PLACE IN RETURN BOX to remove this checkout from your record. TO AVOID FINES return on or before date due. MAY BE RECALLED with earlier due date if requeSted. DATE DUE DATE DUE DATE DUE 2/05 p:/CIRC/DaIeDue.indd-p.1 IMPROVED MODE MATCHING METHOD FOR SCATTERING FROM LARGE CAVITIES By Weiwei Zhang A DISSERTATION Submitted to Michigan State University in partial fulfillment of the requirements for the degree of DOCTOR OF PHILOSOPHY Department of Mathematics 2006 ABSTRACT IMPROVED MODE MATCHING METHOD FOR SCATTERING FROM LARGE CAVITIES By Weiwei Zhang The accurate calculations of electromagnetic fields play a crucial role in aircraft design and antenna design. The methods proposed in the literature for the EM fields calculation of scattering from large cavities are not be fast, accurate and easily implemented enough. In this dissertation, the calculations of EM fields for 2D large open cavities are considered first. By defining two extension operators, the exact solution and the mode matching solution may be formulated under the same framework, which makes it possible to analyze the difference. Using an asymptotic technique, we present a new (improved) mode matching method. In particular, the explicit solution and the error estimate are given. Numerical examples, including EM fields calculations and RCS calculations, are presented. The second part of the dissertation is about the calculations of EM fields for 3D cavities. Following the similar idea, mode matching method and improved mode matching method are presented. Numerical examples are showed to demonstrate the efficiency of the method. ACKNOWLEDGMENTS I wish to thank my advisor, Professor Gang Bao, for his encouragement, sugges- tions, and guidance during the time I worked on this dissertation. It would have been impossible for me to make this dissertation a reality without his sharing the knowledge and discussing various ideas throughout the study. I wish to thank Professor Chichia Chiu, Professor Leo Kempel, Professor Baisheng Yan and Professor Zhengfang Zhou, in my thesis committee, for their suggestions, comments and help in my research and life. I wish to thank my parents and my parents-in-law, for their enduring patience, constant support and encouragement. I wish to thank my friends and all the people who have helped me. Last but not least, I would like to give my thanks to my husband Yang and my son Max, for their love. iii TABLE OF CONTENTS LIST OF FIGURES v INTRODUCTION 1 0.1 Background and Motivation ......................... 1 0.2 Organization ................................. 3 1 TWO DIMENSIONAL CAVITY PROBLEM 5 1.1 Preliminaries ................................. 5 1.2 TM Case ................................... 9 1.3 Mode Matching Method ........................... 11 1.4 Improved Mode Matching Method ..................... 14 1.5 Numerical Experiments ............................ 28 2 THREE DIMENSIONAL CAVITY PROBLEM 34 2.1 Formulation .................................. 34 2.2 Field Representations ............................. 37 2.3 Mode Matching Method ........................... 41 2.4 Numerical Experiments ............................ 43 2.5 Improved Mode Matching Method ..................... 48 2.6 Numerical Experiments ............................ 54 Appendix ...................................... 58 BIBLIOGRAPHY 61 iv LIST OF FIGURES 1.1 Three dimensional open cavity ............................................................. 6 1.2 Two dimensional open cavity ............................................................... 7 1.3 Sine integral and cosine integral .......................................................... 16 1.4 Graph of r(x) ................................................................................. 17 1.5 Magnitude of the aperture field for a moderately wide cavity ......................... 29 1.6 Magnitude of the aperture field for a very wide cavity .................................. 29 1.7 Magnitude of the aperture field for a two-layered filled cavity ...................... 30 1.8 Magnitude of the aperture field for a two-layered filled cavity ...................... 30 1.9 RC8 of a empty cavity ....................................................................... 32 1.10 Monostatic rcs of a filled cavity ........................................................................ 32 1.11 Monostatic rcs of a large cavity. .......................................................... 33 2.1 Mode matching method ((1): 0) ............................................................ 47 2.2 Mode matching method ((15 = 1t/2) ....................................................... 47 2.3 Improved mode matching method ((2) = 1t/2) ............................................ 55 2.4 Improved mode matching method ((0 = 0) .......................................... . 55 2.5 Improved mode matching method ((1) = 1t/2) ........................................... 56 2.6 Improved mode matching method (¢= 0) .......................................... . 56 2.7 Improved mode matching method (6 = 40°) ......................................... . 