THE UNIVERSITY OF MICHIGAN 7741-1-T Notes on Electromagnetic Scattering from Rotationally Symmetric Bodies with an Impedance Boundary Condition Technical Report No. 1 AF 04(694)-834 - - By; -.Dario Cg tellanos July 1966 Prepared for Ballistic Systems Division, AFSC Deputy for Ballistic Missile Re-entry Systems Norton Air Force Base, California 92409

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THE UNIVERSITY OF MICHIGAN 7741-1 -T FOREWORD This report was prepared by the Radiation Laboratory of the Department of Electrical Engineering of The University of Michigan under the direction of Dr. Raymond F. Goodrich, Principal Investigator and Burton A. Harrison, Contract Manager. The work was performed under Contract AF 04(694)-834, "Investigation of Re-entry Vehicle Surface Fields (SURF)". The work was administered under the direction of the Air Force Ballistic Systems Division, Norton Air Force Base, California 92409, by Major A. Aharonian BSYDF and was monitored by Mr. Henry J. Katzman of the Aerospace Corporation. The publication of this report does not constitute Air Force Approval of the report's findings or conclusions. It is published only for the exchange and stimulation of ideas. BSD Approving Authority William J. Schlerf BSYDR Contracting Officer iii

THE UNIVERSITY OF MICHIGAN 7741 -1 -T ABSTRACT An analysis is presented of the problem of the numerical computation of electromagnetic scattering from rotationally symmetric boundaries satisfying an impedance boundary condition. This work extends the work of P. Schweitzer for perfectly conducting boundaries and includes his results as a special case. iv

THE UNIVERSITY OF MICHIGAN 7741-1 -T TABLE OF CONTENTS Page Abstract iv Introduction 1 Section I The Geometry 2 Section II The Incident Electric Field and Polarization Vectors 4 Section III Decomposition of the Incident Electric Field into Cylindrical Modes 6 Section IV The Boundary Value Problem 8 Section V Reduction of the Integral Equations for Numerical Solution 11 Section VI Fourier Decomposition of the Kernel 16 Section VII Fourier Decomposition of the Current 17 Section VIII The Variational Equations 20 Section IX Evaluation of the b's 24 Section X The d-functions 32 Section XI Curvilinear Coordinates on the Scatterer 33 Section XII Explicit Formulae for the T Matrices 34 Section XIII The T' Matrix 43 Section XIV Explicit Choice of the Trial Functions 60 Section XV Classification of the Different Types of Matrix Elements 63 Section XVI Some Comments on the T'-Matrix 70 Section XVII The Case When the Surface Impedance r7 is a Function of the Arc length: rn= r(t) 75 Appendix Evaluation of T-Matrix Elements Involving Integration over a Diagonal Cell and/or Integration over a Cell Bordering a Diagonal Cell 77 v

THE UNIVERSITY OF MICHIGAN 7741-1-T INTRODUCTION This report investigates the scattering of a monochromatic electromagnetic plane wave from a rotationally symmetric object satisfying an impedance boundary condition. It is actually an extension of the work done by Paul Schweitzer at Lincoln Laboratories* and which includes his analysis as a particular case. Many sections of this report have been taken mutatis mutandis from P. Schweitzer's report. It seemed unnecessary to alter his presentation since one could hardly hope to surpass him in clarity. I am indebted to my colleagues Dr. Olov Einarsson, Dr. Vaughan Weston, and Dr. Raymond Goodrich of the Radiation Laboratory for their assistance in formulating the boundary value problem and for many helpful discussions on the physics of the impedance boundary condition. P. Schweitzer, "Electromagnetic Scattering from Rotationally Symmetric Perfect Conductors," Lincoln Laboratory, Project Report PA-88 (BMRS), February 1965.

Section I: The Geometry We consider a body whose axis of symmetry coincides with the z-axis. The body extends from z = O to z = L. The surface is given by x2 + y = f(z) More generally, one could parameterize by arc length instead of z if the surface were re-entry, i.e., if f(z) was a multiple-valued function of z. A point on the surface is given by two coordinates z and 0, where 0 < z < L, and 0 < < 2r. In the usual cylindrical coordinates, a point Y on the surface is given by A A A A r = z z+ p = f(z) cos x + f(z) sin p y + z z (1.1) where z = (0, 0, 1) = unit vector in z direction p=x +y =f(z) (1.2) p = (cos 0, sin 0, 0) = unit vector in x, y plane (1.3) we can construct, at each point on the surface, the triad n, a, t of orthogonal unit vectors A n = (cos 0 a(z), sin 0 a(z), - f'(z) a(z)) = unit outward normal vector to surface (1.4) A A A a = zx p (- sin O, cos p, 0) = unit transverse (azimuthal) vector in direction of increasing p (1.5) A A t = n x a = (f'(z) ar(z) cos 0, f'(z) ao(z) sin 0, a(z) ) = unit longitudinal vector, in direction of increasing arc length. (1.6) 2

where a(z) = = cosine of angle between t and z (1. 7) /T7[rcF(Z For these to exist, f(z) must be a piecewise - differentiable function of z The parameterization by z can be replaced by parameterization by the normalized arc length | dw1 + [f(w)2 t = t(z) = (1.8) where -sL Cj=dw I+[f'(w)] = half of transverse circumference. (1.9) As z runs from 0 to L, t runs from 0 to 1. The surface element is dz = dzdp f(z) 1 +[f'(z) 2 = C dt d f(z) (1.10) The n, a, t triad is related to x, y, z by A A x cos ~ a(z) - sin p f'(z) cr(z) cos 1 n Y l = sin z) c osp f'(z) cos a(z) sin a (1.11) z. - f'(z) ca(z) O Ca() J t n[t rcos a(z) sin0a(z) -f (Z)a(z)l ( x a | |z - sin pz cos p O | (1. 12) t f,(z):c;(z) cos J f'(z) a(z) sin j c11)2) t Z~~~~~~~

These two matrices are transposes of each other because the linear transformation is orthogonal. Section II: The Incident Electric Field and Polarization Vectors We use the convention real [... e ]. The incident radiation will be a plane wave given by (0) (r) = r e (2.1)? points in the direction of polarization, and k0 is the propagation vector. = = w= 27r (2. 2) i0 e~ = 0 (2.3) With no loss of generality, k-0 lies in the x, z plane and makes an angle 80 with the z axis 0 = (k0 sine0, 0, k0 cos 00) (2.4) For objects of rotational symmetry, there are two very special directions of incident polarization, eI and eIi, such that the directly-backscattered electric field has the same polarization as the incident electric field (i. e., no depolarization). These are given by A - eI'~k x unit vector in direction of axis of symmetry (2.5) 0 A - A EII -kO x eI (2. 6) E lies in the plane of incidence, while eI is perpendicular to the plane of incidence. More concretely 4

A A EI=y=(O, 1, 0) (2.7) EII = (-cos 80 0, sin0) (2. 8) The general case can therefore be written A A = aI II II (2. 9) Finally, the incident polarization can be decomposed along triad n, a, t at any point (z, p) on the surface of the object A A a+ aII + similar terms involving aI and eI. (2.10) aI i at] |c~xos p sin p a(z) f'(z) + = a a ] | cos sin -cos 00 co s c (z) f'(z) + sin 00 a(z) (2. 11)

Section III: Decomposition of the Incident Electric Field into Cylindrical Modes In this section, we give expressions for the decomposition of the incident electric field, evaluated on the surface of the conductor, into cylindrical modes. The incident electric field is (0) ik0. E() () ='e On the surface, i = (f(z) cos 0, f(z) sin O, z) (3.1) O = (k0 sin 0, 0, k Cos ) (3.2) Note propagation vector is, with no loss of generality, in the plane y = 0, i. e. o =0. Therefore kO0' = k0 f(z) sin O0 cos + k0 z os0 (3.3) Using the expansion ixcos E m J(X) cos m ( where E =2-6 n mO'

We can write (0) i k0zcos00 e im Jm (k0 f(z) sin 00) cos mO (3.5) r on surface. Inserting (2. 11) into (3. 5) and using the trigonometric identities cos m cos = cos (m+ 1) + cos (m - 1) (3.6) 2 cos m0 sin 0 = sin(m+ 1) - sin (m - 1)0 (3 7) 2 we obtain E)() = a I_= (z) sin mO t+ E (z) cos ma IImIt mll + aII, E (z) cos m ^t+E (z) sinm a: (0r) A) A + E() (i)nf) n, r on surface (3.8) m, IorII, torp E (z) are given by somewhat complicated expressions involving exponentials and Bessel functions. They are given in page 18 of P. Schweitzer's report and it seems pointless to duplicate them here.

Equation (3. 8) contains the desired decomposition of the incident electric field into different polarizations (aI and a; and different modes (different m). Note how, by our special choice of eI and, the various polarizations decouple so nicely. The I mode has an angular dependence (sin m0 t, cos mp a) while the II mode has an angular dependence (cos m0 t, sin m a). Section IV: The Boundary Value Problem We consider the following physical situation: We are given an electromagnetic field E ), (0) incident on a body of revolution. The total field satisfies a Leontovich boundary condition (An xE)xn = r Z n x H (4.1) on the surface of the body. r7 and Z are constants characterizing the body and the medium surrounding the body, respectively. 1 (4. 2) Z=I (4.3) The scattered electric field is given in terms of the surface fields on the body by the integral (see J. A. Stratton, Electromagnetic Theory, page 466).,s= - [iwu (n x H) G + ( x x G+(-E) VG]dS. (4.4) We have assumed that the field vectors contain the time only as a factor exp (- iw t). exp(ik0 f 1) G - G(r, F') = 4rlp T -'I (4.5) 8

The function s defined above is discontinuous across S. The nature of the discontinuity can be investigated by the methods used in potential theory, since G becomes equal to 1/4r - F'l for small values of TI-r'1. Some of the most important results are 1) If A is a vector field tangent to S, and rO is a point on S, then the integral 0 T(Er) = W (?') G (, Yt) dS' is continuous for all'. 2) x t0) x lim JSA(') x V' G(Y, r') dS' 0 =2 A(?o) + n( o) x (r') x V G(Yo0, l ]dS' S In the left hand member, the approach i — 9* is along the normal to S. In the right-hand member, the plus and minus signs correspond respectively, to an approach from the outside and from the inside of S. The integral in the right-hand member can be shown to be convergent. 3) The normal component of js(F') x V' G (, i9')dS' is continuous on S. The term E3 n(.E) VG d S S suffers a discontinuity on transition through S equal to no3 E' where A E3 is the difference of the values outside and inside. The third term in ES, therefore, does not affect the transition of the tangential component, but reduces the normal component of E to zero.

