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ATE IVER ITY LIBRARIES

Illlllllliml \lllll fill l '.| w

3 1293 00908 2110

 

 

 

 

 

 

 

 

This is to certify that the

dissertation entitled

The Scattering and Receiving Characteristics Of
Monopoles and Slots in Tri-Layered Media

presented by

Wang-jie Gesang

has been accepted towards fulfillment
of the requirements for

Ph.D. Electrical

degree in

 

 

Engineering

Mama

Major professor

Date 11/64/9/

M5 U i: an Affirmative Action/Equal Opportunity Institution 0-12771

 

 

 

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Michigan Ci?“
Universixjy

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encirchnS-DJ

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

-:

 

 

 

 

 

 

 

 

 

 

 

 

THI

OF

THE SCATTERING AND RECEIVING CHARACTERISTICS

OF MONOPOLES AND SLOTS IN TRI-LAYERED MEDIA

By

Wang-jie Gesang

A DISSERTATION

Submitted to
Michigan State University
in partial fulfillment of the requirements
for the degree of

DOCTOR OF PHILOSOPHY

Department of Electrical Engineering

1991

THE
OF I

ABSTRACT

THE SCATTERING AND RECEIVING CHARACTERISTICS
OF MONOPOLES AND SLOTS IN TRI-LAYERED MEDIA

by

Wang-jie Gesang

In this dissertation, the scattering and receiving characteristics of monopoles and
slots in tri-layered media have been studied. Two-dimensional Fourier transform tech-
niques are used to derive the dyadic Green’s functions for vector Hertzian potentials
and electromagnetic fields. The electric field integral equation for a thin monopole and
the magnetic field integral equation for a narrow slot are converted to Hallen-type
integral equations. Galerkin’s method is used to solve the integral equations to obtain
induced currents on the antennas. Antenna parameters investigated are input
impedance, radiation pattern, received power, and radar cross section. Various
numeric-analytical techniques are exploited to evaluate the entries of impedance and

admittance matrices accurately and efficiently.

Theoretical results are compared against published data and good agreement is
observed. Antenna current distribution, inth impedance, radiation pattern, received
power, and radar cross section are obtained for a vertical imaged monopole and a slot
in tri—layered media with various substrates and superstrates. Emphasis is placed on the
interaction of a lossy superstrate with an antenna. The theoretical results demonstrate

that, for an antenna in tri-laycred media with a lossy superstrate, the reduction in radar

cross section is ,
this dissertation

capabilities and

cross section is greater than the reduction in received power. The theory developed in
this dissertation can aid in the design of antennas with good transmission and receiving

capabilities and low radar cross section.

To my wife
Wei Cha

First. I wish
and support throui
have had the opp<
also like to expre:
for their generous
and rewarding ex

I am much I
grateful to the fe‘;
their help.

I owe 3 gm
mt km. 1 Hum

mm enmmgen

ACKNOWLEDGMENTS

First. I wish to thank Dr. Kun-Mu Chen, my academic advisor, for his guidance
and support throughout my study at Michigan State University. I feel very fortunate to
have had the opportunity to work under his supervision and to learn from him. I would
also like to express my gratitude to Dr. Dennis P. Nyquist and Dr. Edward J. Rothwell
for their generous advice and help. Working closely with them has been an enjoyable

and rewarding experience.

I am much obliged to Dr. Byron Drachman for his time and direction. I am also
grateful to the fellow graduate students working in the Electromagnetics Laboratory for
their help.

I owe a great deal to my parents for their love and continuous support. Last but
not least, I must acknowledge my wonderful wife, Wei Cha. Her love, sacrifice, con-

stant encouragement and support have been invaluable in the successful completion of

my graduate study.

LIST OF FIGl

CHAPTER 1.]
1.1 Introduct
12 Problem

CHAPTER 2. l
11 Preliminz
22 Boundar;
l3 Integral 1
2-4 Green’s 1

2.4.1 Bou
2.4.2 Scat
14.3 Gre
l5 Gwen’s l
25.1 Scat
2.52 Gre
2'6 Gwen’s 1
2.6.1 cm.
2.62 Gre

CHAPTER 3. 1
MEDIA ..

TABLE OF CONTENTS

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

LIST OF FIGURES iv
CHAPTER 1. INTRODUCTION 1
1.1 Introduction 1
1.2 Problem Description and Decomposition 4
CHAPTER 2. DERIVATION OF GREEN’S FUNCTIONS 13
2.1 Preliminaries 13
2.2 Boundary Conditions for Hertzian Potentials 14
2.3 Integral Representations of Hertzian Potentials 20
2.4 Green’s Functions for Electric Hertzian Potentials 23
2.4.1 Boundary Conditions 23

2.4.2 Scattered Potential Amplitudes 24

2.4.3 Green’s Functions 30

2.5 Green’s Functions for Magnetic Hertzian Potentials 36
2.5.1 Scattered Potential Amplitudes 36

2.5.2 Green’s Functions 51

2.6 Green’s Function for the Fields - 53
2.6.1 Green’s Funcan for the Fields due to an Electric Current .................... 53

2.6.2 Green’s Function for the Fields due to a Magnetic Current .................... 56

CHAPTER 3. PLANE WAVE PROPAGATION IN TRI-LAYERED

 

 

 

 

 

 

 

 

 

 

MEDIA 66

3.1 TM Plane Wave Propagation in Tri-layered Media 66
3.2 TB Plane Wave Propagation in Tri-layered Media 74
CHAPTER 4. FORMULATION OF INTEGRAL EQUATIONS .......................... 82
4.1 Integral Equations for a Monopole 82
4.2 Magnetic Field Integral Equation for a Slot 87
4.3 Hallen-type Integral Equation for a Slot 89
CHAPTER 5. SOLUTIONS OF INTEGRAL EQUATIONS 93
5.1 Method of Moments 93
5.2 Impedance Matrix for a Monopole 94
5.3 Calculaan of Impedance Matrix Elements 96
5.4 Special Consideration on Numerical Integration 103

 

ii

 

5.4.1 Inte g
5.42 Integ
5.4.3 Com
55 Admittanc
5.6 Calculatic

CHAPTER 6. S
61 Scattered
62 Far Field

6.2.1 Inte;
6.22 Stat
6.3 Scattered

CHAPTER 7. l
7.1 Numeric.-
7.1.1 Cor
7.12 Cor
7.1.3 Res
7.2.1 Co]
722 Reg

CHAPTER &
BIBLIOGRA}

5.4.1 Integration through Surface-wave Pele Singularities .................................. 103 '

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

5.4.2 Integration through Branch Point Singularities .............. -- ..... 103

5.4.3 Convergence of Impedance Matrix Entry Integrals .................................... 105

5.5 Admittance Matrix for a Slot - - 107
5.6 Calculation of Admittance Matrix Entries ................................................................ 113
CHAPTER 6. SCA’I'TERED FIELD - -_ ..................... 119
6.1 Scattered Field for a Monopole m---“ -- _ -- - 119

6. 2 Far Field Calculation _ ..................................... 121
6.2.1 Integration along the Real Axis - - - ............................................ 122

6.2.2 Stationary Phase Method ..................................... 124

6.3 Scattered Field for a Slot ...................................................... 128
CHAPTER 7. NUMERICAL RESULTS ........................................................................... 133
7.1 Numerical Results for a Monopole - ................................................. 133
7.1.1 Comparison with Published Results .... 133

7.1.2 Comparison with Experimental Results ........... 134

7.1.3 Results for Lossy Superstrates .................... 136

7.2 Numerical Results for a Slot .............. 138
7.2.1 Comparison with Published Results 138

7.2.2 Results for Lossy Superstrates - - ............ 139
CHAPTER 8. CONCLUSIONS -- 180
BIBLIOGRAPHY ..... 182

 

iii

Figure 1.1 Image

Figure 1.2 Slot it

Figure 1.3 Recei‘
(b) slot. ...............

Figure 1.4 Equiv
Figure 2.1 Hertzi
figure 22 Hertz
HEW 2.3 Hertz
HEW 2.4 Somn
Figure 3.1 Plane

Figure 7.1.1 Imp,

Plates.

.”.00..

511112712 rnp,

F1ture 7.1.3 rnp.
figure 7.1.4 rap.

Esme 7.1.5

LIST OF FIGURES

Figure 1.1 Imaged monopole in tri-layered media. ...................................................

Figure 1.2 Slot in tri-layered media. ...........................................

 

Figme 1.3 Receiving problem decomposition for (a) imaged monopole and

(b) slot. ..... - ..

 

Figure 1.4 Equivalent problems for slot in tri-layered media. ..............................

Figure 2.1 Hertzian potential boundary conditions at interface.

 

Figure 2.2 Hertzian potentials generated by vertical electric current. ................

Figure 2.3 Hertzian potentials generated by horizontal magnetic current. .................

Figure 2.4 Sommerfeld integration path in the complex 2. plane.

 

Figure 3.1 Plane wave propagation in tri-layered media. .......................................

Figure 7.1.1 Input impedance of dipole in free space. .....

 

Figure 7.1.2 Input impedance of dipole between two parallel conducting
plates.

Figure 7.1.3 Input resistance of probe through substrate.

 

Figure 7.1.4 Input reactance of probe through substrate. --

 

Figure 7.1.5 Radar cross section of monopole in tri-layered media with foam

 

substrate and various superstrates versus frequency.

Figure 7.1.6 Radar cross section of monopole in tri-layered media with foam
substrate and various superstrates versus incident angle at 12GHz.

 

Figure 7.1.7 Radar cross section of monopole in tri-layered media with foam
substrate and various superstrates versus incident angle at 156112.

 

Figure 7. 1. 8 Input impedance of imaged monopole 1n tri- -1ayered media
versus number of basis functions.

Figure 7.1.9 Radar cross section of imaged monopole in tri-layered media
versus number of basis functions.

 

iv

12

62

-144

145

146

147

- 148

. 149

150

- 151

r 7.1.10 Rot
vitsguur: numbCT Of

F1 7.1.11 lnp
{flabmm ar

figure 7.1.12 lnp
foam substrate ar

Figure 7.1.13 lnp
17le subsrrate a

Figure 7.1.14 lnp
FIFE substrate a

figure 7.1.15 Rat
with foam substr.

Figure 7.1.16 Rot
foam substrate an

Heme 7.1.17 Ra:
““11 P'TFE substr

Fleur: 7.1.18 Rec
substrate an

Ear: 7.1.19 E-p
With foam

figure 7.1.20 13.
Incdiawith PTFEP

Figure 7.121 Dra

Figure 7. 1.10 Received power of imaged monopole 1n tri-layered media

 

 

 

 

 

 

 

 

 

 

 

versus number of basis functions.-- - ............................... 152
Figure 7.1.11 Input resistance of imaged monopole in tri-layered media with

foam substrate and different superstrates. .............................................................................. 153
Figure 7.1.12 Input reactance of imaged monopole 1n tri- -layered media with

foam substrate and different superstrates. - ................................................ 154
Figure 7.1.13 Input resistance of imaged monopole in iii-layered media with

PTFE substrate and different superstrates. ............................................................................. 155
Figure 7.1.14 Input reactance of imaged monopole 1n tri- -layered media with

PTFE substrate and different superstrates. ....... - 156
Figure 7. 1.15 Radar cross section of imaged monopole in tri- layered media

with foam substrate and different superstrates. - -- - ...................... 157
Figure 7.1.16 Received power of imaged monopole in tri-layered media with

foam substrate and different superstrates. ................ 158
Figure 7. 1.17 Radar cross section of imaged monopole 1n tri- layered media

with PTFE substrate and different superstrates. 159
Figure 7. 1.18 Received power of imaged monopole 1n tri-layered media with

PTFE substrate and different superstrates. 160
Figure 7.1.19 E-plane radiation pattern of imaged monopole in tri-layered

media with foam substrate and different superstrates. - 161
Figure 7.1.20 E-plane radiation pattern of imaged monopole 1n tri-layered -
media with PTFE subsu'ate and different superstrates. - 162
Figure 7.1.21 Drawing of vacuum kayak measurement platform. ................................. 163
Figure 7.2.1 Input impedance of open slot antenna. ............................... 164
Figure 7.2.2 Input impedance of slot on semi-infinite GaAs substrate. ...................... 165
Figure 7.2.3 Input impedance of slot on semi-infinite PTFE substrate. ...................... 166
Figure 7.2.4 Input impedance of slot in tri-layered media with air film and

foam substrate. 167
Figure 7. 2. 5 Input impedance of slot 1n tri- layered media with magnetic coat-

ing and foam substrate. - -- . 168

 

Figure 7.2.6 Input impedance of slot in tri-layered media with resistive sheet

and foam subsrr.

F1 7.2.7 lnp
MMFI‘FE subsu

Figure 7.2.8 lnp
and GaAs substr

Figure 7.2.9 Rat
strate and diffcn

Figure 7.2.10 Rt
state and diffen

figure 7.2.11 R1
sheet and diffen

figure 7.2.10 R1
and different 51.1]

From 7.2.13 E-
Substrate and dig

Fleur 7.2.14 H.
substrate and C111

Bane 7.2.15
“Vi Shéct andfcsi

Bane 7.2 1

and foam substrate. ......................................................................................................................... 169

Figure 7.2.7 Input impedance of slot in tri-layered media with resistive sheet
and PTFE substrate. ........................................................................................................................ 170

Figme 7. 2. 8 Input impedance of slot in tri- -layered media with resistive sheet
and GaAs substrate. ................................................. 171

 

 

Figure 7.2.9 Radar cross seetion of slot in tri-layered media with foam sub-
strate and different superstrates. ................................................................................................. 172

Figure 7.2.10 Received power of slot in tri-layered media with foam sub-
strate and different superstrates. ................................................................................................. 173

Figure 7. 2. 11 Radar cross section of slot in tri- layered media with resistive
sheet and different substrates. -- ....... - .................. 174

 

Figme 7.2.10 Received power of 510t in tri-layered media with resistive sheet
and different superstrates. ............................................................................................................. 175

Figure 7.2.13 E-plane radiation pattern of slot in tri- -layered media with foam
substrate and different superstrates. ......................................................... 176

 

Figme 7.2.14 H-plane radiation pattern of slot 1n tri- -layered media with foam
substrate and different superstrates. ....... 177

 

Figure 7.2.15 E—plane radiation pattern of slot 1n tri- -layered media with resis-
tive sheet and different substrates. ....... 178

 

Figure 7.2.16 H-plane radiation pattern of slot 1n tri- -layered media with resis-
tive sheet and different substrates. ............................. - ......... 179

 

 

vi

1.1 lntroductim

In some ap
such as an aim
that are many
ing antenna sub
of the aircraft 1
an antenna systr
51mm

An efiecti
P7353170 the rec
that While the 1
on“ and cadm-

ami Silffers IWQ

CHAPTER ONE
INTRODUCTION

1.1 Introduction

In some applications, it is necessary to reduce the radar cross section of a system,
such as an aircraft. In order to communication with other airplanes and ground control,
there are many conformal antenna subsystems on board. However an effective receiv-
ing antenna subsystem is also an effective contributor to the overall radar cross section
of the aircraft In other words, the requirement to maintain the receiving capability of
an antenna system conuadicts the requirement to reduce the radar cross section of the

system.

An effective way to decrease the radar cross section of an antenna and to
preserve the receiving ability of the antenna is lossy coating. The physical intuition is
that while the received signal or transmitted signal passes through the lossy coating
once and endures one loss, the scattered signal must go through the lossy layer twice

and suffers two losses.

It is necessary to develop a theoretical model to analyze an antenna with lossy
coating and to provide design guidelines. A practical conformal antenna coated with
radar absorbing material on board an aircraft is too complicated to handle at once. A
simplified model, which highlights the effects of lossy coating on the characteristics of
an antenna, is .established. The geometry is tri-layered media with a ground plane, a
substrate, a superstrate, and a half space. The superstrate can be a lossy coating. This
is a very versatile structure and includes major electromagnetic phenomena. Two
essential antenna elements, a vertical imaged monopole and a slot, in tri-layered media

are studied in the dissertation. A detailed description of the geometry is provided in

the next seetion.

A vertical it
geneous medium
Sommerfeld pior
planarly layered
jett A recent b1
waves and fields
work on the sub;

One of the
problems in elec
functions in la}
Transform techr
[611101111] The

The scatter
“WI! for ye
animus. which
“"0ka on Pfintet

the next section.

A vertical imaged monopole is equivalent to a dipole. A dipole or a slot in homo-
geneous medium is a classical antenna problem and is treated in many books [1]-[4].
Sommerfeld pioneered the study of the propagation of electromagnetic (EM) waves in
planarly layered media [5]. There are extensive research and publications on the sub-
ject. A recent book by Chew [6] presents a comprehensive and updated treatment of
waves and fields in inhomogeneous media. From this book, all the important historical

work on the subject can be traced.

One of the most powerful and commonly used technique to solve boundary value
problems in electromagnetics is the integral equation approach [7]-[9]. Dyadic Green’s
functions in layered media are needed to arrive at appropriate integral equations.
Transform techniques can be used to derive Green’s functions in layered media

[6][10][11] The singularity of dyadic Green’s function has been studied in [12].

The scattering and radiation of apertures in ground plane has been the subject of
research for years [13]-[17]. There is a vast amount of publication on microstrip
antennas, which are closely related to slots, [18]-[26]. Of particular interest are the the
works on printed circuit antenna in a superstrate-substrate configuration [27][28]. Com-

pared with microstrip antenna, printed slot has received less attention [29]-[34].

Sommerfeld integral approach can solve the EM wave propagation in planarly
layered media rigorously. The price to pay for the analytical elegance is that the spec-
tral integrals involved in matrix filling are very difficult to compute numerically. Vari-
ous analytical, asymptotic, and numerical techniques can be used to reduce computa-
tional time [35]-[44].

There are eight chapters in this dissertation. Chapter one gives the motivation for
this research. It also contains a literature survey and describes the problems to be

solved.

Chapter two presents in detail the derivation of dyadic Green’s functions in tri-
layered media. Electric and magnetic Hertzian potentials are used to facilitate the
development of Green’s functions. The planar layers are homogeneous and have arbi-
trary electric and magnetic contrasts. The Green’s functions for Hertzian potentials and
EM fields maintained by a vertical electric current or a horizontal magnetic current in

the substrate are derived.

Plane wave propagation in tri-layered media is investigated in chapter three. This
information is needed to determine the excitation terms of the integral equations

developed in chapter four.

An electric field integral equation (EFIE) and a magnetic field integral equation
(MFIE) are developed in chapter four. Then under certain approximation conditions,

both EFIE and MFIE are converted to Hallen-type integral equations (HTIE).

Chapter five presents solutions of the integral equations developed in the previous
chapter by moment methods. Special effort is made to find accurate and efficient ways
to calculate the Spectral integrals encountered in matrix filling. Induced current on a
monopole or a slot in Iii-layered media illuminated by an incident plane wave is

obtained. Input impedance and received power are computed.

Chapter six deals with the evaluation of scattered field. A stationary phase method
is used to calculate far field. The expressions for radar cross section and radiation pat-

tern are presented.

Numerical results generated by the theory developed in this dissertation are com-
pared with published results and experimental data whenever possible in chapter seven
to validate the theory. Then computer simulation are conducted for several sets of
representative parameters of a monopole or a slot in tri-layered media. Results of input

impedance, received power, radar cross section, and radiation pattern are planed.

In the final
is drawn from 11
mended.

12 Problem De

Two proble
media, have bee:
to analyze slots
layered media is
that the lossy co
more complicate
phenomena of sr
plicatod problem

Consider t1
acOIIdUCting gm
of lhiCkness t, ;
length h and r ad
and ha“? arbitr
respecfivc1y. T111
9. The monopol
magnetically 105

The geOme

111111 Collducdng
length 21 is cut

substrate of 1171c}

the ground Plan

In the final chapter, chapter eight, the work done are summarized. A conclusion
is drawn from the results of chapter seven. Some ideas on further research are recom-

mended.

l.2 Problem Description and Decomposition

Two problems, an imaged monopole in tri-layered media and a slot in tri-layered
media, have been studied. The ultimate goal is to develop a theory and computer codes
to analyze slots in tri-layered media. The main reason to study a vertical monopole in
layered media is that this is the simplist problem in tri-layered media. It is conjectured
that the lossy coating interacts with this simple antenna in much the same way as with
more complicated antenna systems. This simple model keeps all the electromagnetic
phenomena of scattering and radiation in layered media and can lead to the more com-

plicated problem of slots in stratified media.

Consider the geometry pictured in Figure 1.1. The tri-layered media are made of
a conducting ground plane in the z=-d plane, a substrate of thickness d, a superstrate
of thickness t, and a half space on top of the superstrate. A vertical monopole of
length h and radius a is immersed in the substrate. The planar layers are homogeneous
and have arbiuary complex permittivity and permeability £1, £2, 83, til, 11»; and 1.13
respectively. The entire structure is illuminated by a plane wave with an incident angle
0. The monopole has a load ZL attached to it. The superstrate can be a electrically or
magnetically lossy coating.

The geometry of a slot in tri-layered media is shown in Figure 1.2. An infinitely
thin conducting ground is placed in 2:0 plane. A rectangular slot of width 2w and
length 21 is cut in the ground plane. There are three layers above the ground plane, a
substrate of thickness d, a superstrate of thickness t, and a semi-infinite space. Beneath

the ground plane is another semi-infinite space.- All four layers are assumed to be

homogeneous and
is illuminated by
placed at the cen
lossy coating.

In practice, :
to make it radian
in Figure 1.2 is 1
interaction betwet
waves in tri-layer
include backing 2

In the meet
of the energy is
superposition pri
ing Problem and
linear.

Hm consic
figure 1.3. A m

Wave. The curre

homogeneous and can have arbitrary complex permittivity and permeability. The slot
is illuminated by a plane wave with incident angle 9 . A load impedance ZL can be
placed at the center of the slot. The superstrate can be a electrically or magnetically
lossy coating.

In practice, a slot usually is backed by either a cavity or another conducting plate
to make it radiate in only one direction. The reason to choose the structure described
in Figure 1.2 is to simplify the problem and to concentrate on the effects of the EM
interaction between a slot and a lossy coating. Once the radiation and scattering of EM
waves in tri—layered media have been well understood, the research can be extended to

include backing and complicated and practical feeding mechanism for slots.

In the receiving case, an incident plane wave induces current on a antenna. Part
of the energy is delivered to the load and part of it is radiated out in the space. The
superposition principle can be used to decompose the receiving problem into a scatter-
ing problem and a transmitting one as shown in Figure 1.3 because the problems are
linear.

First consider the decomposition of a receiving monopole shown in part a of
figure 1.3. A receiving mode current I is induced on a monopole by an incident plane

wave. The current causes a voltage drop V across the load ZL.

V = "'[ZL (1.21)

In the scattering case, the monopole is shorted to the ground plane and a scatter-
ing mode current I, is induced by an incident plane. There is no voltage drop across
the gap between the monopole and the ground plane. A transmitting mode current I,
is generated by a voltage source V, . There is no incident wave. The input impedance

of the monopole is defined as

Th: receiving

 

current I, and load
I = [3+],
V = V,

From (1.2.1-4

The transmi

current by solvin;

The the TC
“mm by solvi

Where

The receiving mode current I can be expressed in terms of scattering mode

current I: and load and input impedance Zia and ZL by some straight manipulation.

 

I = 13+], (1.2-3)
V = V, (1.2.4)
From (1.2.1-4)
ZL ZL
1‘ = -l Zia = —(I'+ls)-Z‘-—- (1.2.5)

The transmitting mode current can be expressed in terms of scattering mode

current by solving (1.2.5)

 

z
I,=I L

_ 1.2.6
‘ ZL+Z,-,, ( )

The the receiving mode current can be expressed in terms of scattering mode

current by solving (1.2.6).

lit:
1 =ls+l‘ =ISZ—-‘-I'—Z:— (1.2.7)
it:

Finally, the power delivered to the load ZL can be written as

Zia

|2 1.2.8
Zin "I'ZL ( )

 

PL = é-ReWI') = %|1312RL|

where
Zr. = RL +1741.
Then consider the decomposition of a receiving slot shown in part b of Figure
1.3. An aperture electric field E, is induced in the slot by an incident plane wave.

This aperture field generates a voltage V across the slot and an electric current I

flowing along a load impedance ZL. For a narrow slot ( w<l ; w<7c ), the voltage

across the slot is expressed as

v = _j andy (1.2.9)

where the orientation of the slot is shown in Figure 1.2.

1 = V (1.2.10)

”Z

The slot is open and illuminated by a plane wave in the scattering case. An
scattering mode voltage V, can be obtained from (1.2.9). Notice that there is no con—
duction electric current flowing across the slot. A transmitting mode voltage V, is gen-
erated by a electric current source I, placed in the slor There is no incident wave. The
input impedance of a slot is defined as

Vt

=7

Zr. (1.2.11)

This definition of input impedance of a slot depends on the position of the current

source in the slot.

The receiving mode voltage V can be expressed in terms of the scattering mode

voltage V, and input and load impedance Zia , ZL through some straight forward

derivation.
1 = I, (1.2.12)
V = V3+V, (1.2.13)

From (1.2.10—13)

v, = -V—‘"- = —(V_,+v,)— (1.2.14)

Solve (1‘2'14'

u=-H

Substituting 1

V=V5E

1
PL—ZH

According to

by an equivalent m

Mz-fiX

 

hen the problem
shown in Figure 1
Gent’s function in
Green’s function i

development of MI

Solve (1.2.14) to get

V v 2" (1215)
‘— SZL+Zm ..

Substituting (1.2.15) into (1.2.13) leads to

 

V Z" (1 216)
- ‘ ZL+Zin ' ’
The power received by the load ZL can be expressed as
, R
PL = —1-Re(VI ) = l-|V Iz—L———— (1.2.17)
2 2 IzL+z,-,, I2

According to the equivalence principle, the aperture electric field can be replaced

by an equivalent magnetic current defined as:

M = -—;i an (1.2.18)

Then the problem shown in Figure 1.2 can be reduced to two equivalent problems

shown in Figure 1.4. It can been seen that the ground plane makes the derivation of

Green’s function in the upper half space, which has three layers, and the derivation of

Green’s function in the lower half space independent. The coupling occurs in the

development of MFIE.

E1. “1

El 112

63. H3

       

\ ‘ \\
\\\\
\‘I,‘1.\.‘\\\\\\\1 \ \ \ ~ .
\\\~\\\<\\ \\\\\\\‘\\‘~§§\\>\\~\:\ 1
\ \\ x \ \ \ _ 1A
\\\\\\\\‘\\\;

F'
lgul‘e 1.1 Imagfi

 

Az

Hi

81, 111

 

 

 

 

 

cover
z=t
z=0 x
z=.d+h 12(2)
83, “3
2a
substrate
z=~d 21-
s
conducting ground plane

Figure 1.1 Imaged monopole in tri-layered media.

E1. “1

82. 112

83, H3

 

\ _
‘ - ~»\\\\\\&\\§~1 .t~::~a:~-;
\\-\ “-“K‘C' ~
‘\ \\\‘

E4. 114

Figure L2 Slc

10

Az

91, “1

 

z=d+t

 

 

83, #3

 

substrate

2=0 -l M 1
\ = 2 y
\\ \\\\\\\\\\\\\\\\\\
'w
conducting ground plane
64, m

Figure 1.2 Slot in tri-layered media

Receiving

a \\ ""\\\\

 

 

Fig“ ‘3 Pmbl

11

Receiving .-_ Scattering + Transmitting
fl —_ _
\ T 1 \ T 18 \ T It
4..

 

 

 

 

 

 

+
2L; 'v \W-LWV Yt
m \\ x

 

 

 

 

 

 

 

 

(b)

Figure 1.3 Problem decomposition for (a) imaged monopole. (b) slot.