57 Introduction 0.1 Background and Motivation A two dimensional open cavity recessed in an infinite ground plan can serve as a model of duct structures such as jet engine intakes of an aircraft or antenna windows embedded in complicated structures. The phenomena are governed by the Helmholtz equation in an infinite domain along with the radiation condition and Perfect Electric Conductor boundary conditions. The prediction and reduction of Radar Cross Section (or echo area) of this struc- ture are very important and require the information of the fields across a broad range of frequencies. For instance, a short wavelength radar (e. g. missile seekers) and a long wavelength radar (e.g. early warning radar) may both exist at the same time. Thus the accurate calculation of electromagnetic fields is of great importance. When the cavity parameters are small or the frequency is relatively low, a large body of work such as finite element methods, integral equations or hybrid methods have been done [8], [12], [13]. For a large cavity or at high frequencies, the calculation of the fields becomes difficult. I. Babuska pointed out that, for a model problem, the Galerkin solution differs significantly from the best approximation with increasing the wave number. A. K. Aziz showed that “k2h is small” is the sufficient condition to guar- antee that the error of the Galerkin solution has the same magnitude as the error of the best. approximation, where k is the. wave number and h is the mesh size. . . 3“ a. . “FIFE" .F. ~E’LED.J Another major difficulty of the problem is that the mathematical formulations of cavity problems always involve nonlocal boundary conditions and singular integrals [8], [3]. Physically, large cavities involve a large amount of internal reflections. In the engineering literature, modal methods have been adopted to calculate elec- tromagnetic fields at high frequencies or for large cavities because of their efficiency and accuracy [9], [11]. To the best of knowledge, in these methods, the boundary condition at infinity is approximated from some physical intuitions or assumptions and confirmed by numerical experiments in some cases. However, without rigorous error estimates, there is no guarantee that the numerical simulations are correct. In this work, the time harmonic plane wave incidence is considered. The ground plane and the walls of the open cavity are perfect electric conductors and the cavity is filled with magnetic or nonmagnetic layered medium. A bounded domain prob— lem is set up by using Fourier transform, the radiation condition and the continuity condition. The mode matching solution v(:1:,y) and the exact solution u(:r, y) are formulated under the same framework by applying zero extension E0 and periodic extension Ep on the transparent operator T, i.e. M + 9(17) = T(E0('U(17, O»): T + 9(1)) = T(Ep(v(1:v 0») In the case of simple geometries, the mode matching method is easily implemented. By analyzing the error between it and v, and applying some asymptotic techniques, an improved mode matching method is obtained. The new method provides an ex- plicit solution which converges to the exact one at the rate of O(1/ fl), where w is the width of the cavity. Due to the error control, the method is expected to be ex- tremely useful for large dimensions or high wavenumber cavity problems. Numerical experiments of the electromagnetic fields in the aperture and different Radar Cross Section calculations have been studied. The existence and uniqueness of the partial differential equation were proved by Ammari, Bao and Wood in [3] for the case when 8'6 > 0. In this dissertation, a TM case result for 8‘6 < —1/k(2) is given. In three dimensional case, the governing equations are Maxwell’s equations. The application includes metallic cavities RCS study, and electromagnetic penetration and transmission properties of objects consisting of these substructures. It has great importance in industries, including telecommunication, engine and antenna manufac- turing. Besides the same difficulties from two dimensional large open cavity prob— lems, the computational time and memory requirements are sometimes intolerably large. Therefore, the three dimensional cavity problem has been regarded as a grand challenging problem for decades. In the literature, eddy elements, discrete Galerkin methods, integral equations and modal methods were applied to calculate Radar Cross Section [8] But none of them are fast, accurate and easily implemented at the same time. In this dissertation, the continuity conditions, Silver-Mueller Radiation condition and Fourier transform are used to derive the exact PDE model in a bounded region. Numerical experiments are studied. This work could be easily generalized to other applications, such as scattering and transmission through the slot in a conducting plane. Following the idea, Open cavity problem with a canonical shape could be solve similarly using different field expansions. For example, the spherical harmonics. 0.2 Organization In Chapter 1, the two dimensional cavity problem is mainly discussed. The math- ematical formulation, including the Helmholtz equation, boundary conditions, weak formulation are introduced in section 1. The well-posedness of the variational problem is further established for the TM case, when 8(6) < —1/k8 in section 2. In section 3, the mode matching method is introduced by using a periodic extension operator. The idea and fulfillment of improved mode matching method are discussed in sec- tion 4. And the convergence analysis is given in this section as well. The numerical experiments are displayed in section 5. Chapter 2 is focused on 3d open cavity problem. The continuity conditions and Maxwell’s equations are introduced in section 1. In section 2, the field representation based on sine or cosine functions are given. The fulfillment of the mode matching method is given in section 3. Some numerical examples using the mode matching method are shown as well. In section 4, the improved mode matching method is proposed following the idea in 2d case. In section 5, the benchmarks in the literature are studied. Appendix displays some results about the integral involving sin(w;r) /:1:. When the considered function is relatively smooth, the numerical simulations of the integrals are studied in term of a Dirac Delta function. CHAPTER 1 TWO DIMENSIONAL CAVITY PROBLEM 1. 