Since the total electric field E(0) + E is zero inside the scatterer, we obtain from (4.4) 0 =8x E( - lim n(i0) x t tw(n xH) G+ 0rO s +(n x E) x VG +(n E) VG]dS, where the approach is from the inside. Application of 2) gives A -(0) IA - A O n x E + n x E - nxi i i x ) x G + 2 JS +n xE) x V G + ( ) V dS (4.6) For a point just outside the surface of the scatterer n x = n x (r ) + ES).Hence A A o) ^ 0 ( n x E =n x E - lim n x x H) G+ + ( xE) xVG+(n'E) V dS where the approach is from the outside. Application of 2) gives I A A = A ) A F[ A A -2 n xE n-nx Ex I =t xH) G+ + () x x G + (n) VG] dS (4.7)

Adding Eqs. (4.6) and (4. 7) gives n x E + 2 n x [ / (n) G+ (nx E) xVG+( n E) VG]dS 2 nxE (4.8) Use of the definitions A - A ^ A n x E = K,n x H = -K, - and the relations A 1 K - rl Z x K * = i VK ii in Eq. (4.8) gives - Z nxK+Z n xG (nx) xV i_* (V.) VG]dS=- 2 n x o ifo This can be written alternatively as L2 rl ZK+ + iwu K G+ r Z (n xK) xV G + + (v) i~V -(0) = 0 (4.9) tan F on S. This system of integral equations constitutes the mathematical formulation of our boundary value problem. Section V: Reduction of the integral equations for numerical solution Following Schweitzer we will use the method of Galerkin to solve the integral equations (4. 9). We assume the (unknown) induced current is expanded in terms of the complete set fKJ' (r= g Kn(?), r onS. (5.1) n n1 11

If this expression is inserted in (4. 9), we have {1 OD n n= 1 +1 Cn iwpK(r ) G+rZ n x Kn (r'))xVG+ iwe (v.n (ir))vG]dS1 o)} =O 1L0C n tan If both sides are dotted with K (r) and integrated over the surface of the conductor, we obtain (-n Z C }dS(n (d+m(O d~~-~,gz (IdsF (~).~ Zr -X + C dSK dS' w K (r') G+ Z (n xn (') ) x VG n m n n n=JS S S + iE (v.O (( )V) dS= 0 iwE n m That is, adS (r-'j dS'[i MK (Y') G+ rZ (nxK (')h) xVG+

This is of the form oo'Y C T =-b (5.3) n mn m n=1 with T = Z dSK (rI-.K (r) mn 2 t1 d n ( m + jdSK (,|jdSI[iL0 Kn(') G + r Z (n x K (I)) x VG + E (v K (')) vG] (5.4) WE n b = J]E(0) g (r ) dS (5. 5) rn S m We wish to transform the term involving the divergence of K n. We write I- J dSK (r)' adS'(V. K (i)) VG = - jdS K (' V dS (V K (F')) G n =- SK (r.Vh(-) (5. 6) 1TS where the gradient outside the integral sign operates with respect to the unprimed coordinates. By use of a vector identity 13

I = jdS {J V K (r) h ()] -[ V.K () h (Y) Invoking the result VV = n.Vx(n xV) if AVO the first integral becomes JdsV-.[K( h( dS n Vx (n x Kh) =d V(Vx(n x K h))=O 0 Since the first contribution to I vanishes, we have left I dS [V K (i) h () =dS [V. K (i)] dS[V Kn(r')] G S S Use of this result in Eq. (5. 4) gives -mn 2 IZ dSKn(r) -K ) + i W~3 dS K ()-, dS' K (') G n nS + ie dS (VK (r) dS' (V-K (r')) G (5.7) S S Once T and b are available, Eq. (5. 3) constitutes an infinite system of simultaneous linear equations for C1, C2,

In practice one approximates R by a finite sum K (i) 1s Cn RKn (i) (5.8) n=l and solves the N x N set of truncated equations N T C -b m = 1,2,... N. (5.9) mn n m for approximate values of the C's To show the reasoning underlying the Galerkin procedure, we note that the error e () made in the right hand side by replacing the exact current in (4. 9), rewritten as rK () = - r()() tangential components (5.10) by the approximation N (r-) Ci K i(T) (. ll) is W wequi=r that the Ncst er with K (r) be zero: dS T(Y) K i(T = ~ i 1, 2,...., N. (5.13) S These N simultaneous equations are precisely Eqs. (5.9). 15

Section VI: Fourier Decomposition of the Kernel We proceed next to expand the Green's function ik ir r'[ G = G (Tr, fl)~ = 4X|e (6.1) in a Fourier series. R = tF-?I z-_ + (Zf2 +(z)]2 +f(z)]2 - 2 f(z) f(z') cos (B -') (6.2) We write ko G- r,') (Tzr1 z')cos m(G-P') (6.3) 0 4,ff zGZ z)o4s m=O so that. i k R G(z, z') =dO kR cosm (6.4) 0 with =R j(z - z)2 + [f(z2 + (z' - 2 f(z) f(z') cos 0 (6.5) G has the symmetry properties G (z, z) = G (z, z') (6.6) m -m G (z, z')=G (z', z) (6.7) m m1rn 16

Section VII: Fourier Decomposition of the Current It will be convenient to consider the currents as functions of the normalized arc length t instead of the distance z along the axis of symmetry. The transformation from z to t is one-one. We write the fourier decompositions of the current as KI(r- =2K m {tK''(t) sin m + K''(t cosm t m=O K2m -1 I P (t)sinm +K2m (t) cos m a (7. 1) m=O and -II(r) 2 mK2m -, II, t(t) sinm+2m, II, t(t) osm m=O -- 2m-1, II, 2m II.? K ( (t) sin m7n + K t) cos m.a (7.2) m=O The superscripts I and II refer to the two principal directions of polarization discussed in Sections II and III. The trial shapes introduced in Eq. (5. 1) we will assume to be of the form KII) =- III t 2m - 1, IIIK2m, t'R O= n Kn (t) sin mO + 2m, III, t 2m, II, t) cos m t + C K W co n n +~ <{c2m-lIII,OK2m-1,III, 0(t) sinm0+ m nI n n + C2m, III, K2m, m, (t) cos m 7.3) n rlos71

where III is a generic variable denoting I or II. The Km' I or II, t or are known functions, the shapes of which we will discuss in n more detail below. The C's are unknown coefficients, to be obtained by solving the simultaneous equations. For convenience in setting up the equations on the computer, we make the following simplifications: 1) We set N1 = N2 = N. That is, we use the same number N of trial functions for both the longitudinal and transverse current. This seems reasonable because in the sampling approach we would normally sample both Kt and Kp at the same points. 2) We shall insist that the trial functions Km I or II tor be independent of polarization (I or II), mode m, and the direction (t or. This is reasonable since they are usually taken to be say polynomials in t, or exponentials, ej 2 n t which do not depend on polarization, mode, or direction. 3) We shall insist that the Ki(t)'s be dimensionless. The C's will have the dimension of current per unit length, per unit electric field. 4) We shall insist that the K.'s be real. This simplifies the splitting of the quadratures for the b's and T's into real and imaginary parts. The C's will carry the complex behavior of the current. 5) We shall renumber the ordering of the trial shapes so that the t and 0 directions are adjacent. Putting this together, we are going to assume for each mode m exactly N trial shapes K.(t) i= 1,2,... N and construct a 4N-parameter trial function 18

K (ff) mC2i -sin m t + C2i cos mO~'Ki(t) KmI(i)Z CmInmI c K(t) i2i i _ 2i+2N- cos m t+C2+ 2N sinm aK(t) and K C cos mO t + C 2sin a K(t) C2i+2N - 1sin-mt+C2i+2Ncosm a1 3'K(t) (7N5) ~fI~i~l C~mII2i + 2N I' The exact shapes for the Ki(t) will be given later. For the present Eqs. (7.4) and (7. 5) completely describe the trial current. Note that C1, C3,.., C4N_ 1 describe the longitudinal current while C2,.., C4N describe the transverse current. We will also assume that 6) The Ki(t) are independent of the angle of incidence 00: the Cts will carry the dependence on 00. The reason for this is that the T matrix will then be independent of the angle of incidence, and need be calculated only once while investigating several directions of incidence. From the symmetry of the problem and the linearity of Maxwell's equations, we know that if the incident field has azimuthal dependence sin m or cos m, then the components of the induced current will have dependence sin m p and cos m. For a given mode then, the 4N-parameter trial function describing the induced current is given by Eqs. (7. 4) and (7. 5). 19

Section VIII: The Variational Equations Let rF(r,') be the dyadic Green's function which expresses the scattered electric field in terms of the surface electric current E- () ( j) = [i K(r') G(,') + r Z (n x K(r') ) x VG(r r') + +I (V K(?')) VG (', T-)]adS = (r, (= t) K(' ) dS'. (8.1) We assert that the quantity 1-B ~~i~~ 2 +2 s- o(0) [A] = 7 Z dSK +2 dSK(()- (0) ()+ + dSK(r)' dS'r (,')' K(~') (8.2) is stationary when the correct K is used. Proof:: Using the symmetry of r, we have 6 [A] jdS 6 R ().n Z K () + 2 E() (T) + +2 JIdS'r (, rI') K (')] = 0. Since the tangential components of SK may be varied freely 6[A] = 0 if and only if [7 Z R (r) + dS'r(Y, r) KR (r') + E( ) i() = 0 (8.3) tan r on S, which is precisely Eq. (4.9) 20

If the finite approximation (5. 8) is inserted into (8. 2), we obtain [A]= 2 c b + Z C T Cl (8.4) [-I n n m mn n J W n=1 n=1 m=1 If we choose C1 through CN to make [A] stationary, we get the equations a[A] N 0 = 2b + (T +T )C a C m m Z(Tmn nm n m n=l that is nm C =-b m= 1, 2,..., N. (8.5) 2 n m'' If N is sufficiently large, Eqs. (5. 9) and (8. 5) must lead to the same solution for the C's. In other words the two systems of equations must be identical T +T mn nm mn nm T (8.6) 2 mn from which it follows that T = T (8.7) mn nm I have been unable to prove this statement directly for the third term of Eq. (5. 7), but hope to return to it later on. Notice that the symmetric character of the fourth term, when written as in Eq. (5 6), is by no means obvious. -(0) If we insert (7.5) into (8. 2), but replacing E) by the incident tangential electric field for this mode-polarization, which is E III, we obtain [-mm]= nZ dS mK 2+4ds~m~ll.Emm + [A ] ={2 ZJdS Km E + 2iSKm III Em +IdS RmII () ddStr (? m id) 3m III (8.8) 21

4 m b m I+ CN Ia m 4N T. I~ =j j l2 C1 i, j i 1 where b2i -1 =JdS E (r( t sin mp Ki(t) (8.10) b2i+ 2N 1= dS ( tcos m Ki(t) (8.11) b2i dSE (d a cos mI Ki.(t) (8.12) 2i - 1 b2.+ 2N =JdS E (-).m a sin mp Kit) (8.13) with identical expressions for b but with all sin mO replaced by cos mO and viceversa T MZFdin2 K t) KWt) + i dS dS' [sin mO Ki(t) t].r. [sin mO' K.j (tt) ] (8.14) T2i -, 2j2N1 ZdS sin m0 cos m Ki(t) Kj(t) +,jdS dS' [sin mO Ki(t) ] —. [cos mp' Kj (tt) (8.15) T2i - 12j + 2N dS dS [sin mO K. (t) ]'r,'sin m' Kj (t') aJ (8. 16) mI T2i-1, 2j dS dS' [sin mO K.(t) ]'r[cos mi' K. (t') a (8.17) 22

