MflfliFECs'ATIGN 0F
BAKIMSCA‘ITERING L‘F A LQDF’ BY [W’EDANCE LOADING

Thesis for the Degree 0? Ph. D.
{VHCHEGAH STiiiTE UNE“a~‘ERS£‘£‘Y
fiUANG‘LU LlJé

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LIBRAR Y
Michigan C in
University

chblS

This is to certify that the

thesis entitled

Modification of Backseattering

of a Loop by Impedance Loading

presented by
J uan g-Lu Lin
has been accepted towards fulfillment

, ‘ of the requirements {of

Eh, D. degree in___E.L

MM C4

/ Major professor

Date May ‘3; 1957

0-169

ABS TRAC T

MODIFICATION OF BACKSCATTERING OF
A LOOP BY IMPEDANCE LOADING

by Juang- Lu Lin

In this thesis the modification of the backs cattering of a loop
by the impedance loading method is investigated theoretically and
experimentally.

First of all, the resonant phenomena of a loop when illuminated
by an electromagnetic wave is studied. Next, the backscattering of
a circular conducting loop is considered and the impedance loading
method is applied to minimize its backscatter. Finally, the mini-
mization of the backscattering of a rectangular loop is investigated.

In the theoretical analysis, a new method based on a differential
equation for the loop current is developed for the case of a circular
loop. For the case of a rectangular loop an existing method is applied
with slight modification. Throughout the entire analysis, the principle
of superposition is applied to simplify the problem of an illuminated
loaded loop to the combination of an illuminated solid loop and a
radiating loop. The major objectives of the theoretical analysis are
to determine the induced current on a loaded loop, the scattered field
from a loaded loop and the optimum impedance which, when loaded on
the loop, makes the backscatter of the loop vanish.

Extensive experimental study was also conducted. The resonant
phenomena of a solid loop was investigated carefully. The impedance

loading technique was then applied to the circular and the rectangular

JUANG- LU LIN

loops to minimize their backscatters. In the course of this experiment,
the backscatter of a loop was successfully minimized to the noise level.
The major finding of this study is to show theoretically that the
backscatter of a conducting 100p can be eliminated by a properly chosen
impedance loading, and to verify experimentally that indeed the back-
scatter of a loop can be minimized by a practical arrangement of the
impedance loading technique. This investigation should prove
significant in the understanding of the scattering phenomena of a

conducting loop and in the practice of radar camouflage.

MODIFICATION OF BACKSCATTERING OF

A LOOP BY IMPEDANCE LOADING

BY

Juang- Lu Lin

A THESIS

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

DOC TO R OF PHILOSOPHY

Department of Electrical Engineering

1967

ACKNOWLEDGMENT

The author is grateful to his major professor Dr. K. M.
Chen for his guidance and encouragement in the course of this
research. He also wishes to thank the members of his guidance
committee, Dr. J. A. Strelzoff, Dr. L. W. Von Tersch, Dr. Y.
Tokad and Dr. N. Hills, for reading the thesis and valuable
suggestions. The help extended by J. W. Hoffman of the Division
of Engineering Research is also appreciated. The research
reported in this thesis was supported by the Air Force Cambridge

Research Laboratories under contract AFl9(628)- 5732.

ii

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List of Fiat: res

Int 1‘ 0d 11 ct i on

Cha pter I.

1.1.
1.2..

1.3.

Back

smattering of a Solid Loop

Definiticm of Baclascattoring Cross Section

1. .3. 2..
1. (Z. .5.
l. 20%.

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.

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Description of the experinlental
arrangement and“ the nieamlring teclmique
Measurotnent of the backsca ttering cross
section of a circular loop

Measurertmgnt of the backscattering cross
section of LLiLlé’il‘éi loops

Med:-sii.ren'1er1t of the resonance of loops

Beeliscattering le‘OITl a solid circular loop

Geon'ietry of problem

Differential equation for loop cu1 rent
Solution. for loop current

Back :5 catte ring c ros s s e (.ti on
Dots:rrnination of Q S and 13:5

N 1.1.!"HCI‘l cal results

Scattering from a. solid rectangular loop

Formuldtlou of problem

Integr'll equations for loop our “outs;

Zex'otla (studio: :: J.lt.1'ti<.>ns for loop current:

Eiret mailer solutions for loop currents;

Evoluatl ,4 if: ‘;":;:rti<;‘._v..l>:zr int: gra l :1 141 (y)

and 1'.) 1()
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vi

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L)?

TABLE OF CONTENTS (continued)

Page
Chapter II. Minimization of Backscattering of a Circular

Loop 7 R)

Z. 1. Introduction Yo‘

2. 2. Theory 79

2. 2.. l. Scattering from a solid circular loop 81

Z. 2. 2.. Radiation from a circular loop 81

(A) Differential equation for loop current 82’

(B) Solution for loop current 8&7

2. 2. 3. Total current and total back scattered field 88
2. 2.4. Optimum impedance for zero back

scattering r r 89

2. 2. 5. Determination of Q and (bi 90

2. 2. 6. Numerical examples 93

2. 3. Experiment 97
Z. 3. 1. Experimental setup and measuring

technique 97

Z. 3. 2. Experimental results 99

2.4. Comparison Between Theory and Experiment 100

Chapter III. Minimization of Backscattering of a Loaded

Rectangular Loop 118
3. 1. Introduction 1 15
3. 2.. Theory 118
3. 2. l. Scattering from a solid rectangular loop 11‘)
3. Z. 2. Radiation from a rectangular loop 1.3.3}.
(A) Geoinetr'y of Problem .1?»
(B) Integral equations for loop currents 1.33
(C) Zeroth order solutions for loop currents l .31
(D) First order solutions for loop currents 13‘;
(E) Evaluation of particular integrals Gil?)
V - ,
and 91“(x) l—l r
(F) Expansion parameter, c) r 12/
3. 3. Total Loop Currents 1365'

iv

TABLE OF CONTENTS (concluded)

Total Backseatte red Field

3.4.
3. 4. 1. Total backscattered field based on first
order currents
3.4. 2.. Total backscattered field based on
sub-first order currents
3.4. 3. Evaluation of K1 and K2
3. 5. Optimum Impedance for Zero Backscattering
3. 5. 1. First order optimum impedance for
zero backscattering
3. 5. Z. Sub-first order optimmn impedance for
zero backscattering
3. 6. Numerical Examples
3. 7. Experiment
3. 7. 1. Experimental arrangement and
measuring technique
3. 7. 2. Experimental results
References

Page

162.

162

165
167

169

169
170
170

172

172
172

187

1.13.

LIST OF FIGURES

Experimental Setup and Circuit Diagram

ACircular Loop if Illuminated by a Plane Wave
at either Normal or Horizontal Incidence

Backscattering Cross Section of Circular Loops
Illuminated by a Plane Wave at Normal Incidence

Backs cattering Cross Section of CircularLoops
Illuminated by a Plane Wave at Horizontal

Incidence

Backscattering Cross Section of Square Loops
Illuminated by a Plane Wave at Normal Incidence

Backscattering Cross Section of Square Loops
Illuminated by a Plane Wave at Horizontal

Incidence

Scattering Cross Sections of Loops as Functions
of Loop Sizes (First Resonance Curves)

Scattering Cross Sections of Loops as Functions
of Loop Sizes (Second Resonance Curves)

Geometry of Solid Circular Loop

Geometry for the Calculation of Radiation
Field of a Solid Circular Loop

2
s , . a _
cbq as a Function of [30b (g2- - 0. 00179)

2
s . a __
(Di as a Function of [30b ( E2- — 0.00179)

Comparison Between Theory and Experiment for
Circular Loops Illuminated by Plane Wave at
Normal Incidence

Geometry for Solid Rectangular Loop

Geometry for the Calculation of Radiation Field
of a Solid Rectangular Loop

Vi

Page

10

13

15

17

19

26

31

32

33

35

69

Figure

2. 8.

2.10.

LIST OF FIGURES (continued)

Comparison Between Theory and Experiment .
for a Square Loop Illuminated by a Plane Wave
at Normal Incidence

A Loaded Loop is the Sum of a Scattering Loop
and a Radiating Loop

Geometry of a Radiating Loop

= 0. 00179)

0““ mN

Q r as a Function of Bob (

Z
r . a _
(Pi as a Function of Bob (—zb - 0.00179)

Optimum Impedance for Zero Backscattering

2
(32 = 0.00179)
b

Experimental Setup and Circuit Diagram

Backscattering Cross Section of a Loaded
Loop as a Function of Loading Impedance
(f = 1. 61 GC)

Backscattering Cross Section of a Loaded
Loop as a Function of Loading Impedance
(f = l. 71 GC)

Backscattering Cross Section of a Loaded
Loop as a Function of Loading Impedance
(f = 1. 75 GC)

Backscattering Cross Section of a Loaded
Loop as a Function of Loading Impedance
(f = l. 80 GC)

Backscattering Cross Section of a Loaded
Loop as a Function of Loading Impedance
(f = l. 85 GC)

vii

Page

72

.80

82

94

95

96

98

102

103

104

105

106

Figure

2011.

2.12.

2.14.

2.15.

2.16.

2.17.

2.18.

2.19.

LIST OF FIGURES (continued)

Backscattering Cross Section of a Loaded
Loop as a Function of Loading Impedance
(f = 1. 90 CC)

Backscatte ring Cross Section of a Loaded
Loop as a Function of Loading Impedance
(f = l. 95 GC)

Backscattering Cross Section of a Loaded
Loop as a Function of Loading Impedance
(f = 2. 00 GC)

Backscattering Cross Section of a Loaded
Loop as a Function of Loading Impedance
(f = 2. 05 GC)

Backscattering Cross Section of a Loaded
Loop as a Function of Loading Impedance
(f = 2. 10 GC)

Backscatte ring Cross Section of a Loaded
Loop as a Function of Loading Impedance
(f = 2. 15 GC)

Backscattering Cross Section of a Loaded
Loop as a Function of Loading Impedance
(f = 2. 20 GC)

Backscattering Cross Section of a Loaded
Loop as a Function of Loading Impedance
(f = 2. 30 GC)

Backscattering Cross Section of a Loaded
Loop as a Function of Loading Impedance
(f = 2.40 GC)

Backscattering Cross Section of a Loaded
Loop as a Function of Loading Impedance
(f = 2. 50 GC)

Optimum Impedance for Minimum
Backscattering from a Circular Loop
Illuminated by a Plane Wave at Normal
Incidence

viii

Page

107

108

109

110

111

112

113

114

115

116

117

Figure

3. 7.

3. 8.

3. 9.

3.10.

3.11.

3.12.

3.13.

LIST OF FIGURES (continued)

A Loaded Loop is the Sum of a Scattering Loop
and a Radiating Loop

Geometry of a Radiating Loop

Geometry for the Calculation of the Radiation
Field of a Radiating Rectangular Loop

. . a _
or as a Function of (30h (h - 0.0388)

Optimum Impedance for Zero Backscattering
a —
(h — O. 0388)

Backscattering Cross Section of a Loaded Square
Loop as a Function of Loading Impedance
(f = 1. 61 GC)

Backscattering Cross Section of a Loaded Square
Loop as a Function of Loading Impedance
(f = 1. 7 GC)

Backscattering Cross Section of a Loaded Square
Loop as a Function of Loading Impedance
(f = 1. 8 GC)

Backscattering Cross Section of a Loaded Square
Loop as a Function of Loading Impedance
(f = l. 9 CC)

Backscattering Cross Section of a Loaded Square
Loop as a Function of Loading Impedance
(f = 2 GC)

Backscattering Cross Section of a Loaded Square
Loop as a Function of Loading Impedance
(f = 2. 1 GC)

Backscattering Cross Section of a Loaded Square
Loop as a Function of Loading Impedance
(f = 2. 2 GC)

Backscattering Cross Section of a Loaded S uare‘

Loop as a Function of Loading Impedance
(f = 2. 3 GC)

ix

Page

120

122

163

171

173

175

176

177

178

179

180

181

182

3.15.

3.16.

3.17.

LIST OF FIGURES (concluded)

Page
Backscattering Cross Section of a Loaded Square
Loop as a Function of Loading Impedance
(f = 2.4 GC) 183
Backscattering Cross Section of a Loaded Square
Loop as a Function of Loading Impedance
(f = 2. 5 GC) 184
Backscattering Cross Section of a Loaded Square
Loop as a Function of Loading Impedance
(f = 2. 55 GC) 185
Optimmn Reactive Impedance for Minimum
Backscattering from a Square Loop 186

INTRODUCTION

In recent years much research has been conducted on reducing
the radar cross sections of metallic objects. The conventional
techniques include the use of radar absorbing material and the method
of reshaping the body. Recently the impedance loading method has been
found to be especially effective in reducing the radar cross sections of
metallic objects with dimensions of the order of a wavelength. In this
thesis, the minimization of the backscatte ring of a conducting loop, both
circular and rectangular, by an impedance loading method is investigated.

Historically, Kouyournjianland Weston2 are the first ones who
investigated the backscattering cross sections of a thin solid circular
loop with a result of fairly good agreement with the experiment. The
methods employed by them are quite complicated and involved tedious
computation. For our purpose, a mathematically simpler method has
been developed. Our theoretical results also agree with experimental
results quite satisfactorily. For the case of the solid rectangular loop,
the method used by Chen and King3 was adopted in obtaining first-order
solution for current distribution. Experimental results and those of
theory appear to be in good agreement.

In solving the problem, the principle of superposition is
employed. Based on the superposition principle, an illuminated loaded
loop can be considered as the combination of an illuminated solid loop
and a radiating loop. In this study, the two cases are solved separately
and then combined together to yield the final solution for an illmninated

loop.

For the minimization of the backscattering of a circular loop,
a consideration is given to the perfectly conducting loop which is loaded
symmetrically with two identical lumped impedances and is assumed to
be illuminated by a plane wave at normal incidence. The induced current
on the loaded loop is determined as a function of loop dimensions and
loading impedance. The backscattered field produced by the loaded
circular loop is calculated as a function of the loading impedance among
other parameters. It is then possible to find an optimum impedance which
makes the back scattered field equal to zero. An explicit expression
for the optimum impedance for zero backscattering is obtained as a
function of loop dimensions. Some numerical examples are included.
The theory is later verified by an experiment.

A theory on the minimization of the backscatter of a conducting
rectangular loop by the impedance loading method is developed in the
last chapter. A rectangular loop loaded symmetrically with two identical
lumped. impedances at the centers of the long sides of the loop is assumed
to be illuminated by a plane wave at normal incidence. The zeroth order
and the first-order solutions of the induced current are evaluated as
functions of the loop dimensions and the loading impedance. Based on
the induced current, the backscattered field can be calculated. An
optimum impedance which leads to zero backscattering from the rectangular
loop is then found as a function of loop dimensions. Some numerical
examples are included. An experiment was performed for the case of
a square loop loaded with two identical reactive impedances and
illuminated by a plane wave with vertical polarization at normal incidence.
The experimental results indicate that it is possible to reduce the back-
scatter of a conducting, square loop to the noise level if the loading

impedance is properly chosen.

CHAPTER I

BACKSCATTERING OF A SOLID LOOP

1.1. Definition of Backscattering Cross Section

An electromagnetic wave incident upon a metallic object will
induce time-varying distributions of oscillating charges and currents
in the object. The induced charges and currents will, in turn, main-
tain an electromagnetic field which is known as the scattered field.
To characterize the reradiating prOperties of an object, total scattering
cross section a is defined as
ps

1

S

0‘ :

Where PS is the total scattered power and S.1 is the scalar magnitude

of the real part of the incident complex Poynting vector at the location
of the object. In radar, transmitter and receiver are installed at the
same location so that only the power scattered by the object in the
direction of the transmitter-receiver is observed. In this sense,
backscattering cross section of an object is defined as a measure to
characterize the quantity of the power scattered back toward the source
by the object. In symbol, the backscattering cross section of an object

is defined as

IESIZ
0B: lim 41rR2 ——
R—wo 11312

where ES is the magnitude of the backscattered field and E1 is that

of the incident electric field.

1. 2. Experiment

A number of methods are available for measuring the back-
scattering cross section of an object. In this research the cancellation
method was used. The prinicple of this method is to cancel the signal
received by the receiving antenna when the scatterer is absent. After
this the scatterer is put in place. The signal received by the receiving
antenna with the scatterer present will then represent the scattered
field by the scatterer.

In this section experimental arrangement and the measuring
technique of the backscattering cross section of rectangular and circular
lOOpS are discussed. The 100ps are illuminated by a plane wave for
the cases of normal and horizontal incidence. Resonant phenomena of
loops are also investigated.

1. 2.1. Description of the Experimental Arrangement and the

Measuring Technique

The experimental setup is shown in Figure 1.1. The experiment
was conducted inside of an anechoic chamber (0. 8mxl, 4mx0. 7m insize)
which is firmly mounted on a wooden frame. A horn antenna (HP
X890A) is projected into the anechoic chamber through one of the
narrow sides of the chamber. A cellular plastic column made of
styrofoam is used as scatterer support. The support is placed in the
far zone of the horn antenna and its height is so chosen that the
scatterer on the support is on the axis of the antenna. The circuit
diagram of the experiment is also shown in Figure 1.1. The klystron
generator (F X R type X760A) modulated by a 1 KC square wave
generator (HP 715A) is used as the R. F. source. The isolator

(Polytechnic Res. and Dev. Co. type 1203) is used to protect the possible

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

SWR

 

 

Figure 1.1.

 

 

 

 

 

 

 

 

 

 

 

 

meter

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Termination

 

 

 

 

 

 

 

 

 

 

Circuit Diagram

absorber horn antenna scatterer
f
scatter :1
support
support 0
> absorber
> horn
antenna
wooden frame
Experimental Setup
horn antenna scatterer
E:I::l support
/./'7/'7///////////////////
Hybrid < ~— 4 Isolator
T
V slotted variable frequenc 1
section attenuator meter \
detector klystron
‘_.__._ E-H \
t Tuner 1 KC
, Mod.

Experimental setup and circuit diagram.