1 Preliminaries Consider an open cavity recessed in an infinite ground plane ( see Figure 1.1 ). The ground plane and cavity walls are perfect electric conductors (PEC) and the cav- ity is invariant along Z direction. The time harmonic electromagnetic fields (time dependence e—Wt) satisfy the Maxwell’s equations V x E - iwuH = 0, V x H + 2'ch = 0, where E and H are the electric and magnetic fields, respectively. Assume the media inside the cavity are not perfect dielectrics and have finite conductivity. In this case the continuity conditions on the Open aperture F are 2x(E0—E1)=0 2x(H0-—H1)=0 expo—01):“,- é-(BO—Bl)=Js 5 (E,, H.) Ereh Href) Figure 1.1. Three dimensional open cavity. where p5, J5 are the surface charge density and surface current density, D and B are the electric and magnetic flux [16]. Here subscript 0 denotes the quantities above the cavity and subscript 1 denotes the quantities inside the cavity. In the case of the TM polarization, the electric field is parallel to the z-axis. Therefore the electric field has the following form, E E0 = (0,0, 110) in W, E1 = (0,0,u) in (I. Let ui, ur, us be the 2 component of the incident field, reflective field and scattering field in U +, respectively. Let k,- = k0,/€.i/12' (i = 0,1) be the wave numbers above and below the ground plane, respectively. Then 2 2 a no +6 no +k2u0=0 in U+, fl+§l+k¥u=0 in n, (11) 8:122 By? u = 0 0n PEC, along with the upward propagating radiation condition [15], that is, the scattering field satisfies - ' 2_ 2 . . way): ’ [:62ka 5 yaaezé‘rda. (1.2) v27r where (Ml?) = “Dig/=0 EinCaHinc y f ( ) (607/10) U+ PEC PEC X PEC Figure 1.2. Two dimensional open cavity It is clear that 11.0 = u" + “7‘ + us above the ground. By assuming a plane wave incidence with the incident angle 0, (with respect to the y axis), the continuity conditions then read 21.S(.r,0) = u(.r,0) .l‘ 6 [0,11'], Bus (911, ‘ ill-Wham — [1.053(.r,0) — g(.r) .r E [0.11.], (1.3) 7 where g(;2:) = —2z’p1k0 cos die—”€03: sin 61' and w is the width of the cavity. Obviously, above is a problem in an unbounded domain U + U S). Next, the continuity condition is used to derive an equivalent form in the bounded domain 9. For convenience, the transparent operator T is defined [12] and the zero extension operator E0 as M) = «a f: we — emaeiéws, (1.4) :1: :1: 0,10, 1300493)) = ”U H 1 (1.5) 0 otherwise, Here '0 is the Fourier transform of 2). Taking the Fourier transform with respect to :c of the Helmholtz equation in (1.1) and solving an ODE for y, a transparent boundary condition is easily derived [12] as 6u3(:r,0) —,,— = T>). (1.6) 3/ Using the continuity conditions (1.3), the exact solution of (1.1) and (1.2) satisfies 2 2 (Luimkguzo me, 8:122 0y2 u = 0 on PEC walls , (1-7) Bu 17,0 #0 (6y ) +g(:z:) = p1T(E0(u(.r,0))) on F. The weak formulation of the problem is : Find 21. E V (a suitable subspace of H1(Q) [12]) such that a(u,v) = (g,v),Vv E V (1.8) where the bilinear form is defined by a.(u,v) : [VU'W—/k¥uf’—/T(E0(IL(.E,O))D Q D F and the well-posedness of the scattering problem was established in [3] and stated as the following Lemma: Lemma 1.1.1. If6 E LOO(Q), 32(6) 2 61 > 0 and 8(6) _>_ 0, the scattering problem (1.8) attains a unique solution in H6 (Q). In the case of the TE polarization, the magnetic field is parallel to z-axis. Assume the 2 component of the magnetic field in the cavity is u. It satisfies 1 V - (5%) + kg. = o in n, 1 @ = on PEC walls, (1'9) 377. u = S(E0(g%)) + 2eik0$ sin Oi on P where r(eeif‘rds 1 00 1 3 (f ) == —- / — i\/27r / —00 k2 _ 62 Here the cavity is considered non magnetic. The variational form of (1.9) is to find u E V (a suitable subspace of H1(Q) [12]) such that such that, b(u,v) = [@01qu E V (1.10) P where b(u, v) = iVu - V’Udatdy - fez/u ~17drdy — fig—US(EO(22))d;r 61 Q 61012 ()n 0 and the well-posedness of the TE case was stated as follow [3]: Lemma 1.1.2. U6 6 LOO(Q), §R(€) _>_ 61 > 0, (3(6) 2 0 and €(lT) is smooth enough (satisfies assumption (A) in [3]), the scattering problem {2.5) attains a unique solution in Hlm). 1.2 TM Case Following the fractional Sobolev space notation as in [18], (adopt a wave number dependent. norm, equivalent to the usual norm), the norm || - [I H3 is defined by . 0 Hull? = [(1% + 62mm 9 In this case, it can be proved that To E0 is a bounded operator from H 1/2 to H _1/ 2 and satisfies [[T 0 E0“ S 1. Theorem 1.2.1. Let 61 = fre + ieim. Then the variational problem {1.7) has a unique solution in V if Cz‘m 6 (—oo, —1/k(2)) U [0, 00). Proof. When Eim Z 0, the uniqueness of the problem was proved in [3]. Therefore the only one needed to prove is the case when 6- S -—1 1:2. From the definition, 2m 0 Im a(u,u) = -— k26- u dedy — k2 — €2IE (U)‘ 2614 0 2m 0 0 Q [€]> = ( ) E l I v(.r — (n. — 1)u..') .1: E [(n — 1)w, nw] The approximate model satisfies 2,, a? . 2+; k¥u=0 inQ, 01:2 8y2 '0 = 0 on PEC walls , (3'14) (9'r.'(.'1:,0) 11.0 (9y +g(.1:) = ulT(Ep(u(.r,0))) on F. In this model (3.14), only the periodic extension is involved. Similar as (2.11), the approximate solution u may be expanded as 00 . t'(.r, y) = Z b"5111(72‘7?”Sinhhnw + d). (3.15) 7121 I and it is easily seen that , 72.71" A in .