12

58. 833-3 a 87. é nausea c8335 3 2:3,.—

 

 

 

 

 

 

 

 

 

OVN Ba 8038; “=25,va OAN Bu 8030.5 829ch 80305 Raw—O
Y1 .vw V1 .3 v1 .vw
#5:: 8% use» 25m 232»
ll . Al . i .i z... .
x 05E 95on «.1 mm x «”68de n: mm x «a n: no
a: do u: .8 a: do
Fl 40 :— .—m 61 J0

 

 

 

 

CHAPTER. TWO
DERIVATION OF GREEN ’8 FUNCTIONS

2.1 Preliminaries

Consider the geometries shown in Figure 1.1 and 1.2, where a monopole and a
slot in tri-layered media are illuminated by an incident electromagnetic (EM) plane
wave. The existence of a perfectly conducting ground plane makes it possible to
separate the upper and lower half spaces in the derivation of the Green’s functions.

The upper half space has three layers, while the lower half space is free space.

If an elecuic current J or a magnetic current M is placed in region 3, EM fields
will be maintained in all three regions. EM fields produced by a vertical electric
current 2!, in the case of an imaged monopole or a horizontal magnetic current M, in
the case of a slot are of particular interest. The EM fields produced by an arbitrarily
oriented current can be readily obtained following the same procedures outlined. '

Because the layered media are invariant in the x-y plane, it is advantageous
analytically to use a two dimensional Fom'ier transform. This is the famous Sommer-
feld integral approach, by which the Green’s function of an arbitrary source can be
derived rigorously. The price paid for this analytical elegance is the computationally
daunting task of the inverse transform. This chapter deals with the aspects of Green’s
function derivation, while the numerical implementation of the inverse Fourier

transform will be handled in chapters 5 and 6.

One way to derive the Green’s function is to express the EM fields in terms of
Hertzian potentials. The EM fields can be expressed in terms of electric Hertzian
potentials, which are produced by an electric current J , or in terms of magnetic Hert-

zian potentials, which are produced by a magnetic cm'rent M. In this dissertation, the

13

source is either a
to represent either
there is a possibili

The represen

[55 ll 56]
1: = 1211
H = jtoe
where k is the war
k2 = (ole
Md the elecuic H(

Vzll+k

“1° representation
55 ll 56 1

 

H=k2n

and the magnetic

 

14

source is either a vertical electric current or a horizontal magnetic current. I] is used
to represent either elecuic or magnetic Hertzian potentials, depending on the source. If

there is a possibility of ambiguity, it will be mentioned explicitly what II means.

The representation of EM fields by an electric Hertzian potential can be written as

I 55 ][ 56 l
E = k2n+V(V~r1) (2.1.1)
H = jooerII (2.1.2)

where k is the wavenumber of the medium,
k2 = (02611 (2.1.3)

and the electric Hertzian potential satisfies an inhomogeneous wave equation

vzn + 1:er = -—.-J— . (2.1.4)
[(1)6

The representation of EM fields by a magnetic Hertzian potential can be written as [
55 ][ 56]
E = _]- coquH (2.1.5)
H = 1811 + V(V-H) 4 (2.1.6)
and the magnetic Hertzian potential satisfies an inhomogeneous wave equation

V211 + 1:211 = "M- . (2.1.7)

10011

2.2 Boundary Conditions for Hertzian Potentials

To determine the Hertzian potentials, it is necessary to invoke the boundary con-
ditions at the dielectric interfaces and at the ground plane. The boundary conditions for

the Hertzian potentials can be deduced from the boundary conditions for the EM fields.

Consider the
electric Hertzian l
deduction of the 1

those for electric 1
Write equatit
E1 = -j(

Ey =-j(

E: :7“

 

Boundary COn

Exiy=0~)

 

 

Who“)
m0flj
”10:03
If regj0n 2 is l

ExiY‘O‘)

 

510$) :

 

15

Consider the geometry shown in Figure 2.1. The boundary conditions for the
elecuic Hertzian potentials have been derived in [ 10 ] and [ 11 ]. Therefore, only the
deduction of the boundary conditions for magnetic Hertzian potentials are outlined and

those for electric Hertzian potentials are quoted from [ 10 ][ 11 ][ 63 ].

Write equations (2.1.5) and (2.1.6) in component form:

 

 

 

. an, an,
Ex =—Jcou( ay az ) (2.2.1)
, an, an,
E, = —1 com 32 3:: ) (2.2.2)
, an, an,
E. =—qu( ax ay ) (2.2.3)
HI = M1,‘ + i—(V-n) (2.2.4)
H, = kzrr, + %-(V-II) (2.2.5)
Hz = 1:211, + -§—z(V-n) . (2.2.6)

Boundary conditions for the EM fields at the interface between region 1 and 2 are

E, (y=0‘) = E,(y=o+) (2.2.7)
5, (y=o-) = E, (y=o+) (2.2.8)
H,(y=o-) = H,(y=o+) (2.2.9)
H, (y=0") = H, (y=o+) . (2.2.10)

If region 2 is perfectly conducting, the boundary conditions become
Ex(y=0‘) = 0 (2.2.11)

E, (y=0') = o . (2.2.12)

16

It is advantageous to study the cases of three orthogonal components of M

separately and then combine the results to arrive at the general boundary conditions.

1. Vertical current M = 2M,
Vertical current M, produces a Hertzian potential with only 2 component.
1'1 = in, (2.2.13)
This II can describe the EM fields completely.

Substituting (2.2.1-6) into (2.2.7-10) gives

 

 

 

 

 

 

 

 

311,, _ 3H2, 2 4
"1 ax ”‘2 Bx <2. .1)

8H,, ' 8112, (2215)
“'1 8y -“'2 3y . .
82111, aznh

8x82 = 8x82 (2.2.16)
82“” _ 321121 2217
Bsz - ayaz ' ( - )

In order to satisfy equations (2.2.14-17) simultaneously, the following boundary condi-

tions on 11 must hold.

 

1111-11, = [121121 (2.2.18)
81-11, _ 3111, 2 219
az - 82 (H )

It is understood that the boundary conditions are valid at the interface, which is the
z=0 plane.

If region 2 is a perfect conductor, (2.2.1-6) and (2211-12) can be used to arrive

at

3111,
”I ax

 

= 0 (2.2.20)

 

8ml:
“13;"

The followin

H1, =0

2. Horizontal cm

11 can tn Silt

ll must have bodi
H = i 11

In Other words, C1

the case of horizo

 

 

SUbStituting

kink.

1&5

8x

SOiVing CQUarjOnS

Erlllllh

82
“111 ‘

lz~

17

31—11,
“I a),

The following boundary condition on 11 can be deduced from (2220-21).

 

= O . 2 (2.2.21)

r1lz = 0 (2.2.22)

2. Horizontal current M = 2M,

It can be shown that in order to describe the EM fields completely for this case,

11 must have both a horizontal component and a vertical one [ 10 ][ 11 ][ 63 ].
II = £11, + £11, (2.2.23)

In other words, coupling between a horizontal component and a vertical one occurs in

the case of horizontal current excitation.

Substituting (2.2.1-6) into (2.2.7-10) gives

 

= ’11— (2.2.24)

 

 

 

- —) (2.2.25)

 

 

 

 

 

 

 

811 11 an
2 1 1x 12 = 2 i 2: 22
k1 1'le + ax( 8x a ) k2 112, + 3x( 3x _82 (2.2.26)
3 31—11, 31—11, - a 31—12x 81-121
3% 8x + 32 ) - By( 31: + az ) . (2.2.27)
Solving equations (2.2.24-27) leads to the following boundary conditions
811111-11, = exflJQl—I?J (2.128)
BI'IlJr 8H2,
1.11 az — #2 82 (2.2.29)

“In” = [121-122 (2.2.30)

—(n12'

The followin

12) in the case th;

anlz
a),
an ,

( l

’37"

Equations (2.
n12 : 0

T =
3' Horiwmal curl

This case is 1

coIlditions Can be ‘

 

an
11¢
1 dz

“1H1: =1

18

a 8
Earn—Hz.) = 733-611. 412.) . (2.2.31)

The following equations can be written by using equations (2.2.1-6) and (2.2.11-

12) in the case that region 2 is a perfect conductor

 

 

a O C
( , , ) . ( . . )

Equations (2.2.32-33) can be solved to obtain the boundary conditions

 

111, = 0 (2.2.34)
an x
1 = o , (2.2.35)
32

3. Horizontal current M = 5M,

This case is the same as the previous one if y and x are exchanged. Boundary

conditions can be written from equations (2.2.28-31)

 

81“.an = wiznzy (2.2.36)
am, 3112,

11} az — 112—52—- (2.2.37)

“ll-ll: = “21121 (2.2.38) .

31m -n )= —i(r1 -r12 ) . (2.239)

32 l: 22 By 1y y '

If region 2 is a perfect conductor, the boundary conditions become
I11, = 0 (2.2.40)

EDD
82

 

= 0 , (2.2.41)

4. General Boun
Combining t
tions on magnetic

can be expressed

 

19

4. General Boundary Conditions on Magnetic Hertzian Potentials

Combining the results of the above three subsections, the general boundary condi-
tions on magnetic Hertzian potentials produced by an arbitrary magnetic current source

can be expressed as

52112
H =—H 2.2.42
M 51111 20 ( )
a _fl_8_
.8711“ _ [11 82 {120 (2.2.43)
nu=£fib, Q2M)
111

an“ 31-12: 621.12 3H2, anzy
32 az —.(elu1 1" ax T ay ) (2'2“)

 

 

wherea =x.y.

If region 2 is a perfect conductor, the boundary conditions become

 

 

nh=o a2%)
3111,
___ 0 (2.2.47)
32
an
1’ = 0 . (2.2.48)
32

5. General Boundary Conditions on Electric Hertzian Potentials

The general boundary conditions for the electric Hertzian potentials are quoted

from[10][11][63].

£2112
Illa _ 81“] n2“ (2.2.49)
a nu, = E” a 112,, (2.250) -

E 275

20

H12 -—- "—sz (2.2.51)

 

 

) (2.2.52)

In the case that region 2 is a perfect conductor, the boundary conditions can be

expressed as

 

r11, = 0 (2.2.53)
I'll, = 0 (2.2.54)
81112
= 0 (2.2.55)
32

2.3 Integral Representations of Hertzian Potentials

The Fourier transform is a very powerful tool for solving differential equations.
The vector Fourier transform, i.e. two dimensional Fourier transfonn, is an extension
of the commonly used one dimensional Fourier transform [ 49 ][ 50 ]. It is advanta-
geous to use the two dimensional Fourier transform because the planarly layered media

are invariant in the x-y plane.

The transform pair can be written as

 

mkxxyz) = j J n(x.y,z>e"""““’”dxdy (2.3.1)
1 ”- . I

n ,, = n 1““‘9” die . 2.3.2

(xyz) (WU. (mane dk. , ( )

Equation (2.3.2) means that any wave can be expressed as a superposition of plane
waves with proper weighting. Equation (2.3.1) gives the weighting function or the spa-

tial frequency spectrum.

21

Use the following notation for brevity
r = xf+yy‘+z£ ; dzr = the dy (2.3.3)
k = kxf+k y ; dzk = dk, dk, . (2.3.4)

Equations (2.3.1-2) can be rewritten as

f1(k,z) = j j r1(r)e-f'"d2r (2.3.5)

110-): —1—2—j[fr(k,z)ei'"d2k . (2.3.6)
(27:) _..

Hertzian potentials can be categorized into two groups. The primary Hertzian
potentials are produced by primary sources in an unbounded homogeneous space. They
satisfy the inhomogeneous Helmholtz wave equation, which, in the rectangular coordi-

nates, can be written as

(v2+k2)nP (r) = -F(r) (2.3.7)
where
_ fl
F(r) _ jme (2.3.8)

in the case of elecuic Hertzian potentials produced by a electric current and

M(r)
F = 2.309
(r) —ijl g ( )

in the case of magnetic Hertzian potentials produced by a magnetic current.

The scattered potentials are generated by secondary sources caused by the pri-
mary potentials in an inhomogeneous region. The scattered potentials satisfy the

homogeneous Helmholtz wave equation

(V2+k2)11’ (r) = 0 . (2.3.10)

22

Equations (2.3.7) and (2.3.10) can be solved by the Fourier transform technique.
The step-by-step procedures have been given in [ 10 ][ 11 ]. The final results from [
10][11][63]are used.

The scattered Hertzian potentials can be expressed as

°° W’(k) . _
1'1; =II (2:02 elk'eipa‘kdzk a=x,y,z . (2.3.11)

 

The primary Hertzian potentials can be written as

n» (r) = {FM Ii 2(2n)2p (k)

jlt-(r—r’) -p(lr)lz-z’l
e e 2k dv’ (2.3.12)

 

where II” represents electric Hertzian potential if

_ i
- jtoe (2.2.13)
and 11” represents magnetic Hertzian potential if
F = Ji— , (2.3.14)
J (011 ~

In addition, wavenumber parameters are defined as
p (Ir) = Virgil-19L]:2 (2.3.15)
k2 = (02211 . (2.3.16)

In order to properly ensure that waves decay as they propagate in a lossy

medium, the appropriate branch of p (It) used must satisfy

Re(p)>0 ; Im(p)>0. (2.3.17)

In
res

£51"

In
reg

comp

23

2.4 Green’s Functions for Electric Hertzian Potentials

The Green’s functions for electric Hertzian potentials maintained by a vertical

electric current in tri-layered media are derived in this section.

2.4.1 Boundary Conditions

Consider the geometry shown in Figure 2.2. A vertical electric current source
placed in region 3 will maintain electric Hertzian potentials in all three regions. The
potentials have only vertical components for reasons explained in 2.2. In region 1, the

potential will be entirely the scattered potential II;

1112 = n; z>t . (2.4.1.1)

In region 2 the potential will be composed of scattered terms propagating in both the
i2 directions.

Hz, = III-+11,“ 0<z<t . (2.4.1.2)

In region 3 the potential will be made up of a primary component plus two scattered

components propagating in the :12 directions.

113, = Hyman; —d<z<0. (2.4.1.3)

The explicit representations of these terms can be obtained from (2.3.11-12).

 

 

 

r1;(r)= I} “2"()k2)ei'"e‘”d2k (2.4.1.4)
n,‘(r)= -jl ‘(k)e1'””’”d2k (2.4.1.5)
r1:(r)= Ii Eye”? ‘1”de (2.4.1.6)
n;(r)= H W500 ei'" erZk (2.4.1.7)

(2102

where

Apr
leads to l

24

W,’ (k)

 

mm = Ii (2102 ej""e—p‘zd2k (2.4.1.8)
where

12.? = k}+k,2—k,3 m=1,2,3 (24.1.9))

kg = (9211,, cm (2.4.1.10)

J,(r') °° ejk-(r-me-ps'z-z'l

11" = dzk d '. 2.4.1.11
20‘) (11083 ILL 2(21t)2p3 v ( )

 

 

Application of equations (2249-55) at the three interfaces z=t, z=O, and z=-d
leads to the following five boundary conditions

 

 

 

 

 

311,, an,,

82 — 32 Z— (2.4.1.12)
2,111, = 12an 2:: (2.4.1.13)
anz, 3113,

32 — 82 2:0 (2.4.1.14)
£11122 = 831-132 2:0 (2.4.1.15)
8H3,

=0 z=—d . (2.4.1.16)
dz

The scattered potential amplitudes Wz’, Wz‘, W}, W;, and W,‘ are unknown. These five
unknowns may be determined by applying the five equations (2.4.1.11-16) derived
from the boundary conditions.

2.4.2 Scattered Potential Amplitudes
a) Employing (2.4.1.1-2) and (2.4.1.12) gives

_a_ t__a_ + — _
82 II, - az [IIZ+II,] at z—t . (2.4.2.1)

Subsrituti

Taking t
equation

for all x

Proceedi

Proceed

”Ocean

25

Substituting (2.4.1.6-8) into (2.4.2.1) gives

:82- { (21)ZII[Wzte-plz-Wz+e-pzz-Wz-e+pzzlejkfdzk } = 0 at 2:: '
1! —oo

 

Taking the derivative inside the integral and performing the differentiation gives an
equation in which the inverse Fourier transform of a function is identically zero ( ie.

for all x and y ). This is possible only if the function is identically zero
p1Wz‘e—p‘t—p2Wz+e-p"+p2Wz‘e+p2' = 0. (2.4.2.2)

b) Substituting (2.4.1-2) into (2.4.1.13) leads to

61H; = 2.2m;+r1;] at z=t . (2.4.2.3)
Proceeding as in a) gives
81W;e_”"-62Wz+e_P"—82Wz'e+”" = 0 . (2.4.2.4)
c) Substituting (2.4.2-3) into (2.4.1.14) gives
8 + - a r ‘ i
5?“ +H,] = -a—z[flf+flz+flz] at z=0 . (2.4.2.5)

Froceeding as in (a), and using lz-z'l = z-z’ for z>z' gives

i 0153 2P3

 

p2W2+-p 2W;-p3w;+p3w;‘ = ml e” dv ’ . (2.4.2.6)

(1) Substituting (2.4.2-3) into (2.4.1.15) leads to

2411,4113: e3mf+ng+rtjl at z=0. (2.4.2.7)

Proceeding as in (c) gives

. J (r') e-jk'r’
+ W’—e W'—e W‘ = e z
8’2sz z 3 z 3 z 3!,1-“3 2P3

 

ep’z'dv’ . (2.4.2.8)

 

Proceedi

Sur

Whfirc

To
calCUIate
Hide fUn C

[hc abOVe

30M

26
e) Substituting (2.4.3) into (2.4.1.16) gives

§;[ng+n;+n;1= 0 at z=—d. (2.4.2.9)

Proceeding as in (a) and using lz—z ’l = z’—z for z’>z gives

_p3d Jz (r’) 8—11?“

 

 

p3wz'eP’d—p3wge‘m = p38 jwe 2p e‘Piz'dv'. (2.4.2.10)
V 3 3
Summary:
((1) g—lwz‘e‘P" -w,+e"’2’ +1476“ = 0 (2.4.2.2)
2
E
(b) El-Wz‘e'P"—Wz+e'p"—Wz‘e+p" = 0 (2.4.2.4)
2
p p .
(c) p—:w,+-iw;-W;+w; = v: (2.4.2.6)
(d) :2—3W,++3W{-W{-W,‘ = v; (2.4.2.8)
(e) Wz’ep’d—Wzie'P’d = e'P’dV; (2.4.2.10)
where
J (r’) —jk-r ,
v} ajj’m 82p em dv’. (2.4.2.11)
V 3 3

To formulate the integral equation, the total potential in region 3 is needed. To
calculate the back scattered field, the potential in region 1 is needed. Thus, the ampli-
tude functions w;,w;' and w; must be determined. This is accomplished by reducing

the above five equations (2.4.2.2)-(2.4.2.10) in the following sections.

Solving (2.4.2.2) for W; and substituting this into (2.4.2.4) yields

- “'1 “2102‘ +
= —- W .4. .
W2 (1+ 1 e z _ (2 2 12)

where

Substitu‘

Solting

where

NCXL. (2

lastly, (

Here (2.4

T0 (
in region
”J , and

The mOst

27

where

E
a = —‘pi. (2.4.2.13)

8W1

Substituting (2.4.2.12) into (2.4.2.6) and (2.4.2.8) gives, respectively

—[1-—:—+1 e’z”2‘]W+-W,'+W,i = v; (2.4.2.14)

p3

E—[1+ 971.: ‘2“ ]Wz+—w,'-W;' = v; . (2.4.2.15)
3

Solving (2.4.2.14) for W} and substituting this into (2.4.2.15) then gives

 

Wm—li-W: n+1] = Vin—11 (2.4.2.16)
where
01—14,.)
1+—e
8
ya 2P3 “+1 . (2.4.2.17)
e3p2 1_a-1 e‘zpz‘
(1+1

Next, (2.4.2.16) is solved for W,’ and this is substituted into (2.4.2.10) to give

. e-P 34 Vz'+ep’d Vz+
W; = [7—1] 4 _ ,4 (2.4.2.18)
[fillep’ -[Y-1]e "

Lastly, (2.4.2.16) is solved for W,i and this is substituted into (2.4.2.10) to give

 

 

e-M [yum-+1141":

w; =
[New -tv—11e“”"’

(2.4.2.19)

Here (2.4.2.18) and (2.4.2.19) are the desired scattered potential amplitudes.

To calculate the scattered field in region 1 it is necessary to calculate the potential
in region 1 due to a vertical current in region 3. The total potential in region 1 is just
l'l,‘ , and thus it is only necessary to determine W,’ to use (2.4.1.8) to calculate I'I,‘ .

The most straightforward method for calculating W,‘ is to solve (2.4.2.2—10) from the

28

start. Solving (2.4.2.10) for W,’ and substituting it into (2.4.2.6) and (2.4.2.8) gives,

respectively
P2 + P2 .- —2.osd + —2psd _
—W, ——W, +W [1=e ]= v +6 V (2.4.2.20)
P3 P3 2 z z
:—W,*-E-W, +W;[1—e 2”3"] = v,++c‘2P3dV,-. (2.4.2.21)
3 3

Solving (2.4.2.20) for W,i and substituting into (2.4.2.21) then gives

PW,++MW,- = 17, (2.4.2.22)
where

V, = V,+ep’d+V,‘e'P’d (2.4.2.23)
31 . P2

P = —s1nhp3d+—coshp 3d (2.4.2.24)
33 P3

M = 2sinhpgat-p—Zcosbp3ai . (2.4.2.25)
53 P3

Note, (2.4.2.22) is solved for W," , which is substituted into (2.4.2.2) and (2.4.2.4) to

give, respectively

P1

——W,‘e ‘Pi‘+w, {-1}:- e ”he“ 1 = if“ (2.4.2.26)
P2 P
8 _
—1W§e'P"+W,'[y—e"’"-ep"] = —5-e’P2‘ . (2.4.2.27)
82 P P

Finally, solving (2.4.2.26) for W,’ and substituting into (2.4.2.27) gives an equation for
Wz‘

8
w;e‘P*‘{;21[Me'—”P=‘+Pep='] p2 [—-Me 7’" -VPe”=‘]}=2, (2.4.2.28)

The terms in brackets in (2.4.2.28) may be evaluated with the help of (2.4.2.24) and

29
(2.4.2.25) as

Met-”H?” = zgsinhp 3d coshp 2t+2:—:coshp 3d sinhp 2t

8
Me‘Pi—Pe” = -28—2sinhp3d sinhpzt-Zz—Zcoshpgd coshpzt .
3

3

With these, (2.4.2.28) can be solved to yield

where

 

£1 . 8W2 .
x = E—smhp 3d coshp 2t+ coshp 3d smhp 2t
3 3

 

+ lsinhpgd sinhp 2t+%coshp3d coshpzt .
3

€3P2

(2.4.2.29)

(2.4.2.30)

(2.4.2.31)

(2.4.2.32)

The potential in region 1 can now be calculated. Substituting (2.4.2.31) into (2.4.1.8)

gives

H: (r) -

 

II—Vz(k)e1°re"l’t(2“)d2k

(21c)2_..X(k)

Here V, can be calculated using (2.4.2.23) with (2.4.2.11), giving

_ 1,3(r') e—jw
= , coshp (d+z ’)dv’.
z t]: 10053 P3 3

 

 

Substituting (2.4.2.34) into (2.4.2.33) then gives

”[42 (NH) coshp3(d+z’)
(21! 1-=)2 1033193

 

 

e-j k-r’]

;=JJ3(r m

ejk'rdzk } dv’ .

(2.4.2.33)

(2.4.2.34)

(2.4.2.35)

30

2.4.3 Green’s Functions

The Green’s function 0,33 describes the vertical component of Hertzian potential
in region 3 produced by a vertically directed elementary current source in region 3. By

superposition, the total potential in region 3 can be expressed in terms of the Green’s

function as

r13, = r1,1’+11,'+n,i .—. 10,3'3(r,r’).lz3(r’)dv’ (2.4.3.1)
V

Thus, 6,23 can be determined by summing up the potentials for region 3. Using

(2.4.2.18-19) in (2.4.1.8) and using (2.4.1.11) allows the total potential in region 3 to

 

 

be written as
I I 0° 1 - ° e-jkT’ - lz-z’l
n (r) = J3(r )dv —[w'e W+w*e”i’+,———e P3
32 ii 2 ISL (21!)2 z z 101532P3 ]
ejk'rdzk (2.4.3.2)
where the lower case w," and w,i are defined through
wg-‘(t-Jo = £1,3(r’)w,'"(r,r’,k)dv’ (2.4.3.3)
so that from (2.4.2.18) and (2.4.2.19)
. e.” 3" v,‘+ep 3‘1 v;
w‘ =[y-1] (2.4.3.4)
’ Willem” -ttt—11e"’"‘
[ l v'+ 1 v”
w' = e-P’d 7+ 1 ’ [Y— ] ’ d . (2.4.3.5)
mile” -[Y-1]e_p’
Here
e-jk'l’ #32:
v = -,————-e 2.4.3.6
z 1W32P3 ( )

so that

*‘I

“he

31

v; = £1,3(r’)v,*dv’ . (2.4.3.7)

By comparing (2.4.3.2) with (2.4.3.1), the Green’s function is seen to be

-j It'r’

 

03.3 = 1 j j [wge‘P3’+w,‘eP3’+ ,8 e‘Pi'z""]ef'”d2/c . (2.4.3.8)

u (21!)2 J (”5321’ 3

Note that this Green’s function is the inverse Fourier transform of a spectral domain

representation of the Green’s function. Symbolically

- " 3.3
0,23 = F 110,, } (2.4.3.9)
or
0.3" = (zit—)7] j ng'3ejk'rd2k (2.4.3.10)
where
~ 3.3 r _ i e_jkr’ I - 31 Z>Z’
G,zz = er P32+erp3z+meipsz 8+}, {ZQI . (2.4.3.11)

Using (2.4.3.6), this can also be written as

G33 = wz'e'p’z -t-w,‘.ep’z+v,*e“p22 (2.4.3.12)

Before substituting the expression for w,’ and w,i into (2.4.3.12), the quantity 7 in

(2.4.2.17) can be written as

N
Y A ( 313)
where
8
B = 30—3 ‘ (2.4.3.14)
33172

N a acoshp 2t+sinhp 2t (2.4.3.15)

 

 

 

W1

32

A E asinhp 2t+coshp 2t . (2.4.3.16)
Then, substituting (2.4.3.4) and (2.4.3.5) into (2.4.3.12) gives

06,33 = (BN —A)e”’3" e” v,-+(BN -A)ep’d e” 3’ v,++(BN +A)e “" 3" e 'P 3’ v,-

+ (BN-A)e"”"e“’”v,++(BN+A)e”’"e*P’Zvf-(BN —A)e"”de+”’zv,i (2.4.3.17)
where
D E ZBNsinhp 3d +2Acoshp 3d . (2.4.3.18)

Equation (2.4.3.17) can be simplified most easily by considering the following two

C3868.
Case I) z>z’ (upper sign)
Grouping terms gives
06,23 = v,+[(BN —A)e—p3depzz+(BN -A)e ‘Pide‘PS’HBN +A)ep’d e‘W
-(BN -A)e ’Pi" e'PS’ ]+v,‘[(BN -A)e"’"’ e” +(BN +A)e ‘W e ‘P 3’] (2.4.3. 19)

Substituting (2.4.3.6) and simplifying then yields

-'k-r’
.. e 1
00,23 = 2—

, 2cos (d+z ’)[BN cos z-Asin z 2.4.3.20
103832“ hP3 hP3 hP3 1 ( )

Case 11) z<z ’ (lower sign)

Proceeding exactly as above gives

_ -j k'r’
00,23 = 2—5——-

2cos ((1 +2 )[BN cosh z ’-Asin z ’] (2.4.3.21
J 0532P3 hP3 P3 hP 3 )

The above results can be combined into a single expression by using the notation

-'lt-r’
- e 1
06,23 = 2—

_ 2coshp (d+z<)[BN cos z>-Asin z’] 2.4.3.22
10232123 3 hP3 hP3 ( )

where

33

z> E max(z ,z') z< 5 min (2,2') . (2-4-3-23)

The Green’s function transform embedded in (2.4.3.22) can be isolated by dividing
through by D. Using (2.4.3.14-16) then gives

 

 

G 3'3 = film, coshp 3(2 <+d )F (2 >) (2.4.3.24)
u 1016 3
where

Qcos z -Zsin z

F(z) = , hm hp3 (2.4.3.25)
Q smhp 3d +Zcoshp 3d

Q = p 382[81p 2coshp 2t +£2p lsinhp 2t] (2.4.3.26)

Z = p 2€3[£1p 23inhp 21+ezp looshp 21] . (2.4.3.27)

Using (2.4.3.10) gives the final form of the Green’s function

G 3 3 .0 ejHH’)

zz - fiIiWCOShPfifMWQ’MZk . (2.4.3.28)

A check on this Green’s function can be performed by letting 1,12 = til and 81 = £1 so

that the three-layer dielectric system reduces to a two-layer system. In this case
Q = p3€22p2[coshp2t+sinhp2t] (2.4.3.29)
Z = 83p2282[COShp2t+Sinhp 2t] (2.4.3.30)
so that

cos — a sin
F(x)= p382, hp” pz 3 hp” . (2.4.3.31)
p gezsmhp 3d +p 283coshp 3d

 

Thus, the Green’s function is

3 3 l I]: ejk'(l'-l’) COShp 3(Z<+d) p3
22 (2702 -oo j (083;) 3 Tm P 2

 

e
coshp 22 >—g3sinhp32>12.4.3.32)

where

34
83 P3 .
Tm = —coshp 3d+——s1nhp 3d . (2.4.3.33)
82 P 2

Equation (2.4.3.32) is identical to the two-layer expression (7.9.51) from [ 10 ].