MI 1 r7 Zl me K (t) K (t) 2i+2N -1, 2j+ 2N- 1 2 T ZJdS cos i(t) K +ji dS dS'[cos m K,(t)]. r-P[cos m' Kj(t)a (8. T2i+2N-1, 2 dS' os K(t) r- [sin m' K (tt) a' (8.19) m2i+ 2N +0 2jN i i r. Tin+ 2 1 =ldS dS [cos mp K (t) t'r'[cos m' K (t') a] (8.20) 2i+ 2N - 1, 2j =2nJJ sn c m 3t Kj Tml 1 T2i+ 2N, ZdS sin m p Kos(t) K.(t) 2i + 2N, J1 2 +J dS dS sin m K.(t) oir*cos in' K (t')'at] (8.21) TiI 1 Z dS sin2 2i+2N, 2j+2N 2 jj +JfdS dS'[sin m Ki(t) a]. r [sin mp' K. (t')'] (8.22) 3 Ti = 1 dS cos2 m K.(t) K (t) 2i, 2j 2 TZ 3 +}dS dS [Qos mp K (t) ^a} r[cos m' K. (t')'] (8.23) with similar formulas for T.. but with all sin mO replaced by cos mp and viceij versa. 23

Section IX: Evaluation of the b's The b's are given by the integrals (8. 10) - (8.13). In order to evaluate the angular integrals involved in calculating the b's, we will need the results 2,r do sin m e1-ixCos5= 0 (9.1) 27r ~i:ocosm= 2ei Jm(-x) 0do cos mo e ixCos = m J = 2r (-i) Jm(x) (9. 2) m integral. In order to calculate the integrals over t involved in evaluation of the b's it is convenient to define three sets of d-functions by d Iwr-10 ~ -I,~~ f') eikz'cos00 di (O1 _ it ik( e K (t) J (k f(z')sin 00). i02 i m0 0 d.' (0) 1/2 ~d'Th~~ ~~(0) f((z'))2 di 3 (o) a (z) i= 1,2,..., N. (9.3) These are all functions of the single angle variable 00. Using the relations sin ( - 00) = sin 00 cos (r-00)=-Cos 00 it is easy to verify that 24

dm, P( r_0) =()m + P i= 1 2,... N 1 0 p =1,2, 3 (94) so that, when evaluating the d's for a range of 0o, it is sufficient to consider only the set 00 E[0, r/2. We now possess the tools for evaluating both the angular integrals and t integrals involved in the calculation of the b's. The b's depend on the angle of incidence 0 of the incident electric field, and will be denoted by b(00). We now proceed to evaluate the b's From equation (8. 10) we have MI ()-~dSmI A b2i -l(eo) =dS U(r)' tsin mO Ki (t) (9.5) Invoking mode orthogonality, = dS Em'I (). t sin mO Ki(t) = dS e 0 t sin mO K. (t) (9.6) From (2. 7) and (1. 12) we have EI, t=f' (z) a(z) sin0 (9.7) so that (9. 6) becomes, upon use of (3. 3) i k zcose0 bMI (O = d d f(z) e f' (z) (z) b2i-1 (e0) C d t m0 0 i k0 f (z) sine0 cos s e sina sin m Ki(t) 25

Inserting sin sin mO = cos (m-1) p - cos (m+ 1) 2 and performing the angle integral via (9. 2), this becomes bin1 (O)=2rC0 1df() ek0 zcos50 f, (z) a(z) b2i 1 (0 2rC dt f e [i 1 (k f(z) sin 00) - i m+ (k f(z) sin )] Ki (t) Evaluating the t integral by (9. 3), we obtain the result mnI ( 47) [d m-1, 2 (0) dm+l, 2 (l (9.8) b2i-1 (I0)- i 0 1 i Similarly, bnmI (e ) = dS EmI o bb+2 asm I (r- tcos moK.(t) (9.9) 0i kA A dS e I *tcos m Ki (t) 27r ik zos 0 =C dt d f(z) e 0f'(z) a(z) K (t) 000 i k0 f(z) sin 00 cos 1 e (sin (m+l) - sin(m-l) ) )-= 0 where we used Eq. (9. 1). We have then the result b2i+ 2N -1 (00)3'0 (9.10) 26

Similarly, b2i+2N ) dS E a sin m0 K.(t) (9.11) i^ ^ dS e EI a sin m0 Ki(t) I'-a = cos 0 (9.12) So that ~m I = 0 27r ik ZCOS b2i+2N ( dt d0 f(z) e K.(t) 2i+ 2N 0 i k0 f(z) sin 0 cos 0 e cos 0 sin mp i k zcosO = C dt d0 f(z) e ~ K (t) 1 ik0 f (z) sin0O cos 2 e [sin(m+1) 0+sin(m-1) ] = 0 where we have again used Eq. (9. 1). We have then the result mI bM (O )MO (9.13) 2i+ 2N 0 27

Similarly, bI (o) = dS EmI A cos mp K. (t) 2i 0 = dS e ~ I.Aa cos mo K.(t) m1 27 i k zcos0 mI ) Co = 2(e) = dt f(z) e 0 0K. (t) 2i 0 d ikJm (kf(z)sin)+i J 1(kf(z)sinoS0) (9.14) b )e id'os(m- )+d os (m)+ (9-15) 2i 0ki0 i e= 2C dt f(z) e Kcos) O. 28 m-n1 (k0 f(z) sineO)+ im 1 I (k0 f(z) sineO) (9 14) or finally, 4 (e + dT + 1d (e ) ]dm+ ( (9. 15) 2i 0 i 0 i 0 Similarly using EII t = sin e0acz czcos -0 fI (c) az) Cos 0 we calculate 28

b2i+ 2N - 1 = J dS Em II. t sin m Ki (t) ik'-F A = dS e I t sin m p Ki (t) 1 2j r ik ZCS0 0 ik0f(z)sin0 cosO = C dt dO f(z) e e (sin 00 a(z) - cos 0f' (z) a(z) cos 0) sin mO Ki(t) = 0 (9.16) we have then bMH (e )=o. (9.17) 2i+ 2N -1 0 (9. 17) b2i1(00) =JdS Eos K i k0'Y A =JdSe (~ (.) cos m Ki(t) -- ik 0.r = dSe Ki( t) [sin 00 a(z) cos m - cos 0 f'(z) a(z) cos(m -1)+ cos(m+) ik z Cos 0 = 2rC dt f(z) e K. (t) - o00 f'(z) () L(z) im-l 1 in 0 ao z) i - (k f (z) sin 0) m+ (ko f() sin (9.18) 29

or finally mlI 47r Mm 3 b2i (- )-0) i W {sin &O di' (0 2i-1 i 0 -Cos[2 i (0o)+ 2 (0o)1} (9.19) Similarly, using A A (9.20) EII. a = cos 00 sin0 (9.20) we calculate mH I I m A b2i (E ) a sin mo K.(t) 2i 0 d O r\n1 =dS e EAaEII) sin mo Ki(t) =cos 00 JdSe K (t) cos(m-1) 0-cos(m+l) 0 =Cos 90 dSeo K.(t) M 1 O =cos 0 27rC dtf(z) e K 0Ki 2 m- J (k f(z) sin0) - + 1 (k f(z) sin00) (9.21) m-I 0 0 m+Al 0 or finally b (0)=cos Oi w4 [dm-' (eO) _ dm (00)] (9. 22) ~2i 0 0i Li 03 i 30

Analogously mII ( )mdSmII ^ bm0 =) dsS mA K.(t) 2i+ 2N 0 a cos 1 =jdS e 0 (]1. ) cos m Ki.(t) = cos e 0 dS e sin 0 cos mO K.(t) = 0 (9.23) and we have 2i+2N (O0) 0 (9.24) 2i+2N Our results, equations (9.8, 9.10, 9.13, 9.15, 9.17, 9.19, 9.22, and 9.24) explicitly show how the b's may be evaluated in terms of the d's. The d's are given by the one-dimensional quadratures in (9.3) and are presumed to be known. Hence, the b's can be easily calculated. Using (9.4), equations (9.8, 9.10, 9.13, 9.15, 9.17, 9.19, 9.22, and 9.24) imply that bm I (m -) - (- _m+ 1 I (0) (9.25) bm II (r-e)( _)m [bII (e)] (9. 26) 11 so that the b(e)'s need be evaluated only for 8 E [0, ir/2] 31

Section X: The d-functions The d-functions have been defined by equation (9. 3). The relation d (-d. (O) (10.1) 1 0L O has been derived, showing the need to evaluate the d's only for the argument lying in [o, 7/2] Using the relation J (x) = ()m J (x) m integral, (10.2) -m rn we deduce di' P (00) = dm' (00) (10. 3) in particular, d P(e 0) di' (00) (10.4) 1 i 0 This relation will be needed when attempting to evaluate dm with m = 0 For Ki(t) of the form to be discussed later - namely piecewise linear in t - the integrals in (9. 3) seem to be completely intractable analytically. Even for the simplest geometries of interest, cones and spheres, there has been no success in obtaining closed expressions for these integrals. Consequently these integrals will have to be done numerically - numerical quadratures, done preferably by a Gaussian quadrature scheme, should cause no difficulty since the integrand is well-behaved. ik z'cos0 The only complex quantity within the integrand in (9. 3) is the e which can be easily split into its real and imaginary parts. Hence, both the real and imaginary parts of d are calculable by one-dimensional quadratures. 32

The numerical quadratures should be fast, except the numerical generation and/or table lookup of exponentials, f(z), and Bessel functions, since Ki(t) will be a highly localized function of t. Section XI: Curvilinear Coordinates on the Scatterer We construct a set of orthogonal curvilinear coordinates which, on the surface of the body S reduces to the 0, t coordinates used so far. The three coordinates would be t1 = f (11. 1) 2= t (11.2) 3 = (in direction of outward normal to S) (11. 3) and the corresponding unit vectors (ul u2, U3 ) = (a, t ) form a right-handed triad. The surface of the conductor is given by g3 = %30 = constant. The element of arc length, using the orthogonality of the coordinate system is (dS)2 = (hI d 1) + (h2 d 2)2 + (h d 32 (11. 4) where, for 3 30 hi P P(2) (11.5) h2= C (11.6) h3 =h3 (3) (11.7) 33

If an arbitrary vector V is decomposed along the triad as V=v a + v t+ v3 n= v a+ vt t+v n (11. where v= Vi (, 2, 3), (11.9) then we can compute its divergence and curl via 1 5 vn div T = hh h2 h3 hh (11. 10) 123 n curl = h h h E u un ) (11.11) 1 2 3 Imn m 231 3 1 2 312 If v =0, (11.12) then on the surface of the conductor we obtain from (11. 10) div V av at (vt) (11.13) Section XII: Explicit formulae for the T matrices The T matrices have been defined by (8. 14 - 8.23, 8.7). Using Eq. (5.7) we have 34

mI 2 T2i-1, 2j-1 z dS sin m Ki(t) Kj(t) A A I5dS dS' (iwt) sin mO sin mo' Ki(t) Kj(t') (t t') G + r Z | dS dS' (sin mb) Ki(t) t [(n x ) x V' G] sin m' Kj (t') + i dS dS'[V' sin mO K(t)]] [' sin m' K. (t9] G (12.1) T2I1, 2j+2N-d1 = dS' (iwou) sin m cos mp' Ki(t) Kj(t') (tt') G + rl Z dS dS' (sin mp Ki (t) ) [(An x ti) xV' G]cos m0' Kj (t') + i w dS l dS dS'[ sin mO Ki (t)]] [V'.[' cos mo' Kj (t')]]G (12. 2) T2i-1, 2j+2N = iw t dS dS' sin m0 sin mp' Ki(t) Kj(t') (, a') G + rZ dS dS' (sin m Ki (t) ) [(n' x a) x V' G] sin mp' Kj (t') iwte dS dS [V{. sin mO Ki (t)]] LVT [ sin m,' Kj (t'i)] G (12.3) 35