- mr - n7r (E[)Sm(_‘)) (E) = (0(€+—)—0(€——)). (3.16) w \/27r u? ‘ w where 6() is the Dirac delta function. Hence T o Ep(sin(:zfl)) = — sin(£££)un, w where 11., = i\/k8 — (n7r/w)2. (3.17) Then the solution satisfies (3.14) may be given by (3.15) with 2i k0 b = _ n [ill/n SlnhWnd) + #0771 COSh(’Ynd )gn- (3.18) It is clear to see that bn decreases exponentially as 11 increases. This is the so called mode matching solution. Remark 1.3.1. In [10], Morgan expanded the field in the local region above the aper- ture using upward propagating waveguide modes bounded by vertical walls and obtained the same formula. Here, the same mode matching formula is derived by an alternative approach based on the use of the periodic extension operator Ep. Next is focused on the convergence rate of the above series. Lemma 1.3.1. Let C]: be the Fourier coefficients of f(.1:). Then 2k cite-2m < 00 if and only iff 6 H17)". Theorem 1.3.1. Let v(.1:, y) be the mode matching solution of (3.14), then 1*(.r.,0) E H6([0,w]). Moreover, 00 2 (bn, Sillll(’)’-n,(l))2 = 0(i), (3.19) n=[w/rr] [-u,1/7r]—1 Z (n—Wb71,Slllll("fnd))2 1" 0(1). (3.20) ”:1 w w 12 Proof. For simplicity, let’s consider the case of empty cavity with incident angle 6,- = 0. Notice that lim sinh(rc) _ 1 “fit—'0 :1:(sinh(:1:) + cosh(;1:)) — 11.1.1200 S‘nhm = lim i r(sinh(a:) + cosh(;r)) III—100 2:1: From (2.12) and (3.18), 1 bn sinh(*md) = O( . nrrn Therefore (3.19) and (3.20) are true. For more general cases, if nrr >> w . w bn smh('yn,d) = O( 2 ). nrr) and if nrr << w bn sinh('ynd) = GUI—g). u} The same estimates could be still derived. From Lemma 1.3.1, it is clear to see that v(;r,0) E H6+5([0,w]), for some 6 > 0. E] For a two-layered medium cavity problem, define the permittivity inside the cavity 611 y E [—d190]! fr(1.’(.17)) = 62, y E [—(l, —(ll]. Expand the field inside the cavity by 00 111(1', y) = Z [671. sinh(,l3n(y + (1.)) + den cosh(,l3n(y + d))] sum?) y E [—d1,0], n=1 I 00 nrrr 112(.r,y) = Z bnsin( 'w‘ )sinh(‘m(y + (1')), y E [—11, —d1]. r121 13 where fin = \/(n7r/w)2 ‘ k8fl1€1e 7n = \/(nrr/w)2 — kgugeg, 111 and u2 are the upper and lower fields inside the cavity, respectively. Using the continuity conditions on the interface, en, [[10 cosh(,6nd),[3n — #1 sinh(/3nd)1/n] +dn[/10 Sinhfflndlfin — #1 COShffindll/n] = ”971, (3-21) on sinhwndz) + dn cosh([3nd2) — bn sinh(7nd2) = 0, (3.22) on 112 cosh(,l3nd2)13n + (1.an sinli(,8-n112)fin — bnul cosh(7nd2)ryn = 0. (3.23) Here the permeability u is assumed a constant everywhere. By solving the above system, a mode matching solution for the two-layered medium cavity problem is easily derived. The same idea could be applied to a general multi-layered medium problem. If consider the TE cases or the medium is vertical layered, c0412?) should . I . . . n7? . . be used in the field expansmns rather than s1n(—). The mode matclnng solution w is again easily derived. 1.4 Improved Mode Matching Method In order to find the difference between the exact solution and the mode matching solution, the difference between two operators acting on one single mode is analyzed 14 first, i.e., T o E0(sin(¥))— T o Ep(sin(n—:)$)) - [co 1 1 = 22? —k0 812:6 k2- {2 5(1_(_1)n cos(w§))(€ + 171137: _ _ %) dé ' 460 1 1 1 + 52% 00+]: e“:6 13-12 —<1-(—1”) eeecwox +....— m.) «it 5 7.7 ‘ .7; . n 7r + i 1223135 k2 _{2 2131110145 Zn?» _ 381110145 — 7;» 2—7r -OO 2 €+ __ 2 g _ _ w —irr6(£ + —) + i7r6(£ —-)]d{ g 11+I2+I3+I4 (4.24) Before proving the main theorem, let us define Si(.1: ) and Ci(.r ) be the sine integral and cosine integral respectively. They have the following definitions and the asymptotic expansions, (I: t 00 t Si(.r) = / 51” 111=:—/ S—in—dt (4.25) 0 t 2 :r t 1' , t—l 00 t 01(1) = 7+ln:1:+ /0 9—57—111:— / $111 (4.26) .1: S" _ 7r cosa: 00 (——1)"(2n)! sina: 00 (—1)"’(2n+1)! 427 [(1‘) _ E— x _ 1.211 — $2 _ $211 ( ) n—0 n—0 . oo _ smx (—1)"(2n)! cos:1: (—1)n(2n+1)! cm) _ :1: 2 I2” — $2 $2” . (4.28) 1120 r1..=0 The graphs of them are shown in Figure 1.3. Lemma 1.4.1. For sine integral and cosine integral, the following estimates hold, 2 2 ISM-r ) - 7r/2| S -, lCifl‘H S - (429) a: .1: Proof. The estimates could be derived by integration by parts. 1:] 15 Figure 1.3. Sine integral and cosine integral. The well known Riemann-Lebesgue Lemma says if f E L1(R), then lim sin(.r/6)f(;1:)d;1: = 0, 6—40 6lino]cos(;r/6)f(:1:)d;1: = 0. The smoother the function f (r) is, the faster the limit converges. But the inte- grands in 12 and 13 are not in L1, it makes the analysis more challenging. N. Bleis- tein and ect. focused on log-like functions and established a generalized version of Riemann-Lebesgue Lemma [19]. Here in this work, an asymptotic expansion and anti-derivatives are used to prove the convergence. Again for the above limits, if f (r) has singularity at .r, = 0, lim sin(.1:/6) = 7rd(.r) 6—+0 should be used. 16 .05 _ _/ ‘-\\ /’/” \\\\ -o.55 e ’/ \ l / \ / \ / /' I.\ -0.6 " / \ '1 / \ l/ \ -o 65 / \\ e 1' \ -o.7 - / e l / \ '075”; .-< l l -03 l _\ Figure 1.4. Graph of r(x). Lemma 1.4.2. H a 1 sin( ) / ' V1 —.'r2d.r = 71' + 0(6). _la: "b l Proof. Define f(.1:) = V 1 — 1:2, .1: 6 [—1,1]. Because f’(0) = 0, therefore r(x) = 1(0) + fire). ft?) - f(0) Here r .1: = —. It can be seen that r, .1: has sin ularity at .r = 1. Notice 1.2 8 . 