computation of the excitation vector. 2.5 EXCITATION VECTOR FORMULATION In this section, a surface integral will be set up that provides for evaluating the elements of the excitation vector according to (2.11). This equation may be written as Vq= J Hq(x = 0) MOqpdpdq (2.80) eed where p, and a are cylindrical coordinates referenced to the center of the feeding aperture, as shown in Figure 2.9.

29 As a first step towards evaluating this integral an expression for MO will be presented. Second, the magnetic field components parallel to the plane of the aperture (i.e. the y - z plane) will be derived based on the Green's function of Section 2.3. Finally, the two dimensional numerical approximation used to carry out the integration of (2.80) will be presented. The integration is set up to allow for an arbitrary placement of the feed's center with respect to the substrate surface (see Figure 2.9). 2.5.1 Coaxial feed modeling by an equivalent magnetic surface current As stated by Chi and Alexopoulos [20], if the radius of the coaxial feed's inner conductor is assumed to be much smaller than the wavelength (kra < 1), and the coaxial feed line is designed to allow only transverse magnetic (TEM) propagation, we can represent the aperture by an equivalent magnetic frill current given by M = 1n V (2.81) (n () P where Vo is the complex voltage present in the coaxial line at the feeding point rb is the radius of the coaxial feed's outer conductor ra is the radius of the coaxial feed's inner conductor p, 4 are cylindrical coordinates referenced to the feed's center Substituting from (2.81) into (2.80) yields (with ds = pdpdo) Vq - Vo l.I Hq(x = 0)dpdq (2.82) In () sfed where the cylindrical coordinates p and X are defined in Figure 2.9, and ojI is the ( component of the magnetic field evaluated in the plane of the aperture (x = 0).

30 2.5.2 Evaluation of the magnetic field at the aperture To evaluate the magnetic field component Hqq we will first determine the y and z components, and then perform a coordinate transformation to the cylindrical coordinates p and b. Determination of y and z components of Hq The magnetic field Hq anywhere inside the cavity is given by (2.21) -i_ 1 - - Hq =-V x A (2.83) 'Io where (2.27) A = I LJJ Jq.-Gd. (2.84) The y and z components are given by (2.83) 1 (A 9A qy x ) (2.85) Hi = (2.86) qZ P~o ay with At = o 'Sq (y')aq(x')G'xds' (287) A = Jo J ' (y')aq(x')Gizds'. (2.88) We now substitute from these equations back into (2.85) and (2.86) to obtain sq ( aZ ) (Y')aq(x)ds' (2.89) Hqy = OX q ozz Ox Hi = - J |.z (y)(y>(2)dsI* (2.90) These components are evaluated in Appendix G (for i = 1,2). Setting x = 0 in the resulting expressions yields: 00 00 H ()(x = ) = Ho CnqcY1)n sin kyy cos k()z (2.91) m=l n=O

31 oo 00 Hqz)(x = 0) = Ho CnqCzmn 1 cos kyy sin kl)z (2.92) m=l n=0 oo 00 H(-)(x =0) = Ho cnqc(y2n sin ky cos2)(z - c) (2.93) m=l n=0 00 00 H(2)(x = ) = Ho E E CnqC 2) cos kyy sin k2)(z -c) (2.94) qz " zm m=l n=O where k Ho= (2.95) ab sin KI (2 Cnq = (q cos kxq sin Rinsin R2n (2.96) C(1) pn tank 2)(h- c ) ([(kz())2 _ ( )] tanktZ)h dlmnd2mnd3n COS kzh -k(1)k(2)r tan k(2)(h - c) } sin(kyyo)o (k ) (2.97) c(1) - kntan(2)(h - c) (k )(k ) (2.98) 4m = sin(k-ylo)Jo (2.98) dlmnd3n cos k()h 2 C(2) -% tan k (1)h (2n - tan h {k(2)kl) tan )h dlmnd2mnd3n COS k(2)(h - c) - [(k2))2E - k (1 -:)] tan k(2) (h - c)} sin(kyYo)Jo ( ) (2.99) (2p Pntan k h kW\ cZmn = d P-tankcl) h sin(kyYo)Jo ( (2.100) dlmndn cos k 2)(h - c) and d3 = K2-k2 (2.101) R = 1 (K + kx)l (2.102) R2n = (K - k)l (2.103) K = WI/KOu0. (2.104) Coordinate transformation and evaluation of Hqo Equations (2.91)-(2.94) give the y and z components of the magnetic field anywhere inside the cavity. To find the q component we will perform the necessary coordinate transformation in two steps:

32 Table 2.1: COORDINATE TRANSFORMATION VARIABLES VARIABLE RELATIONS UNIT VECTOR RELATIONS x" =X x"=x y" = y- Yc = p cos b y" = y = cos i5- sin b z" = z - he = p cos q z" = z = sin $,5 + cos qb 1. move the origin from the corner of the cavity to the center of the coaxial feeding aperture. 2. perform a cartesian to cylindrical coordinate transformation. Referring to Figure 2.9, let us denote a new coordinate system by (x"1, y", z"1) whose origin is at the feed's center (x,y,z) = (0, Y, hc). The relationship between the new and old coordinates are outlined in Table 2.1. Using these relations, we will make the following substitutions in (2.91)-(2.94) to move the origin to the feed's center y - y"+Yc z - z"+ hc This yields: H)(x = 0) = H(l)(x = 0) = H(2)(x = 0) = H(2)(x = 0) = 00 00 Ho E CnqCl)n sin k(y" + Y0) cos k(1)(z" + hc) m=1 n=O Ho A > Cnq(lm cosky(y" +~ Yc)sink(l)(z" + hc) m=1 n=O Ho E E Cn 2) sin ky(y" + Y) cos k(2)(z- c") m=1 n=O Ho E E Cnqc(2)n cos ky(y" + Y) sin k2)(z" - c') m=1 n=O (2.105) (2.106) (2.107) (2.108)

33 where c" = c- h. (2.109) Now, let fti represent the projection of fl onto the plane of the aperture, then q q z= Hzy~P+ Hqz, = H' 00+ Hqpt (2.110) where Hi and Hi are the 3 and b components respectively. From Table 2.1. = coso^-sinob = cosq$1-sinqb Hence, z- sin OH' + cos H,. (2.111) If we substitute from (2.105)-(2.108) into the above, and let Yif -) Pcos4 Z - psin4 we obtain H(1)= 0)= 00 00 H0 [ sin~$ >j C cnqc~,sinky(pcosq$+ Y0)cosk l)(psinq$+ hc) m=1 n=O 00 00 ~cos ( >qC$,J1n cos ky(p cos Y0)sink41)(psinq$+hc)] (2.112) ryn=1 n=0J H(2) = 0)= 00 00 [m=1 si ymn k (P cos b + Yc) cos zk( P sin - m=1 n=O c c,00 00 ] 21 3 -I-Cos bZ ~CnqC 2)j cos k.(p cos q + Yc)sin k(2)(p sin c") (2.113) m=1 n=0

34 2.5.3 Formulation for numerical integration Consider now the surface integral of (2.82). One factor that complicates the integration is that it must be performed in two regions whose boundaries depend on the feed position as can be seen in Figure 2.9. In other words, the integration can be broken up as follows: In (,^ ed ) \ra = - H ( ) dpd> + H (2)dd (2.114) ln JL feed feed where SMled is the portion of the feed surface lying below the substrate (z" = psin 4 < -t) S(2)d is the portion of the feed surface lying above the substrate (z" = p sin > -t) To perform the above integration in the most general form we will make use of a 16 point Product Gauss formula approximation method [21]. Let us define a pair of dummy variables s and u and a function.F(s, u) such that O<(maa =2T r Pmax=rb. Smax==l /Umaxr=l I / Hzdpdq = / (s,u)ds (2.115) mJmin = J Pmin=ra J min=-1 (umin =-1 where the correspondence between (u, s) and (p, q) is given by the following relations = 2p - (Pmax + Pmin) 2p - (rb - ra) (2.116) Pmax - Pmin rb - ra 1 = - [u(rb - ra) + (rb + Ta)] (2.117) = 2~- (Ymax + Omin) = -(/a'- 1 (2.118) Omax - min = 7r( + 1) (2.119) The numerical integration can be carried out by generating a set of 16 pairs of points (uj, sj) and adding up their contributions according to rm max=2l j7Pmax=rb 16 J m Hm dpd ~ Z BjF(uj,sj) (2.120) nmin O a Pmtn =ra j=l where

