MULT‘ - EMPEDANCE LOADiNG
0F UNEAR ANTENNAS FOR
IMPROVED. TRANSMlTTlNG’ AND
RECEIVING CHARACTERISTICS

T hesi's for the Degree»? Ph..D.‘ ‘
MlCHiGAN STATE UNNERSITYV ; _
PHMP LYLE .FANSONF' .
1972

 

Laws. 'Y

5

Michigan tare
UHIVCISIL')’

 

This is to certify that the

thesis entitled
MULTI-IMPBDANCB LOADING OP LINEAR ANTENNAS

FOR IMPROVED TRANSMITTING AND RECEIVING
CHARACTBRI QTICS

presented by

Philip Lvle Panson

has been accepted towards fulfillment
of the requirements for

_P_h..D.__degree in W11 Engr.

7
{xi/A72» [/évx.

Major professor

/

 

Date '72—“30; /972.

 

0-7639

 

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"0A8 & SUNS”
n, 800K BINDERY WC. .

      

Lsa=nav smoms fl ‘3

”mum: mum ‘1. '-i

 

ABSTRACT

MULTI-IMPEDANCE LOADING
OF LINEAR ANTENNAS FOR IMPROVED

TRANSMITTING AND RECEIVING CHARACTERISTICS

By

Philip Lyle Fanson

This thesis presents a method by which the trans-
mitting and receiving characteristics of a linear antenna
may be improved or modified by the use of multi—loading.

The idea of modifying the antenna characteristics
by impedance loading is not new. In recent years many
researchers in the antenna area have studied the technique
both theoretically and experimentally. Most works,
however, are restricted to single or double loading. A
few experimental studies dealing with multi—loading lack
theoretical explanation. In this thesis, the general
case of multi-loaded transmitting and receiving linear
antennas is investigated based on a systematic and rigorous
approach.

The main findings of this thesis are as follows.

For a transmitting antenna if (1) the antenna current is
specified at M + N points, (2) the radiation field is
specified in M + N directions, (3) the input impedance

is specified at M + N different frequencies, or (A) any

Philip Lyle Fanson

linear combinations of case (1), (2), and (3), then M + N
complex impedances can be found at M + N specified
locations. For the case of pure reactive loading with

M + N reactances, only (M + N)/2 values of the antenna
characteristics can be specified. An interesting case of
loadings which lead to the resonance or instability of a
transmitting antenna is discussed. For a receiving
antenna, M + N complex impedances can again be found at

M + N specified locations if (1) the antenna current is
specified at M + N points, (2) the scattered field is
specified in M + N directions, or (3) the combination of
cases (1) and (2) and if the central load is specified.
The case of loading which will cause resonance or

instability in a receiving antenna is also studied.

MULTI-IMPEDANCE LOADING
OF LINEAR ANTENNAS FOR IMPROVED

TRANSMITTING AND RECEIVING CHARACTERISTICS

By

Philip Lyle Fanson

A THESIS

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

DOCTOR OF PHILOSOPHY

Department of Electrical Engineering

1972

ACKNOWLEDGMENTS

The author wishes to express his appreciation to
his major professor, Dr. K. M. Chen, for his guidance and
support through the course of this work. He also wishes
to thank the other members of his guidance committee,
Drs. D. P. Nyquist, J. S. Frame, J. Asmussen, and B. Ho,
for their time and help.

The author wishes to specially thank Dr. Frame for
his help in developing the minimization method for the
non-linear equations.

Finally, the author wishes to thank his wife, Elaine

and children, for their support during this work.

11

TABLE OF CONTENTS

LIST OF TABLES

LIST OF FIGURES

I.

II.

III.

INTRODUCTION

MATHEMATICAL FORMULATION OF A MULTI—LOADED
RADIATING ANTENNA . . . . . . .

2.1
2. 2

Matrix Formulation

 

Resonance Loading . .

 

2. 2.1 Admittance Constraint
2. 2. 2 Resonance Loading .
2. 2. 3 Resonant Loading - Unstable

Conditions . . . .

2.2.3.1 Numerator Of Vl And V2
Zero . .

2.2.3.2 Coefficient Of V2 Zero

.2.“ Voltage Constraint . . . .

2.2.“.1 Field Pattern
Constraint

2.2.A.2 Current Constraint

2.2.4.3 Input Admittance -

Frequency Constraint

.2.5 Combined Admittance And

Voltage Constraint

Reactive Loading -

2.2.5.1
Current Constraint
2.2.5.2 Reactive Loading -

Frequency - Current

CURRENT DISTRIBUTION ON A LOADED
RADIATING ANTENNA . . . . . .

3.1

Current Distribution On A Short

 

3

3.1
3.1

Antenna . . . .
1.1 Vector Potential of
Symmetric Dipole
.2 Difference Equation
.3 Current Distribution

iii

11
11
12

14

l6
17

17
2O

21
23
23
27

31

31

31
3A
35

IV.

VI.

VII.

 

1.“ Evaluation Of Constants

.1. 5 Special Cases . . . . . . . .
3.2 urrent Distribution On A Long Antenna

2.1 Current On An Infinitely

Long Antenna . .
.2.2 Reflected Current At The

End Of A Semi— Infinite Antenna
.2.3 Current On An Asymmetrical
Finite Antenna . .

RADIATION PATTERNS OF SHORT AND LONG
ANTENNAS . . . . . . . . . . . . .

“.1 Radiation Pattern Of A Short Antenna.
A.2 Radiation Pattern Of A Long Antenna

 

 

NUMERICAL SOLUTION OF THE SIMULTANEOUS
NON— LINEAR EQUATIONS AND NUMERICAL
EXAMPLES . . . . . . . . . . .

.1 Method Of Minimization . . . . .
.2 Example Of Minimization Method -
N = 2 . . . . . . . . . . .
Modified Newton's Method
Additional Techniques . . . .
Numerical Examples . . . . .
5.5.1 Admittance Constraint . . . .
Field Pattern Constraint . . .
Current Constraint
Input Admittance - Frequency
Constraint . . .
5.5.5 Reactive Loading - Current
Constraint . . . . . .
5.5.6 Reactive Loading — Frequency
And Current Constraint .
5.5.7 Resonance Loading .

 

 

 

 

U‘IUTU'I mm
U'IJr-‘LJO

 

U1U1U1
UTUTU'I:
tuck)

MATHEMATICAL FORMULATION OF A MULTI-LOADED
RECEIVING ANTENNA . . . . . . . . . .

6.1 Matrix Formulation
6. 2 Constraint Equations .
6'.2. 1 Load Impedance Constraint .
6. 2. 2 Scattered Field Constraint . .
6. 2. 3 Current Constraint . . .
6. 2. A Current And Scattered Field
Constraint . . . . . . .

 

 

CURRENT DISTRIBUTION AND SCATTERED FIELD
PATTERN OF AN UNLOADED ROD

7.1 Vector Potential or An Unloaded Rod

,7.2 Current Distribution On A Short Rod

iv

39
M2
AZ
A6
A7

61
61
63
6A
66
68
68
71
76
83
85
85
92
92
96
97
100

102

107

107
109

7.3 Scattered Field Of A Short Unloaded

 

 

 

 

 

 

 

 

Rod . . . . 111
7.“ Current Distribution On A Load Rod. . 112
7.“'.l Current On An Infinitely
Long Rod . . . . . 112
7.“.2 Reflection At The End Of A
Semi- Infinite Rod . . . . . . 113
7.“.3 Current On A Finite Rod . . . 11“
7.5 Scattered Field Of A Long Rod . . . . 115
VIII. ADDITIONAL NUMERICAL EXAMPLES . . . . . . 118
8.1 Resonance Loading . . . . . . . . 118
8.2 Scattered Field Constraint . . . . . 120
8.3 Current Constraint . . . . . . . 123
8.“ Current And Scattered Field
Constraint . . . . . . . . . . . . . 123
IX. CONCLUSIONS . . . . . . . . . . . . . . . 131
BIBLIOGRAPHY
APPENDIX

5.

1

LIST OF TABLES

Solutions To Input Impedance—Frequency
Constraint (Four Frequency Case)

vi

77

Figure

Figure

Figure
Figure
Figure

Figure

Figure

Figure

Figure

Figure

Figure

Figure

5.3

5.“

5.5

5.6

5.7

At Four

LIST OF FIGURES

Multi-Loaded Transmitting Antenna . .

Voltage Equivalent Of A Multi-
Loaded Transmitting Antenna .

Asymmetrically Driven Antenna . . . .
Symmetrically Driven Antenna . . . . . .
The c Plane And The Contour CO . . .

Current Distribution For A
Transmitting Antenna Which Was
Loaded With 2“OQ at d1 = 3A/8
(a = 0.00635A) . . . . . . . . . . .
Current Distributions For A Transmitting
Antenna Which Was Loaded With -J 363 Ohms
At d = 0.792 Meter (f = 600 MHz

And a = 0.00318) . . . . . . . . . . . .

Field Pattern For A Transmitting
Antenna Loaded At Nine Points
(PT(O) = Sin(3e))o o o o o o o o o o o o 0

Field Pattern For A Transmitting Antenna
Loaded At Four Points (PT(O) = sin(2e)).

Current
Antenna
At Four

Distribution For A Transmitting
On Which The Current Was Specified
Points (Traveling Wave). . . . .

Current
Antenna

Distribution For A Transmitting
On Which The Current Was Specified
Points (Decaying Wave) . . . . . .

Input Impedance vs. Frequency For A

Transmitting Antenna (Four Frequency Case -

509 " SOlUtion I)o o o o o o o o o o o o 0

vii

32
“5

69

7O

72

73

7“

79

U1

.10

.11

.12

.13

.1“

.15

.16

Input Impedance vs. Frequency For A
Transmitting Antenna (Four Frequency Case -
509 "' 80111131011 II) o o o o o o o o o o o o o o 0

Input Impedance vs. Frequency For A Transmitting
Antenna (Two Frequency Case - Frequency
Dependent) . . . . . . . . . . . . . . . . .

Input Impedance vs. Frequency For A Transmitting
Antenna (Two Frequency Case - Frequency
Independent) . . . . . . . . . . . . . . . . . .

Current Distribution For A Transmitting
Antenna On Which The Current Was Specified At
Two Points (Reactive Loading) . . . . . . .

Input Impedance vs. Frequency For A Transmitting
Antenna (Two Frequency - Reactive Loading -
Frequency Independent) . . . . . . . . . . . . .

Input Impedance vs. Frequency For A Transmitting
Antenna (Two Frequency - Reactive Loading -
Frequency Dependent) . . . . . . . . . . . .

Loading Impedance vs. Frequency For A Trans-

mitting Antenna To Maintain A Rbsonance
Instability (Finite Voltage) . . . . . . . . . .

Loading Impedance vs. Frequency For A Trans—
mitting Antenna To Maintain A Resonance
Instability (Zero Coefficient) . . . . . . . .

Graphical Solution Of Resonance Loading
Condition For A Transmitting Antenna

Multi-Loaded Receiving Antenna . .

Voltage Equivalent Of A Multi-Loaded
Receiving Antenna . . . . . . . . .

Unloaded Rod . . . . . . . . . . . . .

Load Impedance vs. Frequency For A Receiving
Antenna To Maintain The Resonance Instability

Graphical Solution Of Resonance Loading
Condition For A Receiving Antenna .

Field Pattern For A Receiving Antenna Specified
At Three Points (PT(O) = sin 20) . . . . . .

viii

80

81

82

8“

86

87

88

89

91
93

9“
108

119

122

8.“

8.5

8.6

8.7

8.8

8.9

Current Distribution For A Receiving Antenna
For Which The Central Load And Current Were
Specified At Four Frequencies

(f = 300 MHz) . . . . . .

Current Distribution For A Receiving Antenna
For Which The Central Load And Current Were
Specified At Four Frequencies

(f = 320 MHz). . . . . . . . . . .

Current Distribution For A Receiving Antenna
For Which The Central Load And Current Were
Specified At Four Frequencies

(f = 3“O MHz) . . . . .

Current Distribution For A Receiving Antenna
For Which The Central Load And Current Were
Specified At Four Frequencies

(f = 360 MHz) . . . . . . .

Scattered Fields For A Receiving Antenna For
Which The Current And Radiation Pattern Are
Specified At Two Frequencies:

(a) f = 300 MHz e- (b) f = “00 MHz

Current Distributions For A Receiving Antenna
For Which The Current And Radiation Pattern
Are Specified At Two Frequencies:

(a) f = 300 MHz And (b) f = “00 MHz

ix

12“

125

126

127

128

129

CHAPTER I

INTRODUCTION

This study will present a method by which the
transmitting and receiving characteristics of a linear
antenna may be improved or modified by the use of multi-
loading.

The idea of modifying the antenna characteristics
by impedance loading is not new. In recent years many
researchers in the antenna area have studied this technique
both the theoretically and experimentally. However, most
works are restricted to single or double loading. A
few experimental studies dealing with multi—loading lack
theoretical explanation. In this study, the general case
of multi-loaded transmitting and receiving linear antennas
is investigated based on a systematic and rigorous approach.

In Chapter 2, the basic antenna characteristics are
GXpressed in matrix form as functions of loading impedances
and antenna parameters. In addition the desired antenna
characteristics are expressed in the form of constraint
eQuations. These constraint equations are combined with
matrix equations to determine the solutions. In Chapters
3 and “, antenna currents and field patterns for both

short and long linear antennas are derived. Chapter 5 iS

1

2

devoted to numerical techniques and examples of transmitting
antennas. Chapter 6 is concerned with the development
of the matrix and constraint equations for a multi-loaded
receiving antenna. In Chapter 7, the current and scattered
field of an unloaded receiving antenna, both short and
long, are presented. Finally, Chapter 8 presents additional
numerical examples of a multi-loaded receiving antenna.

The main findings of this study are as follows. For
a transmitting antenna if (1) the antenna current is
specified at M + N points, (2) its radiation field is
specified in M + N directions, (3) the input impedance
is specified at M + N different frequencies, or (“) any
linear combination of cases (1), (2), and (3), then M + N
complex impedances can be found at M + N specified loca—
tions. For the case of pure reactive loading with M + N
reactances, only (M + N)/2 values of the antenna
characteristics can be specified. An interesting case of
loadings which will lead to the resonance or instability
of the antenna is discussed. For a receiving antenna,
M + N complex impedances can again be found at M + N
specified locations if (1) the antenna current is specified
at M + N points, (2) the scattered field is specified in
M + N directions, or (3) the combination of cases (1) and
(2) and if the central load is specified. The case of
loading which will cause resonance or instability in a

receiving antenna is also studied.

CHAPTER II

MATHEMATICAL FORMULATION

OF A MULTI-LOADED RADIATING ANTENNA

In this chapter the basic equations which mathematical-
ly describe a multi-loaded radiating antenna will be derived.
The functional relationships between the driving voltage,
the loading impedances, and the antenna dimensions will be
developed. Finally constraint equations will be derived

based on the required antenna performance.

2.1. Matrix Formulation

 

The problem of a multi-loaded antenna can be treated
by considering each load as a voltage driver at an arbi-
trary driving point and applying the superposition principle.
Consider a center-driven antenna of length 2h, and radius a,
having N + M arbitrary loads along its axis, as shown in
Figure 2.1. If each load is replaced by the voltage induced
across it, the problem becomes equivalent to the antenna in
Figure 2.2. This second antenna can easily be broken into
M + N asymmetrically driven antennas as shown in Figure 2.3.
Since the solution for the current on an asymmetrically
driven antenna is a function of frequency, w; radius, a; half

length, h; and position of the driver, dj’ the current at
3

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Figure 2.1 Multi-Loaded Transmitting Antenna

 

V-(M-l)

-M

+1

[GIG-6.919] l9] IGIG)

+ I
+——
Q
I A
.3
l—J
a.
l
3

 

 

 

 

 

 

+ I
+
d l dN
_ N-l
+
d1

+ I

 

 

I +

 

+ I

 

 

+ I

 

 

 

 

 

 

 

(9‘6!

Figure 2.2 Voltage Equivalent Of A Multi—Loaded

Transmitting Antenna

 

 

 

 

 

 

 

 

d 4
d3
6
A
2a -> ‘—
d i

 

Figure 2.3 Asymmetrically Driven Antenna

7

any point can be expressed as

IJ(z) = ij(m,a,h,dj,z) 3: -M to N. (2.1)

To find the voltage Vj’ at each load on the original antenna,

the relation VJ = ZJI t<dj) is used where 23 is the
impedance at dJ and It(dj) is the total current at d1. By
superposition,
N
It(dj) = £_Mli(d3).
Thus
N
V = -Z 0 I d .
J J i£~M 1‘ J)
Using Equation 2.1, the result is
N
V = V1 f ,a, h ,d d
or
N
-Y V = V f a, h ,d d
11 £441” 1’1)

where YJ = l/Zi' Collecting all the terms on the right-hand

side, the result is

N
J) + i£_M vir(a,a,n,di,dJ

i¢0.J

0 = Vof(w,a,h,o,d )

+ v (f(w,a h d ,d ) + Y

J , , J J ). (2.2)

J

If J runs from -M to N, the result is M + N equations of the

form of Equation 2.2. Writing these in matrix form,

F-V + GVO = O (2.3)

 

 

 

 

 

 

where
r d 7
f + Y d d d
11 1 0 le f_11 . f-Ml
d d d d
le O O fNN + YN f-lN ..... O f-MN
F =
f1-1 ' fN-l f-l-l -1 f-M-l
d d d d +
fl-M fN-M f-l-M ..... . f-M-M Y-M
__ ._
f01 V1
d
f0N VN
G = 9 V =
d
f0-1 V-1
d l
f V
L.0'H1 L_ -MJ
d -
and fik "‘ f(w’a’h’di’dk)o

Before proceeding, the number of unknowns in the M + N
equations represented by 2.3 needs to be determined. There
are M + N voltages, M + N admittances, and M + N distances
plus a, h, and w. Since a, h, and w are usually determined

by the problem being solved, they will be treated as known

9
quantities. It should also be noted that the voltages and
admittances are complex while the distances are real. Thus
there are 5(M + N) real unknowns; 2(M + N) real voltages,
2(M + N) real conductances and susceptances, and M + N real
distances. Since Equation 2.3 is also complex, it represents
2(M + N) real equations. Therefore, 3(M + N) real constraint
equations must be introduced before Equation 2.3 can be
solved.