17 that _ .r 1 sm(-E—) ]0 a: f(:1:)d.1: (4.30) 1 = f(0 )Si(%)— [O sin(§);rr(a:)da: (4.31) 1 d = f(0)Si(%) — CCOS%T(1)— 6/0 cos(§)a;(a:r(:r))dx (4.32) 1 d = g + 0(6) — 6/0 cos(§)fi(xr(a:))dr (4.33) Use the Riemann-Lebesgue Theorem in finite domain [a,b] and notice that r(l) = -1 and :rr(;r) is a decreasing function on [ 1,.1] Therefore [01 l—( I7“($ ))ld$— — —/ldd—(:1:r(:1:))d .1: = —:1:r(:c)|(1)= 1. Hence 1 ‘5 lim cos(;€-)d(.rr(r)) = 0 6—10 0 and 1 sin(£) / 6 f(.1:)d.1: = 71' + 0(6). _1.r Remark 1.4.1. The above lemma in fact states that lim€_,0 sin(:1:/6) = 716(.1:). Remark 1.4.2. For Riemann-Lebesgue Theorem, it could be further proved that if [(2) is Lipschitz continuous on [a,b] , then b / sin(.1:/6)f(.1:)d.1: = 0(6), (1. b / cos(r/6)f(r)d.r = 0(6) 11. (See Appendia) But .1:r(.r) E L (,[0 1]), )directly using Riemann-Lebesgue Theorem in. (4. 31) could Just guarantee lim€_,0 fol sin (f) .171(;r )dr 2 0. The first order convergent result could not be obtained. Also notice —(.1:r(.r )) has singularity at .r =,1 the dc change of variable could not be used twice in {4.32) to get 6“ term. 18 Lemma 1.4.3. If FI— < 1, then for any a E [0, w], the following is true. w sin( w(_1; — —)) sin(w(.1: ‘1' 71—71:» /11 Bias, /1_ w d1: nrrw 717T :r—— 23+— w w 7171'0, 7171’ = (Si(w - n71) + Si(w -l- n.n))sin(—TU—) l — (— w ) + 0(1/w). Proof. Define sin(a:1:) - 5111(m) l 71(31‘1) = r171 11 x — _— w 1— :c —)2 w f1(‘.13,(1.) — 7171’ x — _— w sin(a:1:) + Sim-TEE) r2(.1:,a) = 1171 w .r + — f2(¢13.0-)= 11—7—1 w 19 Then using the odd and even properties of the integrand, __2_i/_lli (133m sin(u I-—))_sm(w( :F—l) d:1: .1:+— w7r w sin( w( :1: — —)) sin(w(I + E» = /()1( Sin (GI)V1— — n.7rw dI I—— I+— w w 1171 1 mm sin(w(I—- _)) 1171' f ' 2 =_- [sin(w) (1:—Tim 1—(—w—) d:1: w 1 nn /1 2 — — d +/0r1 sin( (.1: w )) _n(_ w) I +/1<: $11 (1 "new/1 f '(1 "fix "he s — sn1 —— r sme—— I—— :1: 0 111(11 112: w :1: 0 11 w w 1 (1+ ml) )sin( 11 — +/( sin( mrw 1—(fl)2dI 0 r+_ 'IU w 1 —/ r2sin( (u'(I+n—7r)) l—(L—lfl)2d.1: 0 w w 1 . +/1 sin(n—w7m )f28111(u(w;17+—))dI—/ 7‘2f28111(w(.1?+E))($+fl)d£ 0 0 w w Similarly as in Lemma 1, the following facts could be proved: First, 1171 . 1171' 1 sin(w(.1: — —)) SlIl(”tL'(.‘lT + —)) w 11' . / nrr + 111 d": 0 I — -— I + --- w w w—nrr sin I w+mr “in I = f ( )dI + / b—LdI —n.7r ~73 nrr .1: = Si(w — nrr) + Si(w + n7r) Second, 1 4 1 / flsin(w(.1: - 7—U—r))d.1: /0+( f2 sin(u'(.1: + H))d.r 0 w 11' 121— 132*“)2) = —l .f' '..' 1.: ( )n /0 1:2 _ (EH)? s1n(111)11 For convenience the following definition is made: F1(I) = (7171 2111—12—9) (U) . ‘. . 1 . $2 _ (1,11? It IS a decreasing functlon and f0 |F1(I)|dI IS w finite. Therefore the above integral is 0(1 / w). Third, I 1 mr /0 r1 sin(w (I — n—J))dI —/0 r2 sin(w(I + -w—))dI n 1 2(sin(aI)%T — Is;n(L;a)) (—1) f0 717T - (—)2 sin(wI)dI For simplicity, consider the case when a = w. The major contribution of the above integral is 2777f 1 (sin(wI) /0 2_ "7,15 (1.1: 117 -( w) 1 , 1 = :1- [log(t — n71) — log(t + mr) — C1(2t— 2n7r) + C1(2t + 2n7r)]8} = 0(5) Similar, the fourth conclusion is derived 1 nw na 1 ma nw / Tlfl sin(w(I -— —))(I — —)dI mm/O— r2f2 sin(w (I —))(I + —)(II 0 lb w w 111 1 (sin(aI) - sin())(\/1 — I2 (n_7_r )2) 1 = (_1)n/ sin(wI) rm (Tl (1.1: = O(—’) 0 we - —> w u) 00 sin Define qn = / (31:1,)dy wheie n _>_ 1. From (4.25), using the integration by 3!" 1/6 parts, it could be easily proved that 1 . 1 qn, = 6” [cos — + m s1n - — 11(11 +1)e2 c e 1 1 cos 2 — 11(11 +1)(n+ 2)€3 sin —] c +71(n + 1)(n + 2)(n + 3)qn+4, (4.34) 21 and further more, an+4| = O(en+3). Lemma 1.4.4. Assume a = 0(62), then Proof. Define g(:1:) = ‘/l — i2, then g(I) = f(l) from the definition in Lemma :1: :1: 1.4.2. Use Taylor’s expansion and (4.34), 23 ()0 sin(—) 2 fol/$2 —1dI 1 )(x—2 + $14)\/ I2 —1dI + O(€2) 8 Lemma 1.4.5. For any a E [0, w), 71 7? TNT 00 ——1 .- {—— sin(111(I — —)) sin(w(i~ + _)) ./ +/ e'a-E I2 - 1 727?“, — mrw d": = 00/1“)- 1 —OO 15 — — I + — ’11) u) / 1 1 Proof. Define g(I) = —2 — 1, then g(I) = f(—) from the definition in Lemma 1: . tE 22 1.4.2. Use Taylor’s expansion, 7171' 71’” [loo [:33 Fatima-E1) sin(w w(a:+—u7)) -1 TM " n7r d:1: = — sin aI T__sin(wI) V113 1132— 13 On w/loo ( )3; (_)2 1d — CE 00 '( )'( )(i+(-Ei)v 2-1d +0((1/ )2) _ 1 sm aI sm MI: 32 x4 33 113 w 00 sin an: in wI = (13/1 ( l: ( )mo1—§r1111dx+000 a: rate degenerates to 0(6). From the above le111mas, {[4 + Vn, [31(111— nn);Si(11-' + WT) — I] sin(ml} = 0(3) where V", is defined in (3.17). 23 For the second integral and the third integral in (4.24), 00 -217: 12 + 13 = 2W— 1 — {2(1— 1(_)n cos wéko) 2 7% 2 cos(I€k0)d{ 1 C *5?) 0 oo nvr 2' 1 1 n ‘27,,— — _ — — _ - - rm 0 _ 27f /(§ 215 +O(€3))(1 ( 1) coswfiko) £2_(—)2 cos(I£k )d5 1 wko oo n7r i '2‘— : E /§(1—(—1)ncosw€k0) 2 7",)” 2 cos(.1:£k0)d€+ 0(1/w) 1 5 — (mi—(1 if I 71$ 0. Now calculate 1 /€(1 —- (_1)n cos wéko) £2 —- (EV cos(I€kO)d§ ka Notice that the Lebesgue dominated convergence theorem could not be used due to the singularity of the kernel. When n is an even number, the anti derivative is, — k 1 k —1 k. AN('JI) = :( 1) Si(x(u€ 0+( ) mr))sin(mmj) 2 w w k=1 + 22: 1C (I(111€k0+ (—1)I‘n7r)) (TUTTI) 5 l w COS —w— k=1 2 (_1)m+1 k w£k0+(—l)mn7r ( 1)k1:+u1 + Z 31((—1) 1+1“) )sin( 7 ) 4 H) w m,k=l 2 , 1- _ , k _1 m, , . + Z _10,((_1)1 1: 11M 0H ) mr)cos( 7r 1) Hi“) 4 w u! m,k'=1 From Lemma 1.4.1, 12 + 13 = 151010.011-1 011/111) = 0133). 