35 ( uj, sj ) is the jth pair of integration points, and Bj is the weighting factor associated with the jth pair of integration points.F(sj, uj) is the transformed integrand found by performing a coordinate transformation on Hqo. Alternatively, once we have chosen the 16 points (sj,uj), we can find the corresponding values of p and b by (2.117) and (2.119) and obtain the same result. That is max =2r fPmax=rb. 16 = / Hdpdo= Bj (pj, j) (2.121) mn t0 Pmin =ra j=1 where pj - [uj(rb - ra) + (rb + ra)] j = 7r(sj + 1) and H(1)(pj, j) for - hc < pj sin j < -t H(Pj,3 ) (pj,) for -t < pj sinj < c" = c-h (2.122) error condition else. 2.6 MODIFICATIONS FOR ANALYSIS OF TWO PORT STRUCTURES Inthe preceeding sections, the theory has been advanced for computing th as ede impedance matrix and excitation vector associated with a one-port network, such as an open circuited transmission line (Figure 2.5). This section will present modifications necessary to extend the theory for treatment of two-port structures. Our approach for computing the network parameters of two-ports, requires simultaneous excitation of the strip conductors from both sides of the cavity. We will refer to this as "dual excitation".

36 J c - t t, ----- -------- --- ly. - A -........1. y I I — 4110 y C Figure 2.9: Geometry used for numerical integration to compute excitation vector. Note: the relative size of the feed is exaggerated for clarity

37 2.6.1 Application of reciprocity for dual excitation In section 2.2.1 an integral equation (2.3) was derived by applying the reciprocity theorem to the one-port network of Figure 2.5. In an analogous fashion, we will now apply the reciprocity theorem to the two- port network of Figure 2.10. In Figure 2.10, both magnetic current sources MO1 and Mp2 are coupled with the electric current source Js on the conducting strips to produce the total electric field Etot, and magnetic field itot inside the cavity. As before, we consider an independent test source Jq and associated fields Eq and Hq as shown in Figure 2.6. Applying reciprocity theorem between these two sets of sources yields | JEq -Hq - Msl - Hq.' M2) dv = J Jq. Ettdv = 0 (2.123') where the volume V is the interior of the cavity and M,1 = Ml1 (2.124) Ms2 = M020. (2.125) The right hand side of (2.123) vanishes as as described by (2.2). Reducing the remaining volume integrals of (2.123) to the appropriate surface integrals gives J, Eq(z = h) * JsdS = J Hq(x = 0) Mslds + f (x Hq( = a) Ms2ds. Sstrip Sfeedl Sfeed2 (2.126) Comparing the above to (2.3) it can be seen as a natural extension to the theory for the case of single excitation (one-port analysis). 2.6.2 Expansion of current and modified matrix equation The current J, is again expanded according to (2.4) NSECT Js = (y) E Ipa(z)x p=1

38 The only difference is that we now must consider the basis function for the x-dependence on the last subsection (i.e. closest to the right-most feed) as a special case. This is necessary since at each end of the cavity only a half sinusoidal basis function is required as illustrated in Figure 2.11. Hence, for the right-most subsection we let sin[K(x-a)] < <a NsinSECT ()=SECT-1 (2.1277) 0 else where the quantities K and lx are as defined in section 2.2.2, and NSECT represents the index for the right-most subsection. The rest of the basis function expansion is the same as given by (2.5)-(2.7). Substituting (2.4) into (2.126) yields NSECT i Eq (x = h). (y) ap (x) xdsIp = || HIq(x=0)*Msids IS feedl + Hq(x = a). M,2ds (2.128) eed2 which can be expressed as [Zqp] [p] = [Vql] + [Vq2] (2.129) where Sp is the surface area of the pth subsection [Zqp] is the impedance matrix, which has the dimensions of NSECT x NSECT [Ip] is the unknown current vector (NSECT x 1) [Vq1] is the excitation vector of the feed on the left (NSECT x 1) [Vq2] is the excitation vector of the feed on the right (NSECT x 1) Vql Hq(x = 0) Mslds (2.130) Sfeedl Vq2 = I Hq(x = a) * M2ds (2.131) feed2

39 We can now solve for the current vector by matrix inversion and multiplication according to [Ip] = [Zqp1 [[Vql] + [Vq2]]. (2.132) 2.6.3 Modifications to impedance matrix The elements of the impedance matrix for the case of dual excitation are given by the same integral equation as for the one port case, namely (2.10). The difference is in the integration over the last subsection (p = NSECT) where ap(x) is now given by (2.127). The integration for Eqx given by (2.65) is also modified for the last subsection (q = NSECT) in a similar way. The resulting modifications to the impedance matrix are very straightforward. This is because it can easily be shown that the surface integration over the last subsection (i.e. closest to the feed on the right) is equivalent to the integration over the first subsection (i.e. closest to the feed on the left). Hence, the elements of the impedance matrix are again given by (2.79) the only change being that Cq and Cp are as redefined below rather than by (2.69) and (2.77) 12 forp= 1 orp=NSECT (p = (2.133) 4 else 2 for q = 1 or q = NSECT Cq = (2.134) 4 else. (2.135) 2.6.4 Modifications to excitation vector We now consider the integrations of (2.130), and (2.131). By analogy with (2.81) we may express the two magnetic currents as follows: Vol Msl = - I 1 (2.136) (ra

40 V02 Ms2 +-I-a (2.137) ra where the positive sign in the second current source indicates that it is taken to be in the opposite sense (Figure 2.10) and Vol is the complex voltage in the coaxial line at the left-hand feed V02 is the complex voltage in the coaxial line at the right-hand feed rb is the radius of the coaxial feed's outer conductor ra is the radius of the coaxial feed's inner conductor p, (b are cylindrical coordinates referenced to the feed's center Substituting from (2.136) and (2.137) into (2.130) and (2.131) yields Vql -- V Hqq(x = O)dpdq (2.138) In (a) Sf eed Now, the integration required for Vql is identical to that carried out in section 2.5 for Vq (single excitation case). The computation of Vq2 is only slightly modified as we need to shift the origin to (x',y',z') = (a,Yc,hc) instead of to (0, Y,hC). After examining the x-dependence of Hq given in Appendix G it becomes obvious that we need only multiply the result for Vql by cos nTr to get the result for Vq2. That is Vol V = V cos nr Vql. (2.140) V02 Let Vqd represent the excitation vector elements for dual excitation such that Vqd = Vql + Vq2 Then Vqd (- V-o sn7r) Vq (2.141) (2.141

41 where Vq represents the excitation vector of section 2.5 with Vo set equal to unity. The next step is to find dual excitation vector for even and odd mode excitations. As will be discussed in section 2.7, we will find the two-port scattering parameters for a network by exciting it with even and odd modes. For the even mode excitation, we let Vol = V02 = 1. In this case from (2.141) we have Vqe = Vq (1-cosn'r) (0 for n even (2.142]) 2Vq for n odd. For the odd mode excitation, we let Vol = -V02 = 1. Now, using (2.141) Vqo = Vl (1 + cosnr) = (2.143) f 2V4 for n even (2.143) 0 for n odd where Vqe represents the elements of the even excitation vector Vqo represents the elements of the odd excitation vector. Using these two excitations, we can compute both the even and odd mode current distributions using the following matrix equations: [Ipe] = [Zqp] [Vqe] (2.144) [Ipo] = [Zqp-1 [Vqo] (2.145) where Ipe represents the elements of the current vector for even excitation Ipo represents the elements of the current vector for odd excitation.

42 Y STRIP CONDUCTOR Figure 2.10: Total fields inside cavity Etot, IItot produced by magnetic currents M,1, M,2, and electric current J,.

43 y b,. ~, \! \# %s %, % % \ X x,,... x..., x,X,X, X 1 2 3 P NS-3 NS-2 NS-1 NS X x = (p-l)l a P x Figure 2.11: Strip geometry for basis function expansion with dual excitation.