A close examination of f(w,a,h,di,dk) will reveal that
di and dk can not be solved for explicitly. Thus the dis-
tances are taken as known variables, the same as a, h, and w.
This reduces the unknowns to “(M + N) real,and the constraint
equations to 2(M + N) real.

The number of unknowns will be reduced further if only
symmetrical problems are considered, i.e., a monopole
antenna on a ground plane. For this case di = d—i’ Vi = V_i
N = M, and Y1 = Y-i' Thus Equation 2.3 reduces to N complex

equations of the form

F'V' + G'VO = O (2.“)

 

 

 

10

 

 

 

where
—-d d d —1
s s s
fll Y1 fil le
, = ds ds ds
F fli fii + Yi . fNi ,
ds ds ds
le le fNN + Ii
rd Fv
01 1
G' fd V' = V and
01 i
d
f V
0N4 _ NJ
ds d
f f
13 '13 1'3,

and the number of unknowns in Equation 2.3 to 2N complex.

Thus only N complex constraint equations are needed for a

SOlution.

11

2.2 Constraint Equations

It has been shown that N + M complex constraint
equations are needed to obtain a solution for the general
case and N complex constraint equations for the symmetrical
case. This number of equations is exactly equal to the
ruumber of unknown voltages or loading admittances. This
leads to three types of constraint equations: those that
cietermine the voltage, those that determine the loading
eadmittances, or those that determine part of the loading
eadmittances and part of the voltages. These three types of

eequations will now be examined.

2.2.1 Admittance Constraint
While the loading admittances can determine the

current and radiation pattern of an antenna, they can not be
explicitly related to the pattern on the current. This
implies that the constraint equations determining the ad-
mittance must be based on other factors, i.e., available
components, etc. This reduces these admittances from un-
knowns to known variables. Once the admittances are given,

Equations 2.3 and 2.“ can be easily solved by inverting F

and F' respectively, i.e.,

V = -F'1Gvo (2.5a)
t...11
v' = -F G VO (2.5b)
- '—
provided F l and F l exist. The case of the inverses not

existing will be covered in sections 2.2.2 and 2.2.3.

12
(Dnce V or V' is found, both the field pattern and the
<3urrent distribution can be determined.

2.2.2 Resonance Loading

In section 2.2.1 it was assumed that F_1 and

I?'-1 existed. This section and the next will investigate
tzhe results if these inverses do not exist. For simplicity,
t:he discussion will be limited to the symmetric case and

IV = 2. The results, however, can be extended to the

ggeneral case.

For N = 2,F' and G' become

 

 

 

 

 

.— ds ds .-
fll + Yl f21
F' =

ds ds

f12 f22 + Ya‘

and

rd-T

01
G' = . If F' had an inverse, by Kramer's
fd
L 051
rule

d ds

f01 f21
d

f Y'

V =V O2 2 (2.6)
1 o

' ds

Y1 f21
ds '

fl2 Y2

 

 

13

 

 

 

 

 

and
' d
Y1 f01
ds d
,f12 f02
V2 = V0 (2.7)
' ds
Yl f21
rds Y'
12 2
' _ ds ' _ ds
tvhere Y1 - fll + Y1 and Y2 — f22 + Y2. If the

(determinants in the numerators of Equations 2.6 and 2.7 are

'-
riot zero, then at any point where F 1 does not exist, V

1
21nd V2 must approach infinity. The input admittance, Yo’
is given by

_ d ds ds
VOYO — VofOO + Vlf10 + V2f20. (2.8)
l
Because IF I = 0, V1 and V2 are related by either the first
or second equation in the matrix 2.5b, thus solving the
first equation for V1
£31 fgl
V=-—rV--—,- V. (2-9)
1 Y 2 Y o
l 1

Combining 2.8 and 2.9 and dividing by V0, Y is given by

0

ds d ds ds
_ d f10f01 V2 ds f21f10
o 00 V 20
Y1 0 Y1

Providing the coefficient of V2 is not zero, then YO must

also approach infinity. The cases of the coefficient of V2

1“
equal to zero and the numerators of Equations 2.6 and 2.7
(equal to zero will be covered in section 2.2.3.

To see the significance of V1, V2 and Y0 going to
:infinity, consider a simple L-C series circuit consisting
<>f two inductors, L1 and L2, and two capacitors, Cl and C2.
Iget V1 and V2 be the voltage across L1 and L2 respectively
sand VO be the driving voltage. The admittance at the

l
ciriving point, YO would be given by

Y; = 1/(Jw(Ll + L2) — 3(0l + C2)/C1C2a).

Vihen w is adjusted so that the system is in resonance, the
V :
cienominator of the expression for YO goes to zero and Y0

zapproaches infinity. In addition since V0, L1 and L2 are

alssumed finite, the current in the system, and thus V and

1
'V2, must also approach infinity. Thus by analogy the

antenna can be thought to be a series resonant circuit with

distributed components.

2.2.3. Resonant Loading - Unstable Conditions

In Section 2.2.2 indeterminate conditions were

deferred, these will now be discussed.
2.2.3.1 Numerator Of V1 and V2 Zero

If the determinant in the numerator of
Equation 2.6 is also zero, then Vl becomes indeterminate.

TO determine the value of V1 a form of 1'Rospita1's rule

1
will be used. Solving the numerator of V1 for Y2 gives

15

I
Y _ fds f.d

d
2R ’ 21 f

02/ 01, (2.11a)

£3ince the denominator of Equation 2.6 must also be zero,

' ' _ ds ds
Y2RY1R ' f21f12' (2’12)
'
[Ising Equations 2.11a and 2.12 and solving for Y1R gives
' _ ds d d
Y1R - f12f01/f02 . (2.11b)

'
IIt should be noted that for N = 2,Y1R also causes Equation
23.7 to become indeterminate. For N > 2, however, this is

riot true, up to N - 2 of the voltages can be indeterminate

Inithout all of them being indeterminate.

To find V1 and V2 let

Y2 = Y2R + e (2.13a)
Y1 = Y1R + as (2.13b)

undere e approaches zero and a is the tangent of the complex
euugle from which the point is approached. Substituting

Ekguations 2.13a-b into Equations 2.6 and 2.7 and neglecting

the second order 5 terms gives

l

d '
Vl — -VOfOl/(Y1R + aY2R) (2.1“)

d ' ' V
v2 — -VOfO2/(Y1R + aY2R). (2.15)

Since a can be any number in the complex plane, the values
of both V1 and V2 are dependent on how the indeterminate

point is approached. This is typical of an instability point.

16

2.2.3.2 Coefficient Of V2 Zero

In Section 2.2.2, the second indeter-

rninate condition was that if the coefficient of V in

2
IEquation 2.10 equals zero, Equation 2.10 became indeter-

rninate. Setting that coefficient to zero and solving for

V
Y1 gives

' _ fds f.ds

10 ' f2150/1423 ' (2°16)

Y

ligain since the denominator of Equation 2.6 is zero, from

 

 

'
IEquations 2.12 and 2.16 Y2C will be
' _ ds f.ds
Y2C — f12f20/f10 (2.17)
Again letting
! l 8
Y2 _ Y2C + e (2.1 a)
1 l 8
Y1 — ch + a. (2.1 b)
811d neglecting second order 5 terms
ds (2.19)
V2 ds f.ds f.ds f20 ' d ds d
__ = __ F _ .
VE)(f20f21f10/Yl) Y' (Y' + Y' )LY10f02 f12f01]
0 20“ 1c
Substituting Equation 2.19 into Equation 2.10 gives
fd fds fds (2°20)
= d _ .21119 20 d ' d8 d
Yo f00 v' ' ' v ' [f02ch ‘ f12f013‘

lC ch(Y20“ + ch)

Since a is again completely arbitrary, this is another in-

stability point where the input impedance is indeterminate

and loading voltages approach infinity. If, however, F is

l7

a.symmetric matrix (long antenna case), it can be shown that

v v 1
= Y and Y = Y and

YlR 10 2c 2R
_ d d '
YO — foo - fOl/YlR . (2.21)

IIn addition, V1 and V2 become indeterminate as in the first

czase instead of infinite.

2.2.4 Voltage Constraint
The voltage can be specified in several ways.
ESince it is explicitly related to both the current distri-
taution and the field pattern of an antenna, these types of

Jeelationships will now be developed.

2.2.U.l Field Pattern Constraint
For an asymmetrically driven antenna, as
Efluown in Figure 2.3, the field pattern can be shown to have

tl1e following form,

PJ(O) = ng(m,a,h,dJ,o).

1H1erefore by superposition, the field pattern of a multi-
loaded antenna is given by

N
1311(0) = i§_MVig(w,a,h,di’e)

Which reduces for the symmetric antenna to

N
PTS(@) = V0g(w,a,h,0,0) + iglvigs(w:aahadi:e)

where

18

gs(w,a,h,di,e) = g(w,a’h,di,e) + g(w’a’h,di,e)o

E3y specifying the field pattern at N + M angles in the

ggeneral case and at N angles for the symmetric antenna,

Dd + N or N complex equations are obtained. Thus the unknown

\Ioltages can be
I<ramer's rule.

czan be found by

Matrix F can be

Where

 

solved for either_by matrix inversion or by
Once the voltages are known, the admittances

using Equation 2.3 in the form
F.V = —GVO. (2.22)

written as the sum of two matrices

F = A + Y (2.23)
f§1 ffill ffMl—-

ng félN ffiMN

fg-l f(31-1 f(EM-1

f§_M fél_M . . . féM_M-

 

19

21nd
- -1
Y1 0 o 0
Y = YN o .
o Y_1. .
_c . . . o o . . . Y_M'.

 

 

Combining Equations 2.22 and 2.23 and transposing AV to the
other side of the equation results in
YV = -(GVO + AV). (2.211)

Examining the left hand side of Equation 2.2“ it can be

seen that YV is given by

—; V o o o -T
1 1
YV,= o . . . YNVN 0 o
o . o Y_1V_l 0
__O . 0 o Y_NV_E'

 

 

20

Thus the admittances are given by

N
vO + 3;-MA13VJ)/Vi for i=-M ... N.

#0

Y = -(G

i i

It should be noted that both the magnitude and the
phase of the field must be specified before a solution can
be obtained. Usually in this type of problem, the magnitude
of the field is the only component of interest. Thus, there
are many solutions of the problem which will satisfy the

desired field pattern.

2.2.“.2 Current Constraint

From Equation 2.1, the current on an

asymmetrically driven antenna is

13(z) = VJf(w,a,h,dJ,z).

Thus by superposition the current at any point on a multi—

loaded antenna would be
N
It(z) = Z MV1f(w,a,h,di,z). (2.25)
(If the current is again specified at N + M points, N + M
Complex equations for the N + M unknown voltages result.

(“Sing the same techniques as in Section 2.2.“.1, both the

\Rbltages and admittances can be found.

2l
2.2.“.3. Input Admittance — Frequency Constraint
Until now it has been assumed that m
was fixed. A desirable capability would be to be able to
specify the input and admittance at several frequencies.
This can be done but it will lead to a set of non-linear
simultaneous equations. To see this, consider Equations 2.25
and 2.3. If It(0) is taken as VoYo’ Equation 2.25 becomes
N
Vo(f(w,a,h,o,o) - Yo) + i§_Mvif(m,a,h,d1,o) = 0 . (2.26)
i¥0
Rewriting Equation 2.3 and adding Equation 2.26 to it,

results in

B-V = 0 (2.27)
where
__d .1
f +Y d d d d
11 1 . le f-11 . f_Ml £01
d ° d c é a
le fNN+YN f-lN ° f—MN f0N
B:
d d d d d
f1-1 ° ° ° fN-1 f-1-1+Y-1 f—M-l f0-1
d d d d d
f1-M ° ° ° fN-M f—l-M ° ° ' f—M-M+Y-M fo—M
d d d d d
Lflo ° ° fNo f—1o ° f—MO foe'ig

 

 

22

and

 

 

B is an M + N + 1 square matrix whose equations have a so-
lution only when the determinant of B is zero. This gives
one complex equation containing M + N complex unknowns,

Y Y --- YN, Y-l’ --- Y If the input impedance is

l’ 2’ -M'
specified at M + N frequencies, a set of M + N complex
equations for the M + N unknown admittances results. The
form of these equations is
N N N N N N
g-MY + £_Mbj( :_MY1)/Y + Z C3k(12_MY1)/YJYK°°°

J 2
J=-Mk=j+l
a Y + B = O- (2.28)

It can be shown that Equation 2.28 has 2N + M terms with

2N + M-l independent coefficients. It can be seen that for
large N + M, Equation 2.28 becomes unmanageable.

In this development it has been tacitly assumed that
the admittances are independent of frequency. While
admittances which are independent of frequency over a finite
bandwidth exist, they are generally a function of frequency.

One method of partially circumventing this problem will be

presented in Section 2.2.5.

23
2.2.5 Combined Admittance And Voltage Constraint
Until now the constraint equations have been
conditions either on the voltage or on the admittances. The
results have been somewhat less than desirable. If the
loads are specified in a convenient form (i.e., capacitors,
inductors, etc.), the current Shape is arbitrary. If the
current is specified, the load forms are arbitrary.. It is
desirable to be able to specify both the form of the

admittance and the shape of the current.

2.2.5.1 Reactive Loading - Current Constraint
If for example all the admittances are
required to be imaginary, the number of unknowns in
Equation 2.3 is reduced from U(M + N) real unknowns to
3(M + N) real unknowns with 2(M + N) real equations. Thus
M + N real constraint equations are needed. If the current
is specified at (M + N)/2 points, M + N even, both the
current and admittance forms can be specified as desired.
For M + N odd, the current is specified completely at
(M + N)/2 points leaving one real constraint.“ The remaining
constraint can be obtained by specifying either the real or
imaginary part of the current at one additional point.
The solution for M + N even is as follows. Using

Equation 2.25 at (M + N)/2 points, the matrix equation

CiV1 = VoD (2.29)

24

 

where
._
Z Z Z Z
f11 fkl f11 fk+21
_ z z z z
Ci ' flJ fkj f1; fk+2J
Z Z Z Z
f1M+N ko+N f1M+N fk+2M+N
_. 2 2 2 '2
Z Z Z
f1—11 fk+11 f1+1l
Z Z Z
fi-lj fk+lj f1+13
Z Z Z
f1-1M+N fk+1M+N f1+1M+N°
2 2 2
= (N + M) (M + N) matrix

 

 

25

 

 

 

 

 

 

V. Z _ — .-
X1 ‘ f01 V1
- _ z .
D - xJ fOJ , Vk
. Z V1
XM+N‘ f0M+N Vk+2
L- 2 2- V1 = I
V1-1
Vk+l
V1+1
LY-M._
k = -(M E N) for M > N
k = N g M for N Z M
and fij = f(m,a,h,di,zj)

where X

results as i I = V X
sum ng T(zj) o J J

is a specified

value. To keep the resulting non-linear equations as simple

as possible, Equations 2.29 and 2.3 are combined as follows:

E G v 1 = -M, -(M-l),---, k + 1

C -D V 1 ¢ 0

(2.30)

26

 

 

 

where
r’d d d d
fll+JBl fkl f11 fk+21 '
E = d d. d d.
i flk fkk+JBk fik fk+2k '
d d d d
+
Lfli ‘ ' ' fki f11 JB1 fk+21 ° ' ‘
d d d d _'
f1—11 fk+11 f1+1 ' ' ' f-Ml
d ' d. d. d.
' ' fi-lk fk+lk f1+1k ' ' ' f-Mk
d d d d
' f1-11 fk+11 f1+11 ° ' ° f-Ml,
(M g N) x (M + N) matrix
—_i
d
f01
G1: .
I.d
0k
d
f01
L..._J

 

 

381 is the imaginary part of Y Also it should be noted

1'

that the point z is not necessarily equal to the point d

J J°
Since VO # 0, Equation 2.30 has a solution only if
E G '
C —D

27

 

Since 1 indexes over N g M terms, Equation 2.31 generates

the N g M equations needed to determine the N + M suscep-

 

tances, 81' The form of the equations generated by
Equation 2.31 is like Equation 2.28 except the leading term

is

N
’B instead of H Y
11 _MJ
0 O

(13
‘Ikuz’x‘

J J
J J
2.2.5.2 Reactive Loading - Frequency —
Current Constraint
The above case still assumes w to be
fixed. For the case of several frequencies, let Nf be the

number of frequencies and Nc be the number of current

points. Then

 

= N - N (2.32)

assuming M + N even and that the same number of current
points is used for each frequency. This somewhat limits
the choice of M, N, Nf and NC.