24 A similar result could be derived for n which is an odd number. For the first integral, k0 _211—7: — 0‘/k(2) — (2(1 — (_1)n coswé) E2——(T%)_§ cos(I§)d€ 2n7r ikO — E n7r = 1._( __ —— ., d 0 —. WOO/OH c)osw€) £2 - (7:77,)2 cos(r£) £ + ( w ) Evaluating the above definite integral, 7T1? [11 — '11—an12: mall-$1] = 0131+ R> d, it becomes a total reflection problem, thus |u| z 2sin(k0d). The phenomenon may be observed from Figure 1.6. In Figure 1.5 and Figure 1.6, the cavity is empty, i.e., 67 = 1. Figure 1.7 shows the magnitudes of the fields of a cavity with layered mediums. In this case, k0 = 2n, w = 50/7r, 0.,- = 0, 1) ye {—S/Waoli Er“) = 62, y E [——10/7r,—5/7r]. The connected line is for 62 = 1, the dashed line is for 62 = 3 + 3i. Figure 1.8 again shows the magnitudes of the fields of a filled cavity. Here k0 = 271, w = 8, 61, y E [—2.4,0], (r(iL‘) = 62, y E [-4.8, —2.4]. The continuous line is for 62 = l, the dashed line is for 62 = 4 + i. Another important quantity in the EM field calculation is radar cross section (or echo area). It is the most common scattering measurement in antenna design. It 28 1-‘ I l I I Y I l 1.2 ~ 4 1 . .1 3 3 0.0 ~ t z 9 0.e~ 2 0.4 0.2 - . o l J L L L 1 l 0 2 4 6 a 10 12 14 16 XI}. Figure 1.5. Magnitude of the aperture field for a moderately wide cavity. 0.25 r I T l I I T 0.21 . g 0.15 1 D '2 z 9 2 0.1 0.05 - o L L L l l L _l 0 20 40 so 00 100 120 140 160 x/l Figure 1.6. Magnitude of the aperture field for a very wide cavity. 29 3.5 0.5 montzc<2 Figure 1.7. l\r’Iagnitude of the aperture field for a two-layered filled cavity. 4.5 XIX Figure 1.8. Magnitude of the aperture field for a two-layered filled cavity. 30 measures the ”size” of an object as seen at a particular wavelength and polarization and defined by - 2 Din rl—l>moo 47rr —Ds where Din and D3 are incident power density and scattering power density, respec- tively. The prediction and reduction of Radar Cross section plays an very important role in aircraft design and antenna design. Mathematically, for the 2—dimensional open cavity, the echo width (Radar Cross Section) is given as [8] 0(cb) = élgcosfib)feio’xE(:1:)d:c|2 (5.39) F Figure 1.9 displays TM case monostatic RCS of a empty cavity with w = 1, d = 0.25. RCS computed using the FEM (continuous line) and IMM (dashed line) agree well. The unit of RC8 is dB/m. Figure 1.10 shows the monostatic RCS of a filled cavity (62 = 4 + i). It follows the same dimension as the previous example. Because of the different choice of time harmonic, the result is compared with the example in [8] for 62 = 4 —— i. IMM result (continuous line) and FEM result are showed in one graph and they agree well with each other. Figure 1.11 show the RC8 of a large cavity. In this case, w = 10.2, d = 5.1, e = 4 and f = 300MHz. The continuous line is for backseattered RCS, the dashed line is for specular RCS. 31 RCS(db/m) é: '3 RCS(db/m) is -50 ~60 Figure 1.9. RC8 of a empty cavity. 10 Figure 1.10. Monostatic res of a filled cavity. 32 3.3381 70- 60 40 10 Figure 1.11. lV‘Ionostatic rcs of a large cavity. 33 CHAPTER 2 THREE DIMENSIONAL CAVITY PROBLEM 2. 1 Formulation Consider the problem of scattering by a cavity embedded in an infinite ground plane (see Fig 1.1). Assume the ground plane and the cavity walls are perfect electric con— ductors (PEC). As discussed in Chapter 1, the governing equations are the Maxwell’s equations VXE—iwuH=0 VXH+iweE=0 where E and H are the electric and magnetic fields, respectively. Assume k0 is the wave number and i‘ = (.1:,y,z). Let a plane wave E2. = Z0" zkllq'l' illuminate on the structure. Here 13 = cosa(sin ¢,—cos¢,0)T + sin a(cosqbcos 6,sind>cos 0,sin 0)T is the polarization vector, where a is the polar- T ization angle. (7 = (coscbsin 6,5inqbsin 6, — cos 6) is the wave propagation direction 34 (E, ”.1 ref! Hm) and Z0 z 1207r is the intrinsic impedance of free space. Then . - :1: Er = _ Z0p*elk0q :1: Hr = _s£*eik0q*z where p-q=0 q* = (q11q2,—q3) s =p x q s** = (—sl,—32,s3) Because this is a unbounded scattering problem and can be regarded as a perturbation of total reflection problem, the scattering fields are set as E3 = Et - El — ET and H s = H t — H i — H T, where Et and H t are the total field. In this case the scattering 35 fields satisfy the Silver-Muller radiation condition lim rES = lim rHS =0 r—ioo r—voo - s__ s = rlmooNE ZOH xr} 0 Further assume the permittivity £7 and permeability pr inside the cavity are invariant along a: and y directions, but piecewise constants along 2 direction, the boundary conditions are nx(E1—E2)=0 z=0 nX(H1—H2)=O 2:0 n-(Dl—DQ)=pS z=0 n-(Hl—H2)=J5 2:0 where p3 and J3 are the surface change density and surface current density. Notice the cavity walls are PECs, therefore the fields satisfy the following bound- ary conditions inside the cavity n x E2 = 0 on the cavity walls n - H2 = 0 on the cavity walls The well-posedness of the variational formulation was discussed in [3] using the Hodge decomposition and the Unique continuation. 36 2.2 Field Representations Use PEC conditions, the B field inside the cavity is expanded as 12y)SIH[/\mn(z + d)],(2.1) Ex: 2: [amnsin(m)+amncos(fl)]sin( w1 w1 m,n=0 oo , mm: ~ , mir m7r , Ey = 2 sm(—) [bmn sm( w2y) + b77777, cos( 10231)] sm[)\mn(z + d)],(2.2) m,n=0 wl 00 7171']: m7r E2 = 2 sin(?) sin( u y) {Cmn sin Amn<2 + d) + Cmn COS[A7n,n(Z + 60]} (2.3) 1 ’2 m,n=0 where A7777; = \/k(2)plel — (nir/wl)2 —(m1r/w2)2. Note that V - CE = 0 and the medium is piecewise constant, 3E3: aEy 8E3 0 = —— + + at 8y dz 00 7171' 13 7771"? 7771' may = Z [émn cos(—E4) ‘ amn Si11(—uj'.—)] 'w_ sin( w ' )Silllbmnlz + dll m,n=0 1 '1 1 12 00 Tl‘fl'I‘ ~ may TIT/Ty 77177 + Z SIIl(—) [bmn COS( ' ) — b77171, Slll( ):| — Slll[/\Tnn(2 + (1)] 1.1/'1 102 1132 11.12 m,n=0 00 nnr mm; + :sin(—) SlIl( ' ){Cnnz COS /\7nn (Z + d) — CNN), sin[/\77777(z “l" d)]}/\7nn. ”(El 102 m,n=0 therefore 2 077111718111“1'771n(3 + d) = 0, n E b77777ms1n 7mn(3 + d) = 07 m Cmnfi'mn = 0 Since the coefficient matrix involving (”177777 is nonsingular, (177777 = 0, b77777 = 0, 577777 = 0 for 777-777 # 0 and 7171' 77171" 0771.71. + -—f)]nn + ATNTIC'ITUZ : 0 (2'4) “)1 “1'2 37 From the Maxwell’s equations, the H field inside the cavity is 00 1 . . .. H]: = Z .