ELECTROMAGNETIC FIELDS INDUCED IN AND SCATTERED BY BIOLOGICAL SYSTEMS EXPOSED TD NONIONIZING ELECTROMAGNETIC RADIATION Dissertation for the Degree of PI]. D. MICHIGAN STATE UNIVERSITY DONALD EDWARD LIV‘ESAY 1975 LIBRARY Michigan State University w—v—V—w— v a . IlllllllllllIIIIIIIIIIIIIIIIIIIIIIIII L 3 1293 00093 9250 This is to certify that the thesis entitled “Electromagnetic Fields Induced in and Scattered by Biological Systems Expos ed to Nonionizing Electromagnetic Radiation" presented by Donald Edward Lives ay has been accepted towards fulfillment of the requirements for Ph. D. degree in EE Major professor Date May 13. 1975 0-7639 ‘5" amome BY ‘5: . IIDAE & SUNS. BOOK BINDERY INC. LIBRARV BINDERS . I w '. I I ABSTRACT ELECTROMAGNETIC FIELDS INDUCED IN AND SCATTERED BY BIOLOGICAL SYSTEMS EXPOSED TO NONIONIZING ELECTROMAGNETIC RADIATION BY Donald Edward Lives ay This thesis presents a technique for calculating the electric field induced in a finite biological body having arbitrary shape and composi- tion, when the body is irradiated by an electromagnetic wave. A know- ledge of the induced field is important to researchers investigating the biological effects of nonionizing radiation. As an introduction to the study of induced electromagnetic fields in biological media, a plane slab model of a human trunk is analyzed. The electromagnetic field induced in the model by a uniform plane wave is obtained by two methods: (1) by a direct application of bomdary condi- tions, and (2) by transmission line techniques. A group of numerical examples illustrates the behavior of the human trunk ‘model at various frequencies from 100 Hz to 10 GHz. The problem Of calculating the electric field induced in a finite body is considered next. An integral equation for the induced electric field is derived using the free-space dyadic Green's function. The method of moments is then us ed to transform the integral equation to a matrix equation for numerical solution. Techniques for calculating the external scattered field, and for using symmetry to reduce the matrix size, are Donald Edward Lives ay included. A variety of numerical examples, along with some experi- mental data, are presented to illustrate the versatility and the accuracy of the 'mO‘ment solution. In addition, the computer program us ed to calculate the numerical examples is described, and instructions for its us e are given. ELECTROMAGNETIC FIELDS INDUCED IN AND SCATTERED BY BIOLOGICAL SYSTEMS EXPCEED TO NONIONIZING ELECTROMAGNETIC RADIATION By Donald Edward Lives ay A DISS ERTATION Submitted to Michigan State University in partial fulfillment of the requirements for the degree of DOCTOR OF PHILOSOPHY Department of Electrical Engineering and Systems Science 1975 This thesis is fondly dedicated to Dr. Ralph Kron. Without his encouragement, kindness, and understanding, it might never have been written. ii ACKNOWLEDGMENTS The author wishes to express his appreciation to his 'major professor, Dr. K. M. Chen, for his guidance and encouragement throughout the course of this work. He also wishes to thank the other members of his guidance com- mittee, Dr. D. P. Nyquist, Dr. M. M. Gordon, Dr. B. HO, and Dr. J. Asmus sen, for their time and assistance. A special note of thanks is extended to the National Science Founda- tion for the Graduate Traineeship which helped the author complete this study. The research reported in this thesis was supported in part by the National Science Foundation under Grant ENG 74- 12603. iii Page LISTOFTABLES.................... vi LISTOFFIGURES................... Vii I. INTRODUCTION.................. 1 II. INTERACTION OF AN ELECTROMAGNETIC PLANE WAVE WITH A PLANE SLAB MODEL OF A PHYSIOLOGICAL SYSTEM . . . . . . . . . . 3 2. l Qualitative Description of the Plane Slab Model . . . . . . . . . 3 2. 2 Maxwell's Equations and Plane Wave Solutions for a General N- Slab System . . . . . . . . . . S 2.3 Boundary Conditions . . . . . . . . . . . . 7 2. 4 Matrix Representation of Boundary Conditions . . 8 2. 5 Solution by Transmission Line Analogy . . . . 10 2. 6 Numerical Results for a Plane Slab Model of theHumanTrunk ............... 15 III. INTERACTION OF AN ELECTROMAGNETIC WAVE WITH AN ARBITRARY PHYSIOLOGICAL SYSTEM 0 O O O O O I O O O O O O O I O O O O O O O 31 3. 1 Description of Problem and Method of Solution . . 31 3. 2 Integral Equation for Internal Electric Field . . 33 3. 3 Moment Solution of Integral Equation. . . . . . 37 3.4 Calculation of Matrix Elements . . . . . . . 42 3. 5 Reduction of Matrix Size for Special Cases of Cross- Sectional Symmetry . . . . . . . . 51 3. 6 Calculation Of External Scattered Field . . . . . 60 3. 7 Numerical and Experimental Results . . . . . . 61 A. Testing and Convergence . . . . . . . . . . 62 B. Internal Electric Field. . . . . . . . . . . 65 C. External Scattered Field . . . . . . . . . 96 D. Symmetric and Antisymmetric Modes . . . . 98 IV. A DESCRIPTION OF THE COMPUTER PROGRAM USED TO DETERMINE THE INDUCED ELECTRIC FIELD IN AN ARBITRARY, FINITE PHYSIOLOGICAL SYSTEM. 0 I O O O O O O O O O O O 0 111 4.1 Description of the Program. . . . . . . . . . . 111 4.2 Structure of the Data File . . . . . . . . . . . 113 TABLE OF CONTENTS iv 4.3 4.5 V. SUMMARY APPENDIX A . . BIB LIOGRA PHY . Description of the Input Variables . 4. 4 Using the Program Listing of Program and Subroutines LIST OF TABLES Frequency Dependence Of Conductivity and Relative Permittivity for Muscle, Skin, Fat, and Bone Induced electric field at the center of the muscle cubes shown in Figure 3. 10 for various numbers of subvolumes. IE1] = 1 V/m The structure of the data file for PROGRAM BLOCK. If any of the variables marked with an asterisk does not 'match its prescribed codes, the program will be aborted vi 1? 67 114 2.6 2.8 2.9 LIST OF FIGURES An arbitrary N-slab system . . . . . . . . . Boundary condition solution of N-slab system Transmission line solution of N- Slab system . . 7- slab model used for calculations . . . . Electric field and power density in 7- slab model at 100 Hz, for 0 < z < 8.5 cm. Incident field is aplanewave,withIEiI=1V/m. . . . . . . Electric field and power density in 7-slab'mode1 at 1 kHz, for O < z < 8.5 cm. Incident field is a plane wave, with IEiI=1 V/m. . . . . . . . Electric field and power density in 7- slab model at 1 MHz, for O < z < 8.5 cm. Incident field is aplane wave, with IEiI: l V/m . . . . . . . Electric field and power density in 7- slab model at 10 MHz, for O _<. z < 8.5 cm. Incident field is aplane wave, withIEilz lV/m. . . . . . . . Electric field and power density in 7- slab model at 100 MHz, for 0 < z < 8.5 cm. Incident field is a plane wave, with IEiI= 1 V/m . . . . . . . Electric field and power density in 7- slab model at 300 MHz, for 0 < z < 8. 5 cm. Incident field is a plane wave, with IE1I=1 V/m . . . . . . . Electric field and power density in 7- slab model at 600 MHz, for 0 < z < 8.5 cm. Incident field is a plane wave, with IE1]: l V/m . . . . . . . Electric field and power density in 7- slab 'model at 900 MHz, for 0 < z < 8. 5 cm. Incident field is a plane wave, with IE1I= 1 V/m . . . . . . . . Electric field and power density in 7-slab model at 1. 5 GHz, for O < z < 8.5 cm. Incident field is aplane wave, with IE1I=1 V/m. . . . . . . vii 12 16 19 20 21 22 23 24 25 26 27 Figure 2. 14 2.16 3.4 Electric field and power density in 7- slab "model at 2.45 GHz, for 0 5 z .<_ 8. 5 cm. Inset shows power density in skin layer. Incident field is a plane wave, with IEII = l V/m . . . -. . . . Electric field and power density in 7- slab model at 5 GHz, for 0 5 z s 8. 5 cm. Inset shows power density in skin layer. Incident field is a plane wave, with IEII = l V/m . . . . . . . Electric field and power density in 7- Slab "model at 10 GHz, for 0 s z S 8. 5 cm. Inset Shows power density in skin layer. Incident field is a plane wave, with IEII =1 V/m . . . . . . An arbitrarily shaped biological body in free space, illuminated by an electromagnetic field. In gen- eral, the body is inhomogeneous and nonmagnetic . A physiological system partitioned into N sub- volumes, showing a typical arrangement of the cells................ (a) Scattered field E8 in subvolume Vm produced by a unit current in subvolume. Vn when cylindri- cal body is perpendicular to E1. (b) Scattered field E in subvolume V produced by a unit cur- rent in subvolume Vn (”111.211 cylindrical body is parallel to E1. E3 in (b) is about twice as large asEsin(a)................. (a) Equivalent sphere Sn, centered about En, used to calculate the diagonal elements of each sub- matrix. (b) Spherical coordinate system defined inS .................... n A cubical body partitioned into symmetrical quadrants. The quadrants are designated by Romannurnerals................ A cubical body partitioned into syannetrical octants, designated by Roman numerals. The origin of the coordinate system is located at thecenterofthecube. . . . . . . . . . . . Cubical body excited by (a) the symmetric mode E18, and (b) the anti- symmetric mode E3. The octants Of the cube are numbered as in Figure 3. 6 . viii 28 29 3O 32 39 44 48 52 56 58 Figure 3. 8 3.9 3.11 3.13 3. 14 3. 16 A cylinder of muscle partitioned into (a) 6, (b) 48, and (c) 162 subvolumes. The cylinder is illuminated by a 2. 45 GHz plane wave, with the incident electric field E1 parallel to the axis of the cylinder. IE1I= l V/m . . . . . . . . . . Electric field intensity along the axis of the cylinder shown in Figure 3. 8, for 'models (a), (b), and (c). IEII = l V/m. One wavelength in thecylinderis 1.76 cm . . . . . . . . . . Two cubes Of muscle illuminated by a 2. 45 GHz plane wave, treated as single cells in (a) and (c), and divided into 27 cubical subvolumes in (b) and (d). The edges of the cubes ‘measure one wave- length and 1/4 wavelength, respectively . . . . . Electric field induced at the center of a dielec- tric cube, for various values Of frequency and dielectric constant. Incident electric field is a planewave.................. Plane conducting layer exposed to a 300 MHz plane wave, with E1 perpendicular to the plane Ofthelayer............. Electric field induced in the layer of Figure 3. 12, with e = 70 so and 0' = 1 mho/m. Only half of the layer is shown above. E and E arenegligible. . . . . . . . . .y. . .12 A 300 MHz plane wave impinges upon a plane conducting layer, with E1 parallel to the plane ofthelayer................ The x-component of the electric field induced in the layer of Figure 3. 14, with E = 70 60 and 0' = l th/m. Only half Of the layer is shown above.................... The z-component Of the electric field induced in the layer of Figure 3. 14, with 6 = 70 60 and 0' = l 'mho/m. Only half of the layer is shown above.................... A block of tissue composed of a fat layer and a muscle layer, illuminated by a plane wave at 100MHz................... ix 63 64 66 68 70 71 72 73 74 75 3.19 3.20 3.21 3.22 3. 23 3.24 3.25 3. 26 The electric field induced in the block of tissue pictured in Figure 3. 17. Only 1/4 of each layer is shown above. For a corresponding plane slab model, E = 0. 197 V/m at the center Of the fat layer, anfi Ex = 0. 210 V/m at the center of the musclelayer................. A saltwater cylinder of half-length h illuminated by a. 9. 45 GHz plane wave. The cross-section Of the cylinder measures 1. 8 'mm x l. 8 'mm . Electric field E along the axis of the saltwater cylinder shown in Figure 3. 19, for h/A = 0.113, 0. 227, 0. 510, and 0. 737. Concentratio% Of the salt solution is 1 Normal, and the frequency is 9.45GHz................... Power density P along the axis of the saltwater cylinder shown in Figure 3. 19, and total absorbed power Pt, for h/xo = 0. 113, 0. 227, 0. 510, and 0. 737. Concentration of the salt solution is 1 Normal, and the frequency is 9. 45 GHz. Electric field E along the axis of the saltwater cylinder shown in Figure 3.19, for h/Ao = 0. 113, 0. 227, 0. 510, and O. 737. Concentration Of the salt solution is 2 Normal, and the frequency is 9.45GHz................... Power density P along the axis of the saltwater cylinder shown in Figure 3. l9, and total absorbed power P ,, for h/Ao = 0. 113, 0.227, 0. 510, and 0. 73 7. oncentration of the salt solution is 2 Normal, and the frequency is 9. 45 GHz. Electric field E along the axis of the saltwater cylinder shown in Figure 3. 19, for h/lLo = 0.113, 0. 227, O. 510, and O. 737. Concentration of the salt solution is 5 Normal, and the frequency is 9.45GHz................... Power density P along the axis of the saltwater cylinder shown in Figure 3. l9, and total absorbed power Pt, for h/>.o = 0. 113, 0. 227, 0. 510, and O. 737. Concentration of the salt solution is 5 Normal, and the frequency is 9. 45 GHz. (a) A homogeneous muscle cylinder, and (b) an inhomogeneous muscle cylinder irradiated by a 2.45 GHz plane wave . . . . . . . . 76 78 79 8O 81 82 83 84 85 Figure 3. 27 3. 29 3. 30 3.31 3. 32 3. 33 3. 34 3.35 3. 36 (a) Induced electric field E, and (b) power density P along the axes of the muscle cylin- ders shown in Fi ure 3. 26. The frequency is 2.45 GHz, and I 1I = l V/m. Each cylinder was partitioned lengthwise into 100 subvolumes ofequalsize.................. (a) A homogeneous fat cylinder, and (b) an in- homogeneous fat cylinder irradiated by a 2. 