The total potential in region 3 is found by substituting (2.4.3.10) into (2.4.3.1)

 

.. 0
H32 1 111111 I 623’3Jz3(r’)dz’ldX’dy’k’k'Wz/c (2.4.3.34)

(2702 —oox’y’ z’=—d

For the special case of a sheath current ( an axially directed current on the sur-

face p = a), the current density function becomes

 

130’) = 5321803221) (2.4.3.35)
21w
so that
1'] _ 1 a 2n 0 123(2’) "3.3 I I I 'k-r 2
3. - (2102 Ii‘.i..i..‘27a“’u (z.a .t) .2 ,k)dz adoqu d k (2.4.3.36)

where (p,¢,z) are the cylindrical coordinate variables.

Because of the symmeuy of the problem, it is most convenient to evaluate the

integrals in (2.4.3.36) using cylindrical coordinates. Let

x = pcos¢ y = psinq) (2.4.3.37)
k, = xcoso k, = asino. (2.4.3.38)
Then
p2 = k,2+k,2—k2 = 33-18 (2.4.3.39)
k-r = kpcos¢cos¢r+lpsin¢sin¢
= 7chos(¢-<D) . (2.4.3.40)
Also, let

62330 .p’.¢’.z ’.k) = T330 .2 ’.7\)e “j k" . (2.4.3.41)

 

 

Ll;

 

Ffl

Xt

fia‘

35

Substituting these into (2.4.3.36) gives

 

 

21:
H3, (p,z) = —22F,3z',3(z z ’,?t.)dz e‘jMCOS“) —¢)d¢’ x
(2111)2 all: 341%. «£0
21:
j efip605<w>doptdt (2.4.3.42)
0:0
Now use
21: 27:
[ acme-mo = j e!‘ 91mm: = zit/Gap) (2.4.3.43)
0 0
and
J 0(—x) = 10(x) (2.4.3.44)
in (2.4.3.42) to give
1 co 0
n3, (p,z) = E j[ [1,30 ')1‘,3,-3(z ,z ',>.)dz 1100.21 )JOOLP)M 7t (24.3.45)
0 —d
where
~ 1
1‘3'3,’,)e= h < >.
zz (2 z ) “Moos 1930 +d)F(z ) (24346)
Equation (2.4.3.45) can also be written as
0
113,(p,z) = I 0,339 ,p,z')1,3(z')dz' (2.4.3.47)
—d
where 0,33 is the Green’s function
3.3 _ _1_°°~ 3.3 ,
0.. (z.p.z') - 2“ 1 1“22 (2.: 31/0021 yooepwx. (2.4.3.48)
0

Now, letting 0,}3 be the Green’s function describing the vertical component of poten-

tial in region 1 produced by a vertical component of current in region 3, the total

36

potential in region 1 can be written as

= 1‘11t = £G,§3(r,r’)tl,3(r’)dv’

where

13: _J.}°e (1’42“) coshp3(d+z')
22 (21102—4. JOJ€3P3

 

This can also be written as

GL3: Gzl.e3 jksrdzk
7'2 (27—)2-1-10

where

-13 I _ (“(1“) coshp3(d+z’) 5].“,
622 (r’k) -
X 10°83P3

 

is the Fourier transform of the Green’s function.

2.5 Green’s Functions for Magnetic Hertzian Potentials

ejk'(r-r')d 2k .

(2.4.3.49)

(2.4.3.50)

(2.4.3.51)

(2.4.3.52)

The Green’s functions for magnetic Hertzian potentials maintained by a horizontal

magnetic current in tri-layered media are derived in this section.

2.5.1 Scattered Potential Amplitudes

Consider the upper half space (z>0) shown in Figure 2.3 first. A horizontal mag-

netic current in region 3, M3 = 121143,, generates the following Hertzian potentials in

the three regions above the ground plane. Attention should be paid to the coupling

between the horizontal and vertical components. The potentials in each of the three

layers can be expressed as

n5=2ng

(2.5.1.1)

37

11,3 = mg, + £11,! ; i=1,2,3 ; 'y=+,- (2.5.1.2)
where
Hf: Primary potential generated by magnetic source in region 3.
l'I,-+: Scattered potential in region i traveling in +2 direction.
Hf: Scattered potential in region i traveling in -z direction.

The superscript m of magnetic Hertzian potential II’" has been dropped for brev-
ity. This should not cause any ambiguity because in most cases in this dissertation it is
quite clear from the context that H means either electric Hertzian potential or magnetic
Hertzian potential. The superscript m will be added, or explicit explanation will be

provided, whenever there is a possibility of confusion.

In region 1, the scattered potential wave can travel to infinity without reflection.

Therefore

11,-, = 111-, = 0 (2.5.1.3)

By using (2.3.11-12), primary and scattered potential can be written as

 

 

(r’) eik-(r-r’) -pa|z-z'|
115,0 r): 1M” [11" : 42k] dv’ (2.5.1.4)
nyp= J‘j— ”Wi—‘i—k ) e‘W ej""'d2k (2.5.1.5)
(21:)2
where
B=x¢; y=+¢; i=133
r=£x+jiy+22 ; k,-2=(02£,-u,- (2.5.1.6)
k=2k,+yk, ; dzk =dk,dk, (2.5.1.7)

p? =k3+k,2—k.-2 ; Rciptl>0 and Im{p.-}>0. (2.5.1.8)

38

The Hertzian potential in each region can be written as

II1 = fI'lf’xH‘Hf', (2.5.1.9)
112 = 1? (112341;, )+2‘ (11;,+r1,,-,) (2.5.1.10)
113 = f(1'lf,,+l'l3+,+11§,)+é‘ (11;,+r13-,) (2.5.1.11)

Using the boundary conditions (2.2.42-48) at the three interfaces, ten equations
result, which will be solved analytically to obtain the ten unknowns fo, Wf’,, W21,
W22, W21, W22, W3}, W32. W3}, and W52. The ten boundary conditions are listed as

follows:

At interface z=d+t,

“1:: = €21H21n2x (25-1-12)
3 3

En“ = “21511,, (2.5.1.13)
11,, = ”2111,, (2.5.1.14)
1(1'11 4122) = —(€a11121'1)-a—nzx (2°5°1°15)
82 z 81

At interface z=d,

“2. = 5321132H3x (2.5.1.16)
a a

5112; = ”3252-1133: (2.5.1.17)
H22 = 1132113: (2.5.1.18)
£0122 -IT3 ) = ’(8321132-1)'§—H3 (2.4.19)
32 2 ax "

At interface 2:0,

113, = 0 (2.5.1.20)

39

—H3,, = O (2.5.1.21)
where

Ht’j a E‘— ; i,j=1.2.3 (25.1.22)
9,-

”m
\-
Ill
Kan.) I vs?)

Substituting (2.5.1.5), (2.5.1.9), and (2.5.1.10) into (2.5.1.12) gives

II—uW1x(k)e-pl(d+l)ejkrd2k
(21:)2

°°W 00
42111211112): 2 WW) ”3’2“
(2n )
°°W k .
j 12‘” epzwmelszk] (2.5.1.23)
(21!)2

For equation (2.5.1.23) to be valid for arbitrary r, the following relationship must

be true.

Wfie’p‘w”) = gluzltwge’h‘d“’+W,;eP2““"1 (2.5.1.24)

Using equations (2.5.1.13)-(2.5.1.15), following the above procedure, and inter-

changing the order of integration and differentiation when l or i is encountered,

82 8x
give the following equations.
plwge‘Pi‘dm = 1121p2[Wf,e-p'(d+‘)—W2}e“(dm] (2.5.1.25)
wge'PlW’ = u2,[W;,e'P2“’+"+W2-,ePz‘d+"] (2.5.1.26)

d - (1+1)- d
plwfze—ptt “’—p2W{,e P2(+ +2!) W eP2( +0

=(821)121-0119. [W’e 7"“) +W'e ”2“”)1 (2.5.1.27)

40

Equation (2.5.1.4) can be rewritten as

11 (——:":2) (21“de (2.5.1.28)
112
where
31“,) _' .' - lz-z'l
WI; (1t,z) — e “”e P3 dv’. (2.5.1.29)
x 4,2 (0113P3

Substituting (2.5.1.5), (2.5.1.10), (2.5.1.11), and (2.5.1.28) into (2.5.1.16) gives

”W21(k___)_e—p2d elkrd2k+ W2x__(__k)ep2d 8.1""de

=€32143zlii nga‘ 'd) ejkirdzk‘i

_.. (2 )2

°°W
II _(___:x:2)e -P:d ejkr d+2k +1]: (_:_)_2_x(k)e Pad ejk'rdzk ] . (25.130)
It 7:)

The following equation can be obtained because equation (2.5.1.30) must be valid for
arbitrary r

WLe’PI“+W2-,eP =832u32[W§’,(d)1-W+ e—p’d+W3’,eP’d] (2.5.1.31)

Using equations (2.5.1.17-21) and (2.5.1.4-11), following the procedure outlined
8

above, and interchanging the order of integration and differentiation when 58; or E-
is involved give
p2(W{.e e'P’d—Waep’d)- - 1132P3(W§’x(d)'1'W3+,e-p’d W3-,,eP3" ) (2.5.1.32)
wge'PihwgePP’ = 1132(W3+,e-p’d+W§,ep’d) (2.5.1.33)
pa(W2*.e""—Wi.eP“>-pstwee’PP—WaeP’d)

=(e321132—1)jk.<wi.(d)+wa e‘Pi"+W;.ePP) (2.5.1.34)

41

W3’;+W§, = 0

—W€.<0)+W3;-W§. = 0

(r’) . - _. _

ng(d) "" I 2 (:L3p3e-jk'r’e P3(d Z )dV’ = V§,(k)e Pad

V3

31“,) -jk-r’ -psz’ ' —

“gm—121,00”) e e dv =v3,(k)

V
Vi: M3,,(r’) 'Jk'lexp’z'd '

V3 ZjWZ’PB

(2.5.1.35)

(2.5. 1.36)

(2.5.1.37)

(2.5.1.38)

(2.5.1.39)

There are ten independent equations (2.5.1.24-27) and (2.5.1.31-36). These equa-

tions can be solved to obtain the ten unknown Hertzian potential components. Equa-

tions (2.5.1.24), (2.5.1.25), (2.5.1.31), (2.5.1.32), and (2.5.1.36) can be solved first to

get the five x components. The other five equations can be solved to obtain the five 2

components. Notice that equations (2.5.1.27) and (2.5.1.34) describe the coupling

between x and 2 components.

To simplify the derivation, the following notation is introduced:

_Pl'd

d,- E e ; t,- E e-p” ; i=1,2,3
Pi di ‘i . .

Pij E — 3 dij E — i ‘1} E — ; l,j=1,2,3 .
P1 ‘11 '1

Now, pzx (2.5.1.24) plus 821x (2.5.1.25) results in
1
W2; = 3111203121? 12)‘12412W1+x -
While, pzx (2.5.1.24) minus 621x (2.5.1.25) gives

- 1
W2: = 3H12(€12-P12)11‘2d1dzwfl

(2.5.1.40)

(2.5.1.41)

(2.5.1.42)

(2.5.1.43)

42
and p3x (2.5.1.31) plus e32x (2.5.1.32) results in
p3(W{,d2+W2’,d2‘1)+e3,p2(W{,d2+W2‘,d§1 )
= 28321432!) 3(Vixda+W3+xd3)
Some algebraic manipulation then gives

1 _
W3} = _V;x+2'pl3d3l [(512+P 12)(€23+P 23)! 1261 1+
(€12'P12X323—P 23)! 112d 11W 1+1 -
Next, p3>< (2.5.1.31) minus E32x (2.5.1.32) produces

_ 1
W3: = ‘4‘1113‘13K312‘W axe/23“!) 23)‘ 124 1+

(€12‘P12)(€23+P 23)‘ 1126’ 11W 1+; -

(2.5.1.44)

(2.5.1.44)

(2.5.1.45)

(2.5.1.46)

Substituting (2.5.1.45) and (2.5.1.46) into (2.5.1.36) gives the solution for W1}

+ V§,+V3’,
W12: = 492331131P2Ps‘21d31—D—

I

where
0.00 = (eipzfizp 1X62}: 3+63p2>+(81p2-€2p 1)(‘52P3-"33172)122
—<elp2+ew ow 3‘63P2W32451P2-92P1X82P 3"‘531’2V22da2
N134") = 452531131P2P3‘21d31(V3§+V3}) -
Now, rewrite equation (2.5.1.46) as

Ni}

wg=D
X

 

while, substituting (2.5.1.50) into (2.5.1.42) gives

N 5;
DZ

 

wig:

(2.5.1.47)

(2.5.1.48)

(2.5.1.49)

(2.5. 1.50)

(2.5.1.51)

43

where
N2+x(k) = 28319 3(E1P 2+€2P 01112113161 32(V3+x+V3-x) - (25-1-52)

Next, substituting (2.5.1.50) into (2.5.1.43), (2.5.1.45), and (2.5.1.46) respectively

 

 

 

gives
w; = If); (2.5.1.53)
where
N2’s(k) = 2€3P3(€1P2-€2P1)11121131‘22dzd3(V$+V3-x) (2.5.1.54)
W5; = 1;? (2.5.1.55)
with
N32 = [(Eip2+82p amps-Esp 94324461112332? 1)(€op3+83p2)t22d32 1V5:
+ [(elp 2+r.~,,p,)(1»:,p 3+e3pz)+(81p 2—e,p 1)(€2p3-e3p2)t22 1V3:r (2.5.1.56)
W3; = 1232‘ (2.5.1.57)
and
N3} = [(EIP2+€2P1)(92P3‘€3P2)‘132+
(alps-22p1)(esps+esp2>t%d§1<vs;+vs;) . (2.5.1.58)
Jext, 11.51251 x (2.5.1.26) minus (2.5.1.33) and then using (2.5.1.35) result in
W2; = -—dzi—[u,2:,:,d,d3w,+,+u32:22(1—d32)wg,] (2.5.1.59)
(1-12)d3
rile, —u.2‘11t2x (2.5.1.26) plus (2.5.1.33) gives
W2; -1 [1112‘1‘2‘11‘13W1"24'11320-(132 )sz] - (2.5.1-60)

 

— (1-522)d2d3

44

Substituting (2.5.1.59) and (2.5.1.60) into (2.5.1.27) and (2.5.1.34) gives the fol-

lowing matrix equation.
+
011 012 W12 ___ b1
021 022 W35, 172

011=‘1d1(1"22)P1+1112‘1d1(1+‘22)P2

where

012 = 21132‘2(1'd32)d3-1P2

021 = 21112‘1‘2‘111’2

a22 = 1.132(1-(132)(1+t22)d3"1p2+(1+d32 )(1—122)d3-1p3
bl = 1k. (6211121-1X1-‘22)(‘2d2W§x+‘21d2-1W2-x)

122 = -jk, (6321132—1)(1—t§)(d3V3+,+d3W3+,+d§‘W3}) .

Equation (2.5.1.61) can be solved readily:
A = 011022 ' 012021
A1”; = b16122 " 192012

+ _
A32 - 011b2 — a21bl

 

Afz
w,+,= A
A;
w;— A

(2.5.1.61)

(2.5.1.62)
(2.5.1.63)
(2.5.1.64)
(2.5.1.65)
(2.5.1.66)

(2.5.1.67)

(2.5. 1.68)
(2.5.1.69)

(2.5.1.70)

(2.5.1.71)

(2.5.1.72)

The next task is to express Wf, W3“z in terms of known parameters and expres-

sions. Substituting (2.5.1.62-65) into (2.5.1.68) gives

A = [t1d1(1—t§)p1+u12tldl(1+t22)p2111132(1-d32)(1+122)431P2+(1+432)

(1—‘22 )ds-lpsl - [21132120‘432 )d51P2](2H12‘1’2d1P2)

(2.5.1.73)

45

= tldlds" 1151110422 112 1u2+(1+t22 )ulp 211(1‘d32 ><1+r§ )p 2113+
(1+ds2 )(1—‘22)1-12P3] - 4u1p22u3t22(1'—d§)}

(1—222)z,d1

flids

 

2

where
0.00 = [111112PW3(1+122)(1+€132) + 1111122113(1-t%)(1-d32)+
2 _ 2 2 2 2
p1u2p3(1 :2 )(1+d3 ) + [21122112113(1+:2 )(1-d3 )1 (2.5.1.74)

Substituting (2.5.1.51-58) into (2.5.1.66) and (2.5.1.67) results in

 

 

 

 

N + N -
bl = jk,(12.211121—1x1—122)(zzd2 2" “511151 2’ ) (2.5.1.75)
DI DX
. V§+V§
= ka (8211121‘1X1‘t 22 )4fl3251Pfi’3‘53‘ 2d 3 Z) x
x
. 2 , N3; _1 N 3}
b2 = —}k,(e32u32—1)(1—t2)(d3V3x+d3 D +d3 D ) (2.5.1.76)
X x
= ‘ka (332032‘1X1‘t 2 )252P3d3131P2U“ 2 W521? 111“ 2 )]'—D_—
1

Next, substituting (2.5.1.62-65), (2.5.1.75), and (2.5.1.76) into (2.5.1.69) and (2.5.1.70)
gives

V§+V§

D 4u31p2p3I2(1-t22)Ll2-2[(1+t§)(1_d32)(53113’51111)€2112P2
x

41*. = jkx

+ (1—r%)(1+d32)(esuz—elu11uzesps+

<1-r%>(1-d§)<esus-ezuz)ezuw11 (2.5.1.77)

+ - V52“?! 2 -2 2
A3: = -ka D 2t1d1d3P3(1-t2 )uz {(Esus-82112)[u1p2(1+t2 )+
x

46

u2p1<1-t%>1121p2<1+t22>+esp1(1-t3>1+4<ezu2-elu1123usr§p%I . (2.5.1.73)

Then, substituting (2.5.1.73), (2.5.1.77), and (2.5.1.78) into (2.5.1.71) and (2.5.1.72)

results in

 

w+ — 'k N‘:
12 JxDxDz

where
NI; = 4H31‘21d31P2P3[(1+t22)(1-d32)(83113—51111)€2H2P2
+ (1“22)(1+d32)(32H2’51U1)£3112P3
+ (Hg>(1-d32xesus-ezu21e2ulp11<vs§+vs§>
and

N33
DXDZ

 

W52 = jkx

where

N 3’; = 41132123[(63u3-62u2)[u1p2(1+t22)+112p1(1-t22)]

[€1P2(1+t22)+€2171(1—122)i‘i'4(€2112'€1111)33113t2217221(V§X+V3_x) .

Next, substituting (2.5.1.81) and (2.5.1.82) into (2.5.1.35) gives

_ . N32
1 Z

 

where
- _ +
N 32 - _N 32

Finally, substituting (2.5.1.79-82) into (2.5.1.59) and (2.5.1.60) results in

N22
DxDz

 

W2"; = jkx

(2.5.1.79)

(2.5.1.80)

(2.5.1.81)

(2.5.1.82)

(2.5.1.83)

(2.5.1.84)

(2.5.1.85)

47

 

 

 

where
211 221124
2‘. = (312 :2) 3P312[(1+‘22)(1—d32)(€3113‘51111)€2112022
" 2
+ (1‘122)(1+d32)(82H2‘51111)€3112P2P3+(1‘122)(1‘d32)(83113—8211932H1P1P2]
— (1'432){(€3113—52H2)[111P2(1+‘22H1121?1(1—122)][€1P2(1+’22)+€2P 1(1—22211
Mews-61110831192522 11(V3’3+V31) (2.5.1.86)
and
I.
w+ = jk (2.5.1.87)
22 X DxDz
where
211 2d
Ni»; = 3 32P312‘22[(1+‘22)(1‘d32)(53113—€1|>11)€211W22

(1—122)
+ (1‘15 )(1+d32 )(82112-51111)€3U2P2P3+(1“22 )(l'dzi )(83113’52112)€2111P 1P2]
- (l-daz){(83u3-62u2)[u1P2(1+t22Hunt)1(1—t22)][£1p2(1+t§)+62p1(1-122)]

”(Esta-elulksustzzpzz l }(V§.+V3',) . (2.5.1.88)

The following summary will be convenient for later use:

 

 

 

 

N i;
w; = (2.5.1.89)
DI
N 2:
W22 = (2.5.1.90)
Dz
N2}
w- = 2.5.1.91
2, DJ: ( )
N-
W3", = 3’ (2.5.1.92)

48

 

 

 

 

 

 

_ N3-
W31: Dx
X
+ ”1+2
W12 —kaD D
I 2
N22
W22=kaDD
I 2
_ N2}
W22 JkXDD
I 2
N;
W3; —JkXDl;
X Z
_. . N3}
W3z=kaDD
X Z

(2.5.1.93)

(2.5.1.94)

(2.5.1.95)

(2.5.1.96)

(2.5.1.97)

(2.5.1.98)

D,(k) = [elemp3(1+e’2“‘ )(1—e‘2P’d )+e,p22e3(1-e'2")(1+e‘2”"’ )

+ p 12.22p3(1—e“2“’2‘ )(l—e-Z‘D’d )+p 1p21=.2.c.3(1+e‘2*"2‘ )(1+e‘2""’ )1

0.09 = tuluspzps<l+e‘2”“><1+e‘2”"’

>+u1p221ts(1—e‘2”2’)(1—e

(2.5.1.99)

‘2P3d)

+pluips(1—e"”*‘x1+e'w)+plpsusus<1+e'2”x1—e‘w>1 (2.5.1.100)

N 130‘) = 452331131P2P3e-(prpm8—(PTPM(Vrix‘tVs-x)
Nam = 2€3P3(€1P2+€2P1)11323-(prp2)d (vs’;+Vs:.)
Nix“) = 233p 3(91P 2'62? 011323 419216 *pfipm (V‘ix‘l‘Vs-x)

Nam = [(91P2'1’52P1)(€2P3‘€3P7)e_2p3d+(31P2—€2P1)(€2P3+83P2)e

(2.5.1.101)
(2.5.1.102)

(2.5.1.103) '

'2P2‘e’2P3‘1]

V52 + [(elp2+€2P 1)(82p3+£spz)+(61p2-€2P Dem-83192)?“1V3}(2.5.1.104)

N 3x 0‘) = [(811) 2+82P 1)(€2P reap 2)6

+ (em-650 1X52!) 3+€3P 283—2102“? -2p3d](V3ix+V3})

"ZP ad

(2.5.1.105)

49

NEAR) = 4uglp2p38-(prlye-(P3—P1M [(14'3—2’02’)(l-e-2p3d)(€3113-€1u1)£2117P2

+ (I-e'z“><1+e‘2”"’Xesuz—elunesuzps
+ (1-e‘2p2‘)(1—e’2”3" )(E3u3—82112)82u1p 1](v3+,,+v3-,) (2.5.1.106)
211328-2772! 8-0»an
(1—e’2”>

 

NEAR) = p3121<1+e‘2“>(1—e‘2“’ )(83H3‘81111)€2112P22

+ (l—e'zpz‘>(l+e'2”"’)(ezuz—eluoesuzpzps

+ (l-e'zp“)(1-e"2‘”"‘ )(83113—52112)€2111P1P2]

 

— (Fe—2W!)1(€3u3-Ezu2)[u1P2(1+e-2M)+112p1(1—e—zp2‘)]
[e1p2(1+e‘2“ )+e,p ,(1-e'2p2‘ )]
+ 4(52H2‘51111)53H3'22P22 11(Vix+V3_:c) (ZS-1°10”
+ _2u3ze"(-P2+p3)d -2102: —2p21 -2P (1 2
N22(k) = _ 193126 [0+8 )(1-8 ’ )(83u3-81u1)€2quz
<1—e 2”)

+ <1-e'2”><1+e‘2”"’><ezuz-elul>esu2pzps
+ (I—e‘m)(I—e‘ZP’dXesus-ezuoenulp 11221
- (143-30“!){(€3113‘€2112)[111P2(1+e_2m‘)+H2P 1(1-8-2’02'H
[£1p2(1+e-2M)+67p1(1-e-2N)]
+ 4(82112-31H1)€3113‘22P22 ”(x/saws.) (2.5.1.108)
Neck) = —2e‘2”"‘p31<esus—ezuz)tulp2(1+e’2"2‘)+u2p 1(1—e‘2Pz‘n
[81p2(1+e-2‘”)+e2p1(1—e-2p")]

+ “$292431!“)!’—:3113‘3-2p2‘1922 1(V§x+V§c) (251-109)

N3’z(k) = -N3+z(k) (2.5.1.110)

50

Vitxa‘) = I M310“)

, e-J'k'r’ew'dv'. (2.5.1.111)
V3 2} (”“3173

Notice that 0,, DZ,N1+,,N§_*,,N{,,N3+,, N§,,Nf,,N§,, N23, N53, and N3], are func-
tions of lkl = Vk ,y2+k2 only. i

The lower half space shown in Figure 2.3 is a much simpler structure to analyze.
A horizontal magnetic current in region 4, M4 = x‘M4,, generates the Hertzian poten-

tials
11,, = f (1'15,+1'I4‘,) (2.5.1.112)

Because there is no interface except the ground plane in region 4, there is no cou-
pling between the horizontal component and vertical component. In other words, the
horizontal component of Hertzian potential can describe the electromagnetic field in
region 4 completely. Using the integral representation of primary and scattered poten-

tial (2.5.1.4-5) gives

 

 

 

 

M x r’ °‘ ik-(r-r') -p4|z-z'|
Hg,(r)=1 ,4 ( )[11e : dzk]dv’ (2.5.1.113)
V, 10114 ... (21:) 2P4
= ,11W ’2 ep‘zejkrdzk. (2.5.1.114)
4421:) ‘
The boundary condition at the interface 2:0 is
an
4" = (2.5.1.115)
32
Substituting (251.113-114) into (2.5.1.115) gives
M r .
w;,=1—5-"—(—’)—e-i'"’e"’"dv’. (2.5.1.116)

V, 1203114114

51

Now, assume that

mm = W410, )8(z) . (2.5.1.117)

Then, equation (2.5.1.116) can be rewritten as

M ,(r’) .
wg, =11—,—4——e'l*‘"’ds’. (2.5.1.118)
5120414194

2.5.2 Green’s Functions

After solving for the scattered magnetic Hertzian potentials in the three regions
produced by the three orthogonal components of an arbitrary magnetic current source
in region 3, the dyadic Green’s function for Hertzian potentials can be determined. The

following notation will be used:

I'I,-”'(r)=16i'j(rlr’)-Mj(r')dv’ ; i,j=1,2,3. (2.5.2.1)

V1

Here l'I,-’"(r) is the magnetic Hertzian potential in region i maintained by a magnetic
current source in region j.
In this dissertation, the case of interest is the one when i=1,3 and j=3. Also, the

source is assumed to be on the z’ = 0 plane and has only an x-directed component.

Thus
M3(r) = fM3, (r, )5(z)

In this case (2.5.1.111) can be written as

V§,(k) = V3',(k) = e-ikflis' (2.5.2.2)

s3 ZjW3p3

where

I IAI

r,= +yy.