—- {mcmn sin(-TE) cos(mfly) cos[)\mn(z + d)] (2.5) iwu wz w1 w2 m,n=0 —/\mnbmn sin(m) cos(mfiy) cos[/\mn(z + d)]} , w w2 00 1 nvrx may Hy = Z 17/; {Amnamn cos(fi) sin( u2 )cos[/\mn(z + d)] (2.6) m,n=0 _ Ecmn cos(?) sin(TZJZy) cos[)\m77(z + d)]} , 00 1 mr mr mm: Hz = Z _— {—'-b77m cos(—y) cos( ' )sin[)\mn(z + d)] (2.7) ltd/J. w1 w1 wg m,n=0 _ Mam.” cos(m) cos(mfiy) sin[)\m77,(z + d)]} . 1132 ml 1.1/'2 Next, the field representations in the region of upper half plan are established . Above the ground, 60 and 110 are constants everywhere. Thus from the Maxwell‘s equations, new + kng = 0 when 2 > 0 Bay = 0 when 2 = 0. and the scattering components Egg satisfy the I\Iaxwell’s equations and the Silver Muller radiation condition. Further more, Egg/(.12, y, 0) = Ex,y(l', y, 0). Take the Fourier transform with respect to :1: and y of the Maxwell’s equations and solve an ODE for z [2]. In this case, the x y scattering components of the B field above the ground is derived as - .2_ 2_ 2 E‘1s.z>=E'-*ezv"0 ‘1 5‘2 (2.8) . ,- / .2_ 2_ 2 , . . ('anvz) = i‘/‘:>O (E0 OEF.y)A(£10)P2 A0 £1 £2 QEI€IE+I€2yd€ (29) 2n ~00 Ely 38 Note that give E§(:c, y, 0) # Ez(;r, y, 0), a similar result could not be derived for Ez. For Ez, . ,/72_ 15‘3(€1.£2,Z)=[130(E3(€162.0))1661 €2Z Hence, *1/12— 6~E§(él,12.z)=i\/18-5 —12 [5301153311113 0))1e 51 32" Therefore 1,/12— . 1 [30(1‘3346162100l6 £1 €2Z= 3253(€1,€2,Z) 1118-61—63 it implies €311,132): E3(,£1 €212) 2V1?)—l Taking the derivative of E}? with respect to z and using V - CE = 0 [2], [22], 1 - .2 2 2 z\/"0“€1‘52 1533({10 = l-(0.rE3)‘(€, 3) - (311531“, 2)] E301, y, z) 1 g. . 1 ’. ' = — 21—16151) (m) — (01153) (5,211 6’31““?ch 271’ R “/1118— for z 2 0. From the radiation condition [E] = 0(1/r), (0,332)‘ = 1111333.. Thus the total B field above the ground is E_17‘y(.l',y,2) (2.10) = Z11111,2€i(k“‘r+kyy h-eikzz + 6"7"”) 1 OO 1. + '2— (530013117111 0) V 0 ’51 5‘2 1174.42.17, (2.11) 7r —oo " 39 where g = Ez(:c,y,z) (2-12) 20])3ei(ky$+kyy)(elkzz+e—ZkzZ) (2.13) 1 . {—1—1E2111z1—12E5e 21e151$+l€2yd5 2 2 W R \[0 51 52 (61162)- From the Maxwell’s equations, the x-y components of the magnetic field above the ground are + W —[2201(1~yp3 + kzp2)e (2.15) 1Wi€28i51:+i—€:: 27f R2[_ €1< EOOELL') (€10)_ {2(E00Ey)( (NW/1— ,i/,,,,13,/1 —£1—£§ EooEy)(€0)l€1$+l£2yd€} Hy(.r, y. 0) (2.16) —{QZ()2'(— —1 .pl — 1;,3713)e’7(k~1‘1+kyy) (2.17) imp 1 lglelélfi‘l’ngy [ 51 (E0051) (5.0) -€2(EooEy)( 3115 27r R2 0)l\/k8_ 7/32 1/10—{1—£2(E00E)(1.1>)1:7€1x+’52yd£}£1 From the continuity conditions and application of (2.1) to the above equations, two equations involving an,b77,,c77, are derived. Combining with (2.4), the electromagnetic fields could be solved. 40 2.3 Mode Matching Method Replacing the zero extension E0 in (2.14), (2.16) by the periodic extension Ep, H$(2:, y, 0) (318) = fi { 2Zoi(kyp3 + 1zp2)ei(kfl+kyy) (3.19) i€2€i€1x+i€2y 1/12 - £2 - 62 711; [112412-1212 (Ep 0 Ey>‘(:,0>e"€1”+i52ydé} +2.17; l1 21—11031, 0 5111101121131, 0 51,111,011 Hy(x, y,0) (3.20) = m{2Zm(—kzp1—k1:p3)€i(k$x+kyy) (3.21) 1 i£1$+i152y ~27 21-61 EpoEI)(€10)“{2(Ep°Ey 01“” 2 d1 R onflo £1 “52 1152/2 2'1/18— ~53 (.EpoE 11130181151111.1211} It is easy to derive (Epcosflf-n‘m = 12—71(6(1+1’1>1«5(1—:—T>>, mt * 1 , 1 - (_1)7‘1e—;’w€ (E cos(— 1) (1) = —(—1a) ,.1 , 0 w 2n (£2 2_(_)2 ) where (5(-) is the Dirac delta function. Lemma 2.3.1. For the periodic extension. Ep, the following statements are hold: 1 . ., .- . -2— R2 2[Ep(sinP-f—Tcosw m7ry M]\/A {(225.1(1614ry52l11g 71' = /\7nnSln'7m—TCOSm17ry “’1 “'2 1 mm: , "my 2 r) i(r£ + g ) — E( — 1 —-,»1E21 271 R2[ 11(cos Sin 1112)]A12M/ -61 £26 (6 : Am” cos mm: sin mm; “(1'2 41 Lemma 2.3.2. For the periodic extension Ep, the following statements are hold: nn 2 1 mm: may” 626 z($€1+y€2) (‘11:) mar may / [Ep( cos— sinu d6 = cos — sin 27r R2 w1]0\/k_€% __ £3 Amn w1 102 i( ) nmir2 1 x£1+y£2 . 1 [EM Sin—— n7r2: 0081.251528 2d = wlwg cos m sin miry 2—71' R2 U11 )]\/k0— Amn 1.01 7.02 mir 1 nmcC Sm_7r_y) ]A €2€:(I€11+y:2) (112—2) mm: miry R[Ep( sin— = 2 sin — cos ( ) mmr2 . ’i $€1+y€2 . , 1 _./ [Ep( cos— mm: sin _miry “ €1€2€ d = _w1w2 sin m cos mny 27f R2 111 11.12 \/kg _ 6% _ 6% Anzn 1111 102 Using the continuity conditions 112.1(1. 110+) + H2111. 1.0+) + H2111. 11, 0+) = Hx,y(171 1,, 0‘) a system equations is derived 1 mn' 171.111 {—Cmn COS()\7nn(l) /\mnbmn 003(Amnd)] (322) 1 , m-mr2 , , 77127r2 , = ——[S.,1,m — amni Sinhnmd) — bmni SlIl(";‘m,n(l) (3.23) 110 w 1 102 /\m n 102 Amn "anbmn SinO‘mndH 1 7171 Ilhmnamn COS(/\mnd) — U—I‘Cmn COSO‘mndll (3-24) 1 . 12271'2 1 , 17171712 , = —l372nn ‘1' “777.71" 5111(7’11211‘1) + bmnl— 811107127111) (3-25) l1 wl mn w 1102 mn +‘13/\mn(lnm Sin(/\m 71(1)] 42 Here 4 w Srlnn = _ 02/11)] eik$$+l€yy sin(—3:) COS(-¥—y)d:rdy “)le U11 [U2 4 = B k A k w1w2 ”(wl’ 5”) moo lElln'C(97'n'Ca(51%“? or people sometimes use IESl2 o = lim 47r7‘2 (9 HH r—roo lEincl2 Q) ESIQ _ - ,2 l a ”HE-THEM“ I Eincl‘Z (I) . 2 1151312 GEE : rlil’éo4m' 1'71c2 IE9 | 215312 ”EH: li1n 47rr 1‘—>oo lEéncl2 44 With the far field patter derived above, the RC8 is given by RCS = rlim 47r'r2-l—— k8cos29 w] 2112 'k S. 6(' (15+ . Q5) 2 = I\/O‘ / E(l‘,y,0)€2 0 m .ECOS ysm dl'dyl 0 7r Let 072:0,sz 8$=0,8y=0,33 =0; on=0,m;£0 , , "NT 8;]; = (1171,71 8111(A77271d)(-2)E X P1Q2, 8y =3 0,83 = 0; on740,1n=0 , 7171' 8y = bynn SlIl(/\n1n,d)(—l);; X PZQI, 81:20,.9320; o 71750,":750 3.1? = (177m Si11(/\mrnd)(—'i)k0 sinBcos 0m X Q1Q2, 11:2 - . , . , mr 3y = bmnSll'l()‘772.7rd)(—3)k0 5111081110; X Q1622, (amn T7}? + 577mg) n7r m7r Sz : — cos()‘mn‘DEEleQQ- A771." 