6

backward energy from damaging the R. F. generator. Frequency is
measured by a frequency meter (HP X532A). A variable attenuator
(HP X375A) and slotted section (HP X810B) are used to control and
probe the wave from the source. A hybrid T (HP X845A) is used to
separate the incident wave to the antenna and the reflected wave back
from the antenna. Two other terminals of the Hybrid T are connected
to a matched load through an E-H tuner (HP X880A) and to a detector.
The output of the detector is then measured by a SWR meter (HP 415B).
The horn antenna in this experiment serves both as the radiating
element and the receiving probe. In the experiment, a series of

measurements have been made with various sizes of loops,

By the arrangement mentioned above, a plane wave with vertical
polarization can be made to illuminate the loop either normally or
horizontally. When the scatterer is absent, the reading of the SWR
meter can be set to zero by adjusting the E—H tuner. After this
balancing process is completed, the scatterer is introduced. The
reading of the SWR meter will then indicated the back scattered field
by the scatterer.

l. 2. 2. Measurement of the Backscattering Cross Section of

Circular L00ps

To measure the backscattering cross section of metallic
circular loops, a total of 35 circular loops with radius ranging from
0. 3 cm to 3. 5 cm were constructed as experimental models. The
diameter of the wire is 0.1 cm. The frequency of 9. 61 CC was used.
The 100ps are mounted upright on the support which is located 26 cm
(-- 8 A) away from the horn antenna. The loops are, therefore,

considered to be in the far zone of the radiating element. Theoretically,

7

the distance between the antenna and the scatterer should be made as
great as possible. However, to obtain a detectable scattered field
from the scatterer one has to compromise for a finite distance. The
radiated wave from the horn antenna at the location of the scatterer
is then approximately a plane wave with a vertical E field.

In the experiment the plane of the loop is first oriented
parallel to the horn aperture so that the loop is illuminated by a
plane wave with vertical polarization at normal incidence. For the
case of horizontal incidence, the loop is turned 900 from the vertical
incidence position making its plane perpendicular to the horn aperture.
The geometries of these two cases are shown in Figure 1. Z. In Figure
1. 3 and l. 4, relative backscattering cross section of the circular
loops 0' are plotted as functions of 100p dimension Bob where Bo is
the wave number and b is the radius of the loop. Figure 1. 3 shows
the backscattering cross section of the circular loop illuminated by
a plane wave at normal incidence and Figure 1. 4 shows that of
horizontal incidence case. It is observed from these two figures
that the first resonances for both cases occur when the half of the
loop periphery is near ~12- wavelength (1. e., when Bob 2 l). The
first resonant peak in the case of normal incidence appears to be
higher than that of the horizontal incidence case.
1. Z. 3. Measurement of the Backscattering Cross Section of

Square Loops

The same technique used for the circular 100ps was adopted
to measure the backscattering cross sections of the square 100ps. A
number of square lOOpS ranging from O. 5 cm x O. 5 cm to 3. 5 cm x
3. 5 cm in size were constructed as experimental models. The

diameter of the metallic wire is O. 06 cm. The experiment was

2a

 

 

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Normal Incidence

:EL

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Horizontal Incidence

 

Figure 1. 2.. A circular loop is illuminated by a plane wave
at either normal or horizontal incidence.

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11
conducted with the same setup described in previous sections and at
the same frequency.

The backscattering cross sections of square 100ps are plotted
as functions of 100p diInensions [30h (where 2h = one side of the square
100p) in Figures 1. 5 and 1. 6. Figure 1. 5 shows the backscattering
cross section of square 100ps illuminated by the plane wave at normal
incidence, and Figure 1. 6 shows the backscattering cross section for
the case of horizontal incidence.

Once again it is observed that the first resonances of both
normal and horizontal cases occur when half of the loop periphery is

near %wavelength (i. e. , when {30h : O. 8).

l. 2.4. Measurement of the Resonances of L00ps

The backscattering cross sections of a circular and a square
100p described in the last two sections have a common nature that the
first resonance tends to occur when the half of the 100p periphery is
near é—wavelength and the second resonance at E wavelength and so
on. At these resonant conditions the induced currents on the 100ps
are maximum and likewise the scattered fields. However, it is
anticipated that the 100ps with the same periphery but different
geometrical shapes may have different resonant sizes. T0 testify
these conjectures five kinds of rectangular 100ps were constructed.
The five kinds of rectangular 100ps have the ratio of hl/hZ equal to

1, l. 5, 2. 0, 2. 5, and 3. 0 where hl and h are the short and the long

2
side of the loop. The loops are illuminated by a plane wave with
vertical polarization at normal incidence. The experiment . was

conducted inside of an anechoic chamber at the frequency of 9. 61

CC with the same experimental setup mentioned previously.

lZ

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14

The experimental results are shown in Figures 1.7 and 1. 8.
Figure l. 7 shows the first resonance curves of the 100ps. The resonant
curve of a straight wire which corresponds to a loop with hl/hz = O
is also included in the figure. The scattering cross sections of the 100ps
are plotted as functions of the loop sizes. From the peak of the curve
the resonant size of the 100p is determined.

It is observed that the first resonances of the loops with various
shapes indeed occur when the half peripheries of loops are near :12-
wavelength. It is also seen that the first resonant size of a 100p is
dependent of its geometrical shape. These phenomena are summar-
ized in Table 1.1. It is true that the first resonant size of a loop tends

to increase as the loop becomes broader or the ratio of hl/hZ becomes

 

 

larger.
Table 1.1.
first resonant
Loop Geometry size*(17 periphery)
straight wire (bl/h2 = 0) 0. 433).
rectangular 100p (hi/hZ : ~13) 0. 513K
rectangular loop (hl/h2 :: T1125.) 0. 516K
rectangular loop (bl/h2 : -12-) O. 53 A
rectangular loop (hi/hZ : -l-1—5) 0. 561K
rectangular loop (hl/h2 = -i—) 0.61 A
circular 100p 0. 705K

 

 

 

 

For the wire é-periphery corresponds to the total length
For rectangular loop? periphery : 2(hl+h2)
For circular loop-E5 periphery : Tl'b where b : radius

Radius of wire of which the 100ps are made of = a = 0. 01).

15

 

loop sizes (first resonance curves).

 

 

 

10

5, u...
o
(I)
f:
o
‘r—'
.4
E)
0‘ 3
O M-A
(D
co
,q m
.3
go 8*
0 3
H '~H
a; 0
“5 m
3 5
I: .E g
o > .
a 33
Q)
,1: m
0. m
~H
a 2
G)
e 3.
n.
o .S
o H
"* 8
N a
“3 \ 3
F4
0 U)
[Q
.2
a)
H
:5
on
LA
M
—1 o
o
l 1 1 I
oo ‘0 7" N
uonoes ssoxo 81111911203 GAIJ‘BIGJ
v 10 '0 :: ext/n ;o snipe: (2/1 : zq/Iq) dooI xemfiueiosx

dooI xetnoxto 1. (g '2/1 : Zq/Iq) dooI .t'ernSuep ex

:: zq/Iq) dooI 19111311191091 9 («g/I : Zq/Iq) dooI .t'eInBueioeJ

: zq/Iq) d00[ 12111311211391 9 (0 : 2"q/Iq) 9.1 pm qqfiiexas

16
The second resonance curves of the loops are shown inFigure
1. 8. It is observed that the second resonance peaks are not clearly
observable in the loops with large ratio of hl/hZ' The second
resonances of the loops occur when the half peripheries of loops are
in the neighborhood of g- wavelength. The second resonant sizes

2.

of the loops are summarized in Table 1. 2.

Table 1. Z.

 

L00p Geometry Second resonant Size

(é— periphery)

 

straight wire (hi/hZ = O) 1. 44k
rectangular loop (hi/hZ 2 é—) 1. 67k
rectangular 100p (hl/hZ : 1.125) 1. 68k
rectangular 100p (hl/h2 : «9 1. 58k

rectangular loop (hi/hZ : T17? ) not clear
1
rectangular 100p (,hl/h2 = T) 1. 73K

circular 100p not clear

 

 

 

1. 3. Theory

A theory is presented here to calculate the backscattering
cross section of a solid circular 100p and a solid rectangular loop
when they are illuminated by a plane wave with vertical polarization
at normal incidence.

Some numerical results are obtained and are compared with

the experimental results in Section 1. 2.

relative Scattering cross section

28

24

4 rectangular loop (hi/hZ

 

 

17
1 straight\vire (bl/h2 = 0)

1/3)

1/2.5) / \\\
1/2.0) // \\\\

1/1.5) ///

2 rectangular loop (hi/112

3 rectangular loop (hl/h2

ll

5 rectangular loop (hi/bl

6 rectangular loop (hl/hz = 1/1) /
7 cicular100p /
/
radius of Wire 7: 0.01 it /

 

1.0 1.3 1.5 1.7 (t)

1/2 loop periphery in wavelength

Figure l. 8. Scattering cross sections of loops as functions
of loop Sizes (second resonance curves)

18
1. 3.1. Backscattering from a Solid Circular Loop

The backscattering cross section of a circular metallic loop
was first explored by Kouyoumjianl who used a variational method to
determine approximate formulas for the backscattering cross sections
of thin wire 100ps. Numerically he was able to calculate the back-
scattering cross sections of a circular loop illuminated by a plane wave
at either normal or horizontal incidence. A good agreement between
the theory and the experiment was obtained in this study. The weakness
of this method is the tedious computation. Later Weston2 investigated
the same case by solving the wave equation in toroidal coordinates to
obtain general expressions for the electromagnetic field. Numerical
values for an arbitrary incident angle were determined and his analysis
is deemed as a generalization of earlier work by Kouyoumjian.

These two methods are too complicated to serve our purpose
of solving the case of a loaded circular 100p developed in the next
chapter. A mathematically simpler method has been developed in

this research.

(A) Geometry of Problem

The geometry of the problem is as shown in Figure 1. 9.
A l00p of radius b is assumed to be made of perfectly conducting
metallic wire of radius a. A plane electromagnetic wave with the E
field parallel to the plane of the l00p is incident normally upon the

loop. The dimensions of interest are

2
a<<b2, [50a <<l

where 50 is the wave number. We assume that the wire is thin

enough 'so that only the 6 component of current is induced.

19

 

 

 

 

 

 

Figure 1. 9. Geometry of solid circular loop.

20

(B) Differential Equation for Loop Current
Since the tangential electric field should vanish at the surface

of the loop, the following equation is valid:

(1.1)

M
:1

E +E =0 for-n59

where E: is the tangential electric field maintained by the current and

the charge on the loop, E l is the tangential electric field of the incident

1:

wave.

The symmetry of the geometry gives

193 (e) = - 16S (1r - e)

(1.2)
8 7T
16 #2)“)

Where 18(0) is the induced current in the loop. The tangential component

0f the incident E field along the loop is
E 2 EO cose (1-3)

Whe re E0 is a constant electric field.
The current and the charge on the 100p maintain a tangential

l . . .
e ectric field at the surface which can be expressed as

Eta: -(V¢S)9 -ij“;, (1.4)

21

where (1)8 is the scalar potential maintained by the charge on the loop

and AS9 is the tangential component of the vector potential maintained

by the current on the loop. cps can be expressed as

10
file) = 1 S qsle')-‘3——R°——-bde' (1.5)

4N6
o

 

 

 

 

and
2 2
_ l
Rum/43mge e +—a— =bN/2—2cos(0-9')+i—
2 b2 b2

(1.6)

Assume (68(9) can be approximated as

$5
¢fiei = If;— <%e> (L?)
O

Whe re (1)5 is the ratio between scalar potential to charge density and

can be defined as

 

a; = 4neo ¢§9> 0.8)
q (9)
With (1.8)
(1)3 s
s _ 1 q (9)
“74> )ejr:b - I‘ll—Tile: b -—a-e—— (1.9)

 

 

 

 

 

 

 

 

22

From the equation of continuity

 

s
.x 81 (0)
s .1 s . 1 0
9(91-JB’V'I - Jag—‘35— (1.10)
it leads to
8 ch: 3’ 3215(9)
(V4; )9 = 2 2 (1.11)
__kr=b 41160 wb 80

 

Writing A S in terms of the induced current 163(9) on the

CD

loop, we have

u -st
[5188(0) = 3% 5 193(9) 53—33—— cos (9 .. e') de' (1.12)

If Aes(6) is assumed to be pr0portional to 188 (9), (1.12) can be
rewritten as
it (PS

0 i

S S
119(9) :7;- 19(9) (1.13)

Whe re (Ills is the ratio between the tangential component of vector

pote ntial to the current and is defined as

 

5
4n A (9)
1 “o 16 (e)

Wlth (1.11) and (1. 13), (1.4) can be written as

 

 

 

__ ___1
a “It: 6210S (0) 2 2 s 5
-Et 2 2 + (3 b 0. I6 (6)

4n€0wb 89

where

 

The substitution of (1. 3) and (1.15) in (1.1) gives

{-97 +sz2)Ies(6) = KSE c039
as S O

for-nSGSTT

 

where
<I>.S
2 2 s 2 1
BS ‘50 a - 50 Es—
q
s 41T€Owb
K =-_)
(1)5
C1
(C)

Solution for L00p Current

The solution for Ies(9) can be expressed as

s s s . 3
I9 (9): C1 cos fisb9+ C2 Sin fisbe +P (6)

 

(1.15)

(1.16)

(1.17)

(1.18)

(1.19)

(1.20)

24

where C1S and C2S are arbitrary constants, and PS(0) is a particular

integral. C2S is zero due to the symmetry and PS(9) can be found to be

 

 

s K E0 '
P (9) = cos 9 (1.21)
2 2
(SS b -1
Thus, (1. 20) can be written as

s 5 KS E0

10(6):Cl cos [38b0+ ‘3sz 1 cosG (1.22)
s

If the boundary condition of ISS (g) = 0 is applied to (1. 22), we

obtain

(31 = 0 (1.23)

 

16(9): 2 2 ° c039 (1.24)

(D) Backscattering Cross Section

With the induced current found in (1. 24), the vector potential

i . . . .
n the direction of the inc1dent wave and at the far zone of the loop can

be Q alculated as

A‘l
U.

25

“IT
H 7: -jl30R1
s _ o s , , e
AY — TV S 219 (6 ) cos 0

 

__ I
-1 R1 bde
2
. S
“O e-JfiORl K O
= 4 R 2 2 (1.25)
1 (as b -1

 

where R1 = ‘J BC;2 + b2 and R0 is the distance between the center

of the 100p and an observation point on the axis of the 100p as shown
in Figure 1.10 .

The back scattered field in the far zone is

 

E = —ij S
Y

to e-J F3oRl KS E0

= -j —— s b . (1.26)
4 0 R1 ‘3 2 b2 _1
s
#0
Where (,0 = E— = 12017 and [is is to be determined in a
0

f0 1 lowing section.

The Poynting power density of the scattered field can be
fol—1 nd to be

 

 

1. 27)
2 2 2 (

 

 

26

 

 

 

Figure 1. 10. Geometry for the calculation of radiation
field of a solid circular loop.

 

 

27

The backscattering cross section is then found to be

 

 

 

 

2
2 IE SI
0' 2 11m 417 R —X-—
1 IE0 I
2 2
Trz"’0 2 KS
= (B b) —————— (1.28)
4 o l3 2 b2 _1
s
Substitution of (1.19) in (1. 28) yields
l 2
3 2 2 2 4 2 1
0' = 417 g e w b ((3 b)
B o o o .435 (5 2 b2 _1)
q s
2
2 6 1
z X it (50 b) s 2 2
<I>ql (135 b -1)
or
o- l2
B 6 1
: TI’ (5 b) (1. 29)
x2 0 <1): (6521324))

Whe re X is the wavelength.

(E) Determination of (I): and if

( i) Using the equation of continuity (1.10) and (1. 24), qs(0) can be

fou nd to be

 

O .
—-—2——— sin e (1.30)

With (1.5) and (1.30), 0: defined in (1.8) is found as

 

 

 

 

s
S ¢ (90)
(Pg = 4060 s
q (90)
TT ejBOR
5 sin 0' R bd0'
-'rr
: (1.31)
sin 6
o
Where
J .2
R=b 2--2cos(£30-9')+—2
b
Choosing 00 = 121. , the point of maximum charge density, 433 becomes
2
17 -jfisb'\/2-23in9'+%§
(1)8 = 5 Sin 9' e (19'
q _.n / a7-
2 - 2 sin 9' + —-2-
b

N

:5 sinG'Kl(9', >d0' (1.32)

0‘"):
N

29

 

 

J 2
a
_. _ . , __
where 2 3138 b 2 2 Sin 9 + b2
K1(9, 3).. J a" (1.33)
2 - 2 sin 9' +7
b

In actual calculation of <1): , (is will be replaced by (30 as an

approximation.

(ii) Similarly, (I)? is obtained by making use of (1. 12)

and (1.14) as

S
¢s= 411' A9 (60)

l “o 198 (60)

5‘” 6' e 0' equsR bde'
cos cos ( o - ) T
‘" (1.34)

cos 0
o

Where

 

\/ -—Z
a

— I

R—b 2-2cos(9O-6)+bZ

Choosing 90 = O, the point of maximum current, (13.15 becomes

3O

/ Z
-j[3$b 2-2c059'+§7
e

 

 

 

 

 

:IT b
(118 =3 cosze' 2 d0'
1
-1, K/Z-Zcose'+%
b’
w 2 a2
=25 cos 9'K2(9',—2-) de'
0 b ,
(1.35)
where
f a2
2 -jfisb 2-cosG'+-—7
K (9' 9—) e e b
2 ’ b2 J as
2 - 2 cos 9' +—-2-
b (1.36)

In actual calculation of (1118 , (is will be replaced by (30 as an approxi-

mation.

(F) Numerical Results

To calculate the theoretical results of backscattering cross

section, (1)3 and (1118 are numerically calculated for the case of

._2 : 0. 00179 as functions of Bob. Numerical results of (I): and (1318

are shown graphically in Figure 1.11 and Figure 1.12. It is observed
that (I): and <1>is vary only weakly over the range of interest. This
agrees reasonably well with our original assumption of CI): and (Ills

be ing constant. The numerical result of backscattering cross section
0-

\2 is plotted as functions of (Bob in Figure 1.13 in comparison with the

X

e xperimental results .

 

31

.1)—
-Ir-

2
. s . a _
Figure 1.11. <1) q as a function of [30b (32 — 0. 00179) .

d.-

32

RC1).

 

 

Figure 1.12. <I>is

as a function of Bob(

2
3.

b2

 

= 0.00179).

relative backscattering cross section or

33

2a

   

oocooo theory

experiment

 

Figure 1. 13. Comparison between theory and experiment for
circular loops illuminated by plane wave at
normal incidence.