52

Now, using (2.5.1.5), (2.5.1.9), (2.5.1.89), (2.5.1.101), and (2.5.2.1) gives

= 1 6,};3(r|r’)M3, (r, ’)5(z ')dv '
V3

 

N + _ (I‘ ) - .
_ de 11 Pt2 ejkr M3xl' ze-jk-rdsi
de H (211020 e [si— 2! (011:1P32 ]}

—oo

=11 11 W” N”
s, (21:)2 _.. 101113193 Die

 

 

)]M3, (r, "-)d5

 

So

G,,§,-3(rlr') = $0511de 3;: (12:, ) (2.5.2.3)
where

le (z .k) = 4.22:531131p,p3e*"rpl’fie*1D35P1’de2’1z (2.5.2.4)
and

r, =£x+yy .

Other components of the dyadic Green’s function can be obtained by following
the same procedure and using (2.5.1.4), (2.5.1.5), (2.5.1.9), (2.5.1.11), (2.5.1.89-110),
(2.5.2.1), and (2.5.2.2). Some algebraic manipulation gives

 

 

 

               

 

 

G,;-3(rlr') = 0 (2.5.2.5)
°° 'k-(r-I’) N
0,1,3 I - 2k e' ' 1’ 2. . .

(r I") (2 21:021.... d jmusps ’DxDz) ( 526)
ik(r- -r’) (N

dis2 8 3" (2.5.2.7)
102113123 (DJ:

Gyifiulr’) = 0 (2.5.2.8)

J'k(r— -I’) N ,
0,3,3o-Ir'):-—1211cW21dkdk2 " (jk 3 ) (2.5.2.9)

jwuaps 1" DID:

53

where

N12 (2 ’k) = 4113117 217 38-002?” Bin—m” [0+6 4” )(1—8 4W1 )(83113—81H1)52112P2

+ (I—e‘m><1+e”P"’)(e2u2—21u1)e3uzps

+ (1-e‘2p2‘ )(l—e 4“" )(e3u3—82112)82u1p 1]e ‘Plz (2.5.2.10)

N31: (2 1") = [(elp 2+€2P 1X52!) 3+€3P 2)+(81P 2‘82P Mew 3‘83P 2)€ 4” ]e ’p 3’

+ [(81P2'H57P 1)(€2P 3-8312 2)+(81p r821) (Key) 3+€3p 2)e‘z‘”162103".«:”3’(2.5.2.11)
N,, = 2e‘2”"’ps((esus—ezu2>1u1p2<1+e‘2P2‘wzpra—e‘z” >11e1p2(1+e"”">

+6W1043-2””#1032112“:1110631136 2MP 2103,7312 -€—p’z) - (25-2-12)

Again using (2.5.1.112), (2.5.1.114), (2.5.1.118), and (2.5.2.1 ) gives 0,354 as

dkz e_-__jk(r- -r’)

42’"2 0. 2. .2.1
(21102QH jwum K (5 3)

 

Gfi'4(rlr’)— -

2.6 Green’s Functions for Fields

After the Green’s functions for the Hertzian potentials have been obtained, the
Green’s functions for EM fields can be derived by using the relationship between the
fields and the potentials. The dyadic Green’s function for the fields due to an electric

current is derived in section 2.6.1 and the dyadic Green’s function for the fields due to

a magnetic current is derived in section 2.6.2.

2.6.1 Green’s Function for the Fields due to an Electric Current

The electric field maintained by an electric current can be written in terms of

dyadic Green’s function

54

13,.(r) = 1g'i(rlr')-Jj(r')dv' ; i,j = 1,2,3 . (2.6.1.1)

The electric field is represented by electric Hertzian potentials (2.1.1) via

E,- = (k,2+VV-)IT,- (2.6.1.2)
n,- = 18” (rlr’)-Jj (r’)dv ' . (2.6.1.3)
V

Substituting (2.6.1.3) into (2.6.1.2) gives the dyadic electric Green’s function

2"" (rlr) = (k.2+VV-)8‘J (rlr) . (2.6.1.4)

Assume the current distribution in region 3 is a sheath current along the z-axis.

Then (2.4.3.35) holds, and (2.4.2.33) becomes

1 "170») ._ _
t = = —- _Z P1(z t)
112 11,, 21:1 x0») e 10(7tp)).d}. (2.6.1.5)

in analogy with (2.4.3.44), where

_ 1,3(2’) coshp (d+z’)
11,0.) = 11,033 [:3 10(M)dz’. (2.6.1.6)

 

The elecuic field in region 1 can be found using (2.6.1.2). For a vertically

directed potential this reduces to the relations

 

2 aznl,
E12 = k1 le+—- (2m617)
822
8%,,
E 1p = apaz (2.6.1.8)

where p is the radial variable in polar coordinates. The necessary derivatives for using

(2.6.1.5) in (2.6.1.7) are

 

31—112 1 a V2 0") -p,(z—r)
__az_ _ 21; g x0") ple 10(xp)Ml (2.6..19)

55

8211 , “V, _(_2) _ .-

 

32111, 1 0° 171(k)

- — -pi(2—t)
828p 21C 0 x0") P1 11(lp)7\d3. .

   

 

Substituting (2.6.1.10) and (2.6.1.5) into (2.6.1.7) gives

 

°°_;"___V( A) ‘P1(Z"‘)) 1
or, using (2.4.3.39)
21 °° 1:0») _ _
E = — J 2. em“ ”2.3212.

Finally, substituting (2.6.1.11) into (2.6.1.8) gives

211:”172110»)

Elp = “(1) "P1(Z—1)p 17»sz

 

 

         

(2.6.1.10)

(2.6.1.11)

(2.6.1.12)

(2.6.1.13)

(2.6.1.14)

To derive an elecuic field integral equation for an imaged monopole, it is neces-

sary to know the 2 component of the scattered field in region 3. From (2.6.1.2), the

electric field in region 3 maintained by a vertical electric current in region 3 can be

 

 

written as
32113
._ 2 2
E32 - k3 1132+ 322
Mg,
E1. = 3.53:.

Then, substituting (2.4.3.47) into (2.6.1.15-16) gives

E... = (k32 +—)10..(z .p.zp )I.3(z )dz'

(2.6.1.15)

(2.6.1.16)

(2.6.1.17)

56

2 6 2 Green’s Function for the Fields due to 3 Magnetic Current
The relationship between the magnetic Hertzian potential and EM field rs used to

construct the dyadic Green’s function for the EM field. The magnetic field Green 5

function can be written as

mm = 1§“'j(r|r’)-Mj(r')dv’ ; 1,} = 1,2,3,4 . (2.6.2.1)
V
Expressing H1 in terms of magnetic Hertzian potential I11" gives
Hi 1' kiznim+V(V'I-Iim)
(2.6.2.2)

= (k,2+VV-)18i'j (rlr')'Mj(r')dV' -
V3

Exchanging the order of integration and differentiation and using (2.6.1), the magnetic

field Green’s function can be expressed as
$8"; '3 (2.6.2.3)

klz

E“ ’(rlr’)= P. V. (k 2+VV)<’f"1(rlr’)—__._

where P.V. stands for principal value and

Z’ = Lnfi+Lyyfi+Lzz 22
is the source dyad. Each term on the right hand side of (2.6.2.3) is dependent on the
shape of principal volume, but the combination of the two terms is independent on the

pnncrpal volume [ 6 ][ 12 ]. The explicit expression of E is not given because mag-

netic field Green’s function in (2.6.2.3) is not used directly.

Carrying out VV-G’ results in

V6 =%[an+6 ny+G 2142-:— y[G ,r+G”y+Gy,2]

a . ,. .
+5;[an+00)’+0222] (2.6.2.4)

md

57

VV-E’ = f-Q—(V-5)+y—a—(V-5)+i—a-(V-8)
8): 8y dz

2 2 2 2 2 2
:fi(30,,+30,,J30,, 30 H30,,+30,,)

.. 13
8x2 Bxay T 3132 )+xy( 8x2 axay 29sz

AA

320,, 820 , 320,, 820,, 320 , 320,,
fi( + y 2 2 ’

3x2 axay T 8x82 ny axay T 3,2 Tayaz )T

 

 

 

 

 

mam, 820yy £220,y 320,, 320,, 320,,
+ HM + + )+
y“3):;in ayz ayaz axay ayz ayaz
73320,, 320,, + 320,, HZ3,1320,y + 320yy + 820,y H
31:82 (9sz 322 8x82 Byaz 322
.. 320,, 1 820,, 1 320,, 2 6 2 5
”(axazTayazT 3.2) ("')

Throughout the dissertation, the magnetic current is assumed to have only a x-

component
M3 = 2M3, (2.6.2.6)
M4 = fM4x . (2.6.2.7)

The components of E” can be expressed in terms of the components of 6111' by

using (2.5.2.3), (2.5.2.5-9), (2.6.2.3), and (2.6.2.5) and they are summarized as

 

 

 

 

 

         

 

 

a2 32 L38(r-r’)
3,3: _ 3.3 3.3 2 3.3}
g,, P.V.[ax20,, Ta 3 ——0,, +1:3 0,, [(32
Ndz
=1".V.dk2 e (k2 -k, ,
((4)2131: In)“ “:1 3 2..“(00. )1)
L38(r—r’)
1:32
3.3 ejkrde Lna(r—r’)
v.1!”2 ——21)11g,, 1:32 (2.6.2.8)

 

58

g3.3=_iz_G33+__ 32 G33
" Byax ayaz

j (r 1’) ~51,
‘7‘222211Ndk2-.——6W 31-k.I< (D—:"-)k.k,ps(-2-2——Dz )1

3.3 '-k 2
(21021 1 g e1 rd k (2.6.2.9)

32 3 3+ 32 3 3
—Gz,
81.3 =8z 8x ___—G” +23 2
jk-(r—r) 213*,

1 2 e . N32
= k,——— k —+
(2102]... 102113123 U "p3 D. p

DD

1 Z

 

 

2
3

]

 

=02 )211g,, 33 eikrd2k (2.6.2.10)

32 32
: k2+__ Géo3+__
<1 22,)

1.3
31.320“

22:“.

 

 

 

= 1 (121.3, [(1.32 —k,
(2102 —oo Imp-@3010:

                  

)1

~ 1.3 -r
=(22)2 ——1 1g,, ej" d2k (2.6.2.11)

1.3=_13
gyx -a—y_ax n ayaz ___sz

Ndz

3),,1—1. k,(-D— N">-k.k,p3<D D2]

= 2k ej I"W
(2102 U." jwsp

 

 

 

 

 

= (211)21"1'3ej""d2k (2.6.2.12)
1t
2 2
1.3 _ .2— 1.3 _a_ 13
gzx - azaxGxx+aZZsz
jk-(r-r') N“, N
ke kaP3 D" 1’ 1

_ 2 T
_ (2102”... 160113123 p320. Dz

59

 

 

 

 

 

(21c)2 ...
44- 2 a 44 L380. r)
gn PV(k4+ax2)Gn k}
'k )
=P.V lzJI Zke-J (kg-3’ L 80- r)
(21;) _.. JwH4P4 1‘42
= 121123-‘e12'd2k a< )
(2n) - “2

 

 

=(211c)2H§yx 44ejkrd2k

g4 82 04.4
”4 =azax
ejk(r— -I’)

(210211de jwum kam)

 

 

1...}? 4.4ejkrd2k

 

          

 

 

 

 

=1:(2 )2 4.
where
-3.r3(r k) = 32[(k 'kx Nd: ‘1'”,
icon 3 D, Dz )]e
gyi3(r’.k)= . 1 1—k,,k(—D—— :ka p3( Ng ”8"“
101131331) D
zx ’ j 3 x 3 D, p3 DxDz e

(2.6.2.13)

(2.6.2.14)

(2.6.2.15)

(2.6.2.16)

(2.6.2.17)

(2.6.2.18)

(2.6.2.19)

 

                 

 

 

 

 

 

Nd
" g”(r’ k)— - EDI—3,71“? -k, D ND )]e-12'“ (2.6.2.20)
.. 1 3 1 Nd? —jk r’
gny' (1' .k) = wk 19,16y (—- :)—k,kyp3( D Dz )]e (2.6.2.21)
- 1 . N"; N _
g;3(r’ k)- - mp3“ kaP3(31—)+p32(F-D—)]e 1"" (2.6.2.22)
X
E34031) = . 1 (kZ-kf)e"'""’ (2.6.2.23)
100214“

.. 44 k _ 1 _ -jk'l"
gyx (r’9 )— jwp4p4( kxky )6 (2.6.2.24)
.. 4 4 1 . -1”;

(r’ k) = k, )e ’ 2.6.2.25
gzx ij4P4 (J P4 ( )

d _ a

N3x(2 ,k) = _N3x (Z ,k) (2..6.226)

82

= p3{-[(elp 24'62P 1)(€Qp3+£3P2H(€1p2—8j71)(£¢p3—£3p 7)e 'ZPZ’ le‘Psz

+ [(81P2+£2P1)(€2P3-83P2H€1P2"92P1)(52P3+€3P2)e-2N]e-2p’dep32}

11151421.) 5 53-2-1113, (2 ,k) (2.6.2.27)

= 2e'2""p§1(e3u3-62u2)1u1p2(1+e’2“)+u2p.(H‘WH

121p211+e‘2”)+ezp1(1-e'2")1+4(22uz-eluoe3u3e 2"" p§1<e””+e 2”)

N‘f,(z,k) s —a—N1,(z,k) (2.6.2.28)

32

= _46283u3lpID2D3e-(prpxke-(prp1Me-pxz

61

a
N‘f,(z ,k) a 32—N1,(z,k) (2.6.2.29)

= —41131p lpzp3e-(prp‘yefi’rp‘w “He-2”" )(l—e-zp’d )(E3H3‘51H1)€2112P2

+ (1_,-2p.:,(1+,-2p.a )(92212—61211)€3Hzp3

+ (l-e‘b" )(l-e’zp’d)(e3u3-ezu2)ezu1p 1]e"’"

62

61, 111 y

 

 

 

Figure 2.1 Hertzian potential boundary conditions at interface

63

cover

 

 

 

 

 

 

 

 

)1 S Hz
1 l
substrate

X
A

 

conducting ground plane

Figure 2.2 Hertzian potentials generated by vertical elecuic current.

 

 

 

 

A z
111+S
el. “1 ' z =d +t cover
. n .
82’ 112 2 g » . n2 g superstrate
z=d
_. +
113 --
€3,1l3 S y 5 113
M3 substrate
z=0 _. >
—I>—> conducting ground plane
64. 114 M. D.
4

Figure 2.3 Hertzian potentials generated by horizontal magnetic current.

65

 

 

 

2 2.-
-)‘p ”Ab

; , T. » 1r

. . 71'1: 72p
Sommerfeld mtegratton path \

branch point
pole
branch cut

 

Figure 2.4 Sommerfeld integration path in the complex 3. plane

CHAPTER THREE
PLANE WAVE PROPAGATION IN TRI-LAYERED MEDIA

Plane wave propagation in layered media is very different from that in free space.
In order to know the excitation field on a antenna in tri-layered media due to a plane
wave illumination, it is necessary to study the transmission and reflection of a plane
wave in the tri-layered media.

Consider a plane wave illuminating a lossy layer above a ground plane with the
wave vector making an angle 90 with the z-axis, as shown in Figure 3.1. A general
incident plane wave can be decomposed into a TB wave and a TM wave. A TE wave
is defined as a wave with the electric field normal to the plane of incidence, which is
taken to be the y-z plane without loss of generality. A TM wave is defined as a wave
with the elecuic field lying in the plane of incidence. The cases of TE wave and TM

waves will be treated separately.

3.1 TM Plane Wave Propagation in Tri-layered Media
The incident magnetic field is given by
Hf = 2111,4222“ (3.1.1)
where
kf = (49.11190466130ka (3.1.2),
The magnetic fields in each of the regions 1-3 can be written in terms of plane
wave terms similar to (3.1) representing waves traveling in either the +z or -z direc-
tion. The t0tal field in region 1 is composed of the incident wave Hf plus a reflected

wave Hf, while the field in region 2 is made up of a transmitted wave H2" and a

reflected wave H5”, and the field in region 3 is composed of a transmitted wave H;

66

67
11,7: mge‘j“"' t = 1,2,3; 7 = +,— (3.1.3)

where
k2 = It,2 + k,2
(ki)2 = (k1? = 03211151: "12
(m2 = (It; >2 = (0221282 = k% (3.1.4)
(5)2 = “(3)2 = (0222353 = 1‘32
It is understood that Hf = H,- and kli = k1".

Hf is assumed to be a known quantity, while Hf”,H§,H2’,H3+, and H3" are to be
determined by applying appropriate boundary conditions on E and H at each of the
interfaces. The electric field in each region can be obtained via using the Maxwell’s
equation:

1

106

E = VXH (3.1.5)

 

The electromagnetic fields in the three regions can be expressed as:

Region 1
HI = Hf+Hf = £1H§,e‘j"1‘"+H1;e'i"f"] (3.1.6)
E1 = —D16—11(2k1,—y‘k1.>H‘i.e"““"+<2kg-yk1.>Ht:e""”1 (3.1.7)
Region 2
H2 = H§+H5 = 2111;,e'1'";"+H,;e‘j“5"] (3.1.8)
E2 = twice-flea>Hae""i"+<z“kr,-ik2>Hae""5"1 (3.1.9)
Region 3

H3 = 11374.11; = 21yge‘i23‘"+H;,e“i"i "1 (3.1.10)

68
1 , _ , _ _ _. -. , .. _- +.
E3 = $3[(z/t_.,,-ylt3,)H3,e “‘3 '+(zk;y-yk;,)H;,e 1": '1 (3.1.11)

Applying the boundary conditions on the tangential E and H at each of the three
interfaces requires immediately

ky =14, = kg = kg, = k; = k5, = k3; = —k1sin60 (3.1.12)

for continuity of the phase terms. With this relationship established, the boundary con-
ditions can be written as:

B.C. 1: Hum continuous at 2:21. (H1, = H21).
Hi ’jk'irzt+H + -jk1+,21 _ H + "112311 - -jk§,21
1,6 1,8 — 2:6 +H2,e (3H113)

B.C. 2: E“in continuous at z=zl. (E U = E2, ).

k‘iz i —jki 2 k1; + -jk*z kZ-z —'k‘z kit -'k*z
e Hue " ‘+-e—H1,e " ‘= ——H§_’,e ’ " ‘+-?—H{,e ’ 7‘ ‘ (3.1.14)
1 l 2

 

B.C. 3: “tan continuous at 2:22. (H 2, = H3,).

_ -‘k‘ —'k+ _ -'k' -' +
Hue J 2’z"+H3_‘,e J 2.22:er 1 3’224-H§,e ”‘3‘“ (3.1.15)

B.C. 4: Em continuous at 2:22. (E 2, = E13,).

kZ-z _ -jk‘z k3; -jk’z 1‘31 —'k'z kg; —'k”z

E-Hzxe a 2-i-—Hi,,e 2' 2 = Kfoe I 3' 2+?H§,e J 3' 2 (3.1.16)
B.C. 5: Em = O at 2:23. (E3, = O)

k;,H3-,e""2”+k;,H;e‘j"i“ = 0 (3.1.17)

Note that through (3.1.4) and (3.1.12) there exists a relationship between
Icz and Icy in each region. However, care must be taken to choose the sign on the

. square root terms to make each wave decay as it propagates. In general

a: k4 3mm

69
Thus, assuming region 1 to be lossless, the sign on the square root must be chosen
such that
Re{kf,} > O
Re{k2’,} < O Im{k2‘,) > O
Re{k3;} > O Im{k§,} < 0 (3.1.19)
Re{k3‘,] < 0 Im{k3‘,} > O

Rc(k;,] > o Im{k3+,} < 0

giving

 

k1“, = —k1, = xjkf-klisinieo
k{, = —k{, = \lkzz—klzsinzeo (3.1.20)
k5; = -k;, = \Ik32-k123in260

 

 

Remember that in a lossy region, the wave number k is complex, due to the complex
permittivity and permeability,

I . II . G
e = e -_]E = sole-1E] (3.1.21)

11 = u’-j 11” (3.1.22)

To formulate the integral equation for the monopole current or the slot current,
the incident fields in region 3 need to be determined. Thus, equations (3.1.13)-(3.1.l7)
must be solved for 113+, and H 3', in terms of the known quantity I-I‘i, . Using (3.1.20),
these can be solved as follows. From equation (3.1.17)

H37; = ng‘jm“ (3.1.23)

Substituting (3.1.23) into (3.1.16) gives

Haejkizz-Hixe.jk£zz+AH§ [e‘jkitzzfi‘juizzsejkitzz] = 0 (3 1 24)
x . .

70

where

52" 32

 

53k 23

Next, substituting (3.1.23) into (3.1.15) gives

._ - -'k’ _ -'k‘ _. _ 'k'
ngjkfi22+H2xe J 222_H3x[e J 3122+e 121(327-36] 3.22] = O .

For simplicity, (3.1.24) and (3.1.26) can be rewritten as
ngjkizz-ng‘jkizquP = 0
ngjki’zz-t-ng—jki'zz-HfQ = 0

where
P .=_ AIefist-tzsejka—tzzqe-jkilz]

Q E [e_j2ki:z3ejk3-122+e-jk3—IZZ] .

Now, adding (3.1.27) and (3.1.28) gives

Zflfiejki'“ = (P+Q)H§,

 

or
H1. = mg.
where
Z a 23:22 .

Also, subtracting (3.1.28) from (3.1.27) gives
Hi. = "13‘.

where

(3.1.25)

(3.1.26)

(3. 1.27)

(3. 1.28)

(3. 1.29)

(3.1.30)

(3.1.31)

(3.1.32)

(3.1.33)

(3.1.34)

(3.1.35)

71

Next, rewrite (3.1.13) as
—e2jki’z‘H1§+e+H?fx+e’Hix = H5,
and (3.1.14) as
erk'iJ‘H fx—Be+H{,+Be'H2} = Hg,
where

e: ej(k‘i.ik2})21

 

Now, substituting (3.1.32) and (3.1.34) into (3.1.36) gives
—H {Ezezm‘z’+e+ZH3’,‘-l~e'YH3'Jr = H'ix

and into (3.1.37) gives
Hfiezjki’z‘-Be+ZH;+Be'YH§I = Hg,

Adding (3.1.40) and (3.1.41) gives

H3}[Ze+(l-B)+Ye'(l+B)]= ‘1,

Equations (3.1.23) and (3.1.42) give the transmission coefficients.

 

 

T“ = .113;— = 2

Hi, Ze+(1—B )+Ye‘(1+B)
T+ 5 H3; = T-e'jsz-czs

H ‘1;

(3.1.36)

(3.1.37)

(3.1.38)

(3.1.39)

(3.1.40)

(3.1.41)

(3. 1.42)

(3.1.43)

(3.1.44)

Knowing transmission coefficients 7'" and T‘ , it is possible to calculate the elec-

tric field in region 3. From (3.1.11), the z component of the electric field can be

expressed as

72

Ii

[nge'j "5 "+11 ge‘j "3‘ "1 . (3.1.45)
(”53

E32 =
Substituting (3.1.43) and (3.1.44) into (3.1.45) gives
k3"y

E3, = EH‘H‘ixe—j k3"'+T+1L1§,e"' "5 "'1 . (3.1.46)
3

The field along the z axis becomes

k' . . _ . _ . _
1:34:22) = i7“ ‘1,[e"“*’+e‘12"3"3e+’"3*’1 . (3.1.47)
0’53
. . k3 113
Finally, from (3.1.2), (3.1.12) and usmg (063 = 71-— and 113 = E—
3 3

E =—k—‘n Hi T’sin902e-jki’z’cos[k‘(z—z )]
32 [C3 3 1x 32 3

 

= Wcos [\lk32—k12sin200 (2 -z3)] (3.1.48)

where

k . . _
W = —2 k—ln3H‘1xT’sin90e’Jk3’“ .
3

Also from (3.1.11) and (3.1.20)

k . -, _. .,
E3 = — 3’ [ng‘mnge “‘3 '1 (3.1.49)
0°63

 

y

Substituting (3.1.43.44) into (3.1.49) gives

k3 . _. -, _. .,
53y = -m: H5,[T-e “‘3 '-T+e “‘3 '1 . (3.1.50)

 

Then, substituting (3.1.20) and (3.1.43-44) into (3.1.10) gives the horizontal magnetic

field in region 3.

73

_ _. - —'k‘ _. _ . _
H3! =H‘lxe 1“ny (e I 3.2+e 12k3¢z3ejk312) .

On the ground plane 2:23, the magnetic field can be written as

H3,(z=z3) = 2115,.2’1“!y T-e‘jkizs

(3.1.51)

(3.1.52)

The important results are summarized and renumbered for convenient use later.

 

E3z(r)=Wcos[\/k32—k,2sin290(z-z3)] ; x=o,y=o

k _ . . _
W = —2 -l fi11(‘1,,7“sin60(3'11‘3’23
[‘3 83

H3x(r)=2H‘ixe-jk’yT‘e_jk5'z’ ; 2:23

 

 

 

T_= H5; _ 2

H5, Ze+(1—B)+Ye-(1+3)
Z Q+P

zejkizz
Y 9””

ze'jkhzz

P =A[e'12kiglsejk3}22_e-jkizz]

Q = [e‘luizzaelkiz22+e“lk§zlz]

 

 

ct _=_ ejtk'iatkzozi

A ___ €2k3-z
Eskfz
EIkZ-z

B = i
e2klz

 

(3.1.53)

(3.1.54)

(3.1.55)

(3.1.56)

(3.1.57)

(3.1.58)

(3.1.59)
(3.1.60)

(3.1.61)

(3.1.62)

(3.1.63)

(3.1.64)

(3.1.65)

74

 

 

k2; = -\/k22-k125in290 (3.1.66)
k3; = —\/k32-k 125111290 (3.1.67)
3.2 TE Plane Wave Propagation in Tri-layered Media
The incident elecuic field is given by
E{ = iEfixe‘jki" ' (3.2.1)
where
(3.2.2)

kli = (‘5; Sineo-f C0890)k 1

The electric fields in each of the regions 1-3 can be written in terms of plane
wave terms similar to (3.2.1) representing waves traveling in either +z or -z direction.
The total field in region 1 is composed of the incident wave Ef plus a reflected wave
Ef, while the field in region 2 is made up of a transmitted wave E5 and a reflected

wave EL and the field in region 3 is composed of a transmitted wave E3‘ and a

reflected wave E; . All of these terms can be written in generic form as

E; = 213,112" "7" i = 1,2,3; 7 = +,- (3.2.3)
where
k2 = kyz + k}
(k‘i )2 =(k1‘)2 = m2u181= k3
(3.2.4)

(k1?)2 = (k; )2 = 01211282 = k22
(k5)2 = (m2 = 03211383 = k? .

It is understood that Ef = E,” and kf = k1“.