45 GHzplanewave.............. (a) Induced electric field E, and (b) power density P along the axes Of the fat cylinders shown in Fi ure 3.28. The frequency is 2. 45 GHz, and I 1I = l V/m. Each cylinder was partitioned lengthwise into 100 subvolumes of equalsize.................... An inhomogeneous fat layer illuminated by a uniform plane wave at 600 MHz. . . . . . . . . . . The x-component of the electric field induced in the layer illustrated in Figure 3. 30. The incident field is polarized along the x- axis. . . . . . The y-component of the electric field induced in the layer illustrated in Figure 3. 30. The incident field is polarized along the x- axis. . . . . . A homogeneous layer of muscle, illuminated only in the shaded area by a 600 MHz plane wave 0 O O I O O C O O C O O O O O O O O O The x-component of the electric field induced in the layer pictured in Figure 3. 33. The in- cident field is polarized along the x-axis, and irradiates only the shaded portion of the layer . . . The y-component of the electric field induced in the layer pictured in Figure 3. 33. The in- cident field is polarized along the x- axis, and irradiates only the shaded portion of the layer . . . . Backscattering from three different homogeneous cylinders exposed to a 10 GHz plane wave. The 'magnitude of 'T was 100 in each case, but the re- spective values of the phase angle of 7 were 0° 45°, and 90°. The observation point was 30 cm from eachcylinder . . . . . . . . . . . . . . . . xi 86 88 89 9O 91 92 93 94 95 97 Figure 3. 37 3. 38 3.39 3.40 3. 41 3. 42 3. 43 3. 44 3. 45 3.46 Comparison of theory and experiment for back- scattering from a saltwater cylinder of 1 Normal concentration at 9. 45 GHz. The observation point was 15 cm from the cylinder . . . . . . . Comparison of theory and experiment for back- scattering from a saltwater cylinder of 5 Normal concentration at 9. 45 GHz. The Observation point was 15 cm from the cylinder . . . . . . . Comparison of theory and experiment for back- scattering from a brass cylinder at 9. 45 GHz. The Observation point was 15 cm from the cylinder.................. Comparison of theory and experiment for rela- tive backscattering from a brass cylinder and a saltwater cylinder of 1 Normal concentration. The frequency was 9. 45 GHz, and the observa- tion point was 15 cm from each cylinder . . . . . The sym'metric mode Of the electric field in a muscle layer exposed to a 1 GHz plane wave. The symmetric mode is induced by the sym- metric component (cos koz Q) of E1. Only 1/4 Ofthelayeris shownabove. . . . . . . . . . . . The antisymmetric mode of the electric field in a muscle layer exposed to a 1 GHz plane wave. The antisymmetric 'mode is induced by the antisym'metric component (- j sin koz ii) of E1. Only 1/4 of the layer is shown above . . The total electric field induced in a muscle layer by a 1 GHz plane wave. Only 1/4 of the layer is shown above . . . . . . . . . . . A square loop of 'muscle irradiated by a 100 MHz plane wave. E1 lies in the plane of the loop, while H1 is normal to the plane of the loop...................... The symmetric mode of the electric field in the loop of ‘muscle shown in Figure 3. 44. The symmetric 'mode is induced by the symmetric component (cos koz A) of the incident field. Only 1/4 of the loop is shown above . . . . . . . . The antisymmetric 'mode of the electric field in the loop Of muscle shown in Figure 3. 44. The antisymmetric 'mode is induced by the antisymmetric component (- j sin koz A) of the incident field. Only 1/4 of the loop is shown here. . . . . . . . . . . . . . . . . . . . . xii 99 100 101 102 104 105 106 107 109 110 Figure Page 4. 1 An electromagnetic plane wave, incident normally upon a rectangular block Of tissue . . . . 112 4. 2 A block of tissue partitioned into rows, columns, and layers of subvolumes along the x-, y-, and z-axes, respectively . . . . . . . 115 4. 3 A block of tissue divided into 4 symmetrical quadrants. If the incident field is likewise symmetrical, calculating the induced field in only one quadrant will determine the induced fieldintheentireblock. . . . . . . . . . . . . 119 4. 4 A spherical polar coordinate system used in calculations of the external scattered field. This system is used unless the program user changesit................... 122 4. 5 An alternate spherical polar coordinate system for scattering calculations. This system uses the y-axis as the polar axis . . . . . . . . . . . 124 4. 6 A block of tissue composed of a homogeneous fat layer and a homogeneous layer of muscle, illuminated by a 915 MHz plane wave . . . . . . . 125 4. 7 An inhomogeneous layer of fat, illuminated only in one corner by a 915 MHz plane wave . . . . 127 xiii CHAPTER I INTROD UC TION In recent years there has been a growing concern over the possible health hazards of nonionizing electromagnetic radiation. A variety of responses to such radiation have been observed in humans and animals. Some of these reactions are caused by an increase in body temperature, while others are triggered directly by the induced electromagnetic field. Before valid safety standards for exposure to nonionizing radiation can be established, the conditions which elicit a particular response must be known. Thus, it is necessary to determine the electric field inten- sity which produces a nonthermal reaction, and to establish the temper- ature at which a heat- induced effect occurs. The temperature, however, can be derived from the intensity of the internal electric field. Cons e- quently, valuable insight into both thermal and nonthermal effects can be gained if the induced electric field can be calculated. The mathematical complexity of the problem is enormous. It is therefore necessary, in a practical theoretical study, to approximate the biological system of interest by a relatively simple model which can be readily analyzed. Some commonly used 'models are the plane slab I 7], [16], the conducting sphere [6 ], [17], and the dielectric cylinder [ 4]. However, these models are Often grossly oversimplified, and the conclusions drawn from their behavior have only limited validity. This thesis presents a technique for calculating the electric field induced inside a finite biological body having arbitrary shape and composition, when the body is irradiated by an electromagnetic wave. The free- space dyadic Green's function [ 1] is used to derive an integral equation for the induced electric field; the method of 'moments is then used to solve the equation numerically. The calculation of the scattered field is also discussed. A plane slab model of a human trunk is studied in Chapter 11. Two methods of calculating the induced field due to a uniform plane wave are presented: (1) solving the system Of linear equations generated by the boundary conditions on the electric and magnetic fields, and (2) using transmission line techniques. A number of numerical exaanples illus- trate the behavior of the human trunk 'model at various frequencies from 100 Hz to 10 GHz. Chapter III is devoted to calculating the induced field in a finite biological body having arbitrary shape and composition. An integral equation for the induced electric field is derived using the free-space dyadic Green's function. The method of moments is then us ed to con- vert the integral equation to a matrix equation for numerical solution. Details for using symmetry to reduce the matrix size, and for calculat- ing the scattered field, are given. The chapter concludes with a variety of examples illustrating the versatility of this numerical technique. Some calculations of the scattered field from saltwater cylinders are compared with experimental data, with good agreement. Chapter IV contains a description and listing of the computer pro- gram used to obtain the numerical results presented in Chapter III. Instructions for its use, along with some illustrative examples, are also included. CHAPTER II INTERACTION OF AN ELECTROMAGNETIC PLANE WAVE WITH A PLANE SLAB MODEL OF A PHYSIOLOGICAL SYSTEM Since humans and animals have such complicated shapes and struc- tures, an attempt to directly analyze their electromagnetic absorption and scattering properties would be a formidable task. It is therefore necessary to approximate these complex physiological systems by simp- ler models for which the scattering problem can be readily solved. In this chapter we will discuss the Often-used plane slab model, which, in some cases, can provide useful data about heating patterns in the ori- ginal system. 2. 1. Qualitative DescriLtion of the Plane Slab Model A general N-slab system, shown in Figure 2. 1, consists of N con- tiguous plane layers of tissue, each having uniform thickness and infinite cross-section. The tissue composing each layer is linear, homogene- ous, and isotropic, although its electrical properties do vary with frequency. We assume that the entire array contains no sources, and that the incoming electromagnetic field is a uniform plane wave, inci- dent normally upon the first slab. The coordinates are defined so that one axis is perpendicular to all of the boundaries in the model. Both of the vacuum regions enclosing the system extend to infinity along this axis, so that no reflected wave exists in the vacuum on the right. Be- fore considering a specific example, we shall discuss two methods of .EOumam £2ng humufinum :< .H .N oudwfim .— um . b T EDHKVNK/ H ggumk/ .U zq AA NA in mm m“ 3 11.14 . va zfiv my Hmv MU Nmu am?" N determining the electromagnetic field inside an arbitrary N- slab configuration. 2. 2. Maxwell's Equations and Plane Wave Solutions for a General N- Slab System For a mathematical description of the model, we refer again to Figure 2. 1. We have chosen the coordinates so that the z- axis is per- pendicular to all of the boundaries in the system. The incident wave impinges on the array from the left, traveling in the +z direction, with the electric field vector linearly polarized along the x— axis. At any point 1", E(?) denotes the electric field while PM?) represents the mag- netic field. The space occupied by the ntil layer will be indicated by Ln' In the discussion which follows, we can economize on our descriptions by regarding the vacuum regions as additional ”layers" Of the system; we will refer to the region z < (11 as Lo’ while we denote the region z > dN+l by LN+1' With this notation in mind, let us select a layer at random, say Li' and investigate the behavior of the electromagnetic field within it. As indicated in Figure 2. l, the i2 layer has permittivity 6i Farad/ meter, permeability pi Henry/meter, and conductivity (Ti mho/meter. Its input plane is located at z = di’ where "do” is at - OO. It will be convenient to represent the electric and magnetic fields inside Li by EN?) and Hifi"), respectively. Thus, for I" 6 Li’ E5) = Eifi’) (2. 2. la) m?) fiifi?) (2. 2.1b) i = 0,1,. .. ,N+l. We assume a harmonic time variation of ejwt, which we shall henceforth suppress. Since each layer is linear, homogeneous, isotropic, and source-free, Maxwell's equations for Li are Vx E15) = - jwpifii(?) (2.2.2a) vxfiifi’) = (ei+jwei)Ei(I-’) (2.2.2b) v-EiG) = 0 (2.2.2e) v-fiifi’) = 0 (2.2.2d) i = 0,1,...,N+1. Maxwell's equations may be combined in the usual manner to ob- tain the vector Helmholtz equation for EEG"): v2 E16) + R? E16?) = 0 (2. 2. 3a) where kiz = 02,11 61 - jwuio-i (2. 2. 3b) 1 = 0,1,...,N+1. Since the incoming uniform plane wave is incident normally on the first slab, we expect to Obtain uniform plane waves in each layer of the system. Therefore, we anticipate a solution to Equation (2. 2. 3a) of the form -jk.z jk.z E4?) = Ei(z) = x(Aie 1 + Bie 1 ) (2.2.4) i : 0,1,...,N+l where Ai and Bi are complex constants, as yet undetermined. Ai specifies the amplitude and phase Of a wave traveling in the +z direc- tion, while Bi gives the amplitude and phase of a wave moving in the -z direction. Ao represents the incident wave, and is known in ad- vance. Also, since our model precludes any reflections in LN+1’ we must take BN+1 = 0. From Equation (2. 2. 2a), fiifi') is given by —> —-> _ . —> Hi(r) _ —qui[vx Ei(r )] . (2.2. 5a) Substituting Equation (2. 2. 4) in Equation (2. 2. 5a) yields ll ~ fiifi’) = Hi(z) (2.2.5b) i = 0, 1, . . . , N+1 where L. denotes the characteristic wave im edance of L., and is 1 p 1 given by I1). . §=i=I Ill 1:01 N+1(225c) . k vei'J(Ui/w) , ,oo., o o o Let k1 = Bi - Jai. Then, using Equation (2. 2. 3b), we Obtain 1/2 “161 “i (ii = Re(ki) = w 2 1"(3?) +1 . (2.2.7a) i pi is the wave number in the igl- layer; it is also defined by 2 51 = 755 (2.2.7b) i where Xi is the wavelength in Li' <1i represents the attenuation con- stant in Li’ and it is given by “6. 0' ai = -Im(ki) = w l+(———) - 1 . (2.2.8) 2. 3. Boundary Conditions To completely specify the electromagnetic field in the it—h- slab, we must determine the constants Ai and Bi in Equations (2. 2. 4) and (2. 2. 5b). We can do so by imposing boundary conditions on EH") and HG’). Elec- tromagnetic theory tells us that the component Of E(?) and the compo- nent of HG") which is tangent to an interface between disparate media must be continuous there. A reference to Figure 2. 1 and to Equations (2. 2. 4) and (2. 2. 5b) shows that both Efi’) and m?) are tangent to all boundaries in our model. Therefore, the boundary conditions imply that E (1:) and PM?) must be continuous throughout the system. We may use Equation (2. 