Assuming Equation 2.32 is satisfied, for each frequency

the generating equation would be

' ' ' (2.33)
1 G1 V1

, , 0 i=-M,-(M-l),-°,2 + 1

i O i¢0

E

C

where

 

 

 

f

f

f

28

for M > N
C
for N > M
c _.
d d
f21 f11
d d
+
f 2 332 fig
d d
d fd
2+ll i+ll °
d fd
£+l£ 1—12
d f.d
£+li i+li
X
'
D :-

 

2.*-21 O O 0

H22 ’ ‘

 

 

 

Z Z
f11 le f11 f2+21 ° '
' _ z z z z
Ci ‘ f1; fzj fiJ f2+2j
Z Z Z Z
f f f f
LlNc 2N0 1Nc 2+2NC
z z z z _-
' f1-11 fn+11 f1+11 f-Ml
Z Z Z Z
f1-13 f1+13 f1+1J ' f—MJ
Z Z Z Z
° f1—1N f£+lN f1+1N f-MN
C C C S— o

 

Since Equation 2.33 generates only NC complex equations for

M + N
2

 

each frequency, it takes Nf frequencies to have
equations that are required.

In Equation 2.33 the Bi's are still independent of w.
Since the only elements which have pure reactive parts are
capacitors or inductors, a logical frequency dependence for
B1 would be either w or l/w. Thus in Equation 2.33 the
reactances could become either B1 = chi or B1 = -l/Lin

where J = l to Nf. In order to obtain a reasonable result,

30
the solutions to Equation 2.33 would have to be restricted

to positive values for C1 and Li' In general this

restriction may not be satisfied.

CHAPTER III

CURRENT DISTRIBUTION ON A LOADED RADIATING ANTENNA

In Chapter 2 it was assumed that for an asymmetrically
driven antenna the current distribution is known. This
problem has been treated by many authors and is generally
broken into two approaches based on the electrical length
of the antenna. Current distributions for a short antenna
and for a long antenna will be presented in this chapter.
Since many antennas are monopoles and because of the
formulation of the short antenna problem, only the asym—
metrically driven short antenna current distribution will
be presented. The asymmetrical case can be found in the

literature.

3.1 Current Distribution On A Short Antenna

There are several types of approximate current distri-
butions for a short symmetrically driven antenna. The one
presented here will be based on King and Wu's theory (1)

and is valid for kOh < 3H/2 and Q = 2£n(2h/a) 1 8.

3.1.1. Vector Potential Of Symmetric Dipole
Consider the symmetrically driven antenna in
Figure 3.1 The axial component of the vector field on the

surface can be shown to satisfy the equation
31

32

 

 

 

 

 

 

m Z=h
w
V9“: (3115 Z=2
Z
T z=_o
Vt: (S) Z=-£
28-. ‘-
V’ Z=-h

 

Figure 3.1 Symmetrically Driven Antenna

33

(-d% + 2 _ k6 + + )
5;? ko)Az(Z) _ _ w V1(6(z — 2) 6(z 2) - (3.1)

 

The homogeneous part of Equation 3.1 has solutions in three

regions h > z > z, 2 > z > -2, and -2 > z > -h given by

Az(z) —J/eOpO[Cl cos koz + C Sin koz] h > z > 1 (3.2a)

2
Az(z) = -J/E;H6C3 cos koz 2 > z > -2 (3.2b)
Az(z) = -J/eOuO[Cu cos koz + C5 sin koz].-£ > z > -h(3.2c)

where -J/eopo is included to make the C's dimensional volts.

From symmetry Az(z) = AZ(-z), thus C1 = CA and 02 = -C5.

Since Az(z) is continuous at z = :1, from Equations 3.2a

and 3.2b it can be shown

(Cl - C3) = -02 tan koz. (3.3)
3Az(z)
The scalar potential is given by «(z) = km __—55_
0
Thus
¢(z) = —C1 sin koz + 02 cos koz h > z > 2(3.Aa)
¢(Z) = —C3 sin koz 2 > z > —2 (3.Ab)
¢(z) = -Cl sin koz - C2 cos koz> -2 > z -h(3.Ac)
which satisfy the symmetry condition ¢(—z) = -Q(z). The
driving voltage is defined as
(3.5)

V = MN) - cm') = -(c

l - C3) sin koz + C2 cos kol.

1

3A

Combining Equations 3.3 and 3.5 results in

02 = V2 cos koz , C3 = C1 + VA sin kol.
Th
erefore (3.6)
Az(z) = -J¢eOuO(Cl cos kOZ+WNb/2)(sin kolz+£l + sin kolz-AI)).
All that remains to be evaluated is the constant G1 which
is done by requiring Iz(h) = 0. To this point, the develop-

ment applies to either a short or long antenna.

3.1.2. Difference Equation

In King and Wu's method, since

 

 

110 h I v t
Az(z) = ER [hI(z )K(z,z )dz (3.7)
where
v 'Jkr r
K(z,z ) = e r , r = /Qz-Z)2 + a2 (3.8)

a difference equation of Az(z) — Az(h) can be defined. Using

Equation 3.6, it follows that

{:I(z')Kd(z,z')dz' = -333 [C1 cos kozifiVb/2)(sin k0|z+£l
+ sin kolz-zl) + U] (3.9)
where
-Jco h . . .
U = Tn“ [huz )K(h,z )dz . (3.10)

The difference kernel is

Kd(z,z') = K(z,z') - K(h,z'). (3.11)

35
Since the right hand side of Equation 3.9 is zero at

z = h, the constant C1 is

U + V2 cos koz sin koh

C = -[ cos k h
o

 

J (3.l2)

and Equation 3.9 becomes

 

h t t ' 3A
_ I1
{h1(z )Kd(Z,z )dz - Co COS koh [V£(cos koz cos koz sin koh

-CD2)cos koh [sin kolz—2|+ sin kO|z+2|])+ U(cos koz - coskoh)].

(3.13)
3.1.3. Current Distribution
It can be shown that
h . . ,
f Iz(z )KdR(Z’z )dz m Iz(z) (3.1Aa)
-h
h ' t !
IhIz(z )KdI(z,z )dz m cos(l/2)koz -cos(l/2)koh (3.1Ab)

where

Kd(z,z') = KdR(z,z') + JKdI(z,z').

Thus Equations 3.13, 3.1Aa, and 3.1Ab suggest that the

approximate form of the current distribution should be
1(2) = IV(Z) + Iu(z) + ID<z> (3.15)
where
Iv(z) = Iv(cos koz sin koh cos koz -(l/2)cos koh[sin kolz+£|

+ sin kOIz-2IJ) = IVMOZ (3.16a)

36

Iu(z) I F = Iu(cos koz - cos koh) (3

1102

ID(z) IDFOZ' = ID(cos(l/2)koz - cos(l/2)koh).(3

Substituting Equation 3.15 into Equations 3.1Ua-b the

following definitions can be made:

h i v t
[hIV(Z )KdR(Z’Z )dz = IV(Z)¢dR(Z) é Iv(z)"’dR (3°
h , , ' IV Iv
[hIV(z )KdI(Z’Z )dz 2 ID(Z)IdeI(Z) a ID(Z):Ede (3.
h ' ' .
{hlu(z')KdR(z,z )dz = Iu(z)deR(z) = Iu(z)1pdUR (3
, v . . Iu . Iu
{th(z )KdI)z’Z )dz = ID(Z)IBYdUI(Z) = IU(Z)‘1‘3¢dUI (3'
h , ,
IhID(Z )Kd(z,z )dz = ID(z)wdD(z) é ID(z)wdD. (3.

.16b)

.160)

17a)

17b)

.17c)

17d)

l7e)

Noting that the integral on the right hand side of Equation

3.13 equals AHpO-IEAZ(2) — Az(h)] and using Equations 3.15

and 3.17a-e, the result is

-1 _
“Huo [Az(z) - Az(h)] - IdeRMoz + IudeRFoz

 

1
+ (IDwdD + JIdeI + JludeI)Foz ° (3'18)
If this expression is substituted into the original
differential equation (3.1) in the form
_ 2 JAHk
Anuo 1[—9§ + k 2][A (z) — A (h)] = Orv [5(2—2) + 6(z+£)]
dz 0 z 2 Co A

-JwAZ(h)] (3.19)

37
then

2
-IV¢deo cos koh[5(z-g) + 5(z+2)] - IudeRko cos koh

2
+ k0 (IDWdD + JIdeI + JIquUI)((3/“)COS(1/2)koz (3.20)

Junko
- cos(l/2)koh) = — Co (V£[6(z-2) + 6(z+£)] - JmAZ(h)).

 

The coefficients of the 5-function terms must be equal, so

that

Junvk
IV = (3.21)
CowdR cos koh

 

which leaves

-1
IudeR cos koh - Anpo Az(h) - (I + jI

DwdD deI + JIu‘deI)'

((3/A)cos koz - cos(l/2)k0h) = O. (3.22)

Since Equation 3.22 must be true for all z, the coefficient
of((3/A)cos koz — cos(l/2)koh) must be zero.

Thus

ID‘pdD + JIV‘PdI + 31ude1 = O (3°23a)

and

_ -l
IudeR cos koh — Anuo Az(h). (3.23b)

To find Az(h) Equations 3.15 and 3.7 are needed. Using the
same type of approximation as used for Equations 3.17a-e,

the result is

Anuo'lAz<h) = vav<h> + Iuwu(h) + IDwD(h) (3.2a)

38
where

-2
I ' I
w (h) = f cos(2k )sin k (h+z )K(h,z )dz + sin k (h-A)
V —h o o o (3.25a)
h

2' 7 ' ' ' ' '
f cos koz K(h,z )dz + f cos koz sin kO(h-z )K(h,z )dz
2.
-1

h 1 Y '
wu(h) = £h(cos koz - cos koh)K(h,z )dz (3.25b)
311d
h t . v
wD(h) = £h(cos koz /2 — cos kOh/2) K(h,z )dz . (3.250)

Ccunbining Equations 3.2M and 3.23b to form one equation and

ussixig Equation 3.23a, two constants, T and Tu, can be

D
de fined as follows:

1. == :2 = '3Ewd1(wduscos koh ' ¢u(h)) + deIwu(h)] (3 26a)
[3 IV wdDTWdURCOS koh ' 1"um”+ JwD(h)deI

 

L1 IV wdedeRcos koh ” “’ua‘”+ MDU‘N’dUI '

 

Timiss using Equations 3.26a, 3.26b, 3.21, and 3.15, the
approximate current distribution along the antenna driven

by Iidentical voltage generators, Vn, at z = :2 is

 

( JAHV2f(z)
I Z) = (3.27)
z cowdRcos koh
where

f(z) = Moz + TuFoz + TDFoz' (3.28)
witflfwoz, Foz’ and Foz' given in Equations 3.16a-c.

39
3.1.A. Evaluation Of Constants
All that remains to be Specified is the w
functions introduced in Equations 3.17a-e. Since the w
functions represent the ratio of a component of the vector
lootential to the current, the desired ratio is obtained at
tide maximum of the appropriate current. For deR’ deI’

“QiD’ and de this maximum is at z = 0. Thus from

Ek1uations 3.17b—e

wciUR = (l-cos koh)-l f:(cos kaLcos koh)KdR(o,z')dz' (3.29a)
h ' 1 v
chD = (l-cos kOh/2)’1 [ (cos kOZ /2 - cos kOh/2)Kd(O,Z )dz
‘h (3.29b)
-1 h , v v
uuitJI = (l-cos kOh/2) fh(cos ka-cos koh)KdI(o,z )dz
- (3.29c)
-1 -9' i g '
ukiIf = (l—cos kOh/2) [hcos koz sin ko(h+z )KdI(O’Z )dz
9, ' , '
+ sin ko(h-£) lgcos koz KdI(o,z )dz (3.29d)
h ' ' '
+ JACOS koi sin ko(h—z )KdI(O’Z )dz .

Since in the range h i z : O,MOZ may have two relative

nuLXiJna, the definition of wdR is more complicated. The first

rmixiJnum is at z = h — A/U when h—z : A/A and at z = 1 when
h-Z- < A/U. The second is always at z = O. The definition
Inadfé by King and Wu is in terms of an average of the vector

pOtentials .

A0

 

Thus
lMoowdR(O)l + ll‘dozl‘dewl)I
was = (3.30a)
IMOOI + No.1!
where zl = h-(A/A), when h-g > (1/A), and 21 = 2 when
h-z < A/A and where
-1 '9‘ v t v
wdR(z) = (MOZ) [ cos koi sin ko(h+z )KdR(z,z )dz
-h
2' V ' '
+ sin ko(h-£) f cos koz KdR(z,z )dz (3.30b)

"R.

h v c t
+ 1 cos koz sin ko(h-z )KdR(z,z )dz

9.
In particular, MOO = sin ko(h-£) and MOZl = cos koi when
21 = h - A/A, and MOZ - sin ko(h-£) cos koz when 21 = 2.

1
3.1.5. Special Cases
Two special cases need to be considered: 2 = O
and koh = n/2. The first case is that of a center driven
antenna. Since as 1 + O the driving voltage becomes 2V2’

thus V0 is equal to 2V1, and Equation 3.27 becomes

J2HVO

 

Iz(z) = [sin ko(h-|z|) + Tu (cos ka-cos koh)

cowdRcos koh

+ TD(cos ka/2 - cos kOh/2)].
The second case is that of a half wave antenna. The
«expression for Iz(z) becomes indeterminate because of the
cos koh in the denominator. An alternate form of Iz(z) is

‘thus needed when koh is near H/2.

A1

If T and T are redefined as

 

u D
, Tu + sin koh cos koz
Tu = - cos k h (3.32a)
0
TD
To' = 35m (3'32“
0

which are finite at koh = H/2, thus

 

Junv2 _ '
= v _ I _
Iz(z) cowdR [MOz Tu (cos koz cos koh)
(3.33)
+ TD'(cos 1/2 koz - cos kOh/2)]
where (3.3“)
MOZ' = sin koh cos koi - 1/2(sin kolz + 2] + sin kolz-£|)-
From Equations 3.26a-b and 3.32a—b,
Tu' = - {[wdD(wV(h) — wu(h)sin koh cos koz)
+ wdDdeRCOS koh sin koh cos koz — ij(h)o (3.35)

(de - deIsin kbh cos k02)]/[(wdD(deRcos kOh-wu(h))
+ deUIwD(h)) COS kOhJ}
and

- 'JIWdIIWdURcos koh - wu(h)] + deIwV(h)]

T v _ . . . 6
D wdD($dURCOS kOh-¢u(h))¥jdeIwD(h) cos.koh (3'3 )

 

Noting that at koh = H/2, wu(n/U)cos koz = wV(H/A), and

de = cos kol w Equations 3.35 and 3.36 reduce to

dUI’

cos k 2
o

 

T , wdDdeR

u = woD¢u(A/“) _ deUIwD(x/n§ (3.37)

A2

dedeUR
T ' = (3.38)
D wdDwu(A/u) ' deUI¢D(A/U)

which are both finite.

3.2 Current Distribution On A Long Antenna

The approximate solution for a long asymmetrically
driven antenna presented here will be based on papers by
Wu (2) and Shen, Wu and King (3). The solution is valid
for antennas whose shortest arm is Z .15). To follow the

-Jwt

original papers closely, the time dependence of e will

be used.

3.2.1. Current On An Infinitely Long Antenna
Consider an infinitely long center-driven
antenna with its driver at z = O. The axial component

of the vector field on the surface can be shown to satisfy

 

 

( 32 + k 2)A < > - +Jk°2 v s< > ( )
5:2- O 22 - u) 0 Z. 3.39
Thus
“ V Jk IzI
A(z) = 3:0 e o for all z. (3.u0)
0
But
)1 °° v t I
A(z) = 3% [mdz I12 )K(z-z ) (3.A1)
where
H
K(z) = 5% f do[22 + (22 sin (9/2)2:|-1/2 (3 Ala)
'“ 2 2 1/2 .
expfjkofz + (2a sin 9/2) 1 J.

r49

’43
Combining Equations 3.140 and 3.141 gives

2HV 0° ' v I
Tgejkolzl = I dz IOSZ )K(Z-Z ). (30142)

0 '00
To solve Equation 3.112, let Re have a small positive

imaginary part which will eventually be allowed to approach

zero and define the following Fourier transforms

f(;) = fwdz Im(z)e""jtIZ (3.1433)

_<D

k—(c) = Imdz Mme-“Z. (3.143b)

"'CD

Applying the Fourier transforms to Equation 3.142 results

in
(3.1414)
211V

C O [mejkolzleéjczdz = I dze-jcz f dz. 252')K(Z-Z').
0 —00

-‘CD

Using the assumption k0 has a positive imaginary part,

the left hand integral becomes
[we'jko'z'e—Jr’zdz = 2JkO/(k02 - g2). (3.u5)

Interchanging the order of integration of the right hand

integral and using Equations 2.1!3a-b, the integral becomes

[aocizemkqZ foodz' I°°(z')K—(z-z') = Jami-(2;). (3.146)

ombining Equations 3.145, 3.146, and 3.1414, the resulting
>rm for 12(5) is
unjvoko

2 - c2)K‘(c)

 

20:) = . (3.147)

Co(ko

 

All

Tofindlgz),the inverse transform of Equation 3.“? is

 

taken. Thus
3 2 3 (3.118)
_. 1C V :2
1(2) = glfif I(c)eJCch = 4-9- 2 82 _ <12;
oo 00 C
Co 0 Co(kO - c )K(c)

where Co is the contour shown in Figure 3.2.

