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emu”:

BROADBAND ANALYSIS OF RADIATING, RECEIVING
AND SCATTERING CHARACTERISTICS
OF MICROSTRIP ANTENNAS AND ARRAYS

By
Michael Alan Blischke

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

DOCTOR OF PHILOSOPHY

Department of Electrical Engineering

1989

(/001351’]

ABSTRACT
BROADBAND ANALYSIS OF RADIATING, RECEIVING
AND SCATTERING CHARACTERISTICS
OF MICROSTRIP ANTENNAS AND ARRAYS
By

Michael Alan Blischke

Considered are a pin fed circular patch, an array of pin fed circular patches and
an array of microstrip dipoles, all mounted on a dielectric above a ground plane. In
each case, the solution is obtained via an inverse Fourier transform dyadic Green func-
tion Galerkin’s method formulation. The solution is potentially exact, including
higher-order resonances. An attachment current distribution provides for continuity at
the feed pin-patch junction.

In the scattering case, the feed pin is attached to both the patch and the ground
plane, or, for the dipole array, the dipole has no load impedance. The structure is

illuminated by a plane wave.

In the transmitting case, a voltage source is inserted between the feed pin and the

ground plane, or at the center of the dipole to drive the structure.

In the receiving case, an arbitrary load impedance replaces the voltage source and
a plane wave again serves to drive the structure. This case is found as a combination
of the first two cases, with the voltage in the transmitting case chosen so as to be

equal to the voltage across the load impedance due to the current flowing through it.

For the single antenna, the electric field dyadic Green function components relat-
ing current and electric field components parallel to the ground plane on the
dielectric-cover interface and normal to the ground plane within the dielectric region

are obtained.

Michael Alan Blischke

Current on the patch surface is expanded as a summation over a set of smooth
continuous currents. The feed pin current distribution is constant over the feed pin
surface. An additional patch current distribution flows across the patch surface radially
away from the patch-feed pin junction such that current flows continuously from the

feed pin onto the patch with no discontinuous charge build-up.

Current amplitudes are obtained via Galerkin’s method. The current distributions
are chosen so that all spatial integrations arising from Galerkin’s method, including
those involving the patch component of the feed pin current, can be performed in

closed form. Spectral integration is performed via real-line integration.

For the array, the patches are identical and arranged in an infinite and periodic
rectangular array. The inverse Fourier transforms of the single patch dyadic Green
function components are converted to infinite summations, avoiding spectral integra-
tion.

For the microstrip dipole array, the dipoles are arranged in an infinite rectangular

array. Only current flowing along the length of the dipole is assumed.

Copyright by
MICHAEL ALAN BLISCHKE
1989

ACKNOWLEDGEMENTS

I would like to thank my major professors, Dr. K. M. Chen and Dr. Dr. E. J.
Rothwell for their support, assistance and direction during my stay here. I would also
like to thank Dr. D. P. Nyquist for his help. Finally, I would also like to thank Dr. J.

Kovacs.

TABLE OF CONTENTS

 

 

List of Figures.
1. Introduction.
1. Introduction.

 

II. Patch Antenna.

 

2. Problem Description.

 

2.1 Geometry.

 

 

2.2 Problem Decomposition.
3. Scattering Case.

 

3.1 Excitation Field.

 

3.2 Green Functions for Scattered Field.

 

3.3 Derivation of Coupled Inteng Equation.

 

3.4 General Matrix Formulation.

 

3.5 Current Distributions.

 

3.6 Specific Matrix formulation.

 

4. Transmission Case.

 

III. Patch Antenna Array.

 

 

5. Problem Description.
6. Scattering Case.

 

6.1 Green Functions for Infinite Array.

 

6.2 Current Distribution.

 

6.3 Specific Matrix formulation.

 

6.4 Convergence of Feed Pin-Feed Pin Matrix Element.

7. Numerical Results.

 

 

IV. Dipole Antenna Array.

 

8. Problem Description.

 

8.1 Geometry.

 

 

8.2 Problem Decomposition.
9. Scattering Case.

 

9.1 Excitation Field.

 

9.2 Green Functions for Scattered Field.

 

9.3 Derivation of Coupled Integral Equations.

 

9.4 Matrix formulation.

 

vi

10. Transmission Case.

 

11. Numerical Results.

 

V. Excitation field, Green Function and Current Derivations.

 

12. Plane Wave Reflected by Coated Conductor.

 

13. Derivation of Green’s Functions for Sources in the Presence of a
Grounded Dielectric Slab.

 

13.1 Preliminaries.

 

13.2 Representation of Field Quantities using Hertzian Potentials.

 

13.3 Boundary Conditions on Hertzian Potentials.

 

 

13.4 Fourier Integral Representation of Hertz Potentials.
13.5 Solution for the Scattered Hertzian Potentials.

 

 

13.6 Solution for the Primary Hertzian Potentials.

13.7 Green Functions for the Hertzian Potential maintained by a
Horizontal Source in Region 1.

 

13.8 Green Functions for the electric field maintained by a Horizontal

 

Source in Region 1.

13.9 Green Functions for the Hertzian Potential maintained by a
Vertical Source in Region 2.

 

13.10 Green Functions for the electric field maintained by a Vertical

 

Source in Region 2.
13.11 Electric Field Green Function Summary.

 

13.12 Green Functions for an Infinite Antenna Array.

 

14. Current Derivations.

 

14.1 Derivation of Patch Current Distributions.

 

14.2 Feed Pin Current Distributions for Single Patch Antenna. ..................

14.3 Feed Pin Current Distributions for Patch Array.

 

VI. Matrix Element Evaluations.

 

15. Matrix elements.

 

15.1 Patch-Patch Matrix elements.

 

15.2 Patch-Feed Pin Matrix elements.

 

15.3 Feed Pin-Patch Matrix elements.

 

15.4 Feed Pin-Patch Matrix elements.

 

16. Matrix Element Summary.

 

 

16. Matrix Element Integral evaluations.
VII. Programming Details for Single Patch.

 

17. Programming Details for Single Patch.

 

17.1 Expansion of A,” and B,,..

 

17.2 Asymptotic Forms for Spectral Integrands.

 

vii

85
87
120
120

130
130

131
132
140
142
144

150
163
171

179
186
188
197
197
204
214
218
218
218
229
231
239
242
244
259
259
259
269

17.3 Products of Bessel Functions for Argument Approaching Zero.

 

 

 

285

17.4 Asymptotic Forms for Spectral Integrands as Argument
Approaches Zero. 288
VIII. Conclusion. 293
List of References. 295

 

Viii

LIST OF FIGURES

Figure 1. Patch with feed pin.

Figure 2. Excitation field coordinate system.

Figure 3. Geometry of an infinite rectangular array of patch antennas.

Figure 4. Scattered current magnitude for parallel electric field polarization.

Figure 5. Input resistance for each element of array.

Figure 6. Input reactance for each element of array.

Figure 7. Total feed pin current magnitude for parallel electric filed polarization.

Figure 8. Received power for parallel electric field polarization.

Figure 9. Current distribution at 5.2 GHz.

Figure 10. Current distribution at 8.44 GHz.

Figure 11. Current distribution at 9.75 GHz.

Figure 12. Current distribution at 9.85 GHz.

Figure 13. Scattered current magnitude for perpendicular electric field polarization.

Figure 14. Total feed pin current magnitude for perpendicular electric field
polarization.

Figure 15. Received power for perpendicular electric field polarization: ZL = 500 .

Figure 16. Geometry of an infinite array of strip dipoles.

ix

Figure 17.
Figure 18.

Figure 19.

Figure 20.

Figure 21.

Figure 22.
Figure 23.

Figure 24.

Figure 25.
Figure 26.
Figure 27.
Figure 28.
Figure 29.
Figure 30.
Figure 31.
Figure 32.
Figure 33.
Figure 34.
Figure 35.

Figure 36.

Figure 37.
Figure 38.
Figure 39.

Figure 40.

Decomposition of receiving case into transmitting and scattering cases.
Reflection coeficient. a. Pozar et a1. b. Chen et a1.

Reflection coeficient a. Pozar et a1. b. Chen et a1.

Input resistance for each element of array.

Input reactance for each element of array.

Scattered current magnitude at center of dipole.

Received power for ZL = 50 (2.

Received power for ZL = 100 0.

Received power for ZL = Zia.

Received power for ZL complimentary to 2,".

Received power for 2,, = 23,.

Input resistance for each element of array at 17.0 and 18.0 GHz.
Input reactance for each element of array at 17.0 and 18.0 GHz.
Scattered current magnitude at center of dipole at 17.0 and 18.0 GHz.
Received power for ZL = 50 Q at 17.0 and 18.0 GHz.

Received power for ZL = 100 Q at 17.0 and 18.0 GHz.

Received power for ZL = Zia at 17.0 and 18.0 GHz.

Received power for ZL complimentary to Z," at 17.0 and 18.0 GHz.
Received power for ZL = 2,; at 17.0 and 18.0 GHz.

Input resistance for each element of array at 9.8 GHz.

Input reactance for each element of array at 9.8 GHz.

Scattered current magnitude at center of dipole at 9.8 GHz.
Received power for ZL = 50 Q at 9.8 GHz.

Received power for ZL = 100 Q at 9.8 GHz.

X

Figure 41.
Figure 42.
Figure 43.
Figure 44.
Figure 45.
Figure 46.

Figure 47.

Received power for ZL = 2," at 9.8 GHz.

Received power for ZL complimentary to 25,, at 9.8 GHz.
Received power for ZL = 2,; at 9.8 GHz.

Incident plane wave: parallel polarization.

Incident plane wave: perpendicular polarization.
Complex Z-plane.

Feed pin-centered coordinate system.

xi

I. INTRODUCTION.
1. Introduction.

While the transmitting properties of patches and patch arrays have been the object
of much study [1-7], the scattering and receiving modes have had relatively little atten-
tention [8,9]. Further, the majority of the analyses have concentrated on a narrow fre-
quency band near the fundamental resonance, neglecting off-resonance behavior, and

the effects of higher order resonances.

This dissertation presents a theoretical analysis of the receiving characteristics of
a single microstrip patch antenna and of a patch antenna array of infinite extent con-
sisting of a circular patch(es) with offset feed pin(s). With an assumed plane wave
excitation, the power delivered to a load impedance connected to the feedpin(s) is
determined as a function of incidence angle and array parameters, and has been numer-

ically calculated over a frequency range of over three to one.

Since the receiving mode may be viewed as a superposition of transmitting and
scattering cases, an accurate analysis of the array acting as both scatterer and
transmitter is needed over a wide bandwidth. A potentially exact approach is therefore
undertaken, using electric field dyadic Green functions for horizontal current sources
above a grounded dielectric slab, and for vertical current sources within the slab. This
approach allows the coupling between the feed pins and the patches to be included

explicitly and fully, an effect only recently receiving attention of researchers.

The Green’s functions are determined via a two dimensional spatial Fourier
transform over the coordinates transverse to the grounded dielectric slab. Coupled
integral equations are developed for the current induced on the patch(es) and feed
pin(s) both by an applied load voltage for the antenna/array acting as a transmitter, and
by the incident plane wave for the antenna/array acting as a scatterer. Solutions are

undertaken using Galerkin’s method, and are used to calculate the input impedance of

1

2
the transmitting antenna/array, and the current on the feed pin(s) of the scattering

antenna/array. From these, the power delivered to the load of the receiving

antenna/array is deterrrrined.

Mathematical expansion of surface current on a circular patch is rigorously deter-
mined through a taylor series expansion. The current distribution basis functions are
expressed in terms of products of sinusoids (varying azimuthally) and Tchebychef
polynomials (varying radially). This choice of basis functions is prudent in that it
forms a complete set, and in that the spatial integrals arising from the application of
Galerkin’s method can be evaluated in closed form [5]. An additional term is included
to account for the divergent nature of the patch current near the feed pin junction.
Such a "singular" current has been used before [10] in patch current modelling,
although of a different form; its purpose is to provide for continuity of current at the
feed pin junction, and also to accelerate the convergence of the surface current expan-
sion in the vicinity of the feed pin.

Two different singular current distributions are introduced here. The first distri-
bution, used in the case of an infinite array, requires some numerical integration of the
Galerkin’s method integrations. Current flows radially from the feed pin junction to
the edge of the patch, falling to zero there. A second singular current distribution used
for the isolated patch is smooth and continuous, although non-zero over only part of
the patch. The current flows radially away from the feed pin, falling to zero at a con-
stant radius from the feed pin, generally not at the edge. The advantage of this second
distribution is that all Galerkin’s method integrations may be evaluated in closed form,

in terms of simple Bessel functions

For the case of the infinite array, the spectral integrations from the Fourier
transform are converted into an infinite double summation over discrete values of the

transform variables. Each term in the summation represents a single Flouquet mode.

3

For the case of the single patch antenna, the two dimensional inverse Fourier
transform is manipulated into polar form, and the azimuthal spectral variable is
integrated out to obtain a single spectral integration from zero to infinity, which is per-
formed using real-line integration. Large argument asymptotic forms of this integra-
tion are isolated and performed analytically, leaving the remaining integration to be
easily performed numerically to the limit of infinity.

This dissertation also presents a theoretical analysis of the receiving characteris-
tics of a microstrip dipole array of infinite extent. With an assumed plane wave exci—
tation, the power delivered to load impedances centered on the dipoles is determined
as a function of incidence angle and array parameters, and has been numerically calcu-

lated over a frequency range of 3 to 1.

As in the case of the infinite patch array, a potentially exact approach is under-
taken, using electric field dyadic Green functions for horizontal current sources above

a grounded dielectric slab.

The Green functions used in the infinite patch array are conscripted for use here.
Coupled integral equations are developed for the current induced on the dipoles both
by an applied load voltage for the array acting as a transmitter, and by the incident
plane wave for the array acting as a scatterer. Solutions are undertaken using
Galerkin’s method, and are used to calculate the input impedance of the transmitting
array, and the current at the dipole center of the scattering array. From these, the

power delivered to the load of the receiving array is determined.

Mathematical expansion of the surface currents on the strip dipoles is accom-
plished using piecewise sinusoidal basis functions, with no variation across the width

of the dipoles.

II. CIRCULAR PATCH ANTENNA.
2. Problem Description.
2.1 Geometry.

The geometry of the circular microstrip patch antenna to be analyzed is depicted
in Figure 1. A dielectric substrate of perrnitivity 22 , permeability no and thickness d,
(region 2), is located between the z = 0 and z = -d planes. The dielectric is mounted
on a conducting ground plane at z = -d, and is covered by a material with constitutive
parameters a, and u, (region 1). The patch has radius b and is connected to a feed pin
of radius a centered at a point r; = x0)? + y,y‘ running from the patch to a load
impedance 2,, at the ground plane. The patch is located at the dielectric-cover inter-
face in the z = 0 plane, with center located at the origin, and is assumed to be perfectly

conducting and infinitely thin.

Illumination of the structure is taken to be through an incident plane wave of fre-
quency co at an arbitrary incidence angle. The plane wave is expressed in terms of a
coordinate system x", y”, z” rotated with respect to the coordinate system of the patch,
as shown in Figure 2, with 4», the angle between the x and x” axes, and e,- the angle

between the wave vector I?" and the z axis. The 2 and z” axes coincide. The factors

u = sin(9,-) cos(¢.-) (1)
and
v = we.) sin(¢.-) (2)

are the direction cosines for the wave vector 1?" with the -x and —y axes respectively.

 

81

 

 

I j fl z=0

 

 

 

 

/2/ / / / /”“’

Figure 1. Patch with feed pin.

 

 

 

x” = x cos(¢i) + y” sin(¢,~)
y" = —x sin<¢.-) + y" cos<¢.>'

2:2

Figure 2. Excitation field coordinate system.

2.2 Problem Decomposition.

Using the principle of superposition, the antenna acting as a receiver can be
decomposed into scattering and transmitting cases. In the scattering case, the load
impedance is replaced with a short-circuit to the ground plane, the antenna is
illuminated by an incident plane wave, and the feed pin current, I 1, is determined.
For the transmitting case, no illuminating plane wave is present Instead, a driving
voltage V is applied at the base of the feed pin in place of the load impedance,
representing the voltage drop which would exist across the load in the receiving case
due to current flowing on the feed pin. The resulting current on the feed pin, 12, due

to V is found, and the input impedance is then calculated from
V
Zr. - 1—2 (1)

Since V is the voltage drop across the load impedance ZL due to the total current

l=ll+12,

V = 421. = "(1 1+I 2)ZL (2)

But I, = 7V- , so the voltage V can be obtained as

in

 

V=*IlzL—V% (3)
01'

_ ’erL _-llzinZL

‘ 1+ zL/zi, ’ z,,+zL (4)

The total current I is then found as

1 [412... Zr. ]

I=1,+-7— —

Z... Zr. +21. (5)

01’

’12.).
- Z". +21. (6)

 

I

The power received by the structure is then

'2

Zia
2,, +ZL : R" (7)

r 1 '
PL = 312131. =‘inF:

 

where RL is the real part of ZL. For the currents on the patch surface,
Kw“). =K1+Kz (8)

where 1?,” is the total patch current in the receiving case, 1?, is the patch current in

the scattering case, and 1?; is the patch current in the transmitting case.

To determine the current in the post and the power delivered to the load, the
scattering case is solved for I 1, and the transmitting case is solved to obtain Z,,,. Other

quantities of the general receiving case can also be found using superposition of the

scattering and transmitting cases.

3. Scattering Case.
3.1 Excitation Field.

Consider first the case where the patch antenna with feed pin short-circuited to
the ground plane is illuminated by an incident plane wave. The excitation field is that
field generated by the incident plane wave in the presence of the ground plane and

dielectric coating only. The incident plane wave electric field can be written as
ENCP’)=E" Te ”I”? (1)

where -7 describes the incident polarization and I?" is the wave vector of the incident
plane wave. By applying boundary conditions at the conducting plane and at the
dielectric-cover region interface, the total electric field in the dielectric and cover
regions can be found. This is done in chapter 12, and the results are given here. They

are, first for region 2,
E’2‘(r)=§,l(r')+§,|,(r) —ds:so

where the subscripts | | and _|_ stand for polarization (of the incident plane wave) paral-
lel and perpendicular to the plane of incidence, the (x”—z) plane. The two components

of the field are

E; ”(7’ ) = 27‘” in ejkzwme'fime') [2‘ sine, cos(k 2(2 +d )cose,) - 12 ”j c059, sin(kz(z +d )cose,)]
-d s: so (3)
and
1?, 10° ) = .217i 151 e"‘1“""“°'”°°‘°"y‘"sin(k,(z+d)eose,) —dSzSO (4)
with
.1.
E‘}. = E‘ [7." + (ricostr + r§sim1>r)’]2 (5)

10

£1 = 5" [7,‘cos¢.- — mine]

and

jmcose.e""‘"°°'°'

= nzcos9,sin(k2dcose,) — jmcoseicosaczdcoseg

 

Tll

j nzcosei {Ma’w'

T1 = nzcoseisimkzdcose» - j mcos6,cos(k2d c059,)

For region 1, the fields are

EI‘W)=irl(7)+§r||(?) 220

where
E', ”(7” ) = 2 15‘” ej(-;-."+wumei) ijsin(-%-¢l,—klzcos6,-)sin6,
+ x‘”cos(-%-¢l I—klzcos 0,)0059, 220
and
5’, 10’) = 2 Bi eflé‘ifllxn‘inefli'cos(-;-¢l—klzc059,- ) 2.20
with

_ nzcose,sin(lc2dcos9,) + j nlcos9,cos(k2dcos0,)
- nzcose,sin(k2dcos+9,) - jmcosegcosaczdcosea

 

_ nzcoseisin(k2dcose,) + jnlcose,cos(k,dcose,)
i - nzcoseisin(k2dcose,) - jmcose,cos(k2dcose,)

 

In the above, the following relations and definitions hold.

*1 =Wl~|e€r
k2 = (”“1084

x” = xcoso, + ysin¢,-

(6)

(7)

(8)

(9)

(10)

(11)

(12)

(13)

(14a)
(14b)

(14C)

11

II

y = -x sincp, + ycos¢,-
f” = iCOS¢5 + ysm¢i
Y " = 4 sintr + 9008:),-

9; =tan" .3]

16:];1

ki
9.- = cos‘1 [4:]
kl

q)ll =-jl°g¢(r||) or rll = enll

4’1. = -jlog,(1"l) 0" Fl = eml

lie
'01 -- £1
u,
7'12 - £2

_ -1k1 . '
9, "305 (—srn9,)
[‘2

(14d)
(14e)

(140

(14g)

(14h)

(14i)

(14j)

(14k)

(141)

(14m)

(14n)

3.2 GREEN FUNCTIONS FOR SCATTERED FIELD.

The electric field supported by the induced surface currents on the patch antenna
and feed pin is determined using a Fourier transform Green function approach. The
electric field is written as an integral of the product of a dyadic Green function and the
induced currents. The field 173(7) in region a due to a surface current I?(r’ ’) in region b
is

Etrr=IL r‘ttrlr') - rem ds' (1)

where 3’"? | r" ') is the dyadic Green function for electric field in region a due to
currents in region b and where s is the surface in region b where the induced currents

1?? ’) flow. Here, a and b can be either 1 or 2 representing regions 1 and 2.

The tangential electric field over the patch generated by currents induced on the
patch is needed. The Green functions relating horizontal components of electric field
on the dielectric-cover interface to the same components of current have been used by
previous researchers, [2], [11], in various forms. They have been derived in chapter

13, and are given in section 13.11 as

géfi = T2110—2 ” 8030?) e’r'o’”) a=x.y B=x.y (2)
with

1 (1:128, — kx2)P1 +(k12 — k.’)pztanh(pzd)

8:! U7)" jcoe, Pfipzcomtpzdfl [epMszh‘P’d’]

(3)

—. _’ 1 “’58 [P1+P2 tanNPzdil . 4
819(1‘ )5 8,2:(k)5j(m1 [P1+P200t11(Pzd)] [firPr +P2mnh(P2d;‘| ( )

 

r (krzey - 190p: + (k12 - kfmtanhoazd)
jute] {Pr + pzcoth(p2d)] [arm + szmpqu

 

g,,(i?)- (5)

12

13

In the above,
F = m + 155" (6a)
421: = dk, dk, (6b)
p? = 1.3%,er (6c)
p3 = k3+ z-kzz (6“)
k1 = mm}? (66)
k2 = (um (60
e, = 2- (6g)
and
r, = sinh(p2d)+%:-cosh(pzd) (7)
7),, = e,cosh(p2d HZ—jsinhwzd) (8)

The tangential electric field along the length of the feed pin generated by the
currents induced on the patches, as well as the tangential fields over the patch surface
generated by the feed pin current are also needed. The Green functions relating the
vertical component of electric field in the dielectric region to horizontal currents on the
dielectric-cover interface, and relating the horizontal component of electric field on the

dielectric-cover interface to vertical currents in the dielectric region are, from section

 

13.11,
gfie‘ = (2;), I1 3.30? refit-7’ COSh(P2(Z+d)) B=x.y (9)
8°]: = 72%); H 8M?) e’r' ("fl cosh(pz(z’+d)) a = w (10)

with

14

 

 

g... (P) . J—alm Fig-:- (11)
3., (1?); 1.7:;- e"’"'%- (12)
an (7:) .. 1.0;! {Wig-f (13)
8,.(1?)- 1.0161 6'"? (14)

Finally, the tangential electric field along the feed pin due to the current flowing
in the feed pin is needed. The Green function dyad for vertical electric field in region

2 due to vertical current in region 2 is, from section 13.11,

1 17'0””)

2.2 —n____ e x

 

k22+P22 COSh(P2(Z<+d))(P2¢05h(P22>)-€rPrSinh(P22>))

- 8(2-2')+
p2pl Tm

dzk —dSz,z’S0 (15)

where 8(x) is the Dirac delta function, and
z> = max(z,z') (16a)
z< = min(z,z') (16b)
The divergent nature of the above Green function integrals is overcome by the
additional spatial integrations introduced by Galerkin’s method. When these integra-
tions are performed prior to spectral integration, the resulting spectral integrals are

convergent. Equivalent results can be obtained using potential Green functions, which

give convergent spectral integrals prior to application of Galerkin’s method [12].

3.3 DERIVATION OF COUPLED INTEGRAL EQUATIONS.

The induced current on the patch and feed pin surfaces must satisfy a system of
two coupled integral equations constructed by employing the boundary conditions that
the tangential electric field over the patch and feed pin surfaces must be zero. The
scattered electric field on the patch surface results from current on both the patch sur-

face s and feed pin surface p, and is given by

il’(x,y,z=0) = H ?1'1(x,y,z=0 l x’,y',z’=0) ' I?(x',y')d.x'dy’

+ ”E 1'2(Jc,y,z=0 I rp=a ,¢,,,ZP )‘E. (2,. )adzpd¢p (1)

P
where the feed pin current 1?, is assumed independent of op. The integration variables
4),, and z, are from the feed pin centered coordinate system shown in Figure 47, and
described in section 14.2. The scattered field in region 2 due to current on the patch

and feed pin is

52" (x,y,z) = H 311(x,y,z | x’,y ',z'=0)-l?(x’,y ’)dx 'dy’
+ ”3’ 2'20: ,y ,z | r, :0 ,4), .2, )‘E. (2,. )ade d ¢p (2)
’P
The boundary condition on the tangential electric field at the patch surface yields
one integral equation
f X [ir'(x.y,2=0)+ir‘(XJ.2=0)]=0 (3)

where E' 1‘ is the total excitation field in the cover region due to the incident plane
wave. This equation must hold for all points on the patch surface. The boundary con-
dition on the tangential electric field at the feed pin surface yields the second integral
equation. This boundary condition is enforced along the surface of the feed pin, giv-

ing the second integral equation,

15

16
i ' [iz’CfM‘I’Ez‘Vo'IM] =0 —dSZSO OS¢P $21: (4)

where E',‘ is the total excitation field in region 2 due to the incident plane wave, and
where 3(4),) is a vector from the center of the feed pin-patch junction to the pin surface

at the junction.

3.4 General Matrix Formulation.

The induced current on the patch and feed pin surfaces is to be expressed as a
summation over a set of basis function current distributions. The solution for the un-
known amplitudes of the chosen current distributions is to be formulated as a matrix
equation. This will be done using Galerkin’s method--the boundary conditions on the
surface of the patches and feed pins will be forced to hold in an integral sense where
the same set of current distributions will be used as testing functions. The boundary
condition used is the physical constraint that the electric field tangential to the surface

of the patches and feed pins be zero.

Let i ‘ be the electric field due to the incident plane wave and let 5’ ’ be the scat-
tered electric field due to induced current on the conducting surface of the patch and

feed pin. Then

EJ=JI§"I?dS' (1)

Here, 5' is the conducting surface of the antenna structure, I? is the surface current on s

and r is the dyadic Green function. The boundary condition on s requires that
{-(§‘+§‘)=0 (2)
for all arbitrary unit vectors 1“ tangential to the surface 3.

Now let

fi=iC,-I?,- (3)

i=1
where {3,} is an appropriate set of independant and complete current distributions, and

where C, is the corresponding amplitude of 1?} in 1?. Then

mrr- [fzcmjm gut-k:- .1. <4)

81

17

18
By choosing {I?,-} appropriately, the summation can be truncated after a finite

number of terms and the current obtained will approximate 1?. For a finite summation,
it is no longer possible to require 1‘ ~ (5" + E") = 0 at every point on 5'. Instead,
Galerkin’s method is used. For i =1, 2, 3... . ,N, where N is the number of current

distributions used, require

”grim—Harm. (5)

Substituting the expression for E” using the truncated summation,

8

fidslfi-[gcjjjy-fr’jds']=-jj¢slfi-E" i=1,2,3,...,N (6)
01':
jiggle]:.jqums'bjjnz-ri r=r,2,3,...,~ (7)
Making the definitions
Z.,-=]]dsE-]]ds't’-f<3 (8)
vermin-r" (9)
equation (4) becomes
igzyev, r=r,2,3,...,~ (10)

1'1

This can be written in matrix form as

211 Zr: ' " ZIN Cr V1
221 222 ' ' ' zzN C2 V2
: : t ' t = I (11)
ZN! ZN2 . - ' ZNN CN VN

This is the general form of the equation used to obtain the current amplitudes

C1, C2, ..., CN and hence the approximate solution for 1? via (3).

3.5 Current Distributions.

The surface current on the patch is represented using a set of current distributions
based on a two dimensional taylor series expansion expressed in terms of Tchebychef
polynomials radially and complex exponentials azimuthally, along with a singular
current distribution used to model the curent flowing from the feed pin junction onto
the patch surface. A single basis function is used to model the current on the feed pin.
On the patch surface,

M

2:2; f[c,,,,1?,,,+c,,,,l?,,,,]+c,l?, (1)

III-0 lt-L

while on the feed pin,
1? = C: K: (2)

The Tchebychef current distributions are either purely radial or purely azimuthal.
The radial variation of the current distributions are modified by an appropriate factor to
account for the known edge behavior of the two components of current. These current

distributions are, from 14.1(47) and 14.1(48),

1?,“ = 1" K”, = r“ Tm(r/b) e1“ (’1 - i; 1+». odd (3)

KM»! =6Kflm =$Tm(r/b) 8’“ ——1— I+m Odd (4)

1‘ 72
The condition that H»: be odd is implicated by enforcing the patch current, its diver-
gence, and all further derivatives to be continuous at the center of the patch.
The current on the feed pin is modeled as flowing in the i direction uniformly

over the surface of the feed pin.

22:15 (5)

l9

20
The singular patch current distribution, 1?,, is designed to model the current

flowing from the feed pin junction onto the surface of the patch. Its amplitude is
forced to exactly account for the current flowing along the feed pin to the junction, so
that no charge build-up at the patch-feed pin junction occurs. This current distribution

is chosen to flow radially away from the junction. Thus,
1?. = 1*. 1'0.) (6)

where f, is a unit vector directed radially away from the junction and r, is the radial
distance from the center of the junction to a point on the patch surface. From

14.2(12), f (r,) is constructed as
r r 3 f 5
f(r,)=A-r‘-'-+B-ai+c [721]”) [f] aSrpSR (7)
P

where A , B, C and D are constants determined in section 14.2, and where R = b - to is
the minimum distance from the center of the feed pin junction to the edge of the

patch. Outside the region from a to R, f(r,,) is zero.

3.6 Specific Matrix Formulation.

The current distributions used to model the patch current are the Tchebychef dis-
tributions 1?... and 1?“. from equations 3.5(3) and 3.5(4), along with 1?, from 3.5(6).
The feed pin current is modeled by 1?, from 3.5(5). The current amplitudes C, become
the corresponding current amplitudes for the Tchebychef current distributions along
with a common amplitude for 1?, and 1?,. The approximate equation for 1? then

becomes

1?=§ i [Cm1?,,,,+C,,,,I?,,,,]+C,[1?,+1?,] (1)

Mao la-L

where 1?, flows over the feed pin surface, and all other current distributions flow on the

patch surface.

The matrix equation 3.4(11) becomes

I I
2:39"! ' Z 8'75"! ' er'm Cr '1 'm ' vrlm
I ..
2,955?" ’ Z (Win ’ 229 m ' C (V! 'm ’ - Velm (2)
Zrz’l ’m ' Z $7 ’0: ' Zzz' . CZ ' Vz

Letting 7 represent either r or o and letting 8’ represent either r’ or o’, (2) can be

written more compactly as

Zggln'lm' 221m [Cd'l’m'] [V'fim ]

 

3

Zglllml ZZZ! CZ, V2 ( )
The matrix elements 23511,... are

ZWM = “‘13 Kw» 'jjd-Y' Y” 'Eb'l'm’ (4)

where s is the surface of the patches.

The surface current distributions, which are purely radially or azirnuthally

21

22

directed, must be separated into i and y“ components, since §’ is represented using rec-

tangular components. Written component-wise, r is

8.3.1 ngl g...’2

r=gyi1 8);] 8,2: (5)

2,1 2,1
82x gzy 8222

 

where the superscripts on the components are those relevent to the case being con-

sidered. The current distributions are decomposed as
1?,“ =(1?,,,, ‘12)x‘ +(I?,,,, -y‘))‘3
= Kwi + mey)‘:

= Z Kyla-rad (6)

any

and similarly,

I?5lm = 2 KN'M'BB (7)
the

Using (6) and (7), and carrying out the scalar products, (4) becomes

=2 iid‘Kw—ai’ 4"8 ubKa'mr (8)
:a

In (8), the two superscripted "1"s on the Green functions indicate that the field and

source integrals respectively are canied out in region 1, above the dielectric interface.

The matrix elements 23"" are
Zf"=fldsl?yh° [Idsp’g’l-z-K+Hds’g”-1-l?, (9)

where s, is the feed pin surface. The feed pin associated patch current distribution, 1?,

is separated into rectangular components as

1?. = 2 Kart (10)
Ba:

23

The testing function 1?, has only a 2‘ component, and is

1?,=z‘

so (9) becomes

23"": 2 Hds Km,- ”as/g3“ )3 ”arming.
any I ‘P fl’a’y' .r

(11)

(12)

In the above equations, Inn. is the component in the B: :2 or y direction of the

singular current distribution associated with the feed pin. The elements Z31”... are

zi’l’m’ = II dsp z: ' H d3, 2’2'1 ' fib’l'm'

Equation (13) becomes

Zg’l'm' = E II dsp 1 II d5, gzzfll Ko'l’m‘fl
Bay 3' .r

The matrix element 2:. is

Z:'=jjdsp 11' [jjdsp'fi’m'lZi-Hds’gfl'l'la]

This can be written as

2;: n.1,... [Has/3,33.» z ”awn,
’P 3? B'=x’.y’ :

From 3.4(9), the elements V7,, are

VTIM=—,UdYI?TIM.El

V,=-jds,z‘-E’

‘r

 

(13)

(14)

(15)

(16)

(17)

(13)

The patch testing functions I?” in (17) and (18) don’t need to be separated into i

and 9 components. The impressed electric fields 1? 1 and I? 2 are the fields in regions 1

and 2 respectively generated by the incident plane wave, E". This field is separated

24
into two components, 1? 1i and i ‘, parallel and perpendicular to the plane of incidence,

as shown in chapter 14. The fields 1?‘ and E 2 are the also separated into two com-

ponents, and (17) and (18) become
Vm=-_Udsl?7h' II-‘UdSE‘flm'i‘Ll (19)

V,=-]jds,i~i.f-”ds,i-E'f (20)

By setting first one then the other of I? 1" and 172' 1'1 to be non-zero, the two cases of
incident polarization may be solved separately. They may then be appropriately com-

bined to give the solution for any arbitrary polarization.

The matrix elements can be manipulated into a less complicated form. Substitut-

ing 3.2(2) through 3.2(5) into (8), the matrix element becomes

2317:”. = z Hds K1,,“ HdS'IZ—rloi H d2}. gw(F)eIF‘V-”7K,.,.M.B (21)
was.) 8 3 ..
Baa

Rearranging the order of integration, this becomes

 

23%., = (2;), ”all: 2 {[Hds Km, e1?" J [”4" K5,.” e‘Ir'w ] “3023} (22)
-c- My 8 3
B-vw

Define
jjds K,,,., e117" 5 1:,“ (i? ) (23)
”d: Km, etiF'?’ a 15,, (I?) (24)
Hds K...“ at)?" a 1:5,, (I? ) (25)
Has Km, eii‘"? 13..., (I? ) (26)

Using (23) through (26), (22) becomes

25

Zg'tpm' =

 

I] 421. 2 It... (Prime (P)8ap(i;) (27)
~00 M:
h.)

(21:)2

Substituting 3.2(10) and 3.2(2) through 3.2(5) into (12),

 

2.21»: z ”4.x,” ”1.; 1, 1].». gun?).iW-rvcosn(p,(.'.d»
any 8 1' (21:) -

 

l '0 2 I ‘F- -?
d 1: ds I? I 0‘ 7
+31) (2.92 L]. I] gain )e K... (28)
In (28), both integrations over dzk, and the exponential factor contained within
them, are identical. They can thus be combined into a single integration, and (28) can

be rearranged as

=1t(2 )2 U. ”’2" guru... Mir 7]

 

1...