CHAPTER III PRELIMINARY RESULTS The theoretical methodology described in Chapter 2 has been used as the basis for a computer algorithm which was implemented in a fortran program. This program has, so far, been used to obtain numerical results for the open end and series gap discontinuities. This chapter presents some of these results with comparisons to other numerical solutions and preliminary measured results obtained here at the University of Michigan. 3.1 RESULTS FOR OPEN END DISCONTINUITY An open end discontinuity in shielded microstrip can be represented as an effective length extension Lefjf, a shunt capacitance Cop, or by the associated reflection coefficient (Fop) as shown in Figure 3.1. The plot of Figure 3.2 contains the computed effective length for an open end on an alumina substrate (~r = 9.6, W/H = 1), and Figure 3.3 shows the same for a quartz substrate (Cr = 3.82, W/H = 1.57). Our numerical results are shown compared to results obtained by Jansen et. al. [22] and Itoh [23]. The geometrical parameters of the cavity cross section used in the analysis for each of the above cases are (referring to Figure 2.8): b = c =.275", and H =.025". Also, in both cases the distance from the end of the microstrip line to the end of the cavity was fixed at.125". For both the alumina and quartz cases, our results are seen to fall between the other two numerical results. From this we conclude that our results are at least reasonable. 44

45 A comparison with measured results is presented in Figure 3.4, which shows numerical and experimental data for r~p of an open circuit on an alumina (c, = 9.7'1 WIH - 1 ) substrate. Here, our numerical results are compared with numerical data, from Super Compact ' [24] and preliminary measured results. The geometrical parameters for the cavity cross section in this case are: b = c =.250", and H =.025". The measured results are seen to follow our results most closely. 3.2 RESULTS FOR SERIES GAP DISCONTINUITY A series gap in shielded microstrip may be represented as a pi arrangement of capacitances, or alternatively by a set of scattering parameters as shown in Figure 3.5. Numerical results have been obtained for series gaps on an alumina substrate (Cr =9.17) with three different gap spacings (G) 5 mil (i.e..005"), 9 mil, and 15 mil. Results for the scattering parameters of these gaps are shown plotted in Figures 3.6 through 3.13. The geometrical parameters of the cavity cross section are: b = c =.250", and H =.025". For comparison with our numerical results, results obtained using Super Compact, and Touchstone [25] are also shown plotted along with preliminary measured data. W0ith one exception, the measured data best follows our numerical results. The one exception was for the magnitude of S21 of the 5 mil gap. In contrast, the Touchstone analysis had the least agreement with the measurements. This may in part be due to the fact that Touchstone does not allow for taking either the side walls or the shielding cover into account. In summary, our numerical results show reasonable agreement with other numerical solutions and, more importantly, demonstrate very good agreement with the measured It should be noted that Super Compact allows for specifying cover height, but does not take the effect of the side walls into account. However, for the given geometry, the effect of the side walls is minimal.

46 data obtained so far. We are in the process of obtaining another set of improved measurements as the present measured data shown here is not as smooth, as a function of frequency, as it should be.

47 NIL H eff a o, —o I ---o a r op OR _ c op a r op Figure 3.1: Representation for microstrip open end discontinuity

48 END EFFECT FOR ALUMINA LINE (ER=9.6) 0.45 - 0.40 - 0.35 - 0.30 - 0.25 - L, LL L,1 -J -o- JANSEN - - OUR RESULTS -0- ITOH -qp 0.20 -n i -'.,._ _. I., V. vI I a I I I ' I I. I 0 4 8 12 16 20 24 FREQ (GHZ) Figure 3.2: Effective length extension of a microstrip open circuit discontinuity on an alumina substrate (E = 9.6), as compared to other numerical results

49 END EFFECT FOR QUARTZ LINE (ER=3.82) LL UJ 0.50 - 0.45 -0.40 -0.35 0.30 - 0.25 -0.20 -0.15 Ak —.n * — JANSEN -.- OUR RESULTS -- ITOH w I w I w a w I w I -1 I 0 4 0 4 8 12 16 20 24 FREQ (GHZ) Figure 3.3: Effective length extension of a microstrip open circuit discontinuity on a quartz substrate (Er = 3.82), as compared to other numerical results.

50 RESULTS FOR ANGLE OF S11 OF OPEN CKT. 0 - -5 -10 U T(I) v -15 -a- OUR RESULTS -- S.COMP. -a- MEAS. 7/21 -20 -25 -30 0 4 8 12 16 20 FREQ(GHZ) Figure 3.4: Angle of S11 of an open circuit as compared to measurements and Super Conmpact results.

51 a I Ii I C I I g I 6i -- I —6 I P -I p I b OR I I I I I I a b a b Figure 3.5: Representation for microstrip series gap discontinuity.

52 RESULTS FOR GAP G1 -5 m v> -15 - -.- OUR RESULTS -*- SUPER COMPACT -a- TOUCHSTONE -o- MEAS. -25 - -35"f.,..., * A a 0 4 8 12 16 20 FREQ(GHZ) Figure 3.6: Magnitude of S21 for series gap G1 (G = 5 mil) as compared to measurements, Super Compact, and Touchstone.

53 RESULTS FOR SERIES GAP G1 90 - 80 - w 0 %-. Cl) V 70 - -o- OUR RESULTS -*- TOUCHSTONE -a- SUPER COMPACT -o- MEAS B. 60 50 -.........I 0 4 8 12 16 20 FREQ(GHZ) Figure 3.7: Angle of S21 for series gap G1 (G = 5 mil) as compared to measurements, Super Compact, and Touchstone.

54 RESULTS FOR SERIES GAP G1 - -10 - w 0 (1 v -20 - -0- OUR RESULTS -- SUPER COMPACT -a- TOUCHSTONE -o- MEAS. B -30 - -I, _.. — TV 0 I I 8 4 8 12 16 20 20 FREQ(GHZ) Figure 3.8: Angle of Sll for series gap G1 (G = 5 mil) as compared to measurements, Super Compact, and Touchstone.

55 RESULTS FOR SERIES GAP G2 -5 -15 -25 m Trn -o- OUR RESULTS -.- TOUCHSTONE -a- SUPER COMPACT -~- MEAS. 0 4 8 12 16 20 FREQ(GHZ) Figure 3.9: Magnitude of S21 for series gap G2 (G = 9 mil) as compared to measurements, Super Compact, and Touchstone.

56 RESULTS FOR SERIES GAP G2 90 -80 - w 0 (n eJ Cl) V 70 - -0 --a-0 - OUR RESULTS SUPER COMPACT TOUCHSTONE MEAS B 60 - 50 i 0 I 4 8 12 16 I2 20 FREQ(GHZ) Figure 3.10: Angle of S21 for series gap G2 (G = 9 mil) as compared to measurements, Super Compact, and Touchstone.

57 RESULTS FOR SERIES GAP G2 0 --10 - w 1r a Ct) v -20 -:- OUR RESULTS *- SUPER COMPACT -a- TOUCHSTONE -- MEAS. B -30 --40-. -......I... 0 I 4 8 12 16 I 20 FREQ(GHZ) Figure 3.11: Angle of Sll for series gap G2 (G = 9 mil) as compared to measurements, Super Compact, and Touchstone.

58 RESULTS FOR SERIES GAP G3 -10 --20 0-1 m - 1-' cs n -a- OUR RESULTS -+- TOUCHSTONE -a- SUPER COMPACT -0- MEAS. -30 - -40 - I. I... 0 -. I 1 4 8 12 16 20 20 FREQ(GHZ) Figure 3.12: Magnitude of S21 for series gap G3 (G = 15 mil) as compared to measurements, Super Compact, and Touchstone.

59 RESULTS FOR SERIES GAP G3 90 - 80 - w 0 U) vv o- OUR RESULTS +- TOUCHSTONE -a- SUPER COMPACT *o- MEAS. B 70 - 60 -. * i * *. * _ i. 0 I I ' I, I * I 4 8 12 16 20 FREQ(GHZ) Figure 3.13: Angle of S21 for series gap G3 (G = 15 mil) as compared to measurements, Super Compact, and Touchstone.

60 RESULTS FOR SERIES GAP G3 0 - w I) 0 T v) v -c- OUR RESULTS -0- TOUCHSTONE -a- SUPER COMPACT -o- MEAS. B -20 - -30 m w - A I a 9 0 4 8 12 16 2 20 FREQ(GHZ) Figure 3.14: Angle of Sll for series gap G3 (G = 15 mil) as compared to measurements, Super Compact, and Touchstone.