T2i-1 2j dS' [sin mO K. (t) cos mo' Kj (t') (t. a') G + T Z 9idS dS' (sin mO K. (t) ) t.[(' x a') x V' G]cos mo' Kj (t') + ie0 ~{dS dS' V sin m Ki(t)] [. L' cos mo' K (t') G (12.4) TmI dS cos1 2 m K (t) K. 2i+2n-1, 2j+2N-1 2 dS cos + iWs fTdS dS' cos mO cos mo' Ki (t) Kj (t') (') G + r1 ZidS dS' cos mO Ki (t) t HI' x x') xV' G] cos m#' Kj (t') + i IdS dS' IV- (cos m Ki (t)W)] [V'-(cos mo' Kj (t') t')]G (12.5) T2i+2N1 2j+2N - i dS dS' cos mO sin mo' Ki (t) Kj (t'),(t aa') G 2i+2N-1, 2j+2N JJ 1 J + n7 Z~ dS dS' cos mO Ki (t) [(n' x a') x V' G] sin mo' Kj (t') + iw dS dS' [V'(cos m1 Ki (t) t )] -[V'(sin m' K (t') a')] G (12.6) 36

Tm2i+2N-1, 2j i dS dS' cos m cos mp' Ki (t) K. (t') (- aa) G A F~A1 + Tr Z ddS dS mO K.' t) t n'a xV' G cos mK' Kj (t') + i dS dS' [V cos m Ki (t)t V' Os MO' K. (t)'a G (12.7) T2+2N 2j = iC dS dS' sin mp cos m'' K (t) K. (t') (a a') JJ + n Z dS dS' sinmK (t) a n' x a') xV' G] cos m' Kj (t' ) +1 [edS dS' [V in m (t) ]] V [os mt Kj(t') ]G (12.8) mI 1 2 T2i+2N, 2j+2N 2 7Z dS sin m Ki (t) K. (t) + iwjdS dS' sin mO sin mp' Ki (t) Kj (t')a. aa') G + r) Z dS dS' sin mp Ki (t) a [(')x V' G] sin m0' Kj (t') + i~ + iw dSdS dS' V [sin m Ki (t) ]] K'. sin mi' Kj (t') aa] G (12.9) 37

T2i 2j = Z I dS cos m2 K. (t) K. (t) 2i, 2j 2'J1 J + iWAu dS dSt cos mO cos mP' Ki (t) K. (t') (a. a') JJ3 + O Zi dS dSI cos mP K. (t). An' x a') x V' G] cos ms' K. (t') + iWe JJ dS dS' V. (cos mO Ki (t)J [V'.(cos mo' Kj (t')' G (12. 10) Since SdS = C dt f(z) 0do. (12.11) 0 O terms of the form dS sin mO cos mO Ki (t) Kj (t) integrate to zero, and have not been explictly included above. Equations for T I ij follow from those for T. I by replacing all sin m0 by cos m0 and viceversa. We now proceed to simplify these expressions. Let us first note that when t = 0, Eq. (4. 9) reduces to the well known boundary condition for a perfect conductor. The T-matrix elements (12.1 - 12. 10) are made up of terms independent of rl and terms containing r; as a factor. It follows then that the part independent of rl should correspond to the T-matrix elements for the perfectly conducting case. These Tmatrix elements have been calculated explicitly by P. Schweitzer. (See Schweitzer's report, page 68). We shall write here then for future reference: 38

Tmi, p. iw jjdS dS' G sinmO sin mp' (.tt) 2i-1, 2j-1 Ki (t) K. (t') + we jdS dS' G div[ K. (t) sin mO div[t' Kj (t') sin m0']. (12.12) mI, p.C. A A T2i1 2j = i dS dS' G sin m cos m' (t a') Ki (t) Kj (t') 2i-1, 2j JJ + iwe dS dS' G divt Ki (t) sin mp] div a' K. (t') cos m] (12. 13) Ti, 2P = iuSJ dS dS' G cos mp sin mO' (a.t') K. (t) K. (t') 2i, 2j-1 1 j +.' JdS dS' G div[a K. (t) cos mp] div t' K. (t') sin m'] (12. 14) mI, p.C. COS MO.(.A _ A T 2jP. = iwJ dS dS' G cos mcos m' (a a') K. (t) K. (t') T 1 + -i e d dS dS' G div[a K (t) cos m] div a' K (t) cos m'] (12. 15) Similar expressions hold for T i but with all sin mp replaced by cos mO, and viceversa. The superscript "p. c. " denotes "perfect conductor". Use of these results in Eqs. (12.1 - 12.10) gives 39

MI 1 F 2 T i - r Z dS sin m K. (t) K. (t) 2i-1, 2j-1 2 1 J + Z7 Z dS dS' sin mO Ki (t) I i'n' x t') x V' G] sin mo' K. (t') +TmI, p.c. 2i-1, 2j-1 (12.16) T2-1, 2j+2N-1 r7 Z dS dS' (sin mO Ki (t)) t(n' x t') x V' G] cos mp' K. (t') (12.17) The other two terms in (12.17) are identically zero. They are independent of qt and they do not occur in the perfectly conducting case, hence they must vanish identically. This can be checked by direct calculation. 2i-1, 2j+2N S Z=asas (sa mK K (t) n' x') x V' sin mO' K. (t') (12.18) T2i-1 2j = r zj dS dS' (sin mO Ki (Kt))[(' x a') x V' G 2i-1, 2j cos0' K ( TmIs Pc (12.19) 2i+2N-1, 2+2N 1 2 Z dS cos2 m K. (t) K. (t) 2i+2N-1, 2j+2N-1 2 1 J + rl ZIdS dS' cos mp Ki (t) t [(n' x') x V' G] cos m' K. (t')+ Tm IIL p c. i j 2i-1, 2j-1 (12. 20) 40

2i+2N-1 2j+2N r ZjdS S cos m t) S {n' x a') x V' G] sin m0' K. (t') + TMi l, P. c (12.21) sin m 2i-.1, 2j T 2.I2N-1. = 9 r ZjjadS dS' cos m K. (t)'x x V cos m K (t') (12.22) T 2i2N 2j Zds dS' sin m K. (t) a[(' x')x V' G]cos mP' K (t') 2i+-2N, 2j K (12.23) MI 1 2.2 TmI+ =-r1 z dS sin mO K. (t) K. (t) 2i+2N, 2j+2N 2 J J + a ZJdS dS' sin m i (t) x') x V'G] sin m K. (t') + TmII, P.C K1x x sin K 2i, 2j (12.24) ml _ 1 2 T dS cos m0 K. (t) K. (t) 2iTi 2j 2 1 C + rl Z dS dS' cosm m Ki (t a [(') x V' G] cos mp' K. (t') + TmI, p. c. (12.25) 2i, 2j Notice that the T matrix is of the form I, p. Co ITI = + r TI' (12. 6) 0 TmII, p.c. where T arises from the impedance boundary condition. 41

Similarly TII, p.. 0 T= | |+ r TI (12.27) 0 TI, p.C. As pointed out by Schweitzer TmI, p.c. ( )i+j mII, p.c.2.28) ij ii 1] -1.. =. Im>,, IP.C. m>1 (12.29) so that only T P. C. has to be calculated. For m = 0 TI,' P-C. = 0 unless i and j are both even, 1J and 0n1, p.c. T' = 0unless i and j are both odd. Note also that for either polarization the b-column matrix is of the form b2 b b2N (12.30) 0 42

We now proceed to obtain explicit expressions for the elements of T' Section XIII: The T' Matrix We will write down here the expressions for the elements of the T matrix. MI, 1 I T2i-1, 2j Z dS sin mp K. (t) Kj (t) + Z TdS dS' sin mO K. (t) t.(' tt) x V' G] sin mo' K. (t') (13.1) Ti = Z(dS dS sin mO K. (t) t nt x at) x V' G]cos mo' K. (t') 2i-1, 2j ZJdSsi mK(t)'. x' J (13. 2) Tmi' j z 2ZdS cos m K (t) K (t) 2i, 2j 2 1 j + Z JdS dS' cos mO K. (t) a [(n' x a') x V' ] cos mp' K.j (t) (13.3) 2i-1, 2j+2N-1 T 2i-1 2j+2N- = Z dS dS' (sin mO K (t) ) [(n'x Xt') x V' G] cos m.' K. (t') (13.4) 2i - 1, 2j+2N ZJidS dSi (sin m K (t))' x') x V' sin mp' Kj (t') (13.5) 2i+2N, 2j = 1 GZcos (13. 6) 43

TmI = Z1 dS cos mO K. (t) K. (t) 2i+2N-1, 2j+2N-1 2 1 j + ZJTdS dS' cos mO K. (t) t n' x t') x V' G] cos mp' K. (t') (13. 7) 2i+2N-1, 2j+2N Z dS dS cos m K (t) t I a') x V'G]sin mp' Kj (t') (13. 8) T2i+2N-1 2j Z FdS dS' cos mO K. (t)' rn' x V' G]cos mo'Kj(t') (13. 9) 2i+2N, 2j+2N =2 ZdS sin m K. (t) K (t) + Z dS dS' sinm K. (t) a xa') x V G] sin m' Kj (t') (13. 10) miu' The expressions for T.. are given by Eqs. (13. 1) - (13. 10) but replacing all sin m0 by cos mO and viceversa. We have JdS = C dt f (z) d so that dS sin2 m 0 K. (t) K. (t) = C dt f (z) K.(t) K. (t) 2 1 r1 d0 sin m0 =C 7r dt f (z) Ki (t) K. (t) 441 J ~~~~~(13.11) 44

and dS cos m K. (t) K. (t) =C~r dt f (z) K. (t) K (t) (13. 12) Also since t= nxa A x ^ and nxt=- a we have Tm' =- CZ dt f (z) K. (t) K. (t) 2i-1, 2j-1 2 J - ZJTdS dS' sin m0 K. (t) t a' x V' G sin mo' K. (t') (13. 13) Ji T2i-1, 2j Z dS dS' sin mp K (t) [' xV' G cos mp' K. (t') (13. 14) mi' _ 1 T2i, 2j C Zir dt f (z) K. (t) K. (t) 2i, 2j 2 1 j + Z-SdS dS' cos mm Ki (t) () t' x VV' G cos m'' Kj (t') T2i-1 2j+2N-1=: Z dS dS' (sin mp K (t) )t[a1 x' G] cos mp' Kj (t') 2i-1, 2j+2N-1 1 J (13. 16) 2i-1, 2j+2N 1 (13. 17) 45

ImI' T2i+2N 2j = Z dS dS' sin m K (t) [ x V G cos m K (t) (13. 18) T IN Z C7r dt f (z) K (t) K. (t) 2i+2N-1j 2j+2N-1 2 - ZIdS dS' cos Ki (t) i P x G] cos m' K. (t') (13. 19) T2n 1 = ZJsdS dS' cos mO Ki (t) t' x V' G] sin mO' K. (t') 2i+2N-1, 2j+2N J L (13. 20) mI' 2i+2N-1, 2j j1 LJ (13.21) T 2i+2N 2j+2N 2 C Z 7r dt f (z) Ki (t) K. (t) 2i+2N, 2j+2N 2 1 J + Z dS dS' sin mO K. (t) ~a It' xV' G] sin m~' Kj (t') II 1 J (13. 22) The gradient of a scalar function in curvilinear coordinates is U1 af U2 af u3 aj Vf = + (13.23) h a5 h2 a~2 113 a3 In our case we have (u1, u2, u3) (a,t, n) and'1=6 2 =t 3 = (3 (in direction of outward normal) 46