This leaves several integrals to be evaluated. The first, involving Km, is defined

021:
x Ilm[g,,(l?) I! ad¢’,dz’e'f"""'cosh(p,(z'+d))+ Z g,,,(/Z’)H ds'e’jp7'K,p
-d b=x.y :

in (23) through (26). The last, involving Kw is defined in a similar manner as

H ds et’r'r K45, 5 If} B = x . y (30)

The remaining integral can be evaluated immediately. For 7 ’ on the surface of the
feed pin,

r"=r;+a‘+z£ (31)
where z! is a vector from r; to the feed pin surface where it meets the patch. Since 7?

has no 5 component, the 5 component of P ’ doesn’t contribute. Thus,

0 2t 2:

[1 ade', dz’ e*i?"'cosh(P2(z'+d))= et’pfi j dz cosh(p,(z ’+d)) (j d¢', a at". " (32)
.4

26

The integration over 2 is found as

idz’cosh(p2(z’+d)) = M (33)
and the integration over (1:, is obtained using

1‘49} a exit-r = a Tdtb', a?“ com” ‘8) = 21m Jo(lca) (34)
Thus

1? ds’ fir? cosh(P2(2’+d)) = et’r'n in}? 21m Jo(ka) (35)

Using (23) through (26), (30) and (35), (29) becomes

' h _, _. .
L—mgzd) 21m Jo(ka)+ Z g,,,(k )I,-p e ’m (36)

2,1“: :dzkom l a mic

Substituting 3.2(9) into (14),

Zis'r'm' = 2 H (Li, dz 1” ds'Kmimnb x
Bay ‘P .r

 

x (2;), H dzk 13,50?) e-IT‘C” ”’7 cosh(P2(z+d)) (37)

Equation (37) can then be rearranged to become

0 21:

1 ~ 2 ,1"? ’11-?
Z" w = d 1: ° ad d cosh P +d ’
51 (211:): Ii e [ LI ¢p Z ( 2(z )) e

 

x 2 2.105311 We... e-I‘F'?’ (38)
he 8

The integral over the pin surface is given in (35), and the integral over the patch
surface is given in (23) through (26). Thus, (38) becomes

2 3.50? ) Ian... (I?) eW0 (39)

'..,,.= d2}: 2w lea
Zsr )——,- I] o( )———- m

th( ____2_d_)
(21: P2

27
Using 3.2(15) and 3.2(9), (16) becomes

 

2;. =H d5” {,2} ,Hd” (21102 H “'2" 8.5(F)e’r"'"° cosh(p2(2+d)) Km
3' say 3 "’

 

11d: (,—;,.11d=k [5,]e W

 

 

 

 

 

2 2 < > _ -
[—8(z—z’) + 1‘: +P2 005h(Pz(2 +d))(P2¢OSh(P21 ) {Prsmh(1’2z>))]} (40)
Pzpr T...
Rearranging, this becomes
=(2n1—17 U. H {gfwa‘ )x
21:0
x {I add), dz ejp‘rcosh(p2(z+d)) Hds’ e'jr'I'Kw-r
.4 3
+ 1 2“4111(1) eli’"he‘1"'a'ad¢' x
jwfiz g P g P
0 0 2 2 < > __ - >
x I dz dz, [4(er k2 +112 COSh(P2(Z +d))(pzcosh(pzz ) Fprsmhwzz D] (41)
..., p21): T...
Using (35), (30) and (34), (41) becomes
. - __1_ " 2e11‘7’o 21w (2m)2 1.11m
2.1— (21:12 [1 d k [eZ-—$inh(md)pz8.1110(lca)1.11]jm2 I. (42)
where
=} ° .1. else—.3 + k2” +p.’ cosh(p.(z‘+d»(p.oosh(p.z>) _ep,sinn(p,.>» (43)
4.4 P2P1 Tm

When (43) is evaluated in chapter 16, 1, is found to be (16(65))

k22 Bk”
1 = d — - —— sinh(p2d) (44)
' pz’ pz’ T... P2

28
Thus, (42) becomes

1??

WW

J’__(__ka) 1‘22 1»: k2
( M)_ jme pi pinp.”(“fl ( )

The matrix elements for the incident field, (19) and (20) are also evaluated in

chapter 16. For parallel incidence, these are found to be

Vrlm 11 = 5111 e170.- 1‘ 1"“ [31115051 sin(0,-)) + Bl:l,m(k15in(ei))] (46)
Vern. 11 = E111 e170.- 7‘ .I'l [4111,1051 sin(9,-)) 4' A1--1;n(k151“(9i)):| (47)
V = --4M TII jkzixin(e,)+deoe(o,) Sin(9r) (48)

k2 cos(0,)

where 12‘}. and E‘}. are defined in chapter 12, and where 11,3, and 8,3,, are obtained in

section 16 as
“b2 kb kb kb [Cb
min =-.- [JL 2'" [Tilt-Fifi] ’Wi‘fi’mrii7ij

(i1)' 12 o 1 4
O (- il)’ (<0 J ( 9)

I I
31:5. = 7 [211151. - Aria-21 ‘ Art“: ] = '5' [2 A13. - Aria-2 " Art-1+2 ] (50)

and are evaluated there. For the case of perpendicular incidence, the source matrix

elements are
Vrbn l = ’ Ell em" " J" [3111301 sin(0,-)) + 31:1,m(k18in(91)] (51)
V91... 1 = - Ej e170.- 1! I'M P111531 Sifl(91)) — A1:rn(k15ifl(91))] (52)

v, = o (53)

where E 1‘ is defined in chapter 12

29

4. Transmission Case.

The transmission case is arrived at in a manner analogous to that of the scattering
case. The antenna is driven by a voltage at the base of the feed pin which represents
the voltage drop across the load generated by current flowing on the feed pin when the
antenna is acting as a receiver. To find the currents due to this applied voltage, a slice
generator with one volt is assumed to be connected at the base of the feed pin at

z = -d. This gives a tangential electric field along the feed pin
E" ' = z“ 8(z-td) (1)

Since there is no plane wave excitation in this case, the feed pin voltage is the only
excitation present. The integral equations to be satisfied are identical to those for the
scattering case, with zero excitation field on the patch, and 3,12) on the post replaced
with I? '. Thus, from 3.3(3) and 33(4)

2‘ x El‘(x,y,z=0): = 0 (2)

and

i' [€2'W0+fi’,z)+f'] =0 -d 52 SO (3)

Through the application of Galerkin’s method as in the scattering case, these

equations are replaced by the set

U 1?...a(r.¢) - §1’(r.¢.z=0)rdrd¢ = o (4)

l = -L,....,—1,0,1,....,L

m = 0,1,....,M
a = r,¢
and
02x 02!:
Li 5 -§2‘(xo+acos¢,yo+asin¢,z)dz ado, ”it! i -§‘dz amp, = —21I:a v (5)

3O

31

Finally, this is written in matrix form in the same manner as for the scattering
case. In particular, the left hand side of 3.6(2) is unchanged and the right hand side is
replaced with either 0 or -21ta as appropriate, so

zrlm M. . 23%. 2;!“ c M, o
231'»: ' Zwmm W] [CE'I'ZV = 0 (5)
22%", Z¢'l’m’ 2:2 ’ -21ta

Thus, once the scattering case has been solved, the same matrix is used to solve
the transmission case, with only a simple substitution in the driving terms of the
matrix equation. The current on the feed pin is then solved for and the input
impedance obtained through 22(1). The input impedance obtained in this manner is
more accurate than can be obtained using the induced EMF method. If (6) is rear-
ranged into the form commonly used in the induced EMF method, the term which is
solved to obtain 2,, is found to give 2,, - 2,". The input impedance obtained is thus

off by the self-impedance of the feed pin.

III. INFINITE ARRAY OF PATCH ANTENNAS.
5. Problem Description.

The geometry of the infinite array of circular microstrip patch antennas to be
analyzed is depicted in figure 3. A dielectric substrate of permitivity e, , permeability
it, and thickness d, (region 2), is located between the z = 0 and z = -d planes, as in the
case for a single patch antenna. Again, the dielectric is mounted on a conducting
ground plane at z = -d, and is covered by a material with constitutive parameters 61
and u, (region 1). Each patch element has radius b and is connected to a load
impedance 2,, through a feed pin of radius a centered at a point 7:, = x,£ + yd in the
local patch coordinate system. The patches are located at the dielectric-cover interface
in the z = 0 plane, and are assumed to be perfectly conducting and infinitely thin.
They are spaced distances d, and d, apart in the x and y directions respectively, their
centers located at 7;, = p d, i + q d, y“ in the global coordinate system of the array,
with p and q integers.

Illumination of the structure is again taken to be through an incident plane wave
of frequency m at an arbitrary incidence angle. The plane wave is expressed in terms
of a coordinate system x”, y”, z” rotated with respect to the global coordinate system
of the array, as shown in Figure 2, with o,- the angle between the x and x” axes, and 9,-

the angle between the wave vector I?" and the z axis. The 2 and z” axes coincide.

The factors

:4" = sin(6.-) cos(¢;) (1)
and

v = sin(6.-) sin(¢.) (2)

32

 

 

 

 

 

 

 

 

 

81 “0 Region 1.
r F * ‘ T Ff =0
d 20 82 Ho Region 2..
Jr J __— z = —d

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Figure 3. Geometry of an infinite rectangular array of patch antennas.

34

are the same as defined for the single patch antenna.

The decomposition of the receiving case into scattering and receiving cases is

accomplished in the same manner as described in section 2.2 for the isolated patch.

6. Scattering and Transmission Cases.

For the case of an array of patch antennas, the excitation field is the same as that

found for the isolated patch, for both the scattering and transmitting cases.
6.1 Green’s Functions for Infinite Array.

The various Green function components for an isolated current element are
obtained in chapter 13. The corresponding Green functions for an infinite array are
obtained in section 13.12, expressed as a summation of Flouquet modes. For the
infinite array, the Green functions for horizontal electric field at the dielectric interface

due to horizontal currents at the interface are

géi'(7’|7‘")= z 8%,(F)e,p..._., a=x.y B’=x’.y’ (1)
P4“
where
_. (k’e. — k3) + (k2 — k3) tanh )
g;( ) 1 1 P1 ,1 P2 (P24 (2)

 

a jmcldfl, [[21+p2cotl1(p2d)] [8,131 +pztanh(Pzd)]

“kxky [P1+P2 tanh(p2d)]
85(k)'=‘g,2(k)s. * * ' (3)

 

 

,, _. 1 (k?e.—k,’)p.+(k? —k,’)p2tanh<pzd)
g,,( )3 .m d (4)
1 1 *4 [P1+ch°th(Pzd)] [8.p1+pztanh(pzd)]
In the above,
—r . 21! 21: ..
k=k,£+k,y=(k1u+p-d—)£+(k1v+q-d7)y (5)

where u and v are given in equations 5(1) and 5(2). All other definitions needed are

given in section 3.2.
The Green functions relating the vertical component of electric field in the dielec—

tric to the horizontal currents on the dielectric-cover interface, and relating horizontal

35

36

component of electric field on the dielectric-cover interface to vertical currents in the

dielectric for the infinite structure are, from section 13.12,

 

33.34? I r) 1;- g&( i?) JV"M cosh<p2(z’+d» any (6)
where

g; m .. 1.“de d, “’1’ 3f: (7)

g". (1?); m3“, etc}? (8)
and

8:23"? I F”)= i 8254 17) e’r'y’” cosh(p2(z+d)) B’=x’.y’ (9)

P4”

where

g; (I?) a W #475: (10)

3541?): W ““17? (11)

Finally, the tangential electric field along the feed pin due to the current flowing
in the feed pin is needed. The Green function component for vertical electric field in
region 2 due to vertical current in region 2 for an infinite array is obtained in section

13.12 and is found to be,

 

2.2 7-! 7H 1 o- -- e-jP-(‘V-T")
Su( I )—dxd,,§.,§. jars; -5(z-z')

 

k2 2 h < d
2 +P2 Jcos 0’29 + )) [flcosh(p22>)-e,sinh(pzz’) dzk (12)
P2 Tm pl

J
l

where am is the Dirac delta function, and

37
2’: max(z,z') (l3)

2‘: min(z,z') (14)

The spatial integrations introduced by Galerkin’s method are performed prior to

spectral summation.

6.2 Current Distributions.

For the infinite array, the Tchebychef current distributions used in the case of the

single patch antenna are again used.

ii... = f Km = f T...(r/b) a!“ sit - ~32- l+m odd (1)

1?“, = 6 K“, = 6 T,(r/b) e1“ ———1— 1+»: odd (2)

1‘ 7.7
The current on the feed pin is also the same.

1?. =12“ (3)
A singular current distribution is again used to model the current flowing from

the feed pin onto the surface of the patch, but a different distribution is used here. For

the infinite array, from a development in section 15.3,

 

a f cod-37L)
1?.(r’ ) = ' ‘ (4)
P r M
" ““2?

where R, = R,(¢,,) is the distance from the center of the feed pin-patch junction to the
edge of the patch. This distribution has the disadvantage of requiring numerical
integration in one of the Galerkin’s method integrations. It has the advantage of
flowing over the entire patch surface. For a feed pin located very near the edge of the
patch, this distribution would seem to be a more accurate model of the current no the
patch than the other singular current, and fewer of the Tchebychef current distributions

would be needed to accurately model the patch current.

38

6.3 Specific Matrix Formulation.

For the case of the infinite array of patch antennas, the current distributions used
to model the patch and feed pin current are I?“ and 1?“... from equations 7.2( 1) and
7.2(2), along with 1?, from 7.2(4) and K", from 7.2(3). Only 1?, differs from the current
distributions used for the single patch antenna. The current amplitudes C, are again

the current amplitudes for the current distributions. The equation for I? is

mi 5': [c..z?...+c... r...]+c.[r.+z] <1)

nu=0 III-L

where, as for the single patch antenna, 1?, flows over the feed pin surface and all other

current distributions flow over the patch surface.

The matrix equation becomes

1 I
2:353" ' 26,72" ' 2;}?! Cr '1 'm ’ Vrlm
I _
219;”; ' Z $3172; ’ ZzQ m . C (W ’m ’ " Volm (2)
Zrzlllml 23:11,": 2221 C2, V2

Letting 7 represent either r or o and letting 8’ represent either r’ or o’, (2) can be

written more compactly as
23% 29"" [68...] [mm ] 3
Zg'l'm’ 222' C2’ — V2 ( )

The matrix elements 2335'... are

 

men“... ttw” raw (4)

where s is the surface of the patches.

The surface current distributions, which are purely radially or azimuthally
directed, must be separated into i and y“ components, since §’ is represented using rec-

tangular components. Written component-wise, E is

39

40

8.3." 31;" a»?
?= 8,3." 8,;‘1 8,13 (5)
33:1 8.3" 832

where the superscripts on the components are those relevent to the case being con-
sidered. The two superscripts on the Green function dyadic components indicate
whether the field and source integrals respectively are carried out in region 1, above
the dielectric interface, or in region 2, within the dielectric. The current distributions

are decomposed as

Km=(K,,,, ~i)x‘+(K.,,,,. -y‘)y

= 2 Kytnra a (6)
m0
and similarly,
f5’l'm' = 2 K5'I'M’B’B’ (7)
B’u’a'

Using (6) and (7), and carrying out the scalar products, (4) becomes

233%.: 2‘. Hemmfids'gts Ks'z'm'p' (8)
any 3 3

pin I”!
The matrix elements 2,?" are

w=jjar,.,- n dz'w¢',g12-z+;;d.'gx-l-zl (9)

where s, is the feed pin surface. The feed pin associated patch current distribution, K,

is separated into rectangular components as

fé= z 1935' (10)

plaid!
The testing function K, has only a 5 component, and is

R’,=rz* (11)

41
so (9) becomes

I I l I
wig” desxm- I‘tdz admgtfi 3533.0, {Ids slim (12)

In the above equations, Kw is the component in the B’ = 2’ or y' direction of the

singular current distribution associated with the feed pin. The elements 237.... are
0
22... =2na I dz 1?. - H dw'r” is... (13)
-d 3

Testing is done along a line on the surface of the feed pin, hence the single testing
integral. The factor 21m in (13) is solely for dimensional consistancy with the rest of

the formulation. Equation (13) becomes
0
2t... = 2m 2 I dz H ds' g3; Ka'z'm'p' (14)
B’u'a' -d 8

The matrix element Z} is
0

z:..-_2najdzl?,-[jjdz'drpg’m-Ifi+fids'y2-‘41] (15)
-d r, r

This can be written as

 

0

2;. =21ta for: [j] dz’addr’, g3? + Z Hds'gzzfll 1gp. (16)
-d a, B'Iu’a’ :

From 3.4(9), the elements V”, are

VW=-HdsKW-il (17)

and the element V,

0
,=-2...1m.52 (18)
.4

42
The patch testing functions K1,... here don’t need to be separated into it and y“

components. The impressed electric fields K ‘ and E" 2 are the fields in regions 1 and 2
respectively generated by the incident plane wave, [5" . This field is separated into two
components, KI", and E" , parallel and perpendicular to the plane of incidence, as
shown in chapter 13. The fields I? 1 and 52 are the also separated into two com-

ponents, and (17) and (18) become
VW=-HdsKm-illl—HdsKW-§ll (19)

0 0
v,=_2najdzi-I?.2.—2mjdzi-§f (20)
«r -4

The matrix elements can be manipulated into a less complicated form Substitut-

ing 7.1(1) through 7.1(4) into (8), the matrix element becomes

23%,: 2: IIdsK....J ds' i gatheiFV-“Katr (21)
my : :

Pg“
15“.)

Rearranging the order of integration and summation, this becomes

2317;. = i Z {[Hds K7,,“ e17"] [JIJdS’Ka'm'B e-IT’J“ ] g;p(l?)} (22)
In“ Wd 3 3
Define
”d: Km etfi’r a 1,3,, (I?) (23)
lids Km. e*"'" .. It... (I? > (24)
”d: K“, 61"" . 13,“, (I?) (25)
”d: Km, an?" a 13,, (I?) (26)

8

Us ing (23) through (26), (22) becomes

 

43

Zita.“= i 2‘. 1;... (£315.... (Fund?) (27)

P4” W0

ha

Substituting 7.1(12) and 7.1(13) into (14),

o -
23”,. =21ta z j dz 1 j] ds'K&:,,.p 2 5,563) eff-7’7 cosh(p2(z+d)) (28)
the -d 3

P4”

In the above equation, 7 is constant since the z integral takes place along a single
line on the feed pin at r= r; + 21;, where r; is the position vector of the center of the
feed pin junction at the patch, and a; is the vector in the plane of the patches from the
feed pin center to the point on the circumference where the electric field along the feed
pin is forced to zero. Equation (3.7.28) can then be rearranged to become

Zip,“ =21ta 2 e

If" (7. +EL) [
Po!"

0
I dz cosh(p2(z+d)) ] Z g,'5(ic") H 4:1,,” if?” (29)
-d the I

The integral over 2 is evaluated immediately as,

0
j dz cosh(p2(z+d)) = 4- (sinh(p2d) — sinh(0)) = 1- sinh(P2d) (30)
.4 P2 P2

and the second integral, involving [(4, is defined the same as for the isolated patch, the

only difference being the current distribution K,.

Hds’efip'wasIfi i5=x,y (31)

Using (30), (31) and (23) through (26), (29) becomes

- . 'nh
2:... = 2m 2 .fi’ "- M 2: grams... a?) (32)
“7"“ p2 fl-w

Substituting 7.1(1) through 7.1(4) and 7.1(9) and 7.1(10) into (12),

2'1”": 2 HdZ'ad¢'pK1|tt-ru' H ds’ f; g;(E’)efi"('-'°cosh(p2(z'+d))+
my 3 8, Pl"

+ 2‘. IIds’ )3 sat?)e""'<"7°x.p (33)

he I par--

 

44
In (33), both summations over p and q , and the exponential factor contained
within them, are identical. They can thus be combined into a single summation, and
(33) can be rearranged as

7"“ i ’3 “tame"?

P4“ 0‘0

7r COSMP 2(3 I'Hi ))
21m

X

 

“mundane? +BZ 2.5%]st effing, )
a, ’n’J' .r

This leaves three integrals to be evaluated. The first, involving Ky,“ is defined in
(23) through (26). The last has been defined in equation (31) and the remaining
integral can be evaluated immediately. For 7' ' on the surface of the feed pin,

?’=F§,+?+zz‘ (35)

where a? is a vector from 7; to the surface of the feed pin at the patch-feed pin junc-

tion. The magnitude of 3 is simply a. Thus,

a:
U dsIe -jI’- F" cosh(p2(z ’+d))- — e ‘1 dz ’cosh(p2(z '+d)) 1! (up a e-j‘i’ at
’P
_ . . ' h d
= 2M e I? 7’ o (3%)100‘0) (36)
2

where the integral over 2’ is the same as done in (7.3.30), and the integral over 4),, has
been evaluated to Jo, the ordinary Bessel function of zero order, using

2:

2!
i av?" cur, = i i’“ ”'"r ’°’ dc), = 2n: J,(ka) (37)

Using (23) through (26), (31) and (36), (34) becomes

Sin 10’2” )ZM Jo(ka)+ Z swab} (38)

he

Zlh= z 2 ’10-“: [Sara )(——

PR" “‘3"

Using 7.1(12), 7.1(13) and 7.1(14), (16) becomes

45

0 ..
Z,‘=21ta ta: {2; Mr 2 g.r'(i’)e""""°coshtp.(z+d»K.a'
-d ,y 1

Pa“

a: o ,,
+£ Idz’ado’, z [—-— 1 Je’P‘V‘r’x

 

—d pan—o. szdldy
2 2 < > _ - >
x PM.)+ hp 22?: 008h(P2(z +d))(P2005;(P21) Eprsmhtpzz n” (39)
1 In

Rearranging, this becomes

u . ' o ’
ZI=2M X { Z 8.3(?)e’r fichomtpzmd»dzflds'e'fi" Kn:+
pa“ 5“.) -d a

0 0
-'?-? I
+a__—ju)62d,d, {e’ d¢piidz dz x
2 2 < > _ - >
x [4(2-2’) + k2p+P2 COSh(P2(Z +d))(P2€OS;(P22 ) {PrsmMPzZ )) ]} (40)

2P1
Using (30), (31) and (37), (40) becomes

11'"?

. h d -D . ' - ' 0
2 sm (:2. ) g..(k )8]? a; ,1" 7‘3 1,7,] + [f—mmn] (41)

Z,‘=21ta Z

Pfl“ [My p; :B “2‘13“!

where

1:22 + p22 cosh(P2(2‘+d))(P2008h(Pzz>) - @rsinhwzzm ] (42)

o o
I, = dz dz' —8(z—z +
L... [ ') M): T...

Equation (42) has been evaluated in chapter 17 to be

k22 rt”
1, = d — - smh d 43

pr .31.. r. "’2 ’ ‘ ’

Thus, (41) becomes
to ' h d _’ .—9. . .
z:=2na >3 {EM-32.3% >e" “char
PAH Bay p;
j”; :22 n2 (44)

k
e .
+—"'_J k0 d—-——srnh(pd
imzd’d, 0( )[ p 2 T 2 )

46

The matrix elements for the incident field, (19) and (20) are the same as given in
section 3.6 for the case of the single patch. The difference in the singular current dis-

tribution used in each case doesn’t affect the source elements. Copying those here,

Vrbn n = sin cf“.- “I'M [At-$15051 Sin(9t)) + Al:lm(k15in(ei))] (45)
= -1 if... .1 - k . . .. k . . 46
Vol». 1| E” e 1‘] l+l,rn( r 5111(9.» +Al—1n( rsm(9r)) ( )

agrarian + den-(0,) Sin(9r)

V, =47“! TH EH 6 k2 COS(9,) (47)
where Affi, and 8,3,, are obtained in section 17 as

nbz ch kb kb kb

Aim = T [’— ['2‘] L 2"” H1137...” [7] ’—L 2"“ [‘2'] 1
(il)' 12 o
' {(— :tl)’ l<0 } (48)

1 1

31%.. = '4 [2 Alta — Alia-2| " Atfmz ] = '3' [2 Alf». “ Alta-2 - Altn+2 ] (49)

and are evaluated there. For the case of perpendicular incidence, the source matrix

elements are
Vrinr 1 = - Ell e179.- 7‘ J" [3111...(151 sin(6,-)) 4* 31:1.m(k15in(9r)] (50)
V... l = — if e’"" n 1'“ [4111,1051 sin(0.- » — A.:1..(k.sin(e.-»] (51)

v, = o (52)

6.4 Convergence of Feed pin-Feed pin Matrix Element.

Equation 6.3(44) is rewritten here.

~ sinh d _, . .
Z:=2m 2‘. 2_‘P2_)&-5(,,,,W%,Wn .r
IJH the p;
1P? k2 2
e " 2 El: .
+ -.——-—J,(lca) d — — ——-— srnh(p d)
Imeldxdv [ [’22 P22 Tm P2 2

(1)

Looking at the second term in equation (1) as k —) ..., it would seem that the summa-

tion wouldn’t converge. This is seen most easily by considering that

1 ..
d. d, ”2....

 

crudely approximates an integration of the form

~21

jjd2k=dekde

Performing the integral over d6, the exponential e’r‘a’ becomes Jo(ka). The integrand

is then asymptotically proportional to 130cc), and the integral diverges. The solution to

this difficulty lies in the term 1,}, which implicitly is of the form close enough to Jo(ka)

capable of canceling this divergent term. To see this more clearly requires some

investigation.

From 6100) and 6.1(11)

- ~ ._1_ we.
3,, (k )‘ jareld,d, e Tan

8:} ( ) ‘ jmldxdy e T»:

and from 14.3(13)

r-r 2" r
1.3 = e’ ° i d» r“, -B’ I;

47

(2)

(3)

(4)

48
The dot product above becomes for the two cases of [3

f, -2 = cos(¢,) f, -i = sin(¢,,) (5a.b)

A Bessel identity [16] gives,

23
10(k0)= 31n- [archww (6)

Using equations (2) through (6), using k3 = k2 - p22 and changing the dummy vari-

able ¢, in (4) and (5) to o, (1) becomes

 

 

 

- . jp'?’ 2g . . . -
Z’ = 2 21m srnh(p2d) e ¢ 1k (cos¢ cose 4» SIM) srne) I, _ _1- chm-(H) +
z pas-co jmldxdy p2 TM M 21‘
cos —
2R.
'P- a’
- 2m e’ ° 1:} e, sinh(p d)
+ )3 . - —’—- 100(0) (7)
,N,... 10382de P2 Tm P2

 

The last summation in (7) is convergent. Using cos¢ cose + sincb sine = cos(¢ - e),

the integral in the first summation becomes

 

2" 'k -9 1- jkacosQ-O)
i“ Jcos(¢ )R_e 2n (8)
5[ ]
2R,

From 15302), with o, -> 4>

 

_ -J'R. [ H325 - moo-(c - 6»

it-g- -2. cosm- 9»]
I = '
" 21:0: — 2R,k cos(¢ - 0))

fit, [ ~11; + moose — e»

«3%- +2. canto-err]
— 21t(1t + 2R,Ic cos(¢ — 0))

 

"C

(9)

SOince the dependance on o - 0 is cyclic, and since the integration covers a complete
cYcle, let q) - 9 —> o in (8) and (9). Bringing both terms in (8) over a common denomi-

nator, (8) becomes

49

,-_u_
2* kR,cos(¢) [fa-N. m ' e 211, [mm] ' W2 - kR.cos¢) 51“” cos %]
1dr .

 

+
21: (1t: - 2R,kcos¢) cos [fie]

. I"
—. -P o a
2.. —kR,cos(¢) {-je M' ‘°" — e "1 NW" - («12 + kR,cos¢) rim” cos [3%
+ {do ‘

 

] (10)

21: (1: + M,kcos¢) cos [2%]

Using Euler’s identity,

3’35 = .0. [2%] 1).... [5.] (11)

. . . air
in (10), the cosme portron of e ‘ cancels.

 

{do . c , , M +

1w
-— k _
21: (1: 2R, cos¢ ) cos 2R,

h

.. 1......) [,N, .. _ is, [gt] Mm] _ ,2. .. [ ,, ]

 

.-ij,¢06¢ .. 1w -- ._£ --w,, 1m
2* kR.COS(¢) [1e -Jsrn {—21% ]e ”M“ ] 2 e "‘ cos {—ZR, ]
+ £1141

 

(12)

21: (1: + ZR,kcos¢ ) cos [2%]

Putting each integral in (12) over a common denominator, and combining the two

integrands together, (12) becomes

2!
c c

.-ij,coo¢_ .. _RL 4qu _ about _1ta_
21th,cos(¢) [1e Jsrn[2R ]e ] nze cos[2R ]

 

 

£d¢ (13)
211: (1r2 - 41:2 Rfcoszo ) cos [£2-
or
2: . '1'”. cos.
i“ JkR.cos(¢) e + (14)
(1|:2 — 4k2 Rfcoszo ) cos [3%]

50
r
- ammo 1 .121; _ - .11
ij,cos(¢)e [2 cos [2R, ] srn L2R, ] ]
+ ,

 

 

 

21: (1t2 - 41:2 R,2cos2¢ ) cos [£
2R.

J

Looking at (14) as k -1 co, the term with e“"*‘°°“ will integrate to give something

similar to 7‘1- Jo(ka). This will multiply coherently with e’r ' 7° in the Flouquet summa-
tion, but the factor of % will ensure convergence. The other term integrates to give

something similar to i100: R, ) as k —> oo, where R, is some value in the range of

R, (o), and so converges also. Thus, the summation in equation (7) and hence (1) con-

verges, so the matrix element exists.

7. NUMERICAL RESULTS.

Galerkin’s method solutions to the integral equations of both the scattering and
transmitting cases have been implemented on a Cray X-MP/48 supercomputer at Pitts-
burgh Supercomputing Center. The matrix elements are computed by evaluating the
summation in 6.3(27), 6.3(32), 6.3(38) and 6.3(44) over a range of F" sufficiently
large to ensure convergence. Filling the Z-matrix for 6.2(44) takes by far the most
computation time. Total C.P.U. time for each frequency is about 150 seconds, over
140 of which are spent filling the Z-matrix. Once it is filled, the scattering case is
solved with electric field polarization both parallel and perpendicular to the plane of
incidence. The current I, and the patch currents K1 are obtained for both of these
cases. The transmission case is then solved and the input impedance and K2 are
obtained. For an assumed load impedance, the total currents on both the feed pin and

patch surface are computed using 6.3(3).

Numerical results are obtained for input impedance, feed pin current, and power
received over the frequency range 3 to 10 GHz, for both cases of electric field polari-

zation. The dimensions and parameters of the antenna array are chosen as follows.

b = 1.0 cm.

d = 0.15875 cm.
a, = e,

82 = 2.52,

a = 0.05 cm.

d, = d, = 3.0 cm.

'r",=x,£+yoi=0.3cmi

51

52
Nearly normal incidence is chosen, with 9,- = 0.0001 and ¢.- = 0 ( 0,- =0 can’t be

chosen due to indeterrninant forms in the solution). For parallel polarization, the elec-
tric field is parallel to the patch diameter containing the feed pin center. The incident

field strength is taken to be 1 Wm. For matching the electric field to zero on the feed
pin, the line along 4), = {15 is chosen. In 6.3(44), the term e’r'a" interferes coherently
with Java) in those terms in the summation over p and q for which |P~ 1m 2 ka.
Choosing phis, = § puts these terms on the diagonal of the square region of 1’ space

covered by p and q both ranging over :tM . A range on the expansion functions of
I = -5 to 5, m = 0 to 5, and on p and q of -25 to 25 has been found to give adequate
convergence of the matrix elements over the frequency range covered. For the feed
pin-feed pin matrix element, this range is increased to :70 Results are found for two
cases of load impedance, 2,, = 50 o and 2,_ = 25., the complex conjugate of 2,, (2,,
matched).

The scattered feed pin current, input resistance and input reactance are calculated
by solving 6.3(2) for each of the scatering and transmitting cases and using 2.2(4).
For the case of parallel polarization, the current magnitude on the feed pin for the
scattering case, I! 1|, is shown in Figure 4. The input resistance and reactance from
solving the transmission case are shown in Figures 5 and 6 respectively. The feed pin
current for the receiving case is found via 2.2(6) and is shown in Figure 7 for
Z, = 50 Q. The power received for both 2,, = 50 Q and 2,, = 2,; is found using 2.2(7)

and is shown in Figure 8.

The lowest order resonance appears between 5 and 6 GHz as a peak in the feed
pin current and input resistance. The resonance peaks for the total feed pin current
and for the input resistance both occur near 5.2 GHz, and the input reactance passes
through zero near these points. For 2,, = 50 o, the power absorbed also peaks at 5.2

GHz. This resonance behavior is very similar to that reported for a single circular

53
patch [10]. For A matched, the power received is nearly constant for frequencies

below about 9.7 GHz. The current flow on the patch for the dominant mode, near 5.2
GHz, is shown in Figure 9 for Z, = 50 o, and is similar to that expected from the sim-
ple cavity model for a single element [13]. Current is flowing from the patch onto the
feed pin. The simple cavity model gives 5.56 GI-lz as the resonant frequency of the
lowest order mode, slightly higher than that obtained here. It is found that resonances
occur consistantly at lower frequencies than given by the cavity model for other reso-
nances also, and that the difference becomes more pronounced at the higher order
resonances. Due to mutual interaction between the elements and the presence of the
feed pin, the simple cavity model is not expected to give more than qualitative results.
For the scattering case, however, the frequency for peak feed pin current is extremely

close, occurring between 5.5 and 5.6 GHz

Higher order resonance behavior can be observed above 8 GHz. The first
incidence of this occurs around 8.4 GHz, where the feed pin current goes through a
null followed by a sharp peak. The input reactance becomes very large in magnitude,
and changes sign, at the frequency where the null in the feed pin current occurs. This
type of behavior is well known as an anti-resonance in the study of various antenna
structures. At 8.44 GHz, the feed pin current and power received for Z, = 50 o are
near maximum. At this frequency, current flow on the patch surface is as shown in
Figure 10 for Z, = son, with current flowing out of the feed pin. From the cavity
model, current flow such as that shown in Figure 10 is expected at the second reso—
nance, though the cavity model gives 9.22 6112 as the resonant frequency of the

second mode.

At 9.45 GHz. the feed pin current goes through another null. The input reactance
again becomes very large in magnitude and changes sign. This anti-resonance is less
sharp than the one at 8.44 0112, both in the current null and in the approach of the

reactance towards infinity. For both of these cases, the input resistance remains almost

54
constant in spite of the large fluctuations of the input reactance.

Between 9.6 and 9.9 GHz, the feed pin current and power absorbed reach max-
imums while the current distribution on the patch changes form. At 9.75 GHz, the
scattered feed pin current rises to a large peak, greater even than that of the principle
resonance. The patch current is shown in Figure 11, and again, current is flowing out
of the feed pin. Its overall magnitude is smaller than the current at either of the two
lower resonances. The patch current distribution in Figure 11 can be separated into
two main components--a purely radial component which is expected for the third reso-
nance from the cavity model at 11.6 GHz, and a smaller component similar to the

form shown in Figure 9.

At 9.85 GHz, the input resistance encounters a jump in magnitude. The current
distribution on the patch surface is shown in Figure 12, with current flowing out of the
feed pin. This distribution is a combination of current similar to that shown in Figure
9 with a current distribution with 3-fold rotational symmetry. The cavity model
predicts a current distribution with this 3-fold symmetry for the fourth resonance, at

12.7 GHz.

Between 9.75 GHz and 9.85 GHz, the numerical solution breaks down. For the
dimensions we used, the lowest order TM, surface wave mode has a wave length of 3
cm (the spacing between the feed pins) at 9.8 GHz. As noted in [14], the zeros of T,
correspond to surface wave poles. When the Poisson summation is carried out near
this frequency, ,, = 0 for the term with p = q = 0. Since T,, occurs in the denominator

of the Green functions, the solution becomes numerically unstable.

To understand physically why an instability results at this frequency, consider the
current on each feed pin and patch as a source for a surface wave. For frequencies
away from 9.8 GHz, the surface waves from each element in the array add out of
phase and tend to cancel. At 9.8 GHz, these surface waves add in phase along the

array axes, and would radiate strongly in these directions for a finite array. Since the

55

analysis is for an infinite array, the level of this radiation would be unbounded, so no

steady state solution exists.

56

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3 0 ..0 .0
* ‘ “‘ ‘ ‘
.0 .0 00 . .0 .0 .0
0 0.0 .0
0 .0 .00 .00. .0. .0
.0 .o .0. .0 .0
0 .0
n. .0 .0 .9 Av
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o .9 0 § ‘
‘ o . I. ‘4
4 .0

J

62

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Iolsl“...0.0ov Iv...

\'
I. l'. \‘
,Ibllo Io 1000.0.v .0 \0 \Q\1
I I? \0 \c \.
I. I 4.0.0.. L0 \
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.wa m; a 8:335 .550 .: 2:me

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.vaIOATT-Iol 00610:!!04106
3/7K0014l
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f ‘Ku l to"
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K a
s . e (a
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0 g x 0 e r J
0 x o e I
6 0 K O O
.