34

The theory predicts fairly well for the loops with Bob
smaller than 1. 5 which is the range of interest. For a larger loop
the present theory fails to give accurate solution. A refined theory
would give better results but we find the present theory adequate for

further theoretical deve10pment.

1. 3. 2. Scattering from a Solid Rectangular L00p

To study the backscattering from a solid rectangular 100p
an approach which is different from that used for a circular 100p
will be used. The first order solution will be obtained by using the

method developed by Chen and King?

(A) Formulation of Problem

The geometry of the problem is as shown in Figure 1.14.
A rectangular 100p with short side 2hl and long side 2h2 is
assumed to be made of perfectly conducting metallic wire of radius
a. A plane electromagnetic wave with the E field paralléto the
plane of the loop is incident normally to the loop. The dimensions
of interest are
2 2 2

2 2
a << h1 and h2 , [30 a <<1

Where 50 is the wave number. We assume that the wire is thin enough

So that only the tangential component of current is induced.

(B) Integral Equations for Loop Currents

Since the tangential electric field should be continuous at

the surface of the 100p, we obtain the following equation:

 

 

 

 

 

 

 

23—.-

 

 

 

 

 

 

S
IZY
S S
/ 13x I3x
+— -£ ———a-

 

de

 

 

Figure 1. 14. Geometry for solid rectangular loop.

36

For side 4
E43+quI = 0 (1.37)
E4: = EO (1.38)
where E4: is the tangential electric field of the incident wave and

E4: is the tangential E field at the surface of side 4 maintained by

the induced current and charge on the loop.

Due to symmetrical configuration, it follows that

Izy(y) = I4yly) (1.39)
S S

11X (X) 2: - 13X (X) (1°40)

(1280’) = q: (y) (1.41)

qls (X) - - q: (x) (1.42)

where I S (y) is the induced current in side 4 and I

s . .
4y x (x) is that in

1

side 1, etc., whereas qiS denotes the induced charge on side i,

i: 1, 2, 3, and 4. E4: can be expressed as
a 8 _AS
E4t = -(V<b4)y-Ji.1(A4)y (1.43)
_‘\s
where A4 is the vector potential at the surface of side 4 contributed

by the induced current in the 100p; 4348 is the scalar potential at the

37

surface of side 4 maintained by the induced charges on the loop. In
——_.\

symbols, A45 and 048 can be eXpressed as

E52A8§+A89 (1.44)
4 4x 4y
S S S S S

C1’4 ‘ ¢41+¢42+¢43+¢44 (1'45)

where A4: is the vector potential at the surface of side 4 contributed
by the induced currents in sides 1 and 3. The induced currents in

sides 2 and 4 do not contribute to A4: because they do not have x

s . .
component. A carries the same meaning as A

4y except that the

3
4x
contribution is made by the induced currents in sides 2 and 4.

s . . . . .
((94.1 is the scalar potential at the surface of Side 4 maintained

by the induced charges on side i, i: 1, 2, 3, and 4. A4: can be

expressed in terms of current by Helmholtz integrals as

S no h2 S P‘Jfio r44
A (y)=—- S I (y') -’-——-—— d)"
4y 411’ 4y r -
.hZ 44
“o 5312 S 615614:
+ —- I (y') CW
40 -h 2 r42
2
u h3
= ~53 5 I S (y‘) K (y 3") CW
4w -h 4y 1A ’
2

(1.46)

38

 

 

 

 

 

where . .
-J [30 r44 -J ‘30 r42
K (v. y') = + (1.47)
M r44 r42
2
r44 = \/ (y' - y) + a (1.48)
2 2
_ I __ ‘
r42 - ’\/ (y y) + 4h1 (1.49)
With (1. 44) and (1. 45) , (1.43) can be rewritten as
Ea=-—§-(¢S+¢S)-iwAS-—?—(¢S+¢S) (1.50)
4t By 42 44 4y ay 41 43
By the Lorentz condition, it follows that
8 A S (Y)
S s _ 'w “A s ‘3 s _ its 4y
¢42 + 4’44 ‘ 2 v (A42 + A44) “’ 2 (1'51)
(30 60 By

After substituting (l. 51) in (l. 50) and after rearranging, (1. 50)

becomes
a fl .332 2- S 8 s s
E4t : F32(ay2 +50 )A4y(y)--5; (¢4l+¢43) (1.52)
o

Substitution of (l. 38) and (l. 52) in (1. 37) gives

 

2 . .
8 2 s ”30 ”30 8 s s

39

Similarly, the differential equation for the vector potential at
side 1 can be obtained as follows:

For side 1

Elt+Elt: 0 (1.54)
i _
E1t — 0 (1. 55)
where E1; is the tangential E field of the incident wave and E1: is the

tangential E field at the surface of side 1 maintained by the induced
current and charge on the loop.

Eli can be expressed as

E1: = - (WIS 1X -iw(A15)x (1.56)

where $18 is the scalar potential at the surface of side 1 maintined by

s
1

surface of side 1 contributed by the induced currents in the loop. In

_A
the induced charges on the loop and A is the vector potential at the

“is 5
symbols, Al and 01 can be expressed as
_i
s S A S A
A1 -. Alx x-lA1y y (1.57)
s s s s s
4’1 ‘ ct’11 +812 J“ 4’13+ 4’14 “'58)

where A1: is the vector potential at the surface of side 1 contributed

by the induced currents in sides 1 and 3; while A S

1y

by the induced currents in sides 2 and 4. $1? is the scalar potential

is that c ont r ibuted

40

at the surface of side 1 maintained by the induced charges on side i,
i=1,2,3and4.

s .
A can be expressed in terms of current as

1x
s #0 hl s e-J BOTH
_. __ ' ____.___ '
A1x(x) _ 4w 5 le (X) r11 dx
-h
l
110 h1 s e JBori3
._ l I
+ 4" S I3X(x) r dx
13
-hl

ll
dbl ‘2
=1 0
LIED-l
.—a

 

 

S l I I

le(X)K2A (x, x)dx (1.59)
'hl
where
e'Jf‘orii e'J F361'13

K (x, x') 2' —— - —-—————— (1.60)

2A r11 r13

.7
r11 2 N/(x-x')2+a” (1.61)
r13 = \/(x' -x)2+4h2‘2 (1.62)
With (1. 57) and (1. 58), (1. 56) can be rewritten as
Ea---§-(¢S+¢S) wig—105+ 5) (163)
It ‘ 8x 11 13 'J“) 1x‘ax 12 4’14 °

By the Lorentz condition, we have

. 8A S
_b. ._\ 1
¢1f+¢138= LEV. (A118+A13S) 1' u; ‘——)‘{_ (1'64)
(30 (30 8x

Substituting (1. 64) in (1. 63), we can rearrange (1. 63) as

. 2
a w 8 2 s 3 s s
Elt = "L26 (52 +50 >A1x ‘5; (4’12 +4’14) (1'65)
0

With (1. 55) and (1. 65), (1. 54) can be rewritten as

 

2 15
<_6_2 +502 A 8“" = O 5%? (¢1:+¢1:) (1'66)

8x 1x to

Due to symmetry, we note that

A4;(y) = A4;(-y) (1.67)

and

S

Alx

(x) =_A1:(-x) (1.68)

With (1. 67), the general solution for A4:(y) in (1. 53) is then

A4:(y) : 6: E48 cos fioy+ 945(yZ| (1.69)

where C4S is an arbitrary constant, V0 is the velocity of light in free

space and 645(y) is a particular integral which can be found as

42

68(y) 25:9 .5), 3- ¢ S(s)+c)> S(s) sin{3(y-s)ds (1.70)
4 L30 0 as 41 43 o

8(x) in (1. 66) can be

Similarly, with (1. 68) the general solution for Alx

expressed as

Alibi) = ‘72-1— [of sin pox + 913mg (1.71)

where C18 is an arbitrary constant and 913(x) is a particular integral

which can be found as

x
618(x) : -31) 3%[¢128(S) + ¢14S(SZJ sin [30(x - 3) ds (1.72)

With (1.46) and (1. 69), it follows that
4w 3 2 s

= - Jig—EEC: cos 50 y + 94553]
(1.73)

Similarly, with (1. 59) and (1. 66), we obtain

43

1
—- A (x) : S les(x') K2A(x,x') dx'

= - L23:- [Cls sin (30x + 918(x]
o

(C) Zeroth Order Solutions for Loop Currents

For the first approximation, we assume that

A4S(y) “o
—é——= 1:,- w
I (Y)

4y

and

A 8(x) u

s 417 s
le (X)

 

4w A4y (y)

(195(1') = u S
0 I4y(y)

and s
A (X)

4

(1)5(X) : [:1—1T 1:

o le(x)

Substituting (1. 75) in (l. 73), we obtain

3 -'4Tr s s
I4y(y) = WEC‘} COS fioy+64 (ya

(1.

(1.

(1.

(1.

(l.

74)

75)

76)

77)

78)

.79)

-1 ,1

I S'X‘ __ ”—147? i“ 5 (1 J“ ’11.) b’xh‘l1
1x1) __ £03169) '11 b' 1711“ 1‘ 3

By looking clos‘ely at (1. 46) and (1. 5'7}. we note that. 1.1;.-

D,

. . 5
contributions to A‘ly (y) and Alx

(x) are (111:: to the curren‘ (“Mr

1‘)
, .

located at y' 2 Y and X' I: x 1'1;sr.)t—w,i'im-_.1y. '[t is then reasonsxlrdx

assume that (I) (y) and (Fahd are constant over the.~ entire (LUNEZIV
s ..

except at the corners. For simplicity. we assuring that

r ‘ 7 ' '\' S -‘ 5 ' u
.10 dete rniine (5 and (_ , the: inilowmg boundary a a»:

4 ”l
wi’il be used.
(a) Current at the corners be ing ror:t§r=.um1s.
1.9., '14V5{y and} 14113.31 111)
(b) Charge around the i,'('.>;':r.(_-.1‘s 116: v1; 1,,ontim_1<“ms.
1. 11,71 h > <1, h >

.‘ 1..

The substitutiirm of (1. 77“}, (1.50} (1:111 (1.81) Ert (1.5L;- 5%.

. s ,
C, cos 13(th + B

4 4

s s L L, . - . .
q4 (y) and ql(x) can be {round as follows by 11:3 1:13,; UH; minimum

continuity.

1
. 1 1

L ~1'47T S . 8845“,) r

: b.) god’s ~C4i3031ni30y "i" T (1.8?)
s _ 1. _a_ s
ql (X) ‘ U 8X 11K ()5)

. _.4n "7' q 8018(x)

= ‘2; —J—-— Cl‘ (3 cos Bx+ ———,—-— (1.86)
C, (1)3 o 0 dx
0

Combining (1. 85) and (1. 86) with boundary condition (1. 83), we

 

 

 

have
8 ' 3943(y) S 3615(XV
- C4 130 8m 130 hZ + ”71y : C1 F3O COS BO hl + 8x
): hz X:h1
(1. 87)
. . ,. s s ,
Solvmg (1. 84) and (1. 87) 101‘ C4 and Cl’ we obtain
5 l
60 (x)
C S 2: 1 _ __l____
4 130 [:c ot (30111 cos [30112 -- sln (30112] - dx
X~111
39 SW)
_ 4 + .1 X
By tan [3011l sin (3th - cos (30112
IY: hZ

s 8
E1011) + 94 (112—{J (1.88)

46

 

s 1
C = . . - X
1 130 Sin 130111 tan fiohz - cos £10113
8 s
891 (x) 864 (y)
8x - 8y

X:h1 y: h2

_ 50 tan {30112 E15011) + 645012]

 

For the zeroth order solutions, we assume that 948w)

(1. 70) and 618(x) in (l. 72) can be approximated respectively as

01‘]

:1

Y)
”151;:—
S

C1
1” o

 

(all

With (1. 90) and (1. 91) substituted in (1. 88) and (1. 89),

. E cos 13h
[C 5‘1 2 _ __(_)_ o 1
4 «k 60 cos (30(111 +112)

 

 

[:C 8:] -- 1:32 Sin (5th
1 0 (30 cos 130(h1 + 112)

(1.89)

in

(1. 90)

(1.91)

we obtain

(1. 92)

(1.93)

Consequently, the zeroth order solutions for the currents are

obta. ine d a s

 

s ~1 -'477 E0 cos [3111 s \
E4Y(YZ_L 1‘— a?; -1;— COS ‘30 (111 +112) COS BOY '1' 94 (Y)j

(1.94)

47

 

[ s ‘1 -'471' E0 sin 60112 . 9 s 1 ..
I (X)... : C { 8; cos 130(111 + 112) Sm 130x+ l (flj (1'93)

If (1. 90) and (1. 91) are applied to (1. 94) and (1. 95), we have

 

[:14S(y)—l : .4? 143—0 cos Bohl cos (Soy —hcos :0 (111 + hZ) (1. 9(3)
y —~‘OO C’o s 130 COS fio( 1+ 2)

r“ s J _.41T E0 sin (3th
E1X(Y) 00 _. Zo$s 6; cos (30 (h1 + h

 

2) sin 50x. (1.97)

(D) First Order Solutions for LOOp Currents

To obtain the first order solutions for the currents, the
method used by Chen and King is adopted.
The following equations are needed:

Fron1(l. 69) and (l. 71)

s __ _-_l -. s __ s
A4y1y1~ V [9, mm BUY 1 94 M]

O ..

s :— L11- 5 . s
A1X(x) - V0 [C1 8111 Box + 01(x):|

From (1. 75) and (1. 76), we have

—'4w 8 s
[I4Y(Y[]O : 3:5.— [C4 COS BOY + ()4 (Y3!

48

' s
s _ -j477 s . (x)
[11): (x):}0 —- («fps [C1 Sin (30x + 61

..—-

From (1. 85) and (1. 86) we have
s —| 477 S . 8848“,)
[44(1’) = W[ - C4 130 S”) 1303’ + Ti]

89 S(x)
s 417 S 1
[:ql (30:10 2 Wéo 811: C1 [30 cos {30x + 8x -I

.al

First order currents are formulated as follows:

)1:
s _ s 477 s _ _2
E4y(y:)1 —[I4y (yilo + H 058 {A4y 417

2
S 1 1 I (’
S [I4y(y):lOK1A(y. NW] (1.98)

1 _.
° Sh [les(x'flo KZA (x, x'). dx') (1. 99)

49

In effect, (1. 98) and (l. 99) can be rewritten as

S

h
1- 5( T’s zri S1 _ 3— SZ I 5(r') K ( ')d' (1100)
4y YJI“ L4), 1),) 0 "' (I, _h2 -4), ) 0 1A Y’Y Y °

h

_. _' l
E1:(X),,il 'f-f 2[le(x) - Els— 39h [Il:(x'):} K2A(X’ x') dx' (1.101)

‘ 1 0

The substitution of the zeroth order solutions in (1.100) and

(1.101) yields

3
l

i%_{ 2 [C45 cos (30y + 648(fl
o s

 

(:4S he
_ . 1 1 1
‘13s 5:11 cos 1301’ KIA (Y1Y)dY
2
1 h2 s 1
I 1 I 1
s ~hZ

1 1
W I
11>.
:1
FM
0
L‘-
m
.1"
n
O
in
OD
~<
1
o
9.
+
N
show
“i
1
w

ll
6‘!
11A
:1
m
0
43m
1—4
L
)1—
U)
A
E
+
{\a
CD
”9‘01
\<
l
7
F—U)
A
E
W—I

(1.102)

50

where
S Tca(y) 1
L1 (y) = 2 cos 80y - T (1.103)
S nlsm
1‘2
Tcam = 111 cos aOy'K1A1111r'1dy' (1.1051
- 2
s 112 s
r1111) = S 94 (y') K1A (1%)“) dy' (1.106)
_ 2

T (x) d (x)
- 411' S *d . S 1
'—" if {C1 [2 811’) Box ~—%;—~—] "1' 291 (X) -T—j

#477 s s s s
43— C M (x)+29 (x)-D (x))
40 Si 1 1 1 1

(1.107)

where

51

T>’< (X)
s . 51d
M1(x) = 2 Sin {30x - T (1.108)
s
1 dls<x1
h1
Ts=1<d(x) : 5 sin Box' K2A(x,x')dx' (1.110)
-h
1
hl
dls(x) : 5‘ 018(x') K2A(x,x')dx' (1.111)
-h

The scalar potentials (1)415 and (1:438 in (l. 45) can be expressed in ‘

terms of the charges as

 

 

1 hi 311301.41
s s _ s 1 e ,
<1’41 + ‘1’43 ‘ 4TT€ .) q1 (X) r dx
0 --h1 41
1 Shl S e'Jflor43
+ q (x') ———————-— dx'
4w€0 'hl 3 r43
1 Shl s
= . q (1111K (y,x'1dx'
41160 -11 l 13 (1.112)

where

52

 

 

 

 

 

 

 

8’1801'41 P'Jfior43
K (y1x’) - - ‘ (1.113)
1B r41 r43
_ 2 2
r41= J (hZ — y) + (hl - x') (1.114)
2 2
r43 = «/(h2+y) +(hl -x') (1.115)
Similarly,
h _.
s s 1 2 s 1 e Jfiorlz 1
4’12 + 4’14 = 4115 q2 (X) r dy
o -11 12
2
h _.
1 2 S e 3130114
4'11’6 5‘ q4 (x') r dy'
_h 14
2
h
_ 1 52 S K ( 1 d 1
"hz (1.116)
where
( e'JBor12 e'Jfior14
K x, 1"): + (1.117)
2B r12 r14

 

2 2
12 - (h2 - y') +(h1+ x) (1.118)

'1
l

53

 

r14 : 1J(h2 - y')2+ (h1 -x)2 (1.119)

With (1. 51) and (1.112), (1.45) can be rewritten as

S

 

_ s s 3
‘1’4 ‘ 4’44 “1’42 +4’41 “1’43
h
. 3A r"(11) 1
: )1.) 4y 1 S s
2 3Y + 41m h q1 (x') KlB(y, x')dx'
50 ° '1 (1.120)
Similarly, with (‘1. 64) and (1.116), (1. 58) becomes
8 __ s s s 8
‘1’1 “ 4’11 +‘1’13 “1’12 +4’14