Hence A A VG a aG +G nQE aG VG +.. p a5 c at h3 a53 A A A n aG a 3G t x VG =+ (13.24) o ao h3 a3 A n aG a OG a x VG = C - (13.25) c at h3 a3 We also have ttt = a (z) a (z') + f () f (z') cos (p - (13.26) t a' = a (z) f' (z) sin ( - 0') (13.27) AA a att = cos (p - p) (13. 28) t n' = a ( z) t (z') () os (-')- f (z') (13.29) a = - (z) sin') (13. 30) These results follow easily from Eq. (1. 12). Hence ^ [a I ] t.V' aG ttl' aG - C at - I3 a h3 a~3 1C (z) a (z,)[f' (z) cos (0 p t)- f (z)] aI G - a (z) a (z') i f+ (z) f (z') os ( - (13.31) 3 a 3 47

A r.' xAV' _t' a-'P + t- aG = -- a (x (z[f' (z) cos( VI0P -) Off U% )a h3 a h3 1 (Z) () I (Z) Cos V- (Z ap' a=+ z1, =.| aZ ntJ a41 RG r a (z) f (z) sin ( - ) (13 32) h3 at3 a _n aG + a. a. aM- I x VI ml, h3 a~3 aG 1 aG = ta(zt) sin(O -,) T? (13.33) h3 a 3 It is convenient at this point to introduce the scalar function G (z, z') defined by 2 2r i ko R G (z, z') = d o k R eim(c') (13. 34) m 47 ko' R where R= I-rl =i(zZ-) + [f(z)]2 + [2f(z') f(z) fs(z ) cO* ((1 ) (13. 35) More simply, ikoR G (z z d e cos me(13.36) where H =4Iz_,,'2 + [f(z)]2 + [f(z;)f -2 f(z) f(z,) eos e (13.37) 48

Equation (13. 36) is seen to be identical with the Fourier coefficient (6. 4) defined in Section VI. The angular integrals occurring in the T' matrix elements can now be performed dj d d' G cos m0 cos mot = 1 + 6m] 2 (13. 38) jdo do' G cos mO cos mp' cos ( -.') = K1+m 4 Lam- +] (13.39) lido d1' G cos sin m sin m sin ( -')= ko r 1 + 6mOj.Gm+ - G (13.40) fd do' G sin in mp = [1 -6 mO 2 (13. 41) jdo do' G sin mp sin m' cos ( -') = - 6 ] G +- ] (13. 42) d| do' G sin m cos m' sin (0 - ) = L - 6mO Gm + G (13.43) Eqs. (13.38) - (13. 43) have been quoted from Schweitzer. Let us consider the expression 2 2r 2ir ik I d=_ jd' r a e O4 R eim') (13.44) 49

Integrating by parts we have,27r ik0R 2r 0 27r r2r i k0R imJ e a im(p -') 4 0R 7 J27r ek0R i m do dot eikgR eim(O -t') =imG (z,z') m that is jOd'o 4d kR ] ei d = im G (z, z') (13.45) Also 73r r2 r i k R;d:: dU[ a, 4ekoR] er- =, G (z, z') (13.46) and 27r r2i k0 R at'I 47rk R atm we have then from Eqs. (13. 13) - (13. 22) 50

m I' 1 T C Z IT dt f (z) Ki (t) K. (t) 2i-1, 2j-1 2 1'41 C Z K [1 - 6 mdtddt'f (z) f (z) a (z) a (z') f' (z) O O K. (t) K. (t') a t G +m+ ii 1 JCtL Z-i [f+ljj 1 at'J + C Z k 1 -6 Idt dt' f (z) f (z') f' (z) (z) (z) K. (t) K (t') a5 G ] + i C 2Z k I - m dt dt' f (z) f (z') a (z) a (zl) — ( Ki (t) Kj (t1) 4G8 1 51 3 3

mI' _ 2 0 2i-1, 2j 4 l O -6 O dt dt' f (z) f (z) a (z) (z) (z) K. (t) K. (t') j 2 k k [(m+l) Gm+l (m-1) GmL+ZC2m- k [1 6m Fd1 1 dt dt'f (z) f (z') C (z) (z') () (z) K (t) K (t) G ( 1 Z C- ~4 -1mO dt dt' [f(z) f (z') a (z) f (z) Ki (t) Kj (t') Gml +Gm+l] ] (13.49) 3 m~a It 1 C~n a Tm; Mj I C Z 7r dt f (z) Ki (t) K. (t) 2i, 2j C K ( 0 4 p 1 [1 + 6dO[(m+1) Gm+l - (m-) Gm_]] + 2 + 6 | dt it [f(z) f(z') + K. (t) K(t') ~) [G ~ +Gm+l] (13. 50)

T2i-1 2j+2N-1 = dS dS (sin m K. (t) ) t'I' x V' Gcos mo' Kj (t').[~.,a x V a(Z) a] = (z) ~ (z)(Z)] at I, a(z) a(z) [1+ f'(z) f'(z') cos ( -)] I h3 a3 mI' T MI -o (13.51) 2i-1, 2j+2N-1 Ti 2+2N = Z dS dS (sin mo K. (t))'t t' x V' GI sin mO' K. (t') 2[t tV G] 1 ) +\ p a' p +, (Z) f, (z) sin (O -') aG I I h3 8~3 TI'T -0 (13. 52) 2i-1, 2j+2N T 2i+2N = 2j Z dS dS' sin mO Ki (t) A.[t' G] cos mjP Kj (t') A I aG 1 aG a' x V' I = t (Z') sin (O - 0') a + 1,) a h3 a 53 mI' T 0 (13. 53) 2i+2N, 2j (13.53) 53

T2i+2N, 2j+2N1 = 2 ZC dt f (z) K. (t) K. (t) 2i+2N-1, 2j+2N-1 21 - Z C - [1 + dt Ldt' i (t) K. (t') a (z) c (z') f' (z) 4 M o (z) ( f (z' ) - [ G+ + z c + 6 dt dt' f (z) f (z') K. (t) K. (t') 2 m1 j +ZC. 1 + 6 dt' f C(z) f(z') K. (t) K. (t') 2 MO J 1 I- (z% ) a(z') G I h3 5 3 _ k 1 14 + Z C2[11+ 6 J-' dt dt' f(z) f (z') K. (t) K. (t') a (z) a (z) P (z) f' (z) a G + G (13.54) 54 m-1 G+I

mI, 2 1 2 =Z C k l + 6 O|dt dt f (z) f (zl) + (z) a (zt) f' (z) K. (t) Kj (t') [m+l) G 1+ (m-1) m2'11 - Z C m 1 + ] dt dt f(z) f(z), (z) a (zC) ft (z') i j m k11 + zC l1 + 6 J dt dt' f (z) f (zt) 1 () f () K. (t) K. (t') [G+l - G 1 (13.55) ~1 J~a T2mi+2N1 2j ZidS dS' cos Mr Ki (t)' x Vt G]cos mot K. (t') h3 aG3 m -' 2i+2N-1, 2j (13.56) 55

mI' 1 d K T2i+2N, 2j+2N C Z dt f (z) (t) K. (t) 2i+2N, 2j+2N 2 J J + Z C2 k - 6m t dt' f (z) f (z') K. (t) K. (t') a (z) 0 4 JUL1 j P' (m+l) G - (m-l) G ] + Z C k0 [ - 6 dt dt' f (z) f(z') Ki (t) Kj (t') 1 J h3 ae [Gm-+G1] (13.57) In view of equations (13.51, 13.52, 13.53, 13.56) and Eq. (8.7), we see that the T' matrix is of the form TI o TA TI = (13.58) 0 B and TII' 0 TI' 0 A B T = (13.59) O TII_' | | O TI' 0 T B 0 TA What we have in fact shown is that the problem admits of a lower dimensional representation than that assumed by Eqs. (7.4) and (7. 5). In other words we have K(?) 2 sin m +2 cos m aKi (t) (13. 60) 56

and K f ( C~i-Cos MTA+ C2i lsinm a}Ki (t) (13.61) i 2i-1 1 cs 2i A direct proof of this would be desirable. For the T matrix we have TI= TI, p.c. + TI' (13.62) T = TII, Pc. + TII' (13.63) and ( T2i-1, 2j-1 = C Z dt f (z) Ki (t) Kj (t) 2i-1 2j —1 2 J Ki (t) K (t') a-Z (GM_1 + G+l) (13.64) M'I 2 ko T 11C61 C - 6 dt dt' {f (z) a (z) a (zt) 2i-1, 2j 4 L M 3 [f (z) [(m+1) G 1+ + (m-l) G 1] - 2' (z') G m + f(z) f'(z) f' (z') Ca (Gm1 Gm+l Ki (t) Kj (t (13.65) 57

mIl' 1 T 2j C Z tr dt f (z) K. (t) K. (t) 2i, 2j 2 1 J 0 2 k 0F + c2 Z- i + 6 d dt, (z) K (t) Kj(t,) (z) [(m+l) Gm+1 - (m-1) G - f (z') ft (z) a (z').az' m ra a zlz Gm_ + G+ (13.66) muH' 1 5'r 2i-1, 2j-1 2 C Z ir dt f (z) Ki (t) K. (t) 2i- 2j-1 21 0 - c2 z [1 + 6 o dt dt' f (z) f (z) f' (z) a (z) 4 1O Ki (t) Kj (t') a — (G +G ) (13. 67) az 1 2i-1,j C 02 Z - [ + 6m dt dt' {f (z) C (z) a (z') 2i-lj 2j [f (z)[(m+l) Gm+ + (m-l)G 1]- 2 f' (z') G m +f (z') f (z) f' (z') a (G 1 - Gm+l) Ki (t) Kj (t (13.68) 58

1 T In 2 C Z 7r dt f (z) K. (t) K. (t) 2i, 2j 21 C 0 + c 2 L m O- ] dt dt [f (z) a (z') K. (t) K (tl) 4 MO1 {(m+l) G 1 (m-1) G -f(z') f' (z') a(G (13 69) am+ )}1 a zl Yn1 + G(13. 69) Going from Eqs. (13. 48) - (13. 57) to Eqs. (13. 64) - (13. 69), we made use of the results p' = f (z') (13. 70) 2 1 a(z)2 and a= C tV' (13.71) at' a_ = h3 ~V' (13.72) 3 a From Eqs. (1. 12) we have t - f' (z') a (z') cos' x + ft (z') a (z) sin 1' P + a (z') ^ and A cos (z) + sin' a (z) +(-fl (za) a(z')) A this together with h a ^ a ^ a V' x +y —-+z - x a +x' ay' az, and the fact that Gm = G (zJ z'), gives the results 59

aG aG =C c (z) (13.73) at' az' aG aG m rn = -h3 f' (z') )aZ (13.74) a 3 IllI III the b-column matrix is now, of course, made up of the elements b through b 1 uhb2N only. Section XIV: Explicit Choice of the Trial Functions The usual choice of the trial functions Ki(t) in a variational principle is of shapes closely related to the expected shape of the current; in this way only a few shapes are needed and N will be small. The unfortunate part is that each of the matrix elements involves a laborious quadrature over the entire unit square dt dt' These quadratures require an unacceptable amount of time. If N —'20, the T's will 40 x 41 be 40 x 40 symmetric matrices. 2 = 820 of the matrix elements would have to be evaluated for each mode m. We conclude that global trial functions are unacceptable, since the T quadratures are too time-consuming. For this reason we shall pick localized trial functions so that the integrals go, not over the entire unit square, but over a small subportion. 60