APPENDICES 61

62 APPENDIX A REVIEW OF THE METHOD OF MOMENTS The general steps involved for in the computation of surface currents using the method of moments can be summarized as follows: 1. Formulate an integral equation for the electric or magnetic field in terms of the surface current density J, on the conductors. It is generally possible to put this equation in the form Lop (Jis) =g (A.1) where Lop is an integral operator, and g is a vector function of either the electric field E or magnetic field H associated with Js. 2. Expand Js into a series of basis functions Jp so that NSECT s= E IpJp (A.2) p=1 where the Ip's are complex coefficients and NSECT is the number of sections the conductor is divided into. 3. Determine a suitable inner product and define a set of test (or weighting) functions Wq. The result may be expressed as NSECT E Ip(WqLWop(Jp)}) = (Wqg) (A.3) p=1

63 where the inner product is defined as (a, b) =s Ji dS In Galerkin's method, the weighting functions are taken to be test currents Jq which are identical in form to the basis functions Jp. 4. Solve the inner product equation (.3) and form a matrix equation of the form [Zqp] [Ip] = [V] (A.4) where [Zqp] is termed the impedance matrix, and [Vq] is called the excitation vector. 5. Solve for the current coefficient vector by matrix inversion and multiplication according to [Ip] = [zqp]-' [Vq]. (A.5)

64 APPENDIX B DERIVATION OF INTEGRAL EQUATION FOR ELECTRIC FIELD Starting with Maxwell's equations V x E = -jwpH (B.1) V x H = jwCE+J (B.2) ' J = -jwp (B.3) V7(Ep) = p (B.4) V (H) = 0 (B.5) we define A such that H =-V x A. (B.6) Substituting (B.6) into (B.1) yields V x (E + jwA) = O. (B.7) Since V x Vq = 0 for any arbitrary vector function q, we let E + jwA =-Ve. (B.8) Making use of (B.6) and (B.8) in (B.2) yields 1 -V + -V x A = -jw(jwA + q) + J (B.9) or - v2A + V(V * A) = w2IEA - jwIEq,~ + pJ. (B.10)

65 We use the Lorentz condition (V * A) -jWiEVb (B.1I) in (B.10) to obtain v2 A + k2A = — J (B.12) where k2 = W2te. From (B.9) and (B.11) the electric field may be expressed as E = -jwA+ V(V A) jWPE 1 — = -j(1 + -2VV-)A (B.13) We now define a dyadic Green's function G to be a solution of V2 G + k2G = -I6(r - ). (B.14) After some manipulation, the vector dyadic Green's theorem [26] can be put in the following form (an appendix will be added later to give the details of this derivation) I (V2A G -A 2G)dV= JJ {({ x A) vxG +(n' x v x A) G + [A(, G)] -h'.[( A)G] dS (B.15) where, for our shielded microstrip cavity problem, the volume V is the interior of the cavity, and S, and S, are the surface of the cavity walls, and the surface of a small volume enclosing the source region respectively. We will require that the components of A and G satisfy the same boundary conditions on Sw and S,. In this case, it can be shown that the entire surface integral on the right hand side of (B.15) vanishes. If we now substitute from (B.12) and (B.14) for V2A and V2G we obtain I II(V 2A G-AV2G)dV = II v {(-j- k A)*G

66 -A. I6(r - r')k2 * G} dV = -/ l l J J GdV + A(r) = 0. (B.16) Hence, A= | J J - GdV (B.17) Finally, substituting from (B.17) into (B.13) produces the following integral equation for the electric field E = -W(1+ ).) J - GdV = - J [(1+ VV-)(G)] (B.18) where (G)T represents the transpose of G.

67 APPENDIX C EIGENFUNCTION SOLUTION FOR GREEN'S FUNCTION The boundary conditions on the cavity walls are applied here in order to derive the functional form of the Green's function. First, the general solution to the homogeneous differential equations for the components of the Green's function is presented. Then, the boundary conditions on the walls are used to arrive at an eigenfunction expansion for each of the Green's function components. The particular solution for the Green's function is found by integrating the inhomogeneous differential equation across the source region. GENERAL SOLUTION TO HOMOGENEOUS D.E.'s FOR GREEN'S FUNCTION Consider the homogeneous forms of equations (2.23) and (2.25) Ai + 2A = 0 (C..) VG + kG = 0 (C.2) where i = 1, 2 denotes that these equations hold in each region respectively. With G given by (2.30), it can readily be shown that (C.2) implies v2GX + 2Gtl = 0 (C.3) 2 0 V G' + kG' = 0. (C.4) We now apply the method of separation of variables. Let We now apply the method of separation of variables. Let Gt = X(xW)Yx(Y)Z (z) (C (C.5s)

68 Gz = Xi(x)Yz(y)Z(z). (C.6) Substituting from (C.5) into (C.4) produces the following differential equations for Gxx: d2XT+ X = 0 (C.7) dx2 X d2 2 dy + kY = 0 (C.8) d2 2 - r k ZxI = O. (C.9) dz2+k Z Similarly, substituting from (C.6) into (C.4) yields for Gxz d2Xz + k XI = (C.io) 2Y+ki2yt = 0 (C.ll) d2Z +kziZi = 0 (C.12) dz2 where k- = k + k + k2 (C.13) The well known general solution of each of the above differential equations may be put in the form. / = Ai cos ktt + A2 sin k't (C.14) where t = x, y, or z; ' = X',YY, orZs (where s = x or z) and kt is complex in general. The next step is to consider the boundary conditions on the cavity walls. APPLICATION OF BOUNDARY CONDITIONS ON THE CAVITY WALLS In the application of Green's theorem (Appendix B) it is required that the components of Ai and G must satisfy the same boundary conditions. Hence, the following correspondences apply At + GX As A ) Gt x x x I z xC (C.15)

69 That is, A' and G'. must satisfy the same boundary conditions on the waveguide walls and on the substrate/air interface and, hence, must have the same functional form in terms of spatial variation. The same holds true for A' and G/'. In order to establish what conditions A (and correspondingly G ) must satisfy at the walls, we need first to establish more explicit relations between Ai and Ei. From (B.13) Et = -jwA +. V(V * A). (C.16) Now, from (2.26) Az = Ax + Az (C.17) hence, +A (C.18) Ax az and V(V. Ai)= a tA+ )a ( ax + aA' Using (C.17) and (C.19) in (C.16) yields the following expressions for the electric field = - Ax + - z)1 y =- ( a a a+ ) (C.21) z k 9 ax az (C. E: -jw A' + z( +.z) (C.22) [ 10 ~Z( OAx i We now consider the boundary conditions at each of the cavity walls.

70 Boundary conditions at x = 0, a Since the cavity walls are assumed to be perfectly conducting, the tangential components of the electric field must vanish at the walls. Therefore, for the walls at x = 0 and a E(= 0,a) = 0 (C.23) E(x = 0,a) = 0. (C.24) In view of these two equations, (C.20) and (C.22) lead to 0 a) = [ + 9a oa 0. (C.25) Ez(x= 0,a) = -j Ai + z( + ) I (C.26) Now, (C.25) is satisfied if the following condition is imposed: f)Ai /OAi ( + )) Ix=O,a = (C.27) in which case (C.26) leads to A =(x = O,a)= 0. (C.28) If we use the correspondences of (C.15) and the representation of (C.6) we may deduce that Xz(x = 0,a)=0. (C.29) If (C.28) is placed into (C.27) it is seen that OA' ax Ix=O,a =0. (C.30) then (C.15) and the representation of (C.5) leads us to conclude that X Ix=Oa = 0 (C.31) Ox

71 The boundary conditions of (C.29) and (C.31) can be satisfied by choosing the following eigenfunction solutions for the x-dependence: Xt = cos k x Xt = sin kx. z xx (C.32) (C.33) for i = 1,2, where k(l) - k(2) k - n r 4X - a - = for n = 0, 1, 2,... (C.34) Boundary conditions at y = 0, b The tangential component of the electric field must vanish on the walls y = 0 and b; hence, Ex(y=O,b) = 0 E'(y=O,b) = 0. (C.35) (C.36) From (C.20) and (C.21) Ei(y = O,b) Et(y = O,b)., 1 9,QA~ 9aa = - A + a + 9z [ 1OA i aAti If we impose (C.27) at y = 0 and b, we obtain 9A2 aAt ( + ) ly=O,b = 0, and when this is combined with (C.37) and (C.38) the result is At(y=O,b) = 0 Az(y=O,b) = 0 (C.37) (C.38) (C.39) (C.40) (C.41)