2. 4) to express the continuity of the electric field at z=dnby - jkn_ ldn jk d - jkndn j kndn e + B e ) = (Ane + Bne ) (2. 3.1) (An- 1 n- l n = 1,... ,N-I-l. From Equation (2. 2. 5b), we Obtain the expression for the continuity Of the magnetic field: -jk_d jk_d —jkd jkd e n1n_B enln) (Ae nn_Benn) = n n (2.3.2) 4. Q n-l n n =1,...,N+1 . These relationships are illustrated in Figure 2. 2. For convenience, we shall take (11 = 0 with no loss Of generality. 2. 4. Matrix Representation of Boundary Conditions As the index n of the previous section assumes all possible values, Equations (2. 3. l) and (2. 3. 2) generate a set of 2N + 2 simultaneous linear equations relating the unknown A's and B's. This algebraic sys- tem has a solution if the A's and B‘s also number 2N + 2. The N tissue layers contribute 2N unknowns, since there are two for each slab. In addition, we must evaluate Bo and A A0 and N+l' BN+1’ of course, have already been specified. The system of equa- tions therefore contains 2N + 2 unknowns, enabling us to solve it. We define the following column vectors: .539? amHmIZ mo GOEDHOO GOEBGOO Summonsom .N .N ounmmh I'- III III'.. ~+fi fl Hun m m m I I I TI M H..." < < <. A a L 10 {btdlbtfi NI—oI-IO ¢ = : Qnfila) 4% = A a.a1b) .ooOOI—Io—t 11> o -AN+1J -03 The system of 2N + 2 linear equations may then be cast into the form IGNI= R, a.a2) where [C] is the (2N + 2) x (2N + 2) matrix shown on page 11. The vector LIJ may be written as -l q] : [C] 4,0 . (2. 4.3) After the A's and B's have been evaluated, the electromagnetic field in Li can be found by using Equations (2. 2. 4) and (2. 2. 5b) 2. 5. Solutionjj Transmission Line Analogy Equation (2. 4. 2) will be solved by a computer in most cases. Since the boundary conditions relate the coefficients in adjacent layers only, most Of the elements of [C] are zero. Thus, particularly if N is large, considerable computer storage will be wasted, making the matrix method uneconomical to use. We can calculate the coefficients in Equations (2. 2. 4) and (2. 2. 5b) a different way, by exploiting the sim- ilarities between our slab model and a uniform transmission line [12]. The pertinent quantities are shown in Figure 2. 3. As before, the input plane of Li is located at z = di' The input impedance to Li is denoted by Zin' Clearly, Zin is the "load" impedance for Li-l' At 11 H+Zo2xm H+Zo2xn ZoH-an Hn\oo- u T: 12 .8333 nmfimnz mo GOSOHOO OcS GowmmfiEmGOHH on .m .N 33E Hum 13 the boundary 2 = di’ we define a reflection coefficient l"i and a trans- mission coefficient Ti: i-l I‘i — A. (2.5.1a) 1-1 Ai Ti 2 A. (2.5. lb) 1-1 1 = 1, O O . , N+l 0 We will derive expressions for F1 and '7’i by using the boundary condi- tions on E(?) and HG") at z = di’ and by using the definition of impedance. The impedance Zi(z) in Li is given by -jkiz jkiz Eix(z) [Aie + Bie ] 71“”) = W = 4’i -jk.z jk.z 9'5"") 13’ [Aie 1 - Bie 1] where Qi is the characteristic impedance Of Li’ given by Equation (2. 2. 5c). Thus, Zin may be written as -Jkidi Jk d i [Aie + Bie 1] Zin = Zi(z:di) = C’i -Jkid. Jkld (2°5°3a) [A e 1 - B 1] or jZkidi Zi : 1,;[1‘LI‘i-rle ] (2 5 3b) in i jZkidi ' ° [1 " 1“1+1" 1 The boundary conditions at z : di are -jk._ d. jk._ d. -jk.d. jk.d. A. e 111+13. e111=A.e 11+I:3.e11L (2.5.4a) 1-1 1-1 1 1 -jk._ d. jk. d. -jk.d. jk.d. Ai-1e 111_Bi-1e 1-11 Aie ll-Bie 11 z (2.5.4b) C’i- 1 13i 14 By using the relationship expressed in Equation (2. 5. 3a), we re- write Equation (2. 5. 4b) as -jk._ d. jk._ d. -jk.d. jk.d Ai-1e 111'B1-1e 111 Aie 11+Bie 11 L = i (2. 5. 5) i-l Z. In Substituting Equation (2. 5. 4a) into Equation (2. 5. 5) and dividing by Ai—l y1elds -jk. d. jk. . -jk. . jk. e 1-11_1_.ie 1-11 1-11+I..18111 .—_ , (2. 5. 6) gi-l 2? 1n We can readily solve Equation (2. 5. 6) for F1: (2? - g. ) -j2k. d. ri = In 1-1 e 1-1 1 (2. 5.7) (Zin + gi-l) We may rewrite Equation (2. 5. 4a) as -jki- ldi jki- ldi -jkidi jkidi Ai-l[e + I‘ie ] : Ai[e + ri+l e ] (2. 5. 8) Hence, Ti is given by [e'Jki—I i + 1.. eJki-ldi] Ti = -'kT 1 ek d (2. 5. 9a) [e J i i + I‘ eJ i i] i+l By substituting Equation (2. 5. 7) into Equation (2. 5. 9a), we Obtain i 2 z? J(IT‘S-.1)d 7' = ' m e arr 1 1 J . . Zin+§i-l[l+1‘i+le 11] i=1,...,N-l-l . (2. 5. 9b) By examining Equations (2. 5. 3b), (2. 5. 7), and (2. 5. 9b), we see that we can determine Zin’ Pi, and Ti if we know PH Our procedure 1° is motivated by this observation. 15 S1nce BN+1 = 0, 1t follows that FN+2 = 0. Thus, from Equat1on (2. 5. 3b), 25:1 is simply go, the characteristic impedance of free space. Using Equations (2. 2. 5c), (2. 5. 7), and (2. 5. 9b), we calculate FN+1 and TN+1° We then repeat the procedure, evaluatIng ZN, I‘N, TN, ZN- 1, and so on, until I‘ and 'T have been determined at each boundary. Next, we calculate the coefficients, using Equations (2. 5. 1a) and (2. 5.1b): B0 = I‘l A0 (2. 5. 10a) A1 = 71 A0 (2. 5. 10b) B1 = FZA1 = 1727le (2.5.10c) A2 = TZAI = 7271 AC) , (2.5.10d) and so forth. In general, Ai and Bi are given by A1 = TiTi-lTi-2° ° ° ' ° ° ° 7271Ao (2'5'113) Bi = ri+1717i_1 . . ..... 72'7le (2.5.11b) Finally, we use Equations (2. 2. 4) and (2. 2. 5b) to determine the elec- tromagnetic field in Li‘ 2. 6. Numerical Results for a Plane Slab Model of the Human Trunk The model depicted in Figure 2. 4 was chosen to represent a human trunk. The system is 19. 9 cm thick, and comprises 7 layers: two layers of skin, each 0. 2 cm thick; two layers of fat, each 3. 0 cm thick; two layers of muscle, each 5. 0 cm thick; and a layer of bone, 3. 5 cm thick. Table 2. 1 lists the electrical properties of each type of tissue at various frequencies from 100 Hz to 10 GHz. Since physiological tissues are essentially nonmagnetic, it has been assumed that pi = no in each slab. 16 §90m> .mcoflmdpodwo ROM wows HOCOE agate )IWIII/IIIII/Dl/li 80 EU 50 N .o o .m o .m ITIv‘H m 1H Z O H H. m M— < D m rm 2 II/uI\/x/IL .4 .~ onsmE EU 80 80 50 m .m o .m o .m N .o IV? UA 'jl' mm 1H H 0 Z Z w .H. H O D < M m 2 h m 83.6w > In no 17 Frequency Muscle, Skin Fat, Bone (Hz) 6/60 0'(mho/m) I 03/60 0-(mho/m) 102 1,438,039 0.2 71,902 0.04 103 539,265 0.2 21,571 0.04 106 2,000.0 0.4 200.0 0.043 107 160.0 0.625 40.0 0.045 108 71.7 0.889 7.45 0.048 3::108 54.0 1.37 5.7 0.069 61:108 52.47 1.49 5.6 0.086 9::108 51.09 1.59 5.6 0.101 1.5x109 49.0 1.77 5.6 0.121 2.45::109 47.0 2.21 5.5 0.155 ‘5:r109 44.0 3.92 5.5 0.236 1010 39.9 10.3 4.5 0.437 Table 2. 1. Frequency Dependence of Conductivity and Relative Permittivity for Muscle, Skin, Fat, and Bone. 18 The electric field intensity E and the power density P have been calculated in each layer of the system at each of the frequencies given in Table 2. l, and the results are shown in Figures 2. 5 through 2. 16. AC has been taken to be 1. The distributions of E and P at each fre- quency are plotted only for 0 s z s 8. 5 cm, as explained below. At low frequencies, up to about 10 MHz, the electric field is nearly constant throughout the entire model. Therefore, the power density in the second half of the system is merely a mirror image of that in the first half. At higher frequencies (100 MHz and above), the power den- sity is nearly zero beyond the first muscle layer; hence, the latter portions of the system are of little interest. As noted before, the electric field is nearly constant throughout the system for frequencies up to 10 MHz. The heating of each layer is therefore uniform, with the skin and muscle layers heated the most. We also note that E and P are generally larger at frequencies above 100 MHz than they are at lower frequencies. The maximum value Of E occurs in the skin layer at 600 MHz. At 600 MHz and above, the ratio of the power density in the skin to that in the muscle becomes large. Thus, most of the power in the incident wave is dissipated in the skin layer; relatively little penetrates to deeper layers of tissue. Based on this model, the use of high frequencies for diathermy is questionable, since little heating would occur deep in the body. Indeed, a patient would suffer severe burns on the skin before he would experience any significant heating of internal structures. Of course, this model does not take into account the cooling effects of perspiration and circulation of the blood. 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N x Hr HHr . 1' m. [‘3‘ Lug N o .v . _ , ._ m a" m . i H. / m .H_3 pOumnmflBHfi .oommm ovum fl taboo. Rommofiomo. bemoan impugn—Ens c< .H .m 0.”:th 5353/ Mm 33 integral equation for the unknown electric field inside the system. Then, using a pulse-function expansion of the unknown field in conjuncticn with point-matching, we employ the method of moments [ 2 ] to solve the integral equation numerically. Once the internal electric field has been determined, we may, if desired, calculate the field scattered externally by the body. 3. 2. Integral Equation for Internal Electric Field The constituent vectors ER?) and I-I1(?) of the incident electromag- netic field satisfy Maxwell's equations for a source-free region of free space: vinfi') = - jwuofiifi?) (3.2.1a) Vxfi1(?) = jweofilfi") (3.2.1b) v- EN?) = o (3.2. 1c) v- #5“) = o (3.2.ld) p0 and 60 are the permeability and permittivity, respectively, of free space. Eifir) and fii(?), of course, are known functions of I". When the incident electromagnetic field impinges on the physiologi- cal system shown in Figure 3. 1, it creates a distribution of induced charges and currents throughout the system. These charges and cur- rents are the sources for a secondary or scattered field, denoted by Esfi") and HSG"). Thus, E(?) and 35’), representing the total electro- magnetic field at each point, may be written as the sum of the two par- tial fields: E6) EH?) + 1586') (3.2. 2a) m?) fiifi’) + Fifi?) (3. 2. 2b) Since E16?) is known, the problem will be solved if we can obtain an ex- pression for ESG") inside the body. 34 First, we must relate Esfir) to its sources through Maxwell's equa- tions. We begin with Maxwell's equations for the total electromagnetic field inside the physiological system: V): E(?) = - jwuo m?) (3.2.3a) v x in?) = o(?)E(?) + jwe(}’)§(?) (3.2. 3b) v - [a(?)‘E’(i-’) + jw€(?)fi(?)] = o (3.2.3c) v - in?) = o (3.2. 3d) Since Maxwell's equations are linear, we substitute Equations (3. 2. 2a) and (3.2. 2b) into Equations (3.2. 3a) and (3.2. 3b), and obtain V x Eifi'.) + V x ESE.) - jwuo fiifif) - jwuo fisfi") (3. 2. 4a) v x HR?) + v x Ham «(i-Um?) + WM?) - 60] 1‘6?) + jwefiifir’) + 28m] (3. 2. 4b) Subtracting Equations (3. 2. la) and (3. 2. 1b) from Equations (3. 2. 4a) and (3. 2. 4b), respectively, we have v x Esfi’) = - jwuoHsG’) (3. 2.5a) Vxfisd’) = {am +5w[e - 60 mm + jweofisa’) (3.2.sb) Defining an equivalent volume current density jeqfir) by Teqfi’) = T(?)E('r") , (3.2.6a) where n?) = «(5’) + jw[e(i-’) - so] , (3.2. 6b) we may rewrite Equation (3. 2. 5b) as Vx H36?) = 3’ (I?) «+ij ESE) (3.2.7) eq 0 The equivalent current density is non-zero only inside the physiological system, and has two components: 0(F)E(?) represents the conduction current flowing in the body, while jw[€(?) - 60] EG’) gives the polariza- tion current. The equation of continuity for Teqfi?) defines an equivalent volume charge density peqfi"): 35 v-fi’eq(r)+jwpeq(r) = 0 (3.2.8.) from which -> _ .i. . " peq(r) _ wv freq”) (3.2.8b) Taking the divergence of Equation (3. 2. 7) and using Equation (3. 2. 8b) gives 8 _, 98 (F) V°E(r) -—-‘1— (3.2.9) " e 0 Finally, taking the divergence of Equation (3. 2. 5a), we have V' fisfi’) = 0 (3.2.10) Equations (3.2. 5a), (3.2.7), (3. 2. 9), and (3. 2.10) constitute Maxwell's equations for ESG") and HSG’): VxEs(?) = - jwuofis(?) (3.2.11a) Vx $3155) = Teqaf) + 5on E35?) (3. 2. 11b) v- E91?) = sic peqfi’) (3.2.11c) v- fi8(‘r’) = o (3.2.11d) Since peqfi") is related to jeq(?) by Equation (3. 2. 8b), we can ex- press ESG’) as a function of jeq(?) only. Furthermore, we may think of jeqfir) as a current existing in free space, since only “0 and 60 appear in Equations (3. 2. 11a) through (3. 2. 11d). Note, however, that Teqfi’) depends upon the total electric field E(?), which is still unknown. It is this fact which leads us to an integral equation for E(?). The scattered electric field Es(?) can be written in terms of Tqu'") by using the free- space tensor Green's function CG: 3'") [ l ], given by 8&2?) = - jwuorf' + {521]4’3?) , (3.2. 12a) 0 where 36 -ikol?- 1"! (JG-2}") = _, _, . (3.2.12b> 41r|r- r'l ‘f = M+M+Qé , (3.2.12c) and k0 : an!“ 6 , OO Let V denote the volume occupied by the physiological system. If i" is outside of V, the relation between 335:) and Teqfih) is simply Efl?) = 53' ($5) - 61?, r")dv' (3.2. 13) V eq If '1'" is inside V, however, difficulties arise because C(?,?') is singular at 3°" = I". We can remedy the difficulties by using a modified tensor —>—> rr Green's function 55-: 11"), first introduced by Van Bladel [18]. (”u , ') is given by 5(‘1", r") Emmi-2r") - “3512;?" O (3.2. 14) where CG: 3“) is defined by Equation (3. 2. 12a), and the P. V. symbol denotes the Principal Value, to be defined presently. Thus, when '1"> is inside the physiological system, Equation (3. 2. 13) is replaced by E36?) = 5384?!) - [R v. 86-2?!) - T63(J3'w'eolfi]tiv' (3. 2. 15a) 01' 8 -* --> —-p 9 Te (if) E (r) = P. V.Svj'eq(r') - fi(r,r')dv' — 33.