A convenient choice of trial function is obtained by approximating the KmIIIt(t) and KmIHI (t) by piecewise-linear functions of t. We break the unit interval into N points, 0 = t t2 <... < tN = 1, and sample the current at these N points. We define mn III m III t C KI (t.) i=1, 2., N (14.1) 2i-1 j CmIII KmIII (t) i = 1, 2,. N (14.2) 2i 1 = sampled values of the current. and draw straight lines connecting these sampled values. The choice of piecewise-linear trial current is equivalent to the statement mIIIt m III K (t) C2i_1 K (t) (14.3) where t -t t -t ~ = tl < t < t2 K1 (t) f |2 t(14.5) O elsewhere t - t. 1 1-1 ti+ -t< t ti i+ - i 1 +1 O elsewhere 2<i N-1 (14.6)

tN tNi tN_1< t tN = 1 KN (t)= 0 elsewhere (14.7) KI(t) t tl t2 tN=l Ki(t) 1, 1 i_1 Nt ti-l ti ti+l KN(t) t tN1 l-t6

This choice of trial function has the following convenient properties: 1. Most importantly, the K's are localized; the integrals for the T's are only over a small portion of the unit square. 2. The K. (t) are simple to generate and to differentiate. 3. The piecewise linear approximation is a good one: If one wished to approximate a function with 10 maxima a 30th degree polynomial and a 30th degree linear fit are probably equally good - - while the T's are much easier to generate in the latter case. 4. If the piecewise - linear approximation is used the C's have a very simple interpretation via (14. 1) and (14. 2) - - they are the sampled values of the current (and not merely the coefficient of Ki (t) in the sum (14. 3) and (14.4)). Indeed a print-out of the C's, once the variational equations are solved, is precisely a printout of the sampled values of the current, these samplings being carried out at t 1 t2 1'. tN Section XV: Classification of the Different Types of Matrix Elements We define the cell C (i, j) as the rectangular region (t, t') with ti < t <ti+l (15. 1) t t' < tj+l (15.2) Most of the T integrals go over 4 cells. For example T59 goes over the 4 cells C(3,5). C(3,6), C(4,5) and C(4,6). However since KI(t) and KN (t), are special, some of the T's involve integrals over only 2 cells. For example T13 involves integrals over C( 1, 1) and C (1, 2). Finally T1l, T12 T21 and T22 involve integrals over the single cell C (1, 1) whereas T2N-1 2N-1 T2N-1 2N T and T involve integrals over the single cell C(N-1, N-l) 2N, 2N-1 2N, 2N Thus the T.. can be classified by whether they involve integrals over 1, 2, or 4 cells. 63

We next distinguish between integrals which go over non-diagonal cells C(i, j) i i j and integrals which go over diagonal cells C(i. i) The 3 T's listed above which go solely over the cell C(1, 1) involve purely diagonal cells. In addition, diagonal and almost diagonal terms like T33, T34~ T43, and T44 involve integrals over 4 cells, two of which are diagonal (here C(1, 1) and C(2, 2) ) and two of which one non-diagonal (here C(1, 2) and C (2, 1)). Some integrals, say for T13, go over two cells, one of which is diagonal and the other of which is non-diagonal. Some integrals, say for T53, go over 4 cells (here C(1, 2), C (1, 3), C (2, 2), C (2, 3)), one of which is diagonal and 3 of which are non-diagonal. And finally most of the T's involve integrals over 4 non-diagonal cells. Thus the Tij can be classified by whether the cells involved are diagonal or nondiagonal. Putting these classifications together, we find that for N > 4 there are exactly 10 types of T's, which we shall denote as types A, B, C, D, E, F, G, H, I, J. These 10 types are shown in the next figure for the case N = 9, where T is an 18 x 18 matrix. Since T is symmetric, only the elements below or on the main diagonal need be shown. The classification is given in Table I. We assume that N > 4. 64

81 x 81 sT J, Ji sluauIalq xallyI Jo saaOsX, uaj, aq,IL I?~ ~0 DO D O D O 0 0 0 0 U ( TI |f f (3 I2I 2 0I 0 0a 00 0a 1 a aU Hr r 3sa a a a a a a aa a a a Ha a a a a a a a a a a \1a 2 a 2a a a a a a a D a a a a a a a a a a a O H n s s.a a a \H H ~{: 2H O OE ~lc~i c~ c~ c~I 3 U

TABLE I The Ten Types of T Matrix Elements when N > 4. Description of Number of Cells Cell Cell Type Matrix Elements Integrated Over Numbering Diagonal? ~A (T 21 1 C(1, 1) yes T32 C (1, 1) yes B 2 T41 C (2, 1) no 42 2i-C(i-1 1) no 2i-1, 2 C 2 T2i 1 C(i, 1) no 2i, 1 3 < i ( N-1 2N-1, 2 T2N, 1 2N, 2 T 2i-1, 2i-3 C (i-1, i-2) no T nC(i-1 i-2) no 2i-1, 2i-3 E T2i-1, 2i-2 C(i-, i-) yes 2i, 2i-3 C(i, i-2) no 2i,2-2 C (i, i-l) no 3,<i < N-1 66

Description of Number of Cells Cell Cell Type Matrix Elements Integrated Over Numbering Diagonal? Description of Number of Cells Cell Cell Type Matrix Elements Integrated Over Numbering Diagonal? 2i-12j1 C(i-1,j-1) no 2i-1, 2j C(i-1, j) no F 2i-1, 2j-1C(i, ) no _ 2i, 2j 9 C(i, j) no 4Ti- 2N-j F 4 T2i, 2j-1 T2N C(N-i, j1) no 2i, 2j 2 4~ i NN-1 2N-l, 2j-i 2N -1, 2j G 2 T2N 2j-1C (N-l, j) no T2N. 2j 2 < j N-2 2i,2i1 0C(i,- i-) yes T2. 2-C (i-i, i) no H2i, 2i-1 H I 4 T2 2iC(i, i-) no 2( i N-1 C(i,i) yes 2N-1, 2N-1 I T 2N- 1 C (N-l, N-1) yes 2N, 2N 2N-1, 2N-3 T2-, C (N-1, N-2) no 2N-i, 2N-2 T2N 2N3 |C(N-l, N-l) yes 2N, 2N-2 67

Table I must be kept in mind while programming the computer to calculate the matrix elements of T. Diagonal Cells and Cells Bordering a Diagonal Cell The T-matrix elements of types A, B, E, H, I and J involve one or more integrals over diagonal cells. These integrals must be done analytically, due to the integrable singularity in the G. Also T-matrix elements involving integrals over cells bordering a diagonal cell require special treatment since the integrand becomes singular on one of the borders of the cell. This analysis has been carried through in detail by the author for the elements of T P c. (See Appendix "'Evaluation of. T-matrix elements involving integration over a diagonal cell and/or integration over a cell bordering a diagonal cell!'. ) The T -matrix elements involve integrals of the form ti+l i+ I dt Jfdtt' F1 t') F2 (t) Gm(z, Z') (15.3) 1 1 and Cti+1 i1+2 dt dt' Ft') F2(t' ) F Gm (z,) z(15.4) ti ti+l Remembering the equation t=t(z) dw 1+ [f'(w)] 68

and a 11 a 1 + [f' ] at we see that (15. 3) and (15.4) can be put in the form ti+1 tI~ dt tdt F1 (t') F2(t) a G (t, t') (15.5) t at' I 1 and ti+1 i+2 dt dt' F1 (t') F2 (t) at Gm (t, t') (15.6) where F (t) and F2 (t) are different from those of (15. 3) and (15. 4). Eqs. (15. 5) and (15. 6) can be reduced to the form considered in the Appendix by an integration by parts. We have | dt J dt' F1 (t ) F2 (t)- G m (t') i i t.ll tt. 1 1'L+ i+lt dtF (t) F ( ) m(t ti+) - dt F2 (t) Fl (t) G (t, t) +1 dt dt' F2 (t) FI (t') Gm (t, t') (15.7) The first two integrals in the right hand side of (15. 7) can be handled as Eqs. (53) -(55) of the Appendix. The third integral is of the form of Eq. (1) of the Appendix. An identical analysis is applicable to Eq. (15. 4). 69

Section XVI: Some Comments on the T'-matrix For m >1, we compare (13. 64) with (13. 67), (13. 65) with (13.68), and (13. 66) with (13. 69) to conclude TmI' i+j mII' This result also holds for TM p c. as noted in Eq. (12. 28). Hence for the elements of the T-matrix (13. 62) or (13. 63) one has TmI i+j TmII Ti.. T m>l (16.2) ij ij As indicated by Schweitzer, this result could have been anticipated by noting that (8.14) and (8. 18), (8. 17) and (8. 19), (8. 23) and (8. 22) differ only by the replacement of sin mO by cos mO and viceversa. This replacement could be achieved by the substitution, for m.> 1 =0 2- - 0J" which sends sin mO into cos mo", cos mo into sin mo". The same transformation (which is a shift in the angle origin plus a reflection through the plane of symmetry) sends (neglecting the shift in origin) a into - a and t into t. This explains why the T and T which involve a. r- a and t-r-t maintain 2i-1, 2j 2i,-1 2j-1 their sign, whereas T2i-1 2j and T2i 2j-1 which involve /.na and a-r, undergo a change in sign. From Eqs. (13. 64) - (13. 69) we have when m = 0 OI' 1 T2i-1, 2j-1 C Z dt f (z) Ki (t) Kj (t) (16.3) 2i-1, 2j-1 2 J J ToI' T = 0 (16.4) 2i-1, 2j 0I' 1 T C Z ir dt f (z) K. (t) K. (t) +C2 Zk0 dtCdt'ff(z) K.(t)K. (t') a(z')[G -f(z') f' (z')3G1]) (16.5) 70

T2i-1, 2j-1 C Z dt f (z) Ki (t) Kj (t) 2i-1, 2j-1 2 j - 2 Z k0 dt dt'[f (z) f (z') f' (z) a (z) Ki (t) Kj (t') z, G1 (16. 6) T =II (16.7) 2i-1, 2j 1 T2i 2j- C Z -r dt f (z) K. (t) K. (t) (16. 8) 2i.,2j 2 1 We notice that the only elements different from zero are T0H' T0I' T 0' T Oi = T T, and T 0(16. 9) 2i-1, 2j-1 2i, 2j' 2i. 2j 2i-1. 2-19) Hence for m = 0 we have to store only one matrix TO whose entries are'01 _ I' = =T (16. 10) 2i-1, 2j-1 T2i-1, 2j-1 T =T (16.11) 2i, 2j 2i, 2j ~01 ~01'0III T =T =T (16. 12) 2i-1, 2j = T2j, 2i-1 2i-1, 2j-1 from which both T and T may be recovered 71