72 from which it follows that the eigenfunction solution for the y-dependence is given by Y/ = sin ky (C.42) Yz = sin ky (C.43) (for i = 1,2), where k() = k = ky -= mr for m= 1,2,3,... * (C.44) Note that it is easily shown that m = 0 leads to a trivial solution for the y-dependence of both components. Boundary conditions at z = 0, c Similarly at the walls z = 0 and c we have E(z = O,c) = 0 (C.45) =(z 0, c) = 0. (C.46) Making use of (C.20) and (C.21) yields E'(z =,c) = -jw [A + + z ]) Iz=o,c = 0 (C.47) -k ay ( Ox + z) Iz=O, -0. (z=-Oc)= [k A Z)] =o= (Cz48) Again we impose (C.27) this time at z = 0 and c ( + z ) Iz=o = 0 (C.49) which results in At(z = 0,c) = 0. (C.50) If we substitute (C.50) back into (C.49) we deduce Oz lz=O,c = 0 (C.51)

73 Finally, from the correspondence of (C.15), the separation of variables representation of (C.5) and (C.G6), and the conditions imposed by (C.50) and (C.51) the eigenfunction solution for the z-dependence can be written as Z1) sin k(1)z (C.52) Z(1) cos k(1)z (C.53) Z(2) = sin k(2)(z-c) (C.54) Z(2) = cosk(2)(z-C) (C.55) where — from (C.13), (C.34), and (C.44)- k1) and k(2) are given explicitly by ) = )2 M-( )2 (C.56) z 1 a b (2) n7r mr )2 k )2 - )2 (C.57) z a b and kl = wvJT1 (C.58) ko = wv/Jt. (C.59) REPRESENTATION OF GREEN'S FUNCTION BY EIGENFUNCTION SERIES We will now combine the results obtained above, so that the Green's function may be written in series expansion form. Substituting from (C.32), (C.42),(C.52), and (C.54) into (C.5) and taking the summation over all the possible modes, results in the following for Gz: 00 00 G(1) = m A(1) cos kxsin kyysik(1)z (C.60) G(2) B 2)x sinks ky cos k(2)(z - c). (C.61) m=l n=O

74 Similarly, if we substitute from (C.33),(C.43),(C.53), and (C.55) into (C.6) we obtain the following for Gz 00 00 G() = E B) sin kxx sin kyy cosk()z (C.62) m=l n=O oo oo G -() = E A() cos kx sinkyysin k(2)(z - c). (C.63) m=l n=O The complex coefficients A and B (i = 1,2) are determined in Appendix D by the application of boundary conditions at the substrate/air interface (z = h).

75 APPENDIX D BOUNDARY CONDITIONS AT SUBSTRATE/AIR INTERFACE The complex coefficients A, and mn (for i = 1,2) for the Green's function components given by (2.31)-(2.34) are found here by applying boundary conditions at the substrate/air interface (z = h). Figure D.1 shows a cross section of the cavity in the x - z plane. The application of boundary conditions at the interface is made difficult by the presence of the infinitesimal current source on the substrate surface. We will avoid this difficulty by first solving a similar problem with the current source raised a distance Ah above the substrate. After solving for the boundary conditions at z = h and z = h + Ah, the equations required to determine the coefficients An, and Bn are obtained by letting Ah go to zero. FORMULATION From the consideration of the boundary conditions on the waveguide walls the components of the Green's function are given as 00 00 G() = E A(ml cos kxx sin ky sin k(l)z (D.1) m=l n=O G() = B(1) sin kx sin ky cos k)z (D.2) m=l n=0 G(2 = E E Am2cos kx sin ky sin k2)(z -c) (D.3) m=l n=0O 00 00 G(2z) = E Bm2) sin kx sin kyy cos k(2)(z - c) (D.4) m=l n=O

76 z=C - (2) J,,......... -v =...............:.........:.......g.:~::...........:' ~......~~~~~~~~~~~~~ ~~~~~~,.,X............................ x,.....,. -..... —.! x=a a) Actual position of current source z=c z=h' z=h z (2) AL Ah |, ~.~'' l.>..^................ - - - - - -...............................:.......... 1 '.'.... '''.,''' ',' ''. x x=a b) Current source raised above interface Figure D.1: The current source is raised above the substrate/air interface to apply boundary conditions.

77 For region 3, the Green's function must satisfy the same differential equation (2.25) as in the other two regions. We will use the form of the general solution given by G0 = (00 (3)^ ^ (3) 1 G(3) = E cos k~x sin kyy [A() ekz + B(3) -kzJ (D.5) m=1 n=O G(3) sin kx sin ky [C(3(3) ejkz + D(3) ejk (D6) > z sy mn G(~ = ~ sinxxsinkyyC L. + D -Jm (D.6) m=1 n=O where, kX = nr/a (D.7) ky = mwr/b (D.8) kl) = /k2 -k -k2 (D.9) k(2)= 2/k0- k- k (D.10) k() = k(2) (D.11) k = V/J-/ (D.12) ko = wV.oo (D.13) Recall, the electric field solution in terms of the vector potential components (C.20)-(C.22') E} = -,, 4 + k (v.A ) = -j Ai + 2 ( + )] (D.14) -j, a aAx aAt Ey = k ( rx + a z (D.15) tk Z)y O Ez = -j [AtA + ( (D.16) These equations hold in each region respectively (i.e. for i = 1,2,3). The solution for the magnetic field can be written using (2.21) and (2.26) as follows: 1- -i 1 H =-V x A -[ x (A/x + Az\)]. (D.17) i0o I/o

Separating this into x, y, and 78 z components gives 1o Oy 1 (OA 9A\ io oy, z fo 9y (D.18) (D.19) (D.20) APPLICATION OF BOUNDARY CONDITIONS Recall from Appendix C (C.15) that the following direct correspondences can be made as far as boundary conditions are concerned A% +-* GI. As *-+ GI x xx I z xz, (D.21) Boundary conditions at z = h At z = h, the following boundary conditions apply: E(1) E(1) H(1) y H(1) clE(1) = E(3) = E(3) = H(3)y = H(3) "y (D.22) (D.23) (D.24) (D.25) (D.26) (D.27) =- ) (~i3) ==fr, J1) -HZ3 zz We will make use of (D.23)-(D.26) to formulate four of the eight equations needed to solve for the complex coefficients in (D.1)-(D.6). We start with (D.26), then substitute from

79 (D.20), and recall the correspondences of (D.21) to obtain OG(1) 9G_ ~xx IIz=h z=h (D.28) ay -z When (D.1) is placed in (D.5) and orthogonality is applied, the following result is obtained A(') sin kl)h = A(3) e3(z)h + B()e-jk()h (D.29) where k 2) has been substituted for k3) in accordance with (D.11). Next, from (D.24),(D.18), and (D.21) aGy- ay G (D.30) Oy Oy From (D.2) and (D.6) B(1) cos k)h = C(3)n ej)h + D(3) -k)h. (D.31) The combination of (D.25), (D.19) and (D.21) yields /9Gi) G2z) Ih= OGiO2 OG2 t 9 ) - =h (D.32) z x = -z xz Making substitutions from (D.1),(D.2),(D.5) and (D.6) in this expression leads to [(ml)k(l ' — Blk ] ~Osz(2 l (2)'[ [A mnw Bs\) k] cos k = jk(1 A )()ek - Be3e -k ( [ k() h + D e(3)-jk ] (D33) We now substitute for B() cos k)h from (D.31) to reduce the above to A(l) k(l) cos kh =jk 2)[A(3) e h - B )e- (D.34) Now consider (D.23). From (D.15) and (D.21) 1 9 9G0 9G(1) 9 G(3) 9G 9G(3) y(x + z )I z=h ( - + O z) Ih (D.35) c Oy axy Ox Substituting from (D.1),(D.3),(D.5), and (D.6), produces, after simplification, (A () + B k(1 ) () sin k()h (A(3) 2)h + B(3) -j(2)h mn mn mn -jk(2)((3) ek h - (3) e- ). (D.36)