—3?;- (3.2.15b) The P. V. symbol, then, refers to the principal value of the integral in Equation (3. 2. 15b), obtained by excluding from V a small sphere of radius 7? centered at I", then taking the limit as n -> 0. By substituting Equation (3. 2. 15b) into Equation (3. 2. 2a) and re- arranging terms, recalling that Teq(?) = T(?)E(?), we obtain the desired 37 integral equation for EG"): [1 + 4362M )- p. v. S 7(r")§(r") - E’(?,?')dv' = EH?) (3.2.16) 3jw V Equation (3. 2. 16) may be classified as a Fredholm integral equation of the second kind. Eifi") and 7(1'") are, of course, known quantities. EG’) is the unknown total electric field inside the body. A solution to Equation (3. 2. 16), based on the method of moments, will be discussed in the next section. 3. 3. Moment Solution of Integral Equation The inner product of E5“) and 35:, 15) in Equation (3. 2. 16) may be represented as a matrix product: ”(it 39) nyd’. 3“) GXZG’. i5) Ex(?'> Eon-66.?) = cyxu r) cwfifi) 6,26,?) Eva“) sz(r, r') Gzy(r, r') Gzz(r, r') Ez(?') (3. 3. 1) To make the analysis as general as possible, it will be convenient to introduce the following notation: 3 = z (3. 3.2) Physically, Gx x (1:?) is the xp component of the electric field at the observation poilhtqr maintained by a unit xq component of current at the point 1"". If the expression for CH1", 1'") given in Equation (3. 2. 12a) is used to evaluate the inner product of E5") and EG’, in ), and if the resulting ex- pression is compared to Equation (3. 3. 1), we find that 2 -> ~> _ , ' _L a —D --D‘ Gx x(1°.1") — - quo[5pq + k2 m]¢(r.r) (3.3.3) P q o ‘1 P p.q = 1.2.3. 38 where fipq is Kronecker's delta. A straightforward evaluation of Equa- tion (3. 3. 3) (see Appendix A) gives, after some rearranging, . -ja —-> —+ -1wu0koe 2 0 Z G (r,r') = [(0. - l-Ja)6 +cose cosG (3-(1 +3ja)] x x 4 3 pq x x P ‘1 17a P Q (3.3.4) where o. = k R R = ?- r'l o x - x' x - x' C086 -_— ii C086 — J—A x R x R P Cl " _ "’1 _ I I 1 r " (X1! X2, X3) r - (X 9 x2) x3) Equation (3. 2. 16) comprises three coupled scalar integral equations. Using Equation (3. .3. l), we write each scalar component of Equation (3.2.16) as [1 + 3jw€ ]Ex (1‘) - P. V-S 7(r')[ 2: Gx x (1‘,r')Ex (r')]dV' = Ex (r) o p V q—1 p q q p P =19293° (303- 5) We can use the method of moments to transform Equation (3. 3. 5) into a matrix equation for each value of p. The resulting coupled matrix equa- tions may then be combined into a single matrix equation for E(?). First, we partition the body into N cells, or subvolumes, and assume that 7(1?) and E(?) are constant throughout each cell. We denote the rnth subvolume by Vm, and we let Fm denote the position of a repre- sentative interior point of Vm, as shown in Figure 3. 2. Thus, the integral in Equation (3. 3. 5) may be written as X 3 P. v.5 7(r’1)[ >3 ox (?,?')Ex (‘r")]dv' V q=1 P q q II II MW P. my 7(r")cx x (?,?')Ex (1")dV' (3.3.6a) 1 v q p‘q q 39 wcffionm GE. . .330 05 mo “cogomnsuum 303.3 s moEsHoufinn Z 35 poaoflflumm 8393 3033393.“ 4‘. IH (a N .N .m 83E 40 3 N = 2 2: P. v.5 7(‘r")o (r,r‘)}; (‘r*')dv' (3.3.6b) x X x q=1 n=l V p q q 3 N -> -9 -9 —> = Z Z 7(rn)Ex (rn) P. V.S Gx x (r,r')dV' (3.3.6c) q=l 11:1 q V11 p q Hence, Equation (3. 3. 5) becomes [1+ ]E (r) - 2 2‘. 7(r )E (r )P.V.5 G (r, r')dV' = E (r) X n X n V X x x 106 J o p q=1n=1 q n pq P (3. 3. 7) Next, we require that Equation (3. 3. 7) be satisfied at each I’m That is, [l +3'_i-¢3_€—]Ex (rm) - E 23 7(rn)Ex (rn)P.V.S Gx x (rm, r')dV' o p q=1n= q Vn p q i -> - Exp(rm) (3.3.3) Equation (3. 3. 8) can be rewritten in the following form: 3 N n? ) —> -+ -> m -o E Z [7(rn)P.V.S Gx x (rm, 1")dV' - 6pq6mn(l +3-j—hl-E—)]Ex (In) q=l n=l Vn p q o q i —. : - Ex (rm) (3. 3. 9) P m = 1,2, . . ,N P = 1:213- Let [Gx x ] be the NxN matrix whose elements Gin: are given by P q P q mn -b .. ...' ' 7(?m) Gx x - 'T(rn)P. v.5 Cix x (rm,r )dV - qufimn[l +W] P q V11 P q 0 (3.3.10) m,n = 1,2,... ,N. P.q = 1.2.3- 41 Then, Equation (3. 3. 9) is simply 3 N . z: 2: OX x Ex (rn) - - Ex (r ) (3.3.11) q=ln=1 p q q p m: 1,2,...,N P = 19293- We also define the N-dimensional column vectors [Ex ] and [E:{ ], P P given by r. ‘T “ i “P -! Ex (r1) FEX (r1) P P . i . [Ex ] = . . (3.3.12a) [Ex ] = . . (3.3.12b) P o p s —> i _. LEX (rN).l LEx (rN)..J P P P = 1.2.3. The summation over n in Equation (3. 3. 11) represents the inner product of the mth row of [Gx x ] with [Ex ]. As m ranges over all values from P q 1 to N, Equation (3. 3. 11) becomes 3 . 1 2 [(3x x ][Ex] = - [Ex]. p = 1,2,3. (3.3.13) q=1 pq q p After performing the summation over q for each value of p, we obtain the following set of linear equations: [GHHEXHIGXYHEYI+[zeHEz] = - [12:] (3.3.14a) [GYXHEX] +1<2W11EY1 + [cyzn E2] = - 112;] 9.3.141») [sz][Ex] + [Gzy][Ey] + [Gzz][EZ] = - [E12] (3. 3. 14c) Equivalently, we have Wan] new] :[ze1‘ WEXJT PIEill 'fé"i'i122"1":'1'<§"1' '135'1' — - "1132‘? (3 3 is) ---Yi=_-'.--yy.-.'--.v.s- -_.v- ‘ ----.z ' ' L[<3le Hazy] {[Gzz]_ _[Ez]‘ _[E;] _ 42 Equation (3. 3. 15) may be written in compact form as [G][E] = - [Ei] (3.3.16) Equation (3. 3. 16) is the matrix representation of Equation (3. 2. 16). [G] is a 3N x 3N matrix, while [E] and [Ei] each have 3N components. We can find the total electric field in each of the N subvolumes by solv- ing Equation (3. 3. 16) for [E]. 3. 4. Calculation of Matrix Elements In this section we develop explicit expressions for the elements of each N x N submatrix [Gx x ], p,q = 1,2, 3. The m,nth element of P q [Gx x ] is defined by Equation (3. 3. 10): P q _. mn .. —> —> firm) G = 7(r )P.V.SG (r ,r')dV'-6 6 [l+—.—-—] x x n x x m pq mn 33006 P q Vn P q 0 (3.4.1) We will first evaluate the off-diagonal elements; i. e. , m f n. Since Fm f Vn’ the integrand in Equation (3. 4. l) is continuous throughout Vn’ so we need not take the principal value. Thus, the off-diagonal elements of [Gx x ] are given by P q Gm“ = T(?)S G (? ,r")dv' (3.4.2) x X n X X m Pq V,l Pq mfn Poq=11293° The integral in Equation (3. 4. 2) can be evaluated numerically by any convenient method. If the dimensions of Vn are small compared to the free- space wave- length )to, we can approximate Gin: by assuming that the integrand in P q Equation (3. 4. 2) is constant over the region of integration. With this assumption, Equation (3. 4. 2) becomes Gx x = 7(rn)Gx x (rm,rn)AVn (3.4.3a) P q q m {n where AV = S dV' . (3.4.3b) n V n Writing P and 1" as m n -—<> m m m rm = (x1 ,x2 ,x3 ) (3.4.4a) _. n n n In = (XI’XZ' x3) , (3.4.4b) —’ and using Equation (3. 3. 4) to evaluate G (r , F ), we obtain an x x m n approximate expression for the off-diagonal matrix elements. The re- sult is . -> -jamn mn _ - JL‘)"‘okoT(1-n)AVne [( 2 l . )6 x x — 3 amn - - Jamn pq P ‘1 41ro. mn + cos 63m cos Ginn (3 - a2 + 3jamn)] (3. 4. 5a) P q m ;f n where o. =kR , R = |r’ -Fl (3.4.5b) mu 0 mn mn m n m n m n mn x ' x mn x - X cos 9x = —ER—B , cos 9x = __9____9 (3. 4. 5c) p mn q mn If the biological body is cylindrical, experience has shown that using Equation (3. 4. 5a) often produces inaccurate data. When the cylin- der axis is perpendicular to Eifir), numerical integration of Equation (3. 4.2) leads to nearly the same solution as using Equation (3. 4. 5a). The two methods generate different solutions, however, when the axis of the cylinder is parallel to Eifi"). We will soon see why. Figure 3. 3 shows a cylindrical body illuminated by a uniform plane wave at normal incidence, with ER?) linearly polarized along the x- axis. 44 .23 5 mm mm smug mm 033» 3on 3 7: a: mm .H o» Hozmumm mm. 350, Hmowppcfluno can? d> wagon/nan Gm “Cognac find m n5 boon—pom. 8> 2.930». 133m 5 mm 303 monogamom 3v .am 0» umadoflpcomhom 3 canon 33.35313 non? c> oEsHo>ndm Gm “mousse ”as: .0 c3 poodpoum 8.? oESHo>nflm 5 mm 30¢ wouofimom 73 8: :3 .m .m unsure 45 The only significant component of both EG’) and ESQ?) is the x- component. Therefore, it will be instructive to compare the values of Gxx(?m'?n) for the two cases illustrated in Figure 3. 3. In Figure 3. 3a, the axis of the cylinder is perpendicular to EiG-I), so that ?m and Pu have very nearly the same x-coordinate. Thus, according to Equation (3. 4. 5c), 2 (cos 92‘") = o. (3.4.6a) G (i: ,1? ) , the Green's function for perpendicular polarization, is xx m n_L therefore given approximately by . damn G (r ,r = 3 (o. - l - JO. ) (3.4. 6b) xx m n_L 417a mn mn mu 2 2 . . . . If o‘mn << 1, we can neglect the amn term 1n Equat1on (3. 4. 6b), obtaining -ja. -> -» j“Wokoe . 4 Gxx(rm’rn_L — 4" 3 (1+Jamn) (3. .6c) o‘rnn The axis of the body is parallel to Eifi") in Figure 3. 3b, so that Pm and ;n have approximately the same y- and z-coordinates. Hence, 2 (cos 63m) = 1 (3. 4. 7a) ‘Il. -. ' O O O O Gxx(r m' r11).I , the Green 3 function for parallel polarizat1on, 18 approximately . damn -. —» " quo ko e 2 . 2 . (r ,r e [(o -I-Ja. )+(3-a +330. )] xx m n" 41rc13 mn mn mn mn mu (3. 4. 7b) damn j(”“0 R0 8 . = -2 4 a3 (1 +Jamn) (3.4.7C) 11 (1 Therefore, IGxx(rm,rn)“| ~ I H(;m,?n)J-l (3.4. 7d) 1f aim << 1 . Since Gx x ($31") varies as (0.)"3 for small a, the largest matrix ele- ments are those for which “inn << 1. These elements are approximately twice as large for parallel polarization as they are when Eifir) is per- pendicular to the cylinder axis. Thus, for parallel polarization, the matrix elements should be evaluated more accurately to obtain reliable data. The diagonal elements of [Gx x ] may be written as p q F) 'T r :2: = ”in: x - 5Pq[1+ 335%] (3.4.8a) p q p q o where In = P. v.5 G (‘1'? ,?')dv' (3.4. 8b) x X x x n p q Vn p q It is easily verified that ex x ('i’, F), defined in Equation (3. 3. 3), is P q also given by G (E’ i") - -'w [a +-1— 82 spa-"1?» (3 4 9) xx ’ ' J"o pq 25x'5x' ' " P q k0 p q All derivatives in Equation (3. 4. 9) are taken with respect to the variables of integration, so we may set '1? = i-‘n at the outset. Then, since Gx x (E’n, in) is a function of (x11 - x'), (y11 - y'), and (zn - z') only, we can define a coordinate system centered at “r'n, and set r; = U. Thus, M6,?) = w?) = mm = —— (3.4.10) where 47 C ons equently , Z 4, _ ->, _ . l 3 , Gx x (U,r ) ’ Gx x (r ) ’ 'Jw"o|:6pq +17 W] Mr) P ‘1 P q o P ‘1 (3.4.11) To evaluate I: x , we will approximate Vn by a sphere Sn of equal volume centered about the origin of the new coordinate system, as shown in Figure 3. 4a. The radius an of the sphere is given by 32wn 1/3 an = (T) (3.4.12) Thus, 1n = P. v.5 G (?)dv' (3.4.13) X x X x p q 511 p q The variables of integration are merely dummy variables, so we will omit the primes from here on. It is easily shown that Z 2 W =M£lf2f3.,l%lfl[5 his] (3.4.14) xp xq er r r r r pq r r In the spherical coordinate system illustrated in Figure 3. 4b, we have % = sine cos¢ (3.4.15a) § = sine sin¢ (3.4.15b) Z -r- :: C089 (3.4.15c) dV = r2 sinedrdedcb (3.4.15d) Therefore, since the ratio of x,y, or z to r is a function of 6 and (3) only, we write xx quw=$f mq=MA new P q Then, using Equations (3.4. 16) and (3. 4. 14), Equation (3. 4. 11) becomes 48 .sm 5 vocflov 539: oumnmvnooo aeowuonmm 3v .uauuagpan £000 mo 950830 Unowmmp 05 033300 3 p03. .ssm «none 60.3300 .dm 09230 «nonaZu—UH A3 2: 3 .v .m ouvmrw 49 2 prxq(r) ‘ ‘ J“’“o Mr)(A’pq + k2[ er fxpxq(9'4’) 0 d + 11‘ +i'r) (qu fxpxq(9.4>))] (3.4.17) We now substitute Equation (3. 4. 17) into Equation (3. 4. 13) to obtain 11 217 1r 1 = - jwuonlim %q5:11p(r)r ZodrS d4) sinede 0 2n wf +—-2- ang-igilr2 drS d4) (9,4)) sinOdB k0 n dr2 0 o fxpxq a 1 n (r) 2" 1' +3577 9‘41?— rdrS d4) [fipq ' fx x (e,¢)] “made 0 o p q 3.4. 18) Integrating the second term of Equation (3. 4. 18) by parts and combining the result with the third term gives 11’ n e I = - jwu af ¢(r)r 2:.dr5‘1r d¢ sinede xpxq 01171310 qu 0 Zn 1r 1 . +_,Z[,Z 9%12a DIS. d¢ {xx (9,43) s1n6d9 k0 n o 0 P ‘1 a 211' 1r 1 11d r . +7 5' -%-§.—) rdrS' d4, [5pq - 3£x x (9.43)] smede ko n 0 0 P q (3.4.19) In the third term of Equation (3.4.19), 5: ngiliéflrdr becomes in- finite as n+0. However, using Equation (3. 4. "16) and Equations (3.4.15a) through (3. 4. 15c), it is readily verified that 211' 5 d¢ o[1qu - 3fx x (9,4,)1 sinede = 0 (3.4.20) 0 P q Poq = 192:3- The third term of Equation (3. 4. 19) is therefore zero for all finite values of n, and contributes nothing as THO. Thus, Equation (3. 4. 19) becomes 50 n I: x = - jwpo lim 4116 S Mr)rzdr P q ”*0 pq n 3‘n 211' 11 “LP-29192 ]5 d4: f (9.4.) sinede (3.4.21) 2 dr x x k n o o p q It can be readily demonstrated that 211 11 4P5 5 d¢ fx x (9,4)) sinede =63— Pq (3.4.22) 0 o p q We then have n . I = - 3w“ 41161rn:nS. 11;(r)r2 dr xpxq o pq n+0 1 __1_2[ Zildd” an] (3.4.23) 3k r n e-jkor Recalling that Mr): _——Z-1;-;— , we obtain n -Jk r 'Jkor 3'11 I: x = -jwuofi lim 5 e o rdr + -—lz[r2§;(e r ) ] p q pq n+0 n 31:0 n (3.4.24) A straightforward evaluation of Equation (3. 4. 24) gives - iju 6 -jk a 1n = ‘Z’Pq[e °n(l+jk a)-l] (3.4.25) x x o n p q . 3ko The desired expression for the diagonal matrix elements is found by substituting Equation (3. 4. 25) into Equation (3. 4. 8a): Zj n?) - k a 1(E’n) ngx = -5Pq ““02 n [eJon(l+jkoan)-l]+[l “FEED-1%: p q 3ko (3.4.26a) 01' n junoa —> . -> 'jkoan . ox x s —.223 3['r(rn) + sto] - 27(rn)e (1 + Jkoan) p q 3ko (3. 4.26b) 51 If the actual shape of Vn differs appreciably from that of a sphere, Equation (3. 4. 25) can be applied to a small sphere surrounding it; the integration over the remainder of Vn can be done numerically. 