63
For the case of perpendicular polarization, the electric field is perpendicular to the

diameter containing the feed pin. The feed pin current for the scattering case is shown
in Figure 13, and the feed pin current for the receiving case and with 2,, = 509 is
shown in Figure 14. The input impedance is the same as for parallel polarization,

shown in Figures 5 and 6. The power received for a load impedance of 50 a is shown

in Figure 15.

The feed pin current frequency responses for both scattered and receiving cases
are similar to those obtained for parallel polarization, but at a greatly reduced magni-
tude. The magnitudes of the feed pin currents near the first resonance are smaller in
proportion to the current over the rest of the frequency band covered than for the other
polarization case. The major difference beyond this occurs between the third and
fourth resonances, where the feed pin current dips in magnitude for both the scattered
and receiving feed pin currents. This occurs right around 9.8 GHz, where the program
becomes numerically unstable. The received power for 2,, equal to 50 :2 follows the
same trends as the feed pin currents. The power received is much smaller in magni-
tude for perpendicular polarization than for parallel polarization, with a larger drop in
proportion to the rest of the frequencies covered occuring near the first resonance. A

clip in power received also occurs near 9.8 GHz.

The resonance behavior predicted here is qualitatively consistent with published
data. Changes in substrate thickness and dielectric constant give shifts in the position
of the dominant resonance frequency similar to trends described in [8]. In addition,
the higher order resonance behavior seen in Figure 8 has been observed in experiments

performed at Boeing Advanced Systems Company.

 

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0' 0' 0' 0 0 ‘0 o o
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(vn) iNHHHnO GHHHliVOS

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Figure 13. Scattered current magnitude for perpendicular electric field polarization.

65

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67

IV. MICROSTRIP DIPOLE ANTENNA ARRAY
8. Problem Description.
8.1 Geometry.

The geometry of the infinite array of microstrip dipole patch antennas to be
analyzed is depicted in figure 16. A dielectric substrate of permitivity e, , permeability
u, and thickness d, (region 2), is located between the z = 0 and z = -d planes. The
dielectric is mounted on a conducting ground plane at z = -d, and is covered by a
material with constitutive parameters a, and u, (region 1). Each element has half-
length b and half-width a. A load impedance ZL is located at the dipole center. The
microstrip dipoles are located at the dielectric-cover interface in the z = 0 plane, and
are assumed to be perfectly conducting and infinitely thin. They are spaced distances
at, and d, apart in the x and y directions respectively, their centers located at
r’m = p d, 2 + q d, 5» in the global coordinate system of the array, with p and q integers.

Illumination of the structure is taken to be through an incident plane wave of fre-
quency m at an arbitrary incidence angle. The plane wave is expressed in terms of a
coordinate system x”, y”, z” rotated with respect to the global coordinate system of the
array, as shown in Figure 2, with o,- the angle between the x and x" axes, and 6,- the

angle between the wave vector I?" and the z axis. The 2 and z" axes coincide. The

factors

u = sin(9,-) c08(¢.-) (1)
and

v = Sifl(95) Sifl(¢,‘) (2)

are the direction cosines for the wave vector )2" with the —x and —y axes respectively.

68

69

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8.2 Problem Decomposition.

Using the principle of superposition, the array acting as a receiver can be decom-
posed into scattering and transmission cases, as in Figure 17. In the scattering case,
the load impedance for each element of the array is short-circuited at the dipole center,
the array is illuminated by an incident plane wave, and the current I, at the dipole
center is determined. For the transmission case, no illuminating plane wave is present.
Instead, a driving voltage V is applied at the center of the feed pin, representing the
voltage drop which would exist across the load in the receiving case due to current
flowing through the load impedance. The resulting current at the dipole center, 12, due

to V is found, and the input impedance is then calculated from

V
2,. = .1—2' (1)

Following the same development as section 2.2, the total current I is found to be

Zia
= [Zia +21. ]11 (2)

and the voltage V across the load impedance in the receiving case is
“112m
‘ [2,,+z,_ ]z‘ (3)

The power received by the structure is then

 

 

P—IIZR-l PZ‘" '2R 4
L-'2' L'EV! Izmr—Z: ()

where RL is the real part of ZL.

To determine the current at the dipole center and the power delivered to the load,
the scattering case is solved for 1,, and the transmission case is solved to obtain Z“.
Other quantities of the general receiving case can also be found using superposition of

the scattering and transmitting cases.

70

71

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ANV ammo magmfimfiwfi. A: 3.8 wctouaom ommo mfizooom

9. Scattering Case.

9.1 . Excitation Field.

Consider first the case where the short circuited dipole array is illuminated by an

incident plane wave. The total excitation field is the field generated by the incident

plane wave in the presence of the ground plane and dielectric coating. The incident

plane wave electric field can be written as

171"'(7")=Ei Ve-i‘”?

(1)

where V describes the incident polarization. By applying boundary conditions at the

conducting plane and at the dielectric-cover region interface, the total electric field in

the dielectric and cover regions can be found. This is done in chapter 12, and the

results are given here. They are
E,‘(?)=E’,l(r’)+E’,,,(r’) 220
where

. l .
ill|(?)=2E‘||e 2

All

x ijsin(-;-¢,l—klzc059,)sin9,- +x cos(%¢l,—klzcose,)cose,- 220

and

. 1 .
, ](—¢ +k x"nn0l-)
511(7) = 2 E1 3 2 i I y.”008(%¢i.klzcosei) 220

with

_ nzcose,sin(kzdcose,) + jnlcoseicos(k2dcos6,)
- nzcose,sin(kzdcos+6,) — jnlcoseicosaczdcosea

 

_ nzcoseisin(k2dcose,) + jmcose,cos(kzdcos6,)
l - nzcosegsinarzdcosefi - jmcose,cos(k2dcose,)

 

72

(2)

(3)

(4)

(5)

(6)

73
In the above, the following relations and definitions hold.

1‘1 = (Wu. 81 (7a)
k2 = Wu. 62 (7b)
x” = xcosoi + ysin¢,- (7c)
y” = -x sin¢,- + ycos¢,- (7d)
i” = £cos¢,- + ysino, (7e)
y‘” = -£ sincp, + ycoscb, (7f)
4’; = tan'1 [g] (78)
ki = k1 (7h)
k.‘ .
6,- = cos“1 7‘7] (71)
¢.. =-j low”) or n. = e”" (71')
¢l = "j 108: (r1) 0" Pi = 8"” (7k)
111 = ~24; (71)
n: = 8% (7m)

6, = cos'l(-k-1-sin9,) (7n)
k 2

9.2 Green Functions for Scattered Field.

The electric field supported by the induced surface currents on the dipoles is
determined using a Fourier transform Green function approach. The electric field is
written as an integral of the product of a Green function dyadic and the induced
currents. The field 3(7) due to an infinite array of dipoles, where the central element

has surface current I?(r’ ’) is
E’(r)=JLr<rIr3-I?(Mds' (1)

Here 3 (7’ | P ’) is the infinite array Green dyadic for electric field at the interface due
to currents on the interface and s is the surface of the central patch element. The
currents of the other dipoles are identical to the center dipole to within a phase factor.
The phase factor is due to the progression of the phase of the incident plane wave at

each dipole location.

The tangential electric field over the dipole surfaces, generated by currents
induced on the dipoles, is to be set to zero. The Green functions relating horizontal
components of electric field on the dielectric—cover interface to horizontal components

of current for an infinite array used for the patch array are used here as well. They are

 

géifi’l?”)= i sua(?)e’r"'"" mB=xJ (2)
P4”
where

(kit-k3) +(k2-k3) tanh d)
g;(zo)‘_._&_172’_ 7 1 P1 7r P2 (P2 (3)

J 1 . [p.+pzcoch<pzd)] [5P1+P2taflh(P2d)]

-k.k, @wpztanhcpzdr]

35(Fr-g;(F)- (4)

 

1 L _
1mm, [m +p200th(Pzd)] [€P1+P2m"0’2"’]

74

75

.. r (kit — kirp. + (k? — ’92)P2tmh(Pzd) e

 

8;; ( k )3 W [Pr ”2003100sz [Em +p2tanh(p2d)] (5)
In the above,

F=k.£+k,y=(k.u+p%:‘-)£+(k.v+q%)y* (6a)
where u and v are given in equations (9.1.1) and (9.1.2),

d2]. = dk, dk, (6b)

p} = k}+k,2—k.2 (6“)

p22 = k.’+k,’-k2’ (6d)

k1 = (WEI—1 (66)

k2 = ONE—£2— (60

E = e—j (6g)
and

T. = sinh(p2d)+%:-cosh(p2d) (7)

T... =€cosh(p2d)+-£—:sinh(pzd) (8)

The divergent nature of the above summations is overcome by the spatial integra-
tions introduced by Galerkin’s method. When these integrals are performed prior to

spectral summation, the resulting summations are all convergent.

Only x-directed currents are assumed, and only the 1: component of the electric
field will be tested in the application of Galerkin’s method. Thus, the only component
of the dyadic Green function needed is the x—x component, 9.2(3).

9.3 Derivation of Coupled Inteng Equations.

The induced currents on the dipole surfaces must satisfy an integral equation con-
structed by employing the boundary conditions that the tangential electric field over the
dipole surfaces must be zero. The scattered electric field on the dipole surface result-

ing from currents on the dipole surface s is given by
§1'(x.y.z=0) = H ?"’(x.y.z=0 I x’.y’.z’=0) ' l?(x’.y’) dx'dy’ (1)
The boundary condition on the tangential electric field at the strip dipole surface
yields an integral equation
5 X [Er‘(x.y.z=0) + E’r‘(x.y.z=0)] = 0 (2)

where l? 1‘ is the total excitation field in the cover region due to the incident plane

wave. This equation must hold at all points on the dipole surface.

76

9.4 Matrix Formulation.

The induced current on the strip dipole surfaces is to be expressed as a summa-
tion over a set of current distributions. The solution for the unknown amplitudes of
the chosen current distributions is to be formulated as a matrix equation. This will be
done using Galerkin’s method--the boundary conditions on the surface of the dipoles
will be forced to hold in an inteng sense where the same set of current distributions
will be used as testing functions. The boundary condition used is the physical con-

straint that the electric field tangential to the dipole surface be zero.

Let i ‘ be the electric field due to the incident plane wave and let E” be the scat-
tered electric field due to induced currents on the conducting surfaces of the strip

dipoles. Then

E'=jj§’-I‘<’ds' (1)

Here, 5 is the conducting surface of the antenna structure, I? is the surface current on s

and 3' is the dyadic Green function. The boundary condition on s requires that
i-(E‘+E")=o (2)
for all unit vectors, ?, tangential to the surface s.

Now let

1?: i; c. r. (3)

all

where {3.} is an appropriate set of independent and complete current distributions in

1?, and where C, is the corresponding amplitude of 1?. in 1?. Then

n81

E'=Hr- ic.z?.]ds' (4)
or, bringing the summation through the integral,

F=iqfirflm' m

381

77

78

Obviously, the summation over an infinite set of current distributions is an impos-
sibility. By choosing {I?.} appropriatly, the summation can be truncated at some suit-

ably small value, N, and the current obtained will be approximately 11’. That is,

fiquz (m

3.1

Using the above summation to obtain 5' ‘, it is no longer possible to require
i-(E’ " +1? ‘)= 0 at every point on 5'. Instead, Galerkin’s method is used. For

m=l,2,3,...,N require

j]?,.-i*d.:=—fir?,.-E‘ds. (7)

8

Substituting the expression for E” using the truncated summation,

flax... [i C.. ”gt-1?. ds']=—jjdsfi.. f" (8)
r n=1 .3 :
m=l,2,3,...,N
OI‘,
figflar..Hg.r,d.'=_fid.r.-ii (9)
m=l,2,3,...,N

Making the definitions

2..... =Hds1?,.-Hds’§’-I?.. (10)

v... =—jjdsi<*,. 1?" (11)
equation (4) becomes

N

2c.z...=v.. m=l,2,3,...,N (12)

381

This can be written in matrix form as

79

'erzrz"'zuv- 'er "V1”
V

 

 

 

 

 

 

221 222 ‘ ’ ' ZZN C2 2
1 I 2 ' I = j (13)
52"] 2N2 I O ' 2”” 1 1C" In LVN .

This is the equation used to obtain the current amplitudes C1, C2, ..., CN and hence
the approximate solution for 1? via (6). The current decomposition in (6) is con-
structed as a piecewise sinusoidal series with N pieces along the length of the dipole,

and with no variation across the width. The component current distributions are

 

 

 

 

 

I?=K i=1? sin[(b-l-x,|)ko] b..- <x<b..+ —a< <a (14)
a n Sin(bko) 1 l y
where
- _ 21)
b " N+l (15)
b..=5 n-N+1 (16)
2
and where
x, = x — b.. (17)
From (10), using
2.3”: i g..<i£’>ei"""-*° (18)
P4"

where g; (I? ) is given in equation 10.2(3), the matrix element 2..... becomes

a b _
de... Idy'ldx’K. 2; g;(F’)ei""”° (19)
-a -b pat--

&.'-—~Q-

2...=Idy

Rearranging, and using (14), this becomes

b

‘ n+1

no - '4 1 ,5 y . - A x

Zn“ = a k _ d dx b — m k
”2.2-“g ( ) sin(blco) i y e I srn[( II I) o]e

 

n-l

80

b

n+1

 

sin(5k o) q.

n-l

where x..’ = x' — b... This can be written compactly as

2...: i g;(ic")1.t(i’)l:(i’)

 

 

PflH
where
bl+
1312’):- I’d y dew“ )1 dz sin[(5 - Ix I )ko]e 1,1,:
sin m(15k0)-a H
or
1:0? ) = Id y deii‘sr I dz sin[( 5 +1" ) kale 111,1. 3’5”"

sin (bk 0)...

5
«+1de sin[(13 -x.. )ko] e

Evaluating the integral over y, and using
sin(z) = 32L [ej’ - e'fi ]
in the integral over 1, equation (23) becomes

1::(F = l , [eijk’ ‘ - {(th fl] eijk’ b“ x
:tjk, srn(bko)

0 . . .
Idqu — [ej(5+x )koetjk x _e-j(5+x.)ko etjk‘x. ]+
.1,
5 .
+de1[ej(5- x)koetjkwx- -j(5- -x)ko eijkux]
n
2

using (24) again, and evaluating the integral over x, (25) becomes

Idy' 6’" jdx'sinr(5—Ix."I)ko]e”"

:tjkx x. £1:ij b.

(20)

(21)

(22)

(23)

(24)

(25)

81
130?): 2i swig“) 5""- i x

:tjk, sin(Eko) 2
x06 . -iko5 . I 0
e emote». _ e e1(—k°:l: 5):. l
Moi/C.) j(-koik.) I_5
'k -' 5
+ __‘J 05 “40“». _ __e M eiaoflxm : (26)
J'(-koik.) J'(ko=tk.) lo

Evaluating at the limits, this becomes

1 *(P ) = ____-sin(k,a) em‘ b“ x
l k, Sin(5k 0)

x ems [1 _ e-jkob {($12.16) ] _ [ML [1 _ ejkas e-(tjkxE) ] +
(’50 i ks) (“he i ’5)

1:05 . . -ik05 . .
e -jk05 tjkxb _ _ e jkob 1:155 _
+ (+011...) [e e 1] (roux) [‘ e 1]} (27)

Putting the terms over a common denominator gives

I t (I?) = __—sin(k,a) etjk’b" x
" k, sin(koIS)
x (“kc i 1‘3) ejkofi _ [(+186) _ €1,756 + e-jkob ] +
(—k& i kxz)
(k0 i 19:) _ -jk06 4:55) 1,1,6 _ 11:05
+ __(-k& i k,’) [ e + e + e e ] (28)

Using

005(2) = -%- [ei‘ + e"‘ ] (29)
leads to

1 ~--) _ sin(kya) tjk, 5.. ("‘50 i ’9.) - _ - _
1.. (k )— __k, sin(kOE) c [__(—k& + 1:3) [2 cos(kob) 2 cos(i k,b)]

82

(koikx)

_ m [2 cos(kob) — 2 cos(:tk,.b)] ] (30)

collecting terms, equation (30) becomes

4 ' - -k
If (F) = M em‘ b” __9k—2 [cos(lc,.5) —cos(k05)] (31)
- o

k, sin(ko5) k.2

Notice that the I: differ only by a phase factor f": "-. With the definition

4nmga) -ko - -
0 I? =——— __ -
I ( ) k, sin(ko5) [hf-k3 [cos(k,.b) cos(kob) ] ] (32)
equation (31) can be written as
1.30?) =I° thew" ”' (33)

With this, equation (21) can be written as

Z,“ = i g; (I? )1"2 efik’ 5" e—jk’ b“ (34)
PAH

or, using the definition of b.., equation (16),

Z...= i g; (I? )1"2 e“"‘""""" (35)
P4”

From (35), it can easily be seen that

Zn+1,n+l = laws (36)

Once the first row and first column of the Z -matrix are filled, the rest of the elements

correspond to one of these, and can be easily filled in.

The excitation matrix elements (11) now must be evaluated. Since the 1?, only
have an x component, only the x component of the excitation field is needed. For the
scattering case for parallel incidence, from chapter 12, equation 12(45) is specialized to

z = 0 to become

. l .. .

Elllx = 1m ’ i 2 Bill e COS('%'¢| I) 005(9f) (37)

83
From 12(4),

2" - f = cos(¢.-) (38)

and with the definition

.1
E. = 2 5‘}. e150" cos(%<b. .) cos(O.-) cos(¢.-) (39)

equation (37) becomes

5.1. =5. e""""°‘°" (40)
using 12(2) and u and v from equations 8.1(1) and 8.1(2), this becomes

5.1., = 15.8")” 8“” (41)
With this and equation (14), (11) becomes

bani

‘ - ' 5—Ix.|)kol 7....
-E jk‘vy 51”“ - j 1
lie dy oi. sin(kob) e

 

V, dx (42)

01'

sin[( b - IX.I )ko] ejk'u' eihub

sin(k05 )

 

a . -5
V. = .2. I e”‘"’ dy 1! ' dx. (43)

comparing this to equation (22), it can be seen that
V.. = -E1 l..+(k1u £ + klv j") (44)

For the scattering case of perpendicular polarization, the as component of i5" 1‘ is, for

z = 0 from equation 12(75),

' l- + x"sin -
1}: =9" .2 2 Bi e“201 h (0')) cos(-%-d>i) (45)
From 12(5),
9" ° 3 = -sin(¢.-) (46)

Making the definition

84

.1
. o
1.":2 = —sin(¢.-) 2 £1 a]? l cox-$4191)

equation (45 ) becomes

11 =E2 ems" time.)
I

using 13(3) and u and v from equations 9.1(1) and 9.1(2), this becomes

11:52 9"” emu:

With this and equation (14), (11) becomes

bani

a ' . 5"lxnl)k0] 'kux
—E my, srn[( - , I
2 I e dy L sin(kob) e

 

v. dx

-¢ b

01'

a -6 . -
- srn b — .. k - ..., -
V, =—E2 I 81km dy £ [( . Ix-I ) o] e’k‘ . em», dz,
-a srn(kob)

comparing this to equation (22), it can be seen that

v. = -E.I.+(k.u :2 + klv y‘)

(47)

(48)

(49)

(50)

(51)

(52)

10. Transmission Case.

The transmission case is arrived at in a manner analogous to that of the scattering
case. The array is driven by a voltage at the center of the dipole which represents the
voltage drop across the load generated by current flowing at the dipole center when the
array is acting as a receiver. To find the currents on the array due to this applied vol—
tage, a slice gap generator with amplitude one volt is assumed to be located at the

center of each dipole. This gives a tangential electric field along the central dipole
E’ ' = x" 5(x). (1)

At each off-center dipole, the excitation field carries a phase factor due to the incident

wave.

Since there is no plane wave excitation in this case, this voltage is the only exci-
tation present. The integral equation to be satisfied is identical to that of the scattering

case, with zero excitation field on the dipoles, except at their centers. Thus, from

9.3(2)
ixil’(x,y,z=0)=—z‘ x12 8(x) |y| <a. (2)
For this case, the integral in (9.4.11) is trivially evalutated to obtain

{ -20 n = n. a (N+1)/2

V" = 0 else (3)

This is written in matrix form in the same manner as for the scattering case. In
particular, the left hand side of 9.4(13) is unchanged and the right hand side is

replaced with either 0 or (—2a) as appropriate, so

85

86

 

 

 

 

 

 

er' 212’”er7 C1) '0)
z. z' .. ' 1 5° '0
- N1 N2 ‘ ZNN N -

Thus, once the scattering case has been solved, the same matrix is used to solve
the transmission case, with only an easy substitution in the driving terms of the matrix

equation. The current at the dipole center is then solved for and the input impedance
obtained through

 

l
Zia-7;- 20C. (5)

C

ll. NUMERICAL RESULTS.

To test the accuracy of the solution, the results were compared to those obtained
in [4]. The reflection coefficient was obtained for two cases as a function of incidence

angle. The reflection coefficient is defined by

R: 1--— (1)

where P. is the power received by each element of the structure, and P. is the power in

the incident plane wave landing each unit cell of the array.
The results are shown in Figures 18 and 19. The parameters and dimensions of

the array are:

2b=0.39}. , 20:.0021

d = 0.19 l
e. = so , £2 = 2.55 60
d, = 0.5 A. , d, = 0.5 71 in Figure 4.

d, = 0.5155 7. , d, = 0.5 ). in Figure 5.

Figures '18a and 19a are the results obtained by Pozar, while Figures 18b and 19b
are the results obtained here for the E-plane. In both Figures 18 and 19, the load
impedance is matched for normal incidence, Z. = 75 a in Figures 18, while
Z. = 73 + j 4 n in Figures 19. Note that the curves match perfectly in Figures 18, and
in Figures 19 for 9. less than 70°. For 9. greater than 70", a grating lobe exists in Fig-
ure 19. In Figure 18a, the reflection coeficient is defined for transmission from the
array, and power transmitted into the grating lobe is included. In Figure 19b, however,
it is defined for the case of reception of power from a plane wave, and power scattered
into the grating lobe is not included. The difference between plots 19a and 19b

represents power scattered into the grating lobe.

87

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The Galerkin’s method solution was implemented on an LB.M. PC. compatible

personal computer. The number of piecewise sinusoidal basis functions used is 13.
The summation variables p and q were allowed to vary from +50 to —50. The time per
run at each frequency selected is about 5 min. The dimensions and other parameters

were chosen as

£1=£o £2=2.5€o

d = 0.15875 cm. = 1/ 16 in.

b = 0.508 cm. a = 0.0254 cm.
I E" | =1 Wm.
B = 0.0001 ° ¢ = 0.0°

Numerical results for input impedance and scattered current, along with power
received for several load impedances, for a frequency range of 8 to 24 GHz are
obtained. In Figure 20., the input resistance for each element of the array is given. It

peaks at 17 GHz. and has a dip at 21.4 GHz.

In Figure 21., the input reactance is given. It goes through zero near 9.8 GHz,
and shows a standard dipole anti-resonance curve in the vicinity of 17 GHz, where the

resistance peaked, and a small blip near 21.4 GHz, where the resistance dipped.

In Figure 22., the scattered current magnitude is given. It has a large peak at 9.8
Ghz, and a dip at 21.4 GHz. At 21.4 GHz, the guided wavelength of the lowest order
surface wave mode is equal to the dipole spacing. Near this frequency, coupling to
surface waves becomes strong, leading to the observed frequency response in this and

the two previous figures.

For the p = q = 0 term in the Flouquet mode summation, and for nearly normal

incidence, k, = 0 and k, = 0. For this term, 8, p1+p2 tanh(p2d) or equivilently T...

91

becomes zero near 21.4 GHz, and this term of g; (I? ) would seem to diverge. How-
ever, since k, is very small, the numerator also becomes nearly zero. Thus, only very
near the exact location of the zero of T... will the term become large. By way of com-
parison with the patch array, g,,, 3... etc. don’t have such a pole-cancelling term in the

numerator, thus leading to divergent behavior in a larger vicintity of the pole.

With the input impedance and the scattered current, the total power received can
now be obtained for any load impedance. In Figure 23, the received power is shown
as a function of frequency for a load impedance of 50 a. The maximum power
received, near 9.8 GHz, is about 65 % of the incident power. A sharp dip in received
power is located at 21.4 GHz.

In Figure 24, the received power is shown for a load impedance of 100 Q. The
maximum power received in this case occurs near 11 GHz, where about half the

incident power is received.

In Figure 25, the received power is shown for the load impedance set equal to the
input impedance. There are two peaks, located at 9.8 and 17 GHz, the two locations
where the input reactance goes to zero. The power received at these two peaks is

nearly 100 % of the incident power.

In Figure 26, the power received is shown for the load impedance set equal to the
complimentary impedance of the load impedance. If the input impedance Z... was for
an array of dipoles in free space, the complimentary impedance, Z. , would be that due

to an array of slot dipoles with the same dimensions;
Z. = 23 I (42...)

where 20 is the impedance of free space. For this case, the power received again has
two peaks, near 13 and 23.5 GHz. These corresond to the locations where the magni-
tude of Z... is equal to Zn. The power received at these peaks is about 100 % of the

incident power.

92
In Figure 27, the received power for the load impedance matched to the input

impedance is given. For this case, 100 % of the incident power is absorbed for fre-
quencies below about 23.5 GHz. This frequency corresponds to the frequency where
the dipole spacing is equal to a wavelength in the cover region, the cut-off frequency

for the appearance of grating lobes.

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In the remaining figures, numerical results for input impedance and scattered
current, along with power received for several load impedances, are given as the
incidence angle, 6,, is swept from O to 89 degrees in increments of 1 degree. The first

set of figures is for frequencies of 17 and 18 GHz, near the resistance peak.

Figure 28 gives the input resistance of each element of the array. The elements
are driven with a phase progression corresponding to the phase progression of the
incident wave in the receiving case; hence the dependance on incidence angle. The
input resistance is nearly constant with respect to angle, except in the vicinity of the
blind spot, [4], where the resistance peaks and then drops to zero as the angle is
varied. The blind spot occurs when the rate of phase progression of the incident wave
along the array corresponds to the rate of phase progression of a surface wave mode,
so that the fields produced by each strip dipole add in phase, and energy is coupled
into the mode. Note that as the frequency increases, the angular location of the blind
spot approaches zero degrees, occuring at zero degrees at a frequency of 21.4 GHz

(see Figure 18).

Similarly, Figure 29 gives the input reactance as a function of angle. The input
reactance is also nearly constant as a function of angle, except in the vicinity of the
blind spot. Near this angle, the reactance dips then peaks as a function of angle. At
18 GHz, the dip is barely noticeable.

In Figure 30, the scattered current magnitude is given as a function of angle.
Near the blind spot, the scattered current magnitude dips, and then peaks sharply as
incidence angle is increased. Note that the angular location of the blind spots for the
scattering case correspond to those of the input impedances obtained from the

transmission case, as they should.

Figures 31 through 35 give received power as a function of incidence angle 9.- for
the same cases of load impedance as Figures 21 through 25. In Figure 15, the

received power for a load impedance of 50 Q is given. For both frequencies, the

102
received power experiences a sharp drop, followed by a peak as the incidence angle is

increased through the blind spot region. The received power absorbed at zero degrees
is about 10 % of the total incident power. At the peaks, the power absorbed is
between 25 and 30 %.

Figure 32 gives the received power as a function of frequency for a load
impedance of 100 a. The shape of the curves is similar to that of the preceding case,

with about twice the power received.

In Figure 33, the received power is given for the load impedance set equal to the
input impedance. At 17 GHz, where the received power is nearly 100 % at normal
incidence, the received power experiences a drop near the blind spot in an otherwise
smooth curve. At 18 GHz, the plot looks similar to that of the previous figure, a dip

followed by a peak.

In Figure 34, the received power for a load impedance set equal to the compli-
mentary impedance of the array is shown. For both frequencies shown, the received

power drops to zero at the blind spot, followed by a sharp peak in power absorbed.

Finally, in Figure 35, the received power is given for a load impedance matched
to the input impedance. In each case, the power received is about 100 % until the
incidence angle increases to between 20 or 25 degrees. The angle at which the drop
occurs varies with frequency, decreasing as the frequency increases. The drop is due
to phase progression of the excitation of the dipoles becoming large enough that the
radiation into the cover region of each individual dipole can add in phase, so that

power can be scattered into the cover region in the form of grating lobes.

103

 

 

 

 

 

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The final set of figures gives the angular dependance for input impedance, scat-

tered current and received power for a frequency of 9.8 GHz, where the scattered

current experiences a peak, and the input impedance is nearly pure real.

Figure 36 shows the input resistance. It varies smoothly from about 12 a down

to about 8 a at 80° , then drops more quickly, approaching 0 as 6,- approaches 90°.

In Figure 37, the reactance is shown. It varies from -3 Q to -21 n as 6,- varies

from 0 to 89°.

In Figure 38, the scattered current is given. It drops towards O in a smooth arc as

6,- varies from 0 to 90".

In Figures 39 through 43, the received power is shown for the various cases of
load impedance. Note that the units of power in Figure 42 are nW, not uW. The
power received is so much below that of the other cases because the received power
near 13 GHz has not yet peaked and is still very small. In all five Figures, maximum

received power occurs at 9,- : 0°, dropping smoothly towards O as e, approaches 90".

112

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V. EXCITATION FIELD, GREEN FUNCTION AND CURRENT
DERIVATIONS.
12. Plane Wave Reflected by a Coated Conductor

The excitation field for the antenna or array is the field due to a plane wave
reflected by a coated conductor. The total field (incident plane plus scattered) in the

presence of the coated conductor supports the scattering elements.

Consider a plane wave illuminating a coated conductor as shown in Figures 1, 3
and 16, but without the antenna or array elements present The form of the illuminat-

ing electric field is
§i(?)=EiVe"?i'r (1)

where if describes the incident plane wave polarization and 12" describes the incident
direction of propagation. By applying the boundary conditions at the conducting plane
and at the interface between the coating and the surround, the total electric field can be

found.

The case of arbitrary incidence can be handled most easily by superposing waves

parallel to and perpendicular to the plane of incidence.

Case 1: E" parallel to plane of incidence.

The component of I? ‘ parallel to the plane of incidence is the projection of E’ ‘
onto the x’—z plane. As shown in Figure 2 the x’ and y’ axes are rotated through an

angle 0,- from the x and y axes, yielding a coordinate transformation

x” = x cos(¢,) + y sin(oi) (2)
y” = y cos(¢.-) - x sin(¢.-) (3)
f” = i cos(¢.-) + i sin(¢.-) (4)
i” = y“ 00801») - i Sin(¢.-) (5)

The projection of E" ‘ onto the x'—z plane is, in vector form

120

121
SCATTERING CASE

E i
REGION 1 \\/\ if

REGION 2 \ g;

//////////

Figure 44. Incident plane wave: parallel polarization.

 

i.‘,=(E"'-z‘)i+(E"-£”)i" (6)
Substituting (1) gives
if, =E‘ [6} -i) 2 +07 -£”)x”] £17"? ' (7)

The magnitude of if. is then

.1.
5.". =E" [6f -z“)’+(?' -2")’]2 (8)
Using (4) then yields
i
E.'} =E" [752+ (vicos<¢.-)+v,‘sin<¢.-»‘]’ (9)

Figure 44 shows a plane wave incident on a coated conductor, polarized parallel
to the plane of incidence. The direction of propagation makes an angle 6,- with the z-

axis. A reflected wave is also present in region 1, due to the reflection from the

122

dielectric interface and from the conductor. Its direction of propagation also makes an

angle of 0, with the z-axis. Refracted and reflected waves are present in region 2, the

coating, due to refraction at the dielectric interface and reflection from the conductor.

The direction of propagation of these waves makes an angle 9, with the z-axis.

The amplitudes of the reflected and refracted waves can be determined in terms

of the incident wave by applying the boundary conditions at the dielectric interface and

at the conductor. Refering to Figure 44, the incident, reflected and refracted waves

can be written in vector form as

if, = [—2‘ E1, sin(6,)+£”E"H cos(6 )] e ’7" "' (10)
if. = [+2 E'II sin(e,)+£" £1, cos(e )] (1'7"? (11)
:T = [—5 El', sin(9,)+£”El'| cos(e, )]"’ r! '7’ (12)

= [+5 Ef‘. sin(O )+x‘” f, cos(6, )] e"? 7' (13)

17'.- I _ _5,” _If‘lle 8.,7?‘ -? (14)

17.1 -+y" "'1' e-fi” 7’ (15)

17.7 y‘" “2' er" 7’ (16)

17.. - +y‘" -,-"—:- er‘” 7 (17)

where

F" = (— ” sin(9,) - i cos(6,-))k1 (18)

E" = (—£" sin(0,) + 2 cos(6,-))k1 (19)

i?" = (—£" sin(6,) — i cos(9,))k2 (20)

F” = (—£” sin(9,) + f cos(9,))k2 (21)

m = {l 112 = 15 (2221.0

123
k1: Will 81 k2 = (WW 32

(23a,b)

Here,l(14) through (17) have been determined from (10) through (13) and (18) through

(21) using the relationship for plane waves,

fi=PX§
7!

 

(24)

The field amplitudes 51,, Er, , and Er, can be determined in terms of 13‘}, by

applying the following boundary conditions.

a) Tangential E = 0 at z = -d
EI’I,~(z=—d) + Efi,~(z =—d) = 0
or, using (12), (13), (20), and (21)

jkzx”sin(0,) ejkzdcouefi =

E f, cos(6,) ejk’xn'inm') [112400) + E f, cos(9,) e O

which reduces to

_ +2'kdcos(6)
Ell =—EI+I e ’2 '

b) Tangential E" continuous at z = 0
E", l,»(z=0) + Eq,,~(z=0) - Ef,,»(z=0) — Efi,~(z=0) = 0
or, using (10) through (13) and (18) to (21),

it I! "3111(95) I" 1’ "8109.0

i|,cos(6,-) e + E'Hcos(6,-) e

. ~ 0 'k ,, .

— E f, cos(9,) e
For this equation to hold for all x’, the phase terms must be equal, giving
k1 sin(O,1= k2 sin(6,)
which is Snell’s law. Then, (29) reduces to

E‘. I cos(9,-) + 8'” cos(6,-) - Er, cos(9,) — Efi cos(6,) = 0

(25)

(26)

(27)

(28)

(29)

(30)

(31)

124

c) Tangential 1? continuous at z = 0
H". .,»(z=0) + H7.,~(z=0) - Hl-ly"(z=o) - H Fry-(Fm = 0 (32)
Using equations (14) through (17) and (18) to (21) again gives (30), and hence

E‘ E' E’ E”
_II+ II+ H___II_=

711 Th 712 112 (33)

Equations (27), (31), and (33) represent a system of three equations in the unk-
nown field amplitudes E'. I , T. , and 5,]. To solve this system, first substitute (13.27)
into (13.31), giving

 

005(9r) [1 _ e2jkzdcoe(9,)]

 

 

 

Eill 4” E'II = El": 008(6):.) (34)
which eliminates En. Next, substitute (27) into (33) to give
112
Adding (34) and (35) will then eliminate E’.’ . , giving
5*}. =43"? e"‘2“°°“°" [33- cos(k2dcos(9,))+ j “Ne” sin(kzdcos(9,)) (36)
112 cos(6,-)
As a convenient notation, let
El": =Tl|Eil| (37)
where TI, is the transmission coefficient into the coating
TH = . jnzcos(6,-) g'i"2"°°'(°r’ (38)
mooswr) sm(kzdcos(9,)) - 1m cos(9.-) cos(k2dcos(9,))
Substituting (37) into (35) gives E', I
'll =FIIE'iI (39)
where
H nzcos(6,) sin(k,dcos(9,)) + jm cos(9,-) cos(k2dcos(9,)) (40)

= “2008(9r) sin(kzdoosw.» - jm cos(9.-) cos(k2dcos(9,))

is the reflection coefficient from the coating. Note that if k; is real, the magnitude of

125
the reflection coeficient is
This suggests that 1",. be written as

I", = 131.0”

(42)
If k2 is complex, (42) can still be used, but 45,, is then also complex.