Contour(%)is chosen so that the inverse of Equation 3.u5

resultszhithe proper value. Also it can be shown that

K(c)==nJJO[a(kO2 — g2)l/2JHO(1)[a(k02 — c2)1/2]. <3.u9)

Thus Equation 3.A8 is completely defined, but is not in a

usable form. If it is assumed (1) that the antenna is

thin i.e., a/A << 1 and (2) that the characteristic distance

for variations of A(z) is >> a, then Equation 3.A9 can

be approximated by

K<c> = 290 + 3n - 2n[(ko2 - c2)/k02] (3.50)

where §%)== £n(2/koa) - y and y = .57721566.

Using Equations 3.50 and 3.148 and deforming CO so that it

is wrapped around the branch out at z; = k0, it can be shown

that the major contribution to the integral comes from points

near I; = k0. Thus IOSZ) can be approximated by

V J 00 -2k Zn

(z) = —‘?-——e3kolzl[ 9—0—— [(20 — 2m + J3H/2)-l
°° CO 0 n (.0

-(2Cw - inn - JH/2)-l]dn (3'51

'here Cw = 2n(l/koa)- y.

“F

- ‘ "lCl

huh..-

 

 

“5

Imas(c)

Right Branch

cut

 

 

 

 

 

Figureji2

Left
Branch
Cut

 

0 Real (5)

The c Plane And The Contour CO

- r' 1‘; 1i.

 

1&6

Because of e'2koZn in the integral, it can be shown that
thetumerlimit can be replaced by e-Y/2kolzl. Integrating

Emunim13.iland using the new upper limit the result is:

I(z)='VJ§fl§len[l - 2“J J (3 52)
w o :0 20m + £n2kofz| + y + 33n/2 '

for large koz.
Forsmmll ka, the input conductance of the infinitely

long antenna is approximately
(3.53)

2
— £2 11 l. _ '2 _ _ ‘2
Ga, - Co £n(1 + Cw) + 12[Cw £n2) (Cm 2n2 + M) 3-

To obtain a formula for I§z) when koz is small, the term
'1 1/2)

£n2kolz] must be replaced by 2n(k0|z| +-«kbz)2 + D )
- O, the real part

I!
To find D it is required that when koz -

of Equation 3.52 divided by V0 and Equation 3.53 do not

differ by more than l/Cw3 in the limit as ka approaches
V! _
27 Thus for all koz,

 

 

zero. If this is done, D = e .
VOJeJkolzl
I(z) == £n[l - 2HJ/(2C + £n(k |z| +
” CO w 0
+ /?k z)2 + e‘2Y + y + 1.5 HJ)]. (3.5a)
0
.3.2.22 Reflection Current At The End Of A Semi-

Infinite Antenna

IIf i(z) is the normalized current on a semi-
O and the other end

‘.nfinite antenna with one end at z
t z = on and is excited by V = on at z = on such'that the

 

 

“7

amplitude of the incident current is unity, then i(z) for koz

not too small and CO deformed around the branch out at

 

C = k0, can be shown to be
k _ _ CZ
1(2) = fiEL+Hko>J 2 I, 2 e3 2 __ dc (3.5561)
Co(kO - c )K(c)
where
-— 1
in E’( ) = -:l [m-g ' £EELE—l' (3 55b)
4" C 2113—00—5 C C' _ C ° 0

Comparing Equation 3.55 with 3.U8 gives
i(z) = —R;§z)/VO (3.56)

for large kz, where

C _
_ O -2
R — _fi [L+(ko)] .

The approximate value of [Ei(ko)]-2 can be shown to be

[Lnkon — 2% + Jn. (3.57)

From Equation 3.56 it can be seen that -R is the ratio
of the reflected current to the incident current and that
the reflected current away from the end behaves Just like

the current of an infinitely long antenna.

3.2.3. Current On An Asymmetrical Finite Antenna
Section 3.2.2 suggests that the current on a
long finite antenna may be expressed as the superposition of
an outgoing wave, EAZ) and two reflected waves which are

DPOportional to Iéhl + z) and I§h2 - 2), respectively.

1%“ 35-”?

u8
If tflua coordinates are chosen such that the ends are at
z = 'JEL and z = h2, and the driving point is at z = 0,

then
I(z) = I§z) + CdI‘jh1 + z) + CuIéh2 - z). (3.58)

If it is assumed that the reflection coefficient R at the
end of an antenna is the same as that for a semi-infinitely
long antenna, then Cd and Cu can be evaluated. Since at

z = -h1,R is the ratio of reflected to incident waves and

since the amplitude of reflected waves is C V0 and the

d
amplitude of incident wave is Iéhl) + CuI§h2 + hl)’ the

result is

 

 

CdVO
-R = . (3059)
lihl) + Cu££h2 + hl)
Similarly, at z = h2
CuV
-R = O o (3.60)
Téh2) + Cd£§h2 + hl)

Solving Equations 3.59 and 3.60,

C = _R[i§hl)/VO-Ri5n2)/v0;§hl + h2)/VO

 

I (3.61)
1 - R2[;;hl + h2)]2/V'O2

I£h2)/Vo-RI§hI)/VOI§hl + h2)/VO

1 - R2ii§hl + h2)32/VO2

 

C = —R[ ]. (3.62)

This completely defines Equation 3.58 and gives an approxi-
nnte distribution for asymmetrically driven antennas. If a
distribution is desired for the symmetrical antenna, it can be

superimposed using Equation 3.58, km

hit, and h = hit.

1 2

. Aldus“ ij

CHAPTER IV

RADIATION PATTERNS

OF SHORT AND LONG ANTENNAS

Since different current distributions were derived for
the short and long antennas, different radiation pattern

expressions will have to be found.

A.l. Radiation Pattern Of A Short Antenna
Because of the simple form of the current for a short
antenna, the radiation pattern will be found by using

Equation 3.27 and the usual far field approximations,

'
koRo >> 1, RO >> h, R = R0 - z cose (phase terms)

and R = R0 (amplitude terms) where RO is the distance from
the center of'the antenna to the field point. If this is

cknue,.it can be shown that the radiation pattern, Pm(e) is

V c k h

Ifin(e) = m g g sine I I(z)eJkoZ cosedz (A l)
—h

 

where I(z) is given by Equation 3.27 or 3.33. Thus three

integrals must be evaluated

h
‘f Mozejkoz 0086dz = I (4.2)
-h

1

A9

.~c-l:___.fi__’n--An-nu n '-H

50

h .
f (ccns koz - cos koh)e‘]koz COSGdz = I2 (4.3)
-h

h Jk z cose

I (ecu; ka/2 - cos kOh/2)e 0 dz = I3. (A.A)

Eknfli Equations A.3 and A.A can be evaluated easily and can
be shown to equal to

sin koh(l + cose) sin k0h(l - cose)

 

 

 

12: +
k (l + cos a) k (l — cose)
o o
(A.5)
2 cos k h sin(k h cose)
_ o o
k cosO
o

 

 

 

and
sin k h(l/2 + cose) sin k h(l/2 - cose)
I = O + O
3 ko(l/2 + cose) ko(l/2 - cose)
(A.6)
2 cos kOh/2 sin (koh cose)
k cose
0
Using Equation 3.27, Il becomes
h (A.7)
I1 = cos k 2 sin k h f cos k zeJZko COSGdz -(l/2)cos k h.
o o -h o o
h Jzk cose h Jzk cose
[I sin kolz + zle 0 dz + f sin kolz-zle o
-h -h

which can be shown to equal

 

 

dz]

51
sin koh(l + cose)

 

I1 = cos koz sin kohf
ko(1 + cose)

sin k h(l - cose)
+ o

 

] - cos koh (A.8)
ko(l - cose)

sin k 2 sin k 2(1 + cose) sin k zsin k £(l-cose)
[ o o + o o

 

 

k0(l + cose) ko(l - cose)

cos koh(l + cosO) cos kO£(l + cosO)

 

 

 

 

— cos k02( -
ko(l + cose) ko(l + 0050)
cos k h(l - cose) cos k 2(1 — coso)
+ 0 — O )1-
ko(l - cose) ko(l - cose)

 

 

Reducing to a common denominator and simplifying, I1, I2
and I3 are found to be
I2 = 2 [cosesin koh cose(koh cosO) - cos k ho
k cosesin e O
O (u 9)
sin(koh cosO) °
I3 = 2 2 [2 cosGsin kOh/2 cos (koh cosG)
kocosO(l-A cos 0)
(14.10)
— cos kOh/2 sin(koh cose)]
(£1.11)
2
I =-———-——{cos k 2 cos(k h cose) - cos k h cos (k 2 cose)l
l kosin2O o o o o

'- . -31 IL)! mun—M1

 

52

Replacing I1, 12 and I3 in Equation 11.1, the result is

 

 

Vmsine
= r
E3n(9) WdRCOS kohL81n2e(cos koi cos(koh cose)
- cos koh cos(k02 cose)) + u 2 .
cosOsin e
(14.12)
(cosesin koh cos(koh cose) - cos koh sin(koh cose))
TD

+

 

2 (2 cosesin(kOh/2)cos(koh cose)
coso(l-A cos a)

- cos(kOh/2)sin(koh cose))].

If Equation “.12 is inspected, it can be seen that Equation
“.12 is indeterminate if e = O, n/2, H/3, or 2H/3 or if
koh = n/2.

If koh = n/2 then Equation 3.33 must be used in

Equation 4.1. Thus for koh z n/2, I1 is

 

 

h h
I1 = sin k h cos k 2 f eJkoZ COSGdz - 1/2 I (sin k |z +2]
0 o -h -h o
+ sin kolz — £|)ejkoZ COSOdz (H.13)
which reduces to
2 sin koh cos Ron 2 cos(koz 0059)
11 = sin(k0h cose) - 2
k0 cose k0 sin e (A 1A)
2 cos k 2 .
+ —————7?- (cos k h cos(koh cose) + cosesin koh°
k0 sin e O

sin(koh cose))

”a?

.i‘ A...“ ”In.

'_

53
or‘

_ 2 r
Il - 2 Lcosecos koh cos(koh cose)

k0 cosOsin O

 

- (u.15)
+-<xos k02.sin koh sin(koh cose) cos koz
-+ cos@ cos(kO£ cose)].

Thus Pm(e) becomes

Vm sine l
[ [cosecos koh cos kon cos(koh cose)

wdR cosesin e

Pm(O)

+ cos kol sin koh sin(koh cose) - cosecos(kon 0056)]

T V
- u [cosesin koh cos(koh cose) - cos koh

cosesin s

 

v
TD

sin(k h cose)] + 2
O cose(l-A cos 0)

[2 cosesin kOh/2
cos(koh cose) - cos(kOh/2)sin(koh cose)]] (A.l6)

for koh z n/2. It can be seen that Equation “.16 is also
indeterminate if e = O, n/3, H/2, or 2H/3. By appropriately
taken the limit as G + 0 it can be shown that Pm(@) = O

for all koh.

At 0 = H/2 the terms with the form

21m sin koh cose

 

= k h (u.17)
O + H/2 cose 0

become indeterminate.

 

5A

 

Thus
8 m -
Pb(n/2) wdRcos koh [cos koz cos koh + Tu(sin koh
(u.18)
- koh cos koh) + TD(2 sin kOh/2 - koh cos Rob/2)]
koh # H/2
and
Vm ,
Pm(n/2) = EEE{COS koh cos koi - l + koh cos koa - Tu-

7
(sin koh - koh cos koh) + TD (sin kOh/2 -

-koh cos Rob/2] for koh ~ n/2, (u.19)

At a = n/3 or 2n/3 only the last term is indeterminate.
It can be shown by l'Hospital's rule that the last term as
sin koh
o + n/3 or 2n/3 reduces to kOh/2 - ———§———.
Thus Pm(n/3, 2H/3) becomes

Vmsin II/3r

 

Pm(H/3 or 2H/3) = (cos koz cos kOh/2

was Lsin2H/3

u

2

- cos R 2/2 cos koh) +
0 sin n/3

(sin k h cos k h/2
o 0
sin k h

————9—) (14.20)

- 2 cos koh sin Rob/2) + TD(kOh/2 - 2

for koh # n/2. By similarly replacing the last term of
sin k h
Equation A.l2 by TD'(koh/2 - ———§—2—), the expression for

Pm(II/3, 2H/3)koh = “/2 can be obtained.

14.2. Radiation Pattern Of A Long Antenna

 

IBecause of the complex form of the current derived

for a Jinn; antenna, a different approach is needed to find

I W“ “w. j

 

55

the radiation pattern for it. The approach to be used
will follow that used by Wu (2) and Shen (H) and will
be for a symmetrical antenna. The asymmetrical case can
be derived using a similar argument.

iLet 21 spherical coordinate system (r,e,¢) be set up
such that the ends of the antenna are (h, 0,¢) and (h,n,¢).

The radiation pattern can be defined as

Pm(e) = — 21m Ee(r,e)re-Jkor (u.2la)
P+co
which reduces to
_ -l ‘
Pm(0) - Jauo(un) Im(k coso)sin0 (b.2lb)

where

Im(;) = fmI(z)e-chdz

if the usualzfl<0<< l approximation is made. From Equations

3.Al-3.A7 it can be shown that
_ “H ”0-1 A(t)
Im(c) = _ (H.22)
K(c)
where A(;) = I Az(z)e-J§zdz and E(;) is defined by Equation
3.A3b. Substituting Equation 3.22 into Equation 3.2lb

results in

JwA(ko cose)sin0

 

Pm(e) = (n.23)

E(kO cose)

But from Equation 3.50

K(;) = 290 + 3n — £n[(k02 - C2)/k02]

L .L D “-1-“31

56
thus

into 0039) = 2ml - 2n(sine)] (4.24a)

where

01 = 90 + Jn/2, (4.24b)

From Equation 3.6 and correcting for the ert instead

of e'J‘”t time factor, Az(z) becomes
Vm
AZ(Z) = ‘JV/E 11 [Cl COS kOZ - —-2—(sin kO'Z+2’|

+ sin koiz-2|)] h > Z‘> —h.

If it is assumed that the main contribution to the integrals
comes from the vector potential on the antenna surface, then

__ h V
A(k cose) 5 f -Jfibuo[cl cos koz - 7?(sin k0|z+2l
(4.25)
+ sin kolz-2|)]e-JZkocosadz

which is very similar to Equation 4.7. Using the integrals

evaluated to find Equation 4.8

sin k h(l—cose) sin k h(l+cos0)
o + o )
l- cose l+cose

sin k 2 sin k 2(l-cose) sin k 2 sin k 2(l+cose)
_ V ( o o + o o

m l-cose 1+cose

Eo“o
k
0

 

 

A(ko cose) = -J [C

l(

 

 

cos k 2 cos k 2(l-cosa) cos k 2 cos k 2(l+cose)
+ o o + o o

l-cose l+cose

 

 

cos koz cos koh(l-cose) cos ko2 cos koh(l+cose)
l—cose l+cose

 

 

Alltflmt remains is to evaluate Cl'

 

57

IFor a.semi-finite antenna with one end at z = 0,

Wu (2) has shown that

I dz f,(c)[K_(c) - A(O+)/J(c+ko)] = 0 (14.27)
C

on

f dzAz(z)e-JCZ and E+(c) is given by
0
Equation 3.55b.

_ o
where A_(§) =

If again it is assumed that Az(z) = 0

for 2 outside the antenna, then the following functions
can be defined:

'

z
' ' - - -JCZ
T (z ) - é ch+(§)[£dze cos koz + Egfigj (4.28)
. . ° _ 2'
S (2 ) = f ch+(C) I dze'JCZ sin koz , (4.29)
C o
0

Since it has been assumed that z = 0 is at the lower end

of the antenna, Az(z) must be rewritten as

(4.30)

Az(z) = —J/eouo[(Cl cos koh + V cos ko2 sin koh)cos koz-

m

H(2h - z) + (Clsin koh - Vm cos ko2 cos koh)sin koz H(2h-z)
-Vm(sin ko(h+2) cos koz H(h+2—z) - cos ko(h+2)sin kaH(h+2—z))
-Vm H(h-2-z)[sin ko(h-2) cos koz - cos ko(h—2)sin koz)]]

where H(z) = 1 z 1 0
= 0 z < 0
Substituting Equation 4.30 into Equation 4.27 and using

Equation 4.28 and 4.29 the result is

 

58

'
(C cos koh + Vm cos 1402 sin koh)T (2h)

1
i
+ Uh Shikoh - Vm cos ko2 cos koh)S (2h)
! i
- Vm(sin k0(h+2)T (n+2) — cos ko(h+2)s (h+2))

— Vm(sin ko(h—2)T'(h-2) — cos ko(h-2)S'(h-2))= 0.