3. 5. Reduction of Matrix Size for Special Cases of Cross-Sectional Symmetry It is possible in some cases to reduce the number of unknowns in Equation (3. 3. 16), thereby effecting a substantial saving in computer storage space. To illustrate the method, we will consider the cubical body shown in Figure 3. 5. We will assume that the incident electric field is a plane wave, given by ER?) = 41230 e-Jk°z , (3.5.1) and that Eifi") is incident normally upon the face of the cube. The body's cross section is symmetrical about the z-axis, and about the planes x = O and y = 0. We will assume that the electrical properties of the cube are likewise symmetrical. The planes of sym- metry divide the cube into four quadrants, which are indicated by Roman numerals in Figure 3. 5. We could proceed with the calculation of Ed?) inside the body in a purely straightforward manner; that is, we could partition the cube into, say, 4N subvolumes (N in each quadrant), and compute Ea") in each cell. However, under the assumed conditions, we need only determine E(?) in one quadrant; the electric field in the other quadrants can be obtained by utilizing the symmetry of the body. Thus, as we shall show, we can determine E(?) at 4N points in the cube by solving a matrix equation involving the unknown electric field at only N points. 52 mugupmsav 0:8 .30u083s Gogom .5 v0umsmmm0p 0.3 .chuwmsv fifiboflnfigm 33 v0noflmuusm .300. 1.63.90 < \ .m .m 3&3 Mm 53 Following a procedure similar to that us ed in Section 3. 3, we partition the first quadrant into N cells, or subvolumes, and assume that 11?) and E ('1?) are constant throughout each cell. We denote the rnth subvolume by Vm , and its location by i-‘m , where the subscript ”1" refers to the firsthuadrant. By reflectinglel and 3'me about the plane x = O, the z- axis, and the plane y = 0, we obtain their respective images in the other quadrants: Vm and i’m in the second quadrant, 2 Z V and I" in the third, and V and '1'" in the fourth. By m m m m 3 3 4 4 assumption, 7(1‘ ) = 75" ) = 75" ) = 75" ) . (3.5.2) m1 m2 m3 m4 We require that Equation (3. Z. 16) be satisfied at each in . Pro- 1 ceeding as in Section 3. 3, we have 3 N mnl _' mnz _’ mn3 _. Z Z[G E (r )+G E (r )+G E (r) x x x x x x n x x x n q=ln=l pq ql pq q 2 pq q 3 mn4 ... 1 _. + G (r )] = - E (r ) (3. 5. 3a) x x x n x m p 4 l m = 1920- 0 :N i P = 19203: where _. 7(1' ) mnl _. ml ._ I I _ _.__.__ G): x 7(rn )P. V. Gx x (rm ,r )dV 6pq6m[l + 3jw€ :| p q 1 Vn p q l o 1 (3.5. 3b) and mu --> -> ----> G J = T(r )5 G (r ,r')dV' , (3. 5. 3c) x x n x x m P q 1 Vn. P q 1 J j = 2., 3,4 . The principal value has been omitted in Equation (3. 5. 3c) because the integrands are continuous throughout the regions of integration. 54 Under the assumed conditions of symmetry and normal incidence, with ER?) linearly polarized along the x-axis, we can determine by in- spection (and verify by computation) the following relations: Exfi’nl) = Exunz) = Ex(rn3) = Ex(rn4) (3. 5.4a) EY(rn1) = - Ey(rn2) = Ey(rn3) : - Ey(rn4) (3.5.4b) Ez(rnl) = - Ez(rn2) = - Ez(rn3) = E26214) (3.5.4c) Thus, Equation (3. 5. 3a) may be rewritten as N mn1 mn2 Inn3 mn4 _. Z [G + G + G + G ] (r ) x x x x x x x x x n n=1 p 1 N mnl mnZ mn3 mn4 _. +2 ny-ny+ny-ny Ey(rn) n=l p p p p 1 N mnl mnz mn3 mn4 _> 1 + 23[ze -ze -ze +ze]Ez(rn) = -E (rm) (3.5.5) n=1 p p p p 1 p 1 Let wx x] , WK y] , and WK 2] be NxN matrices whose respective P P P elements are given by mn mn1 mn2 mn3 mn4 J4xx = Gxx +Gxx +Gxx +Gxx (3.5.6a) P P P P mu m mn mn fig“; =nyl‘Gx:2+ny3‘ny4 (3.5.6b) P P P P P mn mnl mnz mn3 mn4 fl = G -G -G +G (3.5.6c) x z x z x z x z x 2 P P P P p = 1,2,3. Equation (3. 5. 5) then becomes 2 Z x x Ex (rn ) = - Ex (rm ) (3.5.7) (1:1 n: P q q 1 P 1 55 As explained in Section 3. 3, Equation (3. 5. 7) is equivalent to the following as m and p range over all possible values: FM XXL WK, 15w flHExlfl ’[Eifl : ------ L---- --h ---x.-- ----- --‘—l [flyijlf] “[1123]. [f 11. = " .[F‘yi (3'5'8) L [42x]: [42y1' :- [#22] [Ez]l L. [Ezlll where [Ex ] and [E1 ], p = 1, 2, 3, are defined by Equations (3. 3. 12a) and (3. 3. 12b), respgctively. The subscript "1" again refers to values in the first quadrant. We may write Equation (3. 5. 8) in more compact form: WEE]1 = - [Bi]l (3.5.9) where M] is a 3Nx3N matrix, and both [E]1 and [Ei]l have 3N com- ponents. After we solve Equation (3. 5. 9) for [E]l, we obtain [E]z, [E]3, and [E]4 via Equations (3. 5. 4a) through (3. 5. 4c). Had we solved Equation (3. 3. 16) for [E] in the entire body, we would have had lZN unknowns. Thus, we have reduced the number of unknowns (and the matrix size) by a factor of 4. However, the amount of computation needed to evaluate each matrix element has increased by the same factor. We can sometimes reduce the number of unknowns by an additional factor of 2, at the expense of computation time, by decomposing the incident plane wave into symmetric and anti- symmetric modes. We will again use the cubical body to briefly outline the method. The origin of the coordinate system will be located at the center of the cube, with the axes oriented as shown in Figure 3. 6. The planes x = 0, y = 0, and z = 0 divide the cube into eight octants. We partition the first octant into N subvolumes, as before, and denote the rnt'h 56 .0990 on» mo houses on... an voodoofi 3 532? oumampuooo one no guano can. .u—dnoga 550m .3 vaumammmov .3530 Wownuogctnu 33 venoflfiham 3.03 30.350 < .e .m charm \/ \ x \/M \ > V. / a >H .m \ 57 subvolume and its location by Vm and Fm , respectively. Again, 1 l V and '1? have images in each of the other octants. We assume m1 m1 that n? )='r(? )='r(? )='r(?) m1 m2 m3 m4 = 76’ )= N? )= n? )= n? ), (3.5.10) m5 m6 m7 m8 where the numerical subscript once again refers to the octant in which the point is located. The incident electric field, given by Equation (3. 5. 1), can be written as EN?) = E163” 15:5?) , (3. 5.11a) where Eisfi’) = QED cos koz (3.5.11b) Eiafi’) = - £on sinkoz. (3. 5.11c) Eisfi'.) represents a symmetric mode of exciting the cube, while Eiafi") is an anti- symmetric mode. E185?) and E25?) are illustrated in Figures 3. 7a and 3. 7b, respectively. Let E86?) be the internal electric field induced by E:(?). The components of E8 (3") obey the following relations: Ea? )=E(i-’ )=E(i-’ )=E(‘r’) X8 m1 X8 m2 X8 m3 XS m4 = xs(rm ) = Exs(rm ) = Ex3( m > = Emum ) (3.5. 12a) 5 6 7 8 E -> : - E -> = E -. = _ "’ ys(rm1) ys(rm2) ys(rm3) ys( m4) = Eys(rm5) = - Eys(rm6) = Eys(rm7) = - Eys(rm8) (3,5,1zb) 58 .o .m 0.3th 3 a.» wouonga one 0350 05 mo 03300 2.3. .mM epoch vinegar: 4qu 2.3 fit was .mfl epoch 033550 on» A3 can. vofloxo hponéaogvo .p .m ouawmh 3v 3 am am 1 “ // Ezs(_.ml) = - Ezs(rm2) = ' Ezs(rm3) = Ezsfirm“) --E (i? )=E (5’ )=E ('1‘r )=-E (i? ).(3.5.12c) 28 m5 ZS m6 28 m7 Z8 m8 Similarly, let Eafi'.) be the field induced by E25”. Then, the components of Eafi") satisfy Exauml) = Exa(rm2) = Exa(rm3) = Exa(rm ) = - Exa(rm ) — - Exa(rm ) — Exa(rm ) = - Exa(rm ) (3.5. 13a) 5 6 7 8 E -> = - E —-> = E «b = - E -> ya(rml) ya(rm2) ya( m3) ya(rm4) = — Eya(rm5) = Eya(rm6) = - Eya(rm7) = Eya(rm8) (3. 5. 13b) Eza(rml) : - Eza(rm2) " ' Eza(rm3) — za(rm4) : Eza(rm5) = - Eza(rm6) : - Eza(rm7) = Ezahfma) . (3. 5. 13C) The total electric field inside the cube, with Eifir) given by Equation (3.5. l), is found in three steps. First, using E26.) as the incident field, we require that Equation (3. 2. 16) be satisfied at each of the N points in the first octant. Proceeding as we did in the first part of this section, we use Equations (3. 5. 12a) through (3. 5. 12¢) to reduce . the number of unknowns by a factor of 8. After we solve the resulting matrix equation for EEG.) in the first octant, we can find EB G") in the other octants via Equations (3. 5. 12a) through (3. 5. 12c). Next, we use Eiafir) as the incident field, and compute Eafi") in the first octant. We can again reduce the number of unknowns by using Equations (3. 5. 13a) through (3. 5. 13¢). Once we have found Eafr’) in the first octant, we obtain Eafi") in the other octants by employing Equations (3. 5. 13a) through (3. 5. 13¢). 60 Finally, because of the linearity of Equation (3. Z. 16), the total in- duced electric field in the cube due to ER?) is simply EG") = E8(?)+Ea(?) . (3.5.14) Although we have reduced the number of unknowns by decomposing Eifi") into symmetric and anti- symmetric modes, there are obvious disadvantages. First, we must solve two problems: a body excited by Fifi"), and a body excited by E253. Second, each matrix element will contain eight terms. Finally, we must pay careful attention to the alge- braic sign of the field components in the various octants when we add ESQ.) and Eafi"). Nevertheless, if computer storage space is limited, the decomposition of Eifi") into symmetric and anti- symmetric modes may prove useful. 3. 6. Calculation of External Scattered Field We will frequently be interested in finding the scattered field out- side the physiological system. Once EG") has been determined inside the body, the external scattered field is given by Equations (3. 2. 6a) and (3.2. 13): EN?) = S 7(‘r")E(I~") . §(?,?')dv' . (3.6.1) v Using the notation developed in Section 3. 3, we may write each scalar component of ESQ?) as 3 E; (I?) = SV—r(?i)[qz:lox x (“iii-"m (35)]dV' (3.6.2) x P Pq ‘1 ll I-I ‘ N - W o P Since the volume V has been partitioned into N cells, with '11?) and Ear) assumed to be constant throughout each cell, Equation (3. 6. 2) becomes 61 N 3 E:P(?) = 1121-ng 7(i’n)[(1)21prxq(I-’,?')Exq(?n)]dv' (3.6.3a) or E8 (i’) = I): ;7(?)E (I?) G (i-‘I-‘mv' (3 6 3b) xP n=l q=l n xq n Vn xpxq ' ' ‘ p = 1,2,3. The integral in Equation (3. 6. 3b) has the same form as the integral in Equation (3. 4. 2), so it can be evaluated by the methods outlined in Section 3.4. Therefore, after we have found the N values of EG") in- side the physiological system, we can determine the scattered field at any exterior point by using Equation (3. 6. 3b). 3. 7. Numerical and Experimental Results A number of simple biological models have been studied using the moment solution of Equation (3. Z. 16). The results in this section illus- trate the variety of problems which can be solved by this method. Since techniques for probing the induced field inside a conducting medium are still being perfected, the only experimental results to be presented are those describing the scattering from finite conducting cylinders. The data presented in this section is grouped into 4 general categories: testing and convergence, determination of the internal electric field, the external scattered field, and an investigation of the symmetric and anti- symmetric modes discussed in Section 3. 5. The incident electric field for all of the examples had a magnitude of 1 Volt per ‘meter, and was polarized along the x- axis. Where pos- sible, the symmetry methods of Section 3. 5 were used; in most such cases, an illustration indicates the portion of the body in which the induced field was calculated. The subvolumes in all of the examples were cubes, 62 so that the expression for the diagonal matrix elements would be rea- sonably accurate, and the induced field was calculated at the center of each subvolume. Equation (3. 4. 5a) was used to evaluate the off- diagonal 'matrix elements for the examples shown in Figures 3. 11 through 3. 18. For the other examples, the off-diagonal elements were computed by numerically integrating Equation (3. 4. 2). A. Testing and Convergence This group of calculations was performed to test the convergence of the numerical solution, and to acquire confidence in its accuracy. The following two examples examine the convergence of the solution as the size of each subvolume in the body is decreased. In the first test, a Z. 45 GHz plane wave illuminated a muscle cylinder whose dimen- sions in wavelengths were 3 x 1/2 x 1/2. The incident electric field was parallel to the axis of the cylinder, so that the induced field had essentially only an axial component. The cylinder was divided into a variable number of subvolumes, and the induced field was calculated for each configuration. The models for 6, 48, and 162 subvolumes are shown in Figures 3. 8a, 3. 8b, and 3. 8c, respectively. The edges of each cell 'measured 1/2, 1/4, and 1/6 wavelength, respectively. Figure 3. 9 shows the electric field intensity along the axis of the cylinder for each 'model depicted in Figure 3. 8. Since none of the sub- volumes lie on the axis in Figure 3. 8b, the average of the fields in the front and back of the cylinder have been plotted to facilitate a compari- son with the results from Figures 3. 8a and 3. 8c. All three models are in good agreement, indicating that using subvolumes as large as even 1/2 wavelength ‘may yield useful data in some cases. 63 .E\> H u Tm— .uopfiao 0:» mo Buns 2.3 on. 33.93% «H 30mm 3.3 603 30.305 of saw? .963 053 NmO mv .N m be. poumfiggfi 3 Eopfigo 3:. £0839,an NA: 3 was .3“ 3V .b A3 35 pozofiflumm 30.35 mo .2651? < .w .m onfimmh 885.93.: b 3 3v 3. cm 2. u w 4.4. \ I X X .3: J. a." 3,3. .x. H ,3. :5: Mr: W: . "a“; g m u i Y/L . ““uu ©\x ‘ m ¢\K \/ N\x on “nun \ Q \ “a" ‘ m / we... ‘0; QONQO . wk... AV N30 ma. .N u kocofiwuuh 64 .50 A: .H 3 uowfiau 05 GM sewage??? ocO .E\> a u TW— .on was .2: A3 3355 new .m .m 0.3th E 859.? .3937? 