With the field amplitudes now known, the total electric field in each region can
be determined. The field in the surround (region 1) is

€d=in+fl1 mm
Substituting (10) and (11) and using (39) gives

. ' ~ ' o. -' e. ‘k e.
I} = z. E. em: nn< .) [1.” e m: coe< .)_ e; ,2 ecu .)]

| I 8111(9,‘ )

- 1 ~ ' e. -‘k e. r 9.
+x" E‘H e’ ‘x m ‘) [I], e ’ "co“ ‘)+e’ " co“ J] cos(9,-) (44)

Now, using (42) allows this to be written as

[1&0] ' + klx" sin(ei»

El} = i 2j E‘H e sin(-%<DH - klz cos(9,-)) sin(9,)

Ira-0, I + grams,»

+ i” 2 EH e cos(-:-<1>ll - klz cos(9,~)) cos(6,~) (45)

which shows 5.} to be a travelling wave in the x" direction and a standing wave in the
z direction.
The field in the coating (region 2) is
§|i=EIT+EIT (46)
Substituting (12) and (13) and using (27) and (37) gives
5’3. = 2‘ 2 T” “i. €0,000” ”M 5m.) cosUcztz + d) cos(9.»

— 2" 21' TH 151, ej‘z“"“”‘°" " " “‘°"’ cos(9,) sin(k2(z + d) eos(6,)) (47)

which is, again, a travelling wave in the x” direction and a standing wave in the z

direction.

126
SCATTERING CASE

Hi

 

\05...
REGION 1 \ _ \1‘71‘ 0 :31.
REGION 2 \Fi; e E;

 

Figure 45. Incident plane wave: perpendicular polarization.

Case 2: E’ perpendicular to plane of incidence

The component of E" ‘ perpendicular to the plane of incidence is merely the y”

component, as seen in figure 45. Thus,
if = (13" " ~y‘") y‘" (48)
Substituting (13.1) yields
5‘ = " 9'1? 0'3 e-fl‘” r (49)
The magnitude of E 1" is thus
Ef = E" t? - y"? (50)
which, upon using (5) becomes

1 = E“ [7; com.) - v: sin(d». )1 (51)

127
Figure 45 shows a plane wave incident on a coated conductor, polarized perpen-
dicular to the plane of incidence. As with the parallel case, there is a reflected wave
in region 1 and reflected and transmitted waves in region 2. Referring to Figure 45,

these waves can be written in vector form as

 

Eli = in El e-jE" '7 (52)
if =53"E1 fir"? (53)
if = 9"5; e-II’"? (54)
i+ :9” E1- e-jE’*-r’ (55)
. Ei I 1‘
fig = [—zH-1Lsin(9,- )+:2” % cos(9) e"? '7 (56)
E' E' l . .
171' = [—z‘ —1- sin(6,) — i” 71—? cos(6,) e"? '7 (57)
E- - " . -
17f = [4 T1:- sin(O, )+ i” —i- cos(9,) e"F '7' (58)
[71* = [4‘ 5% sin(9,) — i” % cos(6,) (’10., (59)

 

where the I? vectors are again given by equations (18) through (21).

The wave amplitudes are determined by applying the following boundary condi-

fion&

a) Tangential E = 0 at z = —d
EB~(z=—d)+Ef,»(z =_d)=o (60)

or, using (54), (55), (20), and (21)

BI 1‘28"“!1I9) EM) Efe in: mm iekzdcafi)_ _0 (61)
which reduces to

El = -E+e +21kzdcos(0 ) (62)

128
b) Tangential E continuous at z = o

Ej,~(z=0)+51,~(z=0)-Ef,~(z=0)—Ef,~(z=0) = o (63)
Using (52) through (55) and (18) through (21) gives Snell’s law again, along with

Ei+E1~Ef—Ef=0 (64)

c) Tangential I? continuous at z = 0
1,420) + H',~(z=0) - H 3.42:0) — H I.» (z=0) = 0 (65)

Using (56) through (59) and (18) to (21) again gives (30), and hence

E‘ E' E’ 15*
—-L cos(9,-) — —-'L cos(9,-) — —J- cos(6,) + —-L cos(9,) = 0 (66)
111 Ill 112 1'12

To solve for E' , E I and E f in terms of 15" , first substitute (62) into (64) and (67),

giving
51 + £1 = E; [1 - amid °°s(°"] (67)
. cos 6, °
5'1 — 131 = 45; ( ) 31 [1 + em’d Mm] (68)

 

cos(9.-) 1]:

Now, these two equations are of the same form as (6.34) and (6.35). Thus, using the

 

replacements
005(9r) Th 111 c053(9r)
005(91) -—> 1 1'12 —) E 005(91) (693,13)
gives
51 = 1‘1 5'1 (70)
E f = Ti 5‘1 (71)

where, following equations (40) and (38),

I‘ _ n2 cos(9,-) sin(kzd cos(9,)) + jm cos(9,) cos(k2d cos(9,)) (72)
i _ n2 608(6).) sin(kzd cos(6,)) - jm cos(6,) cos(kzd 608(6),»

 

in: 008(6).) e'fl‘”ll “(9')

T1 = 112 cos(9,) sin(kzd 608(9r)) - j '1: 005(9r) 005(k24 °°5(er))

 

(73)

129

The electric field in region 1 is now found to be
i 1 = if ‘1' El,
which, upon substitution of (52), (53), (18), and (19) gives

. 1 .

, “—0 +k1x"stn(9..))

i5" =y"2E‘ e 2 1
I I

cos(%<bl - k1: 608(91»
where «bl is determined from
I‘l = 1 e’°l
Lastly, the electric field in region 2 is found from
E 2 = E f + E 1’
which, using (54), (55), (20), and (21) gives

jk2(x”sin(9,) + d cos(6,))

if = .9" 2) Ti 51 e sin(k2(z + d) cos(e,))

An interesting special case is normal incidence. Letting 9,- = 0, = 0 gives
ilo

if, =£"2E‘.. e 2 llcos(-;-<1>H—klz)

if, = 4" 2) 15*}. TH ej‘z‘ sin(k2(z + d))

if = —y‘” 2151 11 em" sin(k2(z + 4))

 

 

where
T — T = in: [Mi
" - i nzsin(k2d)-j 111 (WM)
112 sin(kzd) +1 111005004)
1‘” = 1:

112 Sin(k2d) ‘j 111 c05(kzd)

(74)

(75)

(76)

(77)

(78)

(79)

(80)

(81)

' (32)

(83)

(84)

l3. Derivation of Green Functions for sources in the presence of a
grounded dielectric slab.
13.1 Preliminaries.

Consider a grounded dielectric slab of thickness d as shown in figures 1, 3 and
16, but without the patches, feed pins or dipoles present. The region above the slab
(region 1, z 2 0) is assumed to have the constitutive parameters 2, and u], while the

slab (region 2, —d < z < 0) has parameters a; and #2.

If a current source .7 is placed in region 1, an electric field will be maintained in
both regions 1 and 2. The field in region 1 can be decomposed into a primary wave
E 1” and a reflected wave E 1'. The primary wave is equivalent to the field produced by
the current source in unbounded space filled with the material composing region 1,
while E 1' represents the reflection of the primary wave from the dielectric interface.
The field in region 2 can be decomposed into a wave E 1’ propagating in the —z direc-
tion due to transmittance of the primary wave at the interface, and a wave E 1“ pro-

pagating in the +2 direction resulting from reflection at the conductor.

If a current source is placed in region 2, an electric field will also be maintained
in both regions 1 and 2. The field in region 2 is now decomposed into a primary
wave E2”, a wave E; resulting from reflection at the conductor, and a wave E; due to
reflection at the dielectric interface. The field in region 1 now consists merely of a
wave E 1‘ representing transmittance through the interface of the primary and upward

travelling reflected waves.

130

13.2 Representation of Field Quantities using Hertzian Potentials.

Each of the electric fields given above, and their associated magnetic fields, can

be represented in terms of electric Hertzian potentials 1‘1 as

E=k2fi+ wv-fi) (1)

I? = jtoerfI (2)
where k is the propagation constant of the medium,

k2 = (0211:: (3)
and the potentials satisfy the inhomogeneous wave equation

V2171 +k21=1 =- i (4)
.1035

In component form, equations (1) and (2) can be written as

 

 

 

E, = 1:211, + £(V-fl) (5)
E, = 1:211, + %(V-l"1) (6)
E, = kzn, + £(V-fi) (7)
H.=jwe[a—;I-y'--3§L] (8)
H,=jme 831’ -aTr:i-] (9)
H, =jme 331’ -%] (10)

131

13.3 Boundary Conditions on Hertzian Potentials

To determine the Hertzian potentials which describe the fields produced by

current sources in the presence of the grounded slab, it is necessary to employ the

boundary conditions at the dielectric interface and at the conductor. The boundary

conditions on the potentials may be deduced by examining the boundary conditions on

the fields. Letting E 1 represent the total field in region 1 and E2 represent the total

field in region 2, the boundary conditions are

01°

01'

01'

01'

01'

01'

a)

b)

d)

6)

Elx(z=0+) = sz(z=0')

kzl'l + j-(V-1"1 )=k21'l + -a—(V-I'i)
I I! ax I 2 21 ax 2

E1, (Z=O+) = E2y(2=O—)

a
1:311], + %(V'fi1)= [(22112) + 'a—y'(V'fi2)

”13(Z=O+)= H2,(2=0-)

8 31'1“ _ 3111, _e 3112, _ 3112,
I By 32 — 2 By 82

H,,(z=o+) = E2,(z=0')

[3111; an]: ]
51 " — = 82

32 3x

 

 

32 8x

 

am. am, ]

Eh(Z=—d) = 0

.3.

a1(V’flz) = 0

kzznzx +

E2,(Z=—d) = 0

2.

3y (V’fiz) = 0

b32111, +

132

(1)

(2)

(3)

(4)

(5)

(6)

(7)

(8)

(9)

(10)

(11)

(12)

133

Each rectangular component of current gives rise to one or two components of
Hertzian potential. It is convenient to examine the current components individully, and

catalog the boundary conditions on the potentials which result from each component

1. Horizontally directed source: 2’ = i J,
The fields maintained by a current distribution directed along x can be described
completely by a Hertzian potential with x and 2 components but no y component

fi=£ r1,+2rr, (13)

Both x and 2 components of potential are needed to satisfy the boundary conditions on
the fields. Any other combination of components results in a contradiction between

boundary conditions at the dielectric interface.
The boundary conditions on 1‘1 for an x-directed current source are found as spe-
cializations of (1-12) with 1'1, = 0. Equation (4) gives

_Q.
3)

.9.

(V'fir) = By

(V'fiz) z = 0 (14)
The spatial invariance with respect to y of the grounded slab guide allows this equa-
tion to be simplified to

V'fi1 = V'fiz Z = O (15)

A mathematical justification for this step is easily obtained using the Fourier transform

representations of the potentials, which is done later on. Using (15), equation (12)

reduces to

10211,, = kgrrz, z = o (16)
01‘

1'11, = 8, 1'12, 2 = 0 (17)

where

134

e, a — (18)

 

 

 

 

311,, 8112, an), an); _
[az ‘ 32]" 8x ’ 3x] "0 (19)
Using (17) then yields
am, am, _ 8H2. _
[32 - 32 ]--(€'-l) ax 2'0 (20)
Next, equation (6) reduces to
II
81—32; = 28a; 2 =0 (21)
or
n]; = 8, r12, 2 = 0 (22)

where the same argument used to obtain (15) has been used. Continuing, (8) can be

written as
an , an 311 , 81'!
[ex—.5420} [elven-3%] <2»

which, with the help of (14.3.22), reduces to

am, 8112,

32 8' a:

 

 

z = 0 (24)

Note that this equation does not reduce to 11,, = e, nun, since the variation of the
geometry of the slab in the z-direction does not allow the derivatives to be removed.

Next, equation (12) gives

a 8112,, 8112, _ _
5y'[ax + a2 ]’° ”‘d (25)

 

 

or,

3H2, + 3H,,

3): 32 =0

 

 

while (10) yields

2 iflfl-
kznk+31[3x + 32 -0

Using (26) in (27) merely gives
Hz, = 0
and using (28) back in (26) gives

3112,
32

135

2=-d
2=—d

2=—d

z=-d

(26)

(27)

(28)

(29)

In summary, then, the boundary conditions on the Hertzian potentials for an x-

directed current source are

 

 

 

3) nix = 8r Hz:
3111, 3112, __

b) [‘37-'37]"<€r'”
C) HI: = 8r “22

3111, 3I'I2,
d) 32 - 8' 32
3) Hz; = 0

3112,
O 32 - O

 

z=0
2:0
2:0

2:0
2=—d
z=-d

2. Horizontally directed source: f’ = y“ J,

(17)

(20)

(22)

(24)

(23)

(29)

For a current source directed along y, the electric and magnetic fields can be

completely described by a Hertzian potential with y and 2 components and with no x

component.

fi=yrr,+sn,

(30)

136

The boundary conditions on [‘1 for a y-directed current source are found as spe-

cializations of (1) through (12) with 1'1, = 0. Beginning with (2) gives

'aa';(V'fil) '—" £(V'fi2) Z = O

or

V-fll = V1112 z = 0
Using this in (14.3.4) then gives

kfI'II, =k221'12, 2 =0
or

“1, = 8r 1'12, 2 = 0

Now, rewriting (32) gives

 

 

 

 

 

"a.“ "a7' '7);— - Ty- : = 0
and (34) then simplifies this to
Next, equation (8) reduces to
81 am, = e; 8112, z = 0
ax 3x
or
11,, = 8, I12, 2 = 0
Continuing, (6) can be written as
6131-13- _828112, ]= [81231-82325] 2 =0
32 32 3y 3y

 

which, using (38), reduces to

(31)

(32)

(33)

(34)

(35)

(36)

(37)

(38)

(39)

 

Next, equation (10) gives

_a_ [3112, + 3112,]=0

 

 

3): 3y 32
01'

81-12, 3112,
+

 

 

3y 32 = 0
while (12) yields
2 i 3112,
[(2 1'12, 4’ ay ay

 

Using (42) in (43) gives

11;, =0

3112,
+

 

32

]=

and using (44) back in (42) gives

3H2,
32

 

=0

0

137

:L

is

=0
2=-d
z=—d

z=—d

(40)

(41)

(42)

(43)

(44)

(45 )

In summary, the boundary conditions on the Hertzian potentials for a y-directed

current SOUI'CC are

 

 

82
a) r11, = '8: 112, = 6, H2,
3111, 3112, 31'12y
b) [32 ' a: J”‘€"” 3y
C) ”l: = 8r n21
3H1, 31'12y
d) 32 — r 32

8) 11;, = 0

N
II
0

N
II
C

(34)

(36)

(38)

(40)

(44)

138

3112,
32

 

f) =0 2

II
I
A

(45)

3. Vertically directed source: J" = 2‘ J,

For a current source directed along 2, the electric and magnetic fields can be
completely described by a Hertzian potential with a 2 component, but with no x or y

component.
fl = i n, (46)

The boundary conditions for a z-directed current source are found as specializa-

tions of (1) through (12) with n, = 1'], = 0. Beginning with (4) gives

%(V.fi1) = _aa7(v-fi2) z = 0 (47)
or

V’fil = V‘fiz Z = 0 (48)
which reduces to

am, _ anz, _

T: - T ’ ‘ ° ‘49)
Next, (2) gives

Ear-(V1110 = Sax—(V11) 2 = 0 (50)
which again reduces to (49).

Continuing, (6) reduces to

213%: 4231;?"- 2 =0 (51)

HI, = 8, Hz, 2 = O (52)

Continuing, (8) gives

an , 31'!
81 ‘5;- = ‘27:" z = 0 (53)

which again yields (52)

 

a _
E (v 1‘12) _ 0
or,
8112,
32 - 0

while (12) reduces to

3.

3y (V—fiz)=0

139
. Finally, (10) gives

which again yields (54).

In summary, the boundary conditions on ii for a z-directed source are

3111, _ 3112,
a) 32 - 32
b) 1”112 = 8r “22
311
c) 2' = 0

 

32

to
ll
0

(54)

(55)

(56)

(49)

(52)

(55)

13.4 Fourier-Integral Representation of Hertz Potentials

Because of the invariance of the grounded slab geometry along the x and y direc-

tions, it is prudent to represent the Hertzian potentials as a linear superposition of

plane waves propagating along these directions. The two-dimensional spatial Fourier

transform of the a’th scalar component of fl is defined through

mac, .5 ,z) = j] Ha(x,y,2) (””5” dx dy

‘*

where 0: represents x, y, or 2 , and

F=gi+gy

is the two-dimensional spatial frequency, or transform variable.

inverse transform then states

1
(21:

+j (kx'xfiyy)

 

Ua(x.y.2)= )2 H flak, .k, .2) e dkxdky

Using the relations
?’=x x“ + y 55 + 2 2‘
d2}: = dk, dk,
dzr = 4x dy

allows (1) and (3) to be written as

fia(?,2)=I} Hump-mm»

l - - -» r.
natr)=(—2n—),-J_J_na(k .z)e’ ’de

(1)

(2)

The definition of

(3)

(4)
(5)
(6)

(7)

(8)

It is readily seen from (8) that each component of the Hertzian potential can be

represented as a continuous sum of plane waves propagating perpendicular to the 2-

axis, with amplitudes given by the Fourier spectrum of the potential.

140

141
The two dimensional Fourier-integral representation of the potentials provides a

simple approach for solving the wave equation 13.2(4). With the representation given
in (8), the wave equation reduces from a partial differential equation to an ordinary
differential equation with simple solutions. By retaining the variation of the potentials
along the z-direction in the transform representation, the boundary conditions on fl at

2 = 0, and 2 = -d can be explicitly enforced.

13.5 Solution for the Scattered Hertzian Potentials

The electric fields, if, if, if, if, and 1?; represent the transmittance or
reflection of the primary fields maintained by the current source distributions in regions
1 and 2. Letting fig fif, fir, 2', and fl; be the Hertzian potential representations of
these fields, respectively, and letting fi’ represent any of these "scattered" potentials,

the wave equation 13.2(4) reduces to the homogeneous form
Vzfi‘+k2fi‘=0 (1)

Thus, each component of the scattered Hertzian potentials must obey the scalar

Helmholtz equation
(v2 + 1:2) n; = o (2)
where or represents x , y, or 2.

The scalar Helmholtz equation can be solved quite readily by using the Fourier-
integral representation of the scattered potentials. Substituting 13.4(8) into (2) and

expanding the Laplacian operator gives

32 82 32 . r
kufifigay’ “Ml—.875] (210211110l (P 2)" 742k: 0 (3)

Bringing the derivitives inside the inversion integral yields

 

" 2 2 2 - _, .
(2:02Il(k2+'§c7+'aiy7+£7’"5(k’zle’r’de=0 (4)

Taking the derivitives, and remembering that I? - r’ is not a function of z, gives

 

II {Egg-(£3141:2 +k,’- k’)fi;(i'.z)}e"’"d2k=o (5)

32

(21:)2

This integral merely represents the inverse transform of the function. If the inverse
transform of the function is identically zero, the function being transformed must also

be zero. Thus,

142

143

[332:5— z]fi3(F.z)=0 (6)
where
p Judah/$47 (7)

Equation (6) is a second order ordinary differential equation for the transform

domain representation of the scattered Hertzian potential. It has obvious solutions
flaw. 2) = W3 e’” (3)
Potentials corresponding to waves traveling in the +2 direction will assume the nega-

tive sign in the exponential while potentials corresponding to waves travelling in the

—2 direction will assume the positive sign in the expontial.

Using the transform solution (8), the scattered Hertzian potentials are recovered

from the inverse transform (8) as

a k
nap): Ii (__;___()2) e]? '? chm): dzk (9)

13.6 Solution for the Primary Hertzian Potentials

The electric fields E" 1' and if are the primary fields maintained by the current
sources in regions 1 and 2, respectively. Letting fir and fl; be the Hertzian potential
representations of these fields, and letting fi" represent either of these primary poten-

tials, the wave equation 13.2(4) requires

VZfiP+k2fiP=-,—J— (1)
me

This equation is most readily solved using a Green function technique.

From standard potential theory, the Green function for the primary Hertzian

potential in unbounded space, GMT" | r ’), must satisfy
(V’+k’)G'(7’lP’)=-5(7‘-F”) (2)

That is, 6' represents the potential due to a point source excitation. Since the medium
is unbounded, the Green function depends only on the vector difference between the

source and field point position vectors
G’WIV’FGPCV-f”) (3)

Thus, for convenience, the condition 7’ ’ = 0 can be taken. At the end of the solution,

the point charge position will be returned to an arbitrary point. With this, (2) becomes

(V2 + 1:2) 0%? > = —8('r*> (4)

For compatibility with the geometry of the grounded slab, the Fourier transform

of the Green function is utilized

 

GP(E'.2)=H Grmrfi’N’r (5)
_ l u —9 r.
GP(r’)— (2“), 116m: ,2)e’ 7’de (6)

144

145
Using (6), equation (4) becomes

 

(V2442) (21): II G'(E’.z)e"’"d2k =45?) (7)

Next, the Fourier-integral representation of the two dimensional delta function

arx-xvso—y3= (2 -72 II ’T "' mm. (8)
is employed to give
sm=s<x>sma<z>=- 752"} H ”7612* (9)

Substituting this into (7) and taking the derivitives yields

71—21:?” {[3222 —,2—(kz+k, k)](;P (P,z)+s(z) }el""’dzk =0 (10)

Setting the function being inverted equal to zero resulsts in

2 -) - -§
[gag-102% )JG" (k.2)=-8(2) (11)

This is an inhomogenous second order ordinary differential equaton which can be

solved by transforming the z-dependance of G'fk, 2) as

(Val): j G’(F,2)e”z' dz (12)

GP(P,2)=-Tln— If}? ,Z)e’bdZ (13)

Substituting (13) into (11), along with the Fourier-integral relationship of 5(2)

__l. ,2.
5(2) (21:)... I: :12 (14)

gives

322

—.

2 -§
[L-Pzd‘ )1— (2n) I G” (1? Jul” 42 = ”(217) Jelz‘ d2 (15)

146

Combining results in
1 - 32 2 F 6? -§ [21 ’2, _
72-6-1 :h—z-p() (k2)€ +3 62—0 (16)

Taking the derivitives then gives

1 j{ [22+p2(l?)] 5'(IZ’,Z)-1 }e’z‘ dZ =0 (17)

(51*

Setting the function being inverse transformed to zero yields

= l
Zz+p2

 

5%?2) no

The desired transform representation of the Green function can now be recovered
by inverting (18) as

_, :22
G”(k ,2)=2L I e d2 (19)

1t __ Zz+p2

 

The inversion integral involved in equation (19) is most easily computed using

contour integration. Rewriting (19) as

1 .. e’z‘
— a
27‘ i (Z +jP)(Z “jP)

 

G” (I? ,2) = (20)

reveals that the integrand has poles at Z = ijp. To remove ambiguity, the sign on p is

chosen such that

Re {p}>0 Im{p}>0 (21)

The positions of the resulting poles in the complex Z-plane are shown in figure 47. If
the real line integration given by (20) is closed in the complex plane along an infinite
semicircular contour, C“, as shown in figure 47, then Cauchy’s residue theorem can be

used. For 2 > 0, the line integral is closed in the upper half plane to give

12! ,2:
I —5— a + I -_e__ (Z = 2102 Residues (22)
..., Zz+p2 c4» 22+p2

147

 

 

 

 

 

 

lR:{Z}

Im{Z}
C°°(z>0)
J'P .
< X > i T
—J'p
\ \‘r' / C'°°(z<0)
ll

Figure 46. Complex Z-plane.

148

by Cauchy’s residue theorem. The contribution from the second term is zero, since the
numerator is exponentially decaying for Im {Z} > 0, and since the denominator is

decaying as Z '2. Thus, (22) becomes

 

 

- e’z‘ . an I e'”
-— dz = 21: . l = 1: 23
lzz+p2 J Z+JP|Z=iP p ( )
Thus, the Green’s function becomes
_, ‘P’
Gm: ,z)=-‘—— 2 >0 (24)

2p

For 2 < 0, the line integral is closed in the lower half plane along dashed contour
C'“ to give

1'—e‘_dz+ j Jib— dZ=—21t’2Residues (25)

where the negative sign on the right hand Side is due to the reversed direction of the
path of integration. Again, the contribution from the second term is zero, since the
numerator is exponentially decaying for 1m {2} < 0, and since the denominator is

decaying as Z'z. Thus, (25) becomes

 

 

 

- e’z‘ _ . e’z‘ ' _ f"

i 22+p2 a- 27:] Z-jp:Z-jp—n p (26)
and the Green function is

G"(I?’.z)=‘—2:1 2 <0 (27)
Equations (24) and (27) can be combined to give

GP(F,2)= e2“ —°o<2 <00 (28)

Note that with the restriction on the sign of Re {p} given by (14.6.21), the Green’s

function decays as |2 I—m, as is required.

149
The final step in obtaining the Green function is to perform the inverse Fourier

transform of (28) to give

" a)?" €10.)le

p = 2
G ‘7’ U. 202050?) N (29)

 

and to return the source point to an arbitrary position determined by r ’, resulting in

" Err-1") -p<I’Mx-x'l
_, i‘“ _ e e
GP“ I )..Jl 2(21t)2p(l?)

 

d2]: (30)

Lastly, the Hertzian potential for the primary wave can be obtained through the
use of (30) and Green’s theorem as

fipm=jijlal our I Maw (31)

Note that since the primary waves are viewed as radiating into unbounded space, (31)

can also be regarded as a superposition of point source responses.

13.7 Green Function for the Hertzian Potential Maintained by a Horizontal
Source in Region 1.
A current source placed in region 1 will maintain electric and magnetic fields in
both regions 1 and 2. In region 1, the field will be composed of a primary contribu-

tion and a contribution due to reflection from the dielectric interface. Thus, the total

potential in region 1 is

fiy=fif+fif 2>0 0)
In region 2, the field will be made up of a wave transmitted by the interface, propagat-
ing in the —z direction, plus a wave reflected from the conductor, propagating in the +2
direction. Thus, the total potential in region 2 is

fi2=fi2++fi,- —d <2 <0 (2)

The total potentials in each region can be determined by employing the appropriate
boundary conditions from section 13.3. This is most easily done by considering the
effects of each component of the horizontal current distribution separately.

L fi=21u

For an run-directed current, there will be only x and 2 components of the
corresponding scattered Hertzian potentials, and, as seen from 13.6(31), only an x

component of primary Hertzian Potential. The required boundary conditions are

employed as follows.
a) Employ 13.3(17)

11f; + 111', = 8,012+, + 1'15.) at z = 0 (3)
Substituting 13.5(9) and 13.6(31) gives

1} d’kl “30,3 ”TV.” fl" dv'+I.J: __W&(k ) e’r'flrzk

 

150

151

 

 

 

 

 

= e, [I viii) 2’?"de + 8,]1 Wag) 2’1”.de (4)

Since 2 = 2' when employing the boundary conditions, the condition

lz-Z’l = 2’-z (5)
has been used to obtain the first term of (4). Grouping terms now gives

(2;); I1 dzk {I fit: if Pgle-P'x' dv’ + W. — e. [W51 + wand?” = o (6)
Setting the function being inverted in (6) to zero yields

-W. + a. [W2 + W] = V... (7)
where

mu?) .. i 7;: 1') "”25", aw (8)
b) Employ 13.3(21)

Hf, + 111', = 8,015} + 1'15) at z = 0 (9)
A procedure identical to that used to derive (7) yields

—W{, + 8, [W2*, + W23] = V1, = O (10)

Note that V, is zero since there is no 2 component of current (or, as a result, no 2

component of primary potential.)

c) Employ 13.3(23)
3a; (Hf; + 1112.): 8,015; + 115,) at 2 = 0

Substituting gives

11::(7') e,P-(r-r') {my-x)

 

3 o- , .- 120?) 1". —px
3’4”. 4:22! 1'08: 2(2n)’p1 dv +U. (21:)2 e, N ‘ dzk

(11)

l

152

a “W21(k) 2 _Wb(k) 2
=8'32-{Ii (2102 17”": 2212 +2” (1:2)2 1""1 212/2} at2=0 (12)

 

 

Performing the indicated differentiation inside the integrals and then setting 2 = 0

results in

P d .
U ‘12,, {V2.(Pi )- — .eflLl [W22 - W£]}m(k )e!""’= o (13)
P100

 

(21:)2
Lastly, setting the function being inverted to zero yields

a. + 2 ? [W22 - w2+.]= v.. (14)

1
d) Employ 13.3(19)

Sal-(1'11”, + m, — my, - 112'.) = -(e. -1)%(U2*. + r123) at z = 0 (15)

Substituting gives, with Hf, = 0 for an x.directed current source,

;,(E’) , ”W (k) - . — .
a: _{J‘i_ (2102 e]? 7 eHP d-zk _J'l (:n)2e e}? 'r’ e P2 d2k

 

 

‘ H W21“ ) ef’e’” d2).
_. (2n)2

 

{.(k) " W (k) - . 2
_(e,-l)— ‘1{ji— (21:? are" "’2' d2}: H (‘2‘): ‘1???" 21212} at2=0 (16)

or, taking the derivative and setting 2 to zero,

 

(2;), H dk {-plwr.-p2[w2'.—wr.]+ +(e -1)jk. [W2;+W2.]}e “20 (17)

Setting the transform to zero then yields

     

..+—[W2.—W2.]= (2,- ::[W2'.+W2*.] (18)

e) Employ 13.3(27)

IIL+II21=0 atz=-d (19)

153
Substituting

 

W (k) " W (P) 2
I1 (2202‘ e’lr 7 e-PZ' d2k+ +J‘i (__22’n )2 5']? 7‘ (’2 dzk— o

Combining gives

 

(2,102 1? [W£(F) 6'" + W50?) ['2‘] cit-1' d2]: = 0

Thus
Wm?) 2'2“ + Wz’,(17)e-Pzd = 0
f) Employ (13.3.28)
%m2+n2:)=o atz=-d

Substituting gives

W210? )
(21:)2

- W_2_2___(k )

.. -. a "
-a—-z 8J1: (2—1.;__)2e1‘£h’;'ep2 d2k+$jl

217‘?!” 32k =0 atz =—d

Performing the derivitives and combining gives

 

(21102 I} [33,24]?) e-Pzd — WM?) EPZd] P2 elf-7’ dzk = 0

Setting the function being transformed equal to zero yields

W50?) 2"" - W{,(I?) a", = O

(20)

(21)

(22)

(23)

(24)

(25)

(26)

In summary, then, the boundary conditions for an x-directed source current yield

a) -W'Ix + 67 [W23 + W22] = V12

b) -W;, +e, [wg + W;] = V1, =0

C) 12 + 8r"'- [W22 - W21]: V12

(7)

(10)

(14)

154

. a - . _ _ 1'52 - .
d) 1: + p; [W22 " W22] - (8r 1) p1 [W21 + W22] (18)
e) W530?) e" 1" + W50?) 2"" = 0 (22)
f) W54?) {'2‘ — W230?) eh" = 0 (26)

The equations labeled (a) through (f) above represent a system of six equations in
the six unknown transform amplitudes Win 1,, Wig, W21, WL, and W2]. These ampli-

tudes can be solved for as follows.

Solving (a) for W, and substituting into (c), and solving (b) for W, and substitut-

ing into ((1) gives the reduced set of equations

c’) 8, [1+ £2-]W,:, + e, [1- £5]fo = 2v” (27)
P1 P1
2 7‘2 +
d) e, [W{,+W{, +£[W5—WL]=(e,—l)-J;l- [W2‘,+W2,] (28)
e) Wm?) 23"" + W50?) {”2" = 0 (22)
f) W50?) 2'” - wg,(12’)e’2‘ = 0 (26)

Now, solving (7.7.22) for W; gives
W210? ) = -W2‘.(I? ) 22”" (29)

Substituting this into (27) yields

W5; {e,[ -££]-e, [1+QJezpzd}=2V1, (30)
P1 P1
or,
Wf, {6, 2,1407", — and) - 6,? chalk-P" + and) }= 2V“, (31)
1

Using the definitions of sinh and cosh,

sinh(x) = % [2" - e"] eosh(x) = -;- [2" + 2"] (32a,b)

155
equation (31) becomes

W2: {2. e’”sinh<p2d) - 2.1-ii e'z‘coshmd) }= v2. (33)
1
Now, defining
T, = sinh(p2d) + £3 cosh(p2d) (34)
1

allows W21 to be solved for from (33) as

 

 

+ 8,-1le e-P’d
W2: - — T, (35)
Also, using (29) in (35) yields
_1V ‘72“
w; = + 8' 1; ‘ (36)

Note that the zeroes of T, correspond to the eigenfrequencies of the TB surface wave

modes of the grounded slab. Next, (26) is solved for W2;
W; = W5; 12”?" (37)

Substituting (37), (35), and (36) into (28) then yields

3 ‘k .
W2: {[8, — fl] + [8, + 15] £21224}: (6, — Di; {8:12le W} (38)
P P] Tc

1

Rearranging this equation gives

 

W; {8, [2’24 + 2724] + _p_2 [and - 2714]}2’4 = (e, - 1)]

12, 1 sinh(p2d)
’ 2V _—
P1 pl {6, 1: } (39)

T.

Using equations (34) and defining
T... = e, cosh(p2d) + £3- sinh(pzd) (4o)
1

allows (39) to be solved as

156

'k
(2:;1 — 1)-’p—" sinh(p2d)
1

 

 

w; = _ V... W" T r (41)
Using (37) in (41) also gives
'1:
(5:1— 1)14 sinh(p2d)
W; = - V... 2”" p‘ (42)

T. T...

Note that the zeroes of T... correspond to the eigenfrequencies of the TM surface wave

modes of the grounded slab.

Now, equations (35) and (36) can be substituted back into (7), yielding

V 2
W11 = + [end - e-Pzd] — V1, (43)

or, using (32a)

 

 

W. = —V.. [1 - 2%] (44)
Similarly, equation (10) gives
'1:
(2;1_1)2. 2 2
W3, :4, T. r... l sinh(p2d) [2"2 +2“ ]V,, (45)
or,
176. .
(e. - 1) —smh(pzd) coshtpzd)
W12 = 2V1: pl (46)

T. T.

The total potentials in region 1 due to an x-directed current source can now be

calculated with the help of 13.3(13), 13.5(9), 13.6(31), and (1) as

I.112 (7') = 1113(7)

'1:
(e,—1)-J—’- sinh(p2d) cosh(pzd)

= 511;)? H 2V,,(I?) m T T agave-W dz" (47)
_. C III

 

157

and

nlx(?)=nfx(?)+n{x(?)

 

 

 

_ 1 2 12(7”) e!‘"<""" e“"""" ,
— (2n)2 ”.de jwe. 2p. d”
‘ (2.1.): II v..u?>I‘ _2_.mh1(,.21] .225... ...,. (48)

By substituting (8), equation (47) can also be written as

112( r’ ’1 21"” <7”) 2")“ “7
1'1 , d ’x
I“? (21‘ 1)2 [Li-dz 1“! 1051 P1 v

 

'k
<a.-03;" sinhtpzd) cosh<p2d)

 

 

x ‘ LT... (49)
while (48) gives
H1.(F’)—— 21:)2 111d, 1‘! 11(1):) err-2:4) X
W [__nn]} (.0.