75

E,‘ is a known quantity, while Ef,E;,E2-,E;, and E; are to be determined by
applying appropriate boundary conditions on E and H at each of the interfaces. The

magnetic field in each region can be obtained by using the Maxwell’s equation:

H: ,1 VxE. (3°25)

’10)“

 

The electromagnetic fields in the three regions can be expressed as:

 

 

 

Region 1
E1: Ef+E= x[E 3e ejki"+5 1316’"? "'1 (3.2.6)
H1: 1 [(z‘k'iy-y‘k‘iz)E‘ixe_j "i"+(2kfy-y*kfz)E ge‘j "3 "1 (3.2.7)
Region 2
E2: 13,412.; = x“[E+e ”‘3 age—"3"? (3.2.8)
H2 = _aipz[awry/cg,)Ege'ik3‘"+(z3k2-y—y3k5,)Ege‘jm] , (3.2.9)
Region 3
E3 = E§+E§ = £[E3ge'j "3' "age" "3 "1 (3.2.10)
H3: -cou [(zk;y —yk.;,)E3-,e “‘3 r+(zk3+y—yk3+z)E+e (“‘3' ] (3.2.11)

Applying the boundary conditions on the tangential E and H at each of the three

interfaces requires immediately
k5, = {3 = kg, = k5, = k5, = k5; = —klsin90 (3.2.12)

for continuity of the phase terms. With this relationship established, the boundary con-

ditions can be written as

B.C. 1:

B.C. 2:

B.C. 3:

B.C. 4:

B.C. 5:

76

Em connnuous at 2:21. (Ell = EZI)'
. _‘ki _‘k+’ _3k4» — _3k-
E1 6 111145138 1121=E2+xe J 2121+E2xe 1 2:21

1::

Hum continuous at 2:21. (H U = H 2y ).

kiz ,3 —jk{ 21 k1: + -jk{zl k2; _ —jk5,zl k2: + -jk5j,zl
——Elxe ’ +—E1,e ’ = —E2,e +—E2,e

111 111 112 112
E continuous at 2:2 . (E = E ).

tan 2 21 3x

Eite-jki'zz-i-Eixe—jkbzz = nge-Jkizz+E;xe—Jksszz
Hm continuous at z=z2. (H 2, = H 3y ).

k; _.- kg _.. k; _._ k; ..
__Eixe [hula—Egg #2122: __LES-xe Jk3n22+ 2 E2326 Jksalz

112 112 113 113

Elan = O at 2:23. (E3x = 0)

_ —'k’, —'k+
53x6 I 323+Eél-xe 1 ”23:0

(3.2.13)

(3.2.14)

(3.2.15)

(3.2.16)

(3.2.17)

Note that through (3.2.4) and (3.2.12) there exists a relationship between

kz and k’ in each region. However, care must be taken to choose the sign on the

square root terms to make each wave decay as it propagates. In general

6:31:57}

(3.2.18)

Thus, assuming region 1 to be lossless, the sign on the square root must be chosen

such that

Re{k1+,) > 0
Re{k§z} < O Im{k2‘z} > 0
Re{k§z} > 0 Im{k§z} < 0

ReIkgz} < O Im{k3‘z] > O

- (3.2.19)

 

 

77

Re{k3+,} > 0 1m(k;,}< 0

giving

 

1e13, = -k5, = «IkE-klzsinzeo
kg, = —k5, = xjkf—kEsinzeo (3.2.20)
k3“, = -k3-, = \lkf-klzsinzeo

 

 

Remember that in a lossy region, the wave number k is complex, due to the complex

permittivity and permeability:
e = e’— 'e" = e [e - '—°—] (3 2 21)
J 0 r .l (080 - -

u = u’-j u” (3.2.22)

To formulate the integral equation for the monopole current or slot current, the
incident fields in region 3 need to be determined. Thus, equations (3.2.13)-(3.2.17)
must be solved for E 3*; and E 3’, in terms of the known quantity E ‘ix . Using (3.2.20),

these can be solved as follows. From equation (3.2.17)
53; = —E;,e‘12"3’3’3 (3.2.23)

Substituting (3.2.23) into (3.2.16) gives

 

Egejkih-nge‘jkihm ’53;[e’j’F33’3+e‘1'2"3"23e”‘5"3] = 0 (3.2.24)
where
k-
A’s “’2 3f . (3.2.25)
113/‘22

Next, substituting (3.2.23) into (3.2.15) gives

.- _ _.- _ _.k_ _. _ ._
Efxe’k"zz+E2,e I‘m-E342» ’ 33’3—e 12*3’3elk33’3] = 0. (3.2.26)

78

For brevity and convenience, (3.1.24) and (3.1.26) can be rewritten as
Egejkih—Ege‘jkih-ngp’ = 0
Eiejkizz+Efxe-jk’;zz—E§,Q ’ = 0

where
P ’ s A ’[e_j2k§'z’ejki‘zzw-jki‘zz]

Q’ E [_e-12k523ejki.22+e-jki.zz] .
Now, adding (3.2.27) and (3.2.28) gives

zEgejki'“ = (P’+Q )53;

 

or
E5; = Z’E3‘Jr
where
2.5 Q’_+I_D’ .
Zeflcnzz

Also, subtracting (3.2.28) from (3.2.27) gives

52; = 1"ng
where
Y, E _Q_;P:._ .
ze’lkzszz

Next, rewrite (3.2.13) and (3.2.14) as
21ki121+ ++ -—_ 1'
_e Elx'i‘e Eh+e sz —E1x
erkL21Ei4~x_Ble+Eifx+BIe-E-2-x = E111

where

(3.2.27)

(3.2.28)

(3.1.29)

(3.1.30)

(3.2.31)

(3.2.32)

(3.2.33)

(3.2.34)

(3.2.35)

(3.2.36)

(3.2.37)

79

 

e: 5 8101.443». (3.2.38)
k-

B’a “1 f’ . (3.2.39)
llzklz

Now, substituting (3.2.32) and (3.2.34) into (3.2.36) gives

-EgeZ’ki3’3+e+Z'E3-,+e-Y'E3; = 51, (3.2.40)
and into (3.2.37) gives

Efiezfl‘iflua'e+Z'E3-,+B'e-Y'E_.; = E‘i, . (3.2.41)
Adding (3.2.40) and (3.2.41) gives

E3;[Z'e+(1—B wry-(1+3 )1 = 251. . (3.1.42)

The transmission coefficients can be obtained from (3.2.23) and (3.2.42).

 

 

E-
rga 3" = , + ,2 , _ (3.2.43)
‘1, Z e (l-B )+Y e (1+B’)
E+ . _
T; a l- : —T;e"2"3"3 (3.2.44)
Elx

Knowing transmission coefficients 7‘: and T; , it is possible to calculate the EM
fields in region 3. Substituting (3.2.20) and (3243-44) into (3.2.10) gives the electric

field in region 3.
53,“) = 3,61" T¢‘(e-jk5‘z—e-j ”3323.3”33’) . (3.2.45)

Substituting (3.2.43) and (3.2.44) into (3.2.11) gives

k" . . -. . _. .,
H3, (r) = —Eu3’_(r;£‘1,e""3 '+T,+E'1,e “‘3 ’1 (3.2.46)
3
H3 (1') = 33-5— 3, [T-e‘jk3'"-T+e""‘3’"] (3247)
Y x e e - . .
C0113

80

Of particular interest is the tangential magnetic field on the ground plane, which

can be written as

k . . . _
173,043) = 2—3—z—E‘lxrge'1’9’e‘1k33’3 . (3.2.48)
(0113

81

Az

 

 

 

 

 

51, 111 cover
Z=Zl
82, 142 ' 5111263118!“ '
* ~ 2:22 “*
83, 113
substrate
F
2:23
.\ \"
ground plane
X

Figure 3.1 Plane wave propagation in tri-layered media

CHAPTER FOUR
FORMULATION OF INTEGRAL EQUATIONS

The dyadic Green’s functions for the EM fields have been derived in chapter 2.
Integral equations are obtained in this chapter by enforcing appropriate boundary con-

ditions. The case of a monopole and that of a slot will be considered separately.

4.1 Integral Equations for a Monopole

Consider the imaged monopole beneath a lossy sheet as shown in Figure 1.1.
When illuminated by a plane wave, a current will be induced on the monopole surface
causing a voltage drop across the load resistance, and thus deliver power to the load.
The current induced on the monopole surface will be solved by using superposition;
the scattering mode current and transmitting mode current are found independently and

then they are combined to get receiving mode current.

Throughout this dissertation the monopole is assumed to be a thin wire. That is,
the radius is much smaller than a wavelength. Then, the monopole surface current dis-

tribution, 1,3, can be assumed angularly invariant.

Electric field integral equations (EFIE) for the monopole current distribution when
the antenna is acting as a scatterer and as a transmitter can be formulated by applying .
the boundary condition that the total elecuic field tangential to the surface must be

zero:

E, =E:+E;' =0 at p=a,-dSzS-d+h (4.1.1)

E;=-E;’ at p=a,-dszs—d+h.

82

83

Here 15: represents the scattered field maintained by the induced current, and E; the
impressed field due to either the incident wave in the scattering case, or the load vol-
tage in the transmitting case.

In the scattering case, the impressed field is the incident electric field in the sub-
strate. A TM incident plane wave is considered explicitly. A TE incident plane wave
can be solved in a similar way. Comparing the coordinate system in Figure 3.1 with

that in Figure 1.1 gives
21=t; 22:0; z3=—d.
Substituting the above into (3.1.48) leads to
E; = Wcos [k3'z(z+d)] -dszs—d+h (4.1.2)

where

 

k5, = \Ikgz—klzsinzeo.

This expression is derived in detail in section 1 of Chapter 3. Note that the quan-
tity W depends on the incidence angle and incident field strength, as well as the thick-
ness and the parameters (electric or magnetic) of the lossy layer. Also note that in
(4.1.1), the impressed field on the surface of the thin wire is approximated to be the
same as the field on the wire axis. This is a good approximation when the wire radius

is much smaller than a wavelength.
In the transmitting case, the impressed field will be modeled using a delta func-

tion (slice-gap) generator

5,3 = V05(z+d) (4.1.3)

where V0 represents a voltage applied to the terminal region at z=-d.

The scattered field produced by the induced current on the monopole can be

represented in terms of a scattered Hertzian potential 1'1; . The axial component of the

84

Hertzian potential produced by an axial current is found using (2.6.1.7) as

82113,

322

E; = [(321132 + (4.1.4)

 

Substituting (4.1.4) into (4.1.1) yields an inhomogeneous ordinary differential equation

(ODE)

2 .
(38—? + k32fl'132 (z) = —Ez‘(z) —d_<.z S—d+h . (4.1.5)
2

The solution to the ODE takes on a slightly different form in the scattering and

transmitting cases, so each case will be considered separately.
A) Transmitting case

Using (4.1.3) in (4.1.5), the ODE becomes

2
(032—2 + [(3)1132 (z) = —V05(z+d) (4.1.6)

which has the general solution [ 11 ],

v
H342) = Clsink3(z+d) + Czcosk3(z+d) - Elem/calzml (4.1.7)
3

where C 1 and C 2 are arbitrary constants. Now, because of the ground plane, currents
on the monopole must image in the same direction. Therefore the current on the mono-
pole is an even function about z=-d. Thus, the vertical electric field must be even, and
because of the relationship (4.1.4) the potential 1'13z (2) must be even. Thus, the first
term in (4.1.7) is not implicated and the expression reduces to

v
113,02) = Czcosk3(z+d) - Ek-(Lsink3(z+d) —dSzS—d+h (4.1.8)
3

B) Scattering case

85

Substituting (4.1.2) into (4.1.5) leads to

2
(015 + [(32 )1132 (2) = —Wcos [k§Z(z +d )] (4.1.9)
2

which has the solution [ 11 ]

H32 (7.) 3C 151M3(Z +61) + C2COSk3<Z +d) "'
1 2
-k— j Wcos (1:330. +d)]sink3(z-u )du (4.1.10)
3 —d

where C1 andC2 are constants. The integral term in (4.1.10) can be evaluated as fol-

lows. Let

U(z) = j cos [k3’z(u+d)]sink3(z—u )du (4.1.11)
—d

and use the change of variables

v = u+d . (4.1.12)

Then

z+d
U(z) = j cos [k3'zv]sink3(z+d—v )dv (4.1.13)
0

Expanding the sine function gives
z+d
U(z) = sink3(z+d) j cos [ka ]cosk3vdv
0

2+1! \ .
+cosk3(z+d) 1’ cos [k5,v1sink3vdv . (4.1.14)
0

The integrals in (4.1.14) can be evaluated in a straight-forward manner. After a little

algebraic manipulation the result becomes

86

U (z) = k3(cos [k 3', (z +d )]-cosk3(z +d )) (4.1.15)

k324i:

 

Substituting (4.1.15) into (4.1.10), and again eliminating the first term due to sym-

metry, gives the solution to the ODE as

k k" d — k d
H3z(z)=Czcosk3(z+d)- [(1 3(a)“ ”(2+ )1 M 3m D (4.1.16)

2 -2
3 k3 "k3z

 

Upon substitution of (2.4.3.47) into (4.1.8) and (4.1.16), the integral equations for
the monopole current 123(2) for the transmitting and scattering cases, respectively, are
obtained.

a) transmitting case

-d +11

 

3,3 3 I I V0 .
I Gzz (z,a,z ')l, (2 )dz = Ccosk3(z+d) — —s1nk3(z+d)
—d ”‘3
—dSzS—d+h (4.1.17)
b) scattering case
—d+h
I 0233(2 ,a ,z’)lz3(z’)dz' = Ccosk3(z +d)
—d
k (cos [k‘ (z+d )]-cosk (z+d )
— ¥ 3 3’ 3 ) —dSzS—d+h (4.1.18)
3 lei-’63":
Here, from (2.4.3.48)
I 1 ”7' I
0,3-3(z,a,z )= fijrgfiz; 33130.11 )M). (4.1.19)
0

where 173-30 ,z',7.) is given in (2.4.3.46).

87

Equations (4.1.17) and (4.1.18) are Hallen-Type Integral Equations (HT IE). The
EFIE is applicable to an arbitrary source and a wire of arbitrary shape while the HTIE
is applicable to the special case of a one dimensional current and a straight thin wire.
The advantage of the HTIE over EFIE is that its kernel is less singular than that of

EFIE. This makes it numerically easier to solve.

4.2 Magnetic Field Integml Equation for a Slot

Consider a slot antenna in tri-layered media shown in figure 2.3. The receiving
characteristics are determined by the receiving mode equivalent magnetic current on
the slot. This receiving mode induced current can be solved by superposition. The
scattering and transmitting modes are found independently and the results are com-

bined to yield the receiving mode result.

Throughout the dissertation the slot is assumed to be a narrow one. That is,
I >w and WW. A good approximation in the case of a narrow slot is Ey >5}. In
other words, the longitudinal aperture field component B, can be ignored. To incor-
porate the well-known edge behavior of electric field, the aperture field 15'y can be

written as

15y (x,y,z=0) = JE— . (4.2.1)

\I 14%?

The equivalent magnetic currents on the slot in regions 3 and 4 can be written as

 

 

M305) = —2‘er, = J's—L91— = M(x .33) (4.2.2)
\/ 1-(-y—)2
W
M4(X.y) = -(-2‘ )><i‘l‘$y = -M(x.y) (4.2.3)

where

88

M(x ,y) = if (x )W (y) (4.2.3)

1
W33:

and f(x) is the unknown function to be determined.

(4.2.4)

 

WO)=

With the help of dyadic Green’s functions derived in chapter 2, the magnetic field

in region 1 generated by a source in region j can be written as

H,(Mj)= jy‘3f(rlr').Mj(r')dv' ; i,j=1,2,3,4 (4.2.5)
V1

The boundary condition on tangential magnetic field is used to obtain the mag-

netic field integral equation (MFIE)
2x( H§°’(M3)—H,{°‘(M4) ) = K (4.2.6)

tot

where H3 (M3) and Hj°‘(M4) are the total magnetic fields in regions 3 and 4 respec-
tively and K is surface electric current density in the aperture. In region 3, the total
magnetic field is composed of a scattered field and an incident field. In region 4, the
total magnetic field is just the scattered field. Using the results in chapter 2, the scat-

tered field can be expressed in terms of Green’s function and magnetic current.

Scattering case:

H3. (M) + H... (M) = -H;'" (427)
Transmitting case:

He. (M) + H... (M) = K; (4.2.8)

The generic form of the MFIE can thus be written as
L(M) = F (4.2.9)

where L is the proper linear operator and F is the excitation term.

89

In this dissertation L is defined as

L(M) = j j [33'3(r|r')+gx§‘4(r|r')lMx(DMZP'
slot

 

=1 11 1 I I [23'3(zlr’.k)+§$°4(z Ir'.k)1e3'"'342k )M.(p’)dzp’ (4.2.10)
slot 0°

(21):)2
where
p = 2x+y3y (4.2.11)
dzp = dxdy . (4.2.12)

4.3 Hallen-Type Integral Equation for a Slot

The MFIE (4.2.9) has a highly singular kernel, which manifests as a slowly con-
verging integral in the spectral domain. A magnetic dyadic Green’s function in the
magnetic source region is the dual of an electric dyadic Green’s function in electric
source region. It is well known that when the source point and the observation point
coincide, special care must be taken to treat the singularity of the the dyadic Green’s
function. There have been extensive discussions on this subject [ 12 ]. One way to cir-
cumvent the singularity problem is to convert the MFIE into a Hallen-Type Integral

Equation (HTIE). This conversion is possible if the source is one dimensional.

Substituting (2.6.2.2), (2.6.2.3-4) into (4.2.9) gives

32
8x 82

 

2
(ks-2 +%)JIG§'3(I' l r')M(r')d 2r '+ [[02330. Ir’)M(r’)d 2’, I
X s S

2
+(k42+5%;)I£Gfi34(r|r’)M(r')d2 ' = F(r) (4.3.1)

This is essentially a different form of MFIE. A discussion on the singularities of the
Green’s funcrion can be found in [ 6 ]. The terms on the left hand side of (4.3.1) con-

verge as improper integrals if M(r) satisfies the Holder’s condition at r, i.e., there exist

90

positive constants c , A , and 01 such that
lM(r)—M(r’)|SA Ir—r’l"L for lr’—r|<c . (4.3.2)

03.33.033.033“ are given in (2.5.2.7), (2.5.2.9), and (2.5.2.13).

 

 

Define
_ 1 o. ejk'(r‘l’) N32
G 333 a dzk , ) . (4.3.3)
”‘ (21:)2 U. 10mins DxDz
It can be seen by comparing (2.5.2.9) with (4.3.3) that
G 333 = it?” . (4.3.4)
21 ax 2:

After adding a few terms and exchanging the order of integration and

differentiation, (4.3.1) can be rewritten as

2
(k.2+;—2)HGL (rlr'w, (r')d2 ' = FUN-”@201;(rlr’)Mx(r’)d2r ' (4.3.5)
It s S
where
k} a 1:32.43 (4.3.6)
363-3
CL = G;3+Gx§4+a—: (4.3.7)
2 2 ’33
k 60
3 44 4 33 2X
- — —G - + . .
GR ksz GI: ksz xx 8 (4 3 8)

Equation (4.3.5) is a differential-integral equation. Solving the ordinary
differential equation first produces a Hallen-Type Integral Equation (HTIE). The steps

are outlined in [ 48 ] and the result is

HGL (rlr’)Mx (r')d2 ’ = Clsinksx+C2cosksx+
S

.kl— I [F (rHHksZGR (I'lr’)Mx (r’)d2r’]|x__.x'sinks Ix —x’|dx' . (4.3.9)
8 —oo S

91

Generally speaking C 1 and C 2 are two unknown functions of y. How to handle
them will be discussed later. Define the operators

LL (MI) a ”0L (r Ir')M, (r')d2r ' (4.3.10)
S
LR (114,) a ijR(rIr’)M,(r')d2r’ (4.3.11)
S
The HTIE (4.3.9) can then be rewritten as

LL(M )-—jLR(M )I .1 Sink lx—x’ldx’

kS—oo

= Clsink x+Czcosk x+— j 13ml”, sink lx—x ldx’. (4.3.12)

kS—oo

Substituting (2.5.2.7), (2.5.2.13), and (4.3.3) into (4.3.7-8) gives

 

 

 

 

 

G dzkejk'L("""P k 4.3.13
1.: (21:21! () < )
G dzkejk'k("")‘1’ (k) 4.3.14
111210211 ( )
where
N, Nd '
311(11): , 3 e, 3’ 4. 1 (4.3.15)
1001331330. 1031131230102 mum
k4 N31: N32 k3 1
t}! k = __2 . +. t’—2 4.3.16
RU (ks) 1031131230. ImuspanD. ‘k. 10414174 ( )
a
N3z(k) = _a_N3z (Z 1k)|z=0 (43-17)

The right hand side of (4.3.12) can be written in a different form, which is more

convenient to use [ 48 ]

I
1 . , ,
LL (M, )-I—£LR (M,r ) Imam/c, (x —x )dx
3

92
1 x

= C ,sink,x+czeosk,x+k—jp (r)I,=,.sink, (x -x ’)dx ' . (4.3.18)
3 0

The kernel of HTIE (4.3.18) is less singular than that of MFIE (4.3.1), which

means that it is easier to solve numerically. The price paid for the numerical stability

is that the kernel of HTIE is more complicated than that of MFIE.

CHAPTER FIVE
SOLUTIONS or INTEGRAL EQUATIONS

5.1 Method of Moments

Method of Moments is a general procedure to solve linear inhomogeneous func-
tional equations [ 7 ]. The basic idea is to convert a functional equation to a matrix
equation, and then to solve the matrix equation by known techniques. Consider the

inhomogeneous equation

L(f) =g (5.1.1)

where L is a linear operator, g is the source (known function), and f is the response
(unknown function to be determined). Let f be expanded in a series of basis functions
in the domain of L.

f = 2011f» (5.1-2)
n
where a,, are expansion coefficients to be determined. Substituting (5.1.2) in (5.1.1)
and using the linearity of L gives
201.120..) = g . (5.1.3)
It

Assume that a suitable inner product <f,g> has been determined for the problem.
Define a set of testing functions w,” in the range of L. The functional equation (5.1.1)
can be reduwd to a matrix equation (5.1.4) by taking the inner product of (5.1.3) with
Wm

[Imllan] = [gm] (5.1.4)

where

93

94

1,,," = <wm,L (fn )> (5.1.5)

8»: = (Wang) - (5.1.6)

The matrix equation (5.1.4) can be solved by known techniques to determine an. The

particular choice Wu = f n is known as Galerkin’s method.

5.2 Impedance Matrix for a Monopole

The integral equations (4.1.17) and (4.1.18) for the transmitting and scattering
mode current distributions can be solved using the method of moments (MoM) with

pulse function expansion and point matching.

Expand the current as

N
123(2) = 2a,P,,(z ) -d$zS—d+h (5.2.1)
n=l

where

1 -d+(n—1)Aszs-d+nA
P" (z) = (5.2.2)
elsewhere

is a rectangular pulse basis function, a,, is the set of unknown complex expansion

coefficients, and

A = (5.2.3)

.11..

N .

Substituting (5.2.1) into (4.1.17) and (4.1.18) gives
—d+nA

N
2a,, j 0,233(z,a,z’)dz’ =Ccosk3(z+d)+ u(z) -dst-d+h (5.2.4)
n=1 -d+(n-1)A

where

95

 

 

V
--lsink3(z +d)
2k3 transmitting case
u (z) :1 (5.2.5)
w k3 (cos [k3-z(z+d)]—C03k 3(2 +d)) scattering case
L k3 k32 *3:

A system of N equations for the N+1 unknowns an and C can be obtained by

matching (4.1.4) at the N discrete points

2,, = —d+(m--;—)A m=l,2,....N (5.2.6)

representing the centers of the pulse functions P". This gives

-d+nA

N
2a,, 1 6,2'3(zm,a,z’)dz’=Ccosk3(zm+d)+u(z,,,) m=1,2,..,N (5.2.7)
n=l —d+(n—1)A

An additional equation can be obtained by applying continuity of current at the
tip of the monopole. Assuming that the monopole is a thin wire, the current should go

to zero at the tip. Using (5.2.1), this implies

0N = 0 . (5.18)

With condition (5.2.8), (5.2.7) represent a system of N equations in the N unknowns

 

 

 

 

 

 

al, . . . ,aN_1, C . In terms of a matrix equation, (5.2.7) can be written as
. 1 . . . .
A11 A12 ALN-l ‘003k351 01 “(21)
A21 A22 . . . AZN-l ’C03k352 02 “(22)
' " ° ° = ° (5.2.9)
. . - - ° . . aN_1 .
(Am ANZ " ' ANN-1‘605k35Nd _ C . _u(zn)‘
where

8,, = (m--;-)A (5.2.10)

96

 

with
—d+nA
A"... = I 02333(z,,,.a.z ’)dz' n=1,2,...,N—1, m=1,2,..,N (5.2.11)
-d+(n-1)A
and
--isink38m
2k3 transmitting case
u (2“) = 1 3 (5.2.12)
W [(3 (COS [k3-z(am+d)]-C08k3(5m+d)) scattering case
[(3 k32-k3-22

 

5.3 Calculation of Impedance Matrix Elements

Because of the simple dependence of the Green’s function on 2’ , the integral in
the matrix entries (5.2.11) can be calculated in closed form. Substituting (2.4.3.48) into
(5.2.11) allows the matrix entries to be written as

1

 

_. 2
Am _ 21:10:83 (EIMWO (12).)de (5.3.1)
where
—d+nA
1,,,(2) = jtoe3 j rg'3(z,,,,z',7t)dz'. (5.3.2)
-d+(n-1)A

The integrals 1,,," will be calculated based on the values of m and n.
A) m>n
In this case, z>z’ holds. Substituting (2.4.3.46) into (5.3.2) and using z>=zm and
2%2’ from (2.4.3.23), the integrals become
1 —d+nA

IMO.)=—F(zm) j coshp3(z’+d)dz'. (5.3.3)
P3 -d+(n-1)A

97

Evaluating the integral yields

 

 

F m
1,,", O.) = (22 )[sinhp3n A—sinhp3(n—1)A] (5.3.4)
P3
Using [ 53 ]
sinhx - sinhy = 2cosh%(x +y) sinh-é—(x -y) (5.3.5)
then gives
F (2...) 1 . A
1,,," 0.) = 2 2 coshp3(n -—)As1nhp3— . (5.3.6)
123 2 7-
B) m<n

In this case z<z’ holds. Substituting (2.4.3.46) into (4.2.2), and using z> = z’ and
z< = z," from (2.4.3.23), the integrals become

-d+nA
1,,," o.) = °°W-30“” j 13(2 ')dz’ . (5.3.7)
P3 -d+(n-1)A

 

Substituting (2.4.3.25-27) in (5.3.7) gives

1 coshpnzmm 7“”

1,,, 1. = ,
( ) P3 QSlnhP3d+ZCOShP3d -d+(rJi-1)A

 

[Qcoshp3z’—Zsinhp3z’]dz’ (5.3.8)

Carrying out the integral in (5.3.8) and using (5.3.5) eventually leads to

F (2..) 1 . A
1,,", (2.) = 2 2 coshp 3(m ——)Asrnhp3— . (5.3.9)
pa 2 2

 

Comparing (5.3.6) and (5.3.9) shows

1,,", =1", . (5.3.10)

98

In this case z>z’ for the lower half of the domain of integration, and z<z’ for the
upper half. Thus it is necessary to split the integrals into two pieces. Using (2.4.3.46)
in (5.3.2) gives

-d+(n—%)A
1,,, = -l—{F(z,,) j coshp3(z’+d)dz’ +
P3 —d+(n—1)A

coshp3(2,,+d) 4}” [Q h , 25' h ’]d ,} (5311)
. COS p32 " tn p32 Z . . .
Qsznhp 3d +Zcoshp 3d -d+(n _% )A

 

Evaluating these integrals yields

Inn = L2{2F(Zn )COShp 3(n —-3—)Asinhp3_A_ +
P3 4 4

2F(z,,+-’:—)coshp3(n--;-)Asinhp3i:- ) . (5.3.12)

Substitution of (5.3.9) and (5.3.12) in (5.3.1) gives the impedance matrix entries.
In their present forrrr, however, involving hyperbolic sine and cosine functions, these
entries are prone to numerical difficulty. As the integration variable in (5.3.1) increases
toward infinity, both the sinh and cosh functions overflow. In addition, it is very
difficult to ascertain the convergence properties of the integral. Both of these problems
can be overcome if the integrand is written in terms of exponentials. This is done as

follows.
A) m>n

Equation (5.3.6) can be written in terms of exponentials as follows. From

(2.4.3.25-27) and (5.2.10)

-g—c05hp3(-d+6m )-sinhp3(-d +8,,,)
F(z,,,) = . (5.3.13)

g-sinhp 3d +coshp 3d

 

99

By the definition of hyperbolic sine and cosine functions in terms of exponentials, this

becomes

p 5 $41+e‘2P3‘d‘5-’]+[1-e'2"3“"5~)]
F(zm) = e' 3 " . (5.3.14)
%[1—e’2P3"]+[1+e‘2P3d1

 

Here the quantity g- can be written using (2.4.3.26) and (2.4.3.27) as

 

 

 

_Q_ = [P352 €1P211+e-2N]+€2P1[1‘€_2M] (5 315)
Z P283 alp2[1-e-2N]+€y)1[1+e—2N]
Also needed in (5.3.6) is the quantity
A 1 Fag
coshp38n sinhp3-2- = Zep’s'e 2 [1+e—2p35'][1-e-p’A] . (5.3.16)
Substituting (5.3.14) and (5.3.16) into (5.3.6) gives
I 1 [1 ‘22035 H1 ‘P3A] -p3(5_-5.-%-)
= — +e " —e e
""' 21232
"3"“ +[211301—234] +11 _e-zpatd-a.)]
x D (5.3.17)
where
D = %[1-e‘2”3"1+[1+e'2”3"] . (5.3.18)

Multiplying the exponentials together and using (5.2.10) gives

A 3A
- -P3[(m—n)A——] —p [(m+n)A__]
1”“ = 12 [1'8 “Allie 2 +e 3 2
2p3D

 

‘Pslu-(m-n )A-%] -p3[2d-(m+n)A--32A] Q -p3[(m-n)A-%]
—e -e H7 e

100

—pit(m+n)A-37A1 —p.[24—<m-n)A-%1 -P3[Zd-(M+n)A-§2A]
+e +e +e ]} . (5.3.19)

This expression can be written more compactly by letting

 

 

-p.<kA——A—)
e1(k) = e 2 (5.3.20)
wad—MA)
e2(k) = e 2 (5.3.21)
Then
m = 2p320 {—g-[e1(m-n )+e2(m—n )]+[e1(m—n )—e2(m—n )] +
%[e1(m+n-1)+e2(m+n-1)]+[e1(m+n—1)—e2(m+n-1)]} . (5.3.22)
Thus, letting
f(k) = —g—[e1(k)+e2(k )]+[e1(k)—e2(k)] (5.3.23)
gives
1,,, = 12 [1—e‘P3A][f(m+n-1)+f(m—n)] . (5.3.24)
2p3D

Using the form of the integrand given in (5.3.24) allows a dramatic reduction in

the amount of effort needed to fill the moment method matrix. Letting

A(k)= l I 1

, l-e_P’A k Hard). zskszzv-r 5.3.25
21:1(0630 2“le 1f( )Jo( ) ( )

 

allows the matrix entries (5.3.1) to be written as

Am = A (m +n -1) + A (m —n) m >n . (5.3.26)

Thus, only 2N-2 integral evaluations are needed for matn , as opposed to the
N(ZN-1)/2 which would be required if (5.3.6) were used. This is a reduction in

101

computational effort by a factor of N/4.