“ ”2‘ MI: + I )h2 81 '1K 1 '1d '
' ax 4m.- h q4 Y 213 x')’ Y
130 ° ' 2 (1.121)

 

 

If we substitute the zeroth order solutions in (1. 120) and (1. 121)

we obtain

 

 

S . 1 3948(y) C18 hl 1 1 1
= .C34 8111 fioy+5— -—-ay—— + <I> cosfiox KIB(y,x)dx
O s -h
1
h
1 1 8618(x')
‘. I I
+ 5 fl 5‘ ax! K1B(Y1x ) dx
8 o -h
1
as 3(1) T (y) f 3(1)
= -C Esme/144- 4 + C 5 cd + 1
4 I o '56 8y 1 s s
39 3(1)
8 . 1 4 s s s
'-' '04151n50Y+‘3—— T +C1 L2 (Y)+F1 (Y)
o
(1.122)
where
T (Y)
d
LZS(Y) - -%— (1.123)
s
s
8 f1 (Y)
F1(Y) = T (1.124)
s
h1
Tcd(y) = y cos Box' K1B(y,x') dx' (1.125)

-hl

55

 

 

S 1 hi 8918(x')
f1 (Y) = 13— 5 -——————ax, K1B(y, x )dx
0 -h1
s _ s s s s
‘1’11’9 ‘ 4’11 + 4’13 + 4’12 + 4’14
La 3A1: 1 ha s , ,
z 2 8x + 4116 cl4 9") K2139“ V )d”
(‘30 o -h2
'w _. 3918 (x)
= ‘L'Z— 'Q‘J' { C13 ‘30 C08 BO X '1' .732...”—
130 0
T—Ffleo 5:228E14 (Y :1 K2B(x. Y ")dY
s 1 39181") C43 289‘)
: C1 C08 (30X ‘1'? T - T TS *a(x) +T
o s
s 1 3618(x) s s s
= C1 cos Box+—6;- ———a-x—-——— -C4 M2(x)+F2(x)
where
T (X)
3::
M23(x) : .15:
f 51x)

S
F2 (X) :T

(1.126)

(1.127)

(1.128)

(1.129)

56

1
{/3
:3. N

Ts*a(x) — sin fioy' KZB(X" y')dy' (1.130)

1 3948(Y')
T K2B(x, y')dy' (1.131)

m
N U)
E
1
owl"
:r I

The boundary conditions to be used for the first order solutions

are:

(a) Current at the corners being continuous

s _ __ _ s _
I4y (11-112) _ le(x_.h1) (1.132)

(b) Scalar potential at the corners being continuous

s __ s _
414 (yzhz) — <1>1(x—h1) (1.133)
Substitution of (1.102) and (1.107) in (1.132) gives
3 s s 8
C4 L1 (hZ) + 264 (hZ) - Nl (hz)
_. _ s s. s _ s
— I: 1 M1 (hi) + 291(h1) D1 (111)] (1.134)

With (1.122) and (1.127), (1. 133) becomes

57

89 (Y)
s . 1 4 s s s
v- C4 81115th +73; T '1' C1 L2 (112) + F1 (1'12)
Y:h2
as 8(x)
_ s l l s s s
- C1 C08 {30111 +3; T -C4 M2 (hl)+F2 (1’11)
xzh

1 (1.135)

Solving (1.134) and (1.135) for c4S and CIS , we obtain

r——_ M

1:29:92) + 2918911) ‘ le(h2) " D18(h1):1[C°S 13obi ' L592]

 

 

894 S(y) 1 8015(x) S S
Mi‘h’ié a-T "a: 711—- ”1““2"F2“‘111

y=h x=h1
4 - s . s s s
Ml(h1)l:sm (5th — M2 (111)] - 111(112) cos 130111 — L2 (1123

 

(1.136)

r—.

—‘—_1
@:(h2)+ 2915(111) - le(n2) -Dls(nlfl[:sin sonZ-Mzsmlfj

395(Y) 3980‘)
s 1 4 1 1 s s
Mutual-1157,,— ‘Eg—sz— +F11h21'F2<h1’

~— y=h xzh ----*

s 2 l

M15011) sin 80112 — M23011] - 1.13012) cos (30111 — 1125012]

 

 

(1.137)

For the first approximation, we assume that

58

E
3 ~ 3 _ __2_
94 (y) — @(a 0 - Bo (1.138)
913(x) z Elsbg] = o (1.139)
0

Then, (1.106), (1.109), (1.124) and (1.129) can be rewritten as

51. 11
3 5° S 1 d 1 E0 (1 140)
N1 (Y) - T K1A(Y1 Y) Y - 6373—0 Tea(Y) -
-hz
where
h2
Tea(Y) = S K1A(Y: Y') dY' (1. 141)
-hz
D1381) _ o (1.142)
Flsw): o (1.143)
Fzs(x) = o (1.144)
With (1.138) and (1.139), (1.136) and (1.137) can be simplified
respectively as follows if only the terms with the order of —1— are

<I>

S

retained in both nmnerator and denominator:

59

 

l s
E cos Bohl + E)— B
C34s : "8'?“ s 7T
l 0 cos (30(hl + 112) +3; G
1 s

 

S - E0 Sin B0112 "1' as— H
C1 1 _ 15; 1
cos (30(h1 + hz) +3— G

whe re

8

l
_ - 2 [2 Tcd(h2) + Tea(h2) COS Bohlj

1 . .
~2- [Ts*d(h1)81n (3th + 2Ts*a(h1)51n fiohl

- Tca(h2) cos 130111 - 2 redunz) cos (30112:)

1 .
_ .2 [Tea(h2) 8111 (30112 + 2TS*a(h1)]

(1.

(1.

(1.

(1.

(1.

145)

146)

147)

148)

149)

Substituting (1. 81), (1.145) and (1.146) in (1. 79) and (l. 80), we obtain

1
1

cos fiohl +7;— Bs

 

 

: -J'41T {- 2:2 5 cosp y+98(yj(1 150)
é04’s 13o 1 O 4

COS (30031 ‘1' ha) +5— G
S

 

 

. E Sin Bohz "'1' 31— HS
: £114" { T32 1 S 1— sin Box+9f(x)
0% 0 cos 130(h1 + 112’ +——

(I)
5

)(1.151)

(E) Evaluation of Particular Integrals 848(y) and 618(x).

Integration by parts enables us to rewrite 943(y) and 018(x)

in (1. 70) and (1.72) respectively as

60

Y
63(y) 2 E2 -5 i E) S(s) + 4) S(3] sin {3 (y - s)ds
4 so 0 as 41 43 o
E
_ o s s .
~ E; + E41 (0) + ¢43 (OJ Sin [Boy
+ Bo S: E4ls(s) + ¢43S(s)] cos BOW - 3) ds

E
o s s
= E;— + B 5: E41 (3) + (1343 (5):] cos [30(y - s)ds

O
(1.152)

Equation (1.152) is arrived due to the condition of

¢418(0) : ' ¢43s(0) (1.153)

Similarly
X

615(x) : -5; Egg-[4’1 28(5) + ¢14S(s)—‘I sin BO(X - s)ds

H

x
[1913(0) + ¢1:(O)-] sin Box+ [30 5;) E128”) + 4915(0)] cos (30(x - s)ds

X
2¢1:(O)sinfiox+fio So Eblz(s)+¢1:(sEl cosfio(x—s)ds
(1. 154)

61

Equation (1.154) is obtained because

5 s
¢12(0) - ¢14(0) (1. 155)
For the first approximation, we substitute (1.138) and (1.139) in (1. 85).
and (1.86).
4156
qs()=- ° cssinp (1155)
4 y 58 4 oy '

s 4Tr€o s

q1 (x) = T C1 cos BOX (1.157)

The substitution of (1.157) in (1.112) yields
5 8 C18 hl
441 (y) + 443 (y) = TS- Sh cos BOX' K1B1y,x*)dx'
_ 1
GIS
: T Tcdw)
5
(1.1.58)
Similarly, with (1.156), (1.116) can be rewritten as
s s -C4S hz .
¢12(X)+ ¢14 (X) : T Sh Sln BOY' K2B(X9 Y'idY'
b ' 2
-c S
= (1.159)

4
25— T54 a1(x)
S

With (1.155), (1.159) yields

62

.c8

_ 4
2¢12(0) _ T TS}. (0) (1.160)

‘3

Substituting (1.1 58) in (1.1 52), we obtain

8 E0 C1830 Y -
94 (y) = B;— + T o Tcd(s) cos Bo(y - s)ds (1.1.61)

Similarly, with (1.153) and (1.159), (1.1 54) becomes

5 X

CS c

 

 

 

s 4 . 4 o
91 (x) : - 3;- Ts>i=a(0) Sin 60x - (I) 50T5>3<a(3) cos [30(X-S)dS
5
(1.162)
With (1.146), (1.161) becomes
. h + 1 Hs
E 3‘“ 80 2 6‘ s E
s o s o o
1.94”] = ‘6‘+ E," ‘3‘
— 1 0 cos (30(h1 + hZ) + £— 0 S O
s
.5: Ted (3) cos (30(y - s)ds
' h +—1— HS
E [3 8m B0 2 (I) ——7
: "2):1+E)£)' S t (Y) 1'
Bo s cos [30(hl + hZ) + 3}— G Cd __1
8 (1.163)

where

Y

tcd(y) : So Tcd(s) cos 50(y — s)ds (1.164)

63

Similarly, the substitution of (1.145) in (1.162) gives

E cos fiohl + 51- BS
[Bibi = «be S
1 BO

5 cos (30(h1 + hZ) + 51— G
s

x
. [T9, 3(0) sin (30x + (30 50 Tefltés) cos [30 (x - s)dEl

1 3
E0 COS {30111 + a); B

(30 <I>S cos (30011 + 112) + 7121' G]
S

. [Ts>i<a(0) sin (30 x + Bots>:<a(xfl

(1.165)
where

X

ts*éx) = 3;) TS*a(s) cos (30(1): - 3) ds

(1.166)

With (1.163), (1.150) becomes

E‘YSMJU :

 

1

 

<I>

...____. S

 

64

I 1 s . fl...“
Eos pohl cos {30y - cos [30(h1+ h2fl+ E— [B cos (30y - (30 Sin fiohztcdhr) - G]

S

' *2- 50 HS th(Y)

cos 50011 + 112) + 3}- G
S

. 1 s 1 8
cos fiohl Cos [30y - cos [30(111 + hZ) + 5; P41 (y) + $5? P42”)

 

_-———.q .

 

 

where

1:418”)

1342800

cos (30(h1 + hZ) + 51- G
s
s .
B cos (30y — posmfiohztcdhr) -G

- BO HSthW)

Similarly, with (1.65), (1.51) beconrles

 

(1.167)

(1.168)

(1.169)

65

—- .

sinfih sinfix+1{Hssin5x+ T (O)sinfix+fit (xii cosflh
o 2 o 5; o s*a o OS"< ol

 

 

1 s .
+ 37 B €82,250) 5 1n (30x + Bots>1<a(XZl
s

 

 

 

 

—-—— _1
h h 1 G
cos (30(1+ 2) +(I)S
_ 1411 E0
g065 flo
sih' +1PS)+—71Ps()
“8625‘“56)‘ 6' 11(X (1) 12K
S s
1
cos (30(h1 + hz) + 76; C
(1.170)
where
Plls(x) = HS sin (30x +[TS*a(O) sin (30x + (30 ts,:.‘a(x)] cos [30111
(1.171)
s s -
P12 (x) - B [Ts*a(0) Sin (30x + 601:5”?in (1.172)

E4 S(yfl and (:11 S(xfl involve the double integrals which corrlplicate
Y ‘ 11 X 11

the problem.

In order to avoid the double integrals, the following method is

devised at the expense of decreasing the accuracy. First of all,

66

substituting (1.138) to (1.144) in (1.102), we obtain

1 3
cos fiohl +— B

. E <1)
3 _ -_]4Tl' o ' s
[141' (Vii " é ( T3; 1 X

cos 30011; 112) +3;-

S

 

T (y) E E
ca 0 O
[2 COS poy ' ”ES—'1 "L 2 '53??? Tea(Y)}

O S

 

F— .

2 cos Bohl cos (30y - cos {30(h1 + hail+6}— [235 cos (30y - 2G
8

" COS F3ohl Team + COS Boihl + 112) Teawji

 

S

 

1 s
+ "5'2 [GTes‘Y’ ‘ B Tcawj

cos {30(h1 + hZ) @1— G
s

 

_ j4Tr E0
‘ — x
éoés F3o

' l l
2[cos 130111 cos soy - cos 130111, +15%}; U4ls<y) +37 114,56)

S

cos 60(61 + hz) + 61— G

S (1.173)

67

where
s _ s ‘ _ _
U4:1 (y) — 2B cos (30y - 2G- cos fiohl Tca(y) +cos (30(h1+h2)Tea(y)
(1.174)
U S( —GT () 133T ( (1175)
42 Y) - ea y 7 ca y) °
Similarly, the substitution of (1. 138) to (1. 144) in

(1.107) gives

[les(xfl
10

. l 5
Sin (3th +-—— H

"j4 EB (PS [ZSinfi x- ————TS>:6(X)J
{’0qu 50 cos (30(h1+h2)+___l__G O (1)5
(I)

S

 

 

s
1

. . s . . H
2 Sin (3th 8111 (30x +3)— [ZH Sin (30x - BID (3th ng<d(x) - 328T8*d(x)

 

where

glls(xl

. . 1 s 1
Z 5111 B0112 SlnBOX‘l‘E— g1]. (X)+(';—-2- g12 (X)
S

S

.1
<1

S

cos (30(h1 + hZ) + G

S

S

 

cos (30(hl +112) +% G
3

(1.176)

s . .
2H Sln (30x - Sln fiohz Ts’l‘d(x) (1.177)

68

81256:) = - HS Ts*d(x) (1.178)

(F) Backscattering Cross Section ofa Solid Rectangular LOOp

The vector potential, maintained by the induced current in

the solid rectangular 100p, at an arbitrary point on the axis of

the 100p and in the far zone of the 100p (Fig. 1.15) can be expressed

 

 

 

as
h -' R
AS—AS-ig 52 21 s'eJBO d'
_ y 7 4n 4y (Y) R y
sh
2
#0 e—JBoRo h2 s
2’ __ _________ I 1
' 271' R S E437 (VJ dy
o -h 11
2
(1.179)
where
J. 2 2 2
_ l 2.
R _ R0 + y +h1 RO (1.180)
Substituting (1.167) in (1.179), we have
211 “JQORO .1
[AS] = j o E e 1
Y C. F3 0 R . .1 ) 9...]
11 o o 0 (PS cos (30(h1 . h2’ + (1,8

(1.181)

69

 

 

 

 

 

 

 

4v

 

 

 

 

 

 

 

Figure 1. 15. Geometry for the calculation of radiation field
of a solid rectangular loop.

70
where

h
J- 2 h ' (h+h)+—l—BS(')
1— cos Bolcosy -cos (30 1 2 (1’3 41y
.112

1 s
+——z P m] dy'
<I>S 47* (1.182)

J1 will be evaluated later. Consequently the backscatte red

electric field due to the solid rectangular loop is

[12%]

ll

5;] win]

11 V 11

II
N
tr]

CD

p—c

o (1’s [cos [30(h1 + hz) +43%]
(1.183)

The backscattering cross section is found to be

2

8
(TB :- lim 4n RO2 E—lr
R0” CD iEol
J1 2
'2 161T

 

cps cos (30(111 + hz) + £2]

(1.184)

71

“B

the experimental results for the case of the square loop where

is plotted as the function of (30h in Figure 1. 16 to compare with

_ __ _ 5.15 __
hl—hZ—h— 2 cm and a—0.03 cm.

 

The agreement between the theory and the experiment .is fairly good
over the range where [Bob is small.

A silnpler method in obtaining the backscattering cross section
is also presented here. The vector potential at an arbitrary point

on the axis of the loop and in the far zone of the loop is calculated

on the basis of E S(y):l .
43’ 10

 

 

h .' R
AS—As—ffl Szzlswemo d'
" y ‘ 41r 4y Y) R Y
.h2
H -JBOR0 h2
=—"- e I S(') W
1r RO 4y y
-h2 10

(1.185)

Substituting (1.173) in (1.185) , we have

2M 7.130130 J
E s z j o F e 2 __ G
y C (3 7o R
o 0 ° ° ¢SE°Sfiolh1+hzl+o§l

 

 

(1.186)

relative backscattering cross section a

101-

 

Figure 1.16.

 

72

 

 

 

 

 

 

CCL

oooooo theory

experiment

1.0 1.5 2.0

——‘>
(30h

Comparison between theory and experiment
for a square loop illuminated by a plane
wave at normal incidence.

73

where

112
: ' .-
J2 5 (:2 cos fiohl cos (Boy 2 cos 130(h1 + h2)
.h2

 

114186)) + 1142865] dy'
<1: 4,2

S J
S

+

(1.187)

The evaluation of J2 will be presented later.

The backscatte red electric field is, therefore, calculated as

is.) ~ll

10 10 10

E51

~15 R
e o 0 J2

2E G
o <I>S [:cos 130(h1 + hz) +$;j

 

 

(1.188)
Based on (1.188), the backscattering cross section of a solid
rectangular loop is found to be
2 IE3 I.2
0‘ = lim 41TR. —-——Z—
0 o
2
J2
= 1611' (1.189)

 

7 G
(158 cos (30(11l + hz) + 38:]

74
(G) Expansion Parameter p5

From (1.77) we have

 

_ 4v
<I>s(y) —- —

Considering the zeroth order current

 

[-1 3(3)) - j41r E0 cps flphl cos [50y - cos [50(h1 + hz)
4y _ C 73—
l_ 00 o s 0 cos (30(h1 + 112’

we obtain h

I _ l I
-h E03 fiohl cos [30y cos 80(h1 + hail K1A(y, y )dy
. 2

 

(1)8(3') =
cos Bohl cos 80y - cos 130(h1 + hz)

cos fiohl Tca(y) - cos 80011 + 112) Tea(Y)

cos Bohl cos [30y - cos [30(h1 + hZ)

For a loop with 130(h1 + hZ) S 12: , the point of maximum current is

at y = 0. We can then set

(138 =[¢S(Yl:l _ COS pohl Tca(o) - COS F50(hl + hZ) Tea(o) (1.190)
Y=0

cos Bohl - cos (30(h1 + hz)

75
For a loop with 130(h1 + ha) 2 127 , the point of maximum

, so we set

its-I7

current is at y = (h1 + hz) -

X k
- cos Bohl Tca(hl + hZ - Z ) - cos (30(h1 + hZ) Tea(h1 + hZ -174 )

 

cos flohl Sln {50(h1 + hZ) - cos (30(h1 + hZ)

(1.191)

(H) Evaluation of J and J2

1

Since

h
2
_ . .1. 5
J1 - 5 2, [:cos (30111 cos [30y - cos [30(h1 + hZ) + (1)8 P4l(y1),

h
+ 1 P S I d 1
57 42 (y) Y
s
where
s s .
P41 (y) = B cos (30y - (30 Sln (3th tcd(y) - G

P4258) - 80 HS tcdly)

76

_ 3 .
J1 - 3; cos Bohl Sin [3th - th cos (30(h1 + hz)

 

1 2138
+ $— [ 80 $111 flohz - (30 $111 £3th bCd --2h2G]

1
~, —-zq) (30H bCd

8

(1.192)
where
h
2
de : S‘ th (Y') dY'
'h2
h2 y?
2 l [l Tcdisl COS 150 (y' ' S) dil dy'
'h2 0
(1.193)

Jl is difficult to evaluate because of the triple integral involved.