80 Next, we use the expression from (D.29) to replace A(3)ejk) + B(3)e-jk. This yields A(l) (1- c6)k sin k(1)h + BMk(1)) sin k(1)h = -jk2C [C) ejk - )e-]. (D.37) Equations (D29) (31 (D34 andtions.31) (D.34) and (D.37) represent 4 of the 8 equations we need. Boundary conditions at z = h' We now proceed to the boundary conditions at z- h' (see Figure D.1) we have E2) E(2) y E(2) - E(3) H(2) Z X (11(2) - fi(3) H(2) -(H(2) -H (3)) = E(3) = E(3) y = O's = H(3) - z (D.38) (D.39) (D.40) (D.41) = Ji = — = H(3) = Js. (D.42) (D.43) Of the above, we will use (D.38), (D.41), and (D.42) to derive three more equations for the complex coefficients. We start with (D.42) and use (D.18) and (D.21) to obtain GW Iz=h' = z- I\z=h' (D.44) which yields after substituting from (D.4) and (D.6) B(2) cos k2)(h' - c) = C(3)ek ' + D3 )e-n '. (D.45) -- +D~ -j (.45 Next, consider the boundary condition of (D.41). This leads to Iy |z=h'- = |z=h' 9y 49y (D.46)

81 after the use of (D.20) and (D.21). Substitution from (D.3) and (D.5) yields A(2) sin k(2)(h' - c) = A) e jk)h' + B) e- (D.47) This equation, when combined with (D.3) and (D.5), shows that G(2(z = h') = G(3(z = h'). (D.48) With the above equality, we can substitute from (D.14) into the boundary condition of (D.38), and make use of (D.21) to produce 9 a9G(2 9G(2) 0a a9G3 aO(9G ( + XZ) ~ + XZ) (D.49) Substitution in the above from (D.3), (D.4),(D.5), and (D.6) yields [A(2) k + B(2) kk(2)] sin k(2)(h'- c)= mn mn xkn 2)h k [A(3) e ' + B (3) e-(2) h x m mn mnh -jk( [(3) e )- D(3) e-. (D.50) At this point we have 7 independent equations -(D.29), (D.31),(D.34), (DI).37),(D.45),(D.47), and (D.51) — and we have 8 unknown complex coefficients. The other required equation is obtained by integrating the differential equation of (2.25) across the boundary at z = h'. Integration of the differential equation for G across the source region at z=h' From (2.25) we have From (2.25) we have V2 - + Ik = ( - (D.52 )

82 we then substitute from (2.30) to yield (V2 + k) (Gxx + Gxz) -6(r- r')xx (D.53) Hence, (V2 + k?)Gc = -6(- ') = -6(x - ')(y - y')(z - z'). (D.54) We now integrate both sides of this equation over a line passing through the source point r', and then take the limit as the length of this line vanishes lim h (V2 + k2)Gixdz = -b(x - x')(y - y'). (D.55) a'-O J h'-a This may be written as lim [( ++a 2 ) G+ Gdz+I Gzz + = -6(x-' )(y-y') (D.56) If we make use of (D.3) and (D.5), we can show that the first integral vanishes as follows: rh'+ [ ' G dz+ li G dz lim G G( dz + G(2dz -+O h'-t -a -+O h'- ox h "00 00 1 ~2 2 c>C oo \1 + Lm=ln=O k (i A (2 cos x sin kyy cos k(2)(z - c)] z+ m=:ln=0 = 0 (D.57) (since each of the limits on the right hand side vanish individually.) Therefore, (D.56) can be reduced to -yh'+ a lim n 2G dz = -(x - X')6(y - y') (D.58) From which we obtain a-*O z =o ^ | =-^ -x) - 0'^ (D.59) or lG m + - OG-( = -6(x - x')b(y - y'). (D.6 0) - z h 9 z

83 Substitution in the above from (D.3) and (D.5), and simplifying (with the use orthogonality principles) yields ab (2) k(2) cos k(2)(h - c)(3) mn -B() e-k( )z)} -cos kzx' sin y' (D.61) where 2 for n = 0 fr = n (D.62) 4 forn 0 The above represents the final equation needed to evaluate the complex coefficients. EVALUATION OF THE COMPLEX COEFFICIENTS OF THE GREEN'S FUNCTION To evaluate the complex coefficients, we will make use of the equations derived above involving A't Bn (i = 1, 3), and C(3 and D(3). Since we are only interested in AI(, B(), A(2) and B(2) these will be evaluated by eliminating the other complex coefficients along the way. Now, recall that h' = h + Ah. If Ah -+ 0 then h' -* h in equations (I).45), (D.47), (D.51), and (D.61). Starting with (D.45) with h' -+ h we can substitute from (D.31) to obtain B(m) cos k!)h = B2) cos k2)(h -c). (D.63) Similarly, (D.29) and (D.47) yield A() sin k2)(h - c) = A() sin k1)h. (D.64) From (D.37) and (D.51) we get Bk) sin k2)(h - c) = -- A( (- - I. k sin k1) + sin k(1)h. (D.65)

84 From (D.34) and (D.61) b [A(k(2) cos k2)(h' - c) - A()k(1) cos k()h] = - cos kx' sin y. (D.66) kTn The combination of (D.64) and (D.66) yields ab A(') sin _ _)h - [ (mn ) k (2) cos k(2)(h - c) - A(1) (1)cos (1)h =-cos kxx' sin kyy. (D.67) (~n sin k(2)(h - c) Solving for A(') A() = -Pn cos kx' sin kyy' tan k2)(h - c) (D.68) mn (D.68) abdlmn cos ) h where dimn = k(2) tan k)h - kAl) tan k(2)(h - c) (D.69) and Wn is given by (D.62) A(2) is found by substitution from (D.68) into (D.64) (2) _ -(p, cos kxx' sin kyy'tan k)h mn - (D.70) abdlmn cos k(2)(h - c) Next, we combine (D.63) and (D.65) to get B() cosk)h sin k)(h-c) _ 1 [A) (1 cos k(2) (h - c) k A() - + * sin k(1)h (D.71) 6r which, by substituting for A(1) from (D.68), can be rearranged to find B(1) as B(1 - e*)kx cos kx' sin kyy' tan k)h tan k(2)(h - c) mn (D.72) abdmnd 2mn cos z)h where d2mn = k(2)r tan k2)(h - c) - k) tan k()h. (D.73) Finally, place (D.72) in (D.63), and B(2) can be expressed as: B) - c)kx cos k xsin kyy' tan k (1)h tan k (2) (h - c) 2mn ~ r (D.74) dimnd2mn cos kz 2)(h- c) We now have derived explicit relations for the desired complex coefficients A(1), A(2), B(1) B(2) It can be shown that the same relations can be obtained by moving the current source of Figure D.1 into the dielectric region and then bringing it back to the substrate surface.

85 APPENDIX E EVALUATION OF THE MODIFIED DYADIC GREENS FUNCTION The modified dyadic Greens function was defined in (2.53) as I= -JW/oLo 1+ T;V) (G i)T1 L \ i< J (E.1) From (2.30) G = G'xx i+ GXzX:. The dyadic transpose is (G') = GXx + GXZZx -x + (E.2) The divergence of E.2) yields v.(Gt)T(G Oz Forming the gradient of this V * (G )T -x Ox +z x 9 (OGQ, +OG z a y a x az x + a Gtx + Gz + + XXz (Gi+GZ i) ^: az G zOX aZ (E.3) We can now substitute from (E.2) and (E.3) into (E.1) to yield = -o [' + i (Gxa x + Oz ) r jP =x + ko Ox O x + jj

86 ( k 9Gy x +- az Y + [G' z-k2 X+ Xz (EA4) Hence, the xx component of the modified Green's function is given by =S G'. ~ [Xx + T2 X+.z (E.,5) From (2.31) and (2.32) G22-13 A$Q, cos kx sin ky y sinkk') z (E.6) m=1 n=O - Z: Z kxA$7M sin kxx sin kyy sin k$O)z m=1 n=O X - AMZB$ 2coinkxx sinn k.y csin ')z (E.8) m=1 n=O GM Z snkxsin kyy csi k~y sin8 - n m=1 n=O &2G0) _ 00 d - - E S: kO~)BM, csi kxx sin kyy sin kO1z. E9 m=1 n=O 002 __ 1 ik'k2- jwBuo15 5ocosxkxxssinkk y sin k (1z - -(E.9) m=1 noM[A n=Repacetheuio exresson for)(9 ando fE5 romlt (24)ad(.3)nh euti -jwp 0000 00 X X - Eju 5 5cos k~xsin k~ysink ')zAM k kB m=1 n=O 1 FMKabP Cos k xx sin kyy'n s(hn c)z K_ sk__sn__yta k(2 ' sin k- tactnkf k2E.O -d~' Kb1mn-o c~k( co din1n o which can be rearranged in the following form: - WJo 00 k1 m1 n=O ~ dlmn,,d2mn,,cosk 1