45 where eiko sin 9 COS 45w] _ 1 (k0 sin 6 cos (b) if (k0 sin6cos d)) # 0 zwl if (k0 sin 6 cos (1)) = 0 eiko sin 6 sin ¢w2 _ 1 P2 = { (k0 sin 6 sin ¢) if (k0 sin6sin 45) # 0 i102 if (k0 sin6sin d)) = 0 eiko sin6cos $1111 (_1)n _1 if k sin6cos 2 mr w 2 (k0 Sin6cos ¢)2 .. ("W/W1? ( 0 45) # ( / 1) LU? if (k sin6cos ¢)2 = (Tm/w )2 27m 0 1 eiko sin6sin ¢w2(_1)m ._ 1 if (A~.osiné'sin¢>2 aé (ma/w»? Q2 (k0 sin 6 sin ¢)22— (m1r/w2)2 iw 2 if (k0 sin 6 sin d))2 = (m.7r/'wg)2 21mr Figure 2.1 and 2.2 show the RC8 for a long and narrow cavity (wl = 2.5, 102 = 0.25, d = 0.25). The unit of RC8 is dB / m2. Figure 2.1 presents the back scattering RCS for 6 = 900. The upper curve is for H polarization and the lower cure is for E polarization. Figure 2.2 presents the back scattering RCS for 6 = 00. The continuous curve is for H polarization and the dashed curve is for E polarization. They agrees well with the results in [21]. 46 10 ~ 0 :-.--------------~--~~ -10 p “x“ 11) ‘~\ '0 ‘s c. ‘\ N '20 " \\ é \ b \\ -30 1- \ -40 c \\ \ \ \ \ -50 _ “ I \ | $ .60 l l 1 l J; 1 l l ‘L 0 10 20 30 40 50 60 70 80 0, degree Figure 2.1. Mode matching method (g6 = 0). 20 T I V f T I I I -“ 10 ~ \ a \ \ \ ‘\ 0 “ ’I’-‘\\ ‘i | I \ \ l \-v"‘~ b ‘\s p""------~~‘\ -10 — ‘ (D ‘O Né'~ ‘20 ” b -30 » \ -40 - \‘ \ \ \ | -50 _ ‘, a | | | I '60 1 1 1 1 1 1 1 1 ‘1 0 10 20 30 40 50 60 70 80 9, degree Figure 2.2. Mode matching method (ozrr/Q). 47 2.5 Improved Mode Matching Method From (2.14), (2.16) and after some simple manipulations, 1 - . rageflxjéz: d6 \/k0 - <1 - 62 (E0 0 Ey)‘<5,0)ei€1$+i€2yd£ } +21”— [R 21-51% 0 Ems, 0) — 52(E0 o Ey)*) 1 \/—_§ mr - 7mm : 00/2")! 1 — :r x —- — x + — .. w w .. oo ,- ‘ 1 Psin(w(.r -— 72)) sin(u,r(.1: + 121))1 610$ mrw + mrw = 00/11,). 1 v1 — 1:2 :1: — — .2: + — - w w a Proof. The latter two are easily seen from Riemann-Lebesgue theorem. The first two could be similarly proved as Lemma (1.4.3) Cl Note that 00 go 1 1 / f 2 (1.111;; < 00 1 1 \/.I:2 + 3/2 _ 1 l? y 51 when 51 and {2 are large enough, the Lebesgue dominant convergence theorem could be used to (5.31) with (5.33). Another fact used below is that (x) 1 ' _ TNT [a flu — cos u'€)snl(y€)d§ — 0(3) ("w”) for a > 1. This could be seen from Lebesgue dominant convergence theorem and [00 1 ' ( )dé —— sm yé a 62 - (21177:)2 = i“— { [Sid/(Jim’w—“h + say-(131%)] cos(?) 27m _ MM) + wafl “(mg w w w . 'UJ . . . . . . Even though there IS a factor of 2_— 1n the anti-derivative, the following estimate mr could still be derived m 1 . TNT /. mmlww‘i) by the usual taylor’s expansion. Lemma 2.5.4. , mm: may . [E0(sm—cos . f2 wl ”’2 €2(.L‘£1+y€2)d€ R .2 2 2 ' \/"0 ‘ 51 ‘ ‘52 Si w —n7r +527 11) +n7r 2 1 1 -- ( (1 )2 ( 1 l) _ +,__ may) 7r Amn 4"” 2' i , mm mm; mr — . ~ A? k . -— ‘ — + 4mrfn(’€.yl+ 8k07rf"( 0,17)9m.( (Lt/l] 91“ "’1 005 “,2 +0( w) where K. 2 A78 —- (EP and wl [E0(cos ”—73 sin may A _ “’1 “'2 €1(.r{1+y{2)d R2 2 2 2 g \/""0 ‘ 51 ‘ ‘52 Si 11' — 77277 + Si. w + 171% 2 1 = [( ( 2 ) ( 2 )) _ + 7.9111(591.) 27r /\nm 4h7r i i 7277.1: , may n7r ——T—f7,,,(n, .1?) + T'T—fmUVOa y)gm(k0. 1:) ]cos sin I + O(—) 4h7f 8L07r “'1 112 w where K 2 \/k3 — (Ki-7:)? n.7r Proof. Let h.n(1:,w)= (1—(—1)ne-'iw~”C)/(1-2 — (3)2), then [E0(sin m cos may ‘ )1 f “’1 ”"12 ei($€1+y€2)dg = R2 #8 —<¥ -63 = z Tam/62W hm(€2,w2)) +8(hm( £2 w2) ))lezyf2dé2 :flmhnffivwlll+3(hn(€11w1))l l ewfldél a - é RmRn + anIn + ImRn + 1min Among these integrals, using Lemma (2.5.3) 1m In 2 (Si(w — mr) + Si(w + n7r))2 _1 sin PE COS my7r 27" /\mn “'1 “’2 y+R1 Here R1 takes one of the following forms: O(l)a nd O(n— 7r.) It is first order term “’2 w 1 and maybe different in different situation. l mu: m7r R771171.“ — T9771“ y) sin — C05 3/ + R83 4% ml 102 where _ y mr — urn _ + w 7171' — um 9771(K,y) = —2Sl(——(—————)') +Sl((y )(w )) “‘5“ (w - y)(n7r — um)) _ Si( (10 + y)(n7r + 1115)) w w , mr + um , m — mr + 1m»: +251(y£—-'-—)-)+Sz(( may ))]. n ; and RsS(-L‘) are terms involving sin 3 sin mm; .Thev are not essential parts in the wl 2112 sense of L2. ° n7r L‘ may R7117" = 4—_z_.-f-n(K y) sin—cos + RCC u 1 1112 n I: y I and RCC(.’L ) 1s the tel n1 involving cos ——7—r— cos "ml. fn. is defined in Chapter 1 (4.35). 111 1112 , rm (7 may Bran— — 07Tfn(;r)dx+0(e3). —oo 33 —oo Proof Assume supp(q’)) = [—1,1]. 1 sin(f) _ in (2n) 1/00 E ¢(;L‘)dl‘ = /(‘)€ §_y_ [¢(0) + Z ¢(2n)('0)€2ny2n dy n=l ' 2_OOJI y Collect the similar terms of 6i, 60 : gb(0)-72I = é/f; 6(I)qb(r)d.r 1 , °° ¢<2n><01 °° «pm-N0) ecos(;). «1(0) — ”a (2n)! = 1;) (2n)! = —¢(1) 0° (2”) (.2 sin(é) : -(0) + 2: 9b (2n)('0)(2n — 1) n=1 ' 0° WW0) °° ¢<2n>< > , = 12:0 (2”), +71; (2n _1), = —¢(1)+¢ (1) 0° (2”) 63 cos(%) : 296(0) + 2: lb (2n)('0)(2n - 1)(2n — 2) 71:2 ' _ :2 (2n - 2)! "21(211 — 1)! ":0 (271)! I I ll Since supp(¢) E [—1,l], 00 sin(—) 00 f 6 mm = / 6(“)¢(Ild$ + <25”(0)e3cos(-:-) + 0(64). —oo .r -00 Theorem. 2 Suppose ¢(.l') satisfies the following assumptions, (8) 95(1) 6 000W); (1)) (16(1"): 0 for .1: E [—l, 1]; 58 (c) a>cc=<> 1x123: Otn—l—fiforlxm Then 00 5111(5) / 5 qfi(:r)d:1: = 0(6). —00 a: Proof . a: 1 (X) Sln(—) §/—oo $6 ¢($)da:=e/_OO siny Z tn(———— yl)d2n 6 oo 1 1 1 1 = etO cos(—) )+ e Z tn [cosé — + 2716 sin — — 211(2n + 1)62 cos — e 6 n=1 —2n(2n + 1)(2n + 2)::3 sin - :1] + O( 64) = f(1)ecos(1)+2f(11)6251n(:)—e3 cos(:)”([4f +(1)6f’(1)]+0(e4) I Here f(x) E Co0 and f(.1:) = -.-—-n OO__.0tn:rn Further, if f(1)= f (1): f (1) = 0, then 59 BIBLIOGRAPHY 60 BIBLIOGRAPHY [1] A. Altintas, P. H. Pathak and M. Liang, A selective Modal Scheme for the Analysis of EM Coupling Into or Radiation from Large Open-Ended Waveguides, IEEE Trans. Antennas Propagat, Vol. 36, No.1, Jan. 1988, pp. 84-96. [2] H. Ammari G. Bao and A. 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