Equation (1.187) shows that

h

 

 

2
J2 : S [:2 cos Bohl cos fioy' - 2 cos [30(h1 + ha)
7112
s s
U41 (Y') U42 (y'):| '
+ 2 dy
<I> (I)

S S

77

s __ s
U41 (y) - 213 cos [30y - 2G - cos 80111 Tca(y) + cos (30011 +h2) Team

U458!) = G Team - 136 Team

or

_ 4 .
Jz - 132 cos Bohl sm (3th - 4h2 cos 130(h1 + ha)

1 4B8 .
+ 5; [To sm (3th - 4h2 G - cos fiohl tca

1 s
+ COS (30011 + h2) tea]+ <1) 2 [G tea - B 1:ca]

S

whe re

2
= I 1
tea f Tca(y ) dy
.h2
h2
: l I
tea 5‘ Tea“, ) dy
-h

(1.194)

(1.195)

(1.196)

CHAPTER II

MINIMIZATION OF BACKSCATTERING OF A CIRCULAR LOOP

2. 1. Introduction

The backscattering cross section of a metallic circular loop
was studied in the previous chapter. It was found that when the loop
is of resonant size, the induced current on the loop is maximum and
likewise the backscattered field. In radar camouflage it is desirable to
minimize the backscatte ring cross section of a loop, in particular, that
of a resonant loop. Many investigations have been made in the recent
years on the technique of minimizing the radar cross section of a
metallic object. Two conventionally used techniques are to utilize radar
absorbing material and to reshape the body to change the reflection
patterns. Recently, a new method called the impedance loading method
has been deve10ped. This method was found to be especially effective in
reducing the backscattering cross sections of metallic objects with
dimensions of the order of a wavelength. Using this method Chen4-6and

others7’ 8

have investigated the minimization of the backscattering from
a cylindrical object. Liepa andSen’ior9 have applied the same technique
to reduce the radar cross section of a conducting sphere.

The basic principle of the impedance loading method is to control
the amplitude and phase of the induced current on the metallic object by
inseting appropriate impedances at appropriate points on the object in
such a way that the backscatter maintained by the induced current is
minimized.

In this chapter, the minimization of the backscattering of a

conducting, circular 100p by an impedance loading method is investigated.

78

79

A perfectly conducting, circular 100p which is loaded symmetrically
with two identicallumped impedances, is assumed to be illuminated
by a plane wave at normal incidence.

The induced current on the loaded loop is determined as a
function of loop dimensions and loading impedance. The backscattered
field maintained by the induced current in the loaded loop is calculated as
a function of the loading impedance. It is then possible to find an Opti-
mum impedance which makes the back scattered field equalto zero. An
explicit expression for the Optimum impedance for zero backscattering
is obtained as a function of loop dimensions. Some numerical examples
are included.

Analytical study on the radiation of a l00p is rare in the literature.
Storer10 studied a loop antenna using a method of Fourier series ex-
pansion based on a Hallan's integral equation. This method is too ~
complicated for our problem. A new method which is simple enough for
our purpose is deve10ped in this research. This new method is based
on a differential rather than on an integral equation. The theory is
later verified by an experiment.

2. 2. Theory

Based on the principle of superposition, a loaded circular l00p
illuminated by a plane wave at normal incidence can be considered as the
combination of (A) a solid loop illuminated by a normally incident
plane wave, and (B) a loop driven by two identical voltages at 6 = 0
and 9 = 1r. The situation is shown graphically in Figure 2.1.1. Essentially,
case (A) is the scattering of a solid circular loop and case (B) is the
radiation of a circular loop antenna. The case (A) has already been
solved in Chapter 1 and its results are rewritten here for further

deve10pment. The problem of case (B) will be solved in this chapter.

AH:

dog mcmummpmu m was moofl
mcmnofimom a mo 55m ofi ma. moofi popmofl <

.z.z.~ osomrm

 

81
After this the results of case (A) will be combined with that of case

(B) to produce the final solution for the problem of the scattering from

a loaded loop.

2. 2.1. Scattering from a Solid Circular L00p

The following equations from Chapter 1 are needed

 

1a (8): 2 2° cos 8 (1.24)

o 1 KSE
o

 

 

(l. 26)

where 198(9) is the induced current on the solid circular loop and E S
is the backscattered field on the axis and in the far zone of the loop

maintained by the induced current.
2. Z. 2. Radiation from a Circular LOOp

The geometry of the problem is as shown in Figure
2.1.2. The 100p is driven by two identical voltages V at 0 : O and
6 = 11'. The dimensions of interest are
2 2

az<<b2 . (30a <<1

where 80 is the wave number. We assume that the wire is thin enough

so that only the 9 component of current is induced.

82

 

 

 

Figure 2.1. 2. Geometry of a radiating
loop.

83
(A) Differential Equation for Loop Current
The total tangential electric field should vanish at the surface
of the 100p except at the gaps at 0 = 0 and 0 : 1r where voltages of V
are maintained. Assuming that the gaps are very small, we obtain
the following equation:

Ea = —%’- 6(6) "11:— 66: - e) (2.1)

for-1750517

where E: is the tangential electric field at the surface of the loop

and 6(6) is a Dirac delta function. Due to the symmetrical configura.

tion, the following conditions exist:

1‘

r

I6 (9) ---I6 (17 - 0) (2.2)
Ir(:h%) = o (2.3)
where Ier (9) is the current on the 100p. This symmetry condition

simplifies the problem and we only need to consider a half of the

100p. In the right-half of the 100p, (2.1) reduces to

a V
Et = To' 6(8) (2.4)

is 5.".
for-2 0 2

The tangential electric field maintained by the current and the charge

on the 100p can be expressed as

84

a. r . r
Et 2 -(V¢)e -JwA8

(2. 5)

where ¢r is the scalar potential maintained by the charge on the 100p

and A r is the tangential component of the vector potential maintained

0

by the current on the loop. ¢r(0) can be expressed as

 

1T -jB R
r 1 r e 0
<1) (0) _ 4n€0 S- q (9) R bd0

where qr(0') is the charge density induced on the loop at 9', and

 

 

J 2 0 - 0' a2
R = b 4 sin 2 + —-2- as mentioned in Chapter 1.
b

Assume that ¢r(0) can be approximated as

 

(Dr
4179) = 75,—3— 4"(9)
o
where (I); is defined by
(Pr = 4TrE ¢r(9)
q 0 que)

‘1) r is considered to be independent of 9 because ¢r(0) in (2. 6)

is mainly contributed by the induced charge qr(9') in the vicinity

Of 9': 9.

With (2. 7), it leads to

 

(2. 6)

(2. 7)

(2. 8)

(2. 9)

85

By the equation of continuity,

 

 

 

 

r
81 (0)
r . 1 r 1 0
q(9)—J;3\7°I ~1w0 89 (2.10)
(2. 9) can then be rewritten as
r 2 r
"I <I> . 8 I9 (0)
(V4313) = 4 .‘L J, 2 (2.11)
r=b T76o tab 80
Relating A9r to the induced current Ier(0) on the 100p, we have
r ”o Tr r e-JBOR
- _._. _ a _ 1\ I
A6 (0) —. 417 5:” 19(0') R co.: (0 0 )bd9 (2.12)
2
_ 1
where R = bN/4 sin2'<e 29>+ :2—
Assume Aer(0) can be approximated as
Ar6)—‘—J—9<I>r1r(0) (213)
0 ( 7 471’ i 0 '
where (Dir is defined by
A 179)
or = 24.2: -9.—-— (2.14)
l #0 I9 (9)

‘13.: here is presumed to be independent of 0 since (2.12) shows that
main contribution to the vector potential at 0 is due to the current
element located in the vicinity of 0.

With (2.11) and (2.13), (2. 5) can be rewritten as

86

 

 

a -jcbr 82 2 r r
E s q 2 ”—2 + (30b (1 10 (0) (2.15)
t 4weowb 80
where r
r (bi
.. 2.
a — <1>r ( 16)
q

The substitution of (2.15) in (2. 4) gives

2

 

 

a 2 2 r .. rV
< 2 + (3r b>1e (0) .. K ~13- 6(0) (2.17)
00
Tr < <1:
£01 -2-_9_2
where
, or
2 2 r 2 1
Br — so a 2 Bo -7 (218)
<1)
9
r 4w€0wb
K :j r (2.19)
<1)
9

(B) Solution for LOOp Current

The solution for 18170) can be expressed as

19"(0) s of cos 81060 4 c 1' sin (Brbd 4 pr(0) (2. 20)

2
where C1r and C2.r are arbitrary constants, and Pr(0) is a
particular integral. C . r is zero due to the symmetry and 131(0)

2

can be found to be

87

 

pr(0) = 1; sin firblel (2.21)
zero

The application of the boundary condition of 197127.) 2 0 gives

r - KrV 71'
Cl "—' —T tan firb :- (2.22)
28 b
r
Finally, we have
r KrV T1" 17 i
10 (e) = .. :5? sec prb 2- sin or“? - (e I) (2.23)
r

for-TrSG-‘Sw

The above expression for Ier(0) is extended to cover the whole range
of -7r 5 0 $11- by making use of the condition Ier(0) = -Ier(7r -0).
With the current found in (2. 23), the vector potential on the axis of

the lOOp and in the far zone of the 100p can be calculated as

 

r “o :2— r e.J 0R1
~ _ __ 1 l 1
Ay _. 4n 218 (0 ) cos 0 ——————R1 b d 0
‘2
M —JfioR1 r
‘ 417 R 2 2 . (° )
1 Br b - l

 

where R1 = \/ R02 + b‘2 and R0 is the distance between the center

of the loop and an observation point on the axis of the loop.

88

The radiated electric field in the far zone of the loop is

-jl3 R
u o l
r . r . o e K V
E : _ wA : _ w __ ___.____.. 2. 25
y J y 1 2 ( )

Kr and pr are determined in a following section.

2. 2. 3. Total Current and Total Back Scattered Field

Total current can be obtained by superposing 193(6)

upon Ier(0) as follows:

S
19(0) = 19(0) +Ier(6)

 

KSE
o
= “—27—“ COS 9
BS b - 1
r
- KY secpbgsinfirb(%- lei)
28 b“ r
r (2.26)
for - 7r 5 0 S 17
Based on the principle of superposition, V is found to be
s r .
V .. [19 (0) + 19 “£121. (2.27)

With (1. 24) and (2. 23), V is determined as

89

2 (3 152 ZLKSE
r O (2. 28)

 

 

V :
2 r
Zfirb + ZLK tan Brb 2 (38 b -l

[:1
N
N

The substitution of (2. 28) in (2. 26) yields a final expression

for 19(0) as

K871:
I (9) = cos 8
0 ‘35sz _ 1

 

Tl”

r
- ZLK sec Brb 7 sinsb<l- lel>:|
2 prbz + ZLKr tan prb 321 r 2

 

 

(2.29)
for -11 S 0 S 11
Total E field in the far zone of the 100p can be obtained as
E = E S + E r
Y Y Y
.' s
#0 e-JBORl K Eob
: -J(,) —- K
411 R] 1328132 _ 1
r
4 ZLK firb
T' + 2 2 2 r 11
(Br b - 1) (2131b + ZLK tan Brb -Z-)
(2.30)

2. 2.4. Optimum Impedance for Zero Back Scattering

To minimize the backscattering to zero, total field in (2. 30)

is set equal to zero. This leads to

90

4zLKr8 b
r - 0 (2.31)

2

11 + 2 r 11'
(Brb - 1) (ZBrb + ZLK tan Brb -2-)

Solving (2. 31) for ZL, we obtain the optimum impedance for zero back

scattering as

 

(:2 :1 _ - 21181132 (831)2 - 1) (2 32)
L 0 Kr[:4prb + 11(13rb2 - 1) tan firb 71%] °

With (2.19), [:21] can be eXpressed as
O

160 V or ((3er2 — 1) 41> r
[2.1] = 2 2 q (2.33)
0 2E1 orb + 11(13r b - 1) tan pro %]

2.2.5. Determination of <I>qr and (Fir

(A) By making use of (2. 23) and the equation of continuity, we have

r
rm) _ __1__ 810 (9)
q 7' J wb 86

. l KrV 11 V
o...- s.- -.- sass-6 - 1.1)

(2.35)

-for-11$950

+‘F0r 0.59.517

<I>qr is then defined as

91

 

 

 

 

 

 

 

 

490R 1 -100R
_ .11 1 EL..___ 1 IT..- I e________ I
SW cos Brb(2+ 0) R bd0 +§Ocos (3rb(Z 0) R bd0
or —
9
1r
ssls- 9.)
(2.35)
where
J 2 8o - 9' a2
R = b 4 Sin <—-2——) + ~2-
b
Choosing 00 = g , the point of maximum charge, <I>qr becomes)
1‘ 2
I‘ _ 1T. - 1 1 a I ,
(bq -30 cos firbl2 0)K3 (0, Ez)d0 (2.36)
where
J 32
—j(30b 2-251n0'+—-
b
a e
K (6', ) :
3 :2 J a2
2 - 2 sin 0' + -—2-
b
J
1130b 2+2 Sin 0'+-§
e b
J
2 + 2 sin 0’ + --2—
b (2.37)

In the actual calculation of (qu, Br will be replaced by so

as an approximation.

92

(B) Similarly, (Fir can be obtained by making use of (2. 23) as

 

 

r
q." _ 31 __A 9(6)
i -u r
0 I0 (0)
11 -j(30R
S sin Brb(-121- I9' I) cos (0O - 0') E—R—- bd0'
~11"
. 11
em Brb (2 - I 00 I)
(2.38)
where
f 2 - 0' a2
R = b 4 sin ( 2 )+ —-2—
b
Choosing
0 for Brb $1
6 2
0
TT X
s - :11; W firb Z]

. . . r
where 60 [S the pOlnt of nlax1muin current. <1).l then becomes

.
2
l _ _ 1 5 '
““11 mew.)....K,(.,§.)..
(ID-1r: for firbsl
Tr 2
) sinsrsls- l9'|)sin(z%+9') K5(9'.iz)doi
' ' b

for firb 21 (2.39)

93

 

 

 

 

where \/ a2
-jpb 2-2c050'+—-z
a2 8 0 b
K 6', —— = (2.40)
2 - 2 cos 9‘ + —-2-
b
2
_. J . _)(_ , a
2 JISOb 2-231n(4b +0)+b7-
K4953: (2.41)

 

 

b 2

)ze
~/2-2sin(-4)T‘D-+e')+3-Z-
b

In the actual calculation of (Dir , (3r will be replaced by 50 as an

approximation.

2. 2. 6. Numerical Examples

To Show the theoretical results graphically, <I>qr and (Fir
. 2 2
are numerically calculated for the case of a /b = 0. 00179 as

function of Bob. The optimum impedance, [2;] , is then
0

calculated as a function of Bob for the case of aZ/b2 : 0. 00179.

Numerical results of (qu and (Dir are shown graphically in Figure
2. 2 and 2. 3 respectively. The numerical result of [Z1] is

0
shown in Figure 2.4. In Figure 2. 2 and 2. 3, (qu and (Fir vary
only weakly over the range of interest. This agrees reasonably
well with our original assumption of (qu and (1)11. being constant.

In Figure 2.4 the following facts are observed:

94

o
.328 .o u ~o\~£ eon so sososfi o as s e .m .N osowE

 

95

 

.thdoo. n N .m
o N N H
n\ .3 Q a wo c0305; .6 mm a
O C
6 .m . ondwm

 

96

haezoo.o.umcxmsv

mceuofimomxomnfl coon HON ooccpomfifi. SEMHQO

.s.~ osomwm

- com:

 

 

l

l

. com

. cow

.. coo

. cow

. coon
1 com;
. 2:3

1 cos:

-com_

 

97
(1) In general, the Optimum impedance for zero back

scattering needs both a resistive and a reactive component.

(2) It requires a negative resistance around {Bob = 1. 25

and a reactance with negative 510pe around (Bob = 1.

2. 3. Experiment

As mentioned in previous section, the optimum impedance for
zero backscattering, in general, requires a negative resistance.
Practically, it is difficult to implement. To simplify the problem,
an experiment was conducted for the case of a circular 100p loaded
with two purely reactive impedances.

In this section, experimental arrangement and experimental
results are discussed. The comparison between the theory and the

experimental results is also given.

2. 3. 1. Experimental Setup and Measuring Technique

The experimental setup is shown in Figure 2. 5. The
experiment was conducted inside of an anechoic chamber which
is constructed on the tOp of an aluminum ground plane (10' x 12',
0.125" in thickness). In the experiment, the R. F. signal is
radiated from a horn antenna (Scientic Atlanta model 12-1.7)
located at one end of the ground plane and the circular loop is
placed at the other end. With this arrangement, a plane wave is
incident normally upon the 100p. A slot is cut at the central part
of the ground plane. A thin wire probe (Central Res. Lab.

MX-1019/u) protruding out of the ground plane is movable along

 

scatterers
0

slot
probe

L horn
bsorbers antenna

 

 

 

 

98

 

 

 

 

 

scatterers
RF input
h , A w II
. ground
plane
to detecting
system
J .1

variable

Expe rime ntal Se tup

horn
antenna

probe

loading

scatterers

 

 

1.72 GC

 

 

 

 

1 KC stretche r

 

 

square
w. gen.