Similarly, the potentials in region 2 are found using (2) to be

l'12...(7”)=Ui§2(7’)+r1520")

 

= 1 II e’r'y-Vl—x «fife—'2’ +epzdepz’ dzk
(27:): —. Tear

112(7' ') eir' (7'77 2"",
=(2;)2 IIdZ k! jwel pl T: 8r

 

dv' sinh(pz(d+z » (51)

n2.(r)=n2’;(r’)+n2‘.(r)

1 17‘. .
(8: - 1)—smh(Pzd)
P1

T, Tm [c."de-pz‘ + epzdepzz] (1211:

= (21:)2 Ii Cir”? ’le

 

 

158

'k
1 .. (1:1 — 1)-’p—"sinh(p,d)cosh(p,(d+z))
.. 2 1
" (21:)2 H. d k T. Tm x

 

 

XI Jlx(?’) -e’r'(?'w) 8?", dV’ (52)
v ng1 pl

 

2. .71:bi

For a y -directed current, there will be only y and 2 components of the
corresponding scattered Hertzian potentials, and, as seen from (31), only a y com-
ponent of primary Hertzian Potential. The required boundary conditions are employed

in the same manner as for an x-directed source.

a) Employ 13.3(33)

l'If’, + Hf, = 8,015; + 115,) at z = O (53)
Proceeding in a manner analogous to that used to obtain (7) gives

—w;, + 6, [W53 + 111,3] = 12,, (54)

where V, has been defined in (8).

b) Employ 13.3(37)

I'If’, + 111', = 6,015; + 115,) at z = O (55)
This gives
-W;, + e, [wig + W5] = V1, = o (56)

c) Employ 13.3(39)

.33: (Hf, + 1'11") = 54115, + H23) at z = 0 (57)

159
Proceeding as was done to obtain (14) leads to

P2
1) + 8,—p12[W2, - W2,]= V1,
(1) Employ 13.3(35)
a , + _ _ a
3:011: + H1; ‘ n23 " r123) "' -(€r— la) y(n2y + I12y) at Z = 0

This results in, similar to ( 18)
i. + — :[sz — W;]= (a —1)—""[W;, + W23]

c) Employ 13.3(43)

H{,+I'12‘,=O atz=-d
or, as in (22)

W2; a” + W5, [”1" = o
f) Employ 13.3(44)

-a—(n§.+nz)=o atz=-d
32

This yields

- d 4
W2, :72 — W3; e“ =0

(58)

(59)

(60)

(61)

(62)

(63)

(64)

In summary, then, the boundary conditions for a y-directed source current yield

7' -l
a) -W1, 4» 8, [W53 + Wer = V,,
b) -W{,+e, [WL+W;]=V1, =0
, P2 -

C) 1’ + £,— [Wzy — Wig] = V1,

P1
d) {,+%3-[W2’,—W*]=(e,-l)—k’—[W5+W§,]

1

(54)

(56)

(53)

(60)

160

e) W2; and + W5, ['3‘ = 0 (62)
:3 W2: 6"" — W52 ch" = o (64)

The solution of the above equations for W” 3,, W53, W23, Wig, and W2“, is identi-

cal to that obtained for an x-directed current source, with the obvious substitutions

W, -> W, V, —> V, jlc, —) jk, (65)
Thus,
-1 ‘h"
8' V] e
--1 +P24

(.2;l — 1)fl sinh(p2d)
P1

 

 

 

w; = _ v,, {"2“ T. Tm (68)
(a;1 — DIPi sinh(p2d)
d
w; - - v1, e" T‘ ‘7.” (69)
Wu ‘- ‘ Vly [ zsmhfzd) ] (70)
(e, — l) lisinhwzd) cosh(p2d)
w;, = 2v1, p‘ (71)

T, T...

The total potentials in region 1 due to a y-directed current source can now be calcu-

lated as

 

 

 

1113(7) = nix?)
- I -p (1+1) (8,-1)£ Sinh(P2d) COSIKPZd)
= 1 IIde I le(r ) e’r'(7'77e 1 dv, pl (72)
(2102 .. y j (081 P1 T: T»:

and

161
n1,(7')=nfy(7)+niy(7)

 

_ 1 .- 2 J’1(7') eff-(1'4”) -p lx-x'l _ SinW’zd) -p(z+z') ,
-_(21t)2'|1dk!jw81 2p: e‘ —1 2—1‘ e‘ dv (73)

Similarly, the potentials in region 2 are found using (2) to be
H2,(?’)=U2+,(?)+H2',(7)

___l__ " 2 IMF”) e!‘"<”-”" e7!" , -1 Sinh(p2(d+z))
_ (21:)2 H H! jute: p1 dv 6' T, (74)

_-

 

 

1"Izz("")=1'12+x(7’)+Hill")

.. 1c; .
.. (8, -l) lsmh(p2d) cosh(p2(d+z)) 11,5“) eiP-(7-79e'91"

l
(21:)2 U.” T. T... I we. p.

 

 

dv’ (75)

It becomes prudent at this point to write the potentials maintained in regions 1
and 2 by a horizontal current source in region 1 in terms of a dyadic green function.

The following notation will be used:
fi.(r)=16”(rlr’)-mrwv' (76)

Here 11(7) is the potential in region a maintained by a current source in region b,

where a and b represent either 1 or 2, designating regions 1 and 2, respectively.

With this notation, the potentials maintained by a horizontal current source in

region 1 can be represented using

- ,‘E’xr-r') -1’ . 2+:
1). I] 42;. —‘ {31" '- [1-2—smh‘P2d’]fl“ 7} (77)

 

0,33 7 r” = .
( I ) (21: 1033th Tc

6,};‘( r I r ') = o (78)

17‘: .
ef‘" (74,.) e1“, “I, (ix-1); Slnh(p2d) cosh(p2d)

 

G,}."(? I r”) = Ti): J1 42k (79)

faith T. T...

162
G,‘,'l(? | r”)=o

G;R?Ir)=cékrlr)

 

J' .
Cir. (?_7n’ 2.11“ +3') (57-1)??? Slnh(p2d) COShQ’Zd)

 

 

 

 

 

 

 

 

og-‘(r | r“): 1 dzk
(2702 I1 joxlpl Tc Tn
0.”;‘(7’ l r”): 1 I} d2]. e’r'W-V’ 5"" sinh(pz(d+z))
(2702 —. j 0381p; er Tc
r (a:1 1)jk’ sinh(p d) cosh(p (d+ ))
no . . .r_? ‘P f — 2 Z
Gi"(‘r*l?')= 12”,... ‘e’ (. 7“ p‘ 2
(2n) .. 1021p: T. T...
Gfi%?|?)=o
6.5%? | 7') =0
03%? I 7")=G.i-‘(? I r)
1 “ jr-(r-‘P? mr (8:1-1)£’-sinh(p2d)cosh(pz(d+z))
6.3-1(717'): 2” d2]: "‘3 , e p‘
(2n) .. New: T, T...

(80)

(81)

(82)

(83)

(34)

(85)
(86)

(37)

(83)

13.8 Green Function for the Electric Field Maintained by a Horizontal Source
in Region 1.

The electric field maintained by a horizontal source in region 1 can also be writ-

ten in terms of a dyadic Green function. The notation used is
i.(r)=J?‘*(rIr)-fi(r')dv' (1)

Here 5’, is the electric field in region a maintained by a current source in region b ,

where a and b represent either 1 or 2.

The electric field is calculated from the Hertzian potentials by using 14.2(1).

Thus, the fields in region a are given by .
1?, = kffi, + V(V - fl.) (2)

Substituting 13.7(76) gives
i.(r)=1{(k.’+VV-)5"b(rlr”)}-i2(r“) dv' (3)

and thus it is possible to identify
E’""(F’IF")=(k.2+VV-)6”(r|-r“) (4)

as the Green function for the electric field. Using the transform notation introduced in

13.4(8), equation (4) can also be written as

§’"’(?IV’)=(k.2+VV‘)(2_:r)’- (15"(21 7"?)‘W'rdzk (5)
where
8”(z|?’.F).—.jj 8“(7’|?')e'i""d2r (6)

is the Fourier transform of 6’ ‘-"( 7’ | 7’ ').

163

164
At this point further simplification is difficult unless certain assumptions are made

about the convergence properties of the spectral integral. If the integral in (5) con-

verges rapidly enough, the derivitives can be brought inside to operate on the

integrand, giving

 

‘g"""(?|r')= (271:): Il?u(zl r'.1?)e"’"d2k (7)
where
5.6.5 _,, —y _ 2 . 80.5 , -.
(er.k)-(ka+VV) (Zl7.k) (8)

is immediately identified as the Fourier transform of the electric field dyadic Green

function. Expanding out the operations indicated in (8) gives

_ i . ~_§_ . “.3- .
V(V-5’)_£ ax(v 6’)+y ay(v 5')” az(V 6’) (9)

where

+ —[G,, 2 + 6,, y“ + 6,, 2] (10)

For current applications, only the horizontal components of the electric field in
region 1 are required. Using (9) and (10) in (8), the appropriate components of the
electric field dyadic Green function will be found from

 

2.1"(2 u r a I?) = {76: + $63 + kid." (11)

2.;‘(zl r ’. k )= 5%- ',;" + 75251-64“ (12)
2

2,}.‘(zl 7' '. I? ) = 32 6.1-W —a-G..‘.-‘ (13)

165

32
+3yaz

 

- gym 7', r )=3-a;-',;- 6,;v1+k}6,;'1 (14)

In deriving the above equations, it is remembered that 6,1" = 6,1," = O, and that there is
no vertical component of current in region 1. Using equations 13.7(77) through

13.7(81) in (11) to (14) then gives

“ g;1(zl rzF)—-(k.1-k3)G;-1 -jk.prG.l-1 (15)
- g1-1(z| 7,1? )=—k, k,611 —jk,p,6,;-1 (16)
are l r '. r) = —k.k,G.1.'1 - jk,p.G.1.'1 (17)
§,§1(er7’ ) = (k.2 _ k,2)c';,;-1 — jk,p,(';,;-1 (13)

Similarly, only the vertical component of the electric field in region 2 is needed. This

component will be formed from

 

 

2
23%| r“. I? ) = 338262614» 73:26.3" + @613." (19)
~2,l 82 “2,1 82 2.1 2’21
(zlr’, ’k)=ayazG” +-a—-z,,26 +sz,,' (20)

where 6,3," = 6,2," = O has been used. Using equations 13.7(82) through 13.7(88) in (19)
and (20) gives
' 3110] r”, k )= (It2 + p; 2)G,, 2‘1 + jkxpz coth(1’2(d+1))Gm.;2'l (21)
7~1(z I r31? )= (1:21 + p2)G.,’~1+jk,p2 60m(P2(d+2))G,§‘1 (22)
Explicit formulas can be calculated for (15) to (18) by substituting equations
13.7(77) through 13.7(81). However, to employ the boundary conditions on tangential
electric field at the surface of a patch element, knowledge of the field quantities is only

required at z = 0, and only for a horizontal source residing entirely at z' = O. The

proper specializations of (15) through ( 18) are thus found to be

"51(z=0|1' z'=0,x’,,y’k )=

166

'1:
(e.—1)-’p—‘sinh<p.d) coshwn
1

 

 

 

 

 

 

 

 

 

 

_ e—fi..?' 2 . 2 SinhU’zd)
- jmelpl (’51—ka) {—1} J "kxPl LT... (23)
24"(z=0| z’=0.x’.y’.73 ) =
(e -l)&sinh(p d) cosh(p,d)
._ NP", 4, 51M 7, f ”1 2 24
" jump: x ‘ ”I 1P1 T,T.,. ( )
é,§"(z=0| 2’=0.x’.y’.l? ) =
1 ’7‘" ° h d h
_ fly,” k sinmpzd) .k (8.- )zsm (p2 )cos (pzd) 25
"‘ jwelpl "sky T, "Jypl T,T.,. ( )
é,‘,'1(z=0| z'=0.x'.y'.ic" > =
. - (e.-1)i"’—sinh<p.d> cosh<p.d)
_ e-I'E'Jh 02—16 2) smh(p2d) _ P1 (26)
— jmelpl l J y T: k, 1 Tch
The relationships (23) through (26) can be simplified as follows. Let
. - 2 .
F = (k 12 _ k3,) [smhfm ] + (e, 1)k, my?!) cosh(p2d) (27)

Putting both terms over a common denominator and substituting equation 14.7(40) for

 

 

Tm gives
(1:? - k3) sinh(p2d ) [3, cosh(p,d) + %sinh(p2d) + (t:,-l)k,,2 sinh(p2d) cosh(p2d)
l
p = a T. T. ~ (28)
which reduces to
(1:122, - k3) sinh(p2d) cosh(pzd) + (k? — k,2)L2-sinhz(pzd)
r = ”1 (29)

T. T...

Dividing through by sinh(p2d) cosh(pzd) gives

167

(Iris. - k3) + (k? - kx)—tanh(Pzd)
F -.-

 

[SiliMpzd ) cosh(pzd )]T

Finally, substituting for T, from 13.7(34) and for T... from 13.7(40) yields

F 15128,- -kx2)P1 + (k1 2- 15:2)P2 tanh(p2d)
P1
E1 +P2¢0th(P2d)] [€rP1 +P2mhwzdfl

Next, let

-k.k, sinh<p2d) + (2.-1)k.k,sinh(pzd) cosh(p2d)

G = T, T, T...

Putting both terms over a common denominator and substituting for T... yields

-k,k,sinh(p2d) [2, cosh(p2d) + gisinhwzd)] + (e,-1)k,k, sinh(p2d) cosh(p2d)
1
G = *

 

T, T,.
Simplifying gives

—k,k, [fisinhzmzfl + sinh(p2d) cosh(p2d)]

G = T. T.

 

which, upon substitution for T, and T... reduces to

—kxky [P1 + szMPzdfl

G P1 ’ ’ ’
[,.. +122 commo] [c.p. ..., tanhtpzdfl

 

(30)

(31)

(32)

(33)

(34)

(35)

Using (31) in (23) and an analogous formula in (26), and using (35) in (24) and (25),

results in, with the help of (7), the electric field dyadic Green function components are

,,_, e-fi’wtrr')

1,1 I
8:161:0er I z =0.x .y . k) =(——2n)2”‘1 x

(081

(left. - k3»): + (k12 - kx2)P2tanh(Pzd) .
E1 + P200fl1(P2d)] [6m + 102thqu

42k

(36)

168

 

 

 

 

11 , , , .,) ,-fi'- tr-rox
8.; (z=0.x.y I z =0.x .y . )=(2n)2H-‘—7—
‘kxky E91+Pztanh(P24)]
>< * * d2]: (37)
k1 + P2¢0m(Pzd)] [8.101 4' szhwqu
g,l"(z=0.x.y | 2’=0.x’.y’. k )=g.‘,‘(z=0.x.y I z’=0.x’.y’. 5' ) (38)
-—~i£’ tr-ro
gy‘y"(z=0.X.y I z =0.x .y . F)= (2;),H——1—— x
(tie. -k,’)p.+(k.1 -k,’>pztanh(p2d) .121. (39)

[D1 + P2C0WP24)] [8.121 ‘1' szhWflq

Next, explicit formulas can be obtained for (21) and (22) by substituting equa-
tions 13.7(82) through 13.7(88). This gives

 

 

 

 

 

 

 

 

 

(in... ’[ (e, —l)&sinh(p2d) cosh(p2(d+z))
- 2.1 —D I _ 2 2 .Pl'
811 (2 Ir ,k)- jOJ€1p1 (k2 +p2)e l T‘Tm +
+ jkxpz e71" [cosh(p2(d+z )) J > (40)
e, T,
be.» [ (€."-1)jpilsinh(pzd ) cosh(pz(d+z ))
- 2,1 _ 2 2 ‘P1
(2 '7’ 9k)‘ jwfilpl (k2 +p2)e a. T‘Tm +
I
+J'k, p2 8....2’ [WSh(P2(d+Z» ] r (41)
e, T,
I
To simplify these, let
-‘—(e:1-1) sinhtpzd) 1
F = --(k22 + p22) T, T... + mail? (42)

169

Putting these two terms over a common denominator, and substituting for 7'... gives

—(lc22 +p2’)-pl—(e."-l) SinNPzd) + p28." k 008110?de -:-zsinh(pzd)
l 1

 

 

 

 

F = T. T... (43)
which simplifies to
£6.44) sinh(pzd) + p; cosh(p2d) + 523-1111111624)
r = ”1 ”1 (44)
T, T...

Here, note that

[cf 1:,“ = k? (45)
so

—k§ 2:1 + Io} + p22 = 41.2 + 1:; +p22 (46)
But, from 14.5(7)

k22+p22 =k,.2+k,2=k12-1-p.2 (47)
so

-k22 6." + k22 +1222 = 1212 (48)
Using (48) in (44) gives

F = p.sinh(p2d) + pzcosh(p2d) (49)

T, T...

which, using 13.7(34) for T,, reduces to

p = -;—;- (50)
Now, substituting (50) into (40) gives

23%: | 7’ ’.k) = 5i:— ”...,-1. ”smpzmz» (51)

10.31 Tm

A similar set of steps gives

170

7‘ e‘Pt"jk’ COSh(P2(d+z ))
j “1 Tan

 

”31(2|7’.k)=

Thus, the electric field dyadic Green function components are

 

e:fi’(?-7‘) e71"jk'c°sh(pz(d+z»dzk

2.1 I-
g..(f’|f’)— we} T...

..i... —1'1'

1 " eIV'U’q‘? rile, cosh<p2(d+z))
2.,(717’) )2” ...,... e T... a].

(52)

(53)

(54)

13.9 Green Function for the Hertzian Potential Maintained by a Vertical
Source in Region 2.

A current source placed in region 2 will maintain electric and magnetic fields in
both regions 1 and 2. In region 2 the field will be composed of a primary contribu-
tion, a contribution due to reflection from the conductor, and a contribution due to
reflection from the dielectric interface. Thus, the total potential in region 2- can be

written as

fi2=fl5+fig+fig -d<z<0 (1)
In region 1, the field will be composed merely of a wave transmitted by the interface.
Thus, the potential in region 1 is

1"]. = E; z > O (2)
The total potentials in each region can be determined by employing the appropriate
boundary conditions from section 13.3.

For a z—directed current, there will be only a 2 component of both primary and
scattered Hertzian potentials. The required boundary conditions are employed as fol-

lows.
a) Employ 13.3(48)

inf, = 1mg, + 112+. + r15.) at z = o (3)
32 32

Substituting 13.5(9) and 13.6(31) gives

 

 

_. (23)-__2— . 10162 2(211)2 r:
F .
+11%}? +1" ++..+.. +1_'1,“’-—;';.—,—‘ 1. ++.=.} atz=o (4)

Since 2 > 2' when applying the boundary conditions, the condition

|z—z’|=z—z' (5)

171

172

has been used to obtain (4). Differentiating, and grouping terms gives, at z c 0

 

 

 

 

 

- 1M") if": '1' . d’k
W -W .1 dv’ I" 1' =0
ii{:w1+( 2' 1') . jun: 2P2 }P2¢ (210’
Setting the quantity being transformed in (6) to zero then yields
flwix - (W2; "' W23)= V2,:
P2
where
T (r) 4% '2'
V 1"." -‘———-4v
2( )' J]. 2P2
b) Employ 13.3(51)
Uf.='€(l'lf,+n§. +115.) atz=0
Substituting gives
- (2 -I')
I! “W1,(F) elf ‘P {'"d 8,]! d2]! 11;“? ) cl? (7' 77‘72 dv’
.. (21:)2 . me: 2(2102 P2
-w;.( P) " Wat F 1 .
41 (210’ 2'” "'1'd11: 41.—(27):- efi’e'flzr} atz=o

Simplifying this yields

e,'1w;, - (W5 + W5.) = V2.
c) Employ 13.3(54)

%(ng,+n;,+n;,)=o atz=—d
Substituting gives

_3_ - 2 124'") eF-(Y-Vfie'h"""dv
..{11111 ..

 

'W£(F) 'W£(F)
[ii—(211)” A“ Hfidz" II 2...): e"“‘.’e"'111111}-.—.o atz=-d

Performing the indicated derivitives then gives, at z = —d

 

" J r ..ny 73(8"!+d) 2
‘H{W£e"‘-W’e v34 _J-_2_;__( ')e 2011' Pzefiydk ’0

(6)

(7)

(8)

(9)

(10)

(11)

(12)

(13)

(14)

173
Setting the function being transformed in (14) to zero yields

 

W3; eh" — w; {1’2" = .72" 17,, (15)
where

r) _p f 7 -r'?’ 12"

V2(k ),Ii(_2‘ I e dv' (16)

v 1.03: 2P2

In summary, then, the boundary conditions for a z-directed source current yield

p + -

p—iwiz — (W2: - W23) = V2: (7)
e.“Wi. - (W. + Wz’.) = V2. (11)
w; eh“ — w; {1’2" = {1’2“ 17,, (15)

The above system of equations can be easily solved as follows. Equation (15)

can be rearranged to yield

24

W5: = (W5. + V.) 621’ (17)
which can then be substituted into (7) to give

Li-Wi. — W5; [2'2" 472, {2” + W; = V2, (18)
or

fiwi‘ — W5, (62" - 1) = (1., (19)
where

(72 2 V2 + V. [2'34 (20)
Equation (17) can also be substituted into (11) to give

e,“W'., - W5; {2'1“ 47., [2’1" - W; = V2, (21)

Of

174

s.“Wi. — W5. (62'1" + 1) = 02.
which then yields

-1 r
87 W1: _ U23

W5: = ..
e 2’24

 

+1

Substituting (23) into (19) gives

‘szd
P1 e - 1
2 1: 11 21 3.2,”, + l 2:

Factoring out W5, and using 13.7(32a,b) results in
W1. [£1- + e,“tanh(p2d)] = U; [l + tanh(p2d)]
P2

Solving this equation gives

1 + tanh(p2d)

 

1 — U

11 - 2: Pl
— + £,'1tanh(p2d)
P2

The numerator of (26) can be rewritten using

end
1 + tanh(p2d ) = m-
and, from (20)
21924

U2, = V2J + ‘72, e-
First, substituting (8) and (16) into (28) gives

1 r, 'j?‘?’ '0 I
U23 =e-914I 2.( ) e [em ”New" no]

. ch"
11 1‘062 2P2

which simplifies to

 

J 7” #74”
e-pzdj- 2:( )e

U = ,
2’ v 10352 P2

cosh(p 2(2’ + d))dv’

Now, substituting (30) and (27) into (26) gives

(22)

(23)

(24)

(25)

(26)

(27)

(28)

(29)

(30)

175
1,‘—'IJ7((°:,)¢ '7? ,, COSh(p2(Z'+d)) d ,

 

 

 

P1°°5h(Pzd)+e,'lpzsinh(p2d) v (31)
or
((0:33 ”17"” cosh (z +d))dv
=1] 12’ P1 (pg. (32)

Next, solutions for W; and W2: can be obtained quite easily. Subtracting (ll) from (7)

gives

2W2; = W, [8,“ — £1- (33)
P2

which upon substitution of (32) yields

 

 

-1 pl
J r -jI’-r' 6' " '-
W5; =1 31361,) e 2m T p; cosh(p2(z' + d))dv’ (34)
or
E _ a 1
J r’ 45"?” '
W5; =I :fmz’) e £02 p11, cosh(p2(z’ + d))dv’ (35)

Now, substituting (35) and (16) into (17) gives

.j".?' E —e'

w; = cosh(p2(z’ + d)) + [”3" dv’ (36)

 

 

JIM 7' ') e-szd e
v - (082 2P 2 T“

With the transform amplitudes W1” Wig, and W2; determined, it is now possible to
formulate expressions for the total Hertzian potential maintained by a vertical current

source in region 2.

The potential in region 1 is easily determined by using (2). Substituting (32) into
13.5(9) gives

 

 

176
l .. 122(7')ejl'("’7e"' COSh(P2(7-'+d))
n. 7’ = d’k . d ' 37
1( ) (2192!...J {noel p, T... v ( )

The dyadic Green function representation for the potential in region 1 is then easily
identified from 13.7(76) as

GM? l r')=

 

" 4'47-” 1“ '
1 de2k e I e ‘ COSh(p2(z +d)) (38)

(21:? j (081 P 1 Tan

The potential in region 2 is determined by using (1). Using 13.6(31) and substi-
tuting (35) and (36) into 13.5(9) gives

 

 

" 1!"? J (7') -,I’-?' _.
, _ e 2 2.1 e -pz|z 1|
UM? )—'[.‘I. (27‘)2 d I“! jmfiz 2P2 {e

 

 

P2
— _ er
pl I
+ T cosh(p2(z + d)) +
P2
— _ 8'
+ e-Pzz ] e-Pzz :4”, + p17. cosh(p2(z’ + (1))e‘l’2‘i . dv’ (39)

This can be written more succinctly as

 

 

 

" Jr-r J (7') -fi’-”’
_ e 2 28 8 I
n”(P’)_I.L (21:):de 10162 2P2 {F }dv (40)

where F can be simplified by considering the following two cases.

a) z > 2'

For the case 2 > z’, the absolute value in (39) can be replaced by

lz-z’l =z-z’ (41)
Thus, F becomes

P2
— — er
F = e72, eP”, + e12” efl”x e—zpzd + p11, cosh(p2(z’ + (1)) [e-P" e-ZP’d + epzz] (42)
m

 

177
Combining the exponentials in the first two terms and using 13.7(32b) gives

P2
—-87

F = cosh(p2(z’ + d)) e724 2e1" + -&7,-— [e-P" e124 + e”: and] (43)
fl

Putting both terms of (43) over a common denominator, and expanding T... in terms of

exponentials then gives

 

‘92"
e 4 P2 4 2
F = cosh(p2(z' + d)) e, 21" and + e, :72: e-P2 + — :72: e",2 — £- e-P" e724
To: pl pl
P2 4 P2 4 4
+ -p—- e-Pz! 612 + 7’- ep23 6'24 - 6, €121 £12 - 87 ch: ch } (44)
1 1

Canceling, and factoring out common terms results in

 

C-Pzd P2 pd p1 -pz pd p2 “Pf
F=cosh(p2(z'+d)) —e2 (e2 +e 2)-£,e2 (e2 —e )

 

Tn pl
= 2 COMM” + d» [flmshwgfl - e, sinh(pzz)] (45)
Tu pl
b) 2 < 2'

For the case 2 < z', the absolute value in (39) can be replaced by
|z—z’|=z'-z (46)

with this, F becomes

F = e-sz’ epz' + :72“ e-P” e-zhd + flf—cosh(p2(z’ + d)) [KMI {2’24 + eh'] (47)
Ill

combining the exponentials gives

P2
—"€r
P1

T e'“"cosh(p2(z + 4)) (48)

F = 2cosh(p2(z + d)) {'3‘ {’1’} Zoosh(p2(z’ + d)

178

or,

P2
I p— - Er I I
F = cosh(p2(z + d)) e-P’d 2;": + —1T- [e-“' e724 + e” and] (49)

Note that (49) is identical to (43), with the roles of z and z’ reversed. Thus, by anal-
ogy, (49) can be written using (45) as

F=2

 

cosh(p2(z + d)) P2 J (50)

-—cosh z’)-e,sinh 2’)
T. p1 (P2 (P2

Finally, the dyadic Green function representation for the potential in region 2 is
identified from 13.7(76) and (40) as

 

 

 

 

 

1 " eff-(N2 cosh(p2(z+d)) p2
d2" , —cosh z — e, sinh z z < z'
2.2 4 (23):”. x 10362 P2 Tm m (p; ’) (p2 ')
7’ 7' = _
2.”de . —C°Sh(P22) - 8r smh(pzz) z > 2’
_(Zn) .. 10262 p2 T... p1

13.10 Green Function for the Electric Field Maintained by a Vertical

Source in Region 2.

The electric field dyadic Green function can be obtained from the Hertzian poten-
tial dyadic Green function exactly as was done in Section 14.8.

In region 1 only the horizontal components of field are of immediate interest.
Since there is only a z-component of potential in region 1, the appropriate components
of the dyadic greens function spectrum are determined from 13.8(8), 13.8(9) and
13.800) as

 

 

8261.2

~11 I_. - a

3261.2

-l,2 u
,.(z I?.I7)= ayaz (2)

The relationships can be calculated explicitly with the help of 13.9(38). Taking the

derivatives gives
"13(2 I 7’ f):-Jklp161'2 (3)
2.1% 2 I r 'J? ) = —jk,m¢.l’ (4)

Substituting from 13.9(38) and performing the Fourier inversion then gives

 

 

, e’r' v1”) . z ‘305h(P2(Z"**d ))

3.1% r | r )= (7:0,! Jk d’k—— ——(-Ik.)e"‘ T” (5)
-7? z COSh(P2(Z '+d ))

g,.’( 7' I r )= z;(—)—zild’*—— (-jk,)e"‘ T... (6)

In region 2, only the vertical component of field is of immediate interest. Since
there is only a z-component of potential in region 2, the vertical component of field is

given via 13.20) as

3211 ?
EuIrI=k22na<rI+—§;§—’- (7)

The derivatives indicated in (7) must be handled with special care, since the definition

179

180
of 6,? depends on the value of z - 2’. First, use 13.706) to write

 

H2.(r)=jG:-’(r I r'vzxr'wv’ (8)
Then,
8111
3; =33: GEIVIrvaIr )dv (9)

Obviously, the derivatives cannot be brought blindly inside the integral. This is even

more apparent when (9) is written using 13.9(51) as

 

32112, 32 , '
3,2 2,731,! [’LGW(7’IF’V2.(7)dz ]dx dy
+ai” ’[jG‘U’IF’WyhUWdz de’ dy’ (10)
where
G’('f’|‘r"’) z>z’
2.2 I_
Gu(?|P)-{G<(?I?I) z<zl (11)

Since the limits of integration in (10) are variable, the differentiation must be per-

formed using Leibnitz’ rule.

It is possible to bring the derivatives indicated in (9) inside the integral, if 6,3" is

represented properly. Let
032(rlr')=U(z’—z)o<(r|r')+U(z-z')0>(r|r') (12)

where U(x) is the unit step function

1 x > O
U“) = o x < o (13)
With this representation, derivatives on 112, can be safely brought inside the integral.
Thus, the first derivative on 112, is

an 36,?
322‘ =J 32 Jz,(?")dv' (14)

 

The first derivative of (12) is found using the product rule to be

181

2:2 r r < I I_
am. (a: I ') =U(z'—z)301;11) +0.” I rows; 22

aG’(r’|r")
32

8U(z -z')

+U(z-z') a:

+Gflf|73

Now, the derivative of the step function can be written as

.a—Ua—(fl=8(x)
x

where 8(x) is the Dirac 5-function. Thus,

aU(z'—z) _ aU(z’—z)_3_(5_-_z)_ ._
a: ‘ a(z'-z) a: "a“ 2)

30(2 -z') _ 8U(z -z') 3(2 ’2')
32 — 3(2 ’2') 3‘

 

= —8(2 - 2’)

Substituting (15), (17) and (18) into (14) then gives

31124?

32 '1=i1u(?’){U<Z’-z>aa<(?'7')

dz

 

-5(Z'-2)G<(7’ | 7")

BG’U’IT")

+U(z —z’) a:

 

+8(z -z’)G>(f" | f")}dv'

This can also be written as

3112A?)

82 =I dx’dy'IJz’( r’){- 5(2" Z)G<(r I r’) 4» 8(2 _ 2’)G>(r I rl)}dvp

xlyi

\BG<(7 I 7‘“)
z, 32

 

+ U(z -z')aG>(?zI 7”3}dv'

+jJ,,(?'){U(z'- a

The first term of (20) is of particular interest. Let

1(X'.y’ | V)= Jlu(?'){- 5(2’-2)G‘(7’ | 7")+6(2 -z')G’(T’ I V3}dz'

and consider the following two cases.

a) Integration over 2’ does not pass through the observation point z.

(15)

(16)

(17)

(13)

(19)

(20)

(21)

182
In this case, the integral property of the 8-function gives

1122(7’){-5(1'- ZlG‘U’ I r3}d2'= [1230”){50 -z')G’(7 | P')}tlz'=0

Thus,
1(x'.y' l T’)=0
b) Integration over 2 ’ passes through observation point 2.
In this case, the integral property of the 8-function gives
I(x’.y’ I T’)=Jz.(z )[—G‘(x’.y’.z I x.y.z)+G’(x.y.z I x.y.z)]
But, 13.9(51) shows that
G‘(x.y. z I x’.y’.z’=z)=G’(x.y. z I x’.y’.z’=z)
and thus
Iaiy’l?)=0

Thus, the first term of (20) vanishes, and so

 

 

emu—(_r' =J’)Iz.<r’ {Wu ”’aGfig‘zlr‘)+U(z_z,)BG’(’a’zl?')}dv,

V

(22)

(23)

(24)

(25)

(26)

(27)

The second derivative of 11;, can be calculated in a similar manner to above.

Differentiating (27) gives

321—12“ 7’)
az2

z)BG ‘(?[?’)

.ycirlrv
32 ’

+U '—
(z z 822

=Ilzx(7"){- 5(2-

> I 2 >
+5(z —z’)aG (:zl ? l +U(z —z')aG (32' 73}dv,

Substituting (28) into (7) and using (12) gives

.36‘(7’|T”)
3!

Eu(?)=£12.(7’){- 8(2'- 3:

+ U(z'—z)[k22 +%]G‘(? I 7")
Z

+5(z_z,)aG>(rIr')

 

(23)

2
32 + U(z -z')[k22 +$JGV7 l 73}d\" (29)

183
Note that the first and third terms of (29) would not appear if the derivatives were
brought inside the integral of equation (9) without using (12).

From (29), the dyadic Green function component for the 2 component of field in
region 2 produced by a vertical current source in region 2 is easily identified using
13.8(1) as

g:2=—8(z'-z)aG<(;|r’)+U(z'-z)|:k22+ +3322 ]G‘(?|T”’)

32

A3017 | 7")
I dz 3;:

+8(z—z

+U(z— 2') [k2 +— G’(F’ I 7") (30)

Equation (30) can be written in a more explicit form, using 13.9(51), assuming
the derivitives can be passed through the inversion integral. In this case, the required
derivatives become

aG<(7IT’")_ de dkz elf (7' ’3 SinhWZO-M»
(2102...

82 j (.082 T... [':_: —COSh(p 22 ') - 8, sinh(p 22 3] (31)

 

MEL “2 22%? 2’2 coshIp2Iz+dII L200 _
2’ 211(27)" 222. P2 2,. p, 222’20222) esinmpm (32)

 

2 I“ (r- r“) h d
ac (7'1? 2_ ‘(—l'2n)2”d 4.22 22 cos (p2(z’ + )) [:2: mh(pzz)—e,cosh(p,z)] (33)
1

 

82 j 0352 T,
m_ d2 22" (7’ ’7 coshtp2(z +d» _
822 (2;)2_. _Iidk j (.062 p2 T“ 1,—1th 22) 8 rSinhW 22) (34)

with these, equation (30) becomes

 

 

 

 

i? (r- 2") cosh(p (z ’+d))
2,2 _ 2k _e___ 2 _
8n (T’W')’ (21: If.” dk .itoez {RI-117." p —sinh(pzz) ercosh(pzz)
- 5(z’-z) sinh(p;(z+d)) [Pi -cosh(pzz') - e,sinh(pzz')]
2 I
+ U(z-2') [M] cosh(p2(z M» [Q 6051:0222) - ErSinh(P22)]
P2 Tn P1

 

2 2
+ U (z'-z) [#24 008,1?(“d )) [€1cosh(pzz’) " 8' Sinh(P21')] dzk (35)
2 In 1

184
The above expression can be further simplified by using

sinh(x i y) = sinh(x) cosh(y) :l: cosh(x) sinh(y) (36)
and
cosh(x :t y) = cosh(x) cosh(y) i sinh(x) sinh(y) (37)

Since 8(2) = 8(-x), the coefficients multiplying the 8-function can be combined and

written as

cosh(p2(z +d ))
Tan

11 +d
[:—: —Sinh(P22) — E, COSthZ)]- Sin (P720! )) [p—zco COSh(p22 )— 8, 81111109223]

 

 

_2 [[cosmpzz') cosh(pzd) + sinh(pzz') sinh(pzdfl sinh(pzz)

LP
T..pr

— [sinh(pzz) cosh(pzd) + cosh(pzz) sinh(p,d)] cosh(pzz’)]
— ;- [[cosh(p22') cosh(p 2d) + sinh(pzz ’) sinh(pzd)] cosh(pzz)

- [sinh(p 22 ) cosh(p 2d ) + cosh(p 22 ) sinh(p 2d )] sinh(p 22 ')]

= {—Tl—Z—z- [sinh(pzz’ ) sinh(pzz) — cosh(pzz) cosh(pzz ’)] sinh(pzd)
In 1

- ;— cosh(pzz’) cosh(pzz) - sinh(pzz’) sinh(pzz)] cosh(pzd)} (38)

Using (36) and (37) again, this can be written as

= {iflcoshu’ — z) sinh(pzd) - -€—cosh(z’ — z) cosh(pzd)} (39)
Tu: pl TM

Since the above multiplies 8(2 —2 '), set 2’ equal to z and get

= {i-p-z-sinMpzd) - -;'—cosh(p,d)} (40)
Tu pl

From the definition of Tm, equation 13.7(40), this becomes simply

185

=-1

Using the above result and making the definitions
z> = max(z, z')
z< = min(z , 2’)

equation (35) may then be written as

 

" 7412-12)
2.2 2 = 1 2 9’ _
8n (7 I 7' ) (hfjid k jam: {-5tz z')

 

 

22’ + P22 cosh(p2(2‘+d»
+
P2 Tan

(41)

(42)

(43)

[% COSh(p 22 >) - E, srnh(p 22 >) dzk (44)
1

13.11 Electric Field Green Function Summary.

From the development of previous sections, the Green functions for the electric

field which will be needed are

gxi',,1(z=0xy|z’=0’,x’,y,ic" = 3:1?”__ e-J’E’Ir-Xr')

(his. - 2.2». + (k? - k.’)p2tanhIp2d)

 

 

 

 

d2]:
1P1 +P2¢<"1‘(P2‘1)] [ErPr +P2W1h(Pzd)il
11(=0 |I_ollk) IIIej’rzl-y3
' 2 2 - 2 2 —— X
3,, z x y z ,x y =(21t)2_.. 061
—k k, [Pr +p2tanh(Pzd)]
d2}:
[m + p2cothIp2dI] [82m +P2tanh(P2d)]
g,§1(z=0, x, y | z ’=O,x’,y ’,I?)=gx‘y-l(z=0,x,y | z’=0,x’,y’, I?)
" -I’-(?-?‘)
1,] I I I = l e I
8,, (z=0.x.y Iz =0,! .y .17) (21:)2‘U._j0)81
(krzer ' k,2)p1+(k12 " k,2)p2tanh(p2d) d2]:

 

 

[Pr +P2°°“‘(Pzd)d [ErPr +P2mh(Pzd)]

”cf?“ (7- 2”) a.” jk, cosh(p2(d+z))2

d2]:
T...