9.3 m0 mflnm 05. Go? 33:35 3va 3.30on .o .m oufimfim AuxumconZWB :3 .3300 50.5 vacuumfifl m; 0; m5 . o w I_ o um .m oudwrm 803 3mm 0 «V 10 (wt/Am aw .m 0.33m Scum 3mm vvmmnvzw DIII I I ID.. I I I I.D mm .m oudmfim Scum 6.qu :1 O‘ O 65 In the next test, the cube of muscle illustrated in Figure 3. 10a was considered. The cube was exposed to a Z. 45 GHz plane wave, and each of its edges measured 1 wavelength. The cube was treated as a single cell, and the electric field at its center was calculated. Next, the cube was partitioned into 27 subvolumes, as indicated in Figure 3. 10b. The induced field was again determined, and the field intensity in the center cell was compared to the value obtained from the first calculation. The procedure was repeated for the quarter-wavelength cube shown in Figures 3. 10c and 3. 10d, and the results are presented in Table 3. 1.. Noting that the two values agree well for the quarter- wavelength cube, and recalling the results of the previous test, we will take 1/4 wavelength as an upper bound for the largest dimension of any subvolume. The last example was chosen to evidence the accuracy of the method. An electrically small dielectric cube (measuring 4 cm x 4 cm x 4cm) was irradiated by a plane electromagnetic wave, for various values of frequency and dielectric constant, as illustrated in Figure 3. 11. We ex- pect the electric field near the center of the cube to be very nearly equal to the electric field near the center of a sphere with the sane dielectric constant in a uniform electrostatic field. The field E in the sphere is given by E = -é—3Tz- Ei , (3.7.1) 1' where E1 is the externally applied field, and er = 6/60. The table in Figure 3. 11 shows that the numerical solution is consistent with this expectation. B. Internal Electric Field This group of examples is devoted to calculating the induced field in bodies of various shapes and sizes, with both uniform and nonuniform 66 .zozuoomnou .fiwsofiocfimz’ «in was sumac???» odo 0.23.605 manna 05 mo compo 23. .2: and 3V 5 3.5333 3033 S. 35 3336 was .3. can 3 S 53 03¢? on 63.3.3 .233 053 NED m¢ .~ m .3. pouafigam 306.98 .«o 3350 038 .3 .m 0.3th E\onfi 5.... u .0 Au. a /// NED m¢ .N u honoavonh 67 Number Size of [El center Figure of cells each cell (Volts/m) reference 1 A O. 0789 3. 10a 27 x/3 0.0922 3.10b 1 x/4 0.0592 3.10c 27 x/12 0.0556 3.10d Table 3. l. Induced electric field at the center of the muscle cubes shown in Figure 3. 10 for various numbers of subvolumes. lill = l V/m. Ei 68 \\ / Y i'c ' ’32 [El] = l V/m \ f 20 6 )‘e 3 [El center (Hz) (cm) r (cm) 61. 4' Z (Volts/m) 7 3 3 10 3 x 10 5. 0 l. 342 x 10 0. 4286 0. 4172 6 4 4 10 3 x 10 5. 0 l. 342 x 10 0. 4286 0. 4172 3 7 7 10 3 x 10 5. 0 l. 342 x 10 0. 4286 0. 4172 3 7 6 10 3 x10 20. 0 6. 708 x 10 0. 1364 0.1124 3 7 6 10 3 x 10 51. 7 4.172 x10 0. 0559 0. 0503 Figure 3.11. Electric field induced at the center of a dielectric cube, for various values of frequency and dielec- Incident electric field is a plane tric constant. wave. 69 incident fields. The variety of examples presented here underscores the versatility of the moment method. A 300 MHz plane wave iInpinges upon a plane conducting layer in Figure 3. 12, with the incident electric field perpendicular to the plane of the layer. The dimensions of the layer are 6 cm x 4cm x O. 5 cm, and its permittivity and conductivity are 70 60 and l mho/meter, respectively. The induced electric field, which has essentially only an x-component, is shown in Figure 3. 13. The field is nearly uniform near the center of the layer, and is approximately 60/] 6 + a/jwl times the incident field. This result can be anticipated from the boundary conditions on E. In Figure 3. 14, the plane of the layer is parallel to Ei; the x- and z-components of the induced field are presented in Figures 3. 15 and 3. 16, respectively. By is small compared to Ex and E2. We note that Ex is about ten times larger than it was when the layer was perpendicular to ii. In some parts of the layer, E2 is about as large as Ex’ even though the incident field has only an x- component. This example shows that the induced electric field in a conducting body depends very heavily upon the body's orientation with respect to the in- cident field. Figure 3. 1? depicts a system of two tissue layers, fat and muscle, illuminated by a plane wave at 100 MHz. The body measures 160cm x 12 cm x 4cm, and each layer is 2 cm thick. In Figure 3. 18, the com- ponents of the induced field are shown. The magnitude of Ez is com- parable to that of Ex in some portions of the fat layer; this result cannot be predicted by the plane slab model, since it assumes only an x-component in the body. In addition, the two models produce different values for Ex in the system. The plane slab model predicts that 70 £9»: 05 mo 233 05 on negogaomuom am new? .933 £83 £2 com a o» pouomxo uohma waflosvaoo mama 50 m6 b Giana H w 0 won um: com u 4.235695“ .2 .m 989E 6cm C.\ ' 1 I J. Ei . 1? Figure 3. 13. 71 T T I I I .0159;.0161:.0155i.0210 I I----:—---w'-----I----« I .0110I.0112:.0108:.0155 ' I I ----- .t-—--'--—-q—---d I I .0115I.0116:.0112:.0160 .--_-,---1---_1 ..... I .0112;.0114:.0110I.0158 ..... I.--_.'---..I.---__ .0112; .0114I .0110I.0158 .0112:. 0114' .0110:. 0158 I Frequency = 300 MHz Ex(V/m) 1 .0112: 0114I. 0110I. 0158 b---J: ----- r----:L----d .0112:.0114I .0110l. 0158 I I _---'----f----'—---d I I I .0112I. 0114: 0110‘. 0158 I F----r----L-——-:----d .OllSI. 0116: .0112E.0160 I----:----.I-----;.----. | I .0110: .0112' .0108.0155 I I ----- +----4--- - J-----q I | I .0159'.0161: .0155:.0211 J 1 l +-——————chh—————4fi m II 70 e o l mho/m q II Electric field induced in the layer of Figure 3.12, with 6 :70 £0 and o- = 1 mho/m. Only half ofthe layer is shown above. EV and E2 are negligible. 72 #9»: on» mo 093m 2: o» Handgun HE A»; 33.3” moflosvsoo 0539 a noun newsflash obs? 283 3.?” com < :3 .n 0.3th .6 50m .0 E\on§ H W 0 won 832 can u hogauouh '-- I ---¢...l...- I 73 05.00..» gone 3 H083 0:» mo 30: >30 .E\ong — u 6 98 won .I. 0 5m? .3 .m 0.33am mo Henna 05 3 6003.3 30mm 02330 05 mo 30:255on 03H. .3 .m onnmwh E\one g n b O w CF u w T Lumen I. a N‘II . q q 1 . A? d a u q q u u _ n n x n u u u " moo." at." was .m ...S .u 13.“ m2. 2132.12.33." 83.32. . ..... mII- WI---".IIMII..m..IIIIIW ..... “III".IIJHI--...MI-I emo.mmml.na-.mm8~.umml.nola. was.n_ml._so~.mNom.uuo~.uoml. . II+IIwII+II+IIT.-I ..... TII-H.-..--.“..II..."I-I."-I163 . . . . . . go.“ H2." $7.37JS." 2:. 87.2132."$732.32. am . I. u a _ a . u _ . . ..... 1.8.23."58.18.1843? $938."108.32.301.89 - P p p P n p m P m .m AE\>Vum « am: can u 58:03.”... Ufi 74 .o>onm 859? 3 Hubs 05 no as >30 .E\on§ H u b can w on u w 5?? :3 .m 0.2.me Ho 9923 on» 5 @0363 33m 3.34030 93 mo “3239.900 nu 09H. .3 .m uunmwh E\o&§ ~ u b 0 w on. n w Tl Maonw i P N II n u u u d. J.« u n u m u go; 18.. 39:8.“ o8." 2o. to.“ £5.“ NS." m8; 35.. :o. . . _ --..-r.. . . . . _ . . _ . . . . So.m So; 2938; NS." So. m8." So." So." so.“ omo.mo~o. . uluulnllllfilnul.mnnlnLWuualnlulIATllnuwnnulTuulumluulmlInluunuI. GUN m . . wmoLEo; m8."$o.:mo.u$o. ~3.K8..3.o." «8.. $933. fl " u _ . _ - . . u _ IIIIIIIIIIIII IlulLlnlIu Illllllul Illhllllflll InIILIII j .1 u A n .+ u. u " ... u 1 23..3318.38.38.28. 30.38.38."8978.38. _ . p P p — . . . w «M cabana fi NEE com u >0do~50uh 75 Muscle Fat ‘f " I” I” o” I 'I I I” 2 F ' ’ / \ I \‘l : ?\\ | ’o‘ ' ‘s . 1’ ~\ I ‘1‘ I,” l6crn \: : ‘\ x I I k Ei Hi Frequency = 100 MHz / 05° 6 = 7. 45 60 Fat 6 = 71.7 60 o' = 0. 0475 mho/m Muscle 0' = O. 889 mho/m Figure 3. 17. A block of tissue composed of a fat layer and a muscle layer, illuminated by a plane wave at 100 MHz. 76 ‘Fat layer (2cm thick) Muscle layer (2cm thick) 0 u 1 u 4. q 9 _ 5 _ 7 _ 8 _ 8 0 . 0 _ 0 . 0 O O - C O 8 . 8 . 5 _ 9 5 . 9 . 2 . 3 4 — 9 — 8 - 4 0 . .1 . .I. . .1 O P G Ll Oil-H”! O x m E W ( 8 J. 9 . 9 _ 3 9 1.. . 5 _ 9 Z n 3 u 3 . 3 O O O u o l .... u ..... L .... r -II'I 0 . 9 u 4 n 5 Z . 3 . 4 . 5 '-'P'h'-...'L-"O"r--O.-l 0 . 5 8 . 9 3 . 6 9 _ 7 Z n 3 4 u 5 P o r o x 8cm »---- co m h 7 W/ c . 8m 8 l . u 7 0m M = = € 0 6 q 5 d 3 _ 3 3 . .l. . .I. . O 0 . 0 u 0 . 0 O - o I - O '''' *----L'l"-r|-"l 4 . 4 . 3 _ 3 3 _ .1 _ 1.. . 0 O . 0 . 0 u 0 e — o e o llll 4|---‘u'"'-L---IL 3 . 5 . 5 . .I. m _ m . m . m . _ . o + o L elm o v. m E / V I.“ 4 _ 6 . 4. u 8 6 _ 9 . 5 1 l u 0 u 0 n 0 9 . 8 . 8 . 3 8 . 6 . 3 . l 0 u 0 . O _ 0 0 o - O - O T'---‘ ........... J 5 . .2 .q z 4. 4 Z . Z 1 . 0 0 . 0 u 0 . 0 O —y C D O - O O E 5m 5 U/ f. O a 4 0 F 7. mm = = E G - a d 6 3 1 0 . M . 0 n O O u 0 u 0 . 0 "'O-'rll‘O' '—-'-O.--T'-O'l 7 u 6 u 4. . l 0 O 0 . O O u 0 u 0 . 0 O O O - C T---'L """ T""T"'J 8 . 6 . 4 . .I. O _ 0 _ 0 . O O _ 0 . 0 . O o P I P 0]” s z m E / W 6 u 9 u 4 u 3 0 5 . 0 3 3 u l . l u 0 l . 4 n 4. u 6 7 . 6 0 . 3 Z . 1.. _ 1 . 0 O - O .r C - . I"'-+ ......... 1-"J 8 _ o — Z - 6 6 . 7 . 0 . 3 Z . 1 — 1 - o O P O L C \— C = 100 MHz Frequency /m 0. 197 V 10 V/m at the 5’: The electric field induced in the block of tissue pictured Only 1/4 of each layer is shown above. For a corresponding plane slab model, at the center of the fat layer, and Ex center of the muscle layer. in Figure 3. 17. Figure 3. 18. 77 Ex = 0. 197 V/m at the center of the fat layer, and that Ex = O. 210 V/m at the center of the muscle layer. In the next example we examine the induced axial electric field and the power density in cylinders of salt water, having 1 Normal, 2 Normal, and 5 Normal concentrations, and having various lengths. A typical cylinder, along with the coordinate system, is shown in Figure 3. 19. At 9.45 GHz, the lengths of the cylinders are approximately )‘0/4’ Xo/Z, A0, and 3Ao/2, where X0 is the free- space wavelength. The electric field intensity, power density, and total absorbed power for each cylinder are given in Figures 3. 20 through 3. 25. For a given cylinder length, the field distribution has the same general shape for all concentrations, although the total absorbed power is generally greater for the higher salt concentrations. We also note that the total absorbed power Pt reaches a relative maximum when the cylinder is about xo/z in length. The next two examples investigate the effect of an inhomogeneity in a cylinder, as illustrated in Figure 3. 26. Figure 3. 26a shows a muscle cylinder, measuring 10cm x 1mm x lrnrn, illuminated by a Z. 45 GHz plane wave polarized parallel to the axis of the cylinder. In Figure 3. 26b, we have a similar muscle cylinder with a segment of fat 1cm long at its center. The induced axial field and the power density for both cylinders are plotted in Figures 3. 27a and 3. 27b, respectively. Since the electric field in the inhomogeneous cylinder is normal to the muscle-fat boundary, it obeys the boundary condition (a-M + jw€M)EM = (ch + jweF)EF , (3. 7. 2) where the subscripts M and F refer to muscle and fat, respectively. Thus, near the boundary, both the electric field and the power density 78 .d—Hfihw .fi an gm 2" «0.3305 .3937? one no nomaoouunnouo can. .933 £33 £0 a. .o a .3 833.55 a £98732 mo 895.6 3333.. < .S .m 0.3th / Mm “\ .6 N30 mm. .o u hudufiuouh E(v/m) E(V/m) E(V/m) Etvfin) Figure 3. 20. 79 0. 6 ~- 0. 4 0 h = O. 1.13 lo 0. z .. Concentration = l N O ‘ 6 = 50 6 o o. 2 ° x/xo cr = 20. 37 'mho/m O. 8 -- _ Frequency - 9. 45 GHz 0. 6 0 h = O. 227 X0 0, 4 .. 0, 2 .. h=0.510 X0 0 0.2 0.4 0.6 0.8 Electric field E along the axis of the saltwater cylinder shown in Figure 3. 19, for h/).B = 0. 113, 0.227, 0. 510, and 0. 737. Concentration of t e salt solution is l Nor- mal, and the frequency is 9. 45 GHz. 80 2 Concentration = 1 N -3 3 P(x lO mW/cm )1 h=0.ll3Xo P = 4. 97 nW O t 0 0.2 E ___ 50 6 x/Xo ° 0' = 20. 37 mho/m 5T Frequency = 9. 45 GHz 40 h = 0. 227 X0 P 10'3 w/- 3) 3" (x m cm pt = 74. 45 nW 24» 1» o a i : 0 0.3 K x/ o 31 h = 0. 510 X P(x 10- 3‘mW/cm3) o 1.. Pt = 70. 01 nW 0 fl . 0 0.2 0.4 0.6 x/Xo 3-- h=0.737). o P(xlO' 3mW/cm3) Pt = 110. 14 nW L A v A A A 1 T V fl 0 0.2 0:4 0.6 0.8 x/Xo Figure 3. 21. Power density P along the axis of the saltwater cylinder shown in Figure 3. l9, and total absorbed power Pt, for h/Xo = 0. 113, 0.227, 0. 510, and 0. 737. Concentration of the salt solution is 1 Normal, and the frequency is 9. 45 GHz. 0 db 81 0. 6 -- Concentration = 2 N E(V/m) 6:426 O a" = 23.41 mho/m Frequency = 9. 45 GHz 0-6" h=0.227 x0 E(V/m) o. 4 .. h = 0.510 x E(V/m) ° 0.6 h = 0.737 x O E(V/m) 0 4 r : c r. r 4 r 4 o 0.2 0.4 0.6 0.3 x/XO Figure 3. 22. Electric field E along the axis of the saltwater cylinder shown in Figure 3. 19, for h/Xo = 0.113, 0. 227, 0. 510, and 0. 737. Concentration of the salt solution is Z Nor- mal, and the frequency is 9. 45 GHz. 82 1 Concentration : 2 N P(x 10'3mw/cm3) b lh = 0.113 x0 0 p : 6.20 nW o 0.2 t 6:426 x/XO o 23. 41 mho/m q ll Frequency : 9. 45 GHz h:0.227)\ O P = 80.65 nW P(x10-3mW/cm3) t P(xlO- 3mW/cm3) 3 3 h = O. 737 RC P(xlO- mW/cm ) Pt =121.29nW 0 ' 0.2 ' 0T4 0.6 ' 0.8 X x/ 0 Figure 3. 23. Power density P along the axis of the saltwater cylinder shown in Figure 3. l9, and total absorbed power Pt, for h/kO = 0.113, 0. 227, 0.510, and O. 737. Concentration of the salt solution is 2 Normal, and the frequency is 9. 45 GHz. 83 O. 6» Concentration = 5N 0.40 h=0.ll3 X0 E(V/m) 0. 2 ‘N 6 = Z3 6 . o O * 4 0’ = 31.67 mho/m O 0. 2 x/AO Frequency = 9. 45 GHz 0. 8-- 0'6" h=0.227x o E(V/m) o. 4.» 0. 2» 0 - L 4. 0 0. 3 x/Xo O. 6.. 0. h = 0. 