87 * [ab cos kx sin kyy sin k(1)z cos kxt' sin kyy' tan 2)(h - c)] F / k2\ (1 \ kx\ ~ k(2)e (1 - k tan k(2)(h- c) - k(1) tan k' )h (E.11) where the expression for d2mn from (D.73) has been used in combining the terms inside the brackets of (E.10). Evaluation of (E.11) at z = h gives r(l)( = h) J= j OE E {( ml nO dlmnd2mn m=1 n=O * [ab cos kx sin kyy cos kxx' sin kyy'tank(1)htank(2)(h- c)] * k(2)* 1-ki2 tan k2)(h-c) -k(1) (i 2 tan 1)hI. (E.12) Proceeding in a similar fashion for region 2, we have from (2.33) and (2.34) 00 00 G(2) = A(2) cos kx sin kyy sin k(2)(Z ) (E.13) m=1 n=O m=l n=O Q2G(2) 00 00 a2OG = - V XA(2) kcoskXxsinky -c) (E.14) m=l n=O /)2o(2) oo oo - ZE kA(2)B2 sin kx sin kyx sin k sin k2)(z - c) (9 X mn m=l n=O 2 (2) oo0 00 G = - 2)2 cos k x sin kyy sin k(2)(z - c ) (E.15) m=1 n=0 52 5 cos kxx sin k y sin k(2)(z-c) [A( ) (12) - c) GF(2) 2) sin kxx sin kyOy cos k[A ( - ( E.1]5) m=l n=O 2(2) oo oo z - E E k(2)B(2) sin sin kx s sisin k(2)( Z -c ) m=l n=0 Substitution from (E.13)-(E.16) into (E.5) yields () = -jwo cos kxx sin kyy sin kI(2)(z - c) A) 1 - o - 2-o- 2 m=l n=O 0 We now replace A (2) and B(2) with the expressions from (2.42)and (2.44) r 2 o<) r2) = -jW/o yE E cos kx sin kyy s in k2)( z - c) m=l n=O J -Kabcoskxx'sinkyy'tank l)h] ( k) kk_2) (2) k02 J k02 dimn cos kz(2)(h - c) j 0 0

88 [I-b( - ~ O ~'snYY'tanA4~h tan k (2)(h - c)] dimn,,d2mn,,o COS ( - c)II which can be rearranged as follows 00 f____ Fx) - jI 1 Z o dimnd2mn C~k (h -c) K Fab cos kxx sin k~ysin kf2)(Z -_c) cos kx x' sin kyy' t an k 1)h] * k(2),E* ( - tan k 2 h- c) - k0) (i tan kO)h] (E. 17) Evaluation at z=h yields 00 00 / - \ Fr4(2 Z - h) =jwlaoZ (, J m=ln=O '~ \a~lmnaL2mn/ *[rzab cos k.,x sin k~y cos kxx' sin kyy' tan Ohtan k()h- c)] k [k2)EC* ( - x tan k(2) (h -c) -i0) (i -E.k. tan kOBh] Upon comparison of (E.18) with (E.12) we can readily see that rF19(z = h) = r~(2) -Z h) = rxz= h). (E. 18) (E.19)

89 APPENDIX F EVALUATION OF CLOSED FORM INTEGRALS OVER SUBSECTIONAL SURFACES Consider the surface integral given by (2.66) 'qmn = Jf coskxx'sin kyy'(y')aq(x') dx'dy'. (F.1) where from (2.5) 7p (Y,) -2 2 1 — 2(yi-YOl 2 yo- W ~Y'< yo+ (F.2) 0 else. From (2.6) sin[K(xg,+ -x')] sin(Kl x) Yq (X') = sin[K(x'-x-lx) 0 in(KI.,) 0 Xq ~~ X' - Xq+1 Xq-1 ~ X1 ~ Xq else, (F.3) for q $ 1, and from (2.7) sin[K(l, -x')] a, (x') = 0 sin(Kl,,) o <XI' < 1 else (F.4) for q = 1. In the above, lx = Xq21 - Xq = Xq - Xq-1

90 and, for our purposes herel, we let Xq = (q - 1)x Figure F.1 illustrates the strip geometry used to determine the integration limits in (F.1). The boundaries of the qth subsection depend on q as follows: Sq - 0 < x' < I o - y' Yo + 02 - for q = 1 (F.5) Xq-1 < X q < Xq+l Yo- <y <Yo+ W 0 2 - else. With these subsection boundaries, 'qmn may be expressed as qmn - q (F.6) where Yo +W/2 = jY + (y') sin kyy' dy' JYo - W/2 I fo cos kx'a q(x')dx' [xq+l Cos kxx'aq(x')dx',zXq_1 (F.7) (F.8) for q = 1 for q 1. INTEGRATION OVER y' From (F.7) and (F.2) we have v 2 /Yo +W/2 sin kyy' d w1 J Yo - W/2 2(y V L W (F.9) 1 Note that for strip geometries other than an open circuit and a straight thru section of transmission line, the position function Xq will be more complicated in general.

91 y b Y + o W/2 y - w/2 0 I I I I I I 0 1 X X X a x q-1 q q+l Figure F.1: Strip geometry used in evaluation of surface integrals

92 Now, let sin = 2(y'-Y) cos q dq = - dy' (F.10) y = sin q + Yo = dy' = wcos b d with these substitutions IT' =1 / siin [k sin + Yo)] d (F.11) which can be expanded, with the use of trigonometric identities, as follows: y, 1 inf k W T = [ sin(ky- sin q) cos kyYo do + '/ cos(ky- sin >) sin kyYo do] (l)' (2)yF = +. (F.12) Consider the first term: ~ 7*-1 = -cos kYo f sin(ky 2 sin b) do I W d Wl = -cos ky sin(ky sin 0b) do + sin(ky sin >)d [- 2 Jo 2 Now, if in the second integral above we let ' = -; d' = -do then [0 7 = - cos kyYo sin(ky sinm 4)d + sin(-ky-2 sin )(-do") (y 2 [IS s Jo 2 n = 0. Hence, (F.12) becomes Z = ) = -sin kyYo 2 cos(ky 2 sin 4) do 2 1 [ Wsin y sink o cos(ky sin ) d + cos(ky- sin d 7r [T 2

93 If, in the first integral, we let ' = -- and do' = dob we obtain 1 wf I", =-sin kyYo cos(ky- sin 0)ddb. (F.13) 7r 2 By comparison of the Bessel function of the 2i7 order given by [27] 2 fi J27(z) = - cos 27O cos(z sin 0) dO 7r J with (F.13) we may readily see that 1' = sin kyYO Jo (ky 2) (F.14) This completes the y'-portion of the integration. INTEGRATION OVER x' From (F.8), let us first consider the case for q $/ 1 1 = cos kx'aq(x')dx' (q $ 1) (F.15) JXq-1 Substitution from (F.3) yields sin - xq sil K(x' - Xql) cos kxx' dx' js+ | sin (x + xq)cos kx dx']d Xq+ 1 = 1t [I*l)~' -?)'] (F.16) For the first integral we have T(1)' = sin K(x' -Xq-1)coskxxdx' xq-1 = - i {sin [(K + kx)x' - Kxq-1] + sin [(K - kx)x' - Kxq-1]} dx' 2 q-1 -- - K { 1 [cos(KIlx + kxxq) - Coskxxq-l] + Ki1 [cos(Klx - kxxq) - coskxxq-]}. (F.17) A - '