 

327:1(7’lr')= Tit,”—

 

 

 

, I“? (r— 2’2 e11, ,Ijk,cosh(p2(d+z))
33% 7’ I 7’ )= (zit—TJJ— T... dzk

2 8’: (7 7”) -p I c05"‘(P2(Z’+d))
2.1%? I r )= $711422 I——--j Ice.) ' T”

, _e_—’r: 7‘)( -P z cosh(Pz(z "”1 ))
MI? I r' )= $2114" 42" Hint ‘ T...

186

(1)

(2)

(3)

(4)

(5)

(6)

(7)

(8)

187

" JP-(r-r')
guru?) _(MVIid ——jm2 {-sa 23+

2 1 h < d
+ [k2 ”’2 ]°°s (“(2 + » [flcmh(p22>)—e,sinh(p,z>) d2]: (9)
P2 Tn pl

Making the following definitions,

1 (his, - k.’)p1+(k? - k.’)p2tanh(pzd)

 

 

 

 

 

 

 

 

 

en (5’ ) - . (10)
”061 [01+ P2¢°m(Pzd) ErPI +P2mh(Pzd)]
-» ..., 1 -k3ky [pl +p2 mnh(p2d)]
gxy(k)egyx(k)' 'coe * (11)
J 1 TPI'l'PzCOtMPzd) [6,p1+p2tanh(p2d)-
-+ l (kizfir - kyz)P1+ (1‘12 - kyz)p2tanh(p2d)
gyy (k )E .038 (12)
1 I Ewmcothozdfllgpwpzmhozdfl
~ - _1 we.
gm(k)=j(mi1 e T... (13)
" =__1 ”1‘15.
gq(k)-jwele Tn. (14)
-I E 1 711-179
812 (k ) jme, e —T.. (15)
-+ a 1 1:1: "jky
8,; (k) jcoel e _T. (16)
the Green functions involving horizontal components are written compactly as
1 ~ -I r. -
8% = (22:): 113W)” ‘7 7’ a=w B=w (17)
23;} = 1 II 8:30? ) e’r'v'” cosh(pz(z+d)) B=x.y (l8)
(2n)2 ..
g3 = $1]. 8M1?) e’r' ‘7’4" cosh(p2(2’+d)) a= any (19)

188

13.12 Green Functions for an Infinite Antenna Array.

The Geometry of an infinite rectangular array of circular patches on a coated
ground plane is shown in figure 3. This is to be driven by an infinite plane wave of
arbitrary incidence angle. The incident field at each patch is then identical to the
incident field at the patch located at the origin to within a phase factor. For the p,qth

patch, the phase factor is

eik1(sin(0.-) “(95) 1, + “3(95) WK”) y.) = ejkfl'“, + VJ.) (1)
where
u = sin(9,) cos(¢,-) v = sin(9,) sin(oi) (2a,b)

and where, for a patch spacing of d, in the x-direction and d, in the y-direction,
xp = P dz Yq = q dy (3a,b)

For the center, (0,0), element, each component of both the electric field and Hert-

zian potential Green Function dyadics can be written as

ik,(x -x') e150 —n

g<rnr')=jlf(k..k,,z.zve dk.dk, (4)

Each point r ’p 4 on the p,qth patch corresponds to a point on the (0,0) patch such
that

r’p,q=p dx£+qdyy+r’0,0 (5)

or,

I

x M =p d. +x’o.o y’M = q d, + y’o,o (6a.b)

The phase of a current element at r’ '” differs from that of a current element at r‘ ’09

ih(up4.+qu,)

by e . The Green function for field on the (0,0) patch due to current on

the p,qth patch is then

190

8(7’ l VIP‘)=eih(upd,-nqd,)“' f(k..k,,z. z’) ejk,(x-pd,—x') eiso-qg-y') dkxdk, (7)

where x’ = 1’“, , y’ = y’w. The total Green function due to all the elements is then

g.(rlr’3= f: i e”““”‘“‘"

P“ q”

H f<k..k..z.zie”‘*“""“°e’5°"“”dladk, (8)
or,

81(7’I7”): i ejkivqgidky eigO‘qd)"')

q“

Li e"“‘" " } dk. f(k.. k,.z. zie”‘*“"’""°} (9)

Since the choice of the (0,0) patch is arbitrary, the above Green function will hold for

any 7’ and 7’ ’.

Each summation and corresponding integration for both 1: and y can be converted
into solely a summation using the method of Poisson summation [14]. Consider first

the bracketted term. Define
Sx 5 i ei‘lupdx I (1k,f(k,,k,.z,z')ejk’(x-Pd’—n
PM —.

d

= i ejkll‘pd, I dk, fx(kxtky,zr ZI) e-jkxp x (10)
PH —.
where

f'.(k..k,,z.z')sf(k..k,,z.z')2‘4"” (11)

Recalling standard Fourier transform theory,

F(w)= mee'i‘" 4: (12)

191

f(t) = 71; JFuo) em do) (13)

Making the correspondance k, —> t and p d, a 0),

mp d..k,,z.z'>=ldk. i.<k..k,.z,.r).-i*.~. (14)
where

mm,19,2,zu=l},a,k,,z,z')e-iw a: (15)
Then

s. = i e"“‘"‘*r'.<p d..k,,z.z') (16)

p”

The method of Poisson summation is now applied. For any function f(¢) and its

transform F(w), [15],

i f(t+pT)=-71: i e’P“°'F(p (no) (17)
PM P”
where
_ 21:
T -— T). (18)

making the correspondance d, —) m, and k, u —> t, equation (16) becomes

21:

5,:—
dz»

“‘4'

fx(klu+p%9ky,zvz') (19)

From (1 1), this becomes

. 21c
1(k‘u+p—)(x-x’)
5x: f(k1u+p-§-1-t-,ky'z,z')e 4'

I

(20)

Mg:
m

P

Using (20), equation (9) then becomes

83(7’I7')= i efiivqgj'dk’ejso-qs-y')
1" ~00

192

1'0, u +I g:- )(x -x')

-d—n z f(k1u+p-%- ,k, z, z’)e (21)

JI‘11-"-

Bringing the final summation through the integration and first summation, this becomes

.. j(k,u+p -3i‘-)(x-x')

gammy? 2e

1 pa—o

i efllquy Idk,ejk’o-qd’-nf(k1u+p-3-1£,k,,z,z’) (22)
¢""’ "° 1

Now, let

“ kv 21: , 'k - —
Sy=£e e’l qd’iflyf(klu+p7:tgky1,2)e”o 1% 3')

q”

= i e"‘”‘" I dkyfwcluw $412,095“ (23)

q"

where

f‘,(k.u+p-f;;-1t .k,, z z')= “haw-335 k,, .z')e"'5""° (24)

1

Equation (23) is now in the same form as (10), and following a development similar to

that for 5, yields an equation which corresponds to equation (20),

.. 2_1t 21: i(h"+P%)0‘y')
2f(k1u+p— d “1.,“qu .'z)e (25)

.2_“.
’ d...

Substituting this back into equation (22) yields

X

. 2x
a- 1(k1“+P—)(3-1')
g.(rlr')=-:lx[2e “ ]
x P

.. 1(kV+p3'-)0-y‘l
2E2f(k1u+p:-—:t,k1v+q%zz')e ‘ ‘5 (26)
an .

Rearranging the summations gives

193

2

,7? =—l—(2" ' '
3(1’) $4,312..

21: . 21:
1(hu+p—)(x-x‘) 1(k1v+p—)0-y')
f(k1u+p-:—:t ,,k1v+q%zz')e 4’ e d’ (27)

If a single element Green dyadic component is of the form of equation (4), the
corresponding Green dyadic component for an infinite array driven by a plane wave is
of the form of equation (27). By inspection, a simple prescription for obtaining the
infinite array Green function from the single element Green function is

JIdhdk, algl‘L z (28)

d) pas—c-

and

k, —> k1u+pfi1t

 

k, —> h u +p % (29a,b)

From the summary of the previous section, the Green functions for the electric field

which will be needed are

 

1" ’=0 ’ ’7? = l ______e x
gn(z=0.x.y|z .x.y. ) H jam:

("12 8r - k,2)p1+(k12 - k,2)p2tanh(p2d)

 

 

 

d2]: (30)

X1191 + P2C0th(Pzd)] [€rP1 +p2tanh(p2d)]

fi’ (r- 7')
g,,l(z=o, x, y lz’=0,x',y ',=?) (2;)2jj—Tx
-k.k, fin. +P2mh(P2d)]

x * d’k (31)

[121+ momma] arm +pztanh<pfl
gylx’l(2=0, X, y I z'=0,x,y ’,k )=g,;'l(z=0, Xg)’ I z’=orx,1y’a F) (32)

e -iF (r— 1")

TAIT x

(061

#(z=0,,xylz’=0,x’,y 3F):

 

194

("12 8r - ’55P: + 0‘12 " k,2)p2tanh(p2d)
h + P2°°th(Pzd)][€rP1+szh(PszJ

e:f£’ (7-1”) e1.“ jk, cosh(p2(d+z)) d2]:

 

 

 

 

 

8&‘(7 I F”): (211021“! wet T...
g&‘( r I r') = Zulu—FIIL%2 ,w'ikycoshgzwnn 42k
8;.2( 7’ ' 7’ ') ' (220—71 I 42" kit—(Eleanfll' WS“IP;:"+"”

 

2.2 ,_ 1 ~2___7_:). -
gu(?|?)—(2n)21Idk 1.032 {—60 2’)

1:22 +12,2 cosh(p2(z< + d))
4.
P2 T»:

where
z> = max(z ,z’) z< = min(z ,z')
Letting

—. , 2 .
k=(k1u+p %:—‘n+(k1v+q in
and using (28), the above Green functions become, for an infinite array,

3.3;“ =0, | =01?) .. i‘ ”PW-”
z x, 2 Jay = -—1— —..—_—. x
y dxdy P... P... 10181
(klzer - kxz)p 1 + (k12 - k,2)p2tanh(p2d) 7

[P1 + Pzmeflq [firm + szh(P2d)]

 

- ejP- (r- r')

11 I I __ _—
,,'(z=0,,x y lz’--0,x,y, ’k)-d,l Ping-”e [(061 x

[pf COSh(p22>) '- 8 ,Sinh(p22> ) dzk

d2]: (33)

(34)

(35)

(36)

(37)

(33)

(39a,b)

(40)

(41)

195
x , -k.k,Ia1+pzmh<pzd)]
[Pl +pzcoth(pzd)] [arm +P2tanh(Pzd)]

 

eyi"(z=0. x. y I z’=0.x'.y’. ’7 )=g.;-1(z=o, x. y I z'=o.x'.y'. I?)

ZZ-—.——><

p... qua-co 10391

 

3,}‘(z=0.x.y l z’=0.x’.y’.l7)= d
X

We. — mm + (k? - common)
[Pi +P2°0th(P2d)] [3P1 +p2tanh(p2d)]

 

1 .. .. eff-(1L?) _P ,Ijk,cosh(p2(d+z))
.i"(?|7”)=— —-—“
g d! P2 49—0. 1‘08] T”

l .. .. div-7") em,vjk,cosh(pz(d+z))

 

 

8.3"(7 I ?’)=

 

 

 

 

d" d? p=-- q-—~ j 0351 T,
832(7’IV')=-2‘1—P§‘:£;1:’T:;2{—8(2 —z')+

 

 

+ [1‘22 +p22 ] cosh(pz(z‘ + d»

P2 .
—cosh 2’) - e, smh 2’) d2]:
p; T. [p1 (P2 (P2 J

Making the following definitions

1 (1:126, - k3»: + (k12 - khpztanhovzd) ,

g;(F)Ej(0€1dxdy kl+pzcothmd)][€rp1+sz(P2d)I

1 7 -k,k, [P1+P2 tanNPzfl]
jmld‘d’ P1 +P2¢0¢h(P2d)j| [firPl +p2tanh(p2d)]

 

25<FI=g,:(F)=

1 (1:32, - 190p: +(I'c12 - 19012291111026! )

g " ( P) F- J-(m,:ldx(1y [p1+p2cod1(p2d)] [8,111+ pztanh(p2d)]

(42)

(43)

(44)

(45)

(46)

(47)

(48)

(49)

(50)

(51)

(52)

196

no ‘9 5 1 711,1}:-
gu (k ) —jm,d,d, e T... (53)
- . __1 ‘71"!5.
8» (P ’ mm, ‘ r, (54)
g - < I?) :- —-‘— .121". (55)
n jmldxd) Tut
gm? I ——1—— ""'——’k’- (56)

a . c
J (midxdy Tm

the infinite array Green functions involving horizontal components are written com-

pactly as
3.11 = I)?” germ/23 ei‘” 0*?" a = x,y B=x,y . (57)
g. 1 = i 19.3 (I? I e!" ‘7’") cosh(p2(2+d)) B=x.y (58)
”=-
8.13 = i sac?) e114”7 cosh(pz(z'+d» a = m (59)

P4”

14. Current Distribution Derivations.
14.1 Patch Current Distributions.

An arbitrary surface current on the patch is to be modeled as a summation over a
set of current distributions which are independant and which form a complete set. The
current distributions used are based upon a two dimensional Taylor expansion of two

orthogonal current components.

A scalar function w(x,y), continuous and with all derivitives continuous over the

region of interest, can be represented as a power series in x and y. Let

A” 5 v0.0) 02!)
A" = $5!wa A’ 2 €15!wi (1b,c)
A": git-”M A72 = gimw (1d,e)
A" I. 5355!.“ (10

Since w(x,y) and all of its derivatives are continuous,

fl} _L‘ILI (2)

axay lxty-O - ayax lxsy=0
In general then,

I
A,.,. = am}; I (3)

ax'ay” |8=y=0

where the order of differentiation with respect to x and y makes no difference, again

because w(x,y) and all its derivitives are continuous.

The power series expansion of \y(x,y) is then

\l’(x.y) = A" + A’x + A’y + I‘lxzx2 + A"xy + Ayzy2 + + A""”Jc"yP + (4)

197

197a
Converting this to polar form using

x = r cos(¢) , (53)
and

y = r sin(tI (5b)
w(x,y) becomes 9

mm) = A" + A’rcos(¢) + A’rsinmI) + A”r’cos’(¢) + A” r’sin(¢)cos(¢) + A’zrzsinzm)

+ + A""r""cos" (¢)sin’ (o) + (6)

w(r,¢) is to be represented as a summation over terms of the form
Bull r"! C)". (7)

where m is a non-negative integer and l is any integer. 3"" is the amplitude

coefficient and is in general complex. Then

W(r.¢) = 2 f, B”r"'e’“ (8)

03:0 la-c-

Many of the terms in (8) are not needed to represent w(r,¢), and so the
corresponding coefficients are zero. By comparing the two expressions for fine), (6)

and (8), the coefficients 8"" which are zero can be determined.

Since the various powers of r within each representation of w(r,¢) are independent
radial functions, equating expressions (6) and (8) yields independent equations for each

power of r. Thus,

A" = f; 30' e1“ (9)
I”
A‘rcos(¢)+A’rsin(¢) = f; B" r a!“ (10)
I"

A’zrzcoszw) +A"r’sin(¢)cos(¢) +Ar’sin2(¢) = i B” r2 a!“ (11)

la-oo

198

and in general,

gAxl’u-lcosu(¢)sinfl-I(¢)rfl = i Bu! rm a)". (12)

(H

Further simplifications are possible. In each of (9) through (12), the summation
over I needn’t go from minus to plus infinity. By inspection, (9) can be reduced to

B°-°=A° ; 3°“ =0 no (13)

following from the orthogonality of the exponential terms. The left hand side of (10)

can be rewritten as
1.. ' Y 3 ‘ 7
A——21A— (cosm + jsin(¢))r + iii/L (cos(¢I - J'sin(¢))r (14)
Using Euler’s identity, (14) becomes

x_- y . x ' y . " .
A—ZA- e"r + -A—;fl—e”’r = 2 B” r e". (15)
I”

By comparing exponential terms,

31.1: 3.11:1“; 31.-1: M (16a b)
2 2 ’
31" = 0 I ¢2t1 (16¢)

Continuing with (3.5.11), the sin and cos products may be written as

cosz(¢) = EEQZQKI- (17a)

COS(¢)sin(¢) = w (17b)

sin2(¢) = 31% (17c)

By inspection, when these are written as a sum of exponentials, e1“, 1 must be 0, 2, or

-2. Thus

=0 l=0,i2
32.1{=0 else I (18)

199

In general, the coefficients which are non-zero are

at 0 I = —m, —m+2, ...., m—2, m
B " I- o .15. (19>

w(r,¢) then becomes

w(r,¢) = 2 i 8'"-I r" em l+m even (20)

III-0 ln-m

Let I?(r,¢) be the current which is to be modeled on the patch. This can be

represented using two scalar components, K, (r ,4») and K,(r,¢),
I? (r .¢) = K. (r mi + 190.01. (21)

The current components K, (r ,¢) and K, (r ,(to) will both be continuous over the surface of
the patch. Both can then be represented as a sum of the form of equation (20). It is

desired to obtain 1? (r ,o) in the form
I? (r .4» = K.(r.¢Ir“ + K.(r.¢)i. (22)

The two components above, however, can’t be represented using (20) since they are
both discontinuous at the origin due to the discontinuity there of f and 6 in the polar
coordinate system. Thus, K,(r,¢) and K,(r,¢) must be obtained via rectangular com-
ponents. Conversion from rectangular to polar components is made easier through the

use of a third set of current components. Define

fizz-III“ : (rat-Ir (23a,b)
then

mm) = K, (r .4013 + K.(r.¢)¢i (24)
where

K: (f,¢) - jKy (f,¢)
2

 

K, (r AI) = (25)

Kx (r J?) + 179 (r .¢)
2

 

K, (r .0) = (26)

200
The components K, (r ,o) and K, (r ,o) are continuous and can be represented using (20).

Doing this, writing the summation in long form,
Kp(r,¢) = P°'°+ [’1'1 re" + Pl"1 re." + P03 r2 + P23 723‘”. + P2.—2 ,2e-21'0 + . . . (27)

where P“ and Q“ are the amplitude coeficients corresponding to 8"" in (20).

Clearly, the current components in (22) can be found as
K.(r.<I>I=I?<r.¢I - f =K,(r.¢Ip‘ -f +K.(r.¢) <1 I (29)
K.(r.¢I=I?(r.¢I -ci=K,(r.¢Ifi -<‘I+K.(r.¢I 4 -i> (30)

From the definition of p and 4, equations (23a,b),

P 'f = 00801)) + jSin(¢) = 9". (31a)
15 ' 6 = -sin(¢) + jcos(¢) = je" (31b)
13 ° r‘ = cos(¢) — jsin(¢) = 42‘" (31c)
:3 - r‘ = -sin(¢) - jCOS(¢) = -je"" (31d)

The simple exponentials obtained above are the sole reason for using K, and K, rather
than K, and K,, which give sin(o) and cos(¢). Substituting equations (31) into (29) and
(30).

K.(r.¢I = K,(r.¢I e" + K.(r.<II e'" (32)
K .(r 0) = .iKp(r.¢) e" - 1K, (N?) e’" (33)
Substituting the expressions for K, and K, into the above,
K,(r,¢) = P°'° e" + Q” e’" + P"1 r :2" + 1""1 r + Q"1 r + Q1"1r e4"
+ 1,2,0 r2 e" + [’2'2 1'2 :3" + P2"2 r2 e‘"
+ an ,2 e-jo + Q22 ,2 en + Q2.-2 ,2 e-m + . . . (34)

K,(r,¢) = jPo'o e" - on'o e’" +1?” I e2” +jP1"lr -jQ1'1 r — le"l r e'zj‘

201

+ jPz'o r2 e" + jl’z'2 1’2 e3” + 17’2": r2 c‘”
_ 192,0 r2 e-jo _jQZJ ,2 e" _ jQ2.-2 r2 e-ajg + , . , (35)
Combining like terms and renaming coeficients,
K,(r,¢) = R0" e" + R0"1 9’" + R"0 r + R"2 r £2” + I?"2 r £4"
+ [12'1 r2 e" + R""1r2 e'" + Rz'3 r2 :3” + R’"3 r2 (3" + ° - - (36)
K.(r,¢) = ¢°J e" + ¢o,-1 e‘" + <b"° r + CD” r e2" + $1": r (2"
+ c” r2 e" + 49-" r2 e'" + ¢7~3 r2 em «I d?” r2 (31" +- - - (37)

Again, R” and <b”'-' are amplitude coefficients for the various terms in (36) and (37).

By comparison of (36) and (37) with (34) and (35),

o“ = jR°-‘ <b°-" = —jR°-‘1 (38a)

(1)” = jR 1-2 c“ = -jR H (38b)
and generally,

0....“ = th-n“ 4,...“ = —jR"""" (38c)

From the expansions of K, and K., two facts may be observed. First, for odd
powers of r, the azimuthal variation is even, and for even powers of r, the azimuthal

variation is odd. That is, both are a summation of terms of the form,

r’” e!“ (39)
where m+l is odd. The second observation is that, for a given value of m,

|l| - 1 s m (40)

The currents K, (r ,4» and K,(r,¢) then become

0- n+1

K.(r.¢I= 2 2; R" r" cf" (41)

Ill-0 la-m-l

0- n+1
K,(r,¢) = Z Z 47"" r" em (42)

Ill-0 ID-m-l

202
with R“ = 0 and 0"“ = 0 if m+l is even.

It is desirable to modify equations (41) and (42) to a better form for the problem
at hand. In particular, the radial variation will not be modeled as a power series.
First, in place of r", the Tchebychef polynomial T... (r lb), where b is the patch radius, is
used. This has a leading term of r", but also has terms of r"‘2, and 1"“ down to a
term with r1 or a constant term, depending on whether m is odd or even. This
modification is made so that, after a second modification is made, integrals involving

the currents which will be obtained later can be easily evaluated in closed form.

The second modification is a multiplicative factor which forces the known edge
conditions, J, = 0 and J, singular at the edge of the patch. The multiplicative edge fac-

tors are,

1 - {:2- (43)

for the radial current, and

(44)

for the azimuthal currents.

Since the Tchebychef polynomials are even or odd about the origin, and since
both multiplicative factors are even about the origin, their product is even or odd about
the origin as well. When K,(r,¢) and K,(r,¢) were expressed using powers of I, an
even power of r was associated with an odd variation in ¢- This property is preserved,

so that (41) and (42) are replaced with

K,(r,¢)= i i R“ T,(r/b)ei'° rug—2- m+l odd (45)
"‘5‘...

K,(r,¢)= i f; or! T,(r/b)ei't ——1— m+l odd (46)
Ina-01H r

III-H odd b2

203

Since every polynomial T..(r/b) contains either a constant term, or a term which

varies as a single power of r, the restriction II | — 1 s m is no longer valid.

As a further step, the identification is made

KL... = f K”, = T..(r/b) e!“ 1- f,- f
1?». = i K“... = 1.0/12):!" 1 i
r
"7,?
CH,” =Rm'I
C01»! =¢MJ

The expression for the total current I? (r ,o) on the surface of the patch is then

mu (H

KM) = 5: i [C.I... 1?... + Ca... 1?“... }

where the coefficients C». and C,,,,. are to be determined.

(47)

(48)

(49)

(50)

(51)

 

 

14.2 Feed Pin and Singular Current Distributions for Single Patch Antenna.

The current on the feed pin is assumed to be constant over the surface and
flowing in the z direction, so the surface current distribution describing this current is

chosen as

12:12: -ds:s0 (1)

The assumed current on the feed pin leads to a build-up of charge at the patch-
feed pin junction. Physically, the current flowing on the patch from the feed pin will
be continuous, with little charge build-up. Mathematically, it is necessary to have a
patch surface current distribution which cancels the current discontinuity caused by the
feed pin current in order to insure convergent integrals. The singular patch current

associated with the feed pin is written as I?,.

It is convenient to use a feed pin-centered coordinate system to describe the
"singular" patch current distribution. This is shown in figure 47. The feed pin is cen-
tered at 7:, relative to the patch coordinate system, which has its origin at the center of

the patch. A point located at i" is located at
357-055 fr (2)

in the feed pin system. It is desired that this current distribution yield evaluable
integrals upon the application of Galerkin’s method. The integral in question from

equation 3.6(30) is

28 '3
0:1“. I drp r. 3.0.) e*"’""”’°’ (3)
Let
R 3minIR¢(¢p) ] =17 - Vol (4)

and assume current flowing radially away from the patch-feed pin junction, so that

204

205

 

Figure 47. Feed pin-centered coordinate system.

206
K®I=fpmen=eflm (5)

Since f(r,) is not a function of 0,, it must be enforced that
f(r,)=0 r, 2R (6)

Equation (2) can then be written as

21: R
.p . _
I? = et’ 1;, i d¢p fr Idrp rp f(rp) em" I.” a) (7)
C

It will be required that I, f(r,) satisfy four conditions, namely that

f(a)=l (3)
f(RI=0 (9)
V-f,f(r,)r “=0 (10)
V-r‘,f(r,)r =R=0 (11)

The function f(r,) will be constructed as a power series in r,. The dominant behavior

in the vicinity of the junction is expected to be = ,i’ so the lowest power of r, used
P

will be -1. There are four conditions to be satisfied, and only odd powers of r will

P

lead to closed form evaluable integrals, so construct f(r,) as

a

 

3 5
f(r,)=A flu; '7'+C [55] +0 55-] (12)

P
where A, B, C, D are constants to be determined.
From equation (8),

A+B+C+D=1 (13)

and from equation (9),

3 S
A 73-” §+c [1}] +0 [L] =0. (14)

207
Taking the divergence of If (r,) gives

f (r,)

'I

V'(r‘f(r,))=-;L

P

 

Ed: (r, Ha.) I = +f’(r,.)

Using (9) and (15), (10) gives
, - -_1
f (a) - a
while (11) gives
f '(R) = 0
From equation (12),

, —A a 3 3C r,2 SD 1',“
f('p)- r,2 +a+ a3 + a5

 

 

so equations (16) and (17) become

—A +B +3C+5D=-1

2 2 4
—A [1%] +3 + 3C [é] +51) [i] =0.
Solving (20) for A,
A = 1 + B + 3C + SD
Adding (13) to (19) and solving for B gives
B =—2C — SD
so (21) becomes
A = l + C + 20
Using (22) and (23) in (14) and (20) gives

5

R

3
(1+C+ZD)£-(2C+3D)£+C [—] +0 5
R a a

a

 

 

and

=0.

(15)

(16)

(17)

(13)

(19)

(20)

(21)

(22)

(23)

(24)

208

2 4
'] + 50 [L] = o.
a a
Collecting terms, (24) and (25) become

2 4 2 6
[HI-1%] + [is] 16+ [HE] + [£1 IN
2 4 4 6
[-1-2[€-] ”[15] ]C+ [44%] +5[ ]]D=1
Adding equations (26) and (27) gives

[Iii-1141211“I-6{%]‘+6[%I‘ID=°

2
C=’—3 [5] D
a

2
-(1+C+ZD) [1%] —(2C+3D)+3C[

and

clan

01'

2

Using (29) in equation (26) gives
2 4 6 2
Hi] +615] $14-45] ”I
a a a a

l

D: r R I R2
—] —6[— +9—] -4
a aJ a

t

elk:

1°va

01'

 

k

 

 

 

Using (31) in (29),

2
-3 ‘1:
a

ISI‘-6I%I‘+9I%I’-4

Equation (22) then gives

(25)

(26)

(27)

(28)

(29)

(30)

(31)

(32)

209

{-21-631393114

 

 

A=l+ ' '5 ’ '4 2 (34)
If] 4%] ”BI -4

The four coefficients A, B, C, D are now known. Substituting equation (3.106)

into equation (6), 1’:t becomes

_ R 2 4 6 .
frui’mdcp, 1‘, jar, a [A +3 [55-] +C [21] +3 [.51] ]e*""r°°‘""°’ (35)
a a a a

or, interchanging the order of integration,

21:

. R 2 4 6 '
7:: = 8:1??0 J’ dr, [A + B [:5] + C [IiL] +D [LL] ] g a d¢p fp 8:1:le may '0) (36)
a

a

The unit vector f, can be written as

f, = 2 005(4),) + y“ sin(¢,) (37)

or
f, = 11 [cos(¢,—0) cos(9) + sin(o, —9) sin(e) ] + (38)
+ y‘ [cos(¢,-6) sin(0) - sin(o, —6) cos(6) ] (39)

The terms with sin(o, — 6) give zero after integration [16]. Using

21:
I efim‘mcosor 0) d6 = 21:(ii)" 1,,(2) (40)

the terms with cos(¢, - 9) integrate to give

1'}: 3".” ( It cost) +y‘ sine ) x

210

R 2 4 6
xjdr, [A +3 [55-] +C [53—] +3 [1] ](:tj21:a)J,(kr,) (41)

a

The integrations over r, of even powers of r, times J ,(b,) now must be obtained.

Two useful formulae are, from [16] ((11.3.20) and (11.3.21)),

1:

] 11(1) d1 = -Jo(b) + Jo(a) (42)
and

b

[:2 11(1) dt = 32 12(3) - a2 J,(a) (43)
Using (42), the term in (41) with coefficient A becomes

R A a

A I11(kr,)dr, =V [10(ka)—Jo(kR)] (44)

For the feed pin-patch terms, it is useful toelirninate J o in favor of J; and J, with
2
10(1):? 11(1)‘ 12(1) (45)

so (44) can also be written as

3
A I11(k'p)d'p = 32);“. {'22: 11(k0)-Jz(ka) -% 11(kR)+Jz(kR)] (46)

Using (3), the term with coefficient B becomes

R 2

l' B a
3 £23105)“, =

Um I3

 

[k2 3’ 120.3) — k2 a2 12(Ica) ] (47)

For the feed pin-feed pin term, it is desired to have this in terms of Jo and J 1. Using

(45 ).

R 2
3j-::—211(kr,)dr, =71]??? [2m Jl(lcR)—k2RzJo(kR)—2ka Jl(ka)+(ka)zlo(l:a)] (48)

The terms with coefficients C and D will now be obtained via integration by parts.

211

b I”
[1411(1) d: = —:‘ J,(:)

b
+ j 4:3 10(1) dt
a

01'

" r
It‘Jlm d: = —t‘Jo(t)

6
+4]:2 [110(1)]:1:
Another useful identity for Bessel functions from [16] (Eq. (9.130)) gives
d
‘Jo(l)=z [I 11(1)]

so equation (50) becomes

5 b b
I :‘J,(r) d1 = [4410(1) + 412(111(t))] — 8 I 2211(1) d:
Using (43) then gives

b b
[1411(1) d1 = [-t4Jo(t)+4t311(t)— 812120)]

or, in terms of J, and J;

b

a
[1411(1) dz = [2:3J1(I) + (t‘— 812)Jz(t)]

In terms of Jo and 1,, this becomes

11

b
It‘J1(t)dt = [(—z‘+8:2)10(z)+(4r3- 16: )Jl(t) ]

Following essentially the same path,

b lb 5
[1611(1) dt = -t°Jo(r) + 6It’ 10(1) d:

b b b
[1511(1) d1 = [4610(1) + 61‘ ( t 11(1)) ] - 24I1‘11(t)dt

Now equation (55) can be used to obtain

(49)

(50)

(51)

(52)

(53)

(54)

(55)

(56)

(57)

212

b b
[1611(1) d1 = [—I6Jo(t)+ a‘J,(:)+(241‘—19215111049613 — 384: )J,(:) ] (58)

or, using (45)
b b

[1611(1) d1 = [(415 - 4813 )J,(:) + (:6 — 24:4 + 192 :2) 12(1) ] (59)

Using (54), the term with coefficient C becomes

 

R r,‘ C a 3 4 2 R
C [711(krpIdr, = (11.15 [213) lawman) -8(kt) ”2130]“ (60)
or, using (55)
R ’1: C a 4 2 a R
C {:7 11(kr,) dr, = Tia—)5- [(-(kr) + 8k: )Jok(t)+ ( 4(kr) -16kt)11(kl)]‘ (61)
Using (58), the term with coefficient D becomes
R '36 D a 6 I 4 2
D ‘I-a—‘s Jl(kr,) dr, = W [(-(k1) + 24(kt) -192(kt))Jo(kt)+
R
+ (6(kt)’ - 9603? + 384(k1)) ] (62)

or, using (59)

R 5 R
D] £16- Jl(kr,) dr, = (11:21; [(40.05 - 48(3)’)J,(Ia) + ((3)6— 24(3)‘ +192(kt)2)12(kt)] (63)

 

Equations (46), (47), (60) and (63) give all the integrations needed. Equation

(41) can then be written as

. R
2 ”mm cosO + y sine) [f1(kt)11(kt)+fz(kt)12(kt) ] . (64)

7?=:l:j21:a e

where, using (46), (47), (60) and (63),

 

 

 

 

f1(kt)— [m (kt) +2 (ka)5 (kt) + (k0), (4(ktI 480“) I (65)
A 3 C D
f2(kr)=—ka + (kc), (102+ (k0), ((3)4- 8(kt)2)+ (k0), ((kr)‘-24(kz)4+192(k:)2)(66)

213
Equivalently, (41) can be written as

 

 

. _ _ R
i} = :tj21ta2 an?“ ( i cose + y‘" sine) [foam 100:1) +f1(la) 1,0“) ] (67)
where, using (44), (48), (61) and (62), ~
~ _ _ A _ _§_ 2 _ __C_ 4 _ 2 _ D 6 _ 4 2
fo(kt) — kg (ha), (kt) “5((10) 80“) I (“)4 (kt) 24(kt) +192(kt) ) (68)
f',(kz)=—B—2k: +—£— (4(kz)3— 161a )+ D (6(kt)5—96(kr)3+384kt) (69)
(kaI’ (kaI’ (kaI’

14.3 Feed Pin And Singular Current Distribution for Patch Array.

The current on the feed pin is assumed to be constant over the surface and
flowing in the z direction, so the surface current distribution describing this current is

chosen as
1?, = 12 —d s z s o (1)
The assumed current on the feed pin leads to a build-up of charge at the patch-
feed pin junction. Physically, the current flowing on the patch from the feed pin will
be continuous, with little charge build-up. A current distribution term is developed

which has a near singularity at the feed pin junction that matches the known approxi-

mate current flow in the vicinity of the junction.

It is convenient to use a feed pin-centered coordinate system to describe the
singular patch current distribution, denoted K2. This is the same as shown in figure 3
in section 3.6. The feed pin is centered at 7:, relative to the local patch coordinate sys-

tem, which has its origin at the center of the patch. A point located at 7’ is located at
72=7’—75=’p "p (2)
in the feed pin system. In particular, the edge of the patch, at I? | = b, is located at
kiwi-mar; (3)
From (3), R,(¢,) may be determined via
b2 = IR’. + 7:. I2
= (R.cos(¢,) + x. I2 + (R.sin(¢.I + y. I2
= R‘zcoszmrp) + 23,x,cos(¢,) + x,2 + R.’sin’(¢,) + 2R.y.sin(¢,) + y} (4)

01'

R.2 + ZR. moose,» + y. sin(¢p )) - b2 + x} + y.” = o (5)

214

215

80

RI (¢p ) = "(xo COS(¢P) + yo Sin(¢p ))

 

+ V<x.cos(¢,) + y.sin(¢,IIT+ 12’ - x} — y.’ (6)
where the + sign is chosen on the radical since R, > 0.

Five properties are desired in the choice of I?,(r;,) :

1) II?.(r,. =aII = IKI

2) For r, small, l?,(?;,) = :—”

P
3) Forff, =31,1?,(r;)-r* =0
4) Continuous and smooth for a < n; | < R,
5) Integrable in closed form over patch.