Solving for C1

' I
C1 = +Vm[cos k02(cos kOhS (2h) - sin kohT (2h))
Q 1
+ sin‘k0(h+g)T (h+2) - cos ko(h+2)S (h+2)
' 1
+ sin ko(h—2)T (h-g) - cos ko(h-2)S (h-2)]/

V '
[cos kOhT (2h) + sin kOhS (2h)].

(4.31)

(4

V 1 1
Both S (z ) and T (a) have been approximated by Shen (5)

and are given by

S'(z')

-'.[-yi + y (z') - y3(z')]/2

T'(z')

[y; + y2(z'> + y3<z'>l/2J

where

y; = 113/(:21 - 2H2)

 

 

 

 

. a (z') '
y2(z >= zn<a3 ,)+§[—21——.— — -2 1,
2(z) 'n2(z ) 93(2 )
, 23k z
y3(Z )= I 18 0 ’ X [SJ—T— -
koz + /Tkoz)2+(2/n)2 2(z )

 

92(2') = 2 2n(1/koa)+ 2n[|koz'|+/(:Oz')2e-2Y]-Y-Jn/2 (u.

 

(4

. J
)
(4

.32)

.33)

.34)

.35)

.36)

.37)

38)

59
a2<z') + 2n) (14.39)

n3<z')

I
Y

1.64493407

and n is given by Equation 3.24b. Equation 3.32 can be

1
simplified by using Euler's Theorem (e39 = cose + .1 sins)
and Equations 3.33 and 3.34. The result is

(4.40)

eJ2k h

o y2(2h) - y3(2h))+ ejkoz

'
Cl =-4%Jcos kol(yl + (y3(h+2)

- y2(h+2)ej2koh) + e-Jko£[y3(h-2) - y2(h-2)ej2koh]]/A

J2koh

where A = J(yi + y2(2h)e + y3(2h)).

Finally combining Equation 4.24a and a simplified
Equation 4.26 into Equation 4.23
sine Vfi Cl sin koh(l-cose)

= r-..—
Pm(e) 2(2n(sine)-01)L Vm( l-cose

 

 

sin koh(l+cose) l
l+cose ) + 73:33§§)[008(k02 cose)

 

1
cos ko2 cos koh(l+cose)] + (T:EBEET{COS(koz coso)

cos ko2 cos koh(l-cose)]]. (4.41)

It should be noted that like the current distribution, the
pattern function is based on a time dependence of e'Jwt.
.A comparison of Equations 4.40 and 4.41 with the
results obtained by Shen (4)' will reveal two differences.

Instead of cos k02 cos k0h(1icose) in Equation 4.41, he

has cos ko(h:h cose + 2)., However, taking the inverse

Fourier transform of his pattern function will not give the

.I .‘A‘W- -1

r_—-__..__

proper equation for the vector potential.

60

Secondly, his

Cl,lflfleh differs from the one here, will not reduce to

Wu's (2)

result when 2

O.

F

Y fill-1“ ?n#;

I.” _ ..
17-. 1' ..‘.‘

 

CHAPTER V
NUMERICAL SOLUTION OF THE SIMULTANEOUS
NON-LINEAR EQUATIONS AND NUMERICAL EXAMPLES

In Section 2.2.4.3 it was found that in order to get
a solution for the problem at hand, a set of non-linear
equations of the form of Equation 2.28 i.e.,
N N N N

N

aIIY+ b(HY)/Y+ c
=-M 1 ig-M J i=-M i J JE-M k£3+l 3k 1
2.28

N
( g-MY1)/Yk-Yk
N

XaY +8=0

ii

. . .+
1=-M

must be solved. If N = l and M = O, the solution is trival.

For the symmetric case and N = 2, the analytical solution

is straight forward. For N and M greater than 2, however,

the analytical solution of the problem becomes extremely

difficult. To circumvent this difficulty, a numerical

method of solving these equations will be presented.

5.1 Method Of Minimization
A close examination of Equation 2.28 will reveal that

if all but the it}; variable are held constant, Equation 2.28

becomes linear in the ith variable. Thus the system of

N + M non-linear equations reduces to N + M linear equations

will not satisfy all N + M

in the 1311 variable or N + M equations and one unknown, Yi’
Except at a solution point, Yi
61 O

 

62

equations simultaneously. If, however, a new function

N
FM = (a'Y + b')2 (5.1)
J§_M J 1 J

l v
where aJY1 + bJ is one of the linearized M + N equations,

is defined, a value for Yi can be found which will

minimize FM. Since by the way FM is defined, it is either

positive or zero, the least minimum of FM must be ’zero.

If FM equals zero, then all the M + N equations are zero

since each is added to FM as a square. Thus PM = 0 will

occur if and only if FM is evaluated at a solution of the

N + M non—linear equations.

Since FM is a quadratic in Y1, the minimum value of

FM occurs when ggfl = 0. Taking the derivative of FM
1
with respect to Y1 and setting it equal to zero gives
0 Ii12('y ')
8 aa +b
j=—M J J 1 J
or
N I'N '2
Y=-{ab/{a. (5.2)
1 =_M J J =_M J

If FM is not zero, the non-linear equations can be

linearized with respect to another variable using the new

found value of Y1. Again define FM as before and find

the value of the second variable which minimizes FM.

Since the only variable which changed from the first

iteration is Y1 which made FM a minimum, this second minimum

value of FM must be less than or equal to the first. If

the variables are changed one at a time in the above

 

f‘fiti— -

63
Inpcmhne, FM will converge to some minimum value. If
thisxmflue is zero, then the point is a solution to the
desimfliequations. If it is not zero, then the point is
oneiflmt minimizes the set of N + M functions. Because
of Huacomplex nature of the set of non-linear equations,
several minimums and zeros may exist for a given problem.

A procedure to check for other solutions or minimums will

be given in Section 5.4.

5.2 Example 0f Minimization Method - N = 2

To more clearly demonstrate the method outlined in
the last section, consider the case of two real unknowns.

For this case the non-linear equations would be

a XY + le + c Y + d1 0 (5.3a)

l 1

II
C

a2XY + b2X + 02Y + d2 (5.3b)

and FM would be
(5.4)

_ 2 2
FM - (aIXY + le + CIY + d1) + (a2XY + b2X + C2Y + d2) .
Thus if Y is held constant,

a; aJY + bJ (5.5a)

and

bJ cJY + dJ (5.5b)

and if X is held constant,

- a X + c (5.50)

ai J J

‘E

,.. e '

64

and
(5.5d)

bJ = bJX + d3.

From Equation 5.2 for constant Y

[(alY + bl)(clY + d1) + (a2Y + b2)(c2Y + d2)]

X ='- ( Y + b )2 + ( Y + b )2
al 1 a2 2 (5.6a)

and for constant X
[(alX + cl)(b1X + d1) + (a2X + c2)(b2X + (lg—)1

Y = - ( X + )é+ ( X + )2 .
31 Cl a2 C2 (5.6b)

Thus given a Y, Equation 5.6a can be used to find a new X,

and using the new X, Equation 5.6b can be used to find a

new Y and so on. Finally the process will converge to an
X and Y. Substituting this X and Y into Equation 5.4 will
tell whether the point is a non-zero minimum or a solution.

5.3 Modified Newton's Method

While the method outlined in Section 5.1 will rapidly
its rate

converge to the area of a solution or minimum,
Because of

of convergence slows as they are approached.
the nature of FM in the neighborhood of a minimum, the rate
of convergence is adequate to locate the minimum. In the

neighborhood of a solution, however, FM is very slowly

changing, and the convergence becomes extremely slow.

Thus a second method is needed to find the solution once
Because

the neighborhood of the solution has been reached.

of its guaranteed convergence if one is within a small

 

65
enough neighborhood of a solution, the method outlined by
Robinson (6) will be presented here.
Let 3(5) be a vector function of the vector x such

that f(x) = 0. Define the norm of x to be

Hail)2 = lTx (5.7)

and eJ to be the Jth column of the identity matrix. -Then

if x? is chosen close enough to x, the following vector

will converge to zero:

- -l -k
Ak+l = 2Ek+l _ 32k = _ [Fkxklfl ici ) . , (5.8)
where Fk and'Xk are matrices defined by:
‘ k k
Fk = £<§ + XkeJ) ' £(£ ) (5.9)
and ‘. Xk = [ka - xF-lll Hk. Hk is the matrix
k k k k
A1A2 AlAN ’ Al
k 2 ° .
-(Al ) .
Hk = ‘ . . Dk (5.10)
k k
0 AN-lAN ‘AN-l
N-l
O - Z (Ak)2 -Ak
i=1 3‘ N
L. _)

 

 

66
where Dk is diagonal matrix chosen to normalize the columns
in Hk, making Hk an orthonormal matrix. Carrying out the

normalization, Dk becomes

 

 

k .—
dl . . O O
k _ k
D - 0 d1 0
12
__O . . . . O . . . kdflJ
(1+1
where d: = l/[( Z (A? “2 2)1/2] i¢n,l
J= -1 =1
N
a5 = 1/(32 (A3):)1/2
k _ k k 2 1/2
dl - 1/Al (321(AJ) )

Since for the initial point Ak is not given, Hk can be
taken as the unity matrix and Ak can be taken as 50

multiplied by some constant << 1.

5.4 Additional Techniques

 

Totmis point it has been tacitly assumed that the set
of non-linear equations has only one solution or minimum.
This, however, is not the case in general. If, for example,
a set of non-linear equations has several solutions and
several minimums, the solution or minimum found by the
techniques outlined will depend entirely upon the starting

point. Since the location of a. good starting point is no better

67
known than the location of the solution or minimum sought,
an impasse seems to have been reached. The method used
in this study to alleviate this problem was to try a point
at random and find the solution or minimum associated with
this point. Using this solution or minimum as the origin,
look for new solutions or minimums on an N-dimensional sphere
with a radius of at least an order of magnitude larger than
the largest coordinate of the original solution or minimum.
This process can be repeated until the desired solution or
minimum is obtained or until the space has been thoroughly
covered. In using this type of method two things should
be noted: First, only 2(M + N) - 2L real variables need
to be specified where L is the number of constraint
equations since the minimization method uses the first
2(M + N)- 2L to find the remaining variables. Second,
to insure checking all directions each new approach point
should differ from the last by at least one sign change,

i.e., for N = 2, start in a different quadrant each time.

5.5 Numerical Examples

 

In order to consolidate the theory and methods
developed in the first three chapters, several numerical
examples will now be presented. All of these examples
will be for the symmetric case. The computer program used

for the numerical computations is given in the Appendix.

68
5.5.1 Admittance Constraint
The requirements for the admittance constraint

are that the loadings and their position be given. Once
they are given the current distribution and field pattern
for the antenna can be found. Figures 5.1 and 5.2 show
the current distributions of two antennas which are subject
to this constraint. Figure 5.1, which is an example of
pure real loading, shows the very good agreement between
the theory and Altshuler's (7) experimental results.

Figure 5.2, which is an example of pure reactive
loading, compares two theoretical results, the theory of
Nyquist and Chen (8) and the theory presented here.
While the agreement is not as good as in the first case,
the two distributions are quite similar. In addition,
Nyquist and Chen's (8) results are based on a current
distribution which has been forced to be a purely outward
traveling wave. Thus it would seem likely that the actual
solution would lie between the two curves.

While these two examples are for the N = 1 case,
the agreement obtained is such, that it would be safe to
assume the current distributions found by applying the
admittance constraint to the general case would be quite

adequate.

5.5.2 Field Pattern Constraints
The effectiveness of the field pattern

cOnstraint to control the pattern shape is demonstrated

.Aammmoo.o u

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70

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OoH . o . o
.i. .1 . .w .. .. . .. .
02.;
.o.a
com.
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00:1 .o.m
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oos

(AIDA/dWPTIIIw) (2)1

71

in Figures 5.3 and 5.4. Both examples are for an antenna
of half length .71.

The desired field pattern in Figure 5.3 is sin 30
specified at nine points. The resulting pattern is
excellent. There are two results of this solution,
however, which are not desirable. First, six of the nine
loading impedances have negative real parts. Second, the
input impedance is extremely small.

The desired field pattern in Figure 5.4 is sin 20
specified at four points. Again the resulting pattern is
good but not as good as the N = 9 case. In addition to
some of the loading impedances having negative real parts,
the input impedance also has a negative real part. Its
magnitude, however, is four orders of magnitude larger than
the N = 9 case. Thus while the field pattern constraint
can adequately control the shape of the pattern, it
leaves the other parameters of the antenna in less than
desirable form.

5.5.3 Current Constraint

Figures 5.5 and 5.6 show the current distri-
bution for two types of current constraints: 1) traveling
wave and 2) decaying traveling wave, respectively. In both
figures the current was specified at four points as
indicated by the stars on the graphs.

Like the loading impedances for.the field pattern
Constraint, most of the loading impedances for the current

Constraint have negative real parts. Thus in the two cases

72

 

 

 

 

gt

1.0-4
0.8)..
E 0.6 .1-
E-i
P;
0.41.
0.24b
l J I l l I
10 2 30 4b 50 6b 7b 8b
0 (Degrees)
20 = 5.25 x 10"5 + 35.03 x 10'50 a = 0.007021
21 = 12.4 + 33.4 x 1030 n = 0.71
22 = 4.8 + 32.06 x 1030 d1 = 0.063751
23 = 1.85 + 32.08 x 1030 02 = 0.12751
24 = -2.4 + 32.16 x 1030 d3 = 0.191251
25 = -9.6 + 32.37 x 1030 d“ = 0.2551
Z6 = —22.5 + 32.82 x 1030 d5 = 0.318751
27 = -48 + 33.6 x 1030 d6 = 0.38251
28 = -120 + 35.8 x 1030 07 = 0.446251
Z9 = -2.4 x 103 + 34.04 x 1079 d8 = 0.511
d9 = 0.573751
Figure 5.3 Field Pattern For A Transmitting Antenna Loaded

At Nine Points (PT(0) = sin(30)).

73

 

 

/’ \
/ \
// ‘\\
/
/
/
/
/
/
/
/ \
3/
/ \\
/ \
/ \
/ \\
____—-— Desired Pattern \
\
Actual Pattern \
\
\

 

 

1.0 <-
0.8 .L
ECLO 4h
E4
3:
0.4 a.
/
z
O
Z1
Z2
Z3
Z4
Figure 5.4

0 (Degrees)

a = 0.006351
= —.194 + 3.5770 b = 0-7)
= -22.6 + 31.36 x 1030 d1 = 0.11251
= -,12 + 35509 d2 = 0.2251
= 6.03 + 31.16 x 1030 d3 = 0.33751
= -124.8 + 34.2 x 1030 du = 0.451

Field Pattern For A Transmitting Antenna Loaded
At Four Points (PT(0) = sin(20)).

74

 

4.0.. 1 1 t .t : 1
I}
H
8
\\ 3°0‘f
Q.
E -___.___——— ———— -—— ——————— ———-
m
.H
H
H
E 200‘r
1?.
+731,01_ ———————— Actual Current

______ Desired Current

 

  
 

 

 

 

2‘ ‘%
1001
200-
3004
”a? 400-
t
g 500‘[ a = 0.006351
0 h = 2.751
e 6004- d1 = 0.6251
0 00 d2 : l.§51
{c8 7 ‘P 33 : $.515A
i 8004- 4 °
900-1!- \\\
ZO = 351 - 3 1080 \\‘
1000- . 1 . 1 : =
0.51 1.01 1.51 2.01 2.51 2.751
Position
1 = ~88.5 - 3 49.90 Z3 = 205 + 3 36.60
Z2 = -100.9 + 3 66.90 Z“ = -77.1 — 3 279.20

Figure 5.5 Current Distribution For A Transmitting Antenna
On Which The Current Was Specified At Four Points
(Traveling Wave).

75

db

 

cl)-

3.0.3

|It(z)| (milliamp/volt)

Actual Current

1'0 7' _____ Desired Current

 

    
 

 

 

 

4*
100 ..
200 -
A 300 ..
8
m 400 3
a
g 500 - a 0.006351
v h 2.751
m 600 "’ d1 = 0.6251
‘0 d = 1.251
2 700 ‘7- d2 = 1.8751
‘3‘ d3 = 2 51
800 u 4 '
900 ‘7 z0 = 351 - 3 108 \
1000 i i i i : i
0.51 1.01 1.51 2.01 2.51 2.751
Position
Z1 = -41.9 - 3 45.70 Z3 = -53.5 + 3187.60
Z2 = -53-8 + J 35.30 Z4 = 392.7 + 32080

Figure 5.6 Current Distribution For A Transmitting Antenna
On Which The Current Distribution Was Specified
At Four Points (Decaying Wave).

76
where the equations are linear, the resulting impedances
are less than desirable.

The agreement between the desired currents and the
actual currents, however, is good for the decaying traveling
wave (Figure 5.6) but only fair for the traveling wave
(Figure 5.5). Since pure traveling waves are hard to main-
tain for any length on an antenna (even the infinitely
long unloaded antenna decays), the poor agreement in the
traveling wave case should be expected. The good agreement
in the decaying case should also be expected. Altshuler (7)
in his experiments found that an antenna of the same half
length loaded a quarter wavelength from the end with

2400 produces a decaying traveling wave current.