94 We can solve for q 2) in a similar fashion to yield (2)x' = X1 sin [K(xq+l - x')] cos kzx'dx' Xq = 2 { - [cos kzXq+l - cos(KlI + kxxq)] K- Kjk + 1 [cos kxxq+l - cos(Kx - kxxq)]} (F.18) Substitution from (F.17) and (F.18) back into (F.16) yields = sin (K + K k) [2 (cos kq+l + cos kxxqi) - cos Kl, cos kxxq x sin K2lK K(F.19) After some manipulation, this expression may be put in the following form Z -4K cos kxq sin [l(kx + K)l] sin [l(kx - K)l] q sin Kl (K + kx) (K - k) K= - os K qn Sine (+ Sin [(k - K)l] (q $ 1) (F.20) sin Kl L2 J L2 J where in t t 0 Sinc(t) = t (F.21) 1 t=0. Recall now the integral for the case q = 1 from (F.8) If = cos kxXZ'a(x')dx. Substitution from (F.4) for al(x) 1 r1i T = -sin K cos kx'sin(l - x')dx. (F.22) Comparison of this expression to the integral of (F.18) it becomes clear that if we let Xq -- 0 and xq+1 I Ix in (F.18) we can obtain the solution for the integral in (F.22). The result is 1 = 1 1 {[coskzl -cosK +K2 i lxK k K+k 2 s (K + ) [cos kl - cos Kl]. 2 sinA Kl.AK- ka; K + kxj

95 The above can be rearranged to give I 1 -K12 ri 1 K2 Sinc [(+ i) Sinc [ (k-K)lx m - 2sina Kf ts wh (F.2) yl Combination of this with (F.20) yields (for q = 1). (F.23), = (qKi2 cosk kX i n q= K2s'- 2qSinc ri(k + K)lx Sinc[(k2 - KIf] 2 sin Klx L2 2 (for any q) (F.24) where { 2 for q = 1 4 for q 1. (F.25) Finally, substitution from (F.14) and (F.24) back into (F.6) yields 1qmn = / s cos kxx' sin kyy't4(y')aq(x')dxtdy' Kl^2 COSk1Xq fl l -qKl coskxXq Sinc [(kx + K)lx] Sinc [2(kx - K) ] 2 sin kYo Jo(k 2 ) sin kyYo Jo(ky'). 2 (F.26)

96 APPENDIX G EVALUATION OF THE MAGNETIC FIELD COMPONENTS The magnetic field components anywhere inside the cavity are given by the surface integrals of (2.89) and (2.90) Hq = | ( OG z qYz Ox ) (y')aq(X')ds' (G.1) (G.2) Hq = - Gzy') q Z O~~yICQZ) We will evaluate Hy first. EVALUATION OF THE y- COMPONENT OF THE MAGNETIC FIELD From (2.31) and (2.32) OzG 9z OG(1) Ox 00 00 = E kz')A(l cos kx sin kyy cos k(l)z m=l n=O 00 00 - S S k B(1) cos kCx sin kyy cos k(1)z. m=l n=O Using the expressions for A(m and B(1) from (D.68) and (D.72) yields OGz(1) OGX(1) Oz Ox xx xz 09Z 09 ~~ Pn cos kx sin kyycos k) z tan k 2)(h - c) m=l n=O dlmnd2mn cos k )h [k()d2mn - k2(1 - c)) tan kz)h] cos k.x' sin kyy'. Substitution from (2.47) for d2mn in the bracketed term yields (G.3) [k(1)d2mn k - k2(1 - ) tan k(1)h] = k') [k(2) tan k2)(h - c)- kzl) tan kz)h]

97 k'1- c tanklh - k~j)k k(2),C tan k()h- c) - [y4') )2 ~ k2(l _ 6*)] tanklh (G.4) Use of (G.4) in (G.3) results in -GX o - 1T tank (h -c) m=1 nl0 dlmnd2mnco ~ k0) k (2),C* tan k2 h- c) - [(k(1) )2 + k2(1 -E)] tan kO)h} *cos kxx sin kyy cos Ozcos kxx' sin kyy'. (G.5) Similarly, substitution from (2.33) and (2.34) yields after similar algebraic manipulations (2) (2)00 00 - Sk )ABcos k~xsin kyy cos k( )(z - c) m=1 n=O 00 00 tan =ESE(2 m=1 n=O dlmnd2mn C0s kz2 (h - c) * kO)k~2 tan Oh- [(k )r - k~( - ]tan k(')(h - c)} cos kxx sin ky cos k2(z -c) cos kxx'sin ky' (G-6) We are now ready to evaluate the y-component of the magnetic field. Substitution from (G.5) into (G.1) yields - 00 00 S~ntnk2)(H 1 ntn)zzh-c m=l n=o dlmnd2mn cos kl'h k~l) k(2),C tan k(2) (h - c) - [y (1) )2 ~ k2(1 -*)] tan k41)h} *cos kx x sin ky y cos k (1)z [fTqmn] where from Appendix F (F.26) ~Eqmn = Jjcos k~x' sin kyY'I/(Y')aq(x')dx'dy' - qlcOkXq ic[( kx +K)lx] Sinc[-(kx -K)lj *sin kyYo Jo(ky-) 2 (G. 7) (G.8)

98 So q(K) - 1 ol Y2 ntank)(h- c) 2 sin lm=l n=O dmnd2mn Cos h * {k(1)kz2)r* tan k2)(h- c) - [(k1))2 + k2(1- e*)] tan k(1)h} * cos kxxq Sinc -(k + K)i] Sinc [-(ka - K)1i] sin kyYo Jo(kiy-) *cos kzx sin kyy cos k(1)z. (G.9) Substitution from (G.6) into (G.1) and again making use of (F.26) yields H(2) _- -(qK12 ntank(1)h qy ~ 2sin KIx Z Yr2 m=1 n=O dlmnd2mn Cos kz (h- C) *{k(1)k(2) tan k()h- [(k(2))2* - k2(1 - *)] tan k2)(h - c)} * os kxxq Sinc -(k~ + K)l] Sinc [(kx - K)l]x sin kyYo Jo(ky-i-) * cos kx sin kyy cos k2)(z - c). (G.10) We now proceed to the evaluation of Hqz. EVALUATION OF THE z- COMPONENT OF THE MAGNETIC FIELD Substitution from (2.31) aG(l) 00 00 = S = A $kyA1) cos kzx cos kyy sin k(1) z y m=l n=O _ y ynky tan k(2) (h- c) m=l n=O dimn COS kz()h cos kxx cos kyy sin k(1)z cos kxx' sin kyy (G.ll) Similarly, from (2.33) (2) 00 oo G —# = 1y kyAm2) cos kxx cos kyy sin k(2)(z - c) m=l n=O E- x ~~ q nk tankz()h m=1 n=O dimn COS k(2)(h - c) * cos kzx cos kyy sin k(2)(z - c) cos kxx' sin kyy' (G.12)

99 Substitution from (G.11) into (G.2) and using (F.26) yields f(1) ( qKlx E p nky tan k2)(h- c) qz 2 sin XKl 1 di kO()h m=l n=0 dlmn cos k)h * cos kxxq Sinc ( + K)l] Sinc [(k - K)l *sin J(ky ) cs s sinYo o()cos cos sin )z (G.13) Likewise, substitution from (G.12) in (G.2) yields H(2) (q Kl m Pnky tan y kl)h qz 2 sin K (2) 2 sin l m=l n=O dlmn cos k 2)(h c) * cos kxxq Sc (k + Sine [((k + K)lx *sin kyYo Jo(ky )cos kx cosky sink2)(z - c) (G.14) In summary, the y and z components of the magnetic field anywhere in the cavity may be expressed as follows: 00 00 H) = Hqo cnqclmn cos kxx sin kyycos k()z (G.15) m=l n=0 0o oo Hq) = HqO Cnq c1 cos kxx cos kyycos kl)z (G.16) m=l n=O H() = Hqo Cn (2) coskx sinkyycosk2)(z-) (G.17) -y H E nq ymn Z m=l n=O oo oo H2) = HqO Z c 2 c(2, coskxxcoskyysink(2)(z-c) (G.18) m=1 n=O where qCqKl Hqo - q 2 sin Kl Cnq = cos kxzq Sinc [(k + K)lx Sinc [ - K)lx and (1) Cy - yd2= ^ {k(1)k(2)e tank(2)(h-c) )]tan1)h (G.219 - [(k(1) )2 + k 2(1 - 6*)] tan O)h} (G. 19) (1) pky tank(- ) sinkyYo Jo(kyh) (G.20 zmn dimn cos k(1)h (G.0 -2

100 ky Onn{kkV t an k~h - [(k 2))2c6* - k2(l _ C*)] tan 2 h- c)} (G. 21) sitan kohJk~Y y (G.22) dimn cos k(2(h - sin 2ok~

BIBLIOGRAPHY 101

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