Note that each of the exponential terms involved in calculating the integrand of
(5.3.25) go to zero as the integration variable l—m, since Re{p3]—>oo from (2.3.17).
Thus, each of the integrals converge exponentially for m¢n , and little difficulty is anti-

cipated in their numerical computation.
B) m<n
In this case, equation (5.3.10) still holds.
C) m=n
Equation (5.3.12) can be written in terms of exponentials as follows.

.1.

1,,, = 2 [U+V] (5.3.27)
P3
where
i A . A
U = 21%,,)eoshp3(n-7)sinhp3T (5.3.28)
A . A
V = 2F (z,,+7)coshp35,, srnhp3-4— . (5.3.29)

Substituting the definitions of sinh and cosh, and using (5.3.14) gives

1 -2P3(5.-A) wag _ZQ—[l'i‘e-ZP3(d—5')]+[1_e-2P3(d-5.)1
U = 3[1+e 4 ][1-e 2] . (5.3.30)

%[1-e‘2”3‘]+[1+e‘74’3"]

 

Similarly, (5.3.29) becomes

-2). 314—6.-%) -2p.<d—6.—%)

Q
.5 —[l+e ]+[l-e
v =%(1+e‘23’35-][1—e_p 2] 2 (5.3.31)

%[1-e‘2"3" ]+[1+e-2p’d 1

 

102

The entries in the MoM matrix determined in this section are summaried and

renumbered below for convenience.

 

 

 

 

A) m>n
Am = A (m+n—l) + A (m —n) (5.3.32)
_ 1 °° 1 142"“A 2
A(k) _ him (i 2p} D f(k)/O (amen (5.3.33)
f (k) = -g—[e1(k)+e2(k)]+[e1(k)—ez(k)] (5.3.34)
mam-A)
e1(k) = e 2 (5.3.35)
mad—MA)
e2(k) = e 2 (5.3.36)
B) m<n
lnm = I»... _ (5.3.37)
C) m=n
_ 1 .. 2
AM .. 21tj0)63 £1M10(ak)ldk (5.3.38)
1 ‘2P3(5.-A) -p3A 'g—l1H-2p3(d'5')]+[1_e-2pa(d-5.)]
1,,, = —;tl+e 4 NH: 21 4
3 D
- 3(4-5n-A) - std—mi)
1 2,, 5 42.9- %[1+e 2,, 4 ]+[1-e 2,, 4 1
__ " 3 a _ 2
21,32 [1+e )[1 e 1 D (5.3.39)
In the above
6,, = (n-%)A (5.3.40)

D = %[1-e‘2"3"]+[1+e‘2p34] (5.3.41)

103

___ (5.3.42)
P283 £1p2[1-e-2” ]+e,p1[1+e‘2"3‘ 1

 

_Q_ P352 €1P2[1+e-2N]+57P1[1‘e—2N]
Z

5.4 Comments on the Calculation of Impedance Matrix Entries
Each of the impedance matrix entries takes the form of an infinite real line

integral (5.3.1). These integrals have all been done by numerical techniques and

several issues have arisen during their computation. These are discussed below.

5.4.1 Integration through Surface-wave Pole Singularities

In many Sommerfeld-integral type solutions, surface-wave pole singularities of
the integrand appear along the real axis, and are thus within the domain of integration.
In the cases considered in this dissertation, the presence of the loss in region 2 makes
p2 a complex number, causing all the surface-wave poles of the integrand of (5.3.1) to
shift off the real axis. Thus, surface-wave pole singularities are not encountered while

calculating (5.3.1).

5.4.2 Integration through Branch Point Pole Singularities

There are three branch points involved in the calculation of (5.3.32). They are at
[)2 = 0 (5..42.1)

p3 = 0 ; p1 = 0 (5.4.2.2)

Because of the lossy layer, p2 = 0 is not located along the integration path. In
contrast, p3 = O is located along the integration contour, and results in a singularity of
the integrand in (5.3.1). Symbolically, each of the matrix entries (5.3.1) may be written

as

_-J5.’—_k,’§
”10.) 712—1: 2 13 (a MM). (5.4.2.3)
3

 

=-2__7Cj10.)83]:G

Inn

104

Where
A n¢m
Ca A n=m (5.4.2.4)
2

 

Thus there is a first order pole of the integrand at 7. = k3 . Remember , however,
that because of the square root in the exponential, this is also a branch point, and care
must be exercised to ensure (2.3.17) is satisfied. This implies that contributions to the

integral are not symmetric about I. = k3 .

Calculation of (5.4.2.3) is done in a purely numerical fashion, by splitting the
integral into two parts at I. = k3 , and using a routine which does not evaluate the

integrand at the limits of the integration [ 54 ].

It is also instructive to show how the integral in the vicinity of A = k3 can be

done analytically. Isolate the singularity within an interval [k3-7,k3+7] and examine

 

memmgm
”71-03-73; 2
I: 1211:”“0‘1’r242 10(a2)xd).. (5.4.2.5)
3

Assuming 'y is chosen small enough such that

y A < 1 and y < k3 (5.4.2.6)

Then the exponential in (5.4.2.5) can be approximated using the first two terms of its

Taylor series expansion, giving

*3” WM

1~ ~ 3;- Gm(k3)lo(k3a) j—— (5.4.2.7)

Now, using (2.3.17) to determined the sign on the square root, the integral

(5.4.2.7) can be split into two portions

105

 

 

PkrPY k3 .-
. Ad).
1 : 351:0”, (k3)J§(k3a) j —"-4—’”—T — 1 j ——1- (5.4.2.8)
k — k —
3 (AZ-k§)2 377(k32’73)2
Carrying out the integrals, substituting the limits, and using (5.4.2.6) gives
I : 591:0,“ (k3)13(k3a )x/2k3y(1—j) . (5.4.2.9)

5.4.3 Convergence of the MoM Matrix Entry Integrals

Before undertaking the numerical integration of (5.3.1), it is quite helpful to anti-
cipate the rate of convergence of the integrals. Of interest is the behavior of the

integrand as l—wo .

For n at m the integrand has, from (5.3.17), an exponential decay factor

ware—3}) .
e —-) O as k—ioo (5.4.3.1)

and thus the integrals converge quite rapidly.

For m = n the integral has no exponential decay. Since each bracketed term con-

verges to l as k—wo , it is easy to show that, from (5.3.39)

1

1 ___—
"” 2.2-k}

as 1.966 (5.4.3.2)

Thus, the asymptotic form of the integrand in (5.3.1) is, for m=n

13(a 2.) 130: 2.)

~

2.2-k ,2 2» '

 

 

(5.4.3.3)

Numerical integration of a term with the above asymptotic behavior is quite time
consuming. The integral must be computed by summing over periods of the Bessel

function. If the Bessel function is not aided by a strong decay factor, many periods

106

must be summed; more periods are necessary with a thick antenna than with a thinner

one. The additional decay factor of % in (5.4.3.3) is sufficient for convergence, but

the convergence is slow.

To help improve convergence, a term which has the same asymptotic behavior as

the integrand, but can be integrated in closed form, can be added and subtracted as fol-

 

 

lows. Let
_ 1 °°- 2 -

.... _ 27513353 £1” (2)10 (amid). + A (5.4.3.4)
where

Cn=LmO)%;O) (543$
and

"' _ 1 °° a 2

A _ Rimes £50010 (a max. (5.4.3.6)

Here 1:" (7.) is any function which has roughly the same asymptotic behavior as 1",, (7t)
, but also allows (5.4.3.6) to be integrated in closed form. It is easily seen that as
l—no , the two terms in (5.4.3.5) subtract, and the integral in (5.4.3.4) converges at a

more rapid rate than (5.3.1).
A convenient choice for 13(1) is

1

1:" 0‘) = 224432 '

 

(5.4.3.7)

Then,

_ 1 °° 13 (a 2.)
A = , j
27‘10353 () l2+k32

 

MK (5.4.3.8)

which can be integrated in ’closed form [ 51 ] to give

107

- 1
= I 5.4.3.9
A Zfij 3 K 0(k 3a ) 0(k 3a ) ( )

 

where 10 is the modified Bessel function of the first kind and K 0 is the modified

Bessel function of the second kind.

Note that as the radius of the antenna is decreased, the contribution to the total
integral by (5.4.3.9) is increased, and thus the importance of the integral contribution

in (5.4.3.4) is reduced.

Using (5.4.3.7), the integrand in (5.4.3.4) varies as

 

 

r 1
U2 al —
° ( ) [ 2.2-k} 3.24432]

3.1301 2) 1302 7L)
2.4-k; x3

 

(5.4.3.10)

which decays much faster than the original integrand (5.4.3.3).

5.5 Admittance Matrix for a Slot

The integral equation (4.3.12) can be solved by Galerlcin’s method with pulse
basis and testing functions. With the narrow slot approximation, the magnetic current
has a known lateral distribution and an unknown longitudinal distribution. Expand the

magnetic current as

Mx(r) = §anPn(x)W()’) ; xe[—I,I] ;ye[—w,w] (5.5.1)

n=l

Here, the weighting function W(y) is assumed to be

_L._
\I Hi?
W

to account for the edge behavior of the current. A pulse basis function is chosen:

W(y) = (5.5.2)

 

108

1 ; 19,49an

Pn (x) = (5.5.3)
0 ; elsewhere

where

n—l
N

 

x,,_1=( -1)1 ; xn=(-£I——1)l. (5.5.4)

Now, define the inner product as
<f (x.y).g (x.y)> E Hf (x.y) g (x.y)dx dy (5.5.5)

and use Galerkin’s method to reduce the integral equation (4.3.18) to the set of linear

algebraic equations

2N
2<L(P,,W),P,,,W>=<FH,P,,,W> ; m=1,2,...,2N . (5.5.6)

n=1

Using (4.3.10—11) and (4.3.13-16) then gives

L (M1) = LL (Mx )—L§ (Mx) (5°57)
1.5014,) 2 [LR (M,)I,=,,(;1-)sink, (x -x ')dx’ (5.5.8)
0 s
F” (r) = C lsinksx+C2cosk3x+kiIF (r) lxarsinks (x —x ’)dx ’ . (5.5.9)
3 0

Notice that there are 2N equations and 2N+2 unknowns, { an }, C1, and C2.
The boundary condition that the magnetic current is zero at the two ends of the slot

gives two more equations. For pulse basis functions, the two equations can be written

01 = 0 (5.510)

a2” = 0 (5.5.11)

109

Substituting (5.5.10) and (5.5.11) in (5.5.6) gives the matrix equation
lymllvn] = [im] (5.5.12)

where [ym] is a 2N by 2N complex matrix and [vn] and [im] are 2N by 1 complex

VCCIOI'S

<L(P,,W),PmW> ; if n=2,...,2N-1

ymn = <-sinksx,PmW> ; if n=1 (5.5.13)
<—cosk3x,P,,,W> ; if n=2N

[Vn]=[C102 am €le . (5.5.14)

[im]=[i1 izrvlT (5.5.15)

As stated in section 3 of chapter 4, C1 and C2 are unknown functions of y.
Because of the expansion (5.5.1) and the fact that the weighting functions are known,
the explicit forms of C 1 and C 2 will not affect the solution of the matrix equation.

Therefore C1 and C 2 can be assumed to be unknown constants.

The calculation of an admittance matrix entry <L(P,,W),P,,,W> requires a six-
fold integration, four finite spatial integrations and two infinite spectral integrations.
This calculation is very demanding numerically because the integrand is highly oscilla-
tory.

In this dissertation, the matrix entries are calculated by the approach described
below. First, the four spatial integrations are carried out analytically with simple basis
functions. Then, the two spectral integrations are computed numerically. From
(4.3.10-11), (4.3.13-16), and (5.5.5), the admittance matrix entries and the excitation

vector can be written as
<L (P..W).P,,.W> = yin—y; (5.5.16)

)5, = <LL(P,,W),P,,,W>

110

 

M2 211 4le “OF (ky Wk, >Fm<k )I‘J..(k > (5.5.17)

y,§,, = <Lfi(P,,W),PmW>

 

 

=01t 1)2 j I Mark (101‘ (k,)r+(k,)r,;(k,)rm(k,) (5.5.18)
where
l .
13309,) = [Pn(x)e*"‘"dx (5.5.19)
-1
ryiacy) = j W(y)eijk’ydy (5.5.20)
1
Tm(kx) = IPm(x)A(kx,x)dx (5.5.21)
-—l
A(k,,x) =Iejk‘1'(I1-)sinks(x—x’)dx’ (5.5.22)
0 s
jk.x_ -jk.x jk.x_ 11.x
= 211‘ (‘9 19+: 8 k _2 ). (5.5.22)

Substituting (5.5.2) into (5.5.20) and using a known integral identity [ 51 ] give

w cos (k, y)

l"*( )=
yk’ {VI-(1)2
w

Substituting (5.5.3) and (5.5.4) into (5.5.19) results

 

dy = vao(k,w) = 1‘, (ky) . (5.5.23)

 

jkxxn jklxn-l
13:; = e 7" (5.5.24)
.1":

 

_ e-jk'x'—e -J'k.x.-1
F = , (5.5.25)
nx ‘ka

 

111

while substituting (5.5.3) into (5.5.21) leads to

1

— 1
(k1 +k, )(kx —k_, )

2ks (k, —k_,)

 

 

Tm: (k1 ) = r!tl1(kx \fi 13;: (ks )-

1
2ks (k, +ks )

137.106..) .

Then, substituting (5.5.2—5) into (5.5.13) gives
ym.1 = _AyArfr
ym.2N = -AyAr$t

where

Ay = .1 1
“'x/Hlf
W
X—

 

dy=1tw

 

sin (ksxm )-sin (ksxm_1)

 

A; = I cos (k,x)d.x =

 

1.1—r k3

1"" cos (ks x”, )—cos (ks xm _1)
A; = j sin (k,x)dx = k

Ina-1 — 3

The excitation vector can be written as

i", = <J'F (r) lJr=ar ,( -k1—)sinks (x —x ')dx ’,P,,, W >
O S

(5.5.26)

(5.5.27)

(5.5.28)

(5.5.29)

(5.5.30)

(5.5.31)

(5.5.32)

The admittance matrix is independent on the form of excitation while the excita-

tion vector takes different forms for different sources of the slot. In the transmitting

case, a delta gap generator is placed at the center of the slot
K,(r) = 1,800 = no

Substituting (552-5) and (5.5.33) into (5.5.32) gives

(5.5.33)

112

. 1y
1,,, 2k AyA; (5.5.34)
3

 

It is worth noting that because 8(x) is an even function of x, the following result is

obtained:

 

S

x 1 . , 1 .
£1,8(x')(7c-S-)srnks(x—x')dx = 2]: srn(k3x). (5.5.35)

In the scattering case, the source is the tangential incident magnetic field H3";t on
the slot. Plane wave propagation in layered media is studied in chapter 3. Results in

chapter 3 are used to express Hg? in terms of the known incident plane wave field H31".

Because the antenna problem is 3D in nature and has no angular symmetry, it is
necessary to specify an incident plane and the polarization for the incident plane wave
before the scattering case can be solved. In this dissertation, a TM plane wave in the
E—plane (y-z plane) is considered explicitly. Any other orientation and polarization of

the incident wave can be handled by the same procedure.
Comparing Figure 1.2 and Figure 3.1 gives
zl=d+t; z2=d; 23:0 (5.5.36)
Substituting (5.5.36) into (3.1.55) gives I
Hg';(r) = 2Hi';e""*‘i“°°’ T‘ = —F (r) ; xe {-1.1} ; ye [~w,w] ; z=0 (5.5.37)

while substituting (5.5.5) and (5.5.37) into (5.5.32) results in

° ~2H‘fiT-A ’ Ac 5538

where the following approximation is used

Iy 19m).1 —) |k1y|<l —> ej"*““°°’=1 . (5.5.39)

113

5.6 Calculation of Admittance Matrix Entries

It is a daunting task to carry out the numerical integrations of (5.517) and
(5.5.18) because the integrands are highly oscillatory. The 2D infinite spectral integra-

tions can be carried out in either rectangular coordinates or in cylindrical coordinates.

In this dissertation, the 2D spectral integrations are computed in cylindrical coor-

dinates. A generic form of the spectral integrals can be written as

k, = looser
{ ky = Asina (5.6.1)
on on 21!
Hf (k,,k,)dk, dk, = I [ I f mammal (5.6.2)
-°° 0 0

where a is a real variable and 3. is a complex variable. This representation provides
valuable physical insight into the problem. Note that from the results of section 5 of
chapter 2, the branch points and poles of the integrands are independent of the angular
variable. To compute (5.6.2), first the angular integration is canied out numerically.
Then the radial integration is computed. The semi-infinite integral can be converted to

an infinite integral. A generic form of the radial integration can be written as
180M?» = lg ’(WA (5.6.3)
0 —“

There are two methods to do the radial integration. In the complex 2. plane, the

infinite integral can be computed by real line integration or contour integration.

It is necessary to define all the branch cuts and to find all the poles of the
integrand before contour integration can be used. The advantage of the contour integra-
tion method is that the integration is stable and rapidly converging, while the disadvan-
tage is that a lot of analytical work is involved. The existence of three layers above the
ground plane makes the eigen-value equations very complicated. It is very difficult to

find all the eigenvalues (the poles), especially when the layers are lossy.

114

The real line integration involves little analytical effort. But because the integrand
is highly oscillatory, the integration is numerically unstable and converges slowly. If
there are poles on the real axis and their positions are unknown, the real line integra-
tion method might fail. The existence of a lossy superstrate shifts all the poles off the
real axis. Thus real line integration can be used successfully. One drawback of the real

line integration is that it requires extensive computation power.

It is advantageous to explore the symmetry of the integrands to reduce numerical

computation. From (5524-26), the functions can be decomposed into even and parts

1‘3; ac.) = I‘:<k.)ir:(k.) (5.6.4)

 

 

 

F506,) = sin(kxx,, );sin(kxx,,_1) (5.6.5)
13309,) = cos(kxx,, )-.cos(k,x,,-1) (5.6.6)
1k;

r... (k5 = 1;; (k. HTS. (10:) (5.6.7)
T‘(k )- k‘ [1“(k)I“(k)] (568)
mx‘k2-ksznrs 011 °-

0 _ 1 o _ o
Tm(kx) — mlkzrfih) ksrm(kx)] (5.6-9)

Obviously 1": dej are even functions of k, and F: and T; are odd functions of k,.

Substituting (5.6.4-9) into (5.5.17-18) gives

 

 

L = 4 ”dzk‘l’ (k r215, Pm, k, 5.6.10
4 0° 2
m. (2102?“ ) ,(k,)Q ( ) ( )

where

1 15
I’M/(x) = 1‘5.(kx)l‘,f(k.)-l‘,£’.(k,)l‘,‘.’(k,) (5.6.12)
QM (18,) = 1,; (k, )r;(k, )—T:. (k, )r:(k,) . (5.6.13)

The computer can not handle indeterminate forms reliably. All the indeterminate

forms have to be carried out analytically.

When kJr —->0, (5.6.5-6) can be approximated as

 

 

 

I
I“ k -“-’- — . .14
. 2n-1 2 2
1"” k ‘~" -— k I 5.6.15
When lkJr -ks |—>0, (5.6.8-9) can be approximated as
T"(k )2 —i d 1“8 (k) (5.6.16)
m X 2 dkx m S
o .. 1 d o i o
Tm (k1) ~ —-[-k_, —Fm (ks )1-1“,,(k, )] (5.6.17)
st s
where
J; 13508,) = 73—ka (xncoskxxn -x,,_1cosk,x,,_1)-(sink,x,, —sinkxx,,_1)](5.6.18)
x x
-d—I‘°(k )- —-1—-[k (-x sink x +x sink 1: )-
(fit, n x "' 7C2 x n xn n—l xn—l

I

(coskxxn —coskxx,,_1)] (5.6.19)

Now express the spectral integrations in cylindrical coordinates. Substituting
(5.6.1) into (5.6.10-13) gives

 

 

L = 4 “d1 7011 r. L 2. 5.6.20
ymn (2102'; L( )snm( ) ( )
R 4 .. R

,,,,,= d). W 2. m). 5.6.21
y (2102!) R<>s () ( )

116

where

s50.) = I‘y2(}.sina)Pm,-, (Momma (5.6.22)

05-. ”'74

Ml:

5,13,, (7.) = j 13206116.)an (Roscoda . (5.6.23)
0

Integration around singularity points of the integrand needs special treatment. As
mentioned before, integration through surface wave poles is avoided because of the
lossy superstrate. Integration through branch point singularities must be carried out
analytically. All the branch points are contained in ‘PL (7.) and ‘PR (7.). Rewrite
(4.3.15) and (4.3.16)

 

 

‘PL (2.) = ‘1", 0.)+\P,, (75+)}; 0.) (5.6.24)
k4 2 k3 2
\PROJ = (-k—) ‘I’a(7~)+‘1'b(7~)+(k—) “20») (5.625)
where
N3x (Z =0)
‘1', 7. a ,—— 5.6.26
( ) J (0143p 30x ( )
‘1' (7.) - N§,(z=0) (5627)
” ’ jwuapanDz ' °
1
‘1', 7. a , 5.6.28
( ) 10394174 ( )

From (2.5.2.11-12) and (4.3.17)
N31: (2 =0) = [(511) 2+€2P 1X52}? 3+53P 2)+(€1P 2’921’1X82P 3‘33P 93‘2”]
+ [(8117 2+62p 1X82!) 3‘63!) 2)+(€1p2-62D1)(€2p3+83p 98’2” le’zp" (5.6.29)

N§z(z =0) = 4P323-2p’d1(€3113‘€2“2)[“ 1P2(1+€-2p2t)+“2P 1(1-‘3-2’” )]

 

117

[£1p2(1+e—2N )+t-:2p1(1-e—2p‘t)]+4(£/2u2—81u1)£3u3e-2p2‘p22 } (5.6.30)

It can be seen that ‘Pa contains the singularity p3=0 , ‘I’c has singularity p4=0, and
‘Pb does not have a branch point singularity because the factor p32 in N31, cancels the

p3 in the denominator. The branch points can be written as
p3 = O -) 13 = [(3 (5.6.31)

124 = 0 -> 44 = k4. (5.6.32)

If k3 and k4 are real, the branch points 2.3 and 71.4 will be on the integration path.
The integration through them must be carried out analytically. The procedure is out-
lined below. Select a small 7 such that

y<ki and y<l ; i=3,4 (5.6.33)

The semi-infinite integral can be split into three parts and one of them can be

evaluated analytically.

Iii-7’9” 0° .
fl’idt:(j+j+j)—ffl—d7t

Pi o te.—r k.-+r Vii-k?
k'mq a9)f(7t)‘1"Y+\l';5'*'27ki)

:( J + j)—+f(k)1n(k =,34 (5.6.34)

0 4+) 12.- k:"Y+‘j7L2'Ykr)

Semi-adaptive integration subprograms, based on extended Simpson’s rule and Rom-

ch“;

 

berg integration [ 55 ], are used in numerical integration of matrix elements.

After the calculation of the matrix elements, the matrix equation (5.5.12) is solved
to obtain magnetic current in slot Then the aperture tangential electric field can be

obtained via (4.2.2)

Ey (x .y) = M10: 5) (5.6.35)

118

The voltage across a slot is obtained by integration of aperture electric field

W

V(x) = j E,(x,y)dy (5.6.36)
Substituting (5.6.35-36) into (5.5.1) then gives

V(x =0) = (IN I 1 dy = nwaN (5.6.37)
‘" \/1-(—f’7)2

 

 

The input impedance of a slot depends on the location of the current source in the
slot. In this dissertation, a current source is centered in a slot. The input impedance of

a slot is defined as

“W0
2,, = VO‘I‘O) = 1N (5.6.38)

CHAPTER SIX
SCATTERED FIELD

The induced electric current on. an imaged monopole and induced equivalent mag-
netic current in a slot are obtained by solving matrix equations (5.2.9) and (5.5.12). In

this chapter, the scattered electromagnetic fields, radar cross section, and radiation pat-

tern are determined.

6.1. Scattered Field for a Monopole

Once the monopole current has been obtained, the field scattered into region 1
may be determined by using equations (2.6.1.13-14). The term 17,00 given by
(2.6.1.6) is common to both expressions. Substituting (5.2.1) into (2.6.1.6) gives

_ N 74+” cos d +2
V,(>.) = 2a,, , 1 hp3( 342'
n=1 -d+(n-1)A J 0’83 P 3

 

 

(6.1.1)

Carrying out the integration in (6.1.1) analytically leads to

1 . A
V30.) = —— an cos 8,, srnhp — . (6-1-2)
j0)€3 p32 22n§1 hp3 3 2

Substituting (6.1.2) into (2.6.1.13-14) gives

N ” COShp 38!! Sinhp3A 2

E = a,, ,
11 "2:21 i 21‘] me3 X0»)

 

e‘Pl‘H’J 001a )Ioap)——d7. (6.1.3)
p32

N «coshp38, sinhp3-2— e-MH
l

Elp = Earl]

p127.
21tjcoe3 x00 ’1 owylap)—d2 (6.1.4)
n=1 0 .