The first property simply provides that current flowing on the feed pin surface
flows entirely onto the patch. The second property guarantees that V - 3,0,) = 0 for
small r,, which, along with the first property, ensures no charge build-up near the feed
pin junction. This form of current is expected physically. The third property enforces
the condition that normal current flow is zero at the edge. The fourth is demanded by

Maxwell’s equations, which are continous and continuously differentiable to all orders.
For the fifth property, the integral in question is, from Eq. 3.6(30),

2! R —o

w I III. I dr. 381110;)!” B=x.y (7)

where the exponential term comes from the Green function. Closed form integration is

desired since it lessens the computation time needed.

The choice of 11(3) used for the infinite array is

a 1° cox-22%)
1?.(rI= ” —‘ (8)
P r 1ta
” ”St-2W

 

216
Using :13 = r, dr, do, and r" = 17, + 11,, (7) becomes

 

 

28 RC
_ 1 I"? 1 1V - -r
I,p=£d¢,f,-B 1w 2’ °I§cos[2—R%]e’r'dr, (9)
cos — "
Now, define
R. 1 1w
_ -jb com -0)
lk=£dr,§;cos[-2F;-]e ”
R .0 .p
. ,_L ,_L ,
= I dr, —1- [e 23‘ +9 M‘ Je—Ibm“'-°)
a 41:
Re I" {-L—rma -e)] r [AL-rm. -e)]
= I d’p 71,; [9 2R. P +e 2R. P (10)
where
0 = tan“1 [‘21—] (11)
This is evaluated to obtain
1‘ = -jR, e11; - memo, - 0)) _ e112?" ’“ ”‘9; '9’]
R 21:0: — 2R,l: cos(¢, - 6))
JR. [ —j(-’2-‘- + moose, - o» “2% + "‘ ”5“» ‘ 9”] 12
- 21:(1: + ZRJ: cos(¢, - 0)) - e ( )
with (12), (7) becomes
28
P ’ .-
1;, = 2’ 7° 1 34), f, -B 1M I. (13)
COS -2-R:

Since R, is a complicated function of 0,, if 1?, to go to zero only at the edge of
the patch, the o, integral will likely be difficult or impossible to do in closed form.
Property 5) is thus relaxed to performing the azimuthal integral to be obtained

217

numerically.

For most of the range of 0,, (13) can easily be computed numerically. At

 

 

1: = ikcos(0, - 0), 1; becomes indeterminate. Near these values of 0,, make the
definition
1:b
z a 2R, :1: kbcos(0, - 0) (14)

Using the minus sign above, the first term of 1; becomes

 

l [ejk‘ub - 43"] 2 small (15)
Z
'R, R, , 2
:-:— [1+Jbz-(2bzz)+....-l-jz+z7+....] (16)
~ . RI 2 R02
- [,(7_1)_3(?-1)+....] (17)

If the plus sign in (14) is taken, the second term from (13) takes on the form shown in
( 15) through (17) as well. Thus (13) is well behaved for all values of 0,.

VI. MATRIX ELEMENT EVALUATIONS
15. Single Patch Matrix Elements.

15.1 Patch-Patch Matrix Elements.

The matrix elements relating the current distributions on the patch surface to the

tangential electric fields on the patch surface are given in 3.6(27) as

(2;), I] d’k 05.31;)... (I?) 131.11 (02.811?)
fin.)

 

Zwm’ =

where gap (1? ) are given by

1 (kie. - kinI + (k12 - kxz)Pztanh(P2d)
“”81 [Pl +pzcoth(pzd)] [arm +pztanh(pzd)]

 

gn(?)5

 

I: - r - 1 """r [”‘+’”mh(p’d)] ~
gly( )=g,,( )gjwel [pl+p2°0th(P2d)] [8,p1+p2tanh(P2d)]

1 (1:126, - 19031 + (k12 - k,’)pztanh(Pzd)

3” (I?) g 1081 L91+pzcoth(pzd)]F'p‘ +p’mh0’2d)I

Expanding the sum over a and [3 then yields

231%.: 1 [[de 1

(21!)2 _. jmel (P1+ P2 coth(p2 d)) (5r P1 + P2 tanth (1))

 

 

x { [act2 e. — 1,2 )PI + (In2 — kEIpz mnhwzd) Lam/201511.84?)
+ [(1:12 e. — k} )m + (1,2 — 1,2) m tanhIpzd) ] 11.,(F)131.1,(E’)
— 18./WI + p: tanhcpzd» [121?)13'2212’) + 111,? I 181:...(2’) ] }

218

(1)

(2)

(3)

(4)

(5)

219

01'

23157... =

 

111? Ii de 10381 (P1+P2 c01h(P2 ‘01)“; P1+P2 WIMP: d))
x{ [P1+P2 whom] [43 12,1?”me I — 1,2 11,1?)13035’) -
- 1,], 122112013114?) + at, 1,11,? I 132.11? I ]
- k? (8, 31 +32 tanhovzd» [lfim(l?’)la'mv.(l7)+l;‘..fl?)la'..',(?) ] } (6)

Equation (6) will now be expanded for each case of 8 and 7.
CASE I : y=r, 5=r.

expanding 1,3,, and 1,1,, using equations 16.1(15) and 16.1(20)

1,1,, = 1: 1" cl" [ j e19 3,11,, — j w" 3,21,, ] (7)

1,1,, = 1: j' e!“ [e10 3,11,, + r” 3,5,, ] (8)
The four possible products of (7) and (8) are

13...; [J'm'x = 1C2 1"“, ei('+n° {-3219 8111.»: Bit-+1.1!“ - e—2je Bit-1.»: 817-1!“

+ 3111;.- BF-m’ ‘1' 31:1,». BI7+1,m' } (9)
1,3,, 1,7...I, = 1:2 ft” ei('+"’° {em 3,11,. 3,7,“... + e‘m 3:11,, 3.7.1,...

+ 3111,». BF-m’ + 31:1... 3111M } (10)
[rim [J'm'y = 7‘2 J'M' ejtm'IO {1'52” 3111,»: Br‘lm’ — IVE-2’0 31:1,». 317-15'

+ f 3111,». BF-m’ ’1 31:1” 317+11u' } (11)

220
may ”on = "2 I'M, ammo {km 3111,»; 317+11n' ‘18-sz 31:1,»: BF-m’
- .i 3111,». BF-m’ + I 3111... 317+11u' } (12)
Using (11) and (12),
1.1... 1.3.1, +1.1, 1.71.1. = n’ 1”" e"'*’°° {2122” 8.11... 3,7,”, — 2124i" 8111... 3,11,... } (13)

The terms in (6) with coefficient (p1 + p; tanh(p2d) ) are
—k.’ 1.1...(F ) 1.7.241? I - k,’ 1.1.,0? I 1.71.00? ) —

'kxky [ r;m(P)III'm'y(-k-.)+Irzny(P)Ir7'm’x(P) ] (14)

Using equations (9), (10) and (13), these terms become

1:214“ creme {(1,2 - k,2 - 2jk,k, )em Brim 8,7,1,“ +

+ (1,2 — 2,2 + 2jk,k, ) 1).-'2!”9 3,11,, 3,21,, -

- (ka + ’92 ) (3111,». BF—m’ + 31:1,»: 317+1...’ ) } (15)

Noting that
k, = I: cos(0) (16)
k, = I: cos(0) (17)

the following relations hold:

k} - k,’ + 2178.15 = (k. + 1k, I2 = k2 e”0 (18)

k} — 1,2 - 2,1,2, = (k, — jk, )2 = 1:2 e’zje (19)

221
k} + k} = 1:2 (20)

The terms in (14) then become
1:2 jm' ej(l+l’)0 [:2 {3111.21 3'7“”, +3111” 3,74,". — (3,11,, 8,11,”. +B,f.1,,,, 317“.” )} (21)

The terms in (6) with coefficients k1” ( 6, pl + p; tanh(p2d) ) are
Irina III—'5’: + III” Ig’m'y (22)
Using equations (9) and (10), this becomes

7‘2 I'M, ej(l+l‘)6 2 [3111.111 BF-un' + 31:1,». 317+1,m' ] (23)

Using equations (21) and (23), (6) then becomes, for y: r and 8 = r,

nzjrw ej(l+l')9
.0381 (P1+P2 COWPz 4))

 

P1+P2 tanh(pzd)

k2
8r P1 +P2 tanMPH“

41:2
X {hi 2 [3mm BI'-1.m'+ 3H,»: Bl’+1.m') ] "'

X [Bum Br'+1,m' + 3M» Bury '- (BI+1,». Br-m' + 3H,». Bl’+1,m’ ) ] } (24)

Performing the integral with respect to 0 in equation (23), the factor 810*”)9

integrates to zero unless I = -I’, in which case a factor of 21: is obtained.

 

.1.
21:

”die 2}: “251,4'
I 10161 (P1+P2¢Oth(P2d))

 

x k2__l§: P1+Pztanh(Pzd)
l 2 €rP1+Pztanh(Pzd)

J [Bum B-l-lJn' +314.»- B-um' ]

. 1:2 P1 +P2 tanMPfl)
2 6. P1+P2 tanh(pz d)

 

[3mm B-Hm' + 81—1,»: 3443' ) ] } (25)

CASEH: y=0,8=0.

222
expanding 135,...“y using equations 16.1(25) and 16.1(31)

1:1,", = 7C]: e’m [j ejo Alilfll ”j C-je Alf-1,!!! ]
The four possible products of (26) and (27) are

13m lirm', = 1‘2 I“, Jaw” {Vim A111» Allow + 12.2” A111.»- AF-m' +
+1 A111,». AF—m' “j A111,». A17+1,m' }
1:0" “Ian's = “2.1"”, “(hm {32” A111,»: Az7+1.m' + 8-2” Alt-1.»: AF-m’
+ A1311.» AF-m' + A111,». Aflm’ }
1:111!) lo-I'm’y = 7‘2 I'M, eitl+l')0 {—ere A111,» Aflm’ — e-Zje A111,». AF—m'
+ A111,». AF-m’ + «41:1,». Aflm’ }
1:1“, [a'm’x = “2 it”, gimme {482” A111,»: Aflm’ + if“, Alf-1.»: AF-M'

‘j A111,». AF—m' +1 A111,». At7+1.m' }

Using (28) and (31),

(26)

(27)

(28)

(29)

(30)

(31)

1:1." Invite, 1.7% = —2n’ 1“" e’“*’°°{je”° A111... Afmn' -je""° A911,, many} (32)

The terms is (6) with coefficient (p1 + p; mnh(p,d) ) are

-k.? 1;w(?)l;rm'x(F) - k,2 133.00? ) lama? ) -

223

_ kxky [1%(F)l;l’m’y(r)+l;h|y(?)1;l'm’x(F) ] (33)

Using equations (29), (30) and (33), these terms become

—vr2 1“" em“?e {ac} - k,’ — 217:2, ) e2" Ara, Az7+1,..' +
+ (k3 — k} + 2jkxk,’ ) e‘m A111,” AA...“ +

+ (’93 + k, ) ( 14:11.». Ali-15' + Alt-1.1!: AF-rm' )} (34)
Using (18), (19) and (20), the terms in (34) become

_nz J-M' ammo k2 { A111.” Az7+1.m' + A111,... Ag”, + (A1111... 145.1,... + A111,. Aim )} (35)

The terms with coefficients k? ( 8, p1 + p; tanh(p2d) ) are
[an 10.1'01’1 + 1:11” la’m’x (36)
Using equations (29) and (30), this becomes

7‘2 I'M, ej(l+l')0 2 [Altlm AF-m’ 4' A111,... A17+1,m' ] (37)

Using equations (35) and (37), (6) becomes, for y= o and 5 = ¢,

1 “214-w ei(l+l')0
ma: (m +122 comm d»

 

1
2%,. = $1]de j

p1+p2 tanhovzd) k,

x {k‘2 2 [Alfim Ar-m' +Al-1-M A""""") ]— 8 P1+P2 WIMP: d)
r

X [Al-+1.»: Amm' +1414.» Ar-m‘ + (Am... Al’-l.m' + 111-1.». Al'+1,m’ ) ] } (38)

Performing the integral with respect to e in equation (37) gives the factor 21: 6,,”
Thus,

2k 1‘2 51,—1'
jmfit (P1+ P2 COMM d»

 

l-
Z:{7;;=Et-£dk

224

X k2_k_2 Pl+p2mh(p2d)
1 2 e,p1+p2tanh(pzd)

 

] [Anta- A-l-lan’ + Ai-lan A-I‘HJI' ]

_fi p1+p2tanh(pzd)
2 erpl+p2tanh(pzd)

 

[Al-+1.»: A-Hm' + 141-1.»: A-t-m' ] } (39)

CASEIH: 7=r,5=¢.

Using equations (7), (8), (26) and (27),
Irznx I‘l'm'y = “2 I'M, CAN,” {’9sz 31:1.» Alan-v - 3.2” 3111...; AF—m’
+ 31:1,». AF-m’ + 31:1,»: At7+1,:n' } (40)
1,3... 1,7!“ = 421”" HM” {1221‘ 8.11.... Aral,“ — 12'2"" 8:21.... AF-m’
+j 3:11,». AIT-lm’ ‘1' 31:1,». Az7+1,m' } (41)
in) 1&9»? = “2.1.1”, ei(l+l')0 {152” 31:1,». Aflm’ “13-2” 31:1,». AF-tp'
- j 3111,.» AF—m' 4‘} Bit-1,»: At7+1,m' } (42)
1,3,, 1.7%., = 4:2 j”" em“?o {em 3,11,, Ag”, + e4” 35:1,, 145.1,...
+ 3111,». AF-m' + 31:1,»: At7+uu } (43)
Using (40) and (43).
1,1,", 1:15,," «4,12,;y 1.7,,“ = —1I.'2 j”, ej(l+l’)0 2 {e210 31:1,. At7+1,m' 4' [2” 31:1,». AF-lan' } (44)

The terms is (6) with coefficient (p1 + p; tanh(p2d) ) are

—k,’ I'M}? > lama? ) - k,’ 13.x? ) 1am? ) —

225

-k.k, [13...(1?)I;r..',u?)+I,t..,(ic")l;.t..',(i’)] (45)

Using equations (41), (42) and (44), the terms in (45) become

n21” em” {0:} — k,2 - zjk.k, )1 a“ 81:1... Am,“ -
- (1‘12 ' ’92 + zjkxky )j 6.2” 31:1,»: AIT-lml' +

4‘} (’93 + kyz ) ( 3:11,». AF-m' " 31:1,». Az7+t,m' ) } (46)
Using (18), (19) and (20), the terms in (46) become
1‘21”" eja+l’)9 k2 f {3111... Aflm’ - 31:1,». AF-m' + (3111,». AF—m’ - 31:1,». Az7+1.m' )} (47)

The terms in (6) with coefficients 1:} ( e, p, + p; tanh(p2d) ) are
LL: l;l’m'x + [July IO-l’m’x (48)
Using equations (41) and (42), this becomes

'32 I'M, ej(l+l')0 2 1 [311m AF—m’ - 31:1,»: Az7+1.m' ] (49)

Using equations (47) and (49), (6) becomes, for y = r and 8 = m

1 "ZJ'HW ej(l+l')0
we: (p1+p2 who»: 00)

 

2.315;. = 74—11;? ”de j

p: +122 tanhtpzd) k,
8r P1 + P2 WIMP: d)

X {if kiz 2 [Bum Ar—m' ‘31-”: Ann-v) ] + j
X [Bum A:'+1,m' - Bz-m Ar-un' + (3mm Al'—1,m’ " Bz-m Amm' ) ] } (50)

Performing the integral with respect to e in equation (50) yieldss the factor

211: 51 ..1'. Thus,

 

ZJWo=—1--dk 2k 1981,4'
m { 1‘061 (P1+P2C0th(Pzd))

226

. k2 pl 'pZ tanh(p2d) [
— 2 —— B A - '-B- A ']
X { 1 [k1 2 5, l 2 I l( 2 I) (+1.5: -1 1.1!: l 1.»: -l+1,nl

 

° 2 tanh(p2d)
1"— p‘ ”’2 [B A .-B A ] 51
+ 2 8r Pl + P2 1(P2 d) (+1.0: —l+1.m l-LM 4'1!!! ( )

 

CASEIV: 7:4), 8=r.

Using equations (7), (8), (26) and (27),
1;,“ 1;,“ = 4:2 1”” MW” {em 11,11,” 3,7,1,“ + e4” ,4,th 3,21,...
+ 14111.1» BF—m’ + Altlun BIZ-1,45%)
1:0!!! [rt-’m’x = 4:214”, ej(l+">° {jam A111» 8,7,1,“ ‘1 3-2” Alt-1.»: BF—ln’
" j A111,»; BF-m’ +j A111,». 317453)}
Um er'm’y = 7‘21”” ”(M29 {fem A111,»: 3131M - 1.3-2” Altlm BF—m'
+ j A111,». 317-1.:n' - j A111.» 3111654)}
1;,” 1,7,” = 4:2 1“” ei<’+"’° {em 14,1," 3,7,“... + (21° A111,... 3,21,...
- A111,». BF—lfl' “ 141:1... 3:31.45?
Using (52) and (55),
1:,” 1;.” +1;,,,., 1,7...“ = —1c2 1"“ WW9 2 {cm 14,1], 3,7,1...“ + e4“ 11.1,... 3,11,... } (56)

The terms is (6) with coefficient (p1 + p; tanh(p2d) ) are

-kxz land?) 133514?) '— kyz I;hy(P ) IrT'm’yU? ) T

227

-k.k, [Ianti’mmi’ )+13“1,.,?)1r7'..'x(F 1] (57)

Using equations (53), (54) and (56), these terms become

an”? MW {ac} — k} — 2,1,1, )1 e2” Am... 317+1m' —
' (kxz "' kyz + zjkxky ) j e-ZjO Alt-1.»: BF—lan’ '—

‘j (ka + ’92 ) (A111... BF—m’ " A111,». 317+1Ju' ) } (58)
Using (18), (19) and (20), the terms in (58) become
“21"”, 8104436 k2} {Alina 317+1m' — A111,»: BF—m’ - (14:11,». BF—m’ " A111,»: 317+1.m' )} (59)

The terms with coefficients k3 ( 1:, p1 + p; tanh(p2d) ) are
1:101: Iry'm'x + [any Iry’m'x (60)
Using equations (53) and (55), this becomes

7‘21"”, CHM” 2] [Alina BF-m’ — A111,»: 317+1.m' ] (61)

Using equations (59) and (61), (6) becomes, for y = q: and 6 = r,

 

 

= #11121 191”" W”
41172 10’81 (P1+P2 (30‘th (1))
. p1+p2tanh(p2d) 2
[A r. m. -A .. B ' - k
Xjk1{22 (+1.11: Bl 1. z 1,». 1+1,» )]+J e, p1+pztanh(p2d)
X,[Al+1m Bl +1». "Al-1,»: 131—1.». ‘(Amm 31-1». "Al—1,»: 31 +1.». ) ] } (62)

Performing the integral with respect to 6 in equation (62) givess the factor 21: 8,...“
Thus,

zrlm .._1.-dk 2k “28".”
um - 2“ t! fCD€1 (P1+P200th(Pzd))

 

228

x{j [k2_k_2 p1+p2tanh(p2d)

A B - .—A- B .]
1 2 €,P1+Pz tanhwzd) J [ (+1.!!! 41,}! 11,0! 4+1,”

k2 P1 + P2 tanh(pzd) [ ]
+ . — A B ' _ A _ B - O 63
J 2 er P1 | P2 tanth d) Ml» 4+1.» I 1.»: 4 1.»: ( )

15.2 Patch-feed pin Matrix Elements.

The matrix elements relating current distributions on the patch surface to tangen-

tial electric fields on the feed-pin surface, from Eq. 3.6(39), are given by

 

 

 

d

211....» (2; ), H .121 2m Jo(’“1)¢l’L"'—-’T?°lrm(p2 ) 2: 3.30?) 151.115 (1?) (1)

with, from section 13.11,
-9 1 jkx _ k! l

3“" " noel f ' coal T... (2)

and
_L. L"). .. L9. _1_

800‘ )= jml Tm — (1)81 Tax (3)

CASE V: 8 = r
. - " 2 MM - -» - ..
Zr'l’m' " (21!) (.081 Ii d k J0(ka) 3 P2 T,” [k3 lrl'm’x (k) + ky Irl'm'y (k) ] (4)

Using 15.1(7) and 15.1(8), this becomes

0 1t 2 i370 Sinmp 2d) jl'o
z:~,,,.=—-L(2n)(;)ejjdk10(ka)e ——p2Tm e x

X [317.11.111' ejO ka + ky) + BF-ln' e-jO ('jkx + ky) ] (5)
Using equations 16.1(15) and 16.1(16),
jkx + k) =j (k: '- jky) = jk e-jO (6)

_jkx + k) = "'j (k: + jky) = "jk ejO (7)

so (5) becomes

r"'lm

a1: 2 mucous-co) SiHh(Pzd) m . [ _ - ]
=m Ii d k 100(0) e __Pz Tm e’ 1" Bl'+l.m' —Bl'-l.m’ (8)

229

230

Rearranging the integration order,

: 101:1" 7 2 sinh(pzd) - - 2‘ WWW [TO
2,711.7 = (2n) €061 {dk ’5 10050) 7 3174.157 -Bz'-1.m' I 49 e e (9)

A useful identity gives [16]

2:
{ d9 tat/7M“) a!“ = 211: (ij)" cf" J, (z) (10)

Using the identity (10), (9) becomes

 

, ' -1 " ’"° " sinh(pzd)
2,21,,“ = 101!(m8)18 dk k2—— p2]. [3121,» “7'31-13 ] 1141570) 10050) (11)
CASE VI: 8 = o
, 1‘70 sinh(Pzd)
Z.Illml=ijd2k 2M10(ka)e’ 072—";— [k1 l;l"mx(k)+kyl;l"my(k)] (12)

Using 16.1(25) and 16.1(26), this becomes

=—L—M H 42,, 10(ka)e,p7smh(p2d) e”'°><

$1 "‘ (2100161 I): T...

X [AIZHJV ejO (’kx +jky) + Afr-131' e-jO (”k1 - jky) ] (13)
Using equations (6) and (7), equation (13) becomes

3": m' =‘ézfi‘la II ‘12" Jo(’“1)¢’kr0c (Ho) —P2(;—2) "m (‘1‘) [Alan-1' +A17-1.n’ ] (14)

Rearranging the integration order to separate out the O-dependance,

' -

.1 sinh(pzd) _ _ 2" . .

Using the identity (10), (14) becomes

 

3 _
.I’ In! —

-a1t(- 1)”e ’ ° Idk k2 sinh(pzd)

as, p21 [Mm Mam ]JI'(k"o)-’o(k0) (16)

15.3 Feed pin-patch Matrix Elements.

The matrix elements relating the current distribution on the feed-pin surface along
with the associated singular patch current distribution to tangential electric fields on the

patch surface are given in 3.6(36) as

" _, inh _, _ .
1, Hdzk 21;... 3.1.0: )i—(pflzna 10(k0)+ 2 801505 11:13 em (1)

he

 

23"" =

23"" = 23'" + 2.1"" (2)

where p’ refers to the feed-pin current contribution, and s' refers to the associated

singular patch current contribution
CASE VII: 7: r

Looking first at the contribution due to currents on the feed-pin,

4.170 sinh(pzd)

 

 

 

2;!» = 1 2 H 42k 21m Jo(ka) e 2‘, “0:31;“ (I?) (3)
(2“) -oo 2 may

with

11:11:07)5 71‘ (4)
and

-+ E "k,

3,.(k ) m1 T... (5)

Expanding the sum over a,
«__1_“2 2.33912 _ . -._ . ..
2P. _ (210’ we: Ila 1: 2M Jo(ka)e “T” k, 1,,” (k) k, I,,,., (2)] (6)

Using 15.10) and 15.1(8), this becomes

.1 a - ' . hw d) o
’ = a1: 2 ,ryo sm 2 ’10
2;“ —-L(2n) 1 j] d k Jo(ka) e _—r e x

2 an

231

232

x [431:]... c” (it. + k,1- 8111.... e‘” Ht. + 1,) ] (7)
Using equations 152(6) and 15.20), (6) becomes

-a 4 " _' 'nh(p2d) .
rim _ 7‘1 2 IPMH0) 3‘ 19 - _
Zpo — (2n 1 Ii d k 100(0) e p2 T,” e’ jk [81:11”! 31:1,.“ ] (8)

Rearranging the integration order,

”L _.a,, .1 “an: k0 sinh(pzd) + + 2" quorum,» fl, 9
2,7 7-LdL(2n)coel J Jo( )7}:- Bz+1,».-Bz-1.m {”5 e ( )

Using the Bessel identity 14.2(10), then gives

_- 1:90 “ sinh d
2;!"- = 4%5— l[ark 12 723—1 [3111... — 31:1... ]J1(kro1Jo(ka1 (10)
l m

Looking at the associated singular current component,

2:1» = 7211?? H 1121 2 1;,” (F113. (171mm (11)
-. “I,
ha

with 30450?) given in expressions 15.1(2) through 15.1(4).

Expanding equation (11) in the same manner as Eq. 15.1(1),

2;!" =-—1-j} 1122 . 1
41:2 _. JW1(P1+P2°0m(P2d))(€r P1+P2thP2dD

 

x { E1 +122 ammo] [+3 1.1.21? 1 1.;(12’1 -1; 1,1,0? 11.;(78 1 —
— 2,2, 1,1,4}? ) 1,;(12’ ) + 2,2, 1,;(,,12’11,;(1?) ]

_ - 1,2 (a, 121+ p2 tanh<pzd11 [IJm(F')I;(F 1 + 1.1595351? 1 ] } (12)

From equation 14.2(64),

R

_, __-. k
1,;(1 1=-12na2 e 7; [1111:1111«aux/comm ] (13)

233

150? ) = -j21:a2 if?“ 52; [11(k!)f1(kt)+12(kt)f2(kt) ]: (14)
where R is given by

R a b — r0 (15)

The terms in (12) with coefficient (p1 + p; tanh(p2d) ) are

.231,~;.,(2’11,;(2"1 -k,’1,t.,(ic"11.;(E’1-k,k, [13..(E’1I;(E’1+I,tt,7?11;(£'1] (16)
Using equations 15.1(6) and (13),
1101‘;

1,;“(2’ ) 1,;(12’) = —j21ca2 6"?“ nj’ e

R
2 Ji(k1)fi(k’)‘ X

i212
X [leis 81:1,»: ‘18-” 3111...] (17)
and using 15.10) and (14)
+ -’ — -' - 2 ‘17'7’0 -: flaky R 10 + -j0 +
1,1,",(16 )IOUC )=—12M e 1:] 8 T Z 15(k1)P,-(kt) e 814.1” +8 81.1.,” (18)
2:12 ‘

Using 15.1(6), 15.1(7), (13) and (14),

+ "’ - "’ + " - " 2 —jI’-‘Po) -1 “(0 R
rbnxU‘ )IsyU‘ )+lrbny(k )Ia(k )=2M e n] e, E Ji(kt)fi(’“)
i312 ‘

 

{—fy [feje 3:11.». ’ 1?.” 311m]- 3% [fie 3:11,». + 9.” 3111...] } (19)
Rearranging (19) gives

. . R
15.,(2’11,;(2’1+1,;.,(2'11;(2’1= 2m2e”"° m" e’” 2‘. J.-(kz1f.-(Ia1
531,2 0

% {(2, — 112,12” 8111,. - (k, + jkx) e"° 3111...} (20)

Using (17), (18) and (20), (16) becomes

2 ‘ip7’o -1 In R
e TU e 2 Ji(kl)fi(’“)
i=1; ‘

21w

234
1""

-k , .
{7‘23 [Wm -e"°B-m]+-rki [mm +e1e3,_,,]

2, _ . + . _. +
:’ [(k, -sz) e’°Bz+1,.. + (’5 +119) e ”Bx-14}

 

Rearranging (21),

—21ca2e_jr° °1tj 21’“ 2 J(2:)f,-(2:):
i812

{[qu, k’2(—__kk -J"y) JejoBzim

k 'k ' .
_ [2,2 (1,11,) ”five. 11191 Je"°B,tl,,.}

Using 15.2(6) and 15.2(7),

_Zmz e .1}? 01!] ’efle Z J; (kl)fi(kt) :{k2 3111,»: — k2 81:1,!!! }
i=1,2

The terms in (12) with coefficients 2? ( 2,)”, + pztanh(p2d)) are
3.2(F1I;(F1+ 1,2,0?) 150?)

Using (17) and (18), this becomes

I,t,,,,(1?)1,;(12’)+1,z,,(E’)I,;(E’)=-2nja2e “in °1tj cf” 2 J(kt)f,-(kt):
581.2

( +jkx) - ( —jkx) _-
{ k e’eBgtlfl-F‘ky—k—e’oBgi-lfl

or, using 152(6) and 15.20),

tau? 1 1.;(12’1 + 1,3,,(2’1 150?) =

= 2M2 eflrn ° e"° 21:2 J (kl)f. (kl):{3211.m " 31:11.}

Using (23) and (26), (12) becomes

(21)

(22)

(23)

(24)

(25)

(26)

235

 

a. _ f . .
”Hi” 4’2 1 a’t’nm'e’”
‘ (27‘) —. f (051 (P1+P2°°th(P2d))

 

R
. . 2 P1+P2Wlh(Pzd) __ 2 + _ +
X [.212 J,(kl)f,(kt). ] [k 87P1+p2taflh(Pzd) kl Bid-1,»: Bl-ln (27)

separating out the e-dependance,

 

2;!" =_‘_1_{2 dk 1 “2“ Z J,(2:)f,-(2:)R
(21‘) f toe, (P1+P2¢0th(P2d)) mg .

 

x -
5P1 + Pztanmpzd)

 

+ mm d 2" _-. .

Using 15.2(10), this becomes

2

—1 .. l 0 1t R
z'!"'=— kdk M i,“
‘ (21¢); fulfil ([11+P200WP24» [31:3 (‘)f( )‘]x

 

 

x [k2 p1+p2tanh(p2d)

_ I 2 + _ + 1700
ErP1+P2tanh(P2d) k1] {Bum Bz-m}275€ 1:000) (29)

CASE VIII: 7: (I1

Again, looking first at the contribution due to currents on the feed-pin,

 

 

- ' ' “(P20 —b _,
2°!" = 1 2122 kg We.” 2 1-, 2
,, (21:12 we: J1 2mm 1:: p2 £1, g.p( 1 .1...“ 1 (30)
or, with (4) and (5),
0,10: _ a - 2 437° Sillh(pzd) — + ..., - +
Zp " (21:) 0381 lid k 100(0) e __pz Ta 1:, 1,,” (k) k’ 1010963] (31)

Using 15.1(25) and 15.1(26), this becomes

-1 " --. ‘ h(p d) .
¢h_ a1: 2 Ir’bsm 2 to
2,. _mjldkmka): p27”. e’ x

x [A111, ei° (2, — 1191 +1221... e‘" (2, +179) ] (32)

236
Using equations 152(6) and 15.20), (32) becomes

" " -- inh(p d) .
a». _ “7‘1 2 Ibomw s 2 la
2,, _ (2n) 1 j] d2 Jo(lca)e ————p2T~ e’ 2 [A,:,,,. +A,+_,,,, ]

Rearranging the integration order,

4 " sinh 2" _- .
22"“ = 411- die 22 Joan) —M [A111, + All... ] d6 e ""°°°"“°’ e"°
(21'!) (.081 2 Tu

Using the identity 15.2(10), then gives

1'90 " 'nh
In: _ an C 2 S! (Pzd)

p; T [A1113 + 141:1,» ] 12070) 100‘“)

Looking at the associated singular current component,

 

23"" = (2;), U 422 )3 12..“ (P115 (1?) saga?)
-00 m.’
fin.)

with 30,30?) given in expressions (2) through (4).