5.5.4 Input Admittance - Frequency Constraint

Up to this point each example was the solution
of linear equations having a unique solution. In this and
the next two sections, the examples are solutions to sets
of non-linear equations having several solutions. To
demonstrate the multiple solutions of these equations, the
input admittance-frequency constraint was applied to
an antenna such that it would have an input impedance of
500 at four frequencies: 400, 500, 600, and 700 MHz.
Five solutions to the problem are listed in Table 5.1. In
addition to the frequency constraint, the imaginary part
of the loading admittances were required to be frequency

dependent. Thus all the loading impedances in Table 5.1

Ammmo mocmzvopm asomv
pCHMmecoo zocosvopmimocmmeEH psch o9 mcoHpSHom H.m oHnt

 

 

 

77

 

 

 

 

 

 

 

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emwm- n :m swam- 1 mm emma u we emoms- u H

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essaa- ".32 emma- 1 mm eoma 1 mm comm- u a
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assam u as as.as 1 mm emma- n as am.:. 1 a

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cmmml u :m Gama u mm Gama 0 mm cmfisa u H

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cmmmu u :m emmml u mm emozu n mm cmmoa n H
sm.m u so so.m u mo sm.a a mo so.a a He

smmmoo.o u a sms.m u e

 

 

78

are parallel combinations of a resistor and either an
inductor or capacitor.

Figures 5.7 and 5.8 show the behavior of the input
impedance for the antenna as a function of frequency for
two of the solutions. As can be seen, the input impedance
is 500 only at the specified frequencies and varies
widely for the other frequencies. In some regions of
Figure 5.8 the real part even becomes negative.

A second example of the input admittance—frequency
constraint is shown in Figures 5£9and 5.10. The input
impedance of 500 was specified at two frequencies, 800
and 1200 MHz, and the imaginary part of the loading
admittances were made frequency dependent (Figure 5.9)
and frequency independent (Figure 5.10). In both cases, the
resulting input impedances as a function of frequency are
quite similar. This indicates that whenever frequency
dependent loading exists, frequency independent loading
may also exist. In addition, unlike the first example,
there are two regions, 800 to 900 MHz and 1200 to 1300 MHz,
in which the input impedance is fairly constant indicating
possible broadband operation, a desired goal.

In all the solutions to the input admittance-frequency
constraint, however, most or all the real parts of the
loading impedances were negative. Thus in the last three
constraints considered, the resulting antenna which

satisfies these constraints, has had one common factor,

 

 

 

 

 

 

 

 

 

 

 

n a = 0.00635m
800" h = 2 75m
d1 = 1.0m R1 = 17159 1L1 = 0.767 uh
d2 = 1.5m R2 = -1610 L2 = 0.268 uh
700-1- d3 = 2.0m R3 = 181:2 L3 = 0.268 uh
d“ = 2.51“ Ru = “5959 Lu = 0.215 Uh
600*— Re(Z )
R L
1 i Zi
500-1-
400._
3004)-
:; 200-“. \‘
N0 \‘
100.. ‘
‘ P
I)
J— 1, ‘1 71/ \ 4"
) / ‘ {‘0’ \ /
\ / \ l \ /
\ I \ l \ /
4003, 1 / \ 1' \ ,’
| I \ \ I’
\ / \ ( ‘1,
\ / \ : b
-200.. ‘ / \ l Im(ZO)
\ / \ p
\ l l
\/ \ l
-300., J \,
. \l
1 .t 1 1 1 1
300 400 500 600 700 800

Frequency (MHz)

Figure 5.7 Input Impedance vs. Frequency For A Transmitting
Antenna (Four Frequency Case - 500 - Solution 1).

80

 

 

 

1600 11 1 1 ~1 1 1
W a = 0.00635m
h = 2.75m
1u001. d1 = 1.0m R1 = -u13.20 L1 = 0.138 uh
d2 = 1.5m R2 = -1330 L2 = 0.1uu uh
d3 = 2.0m R3 = 79.19 L3 = 0.592 uh
1200.. du = 2.5m Ru = 21610 Cu = 1.08 pf
1000 .. fi—R (z )
e 0 R1 3X1 Z1
8001F
600 ‘1-
u001-
200.. §.J

 

 

 

 

 

3
o
N
-200..
-UOOdr
-600 Jr-
—800-+
-1000 1 1 71 1 1

300 ”00 500 600 700 800
Frequency (MHz)

Figure 5.8 Input Impedance vs. Frequency For1\ Transmitting
Antenna (Four Frequency Case - 509 - Solution II).

81

1000

 

900 «-
800 .1
700 .0
600 ..

500 .-

400 i.
300 ._
200 ..

100 q-

(9)

 

-100 d-

-200 ‘-

-300 d-

 

-u00 .. \ IIm(ZO)

-500 up \

 

 

-600 l A l L l 1 1

I U 1 V I U 1

700 ' 800 900 1000 1100 1200 1300
Frequency (MHz)

 

Figure 5.9 Input Impedance vs. Frequency For A Transmitting
Antenna (Two Frequency Case — Frequency Dependent).

82

“F
+-
q
.4.-
d.

1000

 

q-

900‘

800..

u u

H
- H

B

-67u.9 - 332A70
-85.u + 3117.70

0
700-. 2 0.801 Z

600.).

5001

T

400‘

300.5

Z
O
|._:
O
O
l
1

 

 

—lOO..

-200‘b
-300.-

-HOO¢

U

-500dp

 

 

-600

I l 1 1 l L L
r

I I I ' I
700 800 900 1600 1100 1200 1300
Frequency (MHZ)

 

Figure 5.10 Input Impedance vs. Frequency For A Transmitting
Antenna (Two Frequency Case - Frequency
Independent).

83
negative real loading impedances. The next two sections

will show examples which do not have this problem.

5.5.5 Reactive Loading — Current Constraint
The simplest example of reactive loading with

a current constraint is requiring an antenna to have a 509
input impedance at one frequency. This was done for an
antenna of half length 1.11, radius “.5 x 10-31 and with
loading at .AA and .81. The resulting loadings were
Zl = J “5.39 and Z2 = J 3880 and Z1 = J 1890 and Z2 = J “8.79.
Again there is more than one solution. However, for N = 2
the non-linear equations can be solved exactly and these are
the only solutions.

A second example of this constraint is shown in
Figure 5.11. Using the same conditions as the traveling
wave example in Section 5.5.3 except the current was
specified at two points, a reactive loading solution was
sought. As can be seen, the current distribution does not
satisfy the specified current points and is in fact a
standing wave current. These loadings are a minimum of
the set of non~linear equations rather than a solution.
Because of the requirement that all the loads be reactive,
no solution to the problem exists.

In the first example in this section, it was found
that solutions exist for only a narrow band of input
impedances. Thus while reactive loading removes the
possibility of negative real loading impedances, it clearly

restricts the types of problems that can be solved easily.

8A

 

 

 

  

 

 

 

1 l A l 1 n_
1 f I If U '
5.0..
I;
3 Actual Current
{11.0 .
Q 1 —————— Desired Current
8
H
:1 3 O-{p
H h— — ———— -— _ ——-d—
‘5 _____________
;;2.0..
4..)
I:
1.0-n-
1;:T>. 1e—
‘\
100..
200--
300-0-
1; #004-
m
3 500--
b0
0
£3 600«ra.=
h = .
3 700«-dl = 0.6251
g 02 = 1.251
04 800”"d3 = 1.8751
du = 2.5A
900-1
20 = 86.u - J 152.60
1000‘:— J l J A A A
I I l ' r I
0.51 1.01 1.51 2.01 , 2.51 2.751
Position,'Z
= -J 22H.A0 23 = 3 9880
= J 16980 _ Zu = J 5890

Figure 5.11 Current Distribution For A Transmitting Antenna
On Which The Current Was Specified At Two Points
(Reactive Loading).

85
5.5.6 Reactive Loading - Frequency and Current

Constraint

Figures 5.12 and 5.13 show the frequency
dependence of input impedance of two reactively loaded
antennas. In both cases the input impedance was Specified
to be 500 at 60 and 70 MHz. Figure 5.12 is the case of
frequency independent loading and Figure 5.13 is the case
of frequency dependent loading.

Like the previous example of frequency dependent
loading, the resulting input impedance curves are very
similar. Unlike the previous example, the antenna described
in Figure 5.13 can be easily built from available materials,
four sets of inductors and a metal rod.

It should be noted that like the previous reactive
loading example, the solution was not easily found.

Several combinations of loading points, antenna lengths,
and frequencies were tried before this particular example

was found.

5.5.7 Resonance Loading
In Section 2.2.2 relations were derived for
resonance loading and for the instabilities in that
resonance. For a given set of antenna parameters and
loading points, a family of resonance loadings exist. The
instabilities, however, are unique for each set of parameters.

Figures 5.1M and 5.15 show typical values of Z and Z2

1
as a function of frequency for the two types of insta—

bilities. As can be seen, the real part of either Z or

1

86

 

 

 

 

 

 

 

200 1 ; IL IL
100..
ReIZ I
o
0 I —v i
~100..
~200¢b
C:
\I Ile |
o o
N
—A00-. .
a = 0.00635m
h = 2.75m
d1 = 1.0m Z1 = J 392U0
d2 = 1.5m Z2 = J 8220
.500.. d3 = 2.0m Z3 = J u0530
d = 2.5m Z = J 13760
A u
-700 g : 141 %
50 60 10 80

Frequency (MHz)

Figure 5.12 Input Impedance vs. Frequency For A Transmitting
Antenna (Two Frequency - Reactive Loading -
Frequency Independent).

200

100

-100

-200

g -300

-A00

-500

-600

-700

Figure

87

‘-

 

 

 

<1-

WRE(20)
1-
1. Im(ZO)

 

 

 

a = 0.00635m
h = 2.75m -
d1 = 1.0m L1 = 3.88 uh
02 = 1.5m L2 = 2.00 uh
1- d3 = 2.0m L3 = 6.63 uh
du = 2.5m Lu = 3.78 uh
L 1 1 1
50 60 70 80

Frequency (MHz)

5.13 Input Impedance vs. Frequency For A Transmitting
Antenna (Two Frequency - Reactive Loading -
Frequency Dependent).

 

 

 

 

 

 

 

0.6..
0.“.-
0.2--
3
5
AH
N
(8‘03" 22R
-0.u~~ -- -2.5
-0.6«b 15-3-1
HES
—0.8-p qr- -500:
w
a
up -705
a = 0.00318m
h = 0.25m
(11 = 0.1125m
d2 = 0.1675m
TI. -1000
--—12.5
r 1 1 1 -15.0
60 120 180 2H0

Frequency (MHz)

Figure 5.1“ Loading Impedances vs. Frequency For A Trans-
mitting Antenna To Maintain A Resonance
Instability (Finite Voltage).

89

 

 

 

 

 

 

 

11 1 1 1 1 5.0
0.3-11-
0.2--
+b205
0.1 ‘-
”c;
55
A _O.ldL
or!
N
\d: qD-205
m -0.21-
H
a
-o,3.. HF
din-5.0 V
;:
-O.L1qr 3
a = 0.00318m _7 5
Z2I h = 0.25m 7' '
d1 = 0.1125m
d2 = 0.1675m
~-—10.0
--—12.5
1 1 l 1 3
60 120 180 2b0 300

Frequency (MHz)

Figure 5.15 Loading Impedances vs. Frequency For A Trans-
mitting Antenna To Maintain A Resonance
Instability (Zero Coefficient).

90

Z2 is negative in each case. Thus for the usual passively

loaded antennas, these instabilities are of no consequence,
but if active loading is employed they could become
critical.

Figure 5.16 is a graphical solution of the resonance
equation for f = 2A0 MHz. Because of the complex nature
of the graph, a brief explanation is in order. Each
circle on the line parallel to the real 22 axis, Z2R’

is a curve of constant real Z 2 Similarly each circle

1’ lR'
along the imaginary Z2 axis, 221’ is a curve of constant

imaginary Z Z In addition to the circles there are

1’ 11'
two asymptotes, ZlR = -22.3 and 211

circles approach. Since none of the constant 21R circles

= 6ND, which the

intersect except at the point (-3.9, 1.2“ x 103), any
other intersection on the graph gives the values of Z1
and Z2 for resonance, as shown in the circled example.
The common intersection, however, is not a solution but
equivalent to the condition of Y = 0 for the 2-dimensional
hyperbola XY = 1.

Unlike the instabilities, the resonance loadings
do not have to be active. Because the asymptotes are
skewed with reSpect to the axes, there exists a small region in
which the real parts of both Z

and Z are positive. Thus

1 2
the possibility exists of designing an antenna which has

greatly enhanced radiation using only passive elements.

91

mccmuC¢

wchuHEmcme < pom cofipfiocoo MCaomoq monocommm no COHpsaom awoficampo wfi.m mpzwfim

o.mI

 

 

Aexvmmm
BIS m1 3m 9% mam- o..m- mun
. . 1.1. . .m.mmu . am .
comm w I 83 u mm \c
10R. n+ can u HN ,oam \\

 

 

 

 

It

 

 

 

 

 

 

 

v
n! a —
EmNHH.o u U
Hm N + mmN Emm.o u n mam ucwpmcoo
NH NH I ma . I
Hm MN I N Ewamoo o I a HH
.Nn + .No . am: ozm n .H N pcmumcoo .llllll
. 11 11 1 0 u v

 

u m.»

 

 

CHAPTER VI
MATHEMATICAL FORMULATION OF A

MULTI-LOADED RECEIVING ANTENNA

To this point only multi-loaded transmitting antennas
have been considered. It will be shown that with the
addition of one term, the theory can be extended to

receiving antennas.

6.1 Matrix Formulation

 

Consider a multi-loaded receiving antenna of length
2h, radius a, and with N + M arbitrary loads and a central
load illuminated by a normally incident electric field E0
as shown in Figure 6.1. Replacing the loads by the
voltages across and separating the total antenna current into
a component of current which is induced by the incident
field on the corresponding unloaded antenna and a component
of current which is excited by the voltages across the
loads, the problem becomes equivalent to the antennas shown

in Figure 6.2. It can be shown that

'
Iu(z) = Eofu(z) (6.1)
and from 1.1
N
It(Z) =1§_Mvif(w,a,h,di,z). (6.2)

92

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Figure 6.1 Multi-Loaded Receiving Antenna

9A

 
   
 

 
 

n
1N : 5
VN-l ’Q
A - A
11 :o
V2 .
8 I0 IL

 

 

 

(GIG-GIGIGI

 

[9']

 

 

 

(one:

Figure 6.2 Voltage Equivalent Of A Multi—Loaded
Receiving Antenna

95

Therefore the total current at any point is

It(z)

Again to find the

used. Thus

Eofu(dJ) + (YJ

i=-M

'fids, the relation Y

d N

+
13 1§_M
1%3

+ f )VJ 1 ij

i

N
'
Eofu(z) + X Vif(m,a,h,di,z).

V

(6.3)

i = -It(di) is

(6.“)

Comparing Equation 6.“ to 2.2 it can be seen that Equation

'
6.“ differs only by the extra term Eofu(d

now an unknown rather than a known value.

runs from -M to N,

J) and that’VO is

If J again

including J = O, the result is M + N + 1

equations of the form of Equation 6.“.

matrix form,

FRVR
where
_ d

£11 + Y1
d

le -
d

d

f1-1

d

f1-M

 

+ GREO = 0
d d

' fN1 f01
d d

° fNN + YN f0N
d d

' ° fN0 f00+Y0
d d

- ° fN-1 f0-1
d. d.

- fN-M fo-M

(6 . 5)

d d
f-11 ' f-Ml

d d

f-lN f f—MN

d d

f-10 f--M0

d d

d. d.

f-l—M ° f-M-M+Y-M

Writing these in

 

96

 

 

 

 

F—' --1 — —
fu(d1) V1
'. V.
fu(dN) N
I
GR = fu(0) and VR = VO
'
fu(d_1) V_l
f'(d ) V.
—M -M
L _ L .1
Since FR and VR differ from F and V by the addition of the
fd

10's, V0 and Y0, the number of complex unknowns for the
receiving antenna is two greater than those for the
transmitting antenna and the number of complex equations

is one greater. Thus one additional complex constraint
equation will be needed. Otherwise the analysis of required

constraints follows that of the transmitting antenna.

6.2 Constraint Equations

 

Since the transmitting case requires N + M complex
constraints, the receiving antenna needs N + M + 1 complex
constraints. Also since the field pattern of the unloaded
rod illuminated by an electric field can be shown to have

the form,

Pu(0) = Eogu(w,a,h,0), (6.6)

all the constraints outlined for a transmitting antenna

can be applied to a receiving antenna.

97
6.2.1. Load Impedance Constraint
Since for most practical applications, it is
desirable to specify the central load impedance, one
constraint equation could be Yo equals a constant. This

constraint alters one of the equations in matrix 6.5,

N
d d ' _
(fOO + YO)VO + 2 inVi + Eofu(0) - O (6.7)
=-M
i¢0

where YO is now a known value, and reduces the number of
unknowns by one. Thus there are 2M + 2N + l unknowns and
N + M + 1 equations which reduces the number of additional

constraint equations to M + N.