T.. if both i and j are odd T.. = TO. if both i and j are even (16.13) i jj o if i odd and j even or viceversa T.O j+1 if both i and j are odd T II'I') t Ti3 i-1j T if both i and j are even (16.14) O if i odd and j even or viceversa. Similarly, the result T =. = ( T.. for m > 1 allows a great simplification in storage, since only the T's for the II polarization need be put on tape - - to be later withdrawn and inverted. For the perfect conductor we have the results T.p. c. = 0 unless i and j are both even ii _II, p. c. and T.' = 0 unless i and j are both odd. 0, I, p. C. TOII, p.c Hence for m = O we store on tape the matrix elements T 2i 2 and T2i1, 2j1 2i, 2j 2i-1, 2j-1 (which are the only matrix elements of TO I p.c. and T II p. c. which are non-zero). That is, we construct the matrix TOI, P* c* if i and j are both even T.. (16.15) P~ c0{,j p. c..II pc.' if i and j are both odd 72

h 0, P.C. a T0II, p. C. from which TOI' p-c. and T may be obtained 0T. if i and j are both even ij TOI, p.c. =4 TI. (16.16) i1 O otherwise 1jQ p' c iT if i and j are both odd 1J T p.c.(16.17) ii 0 otherwise In this fashion we can store on tape, for each m, a single matrix T from which both T and T may be recovered. More formally, for each m we shall compute a symmetric matrix T.m iJ T.. T (16. 18) Tm Tp.c. + c' (16. 19) m11 mH, m c m' T. Tm. + r1 T m > 1 (16. 20) 1J iJ iJ from which both TmI and T may be recovered by T0' p.c. + T. if both i and j are odd i jij+1 T T j-1 if both i and j are even 0 if i odd and j even or viceversa T.. 1J 73

and p.~. C.+ ^0' +r T. if both i and j are even ij ij mI.7 T if both i and j are odd T ij 0 if i odd and j even or viceversa (i+j T m l. 1j The result (16. 2) is extremely convenient for two reasons. We have already discussed how it halves, for m 1, the generation and storage of the T matrices for the two polarizations. Second, (16. 2) also implies that a single inversion for each m > 1, gives both [T ]i and [Tml. This cuts the labor in inversion in half. More concretely, it is easy to show that MI -1 i MI iT fij= (-) j [T l-i.m j 1. (16.21) 1i] ij The proof, due to Schweitzer, is: mIi TMnI = m i+k Tm[II k+j TmlI m Ik kj ik kj k k =(_-)ij T mII i =(-) =... (16.22) }J1i 1J Consequently the inversion of T for m >> 1 automatically yields by (16. 21) the inverse of T At this point the reader can turn to Section XXXIII of P. Schweitzer's report, for the consideration of the sequence used in generating the T matrix elements, and the desired storage scheme. 74

Section XVII: The case when the surface impedance r) is a function of the arc length: ri = r (t). It is easy to see, by merely following the steps of the derivation, that Eq. (4. 9) is valid even when rn is a function of the arc length t. Similarly all the statements regarding the variational principle are also valid. The only change brought about is the fact that r1 has to be kept under the sign of integration. The T matrix is now TmIII = TmI, p.c.+ TmI (17' The elements of T are Tm I' 1 I T2i-1 2j-1 =2 C Z r dt f (z) r1 (t) Ki (t) K. (t) 2i-1, 2j- 1 2 1 j C2 Z ~ 1 - 6 ]|dt dt' (z) f (z') f (z) a (z) n (t') MO JI 0 Ki (t) Kj (t') a (Gm + G )] (17.2) 1 a z' i-1i i 1, 2j - - [1 - 6 1 dt dt'f (z) aC(z) a(z') r(t') 2i-1, 2j 4 m 0 0 [f' (z) [(m+l) Gm+ + (m-1)G 1] -2f' (z')m G + f (z') f' () f (z') a (Gm G+) K. (t) Kj (17. 3) z' -1 m 75 75

TmI' it df 1 T2i 2j =2 C Z r dt f(z) r7 (t) K. (t Kj (t) 2iI2j 2 (K(LK( t k0 + C2 Z E [1 + 6m dt dt' (z) K (t) 7 t 4L M (z') (m+l)Gm+ 1 -(m- GM 1] -f(z) (z') ((z') aI z(G 1 +G } 1 (17.4) az' G m+l 1 m II' 1 T2i-1, 2j C Z 7r dt f (z) r (t) K. (t) K. (t) 2i-1Z [ + 1 mO dt dt' (z f (z')f' (z) (z) (t') K. (t) K. (t') a (G + Gm (17.5) 1 j az' r-1 rm+l TiT 2j L1 + 61 5dt i dt' {f (z) a (z) C (z') 0 L nr (t') [f' (z) [(m+l) Gm+l + (m-) G 1] - 2 f' (z') m Gm + f (z') f' (Z) f' (z') a (G - G )j K. (t) Kj a z' Gm-1 - (17.6) mII' 1 2i, 2 = C Z 7T dt f (z) r r(t) Ki (t) K. (t) 2i, 2j 2 1 + C2 Z- 01 m-6 dt dt' f (z) a (z') Ki (t) K (t') rT (t') G 1 O {(m+i) G - (m-1) G - f (z') f' (z') a (G + Gm+ m+ rn-i az rn7676

APPENDIX EVALUATION OF T-MATRIX ELEMENTS INVOLVING INTEGRATION OVER A DIAGONAL CELL AND/OR INTEGRATION OVER A CELL BORDERING A DIAGONAL CELL Introduction. In P. Schweitzer's analysis of electromagnetic scattering from rotationally symmetric bodies the T-matrix elements of types A, B, E, H, I and J (see Schweitzer's report, pages 90-92) involve one or more integrals over diagonal cells, and/or over cells bordering a diagonal cell. These integrals must be treated analytically due to the integrable singularity in G m A detailed treatment of this analysis is presented in the following pages. The method used essentially imitates that used by TRG with only minor modifications dictated by the difference in approach. * The values for the components of the surface field are very sensitive to the values of their diagonal matrix elements, which involve precisely the integral over the singularity. For this reason it is necessary that the analysis here presented be dealt with extreme care, since errors in this part will be strongly reflected in the answers. M. G. Andreasen "Scattering from Rotationally Symmetric Metallic Bodies," TRG Final Report. 77

Evaluation of T-matrix elements involving integration over a diagonal cell and/or integration over a cell bordering a diagonal cell A typical integral over a diagonal cell is of the form t i1 where TT jkR G~m (Z'Z16 k R only the linear part F2(t)" F2(t') + F2 (t') (t - t') (4) 2 2 ti t ti+' ti+

We will treat the integrals ti+1 In=i d(tt-t')n G (t,t'), n=O or 1 (6) i analytically and then do t. t 1 1 by numerical quadratures. From the geometry of the body it is seen that k R can be approximated by k R = [f2(t)- f2(t')] + 4 fl(t') f3(t') sin ( ) If2(t) f2(t+ 4 f1(t') sin (2) (8) The functions fl(t), f2(t), f3(t) and f4(t) are defined in P. Schweitzer's report, page 180. We expand fa(t) in a power series about t = ti and neglect all terms of second or higher order f2(t) f2(ti) + f2 (ti) (t - t.) (9) so that kR =1 [f2(tij2(t-t') +4 fl(t') f3(t') sin2 () f2(ti) (t-t')+4f(t')sin ( 2) (10) 79

From (6) we then have ti+l Im= dt(t - t') G (t, t') n m jk R ii+l jkoR0 = d cos m- t dt(t t)n os m k R 0 z i+1 r t' jkoR n e k R koR = dO cosB mO k (11) "'0 1e 0os ~s-0t' where k R f2(t)]2 s2+4f (t') f(t') f2(ti)sin2 ) s+ 4f (t') sin2 () (12) o 1 3 22 We want to introduce this expression in the amplitude factor of the wave function exp (+ jk R)/k R. In the phase factor, we shall introduce the following expression which is obtained by expanding koR in a power series of s and retaining only the first two terms of the series k R = 2 fl(t') sin( + f 3(t') f2(ti) sin(). (13) 80

We can write jk R j 2fl(t') sin() e 1 2e (l+ j f3(t') f2 (ti) sin(2) s) (14) Let r= f2(t i) (15) \= 4 fl(t') f3(t' f2(ti) sin (2) (16) Q= 2 f (t') sin (-) (17) so that k R =h r2 s2a +As+ (18) 0 and i+l - t' tI - t Let Jn t -t' 81

7f 7r n Jd cos me nn +j f3(t') f2(ti) dO e m o sin ( 2) n+ (21) When I 1 is evaluated 02 can be neglected t +-t' ds i- t' ti+-t' jrlog 2P s+As+d+ 2rF2s+A It -tI. 1 2r r2(t i+1-t') +A'(ti+-t') + + 2r2 (ti+l-t') +A 2Log 2'2(ti-t') +(ti -t') + + 2r2 (ti-t') + A (22) ti+ -t' Jt ds r s +A5 s r r (ti+-t +(ti+l-t)+ %i-t' -r (ti_-t')2 +A(t -t') + - (23) Let us first consider Im 82

Tm r Im eb+J d eej Cos me I =T dO ej fcosmO o + dO e cos 3o 0 7r + j f3(t') f2(t) jdO e cosm sin () (24) where 7r 7r %- 20' =Im 20m m 1, 2,... (25) For Ol0 sin 0r 8 and we have A= f(t) f3(tt) f2(t.) 02 (26) 1 3 Q= fl(t'). (27) Let ti-t'= i fl(t') f3(t') f2(ti) = (28) 1 2 fl(t') f3(t') f2(ti'i + f(t') = i (29) * 2 f (t') f (tl) f'(t f 2 (30) fl(t') f3(t') f2(ti) i+ + f (t') = i+l Then,22 + 2 +f 2 1 |2[ i+ + i+ + 2ri+l + toi 7log 2/2 +z2 2 23g2!(31) 83

Introduce the approximations ej2 =1+jf~tt)(32) ej = 1 + j fl(t') 0 (32) 1 22 cosme = 1 -1 2 2 (33) Define AI= I deej cosme 00 - =g(e) 0 (34) 0 O where g(e) = l+jf (t') e — m2 2 j m f(t') 3 1 2 f1(t')2 Then Inm = +Im+ dOe3ej cosme0 +jf3(t')f2(t) Id ee cos msin( rm (36) To do integral (34) we will consider several cases depending on whether 1i and/or B+l are positive, negative or zero. i ti = t -t -(ti+ -ti) = i+l- (ti+ -ti) ti+1 =+i+1 841 84

Case I i = 0 which implies i+l > 0 We have then 0= log 2 2jg'0+e2 ~t:+l + 2 tI+! e 1 + 2r2 +2 0 log 2 +l 2 - o il i and = t O 0 <) = r3211 dO g(6) 0 -_ Og(O) log 0 4r3 a 3i++1 0 14 4r 2 1 2 I i+I 2r1i+1 + i + M 2 fl1t')' i+ ~rm 1(37) 853 85

Case II hi+l = 0 which implies,i < 0 Then ~2r ~2 oYi 2l 2 2 2 \~I l r |,i |I+ e+e0I and IrfE 2 and ~xo, S [ 2 m, + i fl(t') 3 % ] + 2 m2r3 2Ti +fl(t') 224 18 m 13 86

Case II /3 < 0 which implies i+1. Assume i+l > 0 since /i+ 0 has already been considered 2 |i+1 2 2 2 i+l 2r 1/3 1+ 2 + 22r2 /3.+e2 i+2 12 2 + 2 i+ e 2 1+ 7 i+i /P+1 ee - 4r /3e+-Gl + r 622 1 1i+l1 i+l 2 Afrlog i:1 if | ) + Crlg 1+1 2 2 [)8+ m 2- - "'+1 2+ i+l1 J2+" _24r3 2 and I'Yi+i In, AlI' 0 g(6),de +3 2 8+12 g(e) dO 4r / r g(e) o - de m i+1 + i+l 3 A12r3 2 m i+g 2 1 2 [ f( tt) rTm 1 2 I(39 + L+nf( 4 18 mJ 87