510 X o E(V/m) 0. 0.6 0. 6y 0.4+ h =0.737 X0 E(V/m) 0. 2‘ o 3 7 i ; 7 1‘ ; i O 0.2 0.4 0.6 0. 8 x/Xo Figure 3. 24. Electric field E along the axis of the saltwater cylinder shown in Figure 3.19, for h/K = O. 113, 0. 227, 0. 510, and 0. 737. Concentration of the salt solution is 5 Nor- mal, and the frequency is 9. 45 GHz. 84 1 Concentration = 5 N P(x10'3mW/cm3) | El h : 0'1” X0 0 Pt : 8. 51 nW 0 0.2 x/).() e = 23 60 6.. (r =31.67 mho/m 5.. Frequency : 9. 45 GHz 4'" h : 0. 227 h 0 P(x10- 3mW/c1n3) 3.. Pt = 87. 64 nW _3 3 h = 0. 510 X P(xlO mW/cm ) 0 Pt = 75. 35 nW 3 3 h = 0. 737 he P(xlO- mW/cm ) Pt = 127. 04 nW 0 t f : r . r 0 O. 2 0. 4 O. 6 0. 8 x/Xo Figure 3. 25. Power density P along the axis of the saltwater cylinder shown in Figure 3. 19, and total absorbed power Pt' for h/>.o = o. 113, 0.227, 0. 510, and o. 737. Concentration of the salt solution is 5 Normal, and the frequency is 9. 45 GHz. 85 .963 9.83 ~30 m. .N a .3 60339:“ povfiao gonad msoocomogoncm an 7: can 3365be 30998, mdooaomocuoa < A3 3: I'll ofiomsz \lnnm H8: SUCH nm ‘I oflomflz HM 808 w NED ma. .N u >ocofiwonh A3 1|.II 0H0 max/H .3 .m @3th .m .m 3 z 8 \ 3 LI] 1 o 1.2 MA 5 \ 0.8 B E M I O .. 0.4 2‘, n. 0 Figure 3. 27. 86 ‘ -———-— Homogeneous muscle cylinder J. 'l ------ Inhomogeneous muscle cylinder Distance from center (cm) (a) v LID—- Muscle- fat boundary 1’ I ’ I " l I I I l ., I ' /’ Homogeneous 'muscle I I cylinder 4r a I | I I I 1|- 4 I —————— , / Inhomogeneous muscle I '1 cylinder 1b I 'I r’ I I . . . . __. 0 l 2. 3 4 5 Distance from center (cm) 0)) (a) Induced electric field E, and (b) power density P along the axes of the muscle cylinders shown in Figure 3. 2.6. The frequency is 2.45 GHz, and IE1] = l V/m. Each cylinder was partitioned lengthwise into 100 subvolumes of equal size. 87 in the fat are larger than in the muscle, thereby creating a local ”hot spot". However, the overall heating near the center of the inhomo- geneous cylinder is considerably less than that for the homogeneous muscle cylinder. In Figures 3. 28 and 3. 29, the roles of the fat and muscle have been reversed. We again see a "hot spot” on the fat side of the muscle-fat interface in the inhomogeneous cylinder. However, the heating near the center of the cylinder is markedly greater when the 'muscle segment is present. Thus, the temperature of a cylindrical fat structure exposed to electromagnetic radiation could be significantly increased by the presence of one or more small 'muscle segments. An inhomogeneous layer of fat is exposed to a 600 MHz plane wave in Figure 3. 30. The layer ‘measures 5 cm x 5 cm x 1cm, and has a 1 cm x 1 cm x 1 cm cube of 'muscle imbedded near its center. Figures 3. 31 and 3. 32 show the x- and y-components, respectively, of the in- duced field; E2 is negligible. We note that Ex becomes quite large in the subvolumes immediately above and below the muscle cube, while the field in the ‘muscle is small. Again, this is due to the boundary condition expressed in Equation (3. 7. 2). As the last example in this group, we examine the induced field produced by nonuniform illumination of a homogeneous muscle layer. The layer measures 2. 5 cm x 2. 5 cm x O. 5 cm, as shown in Figure 3. 33, and is illuminated in a small area near one corner by a 600 MHz plane wave. Ex and By are presented in Figures 3. 34 and 3. 35, re- spectively; the z-component is negligible. Ex is largest in the sub- volume which was directly irradiated, and decays rapidly as we move away from the site of incidence. EY varies considerably throughout 88 By All «arm 303.2 \\ \ .53 & H83 ‘lnl warm” .963 053 smO mw .N m >n dogfight nopsmgo an.“ msoosomoccosfi am An: was £09330 «.3 mnoodowgfion < Amy 8.0 Cd .m .m H 9.30 mv .m n anoaosvonh .3 .m charm :3 I «ah Mm 89 3 v ! Homogeneous fat cylinder 1} I l A 2 " l "" " " " "' Inhomogeneous fat cylinder 8 \ > IF E \ V \ m I \\ 1 1' ~_ _ b—— Muscle-fat boundary \ .1.‘\\ I V O l A 44 : j _n O 1 2 3 4 5 Distance from center (cm) (a) O. 6 .. | I M" " I Homogeneous fat cylinder E a { o. 4 0 l E. ~\ l ----- Inhomogeneous fat cylinder 1? \ m \ " I \ I‘ o \ \ H 0. 2 1r \ ' \ 35, J \ Muscle-fat boundary - \ p4 Jb M 0 1‘ 4 0 l 2 3 4 5 Figure 3. 29. Distance from center (cm) (13) (a) Induced electric field E, and (b) power density P along the axes of the fat cylinders shown in Figure 3. 28. The frequency is 2.45 GHz, and IE1] = l V/m. Each cylinder was partitioned lengthwise into 100 subvolumes of equal size. 9O 5 cm = 600 MHz Frequency Fat { 52.47 6 o = 1.49 mho/m 6 Muscle o- o 086 mho/m 6:5.66 :0, 0' An inhomogeneous fat layer illuminated by a uniform plane wave at 600 MHz. Figure 3. 3o. /Fat 91 Ex(V/m) $y - - J 4 . . . . 5 . 7 . 9 . 7 . 5 5 . 6 n 6 H 6 . 5 o - o o o — o . _ u . r llllllll .ulllllilu IIIIIIIII T IIIIIII 4.. llllllll . _ _ . . O/ _ 2 u 4. . 2 . 01 0 . 3 6 . 3 . O 5 6 u 6 . 6 . 5 o - O O - O - O _ . _ . _ . _ . """"""" -'-"-""-"--'-'|'-l""-"J‘-'I""l . . . . 9 . 3 . 7 . 3 . 9 0 2 . l . Z . 0 5 u 6 u 6 _ 6 u 5 O n 0 . O u 0 - O _ ...... &"""--- A11 V.'-"""I-' -"-- -1 - .e n 6 . 9 1 Z 9 . 6 6 _ O c 9 0 . 6 5 u 8 M 0 a 8 . 5 o o . . O - I u M . r'-"|"-l"--"' '--"l-k ........ u . . u 6 . 3 . l u 3 . 6 5 . 6 _ 5 . 6 . 5 5 _ 6 n 6 . 6 . 5 o - O - o . O - O . . . hr F b p T c l 5 1.1 = 52.47 E 5cm o = l. 49 mho/m 6 0’ Muscle { m 0 h om 6% 60 5.0. :: 6.0 Fat { Frequency = 600 MHz The incident field is The x- component of the electric field induced in the layer illustrated in Figure 3. 30. polarized along the x- axis. Figure 3. 3 l. 92 X 1 / Fat E V 'm YI ) / I T 1 I I I I I I : I I I 129 g 049 i 023 : 067 : 121 : 1 I i I. _______ ' _______ I__-_.-_..'_ _______ I _______ T I I .- I | | ' I ' . ' 077 I 020 1 .046 : . 042 i 049 i : I : F’ ------- ' , 7 r. ‘ ———————— -'r ——————— ‘IL ------- -I _ Muscle : : 5 cm .000 ~ .000 .000 g 000 : 000 , . ~ I I .. 1 ' I I- ------- " ’ ' ———————— :-——--—-r—--———--I I I I : . : . . 077 ; 020 I 046 . 042 : 049 : i i ' P. ______ ... ______ .1 ....... l ——————— I ————————— I | I ' : : ' I . 129 l . 049 : 023 i 067 : 121 1 = ' '1 ' l l I ’Y H: 5 cm H 6:5.660 6:52.476 Fat Muscle o 0' = 0. 086 mho/m 0' =1.49'mho/m Frequency = 600 MHz Figure 3. 32. The y-component of the electric field induced in the layer illustrated in Figure 3. 30. The incident field is polarized along the x- axis. 93 0.5cm € : 52.47 6 o 1.49 mho/m 0": = 600 MHZ Frequency ted only in the 1na A homogeneous layer of muscle, illurn shaded area by a 600 MHz plane wave. Figure 3. 33. 94 X T EXIV/m) ' l T I I 5 : : : .0039 : 0041 I .0099 : .0452 I 0107 E I : I I- ------- ' -------- I' ------- M. ‘ e i' 4,5; -------- I I I" E .0081 I .0082 I 0191 I .1006 ; 0207 I I 5: A - h——————-' _______ I ———————————— I'L ‘ " ———————— I I E I 2.5cm : I I I 0074 . .0079 I 0134 I 0650 I 0148 I : I I _______ '—-———-—L—-————l---—-—.—'—_—---- I I l . I , I I I I I .0044 g 0048 I 0081 : .0397 I .0088 I I : I I- ——————— I— —————— L ------- . ------- J -------- I I . I I I I I 0015 I .0017 : .0039 1 0185 I 0041 I 1 I I I I 1 l I l - v [If 2.5cm H 6 = 52. 47 6 O 0' = 1.49 mho/m Frequency = 600 MHz Figure 3. 34. The x- component of the electric field induced in the layer pictured in Figure 3. 33. The incident field is polarized along the x-axis, and irradiates only the shaded portion of the layer. 2. 5 cm- 95 Frequency = 600 MHz Figure 3. 35. X E V m I ,1 7 . I I I ' I I : I 0061 5 .0136 I 0227 . .0065 : .0093 I I ' I ________ §______,J________I I_______‘ I ' iv I I iE‘ 0013 I .0031 I 0048 .0018 .0011 i I * — —————— . ——————— .L ------- .—---——--I I : I I I .0031 : .0068 I .0118 I 0030 : .0055 I l I _ ______ I ______ .I _______ I _______ : ________ I I : I I : I : .0026 E .0060 g 0098 I .0031 ; 0033 I ' I I I ....... .1-----.._-:---_--...I---_--_1'.-_---..-. I : I I I 0027 I 0059 1 0095 I .0030 : 0034 I I I I I I l I 1 I‘" 2.5 cm "I 52.47 s O l. 49 mho/m The y-component of the electric field induced in the layer pictured in Figure 3. 33. The incident field is polarized along the x- axis, and irradiates only the shaded portion of the layer. 96 the layer, but we note that its maximum value is attained in a subvolume other than the one illuminated. C. External Scattered Field In the following group of examples, we examine the backscattering characteristics of various thin cylinders exposed to uniform plane waves. The incident field is parallel to the cylinder axis, so that only an axial field is induced. Equation (3. 6. 3b) is used to compute the backscattered field. The 0 dB reference for all figures in this group is IE8] = 0. 01|Ei|. The first example investigates how the phase angle of 'T affects the scattering behavior of a cylinder. We note that 7', given by Equation (3. 2. 6b), can be thought of as a complex conductivity, having a mag- nitude I?" and a phase angle <1). Teq is purely a conduction current when 4) = 0°, while ¢ = 90° implies that Teq is solely a polarization current; if 4; = 45°, the conduction and polarization components of qu are equal. In Figure 3. 36 we have plotted the backscattering, calculated at a distance of 30cm, as a function of length for three different cylinders. The magnitude of ‘T was 100 in all three cylinders, but (I) was different for each one; the respective values were 0°, 45°, and 90°. We note that the backscattering increases with 4). Although all three cylinders exhibit a resonance at xo/z, the peak grows sharper as 4) increases. This occurs because the losses in the cylinder decrease as 4) grows larger. Hence, for a fixed value of [TI , the scattering will increase with the ratio of polarization current to conduction current. We compare theory and recent experimental results [10] in the next four examples. In each case, the scattered field was determined 97 £06310 some Eoum So on no? ufiom son—«Shown? 03H. .6 mo 3mg woman 05 «o 33.2, 0330mm?” on» «an .oudo homo 3 on: as? b no ovum—mamas“ 03H. .953 383 N30 8 m o» @0893 myopnmgo mdoonomogos «nouomflp 00.23 59G mfinoflmomxomm .ooo can .03. .00 one? .6... .m charm o x\s~ m .N o .N m A o ._ m .o o o u G DID .I ,_ O I U\D\D DID u wag 0m¢ I e 4...-.4 \ /D\.u Em 0H H “N hi ~ r 00 fl Au 1 Id! .fil .0 0 40h. “I . HIL] MN q\ 4a.. r 0 III ..\ . . am \. . . gm . . .o 1w ._. r am a \D 4 o l o .- \U\U\.U xx. TM— 8 I _mM_ .oonouomun mp0 CH \0 Q \0 I . D Inxxdv \ -420. D .4. \ \q 0 INH \0 sq o o o o o... . . £687» . Q . . 3 umO 0H 0 koaodvouh (HP) Sutzaneosno'eg 98 15 cm from the corresponding cylinder. Figures 3. 37 and 3. 38 illus- trate the backscattering from salt water cylinders at 9. 45 GHz. The concentrations of the salt solutions are 1 Normal and 5 Normal, re- spectively. In both figures the agreement between theory and experi- ment is very good. We note the absence of resonances at xo/z, even though, as we recall, the total absorbed power reaches a relative 'maxi- mum at this length (see Figures 3.21, 3. 23, and 3. 25). Although the assumptions we made in developing the moment solu- tion are not valid for a good conductor at microwave frequencies unless the number of subvolumes is extremely large, an attempt has been 'made to calculate the backscattering from a brass cylinder at 9. 45 GHz. We have compared the theoretical results with experiment in Figure 3. 39, and find the agreement to be surprisingly good. Although the current in the brass cylinder is essentially a surface current, the model used here can provide useful data on scattering from metallic cylinders. In Figure 3. 40 we have compared the relative backscattering from a 1 Normal saltwater cylinder and a brass cylinder at 9. 45 GHz. The experimental results agree well with theory. When the cylinder lengths are odd multiples of xo/z, the brass cylinder exhibits resonances, and scatters more than the saltwater cylinder. At lengths which are even multiples of Ito/2, the saline solution scatters ‘more. Thus, it is pos- sible that a cylindrical biological structure, such as 'man, could scatter more microwave energy than a similar metallic cylinder. D. Symmetric and Antisymmetric Modes In Section 3. 5, we discussed the decomposition of an incident plane wave E1 into a symmetric component E; and an antisymmetric component E2, given respectively by Equations (3. 5. 11b) and (3. 5. 11c). When the 99 £09330 0:» Eon.“ Eu 2 003 «smog “339,90 who 0:8 .030 m0 .0 «0 aoflmnflaosoo Wfihoz H .«o H0637? H303?» 0 59¢ m5u0fimomxomo. 90m 3053098 can F305 mo «3382580 og\n~ moN O.N moa 03H moo O P I 5 II - 1 OH I» h I.‘ an. d- 4- I0 {I TM— Ho . H _mm_ ”0oc0n0m0n mp o .a 7... I! 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NED m0 .o u hoaofivouh 1.. on .00 .m 800E (gp) Sutxen'eosnoeg 102 #09330 £000 503 :3 mH 00.3. «52H noHu0>u00no 05 H30 .030 m0 .0 003 930500: 03H. 5030300230 H0§.37H H mo 00051.0 9303:00 0 H30 H0HEHH>0 000.3. 0 895 mGHu0fi0omxo0o. 0>Hu0H0n .3m «0053093 H30 9305 .Ho GO0Hu09hoU .00 .m 0.33m o x\:~ 0 .~ 0 .m m .H o .H m .o o u 0 .. .I v a u n 4 1" 0H _HM_ Ho . .I. _0M_ ”0on0u0m0n mp o 4. I H _ _ INH _ H H .1 0 a ‘II'E‘IN. s m «x imam M. .0 3 AN . I W SH 0 .H u MN m. HI “09510 00093 .308309AM '41.. pH u. F 11 u 009330 000.2H .9393. IIIII a“ .r P H \ as. w o.— " MN .5. 