 

It is important to understand the asymptotic behavior of the integrands in (6.1.3-
4). To write the integrands in terms of exponentials, use
A

pA _ 3__
coshp35 sinhp3-g- =18 ”3% 2[1+e +e'2P3°-][1—e 2" 21 (6.1.5)

119

120

7. - —e’”" e“ 1—e 2"“ 1+ 2” + 1+e—2p’d][1—2p2']+
X()- {:31 ll 6 ] £14m -€

 

2111—62?" 111-e4”) + 5i11+e‘2“" 111+e‘2P2‘1 (6.1.6)
E3P2 P3
to give
=2.) IHO.)e J 00.2: )JOO.p)—d7. (6.1.7)
n=1 p32
N a -P3(d—5'-%) -p2t -p (z-t) p )‘2
Elp= ZanIHQk e e 1 10(71ayl(7.p) d2 (6.1.8)
n=1 0 P3
where
_ 5 ’ZPSA 81d _ 81p (1
Ho.) = [1+e 2“ '][1—e 2 —[1-—e 2“ ][1+e 2W] + —[1+e_2p3 ]
E3 82P3

-1
8
[11‘2”] + fl[1-e‘2f’3"][1—e'2*"2‘1 + 5i[1+e‘2“"][1+e‘2“’] (6.1.9)
E3P2 P3

Since each term in brackets in (6.1.9) converges to unity as 7t-—>oo , the asymp-

totic form of H (2.) is

-1
e e 8

H0.) ~ —1— + _1p_2 + £- + -p—1 = constant . (6.1.10)
83 £2P3 €st P3

Thus, the decay of the integrands of (6.1.7) and (6.1.8) is controlled by the exponential
terms. It is seen that the integrand has the slowest decay when n=N and 2: t, causing
two of the exponential terms to drop out. Then, the integrand behaves asymptotically

as

1009017009)
.. M723 L. (6.1.11)
10090171099)
. J

 

 

121

Since Re{p2)>0 ,(2.3.17), there is always an exponential decay factor, and the

integrals will converge.

Even though the integrals converge, they are still difficult to calculate numeri—

cally. This is due to the oscillatory behavior of both the exponential term and the

Bessel functions in (6.1.11) at large 2.. Care must be taken to integrate over complete

periods of the Bessel function.

6.2 Far Field Calculation

From (6.1.3-4), the scattered electric field in region 1 can be written as

where

 

 

N °° 13
E12 = 20 [H (7.)e W10 Dam—:81).

n=10

N °" _ z pxzd
Elp= za.jH,.(t)e ”11045)”p

n=1 0

_ A
_ 32 3(d-8n-A —)

Hn(l)_ _ mu“ +6—2p36'] [1 ep3 ie—p 2 e’le ePr‘J 00‘”)

E1 —2p3d -2p;l 81p 2 -2p3d -2pzt
( i [ ' e H e ] 3i ][ e ]

5’1"-

[1- e’zp’d][l-e'2p’]+—[1+e’2”’d][1+e_2”]}
33P2 P3

_ _L
5,, —(n 2)A

h
A:—
N

k1: Weill 3 k2 = (WE/2112

z e (t,°°); p E (0.”)

(6.2.1)

(6.2.2)

(6.2.3)

(6.2.4)

(6.2.5)

(6.2.6)

(6.2.7)

122

an — current expansion coefficients
N — number of basis functions

These fields can be calculated either exactly, through direct numerical integration, or
approximately, using the stationary phase method. Both approaches are outlined

below.

6.2.1 Numerical Integration Along the Real Axis

For a lossy superstrate, €22 and/or ttz can be complex. Because of this the zeros
of D0.) are all complex numbers. In other words, the poles of the integrands of
(6.2.1) and (6.2.2) are all off the real axis. Therefore, direct numerical integration can
be used to compute the scattered far field. A real axis integration technique has the
advantage of a wide range of validity in medium, frequency, and spatial parameters.
The major limitation is computation time [ 36 ].

In the far field and radiation pattern calculations, the spatial parameters 2 and p
have a very big dynamic range. Terms like {NZ-0, J0(}.p), and 110p) oscillate
rapidly with large 2 and p. Highly oscillatory integrands make accurate and rapidly
convergent numerical integration difficult to achieve.

The oscillations of the integrands of (6.2.1) and (6.2.2) in the interval 2. e [0,k1]
are due to the terms (NH) , 10(kp), and 11(Ap). The oscillation of (”I") as a
function of it becomes more rapid near the branch point A = k1. Integration of these
oscillatory functions is further complicated by the peak behavior of the integrands near
the branch point. To make the densely packed oscillations more evenly spaced and to
remove the peak behavior of the integrands at the branch point, the nonlinear

transform [ 36 ]

7. = klsinO e e [0%] (6.2.1.1)

123

is used over the interval 2. e [0,kl]. Then let

’51

1,32 = j H" (2.)e ‘P ”1001(3) k
0

3
_th
P3

H, (k 1sin0)e “"1’ °°5910(k1psin0)(k ,sine)3d0 (6.2.1.2)

o~—-.N|=I

and

k1 - z plxz
1,3,, = II-Inak P1 11(2.p)—d2.
0 P3

11,, (k lsin9)e 4‘" 0059J 1(k lpsin0)(k1sin0)2k1cosed 0 (6.2.1.3)

O‘—-. NI?!

After the transform, the branch point is removed and the integrands in (6212-3) have

an almost evenly distributed oscillation, and both approach zero at 0 = g.

A similar transform
2. = klsece 0 e [0,cos‘1(%)] (6.2.1.4)

may be used in the interval 2. e [k1,2k1] to even out the oscillation and remove the

peak behavior of integrands at the branch point 2. = k1. Then

21:
‘ 3
1,3, _ jH,(t)e‘P*’Jo(tp)"—dx
. h 3

P

cos-R?
= 1 ”no;lscc9)e"“m"°]0(klpsec0)(k1sec0)3sec20d0 (6.2.1.5)
0

and

2 2hr _ 2 P17);
[up = (1143).» ”1 11(2.p)—d2.
1‘1 P3

124
cos"(%)

= J Hn(klsec6)e-k‘wmel 1(k1psec0)(klsec0)3tan0d0 (62.1.6)
0

In the interval 2. e [2k1,oo), the exponentially decaying term e7”lz makes the

numerical integration rapidly convergent, so no special transform is needed. Let

°° 3
1,3 = j H, (t)e‘P*’Jo(tp)%—d>. (6.2.1.7)
2k, 3

3 .. _ 2 P112
1,,p = j Hn(2.)e F" mam—d). (6.2.1.8)
21., P3

Romberg integration is performed between the zeros of 10(2tp) and J 1(2.p) and the

results of subsections are summed up to get lull, 1,32, In“, , and 1,3,). A transform 2. = -1—
x

is used to convert (6.2.1.7-8) into proper integrals and then the Romberg method is
used [ 54 ]. The final results for the electric field are

N

1312 = 2a,,[1,,;+1,3,+1,31 (6.2.1.9)
n=l
N

Elp = 2a,, (1,3,, +1:p +139] (6.2.1.10)
n=1

Real axis integration can calculate both the near field and far field. But it is quite
time-consuming. This method can be used to compute the scattered field at a specific
point or to calibrate the results from more efficient approximate methods. It is not

suited for radiation pattern calculation.

6.2.2 Stationary Phase Method

In order to calculate the far field more efficiently, some kind of asymptotic tech-
nique must be used. In this report, a simple and efficient stationary phase method ori-
ginally proposed by Chew [ 38 ] has been used.

125

First the following generic integral is considered
I = j g(0t,2.)d2. (6.2.2.1)

where a is a large parameter. If g(a,2.) becomes rapidly oscillating when or is large,
and if there exists a stationary phase point of g(0t,2.) , a leading-order approximation
can be obtained by the method of stationary phase. Several major steps of the method
are highlighted.

The first step is to factor the integrand g ((1).) into a slowly varying part f (2.) and
a rapidly varying part p (a2).

I = I f (3.)); (amt. (6.2.2.2)
0

Assume p ((1,2) to be of the generic form
p (6.7.) ~ e‘wm a-m (6.2.2.3)
The key in the factorization is to have a function p(a,2.) that can be integrated in
closed form.
The second step is to find the stationary phase point 20 of p ((12), defined by

3.80») _
__37. i=8. _ 0 (6.2.2.4)

Most contribution to the integral in (6.2.2.2) will come from the vicinity of the station-
ary phase point 2. = 2.0 A leading-order asymptotic approximation to (4.6.2.1) can be

written as

I ~ f(2.o) jp(a,).)dt a—wo. (6.2.2.5)

The Sommerfeld identity [ 6 ][ 60 ] is needed in the stationary phase method.

e‘j’”

r

 

IJO(2.p)e’P '“fi-d). (6.2.2.6)
0

126

where
p = \l2.2—k2

The physical interpretation of the Sommerfeld identity is that the spherical wave is

expressed in terms of cylindrical waves.

Now, let

M 3
= JHn(2)e—p‘zJ0(lp)Ldk
0 P3

       

.. P1 2 2t
: IV!" 0" e'p‘zlo(2.p);-]d2. . (6...227)
0 1

The term in the first bracket is slowly varying and the one in the second bracket is

rapidly varying. The next step is to find the stationary phase point.

Express the Bessel functions in terms of Hankel functions [ 53 ]
1.04» = —;—1H,.“><xp)+H.‘2><xp)r . (6228)

Then (6.2.2.7) can be rewritten as
I.z %I(7))[1161’(24))+H(12)(2~1))]""‘—d2.pA
0

= I f... (20116” timer"z £112. (6.2.2.9)
... l
where

12
f... 00 = 11.0.) p ‘
P3

 

(6.2.2.10)

Here the fact that f n, (2.) is an even function of 2. and the following identity have been

used [ 53 ]:

H'flkx) = _e-MHH'S2)(xe-ifl)

127

Note that
kax) ~_\/—%-e-16_22_7) as |x|—>oo (6.2.2.11)
then
H ((2) awe—P12 ~ e-PIZV-Jt—Z—p—e 404,—? ) as lp—aoo (6.2.2.12)

The stationary phase point is given by
a .
5}_\'_[—(p 12 +1 AP” : O . (6.2.2.13)

The solution to (6.2.2.13) is

9" 1 .
7~0 = 1 = klsme (6.2.2.14)
(z 2 +p2)?
where
0 = sin'1[—P—l-] .
(2 2+1?)3

The first order approximation to (6.2.2.7) can then be written using (6.2.2.6) as

 

 

200
1,,, =H 4101’"? je W1 Dam—dz
0
P1102e’m'
... H "(10) r (6.2.2.15)
p3

So the far field asymptotic approximation of E 1, becomes

e-jk 1r

 

=Zanfmao) (6.2.2.16)

where r = Vp2+z 2. The asymptotic approximation of E1p can be obtained in a similar

128

 

 

way.
e-jkr'
=2cznfm0»o)e (6.2.2.17)
where
1043)
1. 1. 1 6.2.2.18
fnr()= H..( )p p3 1.1004?) ( )

Equations (6.2.2.17) and (6.2.2.18) can be used to compute the scattered far field or to

get the radiation pattern.

The radar cross section is defined as

RCS(6,¢)41tr211_r£| -—(_111:(:))Em I2 (6.2.8)

For a monopole illuminated by a TM plane wave, using (6.2.2.16-17), (3.1.1), and

(6.2.8), the radar cross section can be expressed as

Rcs (9.4» ___—“(T101111 2m... (1.11244 )3a..f,., (10)?) (6.2.9)
1x)2n=1 n=1
where
Ef
Tlo = 137' = 120m!) (6.2.10)

1

is the intrinsic impedance of free space.

6.3 Scattered Field for 3 Slot

After the magnetic current in the slot is obtained by solving the matrix equation
(5.5.12), the scattered magnetic field can be computed from (2.6.2.1), (2.6.2.11-13),
and (2.6.2.20-22).

H10“) =Hga1. 3(rlr”)1143..(r)dx ”dy a=x.y.z (6.3.1)

:10!

129

where

 

8.1163“): 1 Héai3(r.k)e“”d2k (6.3.2)

(21:)2 ...

and g3, 12,13, and g2}? are given in (2.6.2.20-22).

Substituting (5.5.1) into (6.3.1) gives

X. W

 

 

WV 13
Hla(r) = 2a,, [[ j dx' jdy' gm; (rlr’)W(y’)] (6.3.3)
n=1 1..-] -w
where
W(y) = —l— y e [—w,w] (6.3.4)
\/ Hi?
W
n
x" = (—-l)l
”_1 (6.3.5)
1: - =(’1 -1)1
n 1 N
Now, define
8.2%“) = S... (106" “’e‘P" (6.3.6)

Substituting (6.3.6) and (6.3.2) into (6.3.3) and carrying out the two spatial integrations
analytically lead to

 

 

2N an °° _ ei(k.x+k,y-k.|z|)
H,,,(r)=n§1 (2102 j _J; P3100 k2 dkxdky ; z>O (6.3.7)
where
101 = J'kz (6.3.8)
13:.(1‘) = 5... (or, (k, mam-1p 1) (6.3.9)

1‘y = anoUcyw) (6.3.10)

130

 

1“,; = j'k1 (e"'"*1‘~-e‘j""'-‘) (6.3.11)
1
Next, define
IV = Isle-Plz (6.3.12)

where N can be Nu, N31,, N,,,N3’, defined in (2.5.2.4), (2.5.2.10), (2.6.2.28), and
(2.6.2.29). Substituting (6.3.6) into (2.6.2.20-22) gives

"d

 

 

 

 

       

~ 1kg 2 N12
Sn = . [(k2 k1?“ —k (— )1 (6.3.13)
101113103 1 :1 D D
§ - 1 [( k XIV”: [V11 )1 (6314)
yx — 1103113P3 1") DJ 1 0x02 1 1
3 = 1 Uk 12(117—1— )+jk (1.2+);2 — (6315)
‘1‘ 1011131030. 1 D D ' '

The integrals in (6.3.7) can be carried out numerically to obtain the scattered
magnetic field. But when the distance r=m becomes large, the integrand in
(6.3.7) becomes highly oscillatory. This makes accurate and efficient numerical
integration almost impossible. This is where asymptotic approximation comes in. A
stationary phase method is used to arrive at the first order approximation to the scat-
tered far field [ 6 ][ 38 ]. The general procedure of this stationary phase method is

outlined in section 6.2.2.
The Weyl identity [ 6O ] makes the approximation of (6.3.7) possible. This iden-
tity is given by

—jb _- °‘ jk.X+ik,y-jk.iz|
‘3 = ‘Ll Idkxdky e k (6.3.16)
—oo 2

 

 

r

where

Ic,3+k,2+k,2=k2 or k, .. k —k, -k, (6.3.17)

1131
To satisfy the radiation condition, the branch cut of k2 is defined by

Im[kz ]<0 and Re [k2 ]>O (6.3.18)

The physical interpretation of (6.3.16) is that a spherical wave can be expressed as an
integral summation of plane waves propagating in all directions, including evanescent

waves.

From (6.2.2.3-4) the stationary phase point k? is given as

R17 = Jimmy-task? (6.3.19)

kg? = k1; = k 1811190084) (6.3.20)
sp = —y- = 1

k, k; r klsrnecoso (6.3.21)

kgp = 21% = klcose (6.3.22)

r = Vx2+y2+22; 0 e [0.1:] ; 6 e [021:] . (6.3.23)

For r—)oo, substituting (6.3.16) and (6.3.19-23) into (6.3.7) leads to the first order
approximation of scattered far field

‘17‘ 1'

2” -
[20,, P2,,(k‘1’1 )] ; a=x,y,z . (6.3.24)

.e
”la = 2} 1V
n=1

 

The second term in brackets on the right hand side of equation (6.3.24) determines the
radiation pattern of a slot in tri-layered media.

The radar cross section is defined as

RCS(0,¢) = 411’2,11_1,11.1£E:'((‘:.)712 (6.3.25)

132

For a slot illuminated by a TM plane wave, using (6324-25) and (3.1.1), the radar
cross section can be written as:

2N ..
[( z 4.18;. (W 112+

RCS (9,4)) = .
we? ..=1

 

7N ~ 2N ..
( )3 21,131 (1.31" ))2+( 2 a, 1310.510 ))2] . (6.3.26)
":1 "=1

CHAPTER SEVEN
NUMERICAL RESULTS

7.1 Numerical Results for a Monopole

FORTRAN programs have been written to implement the MoM solution for the
monopole current and the scattered field described in chapter 5 and 6. These programs
have been run on both IBM PC microcomputers and the Sun workstations of College
of Engineering. The programs are very efficient and it takes a few minutes to run a

case with twenty impedance matrix fillings on a fast 486 PC.

7.1.1 Comparison with Existing Numerical Results

To establish the validity of this analysis it is desirable to make a comparison with
previously published results. The simplest possible comparison is with a dipole in free
space, which is equivalent to an imaged monopole in free space. The input impedance
of a dipole in free space is twice that of an imaged monopole in free space. Free space
is the simplest special case of tri-layered media with both substrate and superstrate
having unit permittivity and permeability. Figure 7.1.1 compares the input impedance
of a dipole in free space obtain by the theory developed in the dissertation with that of
King’s book [ 4]. The two results are in good agreement.

Tesche [ 46 ] analyzed a dipole sandwiched between two perfectly conducting
parallel plates using a Pocklington-type integral equation, the kernel of which was
determined using an infinite image sequence. This situation can be handled by the

present analysis if the superstrate is allowed to become perfectly conducting.

Figure 7.1.2 shows the input impedance of a half-wavelength dipole oriented vert-

ically and centered between two conducting plates, as a function of the plate

133

134

separation. The lossy layer is assumed to have a conductivity which gives
8/2 = (l-j1000)£0 and thus is, for all practical purposes, perfectly conducting. Agree-
ment with Tesche’s results is seen to be good. The discrepancies may be due to

Tesche’s use of the less stable Pocklington-type integral equation.

Comparison have also been made with work done by Chi and Alexopoulos [45 ],
who has studied the radiation of an imaged monopole through a perfect dielectric sub-
strate. This case is handled by assuming the superstrate (region 2) to be nearly free
space. It has been found that to insure the proper convergence of the moment method
matrix entries, the lossy layer must have some small, non-zero conductivity. Best

agreement with [45 ] was obtained by using sinusoidal basis function detailed in [47].

Figures 7.2.3 and 7.2.4 show the input resistance and reactance of an imaged
monopole radiating through a perfect dielectric substrate, for two values of substrate
permittivity, as a function of antenna length. Agreement with [47] is seen to be quite

good for most antenna lengths.

7.1.2 Comparison with Experimental Results

The effect of resistive coverings on the backscattering from a monopole on a con-
ducting surface are studied experimentally by the Boeing Company, the sponsor of the
research project. This experimental work was performed as an aid in confirming the
analytical work presented in this dissertation. Backseatter measurements were made on
a vacuum kayak measurement platform. The experimental setup is shown in Figure
7.1.21. A monopole is short circuited to the aluminum surface of a kayak measure-

ment platform, which means that the load impedance is set to zero

ZL =O(Q).

A 0.23 inch thick foam support and three resistive coverings were used. The

foam is estimated to have near unit relative permittivity and permeability. Throughout

135

the dissertation, a foam substrate is assumed to have unit relative permittivity and per-
meability. The three resistive sheets are believed to have constant surface resistances in
the frequency range from SGHz to 180112. The resistance R and thickness t of the

three resistive sheets are:

R = 75(Q/Cl) ; t = 4.72(mil) = 0.120(mm) (7.1.2.1)
R = 250(QJCD ; t = l.58(mil) = 0.0401(mm) (7.1.2.2)
R = 500(9/El) ; t = O.57(mil) = 0.0145(mm) (7.1.2.3)

With the assumption that the resistance is independent of frequency, the complex
permittivity can be written as

l

—R121tf£0 )50 (7.1.2.4)

82 = (1‘)

where f is the operation frequency and so is the free space permittivity.

The relative complex permeability of the resistive sheets is assumed to be one

Throughout the dissertation, air film, as the name implies, is a superstrate with
unit relative permittivity and permeability.

Theoretical prediction of radar cross section of a shorted monopole in tri-layered
media with foam substrate and four different superstrates versus frequency is compared
with experimental data in Figure 7.1.5. The relevant parameters are specified in the
plot. The complementary incident angle (I) is formed by the incident wave vector and
the ground plane. There is qualitative agreement between the experimental and theoret-
ical results. The biggest discrepancy is 3dB and occurs at the high frequency end. The
relevant parameters are marked in the figure.

136

Radar cross section of a monopole versus complementary incident angle at two
operating frequencies is presented in Figure 7.1.6 and 7 respectively. The trend of the
theoretical data and experimental data are the same. The qualitative agreement between
numerical results and experimental ones in Figure 7.1.6 and 7 is not as good as that in

Figure 7.1.5.

Several factors can possibly cause the discrepancy between the theoretical result
and experimental one. The major factor is that the experimental setup is finite while
the theoretical model is of infinite extend. The contribution to the total radar cross
section from edge scattering can not be ignored. The assumption that the resistance is
independent of frequency and foam substrate has unit relative permittivity and permea-
bility may not hold in the frequency range from 80112 to 186112. Accurate parameter
of the foam and resistive sheets are not available. In the measurement of radar cross
section versus frequency, both the antennas and the kayak platform are fixed in posi-
tion. In the measurement of radar cross section versus incident angle, the antennas are
stationary and the kayak platform is rotated. This can be the reason that the former

measurement is more stable and accurate than the later one.

7.1.3 Results for Lossy Superstrates

It is necessary to check the convergence of algorithms, at least numerically. Fig-
ure 7.1.8 shows the input impedance of a monopole in layered media versus number of
basis functions per wavelength. Two configurations are considered, one with an air
film superstrate and a foam substrate, the other with a resistive sheet of 250 ohm and a
PTFE substrate. The relevant parameters are clearly marked in the plot. The Figure
7.1.9 and Figure 7.1.10 show the radar cross section and received power versus the
number of basis functions per wavelength for the same two configurations. A load
impedance of 50 ohms is located at the center of the slot and a TM plane is illuminat-

ing the entire structure. The angle between the incident wave vector and the ground

137

plane is 20 degrees.

It is observed that the input impedance is quite sensitive to the number of basis
functions used and the radar cross section and received power are less sensitive to the
number of basis functions. In the analysis of monopoles, the density of basis functions

is in the range from 70 to 100 basis functions per wavelength.

In this section, the magnetic coating denotes a fictitious electrically and magneti-

cally lossy layer with the following parameters:

8/2 = (10—j0.5)£0 ; [12 = (5-j4)110; t = 4.72(mil) = 0.12(mm) . (7.1.3.1)

The next ten figures are for the following geometry. A monopole of length 0.216
inch and radius 0.0185 inch is immersed in a substrate of thickness 0.23 inch. The
substrate can be a foam substrate or a PTFE one. The monopole is loaded with a 50
ohm resistor. Five superstrates defined previously are used. The system is illuminated

by a TM plane wave with 20 degree complementary incident angle.

The input resistance and reactance of an imaged monopole in tri-layered media
with foam substrate and five different superstrates are presented in Figure 7.1.11 and
12 respectively. Figure 7.1.13 and 14 show the input resistance and reactance of an
imaged monopole in tri-layered media with four superstrates and a PTFE substrate.
Notice the down-shift of the peak resistance because the the monopole is electrically

longer in PTFE than in foam.

Figure 7.1.15 and 16 give the radar cross section and received power of an
imaged monopole in tri-layered media with a foam substrate and five different super-
strates respectively. The radar cross section and received power of the same monopole
in tri-layered media with a PTFE substrate and four different lossy superstrates are
shown in Figure 7.1.17 and 18 respectively. Tri-layered media with a foam substrate
and an air film is actually a half free space. The case of a monopole in layered media

with a foam substrate and an air film is used as a reference to determine the effects of

138

lossy superstrate on the scattering and receiving characteristics of a monopole in tri-
layered media. Figure 7.1.15-18 demonstrate that the existence of a lossy superstrate
reduces both the received power and radar cross section of a monopole. But the reduc-

tion of radar cross section is more than that of received power.

E-plane (y-z plane) radiation pattern of a monopole in a foam substrate under five
different superstrates is presented in Figure 7.1.19 and E-plane pattern of the same

monopole in a PTFE substrate under four lossy superstrates is shown in Figure 7.1.20.

7.2. Numerical Results for a Slot

The numerical results for slots in tri-layered media based on the theory described

in the dissertation are presented in this section.

7.2.1. Comparison with Published Results

The most convincing way to validate theory and computer code is to compare
experimental results with theoretical ones. The Electromagnetic Laboratory at Michi-
gan State University does not have the capability to do radar cross section measure-

ment. The next best way is compare numerical results with published results.

The simplest case is a slot in free space. S.A. Long [ 32 ] did experimental study
of impedance of an open slot and a slot backed by different cavities. Figure 7.2.1 com-
pares the measured impedance of an Open slot, which radiates freely into the upper and
lower half of the free space separated by a ground plane, with that generated by the
computer code. The slot has a total length of 25cm (21=25cm) and width of 1cm
(2w=1cm). In the measurement, the ground plane is a quarter inch thick and eight

square foot.

To compare with Long’s experimental results, the parameters are set as such

8142434440 ; llr=llz=ll3=fl4=110

21=25(cm); 2w=1(cm); d=1.5(mm); w=0.268(mm) (73°11)

 

139

Figure 7.2.1 shows the input impedance of an open slot. The results from the
theory described in the dissertation show good agreement with that of Long. The
minor discrepancies are caused by the fact that in the dissertation the ground plane is
assumed to be infinite and infinitesimally thin while in Long’s experiment the ground

plane is finite and thick.

M. Kominami et a1. [ 29 ] investigated printed dipole or slot antenna on a semi-
infinite substrate and infinite phased arrays of these elements. The results in [ 29 ] are
compared with the numerical results in the next two figures. Figure 7.2.2 gives the
input impedance of a slot on a PTFE ( 8, =2.55, tan5=0.002; X—band ) semi-infinite
substrate. Figure 7.2.3 gives the input impedance of a slot on a semi-infinite GaAs

substrate ( e,=12.8, tan8=0.002; X -band ).
The rest of parameters are set as

£1=82=e3=£0 , e4=(2.55-j0.0051)£o (PT FE ) or (12.8—j0.0256)£0 (GaAs)
H1=llz=li3=ll4=uo ; w/I =0.02 (7.2.1.2)

The numerical results agree with Kominami’s published results very well.

7.2.2 Results for Lossy Superstrates

This section contains the numerical results of a slot in tri-layered media with
different superstrates and substrates. Terms of interests are input impedance, radar

cross section, received power, and radiation pattern.

Three kinds of superstrates are used. The first superstrate, denoted as air film, is a
vacuum layer with permittivity 62:20 , permeability ufuo, and thickness t=0.12mm.
The second superstrate, denoted as resistive cover, is an electrically lossy sheet with
resistance R=75(Q/Cl), permeability 11qu, and thickness t=0.12mm. The third super-
strate, denoted as magnetic coating, is a fictitious electrically and magnetically lossy

coating with permittivity £q=(10-j 0.5)e0, permeability p2=(5— j 4)).10, and thickness

140

t=0.12mm.

The resistance of the resistive cover is assumed to be constant in the frequency
range of interest and the real part of relative complex permittivity is assumed to be

one. This assumption makes the relative permittivity a function of frequency, which is

written as
1
o = Rt— (7.2.2.1)
82 = 8011—1251:) (7.2.2.2)

In the frequency range of interest, the imaginary part of 84/80 is in the order of one
hundred while the real part is in the order of one. So the above assumption is a good

approximation.

Three substrates are used. The first is a foam substrate with permittivity 83:80 and
permeability 113:110. The second is a reinforced PTFE substrate with permittivity
£3=(2.20—j0.00198)£0 and permeability [13:110. The third is a GaAs substrate with per-
mittivity e3=(12.9- j 0.0258)£0 and permeability 113:110. The last two are commonly
used substrates in microwave and millimeter-wave frequency range [ 18 ]. Another
way to present complex relative permittivity is to use dielectric constant e, and loss

tangent tan5

e = e, 20(1—j tan6) (7.2.2.3)

Figure 7.2.4 gives the input impedance of a slot in tri—layered media with an air
film superstrate and a foam substrate. Figure 7.2.5 shows the input impedance of a slot
in tri-layered media with a magnetic coating and a foam substrate and Figure 7 .2.6
gives the input impedance of a slot in tri-layered media with a resistive sheet super-
strate and a foam substrate. The parameters for the above three figures are set to be

3143:8440 ; “1:113=ll4=110

t=0.12(mm ). d=1.5(mm ), I=5.26(mm ), w=0.268(mm) - (7.2.2.4)

141

The length of the slot is chosen such that at 14GHz, the length of the slot equals to a

quarter of free space wavelength.