From Eqs. (36) and (23) we have m m 1 f341((t') mj 0 = AI-2 j AIm+ dO eJ cos me 0 f2(t)i I 1 m +j f3(t') f2(t) jd {eJ cos m sin( ). [r 2(2i+l 1+1 - tr2(t_ +A(t - t') + J2 1 f3(t') ( 2 j-,dO eJ cos m e sin k (40) f2(ti) where 1= \dO ej os m sin (h() d (41) e 02 1 2 3 1 2 4 he) + j f(t') 1 8 - m fm l(t). (42) Equation (41) can be treated in an identical manner to Eq. (34). Cases I, II and m now give 88

Case I 4 Ai m i+l +~ i+l Em 1 A 1 4;3 12 L. 8 4r3 i+1 1 lo _[1 2 1 3 m r C I A, 6 1 ) m 2 3 1 m m + r [ 8 + j fl(t') 18 (44) Case II 1 3 2 Ai lo m 4 fl (ti) 6 / 2 3 r+ - +j fl (t') - (44) Case I8 m ii+*I + i+l 4 I,32r3 p2 i i+1 +~ + J fl (t') 3 (45) 89

All the integrals in Eq. (40) can be done by numerical quadratures. This fact together with the expressions (37) - (39) and (43) - (45) completes the evaluation of Im(t') Let us now consider I (t'). From Eq. (21) we obtain 1 1 2 d2 ei Cos m {Il2(ti+1 t)2+A(t1 -t')+ 12 2 i fI (t') f3(t') -er (t2 - t') +A(ti - t') + 2 d 2 (ti 2 f2 (t) 3 de e cos m O sin (2 )'0 (46) f2 (ti) m where AI2 = 4 8 de ej 2cos m e sin2 (2 )0 (47) i | de k(O) ~0 where 34 j(t) (48) k(O): = e2 + j (t') - - - We again have to consider three cases. Proceeding as before we obtain 90

Case I Pi ii+ CeI3 4 ITI g -, ~ )n I 1 2 5 - ~il =1 0 1+ 2[_+ j f (t') j r(49) Casel 13 I=0 4 5 m. __. _ t 2 2r 2 L 16 1 201 3 4 F1i~ 2rai 12 + j)fl(t)- 40 2 3 4 + + j f H i (50) +L 36+ ft) 64 200 m rlm Case HI 3. < 0 + 1 543 24 P 64

All the integrals in Eq. (46) can be done by numerical quadratures. This fact together with Eqs. (49) - (51) completes the evaluation of Im (t') 1 Substitution of I (t') and I (t') in Eq. (7) and evaluating the integrals by 0 1 quadratures completes the integration over a diagonal cell. Integration over a cell bordering a diagonal cell A typical such integral is of the form i+ti+2 1 Tm = dt' F (t') Tm =i+2 Ud' F dt F2(t) G (t, t') (52) -i+1 Jti Let m ti) I (t') = dt F2(t) G (t, t') (53) For t' = ti+l the integrand of Im (t') becomes infinite and the integral must be treated i+1 analytically. For t' other than ti+l the integrand is everywhere well behaved and Im(t') i+1 can be done by ordinary numerical methods. T is given by rti+2 T = ti+2dt' F1 (t') It (t'). ti+1 Proceeding as before we obtain I(ttl) = F2(ti+1) (t)+ F2 (ti+1) 1 i+1 92

where m m (ti~l)- 1 f3 (ti+l m I 0 (ti+) =AI 0 (ti+l j I I (ti+l f2 (ti+l) + dO eJ cos mO 0 m +J f3 (ti+l) f2(ti+l) do e cos m sin( ) +jf(t i) +t2 r2 o -r' (ti+l _ t + jf3 (ti+)1 _ i dO e mcos m O sin (2) 0 (56) 2 (ti+ c) m m Q= fl(ti+l) Ein (2) (57) r= f2 (ti+l)j (58) 29 1 2r r P(ti+3+ -t)2 -i+ (ti+ )+ - tit) +A 0 Iarlog......2rl.+ (60) 93

and m 22 1 3 Ai (t II) 8 l+Jfl(ti+l) m 0 i+1 r2 2 L m 3 1 i 1lmJ log F + 2f' 1 il 2 3 2 ( o fl ( i+1 1 f m tn(6i 1 1+ =92 log 2rp ~~Yi+;4 + f (tfi+ t ) 1 1 2 1 3 1r 8 rm + i8 fl ti+2) rm (62) and 1 1(t de) ejt cos me f29 2 i+ 1 r 2 2 2 1 ti) -A.ti+] - t.) f3 (i+ 12 f2 (tt+ 1) 2 i+ 1

where m 1 4 1 5 Ai2 ( i+12 22 rS' 16 rn + j 20 fl (ti+l) nm 3 r _log __ [ j-1 4 1 m2 +r - 1r2 [12 16 1 (ti+1) m( 6 1 3 1 4 1 2 5 + [ 7 + j -l f(t+l) 7 - 200 m u (64) in the above formulas i =ti -ti+l = f (ti+l) f3(ti+l) f2(ti+l) i - fl(ti+l) f3 (ti+l) f2 (ti+l) i + fl (ti+l) i+1 l 1 (ti+l) Eqs. (56) and (63) when substituted in Eq. (55) complete the evaluation of I?(t+l) Eq. (54) can now be done by quadratures. 95

UNCLASSIFIED Security Classification DOCUMENT CONTROL DATA- R&D (Security claeaiflcatton of title, body of abstract and indexing annotation must be entered when the overall report is classified) 1-. ORIGINATING ACTIVITY (Corporate author) 17.. REPORT SECURITY C LASSIFICATION The University of Michigan Radiation Laboratory UNCLASSIFIED Department of Electrical Engineering. ROUP 3. REPORT TITLE Notes on Electromagnetic Scattering from Rotationally Symmetric Bodies with an Impedance Boundary Condition. 4. DESCRIPTIVE NOTES (Type of report and inclusive date.) Technical Report No. 1 5. AUTHOR(S) (.Lst name, first name,!niti.il) Castellanos, Dario 6. REPO RT 9ATg - a 7.. TOTAL NO. OF PAES 7b. NO. OF REFr July 1966 95 3 e8. CONTRACT OR GRANT NO. *4. ORIGINATOR'S REPORT NUMBER(S) b. PROJECT N.A F 04(694)-834 7741-1-T b. PROJEC T N1...-T c. Sb. OTtH R RPORT NO($) (Arr oth,er r nmber thot may be asirn.ed d. 10. AV iL' ABILITY/IMITA'ITON NOTICE' Qualified requestors may obtain copies of this report from DDC. t. SUPP.. EMENTARY NOTES -..P.N$0RiNFG M1 I 1TARY ACT1V1TY Ballistic Systems Division, USAF, AFSC Norton AFB, California 92409 13. ABSTRACT An analysis is presented of the problem of the numerical computation of electromagnetic scattering from rotationally symmetric boundaries satisfying an impedance boundary condition. This work extends the work of P. Schweitzer for perfectly conducting boundaries and includes his results as a special case. D D, ~A~e4~ 1473 UNCLASSIFIED... Security Classification

UNCLASSIFIED Security Classification 14. LINK A LINK B LINK C KEY WORDS | ~ROLE WT ROLE WT ROLE WT Electromagnetic Scattering Rotationally Symmetric Bodies Impedance Boundary Condition Computer Program INSTRUCTIONS 1. ORIGINATING ACTIVITY: Enter the name and address imposed by security classification, using standard statements of the contractor, subcontractor, grantee, Department of De such as: fense activity or other organization (corporate author) issuing (1) "Qualified requesters may obtain copies of this the report. report from DDC" 2a. REPORT SECUMITY CLASSIFICATION: Enter the over- (2) "Foreign announcement and dissemination of this all security classification of the report. Indicate whether report by DDC is not authorized "Restricted Data" is included. Marking is to be in accordance with appropriate security regulations. (3) "U. S. Government agencies may obtain copies of this report directly from DDC. Other qualified DDC 2b. GROUP: Automatic downgrading is specified in DoD Di-users shall request through rective 5200.10 and Armed Forces Industrial Manual. Enter the group number. Also, when applicable, show that optional. markings have been used for Group 3 and Group 4 as author- (4) "U. S. military agencies may obtain copies of this ized. report directly from 1DC. Other qualified users 3. REPORT TITLE: Enter the complete report title in all shall request through capital letters. Titles in all cases should be unclassified.,, If a meaningful title cannot be selected without classific.- tion, show title classification in all capitals in parenthesis (5) "All distribution of this report is controlled. Qualimmediately following the title. ified DDC users shall request through 4. DESCRIPTIVE NOTES: If appropriate, enter the type of, report, e.g., interim, progress, summary, annual, or final. If the report has been furnished to the Office of Technical Give the inclusive dates when a specific reporting period is Services, Department of Commerce, for sale to the public, indicovered. cate this fact and enter the price, if known. 5. AUTHOR(S): Enter the name(s) of author(s) as shown on 11 SUPPLEMENTARY NOTES: Use for additional explanaor in the report. Enter last name, first name, middle initial. tory notes. If:military, show rank and branch of service. The name of the principal author is an absolute minimum requirement. 12. SPONSORING MILITARY ACTIVITY: Enter the name of the departmental project office or laboratory sponsoring (pay6. REPORT DATE Enter the date of the report as day, ing for) the research and development. Include address. month, year; or month, year, If more than one date appears on the report, use date of publication. 13. ABSTRACT: Enter an abstract giving a brief and factual ummary of the document indicative of the report, even though 7a. TOTAL NUMBER OF PAGES: The total page count it may also appear elsewhere in the body of the technical reshould follow normal pagination procedures, i.e., enter the port. If additional space is required, a continuation sheet shall number of pages containing information, be attached. 7b. NUMBER OF REFERENCES: Enter the total number of It is highly desirable that the abstract of classified reports references cited in the report. be unclassified. Each paragraph of the abstract shall end with 8a. CONTRACT OR GRANT NUMBER: If appropriate, enter an indication of the military security classification of the inthe applicable number of the contract or grant under which formation in the paragraph, represented as (TS), (S), (C), or (U). the report was written. There is no limitation on the length of the abstract. How8b, 8c, & 8d. PROJECT NUMBER: Enter the appropriate ever, the suggested length is from 150 to 225 words. military department identification, such as project number, 14. KEY WORDS: Key words are ttechnically meaningful terms subproject number, system numbers, task number, etc. or short phrases that characterize a report and may be used as 9a. ORIGINATOR'S REPORT NUMBER(S): Enter the offi- index entries for cataloging the report. Key words must be cial report number by which the document will be identified selected so that no security classification is required. Identiand controlled by the originating activity. This number must fiers, such as equipment model designation, trade name, military be unique tothis report. project code name, geographic location, may be used as key 9b. OTHER REPORT NUIbBER(S): If the report has been words but will be followed by an indication of technical conassigned any other reprt imbers either by text. The assignment of links, rules, and weights is optional. or by the sponsor), also enter this number(s), 10. AVAILABILITY/LIMITATION NOTICES: Enter any limitations on further dissemination of the report, other than those UNCLASSIFIED Security Classification

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