000506 00300 zH £00520me Isl i: \ 000330 00200 7H H .roonh I from NHHU m0 .0 n twoaofiwouh 103 plane wave impinges upon a biological body, each of these components induces a partial field inside the body; E: excites a symmetric mode 38' while E: induces an antisymmetric mode Ea“ The total electric field E inside the body is simply the sum of Es and Ea. The examples in this group illustrate some of the characteristics of the symmetric and antisymmetric 'modes. In each example, both is and Ea exhibit nearly linear polarization; they differ ‘markedly, however, in their spa- tial distributions. Es is generated 'mainly by the incident electric field, while Ea is due primarily to the incident magnetic field. Figure 3. 41 shows the symmetric 'mode E8 in a muscle layer mea- suring 5 cm x 5 cm x 0. 5 cm, irradiated by a 1 GHz plane wave. Ei is parallel to the plane of the layer, while Hi is perpendicular to it. The magnitude of the symmetric ‘mode is greatest at the center of the layer, and decreases toward the edges. We note that is is roughly parallel to Ei throughout the layer. 3 The antisymmetric 'mode Ea in the same layer is illustrated in Figure 3. 42. Ea is small near the center of the layer, and generally increases toward the edges. In contrast to the symmetric mode, the field lines of the antisym'metric mode circulate about iii. '38! and [Ea] are roughly equal in this example. Since Es and Ea are not in phase, the total electric field in the layer is elliptically polarized. The ellipse traced out by the electric field vec- tor in each subvolume is plotted in Figure 3. 43. The polarization varies throughout the layer from almost linear to nearly circular. Figure 3. 44 depicts a 100 MHz plane wave irradiating a loop of muscle measuring 10 cm x 10 cm x 1 cm. Ei and Hi are parallel and perpendicular, respectively, to the plane of the loop. The symmetric 104 X if 01 U = . 38 . 33 .8” . ' . 34 Ex 1187: 1 |19: 1 |19: 126|21: 1 1.6 + I I / I I I I | / I / .i . . : “132:.01 7 2°I 05 |7,8". 075 1:1 091|11.2°'l .09 |14.4° .21 [17.24. 205|17,8°: .194 [120' .178 RPS-1.561.235- I I I 1 I l f I / I f : I I I 'L.012 7°! 035 |7,5°I 052l8. I 061 07°| 054 ° .256 [16.31‘. 248 [1701' .232|18.6CT 208 21‘, .184 3.9.? T I I . I I I I I I . I f : ' d 008|6°I.02 |6. 6°I 0.7|7.8I 043 9.8d.037 13° r ------------------- 227 21°: 194 4.5 1 I I I I 025|2.2°I .022 ° ------------- fi------- 23 °: .199 4 I Figure 3. 41. I I I Q d .0 008 8.9' .0 007 ZQ3< f = 1 GHz 6 = 50.71 r . -jko z E; = Re(e Gan/I) = coskozé The symmetric "mode of the electric field in a muscle layer exposed to a 1 GHz plane wave. The symmetric mode is induced by the symmetric component (cos k0 2 Q) of ii. Only 1/4 of the layer is shown above. 105 Ex=011 -83‘“: .035 83‘”: .057 820i .079 L81d-E .109 79 I I I | I --——-> : -————.> : ‘\‘fl- : \\‘\ E 1‘ E=.2 .33- .233 §.3°: .189 gfl .124 .4°I .043 [0.8% - --’ - ------ 6r ----------------- ‘ .022 -89 I .068 |-88 : .112 |-87°: . . -84 , 8 I I I - U. ------------ U' . .12 [:92 I .196 -91' .267 |-9o°I .331E89 , I I -“_. : \\~.K : .188l8gg'i .152 M 70%}???- . 154 |-90°: I .136|£._9fI .110L78_.8_d|. +- .11 |-92°: .181 91“, . \ I \i ;l I .082 |77.6°I .066 [153: .044 M .016 I I z .03 . [77° I .027 !77° .022 [76$ .- 0151762“: .. 006 J . z O i E =1GHz = 50.71 1 K H = 1.62 1/52-m -jkz =lrn(e 09) = -jsink 29 o Figure 3. 42. The antisymmetric mode of the electric field in a 'muscle layer exposed to a 1 GHz plane wave. The antisymmetric mode is induced by the antisym- metric component (-j sin koz A) of E1. Only 1/4 of the layer is shown above. 106 Ex=.137h4.3 , .134 [4.3 .133 [11.8" .132 [:14 ., .154|-22 . I Ez-=Z611800 .25 [72.2I .2221643“ 176 .531 .118 [323° ------—--- db----------- inc-“.- --a P’z’é’fi‘fis .197|-17Ir.193|:145. .197 [:27 27.5I. 216 [.386 .134 I54. 6‘I .209I77.6I . ., .079|38,1° 2.48L2i3‘ .237 - 9‘. .233L-_z_gff .24 Lzfin .266|-46° L-.1..L7=‘; 4’ -1.1§._8I .09.|_54_° .055938 1 57-75 w -26'|.-é.._ .258 -23 .269 t38° .295|-§1 W IE0 WC 000 _m '77» “9° LEI? .011 [39.5" Figure 3. 43. The total electric field induced in a muscle layer by a 1 GHz plane wave. Only 1/4 of the layer is shown above. 107 .mooH 05 .«o 0003 05 o» “08.3: 0M um 05:3 .903 05 mo 0:03 05 3 00: a“ .0>0B 0:03 032 o3 .0 .3 0000300.: 03058 mo mooH 000av0.< .00 .m 9303 un O u ~ommao o. ~oou¢n~coo-. coom Noounoeoooaoo o. Nolwoooomo_oo mum and Nw >w .~ I .4400 I zonp2 04w~u J (hawo cum: mhz000 2n ounzanmd «mJOQ thoh otu cacaon I K (hduo NetumnoONOMNo ~¢Iumnon~cONo ~00mo--ooco Nolwcnnmo—Q—c ~01uomosoan~o ooouchommawue noouonco—couo ~00wo~o~hn—mo aeowOMNmooNco x nu ma— Z—ONHQOONQO Nu—oun—n—o—o—a—n > - «1:9 —a¢u Ftvn Al!” x—n—o—oNNNHHfl rozcx I Mpzwzoutou adwuu I K2 _ o~xt cocoocom—o I >uzw30w¢u 130 "DELTA X", "DELTA Y", and "DELTA Z" at the bottom of the page denote the dimensions of each subvolume. The output describing the scattered field from the first example is shown on Page 131. Most of the columns on the page have two headings separated by a ”/”. The first heading in each pair is employed when rectangular coordinates are used; the second heading is used when polar coordinates have been given. Thus, in our example, the first three numbers are the x-, y-, and z-coordinates, respectively, in meters, of the field point. The next three numbers are E2, E3, and E: at that point. The "POLAR AXIS" column is blank when rectangular coordinates are us ed. The magnitude of each field component, as well as the magnitude of the total field, is also given in dB. The 0 dB reference is [ES] 2 0. 01 [Bil . When polar coordinates are used, the first three columns of num- bers give, respectively, the r-coordinate of the field point in meters, and the 0- and ¢-coordinates in degrees. The next three columns show 3 s s Er’ E9, and Ed) at that point. 4. 5. Listing of Program and Subroutines The listing of PROGRAM BLOCK and its subroutines begins on Page 132. 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III II u I! . «mnxu HO“ HUMWGHGQNZU-UH 0“ Pm“ OOHMPHHHOOHHUD HOG HZ QHuHH 1! “HO .- OOHUH K I-(‘DHHCHOGUOOJ'O uni-HU'H "I! N“! ZHN NN ED 2 HHzH £00OuIIII."O£)O¢"JI.A.D:3 omHuHmaamm—mmuo amHHxHBHsz gh(KK:O¥HWW Hmeuuagg wuwumngHu o HN cmOOH ' um OOOH QUH N a 0 0° GOG O U OHM” : I" 3 O m mm m mmm m m mm m U U U UU CHAPTER V SUMMARY This thesis has presented a technique for calculating the electric field induced in a finite biological body having arbitrary shape and com- position, when the body is irradiated by an electromagnetic wave. A knowledge of the induced field is important to researchers investigating the biological effects of nonionizing radiation. As an introduction to the study of induced electromagnetic fields in biological 'media, a plane slab model of a human trunk was analyzed. It was noted that the electromagnetic field induced in the model by a uni- form plane wave can be obtained by two methods: (1) by a direct application of boundary conditions, and (Z) by transmission line tech- niques. A group of numerical examples was presented to illustrate the behavior of the human trunk model at various frequencies from 100 Hz to 10 GHz. The problem of calculating the electric field induced in a finite body was considered next. An integral equation for the induced electric field was derived using the free- space dyadic Green's function. The method of 'moments was then us ed to transform the integral equation to a matrix equation for numerical solution. Techniques for calculating the external scattered field, and for using symmetry to reduce the ma- trix size, were included. A variety of numerical examples, along with some experimental data, were presented to illustrate the versatility and the accuracy of the moment solution. In addition, the computer 139 140 program used to calculate the mnnerical examples was described, and instructions for its use were given. The numerical technique presented in this thesis has a serious drawback: if a researcher wishes to study a model which must be sectioned into a large number of subvolumes, his computer system must be able to invert enormous matrices. Such a case might occur if one or more of the following applies: (1) very high frequencies are used; (2) the 'model is quite large; (3) the model's shape or composition is complicated; (4) a "fine-grained” solution is required. Thus, a use- ful topic for further research would be that of developing an efficient way to invert very large matrices. Also, a great deal of work has yet to be done on probing the induced field in a body of biological material. There is at present a conspicuous lack of experimental data on this subject. APPENDIX APPENDIX A EVALUATION OF THE SCALAR COMPONENTS OF THE FREE-SPACE DYADIC GREEN'S FUNCTION The free-space dyadic Green's function may be written as -jko|?-?'| 563,?!) = -jwuoh’ + VZV] e _. _. , (Al) k0 41rlr-r'l where? z; QQ+A9+ AA and k0 = walpoeo . The scalar components of 86", r") are given by Z -t-*, _ . l a Gx x (r,r) ' -Jw“o[¢ 6pq+-2'Fx_5qx—] ’ (A2) pq k0 q p p9q = 192,3 9 where -jkoR 4. = MR) — 37m.— , (A3) 1/2 —> 2 2 R: Ir-r'l :(ul+u2+u3) , u. : x.-x' 1 = 1,2,3 , 1 1 and xlzx, x2:y, x3=z 5 6R 1’2 1nce R— = R , we have P a”! _ gar: _ d :12 x dR Exp " 3% R ' (A4) Then, ‘141 2 u u u 8 8 d 3 d 8 a—aLx x = qu(€%'fie) = f'a—thfi) + %% qu(‘§) u 2 Bu =_2d 3R+_Li‘1’-[R -u 33]. (A5) R dR 5xq R2 dR xq p 5xq Or, 2 u u 2 u u 3’93 _ 239$ .Léil: -173] A x x ' R2 dRZ+RdR 6pq R ’ (6) Bu since —8—2 = 6 x Pq Substituting Equation (A6) into Equation (A2) and rearranging terms gives G (F r") — -jwu [(¢+._l__di)5 4.3233(fl- iffl’.)] (A7) xpxq o kiR dR pq 1(ng dRZ R dR With 4; given by Equation (A3), it can be readily demonstrated that $5; = -¢[jko+-Ilz] . (A8) Thus, 3—3- =-¢a[jk.+:1- [jko+:]:% = :97 +¢[jko+ lif . (A9) Or, $=¢[fi3+2;2-ki]. (A10) Substituting Equations (A8) and (A10) into Equation (A7) gives Gxx (a?!) : -jw“o [4" +(-lR—+jko)]6 p q k ‘1 ‘1 z ijo 2 1 . )] +§§[¢(-—2+ R -ko)+%(-§-+Jko . (All) 0 143 Equation (All) may be rewritten as _. _. -quo¢ Z Z G x (r,r') = T—Z-[(koR - l- jkoR)6 xp q koR Pq 11233 2 2 + R R (3 - koR +3JkoR)] . (A12) 32 Letting o. : koR, cos 6x = R , P u cos ex = —R3 , and using Equation (A3), q we have —>—» -jw“okoe-Ja Z . Z . Gx x (r,r') = a; [(o. -1-Ja)<‘5pq + cos 9): cos 9x (3-0. +3Ja)] , p q 41m p 9 (A13) which is Equation (3. 3. 4). BIBLIOGRAPHY 10, 11. BIBLIOGRAPHY Collin, R. E. , and F. J. Zucker, Antenna TheorLPart 1, pp. 41- 43, McGraw-Hill, 1969. Harrington, R. F. , Field Computation bl Moment Methods, ch. 1, Macmillan, 1968. Harrington, R. F. , Time- Harmonic Electromagnetic Fields, pp. 123-125, McGraw-Hill, 1961. Ho, H. S., A. W. Guy, R. A. Sigelmann, and J. F. Lehrnann, "Electromagnetic heating patterns in circular cylindrical models of human tissue," in Proc. 8th Ann. Conf. hflfical and Bioflfical Engineering (Chicago, 111.), July 1969, p. 27. Johnson, C. C., and A. W. Guy, "Nonionizing electromagnetic wave effects in biological materials and systems," Proc. IEEE, Vol. 60, No. 6, pp. 692-718, June 1972. Kritikos, H. N. , and H. P. Schwan, "Hot spots generated in con- ducting spheres by electromagnetic waves and biological iznplica- tions," IEEE Trans. B_i£medical Engineering, Vol. BME-l9, No. 1, pp. 53-53, January 1972. Leh'mann, J. F., A. W. Guy, V. C. Johnston, G. D. Brunner, and J. W. Bell, ”Comparison of relative heating patterns pro- duced in tissues by exposure to microwave energy at frequencies of 2450 and 900 megacycles, ” Arch. Phys. Med. Rehabil. , Vol. 43, pp. 69-76, February 1962.. Lepley, L. K. , and W. M. Adams, ”Electromagnetic dispersion curves for natural waters, " Water Rfsources Research, Vol. 7, No. 6, pp. 1538-1547, December T971. Lin, J. C., A. W. Guy, and C. C. Johnson, ”Power deposition in a spherical 'model of man exposed to 1-20 MHz electromagnetic fields, " IEEE Trans. Microwave Theory Tech. , Vol. MTT-Zl, No. 12, pp. 791-797, December 1973. Livesay, D. E., B. S. Guru, and K. M. Chen, "Scattering from finite biological and metallic cylinders," presented at 1975 Inter- national IEEE/AP-S Symposium, Univ. 111. at Urbana-Charnpaign, June 2-4, 1975. Plonsey, R. , Bioelectric Phenomena, pp. 204-206, McGraw- Hill, 1969. 144 12. l3. 14. 15. 16. 17. 18. 19. 145 Ramo, S. , J. R. Whinnery, and T. Van Duzer, Fields and Waves in Communication Electronics, ch. 6, Wiley 8: Sons, 1965. Richmond, J. H. , "Scattering by a dielectric cylinder of arbi- trary cross-section shape, " IEEE Trans. Antennas Propagat. , Vol. AP- 13, No. 3, pp. 334-341, May 1965. Richmond, J. H. , "TE-wave scattering by a dielectric cylinder of arbitrary cross-section shape, ” IEEE Trans. Antennas Propa- gat., Vol. AP- 14, No. 4, pp. 460-464, July 1966. Saxton, J. A., and J. A. Lane, ”Electrical properties of sea water," Wireless Engineer, pp. 269-275, October 1952. Schwan, H. P. , "Radiation biology, medical applications, and radiation hazards, " in Microwave Power EnLineering, Vol. 2, E. C. Okress, Ed., pp. 215-232, Academic Press, 1968. Shapiro, A. R., R. F. Lutomirski, and H. T. Yura, "Induced fields and heating within a cranial structure irradiated by an electromagnetic plane wave, " IEEE Trans. Microwave Theory Tech., Vol. MTT-19, No. 2, pp. 187-196, February 1971. Van Bladel, J. , "Some remarks on Green's dyadic for infinite space," IRE Trans. Antennas Propagat., Vol. AP-9, pp. 563- 566, November 1961. Van Doeren, R. E. , ”An integral equation approach to scattering by dielectric rings," IEEE Trans. Ante_r_1finas Propa at. (Comrnuni- cations), Vol. AP-l7, No. 3, pp. 373-374, May 19 9. HICHIGAN STATE UNIV. LIBRARIES )llWI)"lllWIHIWWIWll)WW\llWIWl 31293000939250