Figure 7.2.7 shows the input impedance of a slot in tri-layered media with a resis-
tive sheet superstrate and a PTFE substrate. The relevant parameters are
E1:593:84450’ 3 111:112=ll3=114=110
£2=(2.20—j0.00198)eo . (7.2.2.5)
t=0.12(mm ), d=1.5(mm ), l=5.26(mm ), w=0.268(mm)
Figure 7.2.8 gives the input impedance of a slot in Iii-layered media with a resistive
sheet superstrate and a GaAs substrate. The relevant parameters are
81:53:84? 0. ; ll1=llz=113=114=110
=(12.9— j 0.0258)£0 (7.2.2.6)
t=0.12(mm ), d =1.5(mm ), I =5.26(mm ), w=0.268(mm)
Throughout this section, a load impedance ZL is placed at the center of the slot

and the slot is illuminated by a TM plane wave (li‘y , E, , H,) with an incident angle 00

ZL = 500 Q
90 = 60. (7.2.2.7)

The radar cross section and received power of a slot in tri-layered media with a
foam substrate and different superstrates, namely air film, resistive sheet, and magnetic
coating, are given in Figure 7.2.9 and 10 respectively. It can be seen from Figure
7.2.9 and 10 that with a resistive sheet or a magnetic coating, the reduction of radar
cross section is more than the reduction of received power. The case of a slot in tri-
layered media with an air film and a foam substrate is used as reference. For example,
at 14GHz the reduction of the radar cross section is 6.79dB for the case of a resistive
sheet and 4.88dB for the case of a magnetic coating. At the same frequency, the reduc-
tion of the received power is 4.53dB for the case of a resistive sheet and 2.57dB for

the case of a magnetic coating.

142

The radar cross section and received power of a slot in tri-layered media with a
resistive sheet and different substrates, namely foam, PTFE, GaAs, are presented in
Figure 7.2.11 and 12 respectively. The relevant parameters are given in (7.2267).
An observation can be made from Figure 7.2.11 and 12. The higher the dielectric con-
stant of the substrate, the more the reduction of both radar cross section and received
power. In other words, a substrate with high dielectric constant will decrease the radia-

tion capability of a slot.

The E-plane (y-z plane) radiation patterns of a slot in layered media with a foam
substrate and different superstrate are presented in Figure 7.2.13. Figure 7.2.14 shows
the H-plane (x-z plane) radiation pattern. Figure 7.2.15 and 16 present the radiation
pattern of a slot in tri-layered media with a resistive sheet and three different substrates
in E-plane and H-plane respectively. For all the radiation patterns, the operating fre-
quency is 14GHz. There are significant changes of E-plane pattern for various super-
strates and substrates. The change of superStrate and substrate does not alter the H-

plane pattern very much.

143

  
  

 

 

I + +
- 4-
500 g — R... +
:1 ---- In .
-. +++++ Rln King
A f; 00000 X." King
E -
5 I
8 100':
8 I
C j 0” ‘\
o .... or, \\
3 - Ox \ °
a—lOO: ,9” ‘.
E -( a, \\a
— .1 0” \s u’
-I I ‘\ ’9’
'H -I I ~~~~~ ‘fi
3 - I, U D D
Q -I D”
C : l’
-300: ,’
-1 ,’
I
I
’l
I h/O=16.56
—500|lrrllllrl[rrrllrrr rrllfi
0.5 1.5

Normalized Length ko

Figure 7.1.1 Input impedance of dipole in free space.

144

 

 

 

 

 

150-1
40 Rln
- ----- in
a 00000 R“, Tesche
A 1 0 xxx” X... Tesche
E ..
o 1
V100 1
Q) -1
O 2 0 0
C
O ..
"C -1
Q) —4
Q -
E ‘
:x
...:
3 -—4
Q. 50 - x x
E " x ,"""‘~----f--_-.¥----z
-l ’X’
.1 "a”,
-I “‘ X ,0”
1 1“‘~ " e”"
- =A/2, h/o=148
O IITIIIIIIIITTIIVUIIIIIIIlllIIlllllllIle
1.0 1.5 2.0 2.5 3.0

Spacing between Plates 2d/A

Figure 7.1.2 Input impedance of dipole between two parallel conducting plates.

145

 

   

 

 

 

:1 0000083=1.9€o
1 _ x x x x x 83:3.980 .
50 '1 85:1.960 Chi; / \

A : _ _ 83:3.980 Chl 14 \
E d

S .1

O -I

v -

8 :

C100:

0 -1

4.:

.‘2 Z

(I) q

0) ..

0: q

*5 l

o. 50q

E :

O 11 rrrlrrrrrfirrrlrrrrrrrrrITr
0.0 0.5 1.0 1.5

Normalized Length koh

Figure 7.1.3 Input resistance of probe through substrate.

146

 
  
   
 

 

 
 

 

100-
‘ o
I 00000 83:1.980
- x x x 1" x 83:3.980
- 83:1.980 {Chi
A ‘1 _ __ 85:3.980 Chl / \
E .1
_C .1
3 0: \
o _
LC) - l
.9 I 1
8 - . \
0 I / \
O: - \x
— —'1 X
*5 100‘ \ I
Q q
E - 0
~ d=0.3)\o
“-200 T1IrrrrrrlrrrrTrrrIITIIIIIIIT[r1

 

0.0 0.5 1.0 1.5
Normalized Length koh

Figure 7.1.4 Input reactance of probe through substrate.

RCS (stm)

_20_

J 1 l

1111

l

1

l
(A
‘11

|
4:
O

1 l

l

l

—50

147

— Air film AAAAA Air film, Experiment

— R=5000 +++++ R=5000, Experiment
- —- R=2500 1* x x x x R=2500, Experiment
---------- R=750 t t o- t 0- R=750, Experiment

11:0. 21
d=0. 23(in
zL= 0.00

6(in). ¢a= 0.0185(in)
=20°

 

‘
s
e
h
‘\
5
.
.
‘
a
Q
‘
s
‘
\
‘
Q
s
‘
m
‘
Q
§
Ԥ
Ԥ

 

 

I I I
12
Frequency (GHz)

Figure 7.1.5 Radar cross section of monopole in tri-layered media with foam substrate
and various superstrates versus frequency.

148

 

 

 

 

-20:
i
-305
A :-
E-40:
tn 2
% -
V :1 Air film 1‘. \
m - — R=5000 \‘\ x \
o : - — R=2500 ‘ , ‘\
a: -50- ----- R=7sn x
3 44648 Air film, Experiment 1 11‘.
:1 +++++ R=500Q Experiment \
_. x x x x x R=2500. Experiment \\\
Z t . o- t t R=750, Experiment . ‘\ 1'
—60§ ‘\
: f=12 GHz). d=0.23(in)
_ h= . 16(m), a-0.0185(in)
:1 ZL=0.0Q
—7O TUITIIIIIITTTTIIIIIITrITTIIITIIIIIIIIII]
0 10 20 30 40

Incident Angle (degree)

Figure 7.1.6 Radar cross section of monopole in tri-layered media with foam substrate
and various superstrates versus incident angle at lZGHz.

I‘V

_ _
€5ch mom

 

figure 7.1.

 

149

 

 

 

-20:
-305
A 3i
E —40j{
m .i
% 1
V j —— Air film ‘\ \ 4' \
m - '— R=5000 \\\ X
o : - — R=2500 \ \ t
05 -5O: """ R=759 " \‘t \
E MAM Air film, Experiment " \‘s \
_ +++++ R=soon. Experiment
2 x x x x x R=2500, Experiment ‘\
- . . o- :- . R=750, Experiment t ‘\
-50: \‘t
: *
I f=15(GHz , d=0.23(in) \\
: h=0.216(m). o=0.0185(in) \
_ ZL=O.00 '
—7O illIifilllrITI—TIIWIIITTIIllll'f‘IIIllle
O 10 20 30 4O

Incident Angle (degree)

Figure 7.1.7 Radar cross section of in tri-layered media with foam substrate and
various superstrates versus incident angle at lSGHz.

150

200

M R5... Air fllm, 83:80
era-aha Xi... R=2SOO. e5=£o
...-...... Rh, R=2500, €3=E2.2-j0.0044geo
Ht“ XI“, R=2500. £5:

2.2—10.0044 so

     

 

l-ll4llll-lllllllll

 

 

A
E
o 100 h/A= 0.256 h/o=16.56
v =.o 23(in) f=14(GHz)
Q)
U
C
C e ‘ A 3.. —.‘.
U I g :- _‘_ :- i- " i "'
Q.) \
Q \
g \‘t
...; "‘ g):*--.¢.--¢---¢---e---A---¢---o----o---¢---.-_-A
c3; O-r \\
E ‘ ‘.
_ \K‘
.1 \‘k‘
: ~““ ~.
- "~o---.~~~*-_+--*--‘
-100 I I I I I T I T I I I I I I I I I fiI I I I I I I
20 7O 12'0

Bosis Functions per Wavelength

Figure 7.1.8 Input impedance of imaged monopole in tri-layered media versus number
of basis functions.

151

 

 

 

 

_30—
d mu Air film, 53::
‘ ...—...... R=2500, 83= 2.2-j0.0044)co
a . A :. —H—a—¢—e—+—-H—A
r“? ‘ "
-40-
E 2
CD
'0 —I
V - h/A=0.256, h/0=16.56
(f) - d=0.23(in), f=l4(GHz)
O .. ZL=5OQ. ¢=20°
m C!
_50—
:N : i A L
—60 I I I I I I I I I l I I I I I I I I I l I I I I I I
20 70 120

Bosis Functions per Wavelength

Figure 7.1.9 Radar cross section of imaged monopole in tri-layered media versus
number of basis functions.

152

“rt-era Air film, 83=8

”...... R=2son, c;=€2.2-j0.0044)eo

ID

A A A A A
a u u a i

1»

A A A A A A
i i u —' a ‘

 

h/A=O.256, h/o=16.56
d=0.23(in), f=o14(GHz)

Received Power (dBm)
I
01

 

rill:1111J1111111L11111Li11111111111441]

 

 

ZL=500. ¢=2O
_1 O X
—15 I I I Ifi I I I I I I I I I I I I I I I I I I I I I
20 7O 1 20
Bosis Functions per Wavelength
Figure 7.1.10 Received power of imaged monopole in tri-layered media versus

number of basis functions.

153

150—
d _— le All' fllm
" ""'"" le R=5000
"‘ ' — le R=2500
- - - - Rs... R=750

----- R.,,, Magnetic coating

Input Impedance (ohm)

 

vi
I
'd
p

- h=0.216(in), a=0.0185(in), d=0.23(in)
- EFF-‘0. fls=flo

OIIWTIIIIIIIIjIIIIIIIIIIII
8 12 16

 

 

Frequency (CH2)

Figure 7 .1.11 Input resistance of imaged monopole in tri-layered media with
foam substrate and different superstrates.

SO-

/'\
E
.C
0
v
a)
o
c
O
.0
Q)
O.
E
...:
D
O.
E
— 1 00
Figure 7.1.12

 

154

— X3", Air film
— X3," R=SOOQ
- — X... R=2500
- - — X.,,, R=750
----- X.,., Magnetic coating

 

h=O.216(in), a=0.0185(in), d=0.2.'5(in)
33:50: fis=flo

 

IIIIIIIT IIIIIIIIIITIIII

'1'2 16
Frequency (GHz)

Input reactance of imaged monopole in tri-layered media with
foam substrate and different superstrates.

155

 

 

1 50 e
Z — R... n=soon
d - — Rh. R=250Q
’ "' - Rim R=750
: ----- Rh, Magnetic coating
A A {I \\\
E —. I, \\
'8 100 ~ x
v ... ’1’ / \\
Q) I II, ,1 K \ \\
g d / f / \ \ \ \\
_I I \ \
.8 u , ”j \ \\\
Q) -4 / / \ \\
0' / II \ \ \\
E -‘ / /, \\ §X
_ _ / \
44 - / / l/ \ \
3 50 “l " \\\‘
Q. //I \
E d ” \ \\
2 \
: ,” {\“\\
,’ g\
"r/ ‘éhh:
I h=0.216(in), a=0.0185(in), d=0.23(in)
: a;=(2.2-j0.0044)eo. m=n.)
O I I I I I I I I I I I I I I I I I I I I I I I I I
8 12 1 6
Frequency (GHz)
Figure 7.1.13 Input resistance of imaged monopole in tri-layered media with

PTFE substrate and different superstrates.

156

 

 

50-
- -— X... R=5000
- — — X... R=2500
q - - - Xm, R=750
. ----- X,,,, Magnetic coating
.
A _ ,”’ \\\
E o -
8/ _I” / \ \\
" / “F ‘.
8 r " ‘x \ \ .\
C _I \ \ t
O _ \ g
8 - \ \\\
E - \ \‘~
_ _ \ .\
\ \
*5 ‘ x \
0. —50— \ \‘\\ z / '
E _ \\\\ ’/ /,/
—I / ’ / 1’
— \‘\\\\ .... I , ’ / 4’
I h=0.216(in), a=0.0185(in), d=0.23(in)
_ £3=(2.2—j0.0044)eo, p3=uo
-100 I I 1 I I I I I I I I I I I I I I I I [T I I I 1
8 12 16
Frequency (GHz)
Figure 7.1.14 Input reactance of imaged monopole in tri-layered media with

PTFE substrate and different superstrates.

157

 

 

 

“25" — 52..., Air film
I _' Rim R=SOOQ
‘ - — R... R=2500
‘ - - - Rs... R=750
d ----- RI... Magnetic coating
_I
_35_
A ..
E -r
(n .-
03 -
.0
v _
(n -I
o q
o: . .. _____
_45- ‘ ‘ - - -
‘ h=0.216(in), a=0.01°85 in), d=0.23(in)
- 83:80. us=uo. <o=20. 650(0)
_55 I I I I I I I I I I r I I I I I I I I I I I I I I
8 12 16
Frequency (GHz)
Figure 7.1.15 Radar cross section of imaged monopole in tri-layered media with

foam substrate and different superstrates.

158

 

 

 

15‘ — R... Air film
" — RI... R=5000
‘ - '— R3", R=2500
"1 - ‘- - le R=750
.. ----- R3... Magnetic coating
E a
m 5..
3 ..
L ..
a) .1
g ..
O _
0- .1
'0 _
Q)
.2 ‘
Q) —
s —5-
CK —.
i h=0.216(in), a=0.01851n). d=0.23(in) \
- 63:50. u3=um cp=20°, L=50(0) \
—15 I I I I I I I I FT I I I fI r I I I I I I I I I
8 12 16
Frequency (GHz)
Figure 7.1.16 Received power of imaged monopole in tri-layered media with

foam substrate and different superstrates.

RCS (stm)

l
U"
(I3

|
03
O
111

1

l l

1

-70

159

R=5000
R=2500
R=750
MogneUc coofing

 

h,

h=0.216(in), a=0.0185(in), d=o.23(in) "
£5: 2.2—10.0044)“. m=n.)
2.: 0(0). ¢=zo

 

 

8

Figure 7.1.17

IIIIIIIII IIIIIITIIIIIII

1'2' 16

Frequency (GHz)

Radar cross section of imaged monopole in tri-layered media with
PTFE substrate and different superstrates.

160

 

 

 

- —— R... R=5OOQ
"‘ \\‘ _ I"! R=2500
.\ x. - — R... R=7sn
J “. ----- R... Magnetic coating
- \
q
E _
CD —5-
3 .
L a
a)
3
0 _
a. —i
"O ..
Q)
.2 ‘
a) —
8 —15—
CE -1
q
I \
‘ h= 0.216(in). o=0.0185(in). d= O.25(in) \
I 85:22 .2- HOOO44 280. fl3=fla \
" ZL= 0(0)] ¢= -02 \
_25 I Ifij I I I I I I I I I I I I I I I I I I I I I
8 12 16
Frequency (GHz)
Figure 7.1.18 Received power of imaged monopole in tri-layered media with

PTFE substrate and different superstrates.

161

— Air film
— R=SOOQ
- —- R=2500
- - - R=750
----- Mognefic coofing

h=0.216(in), o=0.0185(in), d=o.25(in)
82:80, #2:”), f=15(CHz)

 

 

 

Figure 7.1.19 E-plane radiation pattern of imaged monopole in tri-layered media
with foam substrate and different superstrates.

162

— R=5OOD
- — R=2500
- — — R=750
----- Mognefic coofing

h=0.216(in), o=0.0185(in), d=0.25(ing
85=(2.2—j0.0044)80, [1.3:].10, f=15(GHZ

/fl§\\\ / fi\

~~~~~~~~ // ”---“;

i’ \\ / ‘
\\\ / I”

I. .\\\ /, //’,\

 

 

\ \ I
\ \ ,’ /// /
\ \ \\ /’ ‘// /
\ ~ I /
\ N. # "I- I
Figure 7.1.20 E—plane radiation pattern of imaged monopole in tri-layered media

with PTFE substrate and different superstrates.

163

 

 

 

2°“: i

 

Zee DETAIL A-A 15 cm
PLAN VIEW
‘“—- 0037' did
Silver point
(short circuit) i

   

0.216” monopole
l p

 

 

 

‘ $3.:.:.;.:.:...:...:.:.:.3.3:.j
°-“" “°
3 8 0.020’ aluminum
(3 (tips tapered)
...... >438
05““ A" our die rigid coaxial sgc‘r 3-3

Figure 7.1.21 Drawing of vacuum kayak measurement platform.

164

 

 

600.00 1
: — F2in
: ----- Xln
_q +++++ Rim Long's
-i AAAAA Xin. Long's
A "‘
E 400.00 j
.C A
O _
V —1
Q) I
o -1
C —1
O ‘1
'0 200.00 -*
(l) "i
Q ‘i
g a
_.
*5 1. \ +
O. :1 \\
E 0.00 — A X
: ‘x
\
... \
_. A ‘\ A
d \\ Al'-..
d \x A f””
‘ 2|=25cm, 2w=1cm a \“A """
- A
—200.00 Ilrfirirll[IIIIIIIIIIIIIIIIIIfi—l

 

450.00 550.00 650.00 750.00
Frequency (MHZ)

Figure 7.2.1 Input impedance of open slot antenna.

165

C500

 

200

+++++Rm Kominami
00000 X.,,, Kominomi

 
 

100

1111111111111114111111]

Input Impedance (ohm)

-1 9
J
j +
0—0 \ D .... ’
.1 \ n "‘
- \\ U .."’
_I \\ a a”"’pf
q
q
" 84r=12.8, W/|=0.02
-

 

 

—100 IIIIIIIIIIIIIIIIjIIIIIITIIIIIIIIITIIIII]

0.10 0.20 0.30 0.40 0.50
Normalized Length

Figure 7.2.2 Input impedance of slot on semi-infinite GaAs substrate.

166

   
     

 

 

 

400.00 —
1 — Rin
“ ----- Xin
1 +++++ Ran Kominami
a AAAAA X-m Kominami
A ...
E a
_C a
8, 200.00 -
.4
Q) —
o _
E . .
U ‘ ""’ I
(D “ ‘--’-’ I
E g . . . ': +
_ 1 + =
45 0.00 " - ’ :
“ l
O. u
E -* 1 ’4’".
-4 ‘ All
—4 \ A A ’1’
—¢ \\ Ali”
2 s..=2.55, tan6=0.002, w/l=0.02
—200.00 IflTTIIITIITTIIIFIIIITIIIIIIIIIIIIIIIII]
0.10 0.20 0.30 0.40 0.50

Nomalized Length

Figure 7.2.3 Input impedance of slot on semi-infinite PTFE substrate.

167

600

400

200

 

oir film
I 83:80. fl5=flo. d=1.5mm
‘. |=5.26mm, w-=O.268mm

\

Input Impedance (ohm)

111111111111111111111111111411111111111]

 

 

—200 IllllltltrrFrIIIIII]IIIIIIFrIIIIIIIIlilj

10 12 14 16 18
Frequency (CH2)

Figure 7.2.4 Input impedance of slot in tri—layered media with air film and foam
substrate.

168

O)

O

O
1

4s
0
O

 

i magnetic coating
\ m=n). u3=uo, d=1.5mm
\‘ l=5.36mm, w=0.268mm

Input Impedance (ohm)
N
o
o
[14111114111111llllLLllllJJJJlllllllll

 

 

-200 IIIIIIIIIIIIIIIIITIITIIITTIIIIIIllIIIle

1O 12 14 16 18
Frequency (GHz)

Figure 7.2.5 Input impedance of slot in tri-layered media with magnetic coating and
foam substrate.

169

N (A

O O

O O
1

Input Impedance (ohm)
5
o

 

resistive sheet x ,,,,,,
83:80. Lia-1M0. d=l.5mm ""’
|=5.36mm. w=0.268mm

ii1111111114111111111111111111111111111

 

 

—1OO IIIIIIIIIIIIIIITIIIIIIIIIIITT'IIIIIIIIII

1O 12 14 16 18
Frequency (GHz)

Figure 7.2.6 Input impedance of slot in tri-layered media with resistive sheet and foam
substrate.

170

I\) (N
O O
O O

1

Input Impedance (ohm)
5
o

 

111111111lllllIllllllJllllllllllllLLLLl

 

 

0
resistive sheet
83,-2.2. tan6=0.002. mayo
d=1.5mm, l=5.36mm, w=0.268mm
—1OO IIIIIIIIIIIIIIIIIIIITTIIIIIIIIIIIIIIIII]
IO 12 14 16 18

Frequency (GHz)

Figure 7.2.7 Input impedance of slot in tri-layered media with resistive sheet and PTFE
substrate.

171

(14

O

O
I

200

 

Input Impedance (ohm)
S
o

 

Resistive sheet
car=12.9. tan6=0.002. u3=ua
d=1.5mm, I=5.36mm, w=0.268mm

—1OO IIIIIIIIIIIIIIIIIIIIIIIIIIIIjIIIIIrTIIIl

10 12 14 16 18
Frequency (GHZ)

LllllIIIJIIIlllllllllllllllllllLlllllll

 

 

Figure 7.2.8 Input impedance of slot in tri-layered media with resistive sheet and GaAs
substrate.

172

 

 

 

-30-
: ____.Ak Hun
_ .. _ _ Resistive Sheet
- ....... Magnehc Coahng
,\-40:
E a
8 "I
U -I
v -I
(n I
C) -
m e-I
-502
: 83=80, fl3=fitm d=1.5mm
d l=5..'56mm, w=0.268mm
_60 IIIIIIITITTTTIIIIIIIHIIIIITrjIIIIIIIII—l
10 12 14 16 18

Frequency (GHz)

Figure 7.2.9 Radar cross section of slot in tri-layered media with foam substrate and
different superstrates.

173

 

 

 

20 —
I __ Air Film
- _ _. _ Resistive Sheet
0 ..... Mognefic Coofing
A _
0% :
3 10 :
L _
a) ..
g d, /
0 q /
a. _, ’
.0 -
Q) —l
.2 -
(D _.
o 0 q
a)
m —l
: 83=8m Ms=llol d=1.5mm
- l=5.36mm, w=0.268mm
-1O TIIIIIIIWIIIIIIIIII]IIIIIIIIIIIIIIIIIII]
10 I2 14 16 18
Frequency (GHz)
Figure 7.2.10 Received power of slot in tri-layered media with foam substrate

and different superstrates.

174

 

 

 

-45-
I ag,=1.0, tan6=0.0. m=n.
u _ _ _ 53512.9, tan6=0.002, [13:11.0
.. ..... Cy=2.2, t0n6=0.002, #3:”
A —55:
E ... ........
m - ~~~~~
CD ““““
'0 " ....
v -l IIIII
m : ‘~~.\
o q ~~~~~~~
01 _ ~~~~~
—65- . .
- ReSlstlve sheet, d=1.5mm
~ l=5.36mm, w=0.268mm
:
_75 IIrIIIIIIIWIIIIIIIIIIIIIIIIIITII—FIIIIIh
10 1 2 14 16 18
Frequency (GHZ)
Figure 7.2.11 Radar cross section of slot in tri-layered media with resistive sheet

and different substrates.

175

 

 

 

 

10 1
-I \\
A -
.g : ------------------------
3 0 j ................
L _
<1) 4
B 1 8351.0, tan6=0.0. Ila=flo
0? ~ _ _ _ a;,=12.9, tan6=0.002, pal-mo
a ..... c;,=2.2, tand=0.002, #Fflo
.0 .
Q) -I
.2 a ——————————————— s \
<1) _ \
8 10:I \ \
c: a \ \
'I \
" \
.. \
- Resistive sheet. d=1.5mm \
- |=5.36mm, w=0.268mm \ q
—20 IIIlIIII—IIIIIIIIIITITIIIIIIIIITTIIIIIII]
10 1 2 I 4 16 18
Frequency (GHz)
Figure 7.2.12 Received power of slot in tri-layered media with resistive sheet and

different superstrates.

176

—— Air Film
----- Magnehc Coafing
— - — Resistive Sheet

83:80, 1.1.3:”0, d=1.5mm
I=5.36mm. w=0.268mm

 

 

 

Figure 7.2.13 B—plane radiation pattern of slot in tri-Iayered media with foam
substrate and different superstrates.

177

—— Air Film
----- Magnefic Coafing
- — — Resistive Sheet

83:80, “3:00, d=1.5mm
I=5.36mm, w=0.268mm

 

 

 

 

Figure 7.2.14 H-plane radiation pattern of slot in tri-layered media with foam
substrate and different superstrates.

178

 

83r=1°0v tan6=0.0
_ _. _ £3,529, tan6=0.002
..... 85'22.2, t0n6=0.002

Resistive sheet
d=1.5mm, l/w=20

 

 

 

 

Figure 7 .2.15 E-plane radiation pattern of slot in tri-Iayered media with resistive
sheet and different substrates.

179

 

83":1'0! ton6=o'o
_ _ _. £3,=12.9, tan6=0.002
..... 85r=2°2' t0n6=0.002

Resistive sheet
d=1.5mm, I/w=20

 

 

 

Figure 7.2.16 H-plane radiation pattern of slot in tri-layered media with resistive
sheet and different substrates.

 

CHAPTER EIGHT
CONCLUSIONS

The scattering and receiving characteristics of imaged monopoles and slots in tri-
layered media have been investigated in this dissertation. Emphasis is placed on the
effects of lossy superstrates on the scattering and receiving characteristics. Basic elec-
tromagnetic parameters of monopoles and slots, such as input impedance, radiation pat-
tern, radar cross section, and received power, have been studied by the full-wave
integral equation approach.

Electric and magnetic Hertzian potentials have been used to facilitate the deriva-
tion of electric and magnetic dyadic Green’s functions in tri-layered media. The dyadic
Green’s functions for electric Hertzian potential, magnetic Hertzian potential, electric
field, and magnetic field in tri-layered media have been derived and expressed in terms
of Sommerfeld integrals. An electric field integral equation (EFIE) and a magnetic
field integral equation (MFIE) are converted to Hallen-type integral equations (HTIE)
and the HTIEs are solved by the method of moments to obtain unknown electric and

equivalent magnetic currents.

The existence of a lossy superstrate shifts all the surface wave poles of Sommer-
feld integrals off the real axis of the complex l—plane. This fact makes it possible to
evaluate the impedance and admittance matrix entries via real axis spectral integration.
The stationary phase method is used to compute the scattered far field.

Two representative antennas, an imaged vertical monopole and a narrow rectangu-
lar slot, in tri-layered media have been investigated numerically. The results are com-
pared with published data whenever possible. In the case of a monopole shorted to the
ground plane in tri-layered media, theoretical results are compared with experimental

ones. The numerical results demonstrate that, for an antenna in tri-layered media with

180

 

181

a lossy superstrate, the reduction in radar cross section is greater than the reduction in
received power. The theory developed in this research can aid in the design of anten-

nas with good transmitting and receiving capabilities and low radar cross sections.

In the case of a slot in iii-layered media, it is a very demanding computational
task to fill the admittance matrix. Further research is needed to find efficient and robust

analytical and numerical techniques for the evaluation of admittance matrix elements.

In most applications, another ground plane or a cavity is placed under the slot to
make it unidirectional and to provide more practical feeding mechanisms. The current
theory can be extended to analyze a cavity backed or microstrip fed slot. The kernel of
the integral equation for such an antenna system will be even more complicated. The

challenge is find efficient matrix filling methods to keep the computer cost in check.

 

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"Illlllllllllllllllllllli