 

 

 

////////7/7//////// ,

mixer

te rmi nation

RF 1) directional
gen. ' - - coupler
l

ine attenuator

 

l

 

m

///// //f/ ////// “f/V/

variable
loading

 

 

'

 

 

 

detector

 

 

 

 

Circuit Diagram

\(

 

 

 

SWR
meter

Figure 2. 5. Experimental setup and circuit diagram.

99

the slot. The loop is loaded with a pair of identical coaxial lines

(characteristic impedance ZC = 50.1) underneath the ground plane.
This simple device provides a purely reactive loading to the loop.
The approximate impedance of the coaxial cavity can be calculated

from the well known expression Z = jZC tan [501 where ZC is the

L
characteristic impedance of the coaxial cavity, [30 is the wave
number and 1 is the length of the coaxial cavity.

The circuit diagram is also shown in Figure 2. 5. The
method of cancellation is employed in the experiment. The R. F.
signal is generated from a microwave oscillator (GR 1360),
modulated by a 1 KC square wave generator. The output of the
R. F. oscillator is connected to the horn antenna. When the scatterer
is absent, the signal received by the probe is cancelled by a refer-
ence signal from the R. F. oscillator through a line stretcher
(GR874-LK20L) and an antenuator (ARRA 2414-20). The cancal-
lation of the probe signal by the reference signal is accomplished
by the combination of these two signals in a mixer after going through
a directional coupler. After the cancellation process is completed,

the loop is introduced. The output of the mixer or the reading of

SWR meter will then represent the backscattered field by the loop.

2. 3. 2. Experimental Results

A circular loop (diameter : 5. 15 cm) was constructed as as
an experimental model using cylindrical wire of 0. 1 cm radius.
The experiment was performed at various frequencies (1. 6 GC -

2. 5 GC). Experimental results are shown in Figures 2. 6 - 2. 20
in which the backscattering cross sections of the loop wire plotted

as functions of loading impedances at each specified frequency.

100

The backscattering cross section of the loaded loop is represented
by a solid curve and that of the solid loop by a solid straight line.
It is observed that if the loading impedance (or the length Of the
coaxial line) is properly adjusted, the backscatters of the loop can
be minimized to the noise level. At each frequency about 15 db
reduction in the back scattering cross section was obtained. It
was also observed that the scattering in the off broadside direction
was reduced when the backscatter in the broadside direction was

minimized.

2.4. Comparison Between Theory and Experiment

To check the accuracy of the theory, the Optimum reactive
impedance for the minimum backscattering was numerically cal-
culated and then compared with the experimental results.

The theoretical value for the optimum reactive impedance

is calculated as follows:

(1) In the expression for the total back-scattered field,

equation (2. 30), the loading impedance ZL is replaced by jXL,

(2) An expression for the total backscattered field is
numerically obtained for a particular frequency as the function

of XL'

(3) A computer program is then set to calculate the

particular value of X which gives the minimum value of the total

L
backs catte red field,

(4) Thus the Optimum reactive impedance, E1] , is
---0

obtained as a function of Bob.

101

The experimental value for the Optimum reactive impedance
was obtained by taking into account of the stray capacitance which
exists at the end of the coaxial line.

The experimental and theoretical results of the Optimum
reactive impedance for the minimum backscattering are shown
graphically in Figure 2. 21. The dotted curve represents the ex-
perimental results and the solid curve represents that of the theory.

The agreement between theory and experiment is reasonably good.

.60 S .H u a 8.8895 mseosos
mo COSOCUM m mm mood popmoa .m mo sofloom mmono wswnoflmomvfiomm .o .N onsmwh

1a

Son Emu 0N... n N OOGMUOQF: wcwpmod on: 33300 HO gums“: A:

 

102

m; MN mN mm 5N am am

_ _ . p r . p

_ _ a . _ a \ q CAI

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CHAPTER III

MINIMIZATION OF BACKSCATTERING OF A
LOADED RECTANGULAR LOOP

3.1. Introduction

A theory of the minimization of the backscatter of a metallic
rectangular loop by impedance loading method is developed in this
chapter. A rectangular 100p loaded symmetrically with two identical
lumped impedances at the centers of the long sides of the 100p is
assumed to be illuminated by a plane wave at normal incidence. The
induced current on the loaded loop is determined as a function of the
loop dimensions and the loading impedance. Based on the induced
current, the backscattered field can be calculated. It is then possible
to find an optimum impedance which leads to zero backscattering from
the loOp as a function of loop dimensions. Some numerical examples

are included.

3. 2. Theory

Based on the principle of superposition, the problem of the
scattering from a loaded loop illuminated by a plane wave at normal
incidence can be considered as the combination of the scattering of
a solid loop and the radiation of a driven 100p as described in Chapter
H. Thus the following two cases will be considered separately and
the results will then be combined to yield the final solution for the

scattering from loaded loop:

(A) A solid rectangular loop illuminated by a normally incident plane

wave.

118

119

(B) A rectangular 100p driven by two identical voltages at the centers
of the long sides.

The situation is illustrated in Figure 3.1. In fact, case (A)
is the scattering of a solid rectangular 100p and case (B) is the
radiation of a rectangular 100p antenna. Case (A) has already been
solved in Chapter 1 and its results are quoted here for further theoretical
development. Case (B) will be considered in this section. The results
of Case (B) will be superposed upon those of case (A) to obtain the

solution for an illuminated, loaded rectangular loop.

3. 2. 1. Scattering from a Solid Rectangular Loop

The following expressions from Chapter 1 are needed for

further development:
I E
s 34w 0
I :

cos Bohl cos [30y - cos [30(h1 + hZ) + E:- P418(y) + 31-? P4Za(y)
s

 

cos (30(h1 + hz) + 51— G
s

 

 

‘_ (1.167)"—‘

s _ -°411’ E0
le(x) — ZJT IX

11 o

 

 

_._ . 1 s 1 8
sm [‘3th 811'). Box +3); Pll (x) + 3—2 P12 (x)
s
(h +h )+ —1— G
cos [30 1 2 (1’5

 

 

"_” —_ (1.1m)

120

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121

 

S : 3411* o
E4HYBIO ’5 (I) 50 X

OS

-———

1 1 s
2Ecos fiohl cos [30y - cos {30(hl + h2):l + E; U418(y) + g—z- U42 (Y)
s

 

cos [30(hl + hz) + $— G
s

 

 

 

 

 

 

 

(1.173)
E ski] : 1'4“ Eo X
1x 10 goq)s £3o
—. . 1 s 1 s ——‘
2 Sin (3th SH] (30x + @— g11 (x) +Ezg12 (x)
s
s
l
-—-—. cos 130(h1 + hz) + 5: C
(1.176)
[ 3:] e'jBoRo J1
E = 2 E
11 0 R0 <I>S cos (30(hl + hz) + 51— G]
8
(1.183)
s e-jBoRo J2
13:] = 2E
l: 10 0 R0 cps cos 50011 + hz) + g1— G]
5
(1.188)

where I4ys(y) and les(x) are the induced current on side 4 and side 1
of the rectangular 100p reSpectively. Es is the backscattered field

on the axis and in the far zone of the 100p maintained by the induced

12.2

 

 

 

 

2a H.» .—

 

 

 

 

 

 

 

 

1 12"

 

3x

 

 

 

 

 

 

 

 

Figure 3. 2. Geometry of a radiating loop.

123

currents. The subscripts outside the brackets in the above expressions

are referred to the order of solutions.

3. Z. 2. Radiation from a Rectangular LOOp

(A) Geometry of Problem

The geometry of the problem is as shown in Figure 3. 2. A
rectangular 100p with short side 2h1 and large side th is assumed to
be made of perfectly conducting wire of radius a. Two identical
voltages V are connected across the gaps at the centers of the long

sides of the loop. The dimensions of interest are

2 2 Z
a << hl amdh2 , [30a <<1

where B0 is the wave number. We assume that the wire is thin enough

so that only the tangential component of current is induced.

(B) Integral Equations for Loop Currents

The tangential electric field should vanish at the surface of
the 100p except at the gaps at the centers of the long sides where
voltages of V are maintained. Assuming that the gaps are very small,
we obtain the following equation:

For side 4

a
E“ =-V6(y) (3.1)

where E4“;81 is the tangential electric field at the surface of side 4
and 6(y) is a Dirac delta function. Due to symmetrical configuration,

if follows that

124

Izyrm = I4yrm (3.2)
ler(x) =-I3Xr(x> (3.3)
qzrm = qfly) (3.4)
qflx) = -q3r(x> (3.5)

where I4yr(y) is the induced current on side 4 and ler(x) is that on

side 1, etc. qir denotes the induced charge on side i, i = 1, 2, 3, and 4.

E4: can be expressed as

_\
E45 = - (v 45f) Y - waif) Y (3. 6)

where A41. is the vector potential at the surface of side 4 contributed
by the currents in the loop; ¢4r is the scalar potential at the surface of
side 4 maintained by the induced charges on the loop.

In symbols, A r and ¢4r can be expressed as

4
(3.?)
‘1’4 = ¢41r + ¢42r + ¢43r + ¢44r (3'8)

where A4}: is the vector potential at the surface of side 4 contri-

buted by the currents in sides 1 and 3. The currents in sides 2 and

4 do not contribute to A4xr because they do not have x component.

125
Similarly, A4},r has the same meaning as A4xr except that the contri-
bution is made by the currents in sides 2 and 4. ¢4ir is the scalar
potential at the surface of side 4 maintained by the charges on side
i, i=1, 2, 3, and 4.

r .
A can be expressed 1n terms of current as

 

 

 

4y
r #0 hZ r e-Jfior44
_ __ I I
"hz
“o hZ r -Jfior42
+ 2;; 5 12y (Y') r dY'
h 42
- 2
#0 hZ r
.. __ I / t l
'hz
where . .
e-Jfior44 e-mor4z
K (Y. Y') = + (3.10)
U“ r44 r42
r =N/(y' -y)2+a2 (3.11)
44
Z 2
r42='\/ (y' -y) “”1 (3.12)

With (3. 7) and (3. 8), (3. 6) can be rewritten as

126

 

r __ _8_ r r . r _8_ r r

E4t - -8y (4)42 + ((144 ) - JwA4y - 8y (4)41 + 4’43 )
(3.13)

By the Lorentz condition, it follows that
r r 'w “J r —') r jw 8A4yrw)~
4°42 +4’44 = Z V'iA42 “‘44 )= 2 8y
(3., 5.

(3.14)

After substituting (3.14) in (3.13) and after rearranging, (3.13)

becomes

Ea_ __j_w_( 82 2)A 1' _8.( r+ r) (315)
4t " 2 WE +50 4y (Y)'ay ¢41 4’43 '

The substitution of (3.15) in (3.1) gives

 

 

 

 

. z . 2
2 16 V 1B
8 2 r o o 8 r r
(“Tay +130 )A4yiy)"‘—“’“w 5m" w ‘5?(¢41+¢43)
(3.16)
where
¢ r+¢ r’ 1 fl r(')eror4l dx'
41 43 4175 q1 x r
_h 41
1
1 hi r e’Jfior43
l
4Tr€ 5. C13 (x') r dx
0 h 43
" 1
1 hl r
_ I I I
' 4Tr€O .) q1 (X ) K113(y”‘ ) d"
-h

1 (3.17)

127

 

 

where

-jBor41 4501.43

K (y.x')= e - e———— (3.18)
1B r41 r42

2 , 2
r41= N/(hz-y) +(h1-x) (3.19)

2 2
r43: ~/(h2+y) +(h1-x') (3.20)

Similarly, the differential equation for the vector potential at
side 1 can be obtained as follows:
Since the tangential electric field should vanish at the surface of

side 1, we have

a—
Elt -0 (3.21)

where Elta is the tangential E field at the surface of side 1 maintained

by the current and charge on the 100p.

a
can be expressed as

E1t

a

Elt

= -<v¢1r) -jw(A1r> (3-2?->
X

X

where ¢1r is the scalar potential at the surface of side 1 maintained
_;

by the induced charges on the loop and A1r is the vector potential

at the surface of side 1 contributed by the currents in the loop.

-‘ r r
In symbols, A1 and (bl can be expressed as

9 (3.23)

¢1r=¢11r+¢12r+¢13r+¢14r (3'24)

128

where Alxr is the vector potential at the surface of side I contributed

by the currents in sides 1 and 3; while A r is that contributed by the

1y
currents in sides 2 and 4. (bur is the scalar potential at the surface

of side 1 maintained by the induced charges on side i, i = 1, 2, 3 and

r .
4. Alx can be expressed in terms of current as

 

 

r #0 h1 r e'Jfioru
Alx = '4}? _) I1x (x') _r dx'
_h 11
1
no hl r 81.130113
+ T 1 13x1”) dx'
W -h r13
1
“0 hi r
= 7}"? 5‘ 11X (x') KZA (x,x') dx'
'hl (3.25)
where . .
fiflorll 'Jfiorl3
I — E...___._ _ E__.____
K2A(x,x ) — r11 r13 (3.26)
r11:“/ (X' --X)Z+&12 (3-27)
r13 = [(x' - x)2 + 4h22 (3. 28)

With (3. 23) and (3. 24), (3. 22) can be rewritten as
Ea=-—§-¢ r+c1> r>-”wA ‘23-’— ¢ 1L”+4 1' (329)
It 3x 11 13 3 1x ax 12 14 ) '

By the Lorentz condition, we have

._\. __1 . 3A

r r_ jw . r r _ 31.) 1x
4’11 +4’13 “ 2 V (A11 +A13 )" 2 3x (3°30)

 

Substituting (3. 30) in (3. 29), we can rearrange (3. 29) as

 

2
' 2

Ema: ' gm: (3:2 + Bo )A1xr "58; (4’12r + ‘11:) (3'31)
0

where ¢12r and ¢14r can be expressed in terms of charge as

 

 

 

 

 

¢ r+¢ r___1_ hZ r I 9;?11—2 d I
12 14 ‘ 4716 qz (Y) r Y
o -h 12
2
ha r e‘jflorm
4m. 5 <14 (3") r dy'
o -h 14
2
1 ha r
_ l l l
_ 411.60 5 C14 (Y ) KZB (X, Y ) dy
'hz (3.32)
where
-j(30r12 -jfior14
K (x.y')=e————- + e (3.33)
2B r12 r14
. , 2 2
r122Jih2'Y)+(hl+x) (3.34)
2 2
r14 = N/ (h2 - y') + (h1 - x) (3.35)

With (3. 31), (3. 21) can be rewritten as

 

2 '(3
(87' + 502) A1xrix) = on "5'8; ( ¢12r + ‘)’14r) (3'36)

Due to symmetrical configuration, we note that

r r
A4y (y) = A4}, (-y) (3.37)

(x) = - A r(.x) (3.38)

r
A 1x

1x

(3. 37) implies that A r(y) is an even function of y and,

4y

therefore, the general solution for A r(y) in (3.16) can be expressed as

4V

0

where C4:r is an arbitrary constant, and 64r(y) is a particular integral

which can be found to be

64r(Y)=% sin BOiYi - j: 5%[4541r(8)+¢43r(5flsin130(Y-S) ds

(3.40)

Similarly, (3. 38) implies that A Xr(x) is an odd function of x and,

1

therefore, the general solution for Alxr(x) in (3.36) canbe expressed

as

Alxr(x) -_- EZOiEC1r sin (30x + 611' (x):] (3.41)

131

where Clr is an arbitrary constant and 61r(x) is a particular integral

which is found as

91r(x) = — SO 5%[¢12r(5) + ¢14r(sfl sin (30(x - 8) ds (3.42)

With (3. 9) and (3. 39), it follows that

h
41v r 2 r
f,— A4y (y) = 5 1,y (y') K1A (y. y') dy'
o
-hz
- 411' r r
: ZLo—[C‘l cos Boy+ 64 (yj
(3.43)
Meanwhile, with (3. 25) and (3. 41) we obtain
h
4w r l r
”—- A1X(x) =5 I1X (x') KZA (x,x') dx'
0 -h
1
-'41r r . r
= —J§-O—[Cl Sln Box+ 61 (35)]
(3.44)

(C) Zeroth Order Solutions for Loop Currents

At the first approximation, we assume that

132

A r )1

fl = __0_ @(y) (3.45)
I r(y) 411 r

4y

A r [.1

fl 2 _4_° q; (x) (3.46)
r' TT r

I1x(x)

where <I>r(y) and <I>r(x) are defined respectively as

4" A4 r(1r)

<1> (y) = __ __L__ (3.47)
r “0 I4Yr1y)
A r(x)

<I>r(x) = :47 —l-"—r—— (3.48)
0 1x (x)

Substituting (3.45) in (3. 39), we obtain

I4y(y) : fl [C4]: COS BOY + e4r(fl (3.49)
O 1‘

Similarly, the substitution of (3.48) in (3.43) gives

ler(x) = fig-111$?) [Clr sin (30x + 61r(xZI (3. 50)

As we can see from (3. 9) and (3. 25) respectively that A4yr(y) is
primarily contributed by the current elements in the neighborhood of

y and that A xr(x) is mainly due to the current elements located

1

133
nearby x. It is, therefore, quite reasonable to assume that ¢r'(y) in
(3. 47) is independent of y and ¢r(x) in (3. 48) independent of x. For

simplicity, we assume that

<I>r(y) = <I>r(x) = <1> (3.51)

1‘

The boundary conditions for determining C41. and Clr are

(a) Current at the corners being continuous.

. r _ r _ ,
1.8., 14y (y: hz) - -le (x- hl) (3.52)

(b) Charge around the corners being continuous.

i.e., q4r(y = h2) = qlr(x = hl) (3.53)

The substitution of (3. 49),(3. 50) and (3. 51) in (3. 52) yields

r r . r
C cos (3th + 94 (h2) = - C sin Bohl — 91 (hl) (3.54)

4 l

q4r(y) and qlr(x) can be found as follows by making use of the

equation of continuit y.

8 r
ay 14), (Y)

ere-

q4r(y)

(3. 55)

134

qflx) - 1 ~—— I (x)

861%.)