6.2.2. Scattered Field Constraint
It is desirable to be able to control the
scattered field of a receiving antenna. Thus a possible
constraint would be to specify the scattered field in M + N
directions. Using Equation 6.6 and the first equation in

Section 2.2.“.l., the equation for one direction would be

N
PT(GJ) = Eogu(w,a,h,03) + §_Mvigi(w’a’h’93)' (6.8)
Applying Equation 6.8 in M + N directions results in N + M
equations. But there are M + N + l unknowns. The remaining
equation can be generated by specifying one more direction,
or it can be Equation 6.7. If one more direction is used

the resulting equations are like those in Section 2.2.“.1.

and the solution follows that outlined there.

.....-..i

 

that

If Equation 6.7 is used, the resulting set of

equations in matrix form is

77

RV = éSE

 

 

 

 

98

R o
where
_v;;1
77- .
VR - vO , s -
V. '(e )
g +
L_ N_ u M'N
fu(0)
and ‘-
_7 7
' ( > '( >
8_ 9 g 9
R = M J O J
7 ° 7 .
8_M(0m+N) go(eM+N)
d d
f—MO f00 + Yo
7
where g (93) = g (w,a,h,0J).

by 8'1

Multiplying Equation 6.9

- P

gu(ej) - PT(GJ)/EO

T(GM+N)/Eo

' .—
gN(el)

g$(ej)

7
gN(OM+N)

d
fN0

 

 

.1

gives the solutions for all the Vi's.

(6.9)

 

 

 

-
. ‘

99

Since one of the equations in matrix 6.5 has been

used, care must be taken to find the Y 's. If three

 

 

 

 

1
7 7
new matrices Y, FR’ and GR are defined as
r‘8 d d d d -1
f11 ' fN1 f01 f-11 ' f-Ml
d d d d d
F' = le ' fNN f0N f-lN f-MN
R
d d d d d
f1-1 fN-1 f0-1 f-1-1 f—M-l
d. 8 8 d 8
f1-M ' fN—M fo-M f-l-M ° ° ' f-M-M
A —_I
-‘ -1
Y1 . 0 0 0 . 0
0 . Y 0 0 . 0
Y = N
0 0 0 Y—1 0
0 . 0 0 0 . . . Y_M
L. —
and
_.' ._
fu(dl)
' .
' ru<dN>
G =
R 7
r (d_l)
' .
fu(d_M)

 

 

100

then matrix 6.5 less one row (Equation 6.7) may be written as

7

R

(Fé + Y)vR + 0 E0 0 . (6.10)

7
It should be noted that FR

but are M + N by M + N + 1 matrices. Thus they do not

and Y are not square matrices

have inverses. To find the Yi's the same procedure as

outlined in Section 2.2.“.1. can be used, namely:

YVR = -(GREO + FRVR) (6.11)
01”
1 g 1
Y = -(G E + F v )/V . (6.12)
1 R o J=—M HiJ RJ R1
J¢O

6.2.3. Current Constraint
Since on a receiving antenna the current at
the load is of primary concern, it would be desirable to
specify the current at d = 0. This condition, however,
only supplies one constraint equation and M + N + l are
needed. Additional current points could be specified as
was done in Section 2.2.“.2., but a more interesting case

is to specify I(O) at several different frequencies.

From Equation 6.7 and 1(0) = -YOVO
z N z '
fOOVo +1§_Mvif10 + Eofu(0) = 1(0). (6.13)
i#0

But Equation 6.13 is not in a useable form. If CF(w) is

defined as

CE(wl) = le(O)/Eo (6.1“)

101
then Equation 6.13 becomes

N

z ' z _
1; Mviin + EOEfu(0) - CE(1 + foo/YO)J — 0. (6.15)
i#0
Applying 1(0) = -YOVO and Equation 6.1“ to Equation 6.10

and combining the results with Equation 6.15 gives

 

BR . VR = O (6.16)
where
_'d d d
f11+Y1' fN1 f-11
d 8 d
f1N fNN+YN f-lN
B =
R d d d
f1-1 fN-i f-1-1+Y-1
d. a 8
f1-M ° fN-M f-l-M
Z Z Z
L_f10 fN0 f-10
d ' d ‘—
8 . ° d
. . . f_MN (fu(dN) - CEfON/YO)
d I d
f-M-l (fu(d-1) ' CEfO—l/Yo)
rd +Y (f'(d ) - 0 rd /Y )
- - - -M-M -M u -N E O-M o
z ' z
f_MO (fu(0) — 0E(1 + foo/quJ

 

102

and

 

 

Like Equation 2.27, Equation 6.16 has a solution only if

the determinant of B is zero. This produces one equation

R
with N + M + l unknowns. Since, however, YO is usually

known, there are only N + M unknowns in Equation 6.16.
Thus if 1(0) is specified at N + M frequencies, the
required N + M equations will be generated and all the

admittances can be found.

6.2.“. Current And Scattered Field Constraint
The final type of constraint to be considered
will be a combination of the current and scattered field
constraints. Again the current constraint will be to specify
the current at d = 0. If the scattered field is specified
in L directions, then the resulting constraint equations

in matrix form are:

c v =0 (6.17)

where

gi(el)

gimL)

 

10

‘ g1-1

' g1—1

. f

. g_M(01) (gu(01) - (PT(01)/EO + CEgo(el)/Y

. g;(01)

(0

(0

d
1-10

1

L

g};(0L)

f

)

)

2
k0

103

g

f

k+1<0

d
k+10

L

)

)

g1+1(91)

(0 )

g1+1

d
f1+10

L

.1

O)

. g_M(eL) (gu(0L) - (PT(0L)/EO + CEgO(OL)/YO)

d

. f

-MO

' z
(fu<0) - CE(1 + foo/YO)

 

10“
and

 

 

Following the same method used to obtain Equation 2.30,

define the matrix B as

i

 

 

 

105

 

 

d d d d
f11+Y1 ° ° fkl f11 fk+21
Bi = d ° 8 8 d.
flk fkk+Yk fik fk+2k
d d d d
.ufli fki fii+Yi fk+2i .
d d d d I
° f1-11 fk+11 f1+11 ' ' f-Ml (fu(d1) CE f01/Yo )'
d. d. d. d. I °
fi-lk fk+1k fi+lk f-Mk (fu(dk) I CE f0k/Yo )
d d d d I
fi-li fk+11 f1+11 f-Mi (fu(di) ' CE f01/Yqfl
where
k = -(M - L - l) for M > L + l
k = N + M — (L + 1) for L + l 1 M
and i = k + l, . . ., —M. B1 is a M + N - L by M + N + 1

matrix and C is a L + l by M + N + 1 matrix. Thus if they

R
are combined, the result is an M + N + l by M + N + 1

matrix

V = O (6.18)

1 1 = k + 1, . . ., (-M)

whose determinant must be zero. Equation 6.18 generates
L + 1 equations with M + N unknowns. Thus Equation 6.187

IWJSt be used for P frequencies such that

106

(L +1)P = M + N (6.19)

to obtain the required number of equations needed for a

solution.

CHAPTER VII
CURRENT DISTRIBUTION AND SCATTERED FIELD PATTERN

OF AN UNLOADED ROD

In Chapter 6 it was shown that the current and
scattered field of an unloaded rod was needed to solve for
the current and scattered field of a multi-loaded receiving
antenna. In this chapter the induced current for both a
short and long unloaded rod will be derived. The
derivation will be for a normally incident wave. The
derivation of other than normal incidence can be found in

papers by King (59) and Yu and Shen (10).

7.1 Vector Potential Of An Unloaded Rod

 

For an unloaded rod of length 2h and radius a illumi—
nated by a normally incident electric field, E0, as shown

in Figure 7.1, the vector potential is given by

 

 

d2AZ 2 3802
2 + 80 AZ = - w B0 (7'1)
dz
the solution of which is
= _ / — 2
AZ J Eouo [Cl cos koz + EO/ko] h i z i h. (7. )

Equation 7.2 is valid for both short and long antennas.

107

108

 

’m

 

 

251—. <—-

 

 

I
1V:

Figure 7.1 Unloaded Rod

 

2h

 

109
7.2 Current Distribution On A Short Rod

For the short rod the procedure outlined in Section 3.1

will be used. From Equations 3.7 and 7.2, it follows that

h ' ' ' E
fhlrod(z )Kd(z,z )dz = — g53[cl cos koz + E9 + U] (7.3)
.. o 0

where U is defined by Equation 3.10. Since at z = h the

right hand side of Equation 7.3 is zero

cl = - [U + EQ/kOJ/cos koh (7.u)

so that Equation 7.3 becomes
h
[hIrod

 

I t 1 '- Jun
(2 )Kd(z,z )dz — Co cos koh[cos koz - cos koh]

(U + Eo/ko). (7.5)
Equations 7.5, 3.1Aa and 3.1Ab suggest that the approximate

form of the current should be

(7.6)

Irod(z) = IEI(cos koz - cos koh) + IER(cos Roz/2 - cos kOh/2)

Substituting Equation 7.6 into Equation 7.5 and using

Equations 3.17c-e, the result is

 

IEIEUdUR(cos koz - cos koh) + jdeI(cos ka/2 - cos Rob/2)]
+1 _ = 3““ r _ .
ERwdD(cos Roz/2 cos kOh/2) cos kohLcos koz cos koh]

(U + El/ko). (7.7)
Since Equation 7.7 is true for all 2, then

j“’dUIIEI + IERwdD = 0 (7'8)

110

and

 

_ JUN
IEdeUR ‘ Cocos koh(U + Eo/ko)' (7'9)

All that remains is to find U. Substituting Equation 7.6
into Equation 3.10 and using Equations 3.25 b-c results

in
3:0

Combining Equations 7.10 and 7.9 and using Equation 7.8,

IE, and IER can be found. Thus

 

3w hnE
IEI = dD O (7.11)
wdD(deR COS koh ’ wtfh)) + MmedUI Coko
and A
w . HE
- dUI o
IER ‘ .(7.12)

 

wdD(deR COS koh ‘ wu(h))+ 3¢D(h)de1 z:01‘0

If new variables, TE1 and TER’ are defined as
E
_ Mn 0
o
E
An 0
ER ER goko
then Equation 7.6 becomes
(7.15)

['11

um

- __._2 _
Irod(z) - Coko[TEI(cos koz cos koh) + TER(COS ka/2

- cos koh/2)]'

111
7.3 Scattered Field Of A Short Unloaded Rod

 

Once Irod(z) has been found, the scattered field for a
short rod can easily be found. Since Irod(z) has the very
same form as the current of a transmitting antenna, with

the use of Equations “.1, “.9 and “.10 it can be shown

 

 

that
JEO sine TEI
= _ r r
Prod(e) k cose L 2 Lcosesin koh cos(k0h cose)
0 sin 0
TER
— cos koh sin(koh cos®)] + 2
(l—“ cos 0)
[2 cosesin kOh/2 cos(koh 0080) (7.16)

- cos kOh/2 sin(koh cosO)].

Like Equation “.12, Equation 7.16 is indeterminate if
G = 0, H/2, + n/3, or 2H/3. For 0 = 0 it can again be
shown that Prod(0) = 0. If e = H/2, then Equation “.17 is

used and Equation 7.16 becomes

JE
_ _ _43 _
Prod(H/2) - k0[TEI(sin koh koh cos koh)
+ TER(2 sin kOh/2 - koh cos koh/2)]. (7.17)

For e'= H/3 or 2n/3 only the T term is indeterminate.

sin k h
_____2_
2

ER
It has been shown that it reduces to kOh/2 -

112

Thus

. 1E T
P (0):.- O'EEI

r
0d kO cose sine

 

' '
T(cos 0 sin koh cos(koh cos a )

'
- cos koh sin(koh cos 0 )) +

sin k h
o

2 )sin e']. (7.18)

+ TER(kOh/2 -

7.“ Current Distribution On A Long Rod

 

The approximate solution for the long rod will be
based on a paper by Shen (11). Unlike Shen, only the
normally incident wave will be considered.

7.“.1 Current On An Infinitely Long Rod

Jwt time

Applying Equation 7.1, altered for e-
dependence, to an infinitely long rod, the equation's

solution is

E
A (z) = j —9 for all z . (7.19)
z w
Since
U m t v v
Az(z) = E% 1 dz réz )K(z-z ) (7.20)

V

where K(z-z ) is given by Equation 3.“la,

“HJEO m t 1 v
C k f dz Imrod(z )K(z—Z ). (7.21)
O O "°°

 

Applying the Fourier transform to Equation 7.21 and using
Equation 3.“6, the result is
_ “nJE

1mr0d(;)i(;) = Co k: 2n5(;). v (7.22)

 

113

Solving for I(C) and taking the inverse transform,

 

 

3E
I- (z) = “A - 9‘. (7.23)

rod ; —

o k K(o)
0
But from Equation 3.50, K(O) = 290 + JH thus

“n JEo 1

I”. (z) = —— . (7.2“)
rod Co k0 (290+ 3H7

7.“.2. Reflection At The End Of A Semi—Infinite Rod
If the semi-infinite rod is in the upper half
plane with its end at z = 0, and if it is illuminated by a
normal E field such that I rod(z) = l, the reflected current
can be expressed as
k ejczdz

1 (z) = ——9 E (o) _ . (7.25)
r 2“J ' (0(kO—c)cL_<;)

 

If again it is assumed that the contribution to the

integral in Equation 7.25 comes mainly from C = k and

 

 

0
that CO can be deformed around the branch out at z = Cko,
Equation 7.25 becomes

k E (0) 3:2
1r(z) = 21? _ ' 8 dz- (7.26)
L+(ko) C2(ko-c)cK(c)
which is approximately
ir(z) = -RS I§Z)/VO (7.27)
where
R =E_g.__I::_(_9_).
S 2H *
L+(ko)

11“

For large koz,

E (o)
:1———— = 20w + 2n2 + jn . (7.28)
L+(ko)

7.“.3. Current On A Finite Rod

To find the current on a finite rod the
results of the last two sections and Section 3.2.2. must
be used. Away from the end, the finite rod will seem like
an infinitely long rod and this should support a current
like Iwrod(z)' Near the ends Ierd(Z) will produce a
current like Iéz)/VO from both Equations 7.27 and 3.56.
Thus current one finite rod should be of the form

Im(z + h) Im(z - h)

Irec(z) = Iwrod(Z) + CR1 VO + CR2 vO ° (7°29)

 

 

To find C and CR2, Equations 7.27 and 3.56 must be used.

 

 

 

Rl
At 2 = -h, the incident current would be I0° (-h)-
Im(2h) rod
+ CR2 ——V——— which would produce a reflected current at
o
z = -h of
Im(0) Im(2h) Im(0)
Irefl = ”Rslmrod('h) v ’ R032 V V ° (7'30)
0 o o
1m(o)
But by definition Iref = CR1 V0 , thus
Im(2h)
CR1 = ‘ RSI”r0d(-h) ‘ RCR2 ‘“V;"“ (7°31)
Iw(2h)
Similarly at z — h,cR2 — — RS Imrod(h) - Rch-—1V———. (7.32)

O

115

Solving Equations 7.31 and 7.32 for C and C

R1 R2 yields:

R I». (h)
CR1 = — S r°d = CR2. (7.33)
1 + RI..(2h)/vO

Simplifying Equation 7.29, the current distribution for a
finite long rod becomes
(7.3“)
on + co --
S (I (Z h) + I (Z h))]
1 + RIm(2h)/vO vO v

R

 

Irec(z) = I""rod(z)[l -
0

since I0° (z) is a constant.

rod

7.5. Scattered Field Of A Long Rod

 

To find the scattered field of a long rod, the same
type of method as used in Section “.2 will be employed.
From Equation 7.2 the vector potential, corrected for the

e.Jmt time factor, is

AZ = J/quOECl cos koz + EO/kO] - k i z i h

and from Equation “.23

( K(ko cose)
P ) = 3w _ sine
rec G K(kO cose)

 

If again it is assumed that the main contribution to the

integral is the vector potential on the antenna's surface,

then
(7.35)

h
I JVHOSO [Cl cos koz + EO/koje‘JZkOcosedz
-h

Ill

A(kO cose)

116
which using Equation “.5 is

 

 

 

 

_ JC /u a sin k h(l + cose) sin k h(l-cose)
A(k cose) = 1k 0 O( O + O )
o l + cos@ 1 - cose
32E /p e sin(k h 0056)
+ o o o o . (7.36)
k 2 cos@
0
All that remains is to find Cl' This can be done by again

using Equation “.27. If Az(z) = 0 for 2 outside of the
antenna, the functions defined by Equations “.28 and “.29

and a third function

v'(z') = I dcf+<;) Imdze‘JCZ(H(z'-z)-e‘3koz) (7.37)
O O

are needed.
Rewriting Az(z) so that one end is at z = O,

Az(z) becomes
Az(z) = j/pon[Cl(cos koh cos koz + sin koh sin koz)
+ EO/kO]H(2h - z). (7.38)

Using Equations “.28, “.29, “.27, and 7.37, the result is

. . (7.39)
Cl(cos koh T (2n) + sin koh S'(2h) + EO/ko v (2h) = 0

01"
_ ' ’2k h
c - -(EO/ko)2j v (2h)/(yl + y2(2h)eJ o

1 + y3(2h))e'3koh

7
using Equations “.33 and “.3“. Only V (2h) needs to be
' _
evaluated. It can be shown that V (z) is equivalent to

Pl(ei = n/2) as defined by Yu and Shen (10).