= ~15 43—;04: [Clr (30 cos Box+ -——-—-8x :1 (3-56)

Combining (3. 55) and (3. 56) with boundary condition (3. 53), we

 

 

 

have
r . ae4r(y) r 891%.)
.. C4 flOSlnfiOhZ + -———a—);-—— : C1 flocos fiohl-f T
y: 1'12 x=h1
(3.57)
Solving (3.56) and (3. 57) for c4r and clr, we obtain
.- 1 aeflx) 894%)
C' _—_ ' h 'T
4 Botcot 80111 cos (3th - 81171130 2] 8x y
X: h1 yzhz
1 I‘ 1‘
+ [61 (h1)+ 6‘4 (112)]
tan [1 h1 sin (3 h2 - cos (3 h2
0 O 0 (3.58)
r 1
C1 :(3 [sinfihtanph -cos(3h X
o o 1 o 2 o 1:)
I‘ I‘
391 (x) 884 (y)
8X - 8y
X ‘—" 1'11 y: [‘12

)
.. (30 tan Boh2E1r(h1) + 64r(h2fl 2 (3. 59)
J

135

F01 ze ioth order solution, we assume that 94 r(y) in (3. 40)

and 61r(x) in (3. 42) can be approximated respectiver as

[M
0
Es <5)
0

With (3. 60) and (3. 61) substituted in (3. 58) and (3. 59), it follows that

[of] _ :21 tan (30011 + hZ) (3.62)
o

4 V 1
, — (3-63)
I: 1 :lo 2 cos (30(111 + h2)

sin (30 lyl (3.60)

u
N) <

(3.61)

ll
0

 

Consequently, zeroth order solution of currents are obtained as

[My (y‘] 0 %—-[ tan; O111+h2)cos B0y+84y(yEI (3.64)

=-:4: Sinfiox + e r 3 65
erX 2 cos Bo(h1+h;) 1 (X) ( ° )

 

If (3. 60) and (3. 61) are applied to (3. 64) and (3. 65), we have

136

 

 

 

. ' h + h - ly l)
r __ JZTTV sm 50( 1 2 3 66
[14y (yfloo - (JO‘br cos (30(hl + hZ) ( ' )
. sin (3 x
1 r = 12‘” 0 (3.67)
E 1" (X130 éocbr C05 60011 + h2T

(D) First Order Solutions for Loop Currents

To obtain first order solutions of currents, the following

equations are needed:

From (3. 39) and (3. 41)

A4yr(y) : é‘l— E41. cos (Soy-i 64r (yZJ

From (3. 49), (3. 50) and (3. 51), it follows that

__ #411 _._ r y
E48111, ‘ 218164 BM ‘1']

Elxrbc] = if;%— [Cir sin (30x + 91r(x):l
O

137

From (3. 55) and (3. 56)

11 89 r(y)
[4:63 - nz—d «a: s. a.“ 42,—]

O O r

r 411 r 891r(x)
[:q1 (xfl = W0 r C1 (30 cos (30x + ————--8X :1
0

First order currents are formulated as follows:

r _ r 411 r _o
[:14), (yZl —l:I4y (yflo +“To r {A4y - 4n
1

h
51 i: I4yr(y'flo KIA (Y! Y') dY'}
-111 (3.68)
r I‘ 411 “o
[11" (xii :[le m] + “jail AIX' 4?
l 0

r I 1 1
. 5 [11x (XEJOK2A (x, x ) dx i
.111 - (3.69)

In fact, (3. 68) and (3. 69) can be rewritten as

n
2
E4;(Y):i 92E4;(y)] “(ii—S [I4yr(>")] K1A(7.y')dy' (3.70)
1 O I‘ _h2 O

138

h
1
[le11313 -:-’ zfilxflxfl “(1}- S [11;(31'3 K2A(x,x') dx' (3.71)
1 0 r 4.1 0

Substituting zeroth order solutions in (68) and (69) yields

[:14eriil zi-Egr: {2[C4r cos (30y + 64r(y)-l

C: hz
' E— 5 COS Boy. K1A(Y: Y')dY'
1' ~11
2
1 hz r
- q.— ) 94 (y') K1A(Y.Y') dy'}
r -h2

6%?— [C4rL1r(y) + 294r(y) - Nlrlyfl

O r
(3.72)
where
r Tca(Y)
L1 (y) : 2. COS Boy - T— (3.73)
r
r nlrm

139

h

2
Tca(Y) = 5‘ C05 BOY' K1A(y,y') ch“ (3. 75)
.h2
h2
n1r(y) : S 9437“,.) K1A(Y: Y') dY' (3. 76)
‘hz

-'4 .
Emmi, =35;— izEClr 5‘“ p6+91rixfl
1

c r hl
- ‘4'):— 5 sm (3 X' K2A(x,x') dx'
'hl
h
1 1 r
- $— 91 (x') K2A(X’ x') dx'
r -h

' T (X (1 r(x)
'- 411 . *d 1
5:3; {Clrigsm Box-$15 291r(x)_T}

-'4TT r r r r
= 5:3; E31 M1 (x)+ 291 (x) - Dl (xi)

(3. 77)

where

r . Ts*d(x)
M1 (X) = 28111 BOX -——-§——-— (3.77)

r

140

1. d1 (X)
D1 (X) = T
h1
Ts>l<d(X) : 5 sin (Sox' K2A(x, x') dx'
-h

1

h1
d1r(x) 5‘ 91r(x') K2A(x,x‘) dx'

.111

With (3.14) and (3.17), (3. 8) can be rewritten as

 

 

 

h
. 8A r(y) 1
r __ Jw 4y r
434 (Y) "' B 2 8y + 4v€o S‘ ql (x')KlB(y, X')dX'
o -h

1

Similarly, with (3. 30) and (3. 32), (3. 24) can be rewritten as

r '0) 8A1):(X) 1 ha r
¢1(X) = ESL-z T + 411—5— 5 C14 (Y') KZB(X’ Y') dY'
o O -h
2

(3.78)

(3. 79)

(3.80)

(3.81)

(3.82)

If the zeroth order solutions are substituted in (3. 81) and

(3. 82) respectively, we obtain

141

r .w _-.'1 r . 8941's)
(1)4”):ng vO EC4 Bosmfioy+ 8y :1

O

 

E
O

h1
1
+ 411 .) Elrix'fl K11_=.(Y"")dxl
-111 0

r . 1 864r(y) clr 11‘1
-h
1
1 h1 891r(x')
+ EB—o' 5' ax! K1B(Y2 x') dx'
-hl
r 1'
=-c ramp 7+; 894 (Y) +0 1.Tcd()')+ f1()')
4 o (30 By 1 Er 5r
as r( )
_ r . 1 4 Y r r r
_ — C4 SinBOY+;3——'-a-y—— + C1 L2 (”+171 (Y)
0 (3.83)”
where
T (Y)
d
Lzrw) = 435— (3.84)
I‘
r f1r(y)

142

h

1
Tcd(y) = 5 cos Box' K1B(y,x') dx' (3.86)
-h
I
f r 1 h1 891r(x')
1 (Y) = E;- S T K1B(Yox') dx' (3087)
-h

1

. _. 89 r(x)
¢1r(x) =_)_2_w 3L [€11.80 cos (30x + ——-la—x————-]
(3 o

O

 

h
1 2 r I 1 I
+ 411’60 E14 ‘5’] 1‘23“” l 9"
..h2 0

 

ae r(x) c r f r(x)
=Crcosfix+—l— 1 - 4 T (x)+ 2
1 0 50 8x Er s*a 6r
"Crcos x+_1_ ___391r(x) -Cer()+Fr()
‘ 1 50 (30 8x 4 2 X 2 x
(3.88)
where
T (x)

M2r(X) 1' “—3-— (3.89)

143

, farm)
F2 (x) _ T (3.90)
h2
Ts*a(x) = 5 sin Boy' K2B(x, y') dy' (3. 91)
‘hz
f2 (x) = 5: ————,—YT——- Kznlxw' ) dy (3.92)
'hz
The boundary conditions to be used for the first order
solutions are:
(a) Current at the corners is continuous
ie 1 r(=11)=-1 r(x=h) (393)
° ” 4y Y 2 1x 1 °
(b) Scalar potential at the corners is continuous
' r — h — r — h (3 94)
1080: (84 (Y‘ 2) - 9’1 (X— 1) °

The substitution of (3. 72) and (3. 77) in (3. 93) gives

r r r r
04 1.1 (hz) + 294 (112) - Nl (ha)

= - Elerrml) + zelrml) - DEM] (3.95)

144

With (3. 83) and (3. 88), (3. 94) becomes

 

 

39 rm

r . 1 4 r r r

-C4 811']. BOh2+Eg T +C1 L2 (h2)+FI (ha)
y=h2
ae r(x)

_ r _1__ 1 r r r

— C1 cos Boh1+ 80 ————8x - C4 M2 (hz) + F2 (hi)
x=h

(3. 96)

So1ving (3.95) and (3.96) for c4Jr and of, we obtain

——-d

)—
E64r(h2) + 261r(h1) - Nlr(h2) - 131171113] [cos (30h1 - L2r(hzfl

ae r(y) 69 r(x)
r l 4 1 1 r r
+Ml(h1)E3:T "5'; 6x ”“1 (1.2)—17241))

y=h2 x=h

 

 

 

1

h—— _

M1r(h1) [sin (30112 - M29111] - Llr(h2)[:cos pohl — L2r(hzfl

(3.91)

F

Ee4r(n2) + 261r(h1) - Nlr(h2) - 13117111] [sin (3th - M2r(hlfl

1 4

r 1 r r
+ L1 (ha) (30 8y " E: ax' + F1 “‘2’ ’F2 0‘1]

 

h—

 

y=h2 x=h1

 

M1r(hl) Em floh2~M2r(hlfl - 1.117112) cos (30hl - L2r(hzfl

(3. 98)

145

For our first approximation, we assume that

94r(y) ‘2’ [94r(y):] = 1g- sin (30 Iyl (3.99)
0

H
O

(3.100)

81%;) 9.11:8ng
0

Then (3. 74), (3. 78), (3. 85) and (3. 90) can be rewritten as

1'

where

_Y_ hz .Y.
2 .
N1 (1') =‘o—S sm BolY'l K1A<y.y') dy' = —-§— 13am (3.101)
r -h r
2
hz
Tsa(Y) = 5‘ sin (30 lY'l K1A(y’ yr) dy' (3.102)
-hz
r
D1 (’9: 0 (3.103)
r ,
F1 (Y) = 0 (3.104)
Y. hz + for y' > o
2
F2 (X) - T :1: cos [5 y' K2B(x’ y!) dyu
'hz - for y' < O

V
2 32.. Tba(x) (3.105)
r

146

where

hZ + for y' > 0
_ I I I
Tba(x) -. S t cos Boy K2B(x, y ) dy
I
-112 — for y < 0
(3.106)

Using (3. 99) to (3.106), c4r in (3.97) and c r in (3.98) can be

1
obtained as follows if only the terms with the order of 5L are

r
retained in both numerator and denominator:

- (3.107)
1 cos Bo(h1 + hz) + $9
r

 

r
3’2. [sin (30(111 + hz) + 3"]
[Cf] .

1'
1154;]

[C11] = (3.108)

1 cos (30(11l + 112) + if?
r

 

where

r

1 .
B = - 2- [Tsa(h2)cos Bohl + 2 Tcd(h2) Sin {3th

+ Ts*d(h1) cos (3th + 2 Tba(hl) sin (Bohfl

(3.109)

147

1 . .
G = 2 EI‘S*d(h1)sm (3th + 2Ts*a(h1) sm fiohl

- Tca(h2) cos (Sohl - 2Tcd(h2) cos 1301,12]
(3.110)

:1}
n

r _ .12. { sin (3th [Tsa(h2) + ZTs*a(h1)]

+ 9.112 5.4112) + ”ba‘hlfl l <3 I“)

With (3. 51), (3. 107) and (3.108), (3.49) and (3.50) can be

rewritten as

r
1% [s in (30(hl + ha) + g—]
r

 

Eurwil =Zigif ' o C” l3oy + 941.1”)
1 o r cos (50(hl + hZ) + E-

r

(3.112)

.2145)

cos [30(h1 + hZ) +%

 

E1591 4%"; i

sin pox + 911750}

(3.113)

(E) Evaluation of Particular Integrals 64r(y) and 91r(x):

Integrating by parts, we can rewrite 94r(y) and 91r(x) in

(3.40) and (3.42) as

148
Y
e4rly) = 3;. sin solyl - )0 5% E4116) + ¢43r(8):|sin (Bow-s)ds

= X2: sin poly) + E41r(o) + ¢43r(ofl sin (30y
+ 90 S: [(141%) + 443%] cos (sow - s)ds

V . r
= 7 sin poly) + (30 S: E41r(s)+ (1)43 (3E) cos (30(y - s)ds

(3.114)

(3.114) is derived based on the relation of

¢41r(o) = - ¢43:"(o) (3.115)

Similarly,

X

61r(x) - 50 5%- El 2r(s) + ¢14r(sfl sin (30(x - 8) ds

x
r r . r r
1312 (o)+ (1)14 (0)..) Sin fiox+ (30 50 E312 (s)+ 4114 (83

° cos (30(x - 8) ds

x
= 2 ¢12r(o) sin Box+po 50 E31 Zr(s)+¢l4r(x)—J cos {30(x- s)ds

(3.116)

149

based on the relation of
r
412%) = (>14 (o) (3.117)

For the first approximation, we substitute (3. 99) and (3.100) in

(3. 55) and (3. 56). This leads to

4116 + for y > 0
qr(Y)=+-TIO -Crsin(3y:l:lcos(3y
4 4 o 2 o
r - for y < 0
(3.118)
r 41T€o r
ql (X) = + T C1 COS [30X (3.119)
The substitution of (3.119) in (3.17) gives
r h
ct r()+ct> r()=+Cl £1 cospx'K (x')dx’
41 Y 43 Y 745—- o 113 Y:
r -h
1
c1r
: r _ Ted”) (3.120)

Similarly, with (3.118), (3.32) can be rewritten as

150
r r C4r hz
_ ' I I I
(1)12 (X)+¢14 (X) "' ”-3;— S Sln I30), KZBIXsY)dY
'hz
V h
-2- 2
+ E— S t cos (Boy' K2B(x, y')dy'
r
-h
2
+ for y' > 0
- for y' < O
— - Tr Ts*a(x) +6; ba(x)
(3.121)
With (3.117), (3.121) yields
V
26 r() - C4r T (o)+ '3 1' (o) (3 122)
12 ° ‘ ' E; s*a 35' ba °

Substituting (3. 120) in (3.114), we obtain

r V C1r8o
04 (y)=7 sin poly) +T 0 Tcd(s)cos BOW-s)ds (3.123)

1‘

Similarly, with (3.121) and (3.122), (3.116) becomes

151

011’(x) : 51.) l:_ C4r Ts*a(o)+\% Tba(ofl sin (30x

1’

X

r V
- C4 50 Ts*a(s) cos (30(x - 5) ds + 2

X

- SO Tba(s) cos [30(x - s) ds ‘I
(3.124)

With (3.108), (3.123) becomes
8 r :1 V .
[4 .2. We...)

V . 13r

- 2 [Sin 50011 + hz) + 37:] (3 y

r 52 S T d(s) cos (3 (y-s)ds
I‘ O C O

 

cos (30(hl + hz) + £—
r

(3 1‘
sin poly] Eos (30011 +hzl+£fl - 21>? sin (30011 +h2)+§:]tcd(y)

 

NI<1

cos (30(hl + hz) + EG—

r
(3.125)
where

Y
tcd(y) : 50 TCd(s) cos Bo(y - 5) ds (3. 126)

152

Similarly, the substitution of (3.107) in (3.124) gives

1‘

sin (30(h1+ hZ) + £—
Elrg = <1; 2X r Ts*a(o)+Tba(o) Sin pox
1

cos (30(hl + hz) + FE—
r

 

 

Br

 

sin (30(h1 + hz) + $—
+ Gr ts*a(x) + tba(x)
cos (30(h1 + hZ) + E)—
r
(3.127)
where
x
ts*a(x) : 5‘ Ts*a(s) cos (30(x - 5) ds (3.128)
0
x
tba(x) : SO Tba(s) cos (30(x - 3) ds (3.129)

With (3.125), (3.112) becomes

153

r _ '211V
E4)! WE)“ _1_535: x

(_—

sinfi (h +h - lyl)+ 1 BrcosBOy-Gsinfi iyl
o 1 2 E o
r

—

 

 

+ (30 sin (30(h1 + hZ) tcd(y):] + fi (30 Br tcd(y)

——-— r —_—4

 

cos [50(h1 + hz) + EG—
r

sin (30th + h2 - lyl) + .g— P41riy) +£72— P42r(y)
r

 

 

: 1211V 1‘
Locpr 1 G

cos (30(hl + hz) + E—
r

(3.130)
where
r r . .
P41 (y) = B cos Boy - G 8111 poly) + Bo Sin (30(h1+ hZ) tcd(y)

(3.131)

P42r(y) = ()0 Br tcdm (3.132)

Similarly, with (3.127), (3.113) becomes

154

r _-'211V
152911 ‘53— X

)— _.

sin (30x + £—{ Hr sin (30x + sin (30(h1 + hZ) Ts*a(°)
r

+ cos (30(h1 +h2) Tba(o)]sin (30x + sin (30(h1+h2)ts*a(x)

+ cos (30(h1 + hz) thaw);7
1 . r
+ (3.—2. { [BrTs*a(o)+ GTba(ofl Sln Box+ B ts*a(x) + tha(x)}

1'

h——— —

 

 

 

cos flo(h1 + hz) + £-
r

. 1 r l r
SlnfiOX'I'a: P11 (X) ')'E—Z- P12 (X)

=-°211V r
5 5

cos (30011 + hang-
r

 

(3.133)

where

Plll‘(x) : Hr sin (30x + Ein (30(h1 + h2) Tsakaiol + cos (30(h1 + h2)Tba(o)]

sin (30x + sin (30(h1 + hZ) ts>i<a(x) + cos (30(h1+h2)tba(x)

(3.134)

155
r r .
p12 (x) : E Ts*a(0) + G TbawZ) Sin (30x

r

+ B ts*a(x) + tha(x) (3.135)

Eyr (yi' and [II xr(x)] involve the doubel integrals which in
11 11

turn complicate the problem. In order to avoid double integral, the
following method is presented at the expense of sacrificing the
accuracy of the theory:

Substituting (3. 99) to (3.105) in (3. 72), we obtain

 

V . h h Br

7 Em fio( 1 + 2) + Eli] Tca(y)

i- [:2 cos (30y - T—
r

cos (30(h1 + hz) + £—
r

|<1

 

T (Y)
+Vsinfioly)- 2 2a }

1‘

. 1 1
2s1n(30((h1+h2)- lyl)+_<jb U411.(y)+-------—2 U42rIY)

.2 r ‘1’

11V r

 

cos (30(h1 + hz) + £3—
r

(3.136)

156

where
r . .
U41 (y) = ZBr cos (30y - Sln (30(hl + hZ) Tca(y) - G 8111 poly)

+ cos [50(h1 + hz) Ts (y)

a

(3.137)
U42r(y) = - 13‘r Tca(y) + o Tsa(y) (3.138)

Similarly, the substitution of (3. 99) to (3.105) in (3. 77) yields

v 1+1r
.4 7111‘s] T M(x)
Exr(x] ‘21; r c. (“infiox'—%‘—
r cos 00(h1+h2)+§— r
1'

 

 

10 °

. 1 r 1 r
2 Sin flex-1'3): gll (X) +3712- glz (X)

 

_ iZnV r
_ § §
° r cos (30(111 + hz) +£—
r
(3.139)
where
gllr(x) = ZHI‘ sin Box - Ts*d(x) (3.140)

g12r(x) = "' Hr TS*d(x) (3.141.)

157

(F) Expansion Parameter, CIJr

With zeroth order current in (3.66), <I>r(y) in (3.45) can be

rewritten as

h2
( sin (30(h1+ 1‘2." ly"l ) K1A (Y. y') dy'
'hz

 

{3(9)

sin (30(h1+ h2 - IYl)

sin 5‘):th + hz) Tca(y) - cos (30(hl + hz) Tsaw)

 

sin [30(hl +112 - lYl)

For a 100p with (50(hl + hz) 5% , the point of maximum current
is at y = 0 , so
<I>r = Er(y):] = Tca(°) - cot (30(h1 + hz) Tsa(°) (3.142)
y= 0

For a loop with (30(h1 + ha) 2 1T2- , the point of maximum current is

aty=(hl+h2) ->‘- , so

4
r [jg-(3')]

'9'
ll

_ X
y-(hl +112) -71—

-. X A
Slnflo(h1+h2)TCa(h1+h2- 4) - cos (30(h1 +h2) Tsa(h1 +hZ - 4)

(3.143)

158

3. 3. Total Loop Currents

By the principle of superposition we note that

1' S
Iyly) = I4yiy) + I4y (y) (3.144)
Ix(x) = ler(x) + les(x) (3.145)
V: - Iy(O)ZL = - [:I4yr(o) + I4ys(o)] Z_L (3.146)

(A) Considering the first order solutions of currents

[Luigi in (3. 130), [:14ys(y] in (1.167), [ler(x_)] in

11 ll 11

(3.133) andl:I1xS(xZ[ in (1.170)) V in (3.146) can be solved as

E
. o G
- J41T ZL B;- 4’1. E208 (30(1):) 'I' 1'12) '1' 6:]

 

cps cos (10(111 + hz) + Sis—j

[cosfiolh -cosfiol(h +h2)+§-—P431(o)+-—- P428(0)]

¢S

Q—o‘br Eos (30(h1 + hZ) + ill—1+ j211 ZL sin (30(hl + hZ)
r

_-

l r 1 r
+ '6: P41 (0) +32 P42 (OZ)
r

 

m

(3.147)

 

159

With (3.147), (3. 130) can be rewritten as

2
I r(y) = ——-—8n ZL 33 x
4y 11 £0 50

r.
Eos pohl - cos {30(hl + hz) + $— P4ls(o) + gl—Z- P428(o) X
s
s

 

Ein $0011 + h‘2 - lyl) + 31,— P41r(y)+ (biz P42r<yfl
1‘
r

 

 

_

 

F— ——
(be cos (30(hl + hz) + 5%]{goér cos [30 (hl + hZ) + 39:]

 

 

+ 32" ZL Em l3olhl + 1‘2) + 31>— P41r(°) + @12 P42r(o):l}
—' r
1'

Similarly, (3.133) can also be rewritten as

2
817 Z E
r _ _ L o
Elx ”all ’ 40 50 x

 

l:cos fiohl - cos [30(hl + 112) + $1;- P4ls(o) +71%? P428(O‘)—"X
s

[:3 in 50x + 61; Pllr(x) + 3}: Plzrlxfl
r

L—

_»

 

I'— (1)3 [(205 50(h1 + ha) + %j{ £0431. [cos {30(h1+h2) + 33-3 ——l

 

 

—

+j21rZL Em 50011 + hz) + 5}- P4lr(o) + (Pl—2 P42r(oa}_—J
r

r

(3.148)

(3.149)

160
Thus, the total loop currents in (3.144) and (3.145) can be expressed

as the functions of the 100p dimensions and the loading impedance ZL'

(B) If consideration is given to the sub-first order solutions of currents

[I4yr(yfl in (3.136), [:leflxil in (3.139)’[:I4y5(y)] in (1.173)
10 10 10

and [IIXSWEI in (1. 176), V in (3.146) is obtained as
10

E
. o G
- J41TZL T30 é]? COS [30011 ‘1‘ ha) '1' E]

 

(1)8 Eos {30(h1 + hZ) + 39;]

1
E cos pohl - 2 cos 50(h1 + hz) + 5; U418(o)+ ;p—l-z U428(ofl
s

 

god’r cos 130(h1 + hz) + £31

+j21TZL 2 sin 130ml +h2)+31)— U41r(o) + 317 U4zr(o]
r

n __ r __.__J

 

 

'(3. 150)

With (3.150), (3.136) can be rewritten as

 

161

r
‘ r

E COS Bohl - 2 cos [30(h1 + hz) + $;U4ls(o) + $17 [1428(0)] x

 

 

 

 

__ J—:-:l
<I>s cos (30ml + hz) + Egjxigoor cos [50(h1 + ha) + 3):]
. . l r 1 r
+ JZTTZL 2 Sin {30(h1+ hz) + E- U41 (0) + $7 U42 (ofl
i—— r —.
r
(3,151)
Similarly, (3.139) can be rewritten as
81TZZL Eo
Elx bi] = "—2"— “T3" x
10 o o
1 s l '
2cosfih-2cosfi(h +h)+—U (o)+—U (o)X
l: o 1 o 1 Z (I) 41 <I> 2 42
s
. l 1
E Sin [30x + ; U11r(x) + $7 U12r(xa
r

 

 

r

. , 1 r
EJZTTZLEZ Sln 130(h1 + hZ) + 71>: U41

(”+317

1‘

 

 

E (30550011 +h2)+£;:l{§o<br cosfio (hl +h‘2 )+C::J “’

 

U42r(°);L

(3.152)

162

3.4. Total Backscattered Field

The vector potential, maintained by the induced current in the
radiating 100p, at an arbitrary point on the axis of the [00p and in the

far zone of the loop (Fig. 3. 3) can be expressed as

 

Ar-AIU—BX‘2 21 r(1)eo d'
- y — 4n 4y Y R Y
-hz
H -1floRo hZ
:- ._0. e S I r(u)d1
217 R0 4y Y Y
'hz (3.153)
where
_ 2 ,2 2.,
R _ JRO +y +h1 — R0 (3.154)
3.4.1.

Total Backscattered Field Based on First Order Currents

With (3.148), (3. 153) can be rewritten as
(”Ar 1. _ A r _ 41r/JOZL e o o
L J ‘ ‘ "'1" (3 Eo R x
11 Y 11 o o

1 1
E05 (30111 — cos po(h1+h2)+ 21>; p4ls(o) + <37 p425(o)] K1

8

 

 

F '7
G G
<I>s cos (30(h1+h2)+-5;]{ §O<I>r cos Bo(h1+h2)+ 5;]
. . l r 1 r —l
+521rzL Sinflo(hl+h2)+$— 1341 (o) + :2- p42 (oflj
r
L. r __)

 

(3.155)

163

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

3x

Figure 3. 3 Geometry for the calculation of the radiation
field of a radiating rectangular loop.

164

where h r r
2 . P41 (1") P42 (Y')
K1: S [:51nfio(hl+h2- ly'l)+—$r——+——¢—T— dy'
-h r
2

(3.156)

K1 will be evaluated in a following section.

The backscattered electric field due to the radiating 100p is

-jl30Ro

r _ _. r _ _ . e
[EV] — 31.01:ij - J4TTEOZL R0 x
11

ll

 

Eos Bohl - cos (30(h1 + hz) + RID-8.13418“) + g-l-z P428(o) K1
8

 

<I>s cos (30(111 + hz) + $311.04; Eos (30011 + 112) + gr]

 

 

+ jZTI’ZL sin )1'30(h1 + hz) + 3% P4lr(o) + 317- P42r(ofl _J
'— r
r

(3.157)

The total backscattered field from a loaded rectangular 100p

is obtained with (l. 183) and (3.157)

165

E31“ {Bill {Bill

 

 

 

 

 

-jpoRo 1
= 2E e x J -j «z x
l 2 L
0 R0 (PS cos (30(h1 + hZ) + 5%]
1
E05 Bohl - cos (30(h1 + hZ) + 51- P4ls(o) + 7 1:34‘23(o):_l—__K1
§O<I>r cos (30 (h + 112) + %] +j21TZL sin (30 (h + hz)
+ 1 p r l r
E— 41 (O) + —2- p42 (0)
1—— r <I>r __.1.
(3.158)

3. 4. 2. Total Backscattered Field Based on Sub-first Order Currents

With (3.151), (3. 153) can be rewritten as

e-jfiR
[Arj1o:[Ayr‘—lo %_ 0 ° 5:: l:4v(y:ll Ody'

ll

-J'B R
41THOZL e o o

=———— E
4050 o R

 

O

Ecosfiolh -2cosfiol(h +h2)+$_U4$1(0)+q)—-1_2—U 480le
s

__ G l _ _—
lbs cos [50(h1 + hz) + 35;]{c’ocbr [cos (30(hl + hZ) + 5: G__'

 

g

 

+ jZTrZL 2 sin (30(h1 + hz) + 3}- U41r(0) + $17 U42r(°):ll
r
r __

(3.159)

166

where h

2
_ . , 1 r , l r . .
K2 —5‘ [:ZSInfiO(h1+h2-1Yl)+q)—U4l (y)+—@2U42 (y']dy
-h r r
2
(3.160)

Then, the backscattered electric field due to the radiating 100p

is obtained as

-16 R

0 O
r _ r __. __. e
[E] [Y] ' WEE] ’ WEOZL R, x
10 10 10

[:2 cos (30111 - 2 cos (30(111 + ha) + 3}; U413(o) + 317 U428(o)] K2
8

(be [:cos (30(h1 + hz) + %]{§o¢r [:cos [30011 + 112) + 3%] fl

 

+32sz [:2 sin (30011 + hz) + $1— U41r(o) + 3% U42r(o):]}
r
—- r —‘ (3.161)

 

 

The total backscatte red electric field due to the loaded rectangular

100p can be calculated by superposing [Ea in (1.188) uponEEr]
10 10

in (3.160) as

151 =51, +1211.

'15 R
e

O O 1

o (ESL-cos (30(h1 + hZ) + 331: C1]

2153

 

 

o R
.12 - JZTTZLX

1 1 s
[2 cos (30111 - 2 cos (30(111 + hz) + 6; U4ls(o) +3—2— 114‘2 (0):] K2

 

 

 

“m
Q0 <19 r[cos (30 (h1 + 112) +——- 1C3 + jZTTZL [2 sin Bo(h1+ hZ)
(pr
1 r 1 r
+ — U (o) + —— U (o)
<1>r 41 (1)12 42 :l
L_. _—
(3.162)

3.4. 3. Evaluations of K1 and K2

From (3.156) we note that
Z 1
5‘ Emfio(hl+h2-1Y'l)+§—P 41 (Y) (1,]? 42 (dey
-h

where

P41r(y) : Br (:08 BOY - G sin B01371 '1‘ Sin $0011 '1' ha) th(Y)

168

P42r(Y) = Ro Br tcdw)

then

K1 = é [cos Bohl - cos (30(h1 + hag

1 r . '
+ ?r. {2 B Sin flohz - G [1" COS BthJ + Bo Sln F30011+ hZ) de}

1
+—2 fiOBb

‘13 cd (3.163)
r
where 1 -
h2 h2
bCd : Eh th(y')dy' : 5 E 0 Tcd(S) cos 60(y' - s)da dy'
2 , 'hz (3.164)

Similarly, from (3.160) , we have
h
2 . 1 r l r
K2: 5 [28111130011 +112 - ly'l)+§: U41 (Y')-lQ—Z- U42 (Y'E'dY'

-hz 1‘

where

r r . .
U41 (y) 2 2B cos (30y - Sin [30(h1 + hZ) Tca(Y) - G Sln BOW.)

+ cos (30(h1 + hZ) TSa(Y)

114211,.) = - 13r Tea (y) + G TS (y)

a

169
Then

 

 

‘13
r
(3.165)
where
hZ
tca :5 Tca(y') dy' (3.166)
..h2
h2
tsa = 5‘ Tsa(y') dy' (3.167)
-h

3. 5. Optimum Impedance for Zero Backscattering

The optimum impedance for zero backscattering can be

obtained by letting total backscattered field equal to zero.

3. 5.1. First Order Optimum Impedance for Zero Backscattering

Letting Fly] in (3.158) equal to zero and solving for ZL’
' 11

we have

170

-ng<I> [cospai +h2)+-(I)l: o]

 

1 o r o 1
[2.] = _.
012P428(°):1

cos h1 - (hl +h2)+ S(°)+
K11: Ro COS13o 351;}, 41 «D

S
211

 

. 1 1
- .11 sin (30(111 + hZ) + F1»: P41r(o) + 37 P4Zr(ofl
r

L._.

 

 

(3.168)

3. 5. 2. Sub-first Order Optimum Impedance for Zero Backscattering

Letting [1533:] in (3.161) equal to zero and solving for ZL, we

get

[21

. 1
.. Jngocbr cos (30(111 + hz) + If; G]

 

O SE2 cos Bohl - 2cos {30(h1 +h2)+§1— U418(o)+—1-2- U428(o):]
s <I>s
2n
- JZE sin (30(111 +h2)+3}— U41r(o)+ +4711 U42 r(o):]
_ r @1‘ ___.1

 

 

(3.169)

3. 6. Numerical Examples

The numerical results are obtained for a square 100p with
E- : O. 0388 . (Pr is numerically calculated as functions of 50h

and is plotted in Figure 3. 4. The Optimum loading for zero back-

scattering [21] in (3.168) is then calculated as a function of

171

 

O
Hflp.

 

Figure 3.4. <I>r as a function of (30h (a/h = O. 0388).

172

00h for the case ME: 0. 0388. Numerical results of [21:] are
0

shown in Figure 3. 5.

3. 7. Experiment

The Optimum impedance for zero backscattering from a
rectangular loop requires both the resistive part and the reactive
part shown in Figure 3. 5. To simplify the problem, an experiment
was conducted for a conducting square loop loaded with two identical
reactive impedances. Experimentally, this reactive loading method
proved to be quite adequate in reducing the backscattering cross

section of a square loop.
3. 7.1. Experimental Arrangement and Measuring Technique

The experimental setup and measuring technique were
identical to the case of the circular loop mentioned in Chapter 2

and no further description seems to be necessary.
3. 7. 2. Experimental Results

A square 100p (side length = 5.15 cm) were constructed as
an experimental model using cylindrical wire of 0.1 cm radius. The
experiments were conducted at various frequencies. The experimental
results are shown in Figure 3. 6.to Figure 3.16 in which the back-
scattering cross sections of the square 100p were plotted as functions
of loading impedances at each particular frequency. The solid curve

represents the backscattering cross section of a loaded square 100p

zoo
220
200
180
150
T140
120

0 100
(ohms) 80
oo

40

20

-40
-80

-100

-120

 

 

 

 

 

 

 

 

 

Fi ure 3. 5. Optimum impedance for zero backscattering
g from a square loop (a/h = O. 0388).

174
and the solid straight line represents that of the solid quare loop. It
is observed that if the loading impedance (or the length of the coaxial
line) is prOperly adjusted, the backscatters of the square loops can
be minimized to the noise level. About 15 db reduction in the back-
scattering cross section was obtained in the experiments. In Figure
3.17. the optimum reactive impedances for minimum backscattering

are plotted as a function of (30h.

175

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10.

REFERENCES

Kouyomnjian, R. G. , ”Back-Scattering from a Circular
Loop, " Ohio State University Engineering Experiment
Station, Bulletin No. 162 (November 1956).

Weston, V. H. , ”Scattering from a Circular Loop, "
University of Toronto (1957).

Chen, K. M., and. King, R. W. P., "A Loop Antenna
Coupled Electromagnetically toaa Four- Wire Transmission
Line, ” Scientific Report 5, Cruft Lab. , Harvard University,
Cambridge, Mass. (January 1960).

Chen, K. M., and Liepa, V., "Minimization of the back
scattering of a cylinder by a central loading, " IEEE Trans.
Antennas Propagation AP-IZ , 576 (1964).

Chen, K. M. , ”Minimization of backscattering of a cylinder

by double loading, " IEEE Trans. Antennas Propagation

AP-13, 262 (1965).

Chen, K. M., "Reactive loading of arbitrarily illumined
cylinders to minimize microwave backscatter, " J. Res.
Nat'l. Bur. Std. 690, 1481 (1965).

Hu, Y. Y. , ”Backscattering cross section of a center-loaded
cylinder antenna, " IRE Trans. Antennas Propagation AP-6,
140 (1958).

RS, B. 0., and Schmitt, H. J., "Backscattering cross section
of reactively loaded cylindrical antennas, " Scientific Report
18, Cruft Lab., Harvard University, Cambridge, Mass.
(August 1958).

Liepa, V., and Senior, T. B. A., "Modification of the
scattering behavior of a sphere by reactive loading, " Proc.
IEEE _53, 1004 (1965).

Storer, J. E. , ”Impedance of thin-wire loop antennas, " Trans.
AIEE (Communication and Electronics) ]_5, 606 (1956).

187

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