117

Thus
V'(2h) = -ZJ y3(2h) + 2n/(20w + £n2 + 3n) (7.“0)
and (7.“1)

= Jk h
C +Eo/ko2e o

1 (-2J y3(2h) + 2n/(2cw + £n2 + JH))/A

where A is defined by Equation “.“0. Finally combining
Equations 7.36, “.23 and “.2“a

sine C kO sin koh(l+cose)

 

 

I 1 (

P (0) = -(E /k ) .
rec O O - £n(sin0)) 2 E0 1 + cose

(91

sin koh(l - cose)
+ ) +
l - 0030 c050

sin(koh coso)
l. (7.u2)

 

 

Equation 7.“2 has two indeterminate points 9 = O, H/2.
At a = 0 it can be shown that Prec(e) = O as it should be.

At 0

H/2 the last term approaches koh and Equation 7.“2

becomes

C k
_ l o
Prec("/2) — -(EO/k091HF§—E; 2 sin koh + koh]. (7.U3)

CHAPTER VIII

ADDITIONAL NUMERICAL EXAMPLES

To this point all the numerical examples have been
for transmitting antennas. This chapter will give some

examples for symmetrically loaded receiving antennas.

8.1 Resonance Loading

 

In Section 2.2.2 the conditions for resonance loading
on a transmitting antenna was derived. In Section 6.2 it
was found that all the constraints of a transmitting antenna
can be applied to a receiving antenna with only slight
modifications. The modifications for resonance loading are
that the second loading impedance is replaced by the central
load impedance, 20; V2 is replaced by V0, V0 is replaced
by ED, and the zero coefficient instability is dropped.

Figure 8.1 like Figure 5.1“ shows the values of the
load impedance as a function of frequency for the finite
voltage instability. Figure 8.1 shows two things of
interest: 1) the real part of central load is positive and
2) it is near 509, a typical impedance of a recieving
antenna. Since for a receiving antenna, V0 is one of the

unstable voltages, this instability point could prove to

be extremely undesirable.

118

119

 

 

 

 

 

 

 

 

 

l L L I
j Y t T
80 ‘-
df3.0
6O ._.
db2.0
“0 4p
d 1.0
20 .. r
H
i
N
p.
E :9
:3
”Q
N
g
-20‘r
‘b-1.0
"L‘O-I-
1--2.0
-60'P
a = 0.00318m \
—80 h = 0.25m \\ “ 3'0
d1 = 0.1125m \
\
ZOI Z1R
% 1 11 1 -“.0
60 120 180 2“0
Frequency (MHz)
Figure 8.1 Load Impedance vs. Frequency For A Receiving

Antenna To Maintain The Resonance Instability.

120
Figure 8.2 is constructed like Figure 5.16 and gives
the other values of impedance required for resonance.
Like Figure 5.16, it also has an area of passive loadings.
Solving the equation given in Figure 8.2 for 21R = 0,
it can be shown that resonance can occur if ZOR i 209.

Thus with the right reactive load at d the antenna can

1’
be taken in and out of resonance by adjusting the central
loading.

Unlike the transmitting case, the receiving antenna
coefficients are dependent on the direction of the incident
radiation. Thus the receiving antenna can also be taken
in and out of resonance by varying the direction of the
incident radiation. Since resonance enhances the voltage

of the central load, this would indicate that an extremely

sensitive and narrow beamed receiving antenna could be made.

8.2 Scattered Field Constraint

 

Figure 8.3 shows the desired (dashed line) and the
resulting scattered field (solid line) for a receiving
antenna of half length .7). Like the field pattern
constraint for a transmitting antenna, Figure 5.“, the
field was specified at.four points. But unlike the
transmitting case, the central impedance was fixed at 509
for the receiving antenna. This last constraint removes
one of the problems with the transmitting case, uncontrolled
input impedance. Thus the receiving scattered field

contraint can produce the desired scattered pattern as well

121

mcsouc< wcfi>fiooom < pom cofipfivcoo wcfiomoq momeOmmm mo soapSHom HmoHSQMLo m.m opzmfim

 

  

 

 

 

 

 

 

 

 

 

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oom
\
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\\\|\a - 'l'l lllllll l1
3 Ho \ \\ \ A.////
f How I N u .N \\ \ \ / / /
.. m0 m0 \ x / 1!
2.0 + N u .N \ _ . / ./ o.H
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mom 1 N u .N / \ cow /
I
1' / \ // I'moH
Ho mo Ho mo
NHN + NAN . HHN NHN + NHN . mHN HAN pcdpmcoo IIIII
Ho 1 mo I . Ho mo 1 .
.No . .Nn - .Nm + .No. .mHN pcdymcoo r
d u T N d 4 1 Com

 

 

 

 

.l“ . ,2 x
\
//” ‘\
// \
/ \
.12-H- / \\
/ \
/ \
.10.. /
/ \
/ \
\
i: .08-- \
3 \
(LB \
.061. / ——————— Actual Pattern \
\
/ ————— Desired Pattern \
/ \
.ou.. / \
/ \
/ \
, a = 0.0106) \
.02 1 / h = 0’71 \
‘i \
/‘ \
/ \
/ \
1 s 1, 2% : s : 1 4
10 20 30 .“0 50 60 70 80 90
6 (Degrees)
ZO = 509
Zl = -80.8 + J 3119 d1 = 0.11
Z2 = -37.0 + 3 5029 d2 = 0.291
Z3 = 16.6 + 3 6289 d3 = 0.“01
Z“ = 168.9 + J 5829 d“ = 0.55)
Figure 8.3 Field Pattern For A Receiving Antenna Specified

At Three Points (PT(O) = sin 29).

123
as maintaining a reasonable central impedance. However,
like the transmitting case, it also has negative real

loading impedances.

8.3 Current Constraint

 

Unlike the transmitting current constraint, the
receiving current constraint is designed to control only
current at the central loads and the value of the central
load. Figures 8.“, 8.5, 8.6, and 8.7 show the current
distribution for an antenna of half length 2.75 meters
at four frequencies: 300, 320, 3“0 and 360 MHz. The
specified currents and the resulting currents are given on
the figures. As can be seen, the currents outside the
neighborhood of the central load vary widely. Thus good
control of the central current and load is obtained by this
constraint. Therefore the objective of constructing a
frequency rejecting receiving antenna with constant central
loading can be achieved by this constraint if active loading

is acceptable.

8.“ Current And Scattered Field Constraint

 

The final example of the versatility of the methods
outlined in this thesis will be current and scattered field
constraint. Figures 8.8 and 8.9 show the scattered fields
and current distributions, respectively, for a receiving
antenna for which the backscattered field was Specified
to be zero at two frequencies, 300 and “00 MHz, and the

central current was specified to be 10 ma-m/V at 300 MHz and

12“

 

  
  
 

z = 509
R2 = -10959 ' L1 = 1.17 uh
R2 = -2119 L2 = 1.82 uh
“0 R3 = 5759 L3 = l.“9 uh
‘* Ru = -2339 L, = 0.u26 uh
301

*r
:0
H-
I-'
5.1.
N
[.5

   

20‘

10-

 

= 0.00635m
= 2.75m
0.625m
1.25m
1.875
2.5m

-10 1)-

It(Z) (milliamp-meter/volt)

0:00-04:79)
4:"me

I

-20.

Re(Iz(z))

-301r --"--Im(Iz(z))

It(0)
It(0)
-“O f

0.5 1.0 1.5 2.0 2.5 2.75
Position, 2 (Meters)
Figure 8.“ Current Distribution For A Receiving Antenna For
Which The Central Load And Current Were Specified
At Four Frequencies (f = 300 MHz).

1.0 pa - m/V (Desired)
1.8 - 3 0.10 ua - m/V (Acutal)

 

 

AL 1

 

125

 

 

  

 

 

 

  

 

 

0.25 i % : % ‘L
20 = 509
R1 = -10959 L1 = 1.17uuh
0 20 +_ R2 = -2119 L2 = 1.82 uh
R3 = 5759 L3 = 1.“9 uh
R“ = -2339 L“ = 0.“26 uh
O 15 “r a = 0.00635m
h = 2.75m
d1 = 0.625m
d2 = 1.25m
d = 1.875m
0.10 db 3 _
du - 2.5m
S
H
g 0 05 4)-
\.
L.
(D
1.)
(D
7
Q.
s ~~4
5 I
/\ I
N I
F$0.05 .. I
I
I
I
I
-0.10 q- \ /
\ I
L]
—0.15 d- Re<It(Z))
_______ Im(It(Z))
I (0) = 10.0 ma-m/V (Desired) _
IE(0) = 9.98 + J .016 ma-m/V (Actual) f ' 320 MHZ
-0.20 : : “I : :
0.5 1.0 1.5 2.0 2.5
Position Z, (Meters)
Figure 8.5 Current Distribution For A Receiving Antenna For

Which The Central Load And Current Were Specified
At Four Frequencies (f = 320 MHz).

126

 

 

 

    

 

 

 

25 1. 1r : : 'L
13(0) = 1.0 ua-m/V (Desired) /\
13(0) = 1.16 - J 0.38 ua-m/V(Actua1)‘/ \
I
For Values of
20 IF Loads See I
Figure 8.“ l
I
I
15 “P ‘
I
I
I
10 a. ‘
" I
H
O I
>
\ I
p
B 5 "I' ‘
a) I
E .
E /' \ |
:1 76 \
g \\ / I
V \//
g b
\1>-5 I
I4
_10 up
Re(It(Z))
f = 3“0 MHz
-20 . 1 1 L7 1
U I I ' T
0'5 1.0 1.5 2.0 2.0

Position Z (Meters)

Figure 8.6 Current Distribution For A Receiving Antenna For
Which The Central Load And Current Were Specified
At Four Frequencies (f = 3“0 MHz).

127

 

 

 

 

 

 

 

25 : i I i I
For Values of
20 Loads See
1* Figure 8.“

15 «I-
33 10..
H
O
;>
\.
S...
(D
4.)
7 5"
o.
g -~‘
«-I I)-
:I ,,/
'r'I M
E
’13
If -5...

~108r

Re(It(Z))
'15‘P ______ Im(It(Z))
It(0) = 1.0 ua-m/V (Desired) f = 360 MHz
It(0) = 1.01 - 3 0.02 ua-m/V (Actual)
'20 i I i I *I
0.5 1.0 1.5 2.0 2.5

Position Z, (Meters)

Figure 8.7 Current Distribution For A Receiving Antenna For
Which The Central Load And Current Were Specified
At Four Frequencies (f = 360 MHz).

1287

 

 

 

 

 

 

 

 

3.0.. h = .7m f = 300 MHZ
a = .0106m
d1 = 0.1m
III- (12 = 0.29m
(13 = 0.“m
dLI 3 0.55m
2.0.b
T:
m
__ 3_
1.0..
: : t : : : :2 :
0 10 20 30 “0 50 . 6O 70 80 90
0 (Degrees)
(a)
3.0
R
R1 = -95609
gf‘ L1 = 0.373 uh
232 0-- R2 = -1979 f = “00 MHZ
H L2 = 0.106 uh
:; R3 = 13109
Of) Tb L3 = 0.192 uh
__ R“ = -57509
1 0 L“ = 0.179 uh
0 10 2O 30 “0 50 60 70 80 90
0 (Degrees)
(b)
Figure 8.8 Scattered Fields For A Receiving Antenna For

Which The Current And Radiation Pattern Are
Specified At Two Frequencies:

(6) f =

“00 MHZ.

(a) f =

300 MHz

I(Z) (Milliamp-meters/volt)

I(Z) (Milliamp—meters/volt)

60

129

db

1 l
I 1

{I-

 

   

 

————Re(I(Z))
Im(I(Z))

 

 

 

 

 

 
 
 
 

 

 

 

 

 

'60 I I 1 T
001 003 005 0.7
Position Z, (Meters)
(a)
;_
0.8 «I-
For value
of loads see
0.6 ‘- Figure 8.8
O.“
0.2
-0.2
-0.“
-0.6 4 ;
0.1 0.3 0.5 0.7
Position Z, (Meters)
(b)
Figure 8.9 Current Distributions For A Receiving Antenna For

Which Current And Radiation Pattern Are Specified
At Two Frequencies: a) f = 300 MHz And h) f = “OOMHz

130

0.01 ua-m/V at “00 MHz. Again with the use of active
loading, the desired conditions were obtained. Thus if
active loading is allowed, the current distribution, the
scattered fields and the central load can be controlled
as a function of frequency. Since these are all the
important characteristics of a receiving antenna, the
objective of improving the receiving characteristics of a

linear antenna has indeed been achieved.

CHAPTER IX

CONCLUSIONS

In the preceeding chapters a method has been
developed to improve the transmitting and receiving
characteristics of a linear antenna. Matrix equations to
describe multi—loaded transmitting and receiving antennas
were derived. Further constraint equations were developed
which related the loading impedances to the desired
antenna characteristics and the characteristics to the
loading impedances.

In particular, relations were developed for a
transmitting antenna such that:

l) M + N loading impedancescnuibe Specified and the

current and field pattern could be found,

2) M + N current points can be specified and the M + N

complex loading impedances could be found,

3) M + N field pattern points can be specified and

the M + N complex loading impedances could be found,

“) the input impedance can be specified at M + N

frequencies and the M + N complex loading
impedances could be found,

5) any of the above can be specified at (M + Nl/2

and the M + N reactive loading impedances could be

found,
131

132
6) resonance loading can be achieved and no
characteristics could be found.
In addition to the above relations, constraints were
developed for a receiving antenna such that the central
load impedance could be specified and relations 2) and 3)
from above be applied.

The first three relations are linear and presented no
problems in obtaining a solution. The remaining relations
were non-linear and may or may not have had one or more
solutions. Since in general these non-linear equations
could not be solved analytically, two numerical methods
were employed: 1) a minimization method for finding the
neighborhood of a solution and 2) a linear method for
finding the solution.

Examples of each of the constraints were presented.

As a result, it was found that in each case of complex
loading, the desired characteristic could be obtained but
only by active loading. When reactive loading was required,
it was found that if a solution existed, the results were

as good as the case of active loading. However, the
requirement of reactive loading severely limited the chances
for a solution.

The most interesting result of the study, was resonance
loading of a receiving antenna. It was found that a single
reactive load and the proper adjustment of the central load

impedance could create an antenna with high gain and a very

narrow beamwidth. In this resonance loading, however, it

133
was found that certain values of loadings would cause the
antenna characteristics to become unstable.
Thus multi-loaded antennas can achieve improvements

in receiving and transmitting characteristics normally

requiring extensive arrays.

1.

10.

11.

BIBLIOGRAPHY

R. W. P. King and T. T. Wu, "Cylindrical Antenna With
Arbitrary Driving Point," IEEE Trans. Antennas
Propagat., Vol. AP-l3, pp. 711-718, (Sept. 1965).

T. T. Wu, "Theory Of The Dipole Antenna And The
Two-Wire Transmission Line," J. Math. Phys. (New York),
Vol. 2, pp. 550-57“, (July 1961).

L. C. Shen, T. T. Wu, and R. W. P. King, "A Simple
Formula Of Current In Dipole Antennas " IEEE Trans.
Antennas Propagat., Vol. AP-l6, pp. 5“2-5“7, (Sept. 1968).

L. C. Shen, "The Field Pattern Of A Long Antenna With
Multiple Excitations," IEEE Trans. Antennas Propagat.,
Vol. AP-l6, pp. 6“3-6“6, (Nov. 1968).

L. C. Shen, "Current Distribution On A Long Dipole
Antenna," IEEE Trans. Antennas Propagat. (Commun.),
Vol. AP-l6, pp. 353-35“, (May 1968).

S. M. Robinson, "Interpolative Solution Of Systems Of
Nonlinear Equations," SIAM J. Numer. Anal., Vol. 3,
pp. 650-658, (1966).

E. E. Altshuler, "The Traveling-Wave Linear Antenna,"
IRE Trans. Antennas Propagat., Vol. AP-9, pp. 32“-329,
(July 1961).

D. P. Nyquist and K. M. Chen, "Traveling Wave Antenna
With Non-Dissipative Loading," IEEE Trans. Antenna
Propagat., Vol. AP-l6, pp. 21-31, (Jan. 1968).

R. W. P. King, "Current Distribution In Arbitrarily
Oriented Receiving And Scattering Antenna," IEEE Trans.
Antenna Propagat., Vol. AP-20, pp. 152-159, (March 1972).

I. P. Yu and L. C. Shen, "Modifications Of Scattered
Field Of Long Wire By Multiple Impedance Loading,"
IEEE Trans. Antenna Propagat., (Commun.), Vol. AP—19,
pp- 55“-557, (July 1971).

L. C. Shen, "A Simple Theory Of Receiving And Scattering
Antennas," IEEE Trans. Antennas Propagat., (Commun.),
Vol. AP—18, pp. 112—11“, (Jan. 1970).

APPENDIX

A-1

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