Expanding equation (36),

l 1
Z: ’13:, = _ dzk .
f 4,;2 [L [we] (p1+p2 coth(P2 ‘1)) (8r P1+P2 tanh<p2d))

 

x { flax +p2 tanh<p2d1] [4:3 1:120? 11.;(1? 1 - 2,2 1:1,,(1? 1150? 1 —
— kxk, 1:1...(E'1I.;(E’ 1 + kxk, 1.1.01? 1 1.;(12’1 ]

— 2,2 (e, p1+ p2 tanh<p2d11 [1:W(F1I.;(F1+I:§.,P1I.;(F1 ] }

The terms in 15.1(6) with coefficient (p1 +p2 tanh(p2d) ) are

—2,2 1:220? 11.;(1? 1 - 2,2 1:290? 1150? 1 —

- kxk, [12...(5'1 1.;(7?) + 1:2,? 1 1.;(12’1 ]

(33)

(34)

(35)

(36)

(37)

(38)

237
Using equations 15.1(6) and (7),

45%

R
1.9.41? 11:0? 1=2na e n} e"°— 2 2; J9-(k1)f.-(kt)‘ [e’°A9:9,.-e"°A9ti,.] (39)

i=1}
and using 15.10) and (8)
1:1,,(1? 11.;(1? )=—j21ta2e ’T 099,- a“: )3 J(kz1f.-(kr1:[e"°A9:9,. +e ”A9: 1,..] (40)
£312

Using 15.1(25), 15.1(26), (13) and (14),

1:9.“(5’11 50: 1+1,,,,,(2’ 11.;(1? 1=—j2na e “”7 099,491" )3 J.-(kz1f9(kr1:
i812

% {Hey +jk991ei°A91m + (-k, ~jk,1 e'“a A919,} (41)
Using (39), (40) and (41), (38) becomes

j21ta2e ’F° °1tj ’el”6 2 J(kt)f,-(kt):
i=1}

     

x [en/1,11,, + e'” [Minn] + L:’- It; [an A111,». — e'je Alina]

k, - . -' +
" _ikL [(1‘) ’jkx) e10 Alt!» - (k, +ka) e [0 Al-lfl] } (42)

Rearranging (42) gives

--21ta2 [1770 °1tj e"° Z J,(kt)f,-(kt):

i-ll
{ [,9 ‘-'5-,-’k’-) , 192M ],,,m
_ [k3 (’5: 1178) +k,2-——— (kx+ kjk’) J e-jO Alt”. } (43)

Using (41) and (42), (43) becomes

—j21l:a2 e 4"] F"°fl:j e”° Z J,(kt)f,-(kt):[lczA,1m 44211;”, ] (44)
i=l.2

238
The terms in (37) with coefficients 2,2 (e,P, + pztanh(p2d)) are

1:9...(12’1Iai’1 + 1:th 1 1.;(E’ 1

Using (13) and (14), this becomes

—1 -+ —' ' - R
l;m(F)lx;(k 1+I:9..,(I?1I.;(k )=j21ta2e ’mnj' e!” 2‘, Ji(kt)f.-(kt)
i=1) a

(k,-jk, - (k,+jk,) _9
{—E—eleAltln+-_—k_e’°Alt-lm

or, using 152(6) and 15.20),

1:19.990?) 150? 1+ 13,,(2’11,;(2’ ) =

_ - 2 _,r.r° -1 m R + +
- 127‘“ e 7‘1 e 2 JiU‘UfiU‘Ua Am» +1414...
i=1.2

Using (44) and (47), (37) becomes

1 ja2 24% nJ'L' €on

_1 "

29"" = — 9122

S (2")2 ‘U. 27‘} (0 81 (P1 + P2 COdKPz d»
J [22 p1+p2tanh(p2d)

—22 [14+ Aim]
e9pi+pztamh(pzd) ‘] '“""+ '1'

R
X [2 Ji(kt)fi(’“)
i=1; ‘

Separating out the O-dependance,

2,92» = .1 £2 dk 1 “2" )3 1,-(kt)f,-(la)k
(21:) coal (P1+ch°m(P2d)) 9-12 9:

 

 

2‘!
2 P1 + p2tanh(p2d) _ k 2 + + -l '1'??o 1'10
X [k e,p1+p2tanh(p2d) 1 A1+1W+A1_1fl {d0} e e

Using 15.2(10), this becomes

29"": -2212 '1 “2“ X J-(2z)f-(2:)R x
s I! (081 (P1+P200¢h(P2d)) 5,1,2 ' ' a

 

 

P +p2tanh(p2d) re
x [k2 91:1 +p2tanh(p2d) 4‘2 “1"" ”’3’" ‘1 01‘0””)

(45)

(46)

(47)

(48)

(49)

(50)

15.4. Feed pin-feed pin matrix elements

The matrix element relating the current distribution on the feed pin along with the
associated singular patch current distribution to tangential electric field on the feed pin

surface is given by
2;, = 2;. + 2:9 (1)

where s’ refers to the component on the patch surface, and p’ to the component on the

feed pin.
Looking first at the component due to the associated patch current,

['52-‘70

 

 

kL—inhptfl) _
23’: (2;)2 11622 2M 100‘“) g3 82: 1:5 e (2)
From equation 14.2(67),
.9 42’? k2 - - R
1,;(k)=-j21ta2e 0- [10(k1)fo(kl)+11(ki)f1(kt)]a (3)
—¢ . -fi'-r ky - - R
15<k1=-12na2e °T [19<k:1f9(kz1+19(kz1f9 (29) ] . (4)
Using 152(2), 152(3), (3) and (4) equation (2) becomes
, _ _l__ 2. 2 a Sinthd) [2‘70
22’ - (21:) [id k 0081 P2 Tn 100“”
2k1_+__2_)
x(—2nja122° zf9(kt11(kr1:(k‘ (5)
i=1
Integration over 6 gives a factor of 21:.
2' 21w2 SinhU’zd) 1 IR
2;: dkk—,-————Jkalca ,2:J,2z 6
2 10,81 pZTm O( ) §f( ) ( )2“ ()

239

240

From 3.6(45),

 

 

" 2 k2 e, [:2 sinh(p d) ‘
x, = 2122 ," 12 ka d—2-— 2 7
2" U. 1062 °( )[ p22 p} T... P2 J ()

The 6 dependance is integrated out to give

 

 

" 2 k2 e, [:2 sinh
2;. =12 die 27"“ 13(2a) d—-2;- 2 (”2) (8)
1‘05: P2 P2 T». P2

Both 2;. and 2," have a nonconverging (Jo(ka) )2 term, but taken together, the

terms cancel one another. To show this explicitly,

 

 

 

" lc Jo(ka) sinh(pzd) 1 .. IR
25:2 2 dk , ———ka ,2 1,2:
M I stl { p27“, Eof( 1) ( 2:42.
kfd k2 SiDh(P2d)
+J ka - J ka 9
0( ) 8,p22 p22 Tm P2 0( )} ( )

01'

" 2 {13(2a1222d ”a smh(p2d) x

2'9 = 2M2£dk -,——
2 10381 8,. P22 pZTm

 

x [4309» [M99919 “:2, ]+f9(kR119(ka1Jo(kR1+ Jo<ka1f9(k:1J9(kt1{:]} (10)

must approach zero faster than ka'

 

For the integral in (10) to exist, foam) + a :2
2

Examining this term using 14.2(68)

 

 

- 2 -1 8C 240 1920 22
f0(ka)+ap22 ka [A+ +C+ ( )2 ( )2+( )4 p22] (11)

But k2=p22 +222 andA +B +C +D =1, so (11) becomes

~ k -l
f°(ka)+ap9"ka [1-(ka)2-(lca)2 (2914 ,9;

k2
sc 24D+19ZD-—2—1] (12)

01’

k2
2 1 [8C 241) 1920 2 (13)
2

fawn—r“: firm—79:73:
The right hand side of (13) approaches zero like 7:27;, so the integral in (10) exrsts

Equation (10) then becomes
" k k; 2d _ka smh(p2d)
2:9 = 2 2 dk -—,—— 12 lca
M I [(081 { 0( ) [— 6r P2 P272... x

sc+240 192D 1:22
x ——————+— +
29:3 22:5 p221“:

ka inh d J ka - - -
9 9 23’ °( 2 [fo(kR)Jo(kR)-fi(ka)Ji(ka)+f1(kR)Ji(kR)]}(14)

The expressions for f,(kz) from section 14.2 are

- =__é___ 2__C_ 2 ___D__ 6 4 2
fo(kt) ka 29:30“) 2a5 ((kt)- 8(kt)) (2a) ((kt)- 24(kt) +1920“) ) (15)

- B C D
f1(kt)=-k—a-§- 2kt «I»? (4(kt)2- 16kt )+ (ka 7(6(Ia)5-96(kt)3+3841“) (16)

 

Using (15) and (16), the matrix element 2:9 is found to be

 

 

" 2%: ka 'h d 22
Z}=21ta£dk .ka 13(ka) 22_ 8111012 ) 8C+2340_ 1921:+ 22 +
10351 6, P2 Psz (’60) (1“!) P2 ’50
'h d J ka 2 4 6
+sm(p2)o( 210(kR)[— [A +BR2 +CR—4+D£;]+
P21... 0 a
8 R2 R‘ 1920 R2
__ C 30 _ _
+ (“)2 [ a2 a4 ] (“)4 02 ]

 

 

Sinh‘Pzd) 100m) 1
- P27... 11(ka) [Ta—

+ Sinh(P24)-’o(k0) MR) [
P272".

——16 [c£+wR—:]+—3——34‘ZR]} (17)
a a (ka) a

15.5. Matrix element summary.

The matrix elements obtained in the previous sections are here summarized

Zr,en'=__l_-dk 2k “Zak—V
z». 2"! 1.031 (P1+P2¢0¢h(Pzd))

 

k2 P1+P2 WNPzd)
k2—— [B B _ .+B_ B ]
X {[ 1 2 6,. P1 +p2 tanth d) (+1,» -I 1.». 1 1.». 4+1...

[:2 p1+pztanh(pzd) [ ]
+ B B :+B_ B _ ' 1
2 8r p1 +p2 WIMP: d) 1+1,» 4+1.» 1 1,»; -1 1;— ) ( )

2k 1'72 81,-4'

l -
Pad—{cu .
N 7‘ 1081 (P1+P200th(Pzd))

2

 

k2 p1+p2tanh(pzd) [
2 _ _ A__ I A _ A ']
X {[k1 2 8r P1 +P2 ta"MP2 4) AM” I I” + I 1'” 4+1”

1:2 P1 +P2 timmpzd) [ ]
— — A , A- .+ A _ A _ . 2

1 Idk 2k “251,-1'

2391;" = _ -
27‘ 10351 (P1+P2 CO‘h(Pzd))

 

[:2 P1 + P2 thzd) [
-' 2—— B A _ .—B_ A ]
X { 1 [1‘1 2 8' p1 +P2 (31111072 d) 1H,»: —l 1.01 l 1,»: 4+1»

 

' 2 + tanh(p2d)
15- p‘ ”2 [B A .— B A , ] 3
+ 2 E, P1+P2 mh(P2 d) 1+1” -—l+l,m l-l.m -l-1,nl ( )

2'“ = _ dk 2k 1‘2 81,-1'
oz». 2“ I! jmfit (P1+ch°‘h(P2d))

. k2 P +P2 tanh(p2d)

 

X

2 Er P1 +P2 ta“MPH“

 

[:2 P1 +P2 tanh(p2d) [ ]
’ — A B w-A _ B _. 1 4
+1 2 er P1+P2mhw2d) 1+1.» -l+l.m l 1.1!: -l 1.1!: ( )

242

15.5. Matrix element summary.

The matrix elements obtained in the previous sections are here summarized

E 2k 7'52 51.4'

,,,,. 1
I"! = — dk .
Z” 2n 10081 (nun who): am

 

k2 P1+P2 tanh(p,d)
X k2 - — [B B _ I+B _ B 1]
{[ 1 2 at P1 +p2 ““1th d) 1+1... .1 1... 1 1.». 4+1...

 

1:2 p1 +122 tanh(p2d) [ ]
_ B B I B __ m B _ ’ 1
+ 2 8r p1+p2tanh(p2d) 1+1.» 4+1... + 1 1. -1 1... ) ( )

2’3". =-1—-dk 2’: "251,4'
. m 1[ 1.061 (P1+P200'-h(P2d))

 

k2 P1+P2 tanh(p2d)
k2 _ _ [A A-— i A _ A ’]
X {[ 2 8r P1 +p2 tanh(pzd) (+1,» 11.»: + :1... 4+1...

1:2 P1 +P2 tanh(p2d) [ ]
— — A A_ . + A _ A _ . 2
2 8r P1+P2 tanh(p2d) (+1.17: (+1»: I 1.»: -l 1.»: ( )

“2 51,—1'
jwfi (P1+ P2 coth(p2 d))

. l’c2 P1+P2 W‘NPM)
X {-1 [k12 - -2— 8r p1+pz tanh(p2 d) [Bum A-l-l,nl'-Bl-l,m A-mn' ]

 

 

° 2 + tanh(p2d)
[k P1 P2 [B A . _ B A ’ ]
+ 2 8, P1 + P2 tanth d) (+1.01 -l+l.m l—l,m -l-l,m (3)

zrim =_1_-dk 2k “261.4'
“m 27‘ { 10381 (P1+P2C°‘h(Pzd))

p1+pztanh(pzd)

. k2
x {J [k12 ' —2— s, p1+p2 tanh<p2 d) ] [A’*"" B""”'_A”"" B’”""']

 

. [:2 P1+P2 WMPM) [ ]
+1 2 8, P1 +P2 tanh(P2 d) Am» B-Mm' Al-l-m 3.4-1,: (4)

242

243

 

 

 

 

 

 

z:,,.,, = 1““ (2012:?” 1‘“ "2 52%?) [Raw - 3.7.1,... ] mm.) 10001) (5)
3,1,”, = —an: ($251730 14]; k2 3.2% [Afllfl’ +A,7_M. ]J,r(kro) Jo(ka) (6)
Zfi’”=--l—Ilcdk ,1 “z" [21(k1)f,.(k1): ]x
(21:) J to 81 (P1+P2 000100de 1.1,;
x [*2 :grfii‘rzzzx) 4%] {Bav- } <7)
239’" = 1k dk (1)-:1 (p1+pzac201tth(pzd)) 1§31'(kt)fi(kt)e RJX
x [k2 8? 111:” 12:23:?) 1:3] {mm +A,_1,,}e "°°J,(kro) (8)

5,. P22 pZTm

   

2,122,111! (m3 "(m5 pike

“ . 112d ka 'nh d 1:2
$81 {13(ka)[ 2 s: (102) [8C-1-24D 1920 2 ]]+

a‘

' 2 4 6
+s1nh(pzd)Jo(ka) Jo(kR)[- [A+BR—+C5—+Dfl-]+
PZTM a

 

8 122 R‘ 1920 R2
C — 30 — - —— — -
+ (“)2 [ a2 + a4 ] (“)4 a2 ]

_ sinh(pzd) Jo(ka)
psz

 

[28+4C+6D]— -(—klaé)3 (C+ GD)+—

Elv-

3_8__4D
(Ira )’

110‘“) [

sinh(p 24 ) 100“!)
+
P 2T»:

 

3 5
11(kR)[-k1;- [235+4CL+60L]-

16 R R3 3840 R
-—(ka)3 [C:+GD-c1_3]+-(__ka)5 0]} (9)

16.

Matrix Element Integral Evaluation

The integrals to be evaluated are given in equations 3.6(23) through 3.6(26),

3.6(30) and 3.6(43). Rewriting these here,

 

Hds Km e*"'"zl,%... (1? ) (1)
Hds K,,,,., #1?" a 1,3,, (I? ) (2)
[Ids Km, :1!" '7’ a 13,... (I?) (3)
lids K,,,,, M" a 1.1+, 0? > (4)
H d: fir? Kw, a 1,3 |3= x. y (5)
I, = i:dz dz’ [—8(z —z’) + ki;f22 cosh(p2(z<+d))(p2cosh1(:zz>) _ 8yP1Sinh(Pzz>)) (6)

The integrals in (1) through (4) now will be evaluated, beginning with (1). Let

d3 = r dr d4) (7)
Km = 1,,(r/b) e!" (f -£) 1+? (8)
f -2 = 005(4)) = % [c" + e"’] (9)

So, (1) becomes

Let

b 21:
1,?” = 1’; dr Tm(r/b) r \/l - -;—2 {do etjr'7-é [510“). + CHI-1”] (10)

ehrv=etikrcosm01 (11)

244

245

where

e = tan" [72—] (12)

An identity that will be used frequently [16] is

2:
1! eixcuu-O) ej" d¢=ej"°2nj" J..(z) (13)
Thus
b 21:
13m=£dr T... (r/b) r \/1 — { d4) e’”"°°‘(’ °) ; [eiam‘ + ew‘m]
b2
b , .
=£dr 7",,(r/b) r N, - L2- [nj“+1’e"'+"°11+1(:kr) + nj"“’ el“"’° 1(1-1)(ikr) ] (l4)
Rearranging,
Li... = n 1"“ 43’ «+119 3&1... + n 1'H e"""° 815.... (15)
where

b
3.3,, =£dr T...(r/b) r \/1 - é um) (16)

Proceeding similarly for (2),

KM, = T...(r/b) e!" (f -y‘) 1+; (17)

f y=sin(¢)=-21[e”-e"’] (13)

So, (2) becomes

21:

b
1%., =£dr T... (r/b) r ’71 my ¢*i‘='°°'(¢ 9) ..L [ei(l+1)¢ _ ext-1).]

b
=£dr T... (r/b) r

N

r—2[fljl ej(’+1)°J,+1(ikr) + 1tj Iem' l)0 J“- ”(310] (19)

246

01'

1,3... = n j’ PM)" BE...” + 1: j' MM)" 33.... (20)

Continuing for (3)

 

Km. = T...(r/b) a“ (£13 - :2) (21)
r
1+?
«I: - 2 = -sin(¢) = if [cit - e‘”] (22)
So, (3) becomes

1: 2x .
13...... = {(1, T..(r/b) __'_ I! M. etjbcom—e) ..L [€10er _ cut-11¢] (23)

1 — '— 2

b2

Using (13), (23) becomes

b

 

1:1... = 1[air T..(r/b) ' fl [1“ MM” Jmelcr) — j” e"""°Ja-n<:~Jcr)] (24)
r
1 ‘ 7,?
or
13in: = “'7‘ J" CHM” Alina ’ 7‘ J" ei(1-l)9 Alina (25)
where
b
Alf... = 1dr T... (r/b) r J,(:tkr) (26)
r
1‘ ?
Finally, for (4)
K»... = 1.0/b) e’“ (i - y‘) (27)
l+-;-5-

(3'? = cos(¢) = % [e“ + e‘"] (28)

247

So, (4) becomes

b
If...,= {dr T... (r/b)——— Trio emw“"°)-;- [e"’“” + e“""”] (29)

n1

Using (13), (29) becomes

1:
13...: I[air T... (r/b)——— -2—2"j['+‘ aim” J. 1(:tlcr)+j -‘ ei“ 1’9 J(,_1,(:tkr)] (30)

H

or

1:10!) = 7‘ I'M ej(l+l)0 41:1,». + 7‘ 1H 810-110 Alf-1.»: (31)

The integrals in (26) and (16) will now be evaluated. Let 2 = r/b and perform a

change of variable on the integral in (26), giving

1
11.3.. = bzgdz T..(z)

 

’ :11, 32
r—gl _ 2 J:( 2 ) ( )
where T...(z) is the Tchebychef polynomial of order m.

The recurrence relationship for Tchebychef polynomials is [16]

2 Z Tm(z) = Tin-1(2) + Tan“) (33)

If m is O, m-l is negative and Tchebychef polynomials aren’t normally defined for
negative orders. It is desired to use the formula for m = 0, so (33) will be considered

as defining T...(z) for m negative. When this is done, it is found that
T...(z)=T....|(z) all integerm (34)

Thus, (33) may be used for all m if (34) is applied afterwards. Equation (32) then

becomes

 

A13. = Pf-i dz [7.4.121 + 1.42)] mun) (35)

xii—:5

248
A standard integral from [17] is

1

{mm

 

1 -33 ..y. 1

Since the Tchebyshev polynomials in (35) have a non-negative order, I 2 0 satisfies the

restriction on (36). For I < 0, use

J.(:kbz) = (-1)' 1.. (:tkbz) I < o. (37)
Using, in addition, the relation [16]

1.1—kbz) = 1—1)’ mun) (38)
the Bessel function in (37) can be written as

:1 ’ 12 0
( ) } (39)

1((ikb2)=~,”l(kb2)' {(_ i1)! (<0

Using (36) and (34) in (35),

Is

kb kb kb x
2

“b2
Alia = T [1|I|+|2m-ll ['7] Jlll-lzm-ll [_] +J|ll-;m+l [_z—J Jill-znv-l [

N

}

(:tl)' 120
x (-;1:1)' (<0 (40)

By inspection, the absolute values on the m—l subscripts may be dropped, since the

 

bracketed term as a whole will not be affected. Thus, equation (40) becomes

113-121.. 591.. -"9-+JII -"-"-J,, E
2 2 2 2

(:tl)’ 12 o
' (- i1)’ (<0 (41)

 

249
Proceeding similarly for (16), let 2 = rib and perform a change of variable to

 

obtain
1
3.3., = bzgdz 1,,(2) 2 V1 — 22 hard») (42)
or
1 1
Eff... = bzldz T...(z) (2 -z3) W J.(ikbz) (43)

Using the Tchebychef recursion relationship repeatedly,

8 237'... (z) = 4 zzT.....1(z) + 4 22 T...+1(z) (443)
= 2 z T..._2(z) + 4 z T... (2) + 2 Tm+2(z) (44b)
= ..-3121 + 3 Tun-1121+ 3 T...1(z)+ T...3(z) (44c)

With (34),
233.12) =% ( T..-3.(z) + 3 T..-..(z) + 3 Tm+1(2) + mam) (45)

so, using (33),
(z - 231T..(z)= % (4...-..(21 + 7...-..12) + 3.112) - T...3(z)) (46)

Equation (43) is then, using (39) and (46),

1
Bit" = bzgdz [é’ (”Thu—3|“) + TIM-1|(z)+ Tm+l(z) "' Tm+3(z))]

 

1 (111' 12 o

Reworkin g the bracketed term,

1

Bi-b2d1[r T 1 2T T 1
up?! 2; — ..-3.(z)— lm- |(2)+ ( ..-1.(z)+ |m+ 1(2))-

1 (il)' 12 0
- n+1(2)- Tm+3(2) Lil—1?; 1111(kb2)' (_ il)’ (<0 48)

250

01'

b2 1 -1 1 (in,
as. = 7 I” T [Tm-31W T'm-"<‘)]r1—.—:;' WW” {- (in'

l

2 (in'
+2 {(122 -4- [Tlm-1|(z)+T|m+1l(z)]

1
TVA—2'7 1......) {- my

1

+3: {.124 —1 [T,.....(z)+T|m+3|(z)]

2

 

l
W """"’" {- (+1):

Then, using (35),
l 1
31?»: = 7 [2141?» " Azfum—zl " Afmz ] = 2' [2 41?» - 2111—2 ' «413“: ]

Equations (1) through (4) are now fully evaluated. Rewriting (5),

w" ’31, 5:” ds’ejr'7K.5' B’=x’.y’

where, from 14.2(6),

K33 =fpf (rp)
Using d: = r. dr, do, and F; = 'r'”— r3, (5) becomes

2:

I,}=£d¢,fp B’ if (r,)ejr 'rp drp

(in’

120
(<0 +
120
1<o +

$2)" }<49>
(50)
(5 1)
(52)

(53a)

This is evaluated in section 14.2 with the result in equations 14.2(64) and 14.2(67) for

the case of a single patch.

For the case of the infinite array, (51) is evaluated in section 14.3 to obtain

- 7 , l -
135’: e]? °£d¢pr-B IR

with 1; given in Equation 13.3(12).

(53b)

251
The final integral to be evaluated is (6). Rewriting this,

k22 + p:2 cosh(p2(z‘+d))(pzcosh(pzz’) - epxsinhmz’»

 

o
1. = _dez dz’ -8(z—z’) + m T... (6)
Remembering that
z> = max(z, 2') (54a)
z< = min(z, 2’) (54b)

the above integral can be separated according to whether 2 or z’ is larger. Equation

(6) then becomes

0 0 2 2
k +
I: = Id}. {I—su — 2’)d2’+ 2 P2

.4 P

.4

 

I
T [pzcosh(pzz) — e,p1 sinh(pzz)] I cosh(p2(z’ + d))dz’
2pl 1n .4

o
+ cosh(p2(z + :1» I022 cosh(pzz’) — 6.17. sinh(pzz ’))dz’} (55)

The hyperbolic integrals in (55) are evaluated as

 

 

Icosh(p2(z’ + d)) dz’ = J- [sinh(p2(z + d)) - sinh(0)] = s1nh(p2(z + d» (56)
—a P2 P2
isinhwzz) dz’ = -l— [cosh(0) — cosh(p22)] = i [l - cosh(pzz)] (57)
2 P2 P2
0 .
Icosh(p22') dz' = -1— [sinh(O) — sinh(pzz)] = smh(pzz) (58)
. P2 P2
Using (56) through (58) and the S-function, equation (55) becomes
0 2 2
1. = ] dz {-1+i2’—+—pz— E2 cosh(p22) - £,plsinh(pzz)] sinh(p2(z + d))
-d P2 pl T1»
1‘22 + 1022 .
+ -—2——— [—p2 sinh(pzz) + £,p1 cosh(pzz) - 8,] cosh(p2(z + d)) (59)
n: m T...

Using 13.10(36) and l3.10(37), this becomes

k2+ 2 0
1,=—d+—2’—P2—j
P2 ple-d

252

{ choshcpzz) - e.plsinh(pzz)] [sinh(pzz) cosh(pzd) + sinh(pzd) cosh(pzz)] +

+ [cosh(pzz) cosh(pzd) + sinh(pzd) sinh(pzz)] [—p2sinh(pzz) + e.p1cosh(pzz) — e.p1]}dz (60)

With cancellation of several terms, this becomes

12 2 °
2 + p” l {P2[Cosh2(P22) sinh(pzd1- sinh’W) sinh‘Pzd)]

I =—d+————
8 p22plTu-d

+ e.p. [cosh2(pzz) cosh(pzd) - sinh2(pzz) cosh(pzd)]

 

- 8,p1 [COSh(p22) COSh(p2d) + sinh(pzd) sinh(pzz)]}dz (61)
Using cosh2(x )—-sinh2(x) = 1, and l3.10(37) again gives
k22 +p2’ ° .
I, = —d + 2 j p2 51111101211) + e,p1cosh(p2d)— e.p.cosh(p.(z + d)) (62)
P2 P1 Tm -4
Using the definiion of T... equation (14.7.40), this is written as
1‘22 W? ° 8,
I, = —d + 2 j' 1- — cosh(p.(z + d)) (63)
P2 -4 Tm
Performing the integral, with the use of 3.6(33) gives
k2 + 2 e. k2
I. = d {—2 2“ - ]- —, Sinhled) (64)
pz 102 Tn 172
or
2
(65)

1 d I": 8' k inh(p d)
= — — —— s 2
' m” 102’ T... p2

The matrix elements for the incident field, equations 3.6(19) and 3.6(20), will
now also be evaluated. Rewriting these,

(66)

V11... =-Hds E11... ‘Elll "IIdS KW 'ill

CASE 1 E" ‘ parallel to plane of incidence

For this case, equation (66) reduces to

V71..=Vy1.. H=-l!d-i E11»- 'Ell
For 7 = r, this becomes

V... II =-.qu 1?” 'iflEIlIx"
since E" 1‘. has no y" component. Let
5.1..» = E1. M"
where, from 13(45),

5‘}. = 2 El. e’lumn'cosU/Zm,l - klz cos(9.)) cos(9.~)
Now,

x” sin(O.) = x sin(G.) cos(¢.-) + ysin(6.-) sin(o.)

= sin(e.) [rcos(¢) cos(¢.-) + r sin(cp) sin(¢.-)]

= r sin(e.) cos(¢ - ¢1)

so (71) becomes
5'1". :53. eihrsin(0.-)cos(¢-¢.)
From 13(4),

f 'f” = cos(¢) cos(¢.-) + sin(¢) sin(¢.-) = cos(¢ - ¢.-)

Equation (69) then becomes

(67)

'(53)

(69)

(70)

(71)

(72)

(73)

(74)

254
V... n = H 13” 1‘. 3"" m" °°‘" "" e’“ cos<¢ - 41) 11.1 r/b ) 11 - 1 Nb )’ rdrd¢ (75)

Using Euler’s identity, and the identities
2 cos(a) cos(b) = cos(a +b) + cos(a —b) 2 sin(a) cos(b) = sin(a-1-b) + sin(a —b) (76a,b)

equation (75) becomes

21 bu ' . ' r ' . — .
V”... II = #11! rdrdtp em' em “(6‘)“. M T,.( r/b )‘Jl — ( r/b )1

[008((l+1) (¢ - 4%)) + 008((1-1) (‘1’ - (M) +j Sin((l+1) (¢ - ‘21)) + j sin((I—l) (4’ - ¢1))] (77)

Using the Bessel identities [16]

21:
l e‘ ' We) cos(nB) d9 = 2111'" J..( z ) (78a)
21:
l e" ‘ °°5<°> sin(n 9) d6 = o . (78b)

equation (77) becomes

1:

V..... H = Ell, em" 11th+1 I [Jmaclr sin(O.)) —J,_1(l:1r sin(G.))]T...(r/b) V1 — (r/b)I rdr (79)

Using (16), this becomes
Vrlm 1| = 5111 8179.- 75 I'm [31111.1(1‘1 sin(G.)) - 31:1,m(k15in(91)] (80)
For 7: o, equation (68) becomes

an="”d51?¢bu'£”Elll (81)

111 - 2” = cos(¢) sin(o') - sin(o) cos(o’) = -sin(¢ — 41’) (82)
Using (73), (82) and 15.1(44), equation (81) becomes

1

—— drd 83
\ll—(r/bfr ¢ ( )

V». n =1! 531‘: e"" M" °°"" "’ e!“ sin(¢ - <1.) T..( r/b )
3

255
Using Euler’s identity, (78b) and

sin(a) sin(b) = -;- cos(a - b) — % cos(a + b) (84)

equation (83) becomes

£222

3.1 ' . ' r' . —.
thmll= 2ll Hrdrdwm‘e’k‘ un(0.)cos(¢ .')T...(r/b) 1-(lr/b)

[‘1 cos(a-+1) (¢ - 4%)) + f 608((1-1) 0) - 4%)) + Sin((l+1) (<9 - t21)) - Sinai-1) (¢ ‘- ¢1))] (85)

Using (78a) and (78b), this becomes

b

V :1:1 11¢. " [1.1: 'e. 1.11 '6-]T lb—1——-d
.1”. ll ll 8 1t] g (1(17 sm( .))+ (1(17 Sln( .)) ”(r )W 7 r (86)
Using (32), this becomes
Va». 11 = 5111 em" “ J'm [Alina-(k1 Sin(91)) + A111,m(k15in(91))] (87)
For E’ ‘ parallel to the plane of incidence, (68) reduces to
0
V,|l=-Jd221taK,f'E—"fi (88)
.4
From 13(47),
E .2” = 2 TH E‘H e’*2"""“‘°"*‘°°“°"sin(e,) cos(k2(z + d)cos(0,)) (89)
Equation (88) then becomes
'2 "' 0 d 0 0
v,ll = -2111: 2 TH 15"” e’ 1" ““‘ "* °°“ " sin(e.) j dz cos(k2(z + d) cos(G.)) (90)
.4

Evaluating the integral, this becomes

jk 2:”ein(0,) + dcos(0,) sin(B.)

V111 =47” Tll H e kzcos(6.)

(91)

Case 2 E" ‘ perpendicular to plane of incidence

256
For this case, equation (66) reduces to

V.,.=V...l=-j!dsl?...-il‘ (92)

For 7: r, this becomes

Vrlmi="IIds E71»: 'qully" (93)

since 1731‘ has no component in the x” direction. Let
E11)" = 2.12”” (94)
where, from 12(75),

-1_ .- le0
El. —2Ele

”emu/26,, - 1:12 cos(a.» (95)
Using (72), equation (94) becomes

511:" = if eikgr magma-1.) 3 (96)
Using, from equation 13(5)

7‘ ' y‘” = sin(¢) cos(¢.-) - cos(¢) sin(¢.~) = sin(¢ - <11.) (97)
equation (93) then becomes
1w sin(e.) com - 1.)

V... 1='ll if, e e“ sin(¢— 11,-) T...( r/b )111 - ( r/b )1 was (98)

8

Using Euler’s identity, (76b), and (86), this becomes

El " b2! ‘ . ' r ' . - .
14...: —2L 1 1! mm»; a” e"‘ “"°"°°‘" "’T..( r/b )N’l -(r/b )2

[608((l+1) (¢ - ¢1)) + c08((l-l) ((1 - (11)) + 1' sin((1+l) (¢ - ¢1)) -J' sin((l-l) (¢ - 4W] (99)
Using the Bessel identities (78a) and (78b), this becomes
. b
V... 1 = -£‘ 112'" n: j’+1 l [J...(k.r sin(O.)) +1.-,(k.r sin(O.))]T..(r/b) ~11 — ( r/b )zrdr (100)

Using ( 16), this becomes

257

VA». 1 = — E: 9i,” 7‘ J.l [3111.1110‘1 sin(G.)) + 31:1,m(k15ifl(91)] (101)
For 7: «1, equation (66) becomes
V.hl=—JJdSE.bn'y”Ef (102)

ii - y'” = cos(¢) cosm + sinm» sinm = cos(¢ - 2') (103)

Using (73), (103) and 14.1(44), equation (102) becomes

 

V». 1 = _ ”121 12"" "’9" ““9"" e!“ cos(¢ - 1.1m r/b ) W rdrd¢ (104)

Using Euler’s identity, and (78a,b) equation (104) becomes

21:

E" b 11¢. ,1; r 1111(9) cos(9-9-) 1
v =- rdrd¢e 'e ‘ ' 'T...(r/b)
“Mi 2 {g Vl—(r/b):

[008((1+1) (9 - 91)) + 605((1-1) (9 - 91)) + jsin((l+l) (9 - 91)) - jSifl((1-1) (9 - 90)] (105)

Using (78a) and (78b), this becomes

b

V... 1 = -E.‘e”°‘ n j’” 1 [1...(k.r sin(9.))-J.-1(k,r sin(9.))]T...(r/b) Wm (106)
Using (26), this becomes

V... 1 = - E .1 em‘ 1: j"" [Ag-Mk. sin(e.» - A.:.,.(k.sin(e.))] (107)
For perpendicular incidence, equation (68) reduces to

V.l=—:Ldz 2mg; 15.2 (108)
From 12(82), if has no component in the z direction, so

V.l = 0 (109)

Below is a summary of the integrals evaluated.

[rim = 7‘ 1.1+] ei(l+l)9 81:1,»: + TC jl-I €1.04” 81:51.»: (15)

258

1.3., = 1: j' eimm 13.3.... + 1: 1" MM)“ 3.3....
1:101: = ‘7‘ J" ei(l+1)0 41:111. ‘ 1‘ .il eiU—DO Alix...

, '1 . _ ’ ..
131..., ___ “114-1 el( +1)0 A141,.» + 1:}, 1 e’“ ”GAE-1,».

A* —£L2 J 59 J 5’1 +J 1‘3 J 1‘3
1.»: - 4 [I'm-1 2 Ill-n+1 2 Ill-HIH-l 2 |l|-m-l 2
2 2 2 2
(:1)’ 120
' (-¢1)' I<0

1 1
31?». = '4' [2 41f». ’Azfim-zl — Alina ] = '3' [2411}. -Alfm-Z — 415m ]

I .1"22 2.1.2 'h(pd)
= —-———-sm 2
' p22 PzszPz

V... .. = if. e""‘ nj’“ [B.‘....(k. sin(e.» —B.:.,..(k.sin(e.-)]
V... I. = if. e”"’ 11"“ [A.‘..,.(k. mm.» + A.:....(k.sin(e.- 1)]

e jk fame.) + 4mm.) Sin(9,)

Vrlm 1 = — Ell em" 1|51'l [31111.1(k1 sin(6.)) + 31:1,m(k18in(91)]

- 'I - . - . - .
V... . = — E .1 e’ " n 1’“ [2,...(2, sm(e.- )) — A.-...(k.sm(e.- 1)]

th=0

(20)
(25 )

(31)

(41)

(50)

(65)

(80)

(37)

(91)

(101)

(107)

(109)

VII. PROGRAMMING DETAILS FOR SINGLE PATCH
17. Programming details for single patch.
17.1 Expansion of Afi, and 8,5...

The matrix elements of section 15.1 are obtained in terms of products of A,” and
B”... These products are here expressed in terms of quadrupole products of Bessel

functions. Also, 8,, is obtained in terms of A”... From equation 16(41),

2
Azfm='n:— J|l|+m-l £22 J|l]-m+1 £22 +J|l|+m+l 'k—Zb' J|l|-m-l %
2 2 2 2
(:1)’ 120
' (-¢1)’ (<0 (1)

In practice, Af, is needed only where m+l is even. This implies the Bessel func-
tion orders are all half-odd-integer. Consequently, the use of spherical Bessel func-

tions is indicated.

Jam (2) = \I-ff mu (2)

where j,(z) is the spherical Bessel function of order n [16]

Rewriting Afi, then gives

 

AF =£ 21'. 491]}. [1‘2]+&,-, [51,-sz 2[-"5’-]
-"' 2 2 LL22- 2 4421 2 2 Ufa 2 Lb;— 2
(il)' 120
(~il)' (<0 (3)
The products needed are in the forms
Alina AJ—ln' i Alt-1,»: A:I+l.m’ (4)

259

260

and
A111,». A-Tlflpn’ i Alt-1,0: 143-15' (5)
Equation (4) is expanded using equation (3) to give

+ — + - _
Al+l,m A—l-ln' i Al-l,m A-H-lwl' '-

2
= 2: 9;. - (52) - (2, . (2, - .(2)
4 2 j Il+1|2+m-2 2 J II‘HZI-ll 2 J I'd-1|? -2 2 J II'HZI-fl 2

+ a - (2.,- (122,. (a, (1:2,
2 J|l+l|2+nI-2 2 J|l+12I-m 2 J|l+12|+m 2 JIM-1|;m-2 2
1:2 . a - 1:2 - 12 . 12.
+ 2 J|l+l|+m(2 )JI:+1I-m-2(2)JI:+1I+m'-2(2)JI:+1I-m'(2)
2 2 2 2
+ [
i [

i .12 . .52. (2, (2- .4».
2 JIz-IzIm(2)JIz-1I2-m-2 2 JII-lIrLZ 2)J|l-12|-m'(2)

IS:

2
]JI1+121+..kb (— 2)j|1+1|2—M-2 (—)JI1+1I+». (—)jIl+1|2-m -2'-(2 )

N

|+
~|gz

MIG:

2. kb .
]J|l-1I+m—2(-2-')JII-l|-m(_2 _)j|l-12|+m "'—( 2)j|l;m-l| -2—'(2 )
2 2

kb

—2'. I (59V. (Er. I «lit,- .(-"9)
2 Jl-lzm 2 Jl-llz-m-Z 2 1142+». 2 JIl-lIz-m-Z 2

2

]j]!-l|+m-2(L2b')jIl-lI-m(k:—)jll;m—l| .2—(2)JIz-1I-m (—‘k22)]

2 2

] (6)

H-
r———fi

261
Equation (5) is expanded using equation (3) to give

4» - + - _
Am»: A-H-Un’ it 141-1» A-z-m' -

+[

+[

+[

i[
[

2
i 7 JIt-1I+m('3')JIt-tI-m-2() J|l+l|+m'-2(—{)J|l+l|-m’(_2')
2 2 2 2

b‘ ch 2. kb . ch . kb . lcb
= — '5' J|l+1|+m-2('7)J “HI-"(7)1 Il—lIm'—2(—2")J Il-lI-m’('§-)
2 2 2 2

Is:

2
] J |l+l|+m-2 (‘2‘) J Inn-M7) J |l—l|+m'(—2—) J |l-l|-m'-—2 ('2‘)
2 2 2 2

N

MIG:

2. kb . kb . kb . kb
JI1+1I+m(—2")JI1+1I-m-2("i')JIt-tIm'-2(‘2—)JI:-1I-m'(—2')
2 2 2 2

MIG:

~|§

2. kb . kb . ch . ch
J |l~l|+m-2 ('5') J Il—I |-m(—2') J |l+l|+m'-2 (‘2‘) J |l+lI-m’(—2-)
2 2 2 2

H
I;

2. kb . kb . kb . kb
J ”AIM-2(7) J “AI-M?) J |l+1|+m'(—2-) J Il+l|-m'—2(?)
2 2 2 2

N

. kb . kb . kb . kb
] J |l+lI+m(—2-) J |l+l|-m-2 ('2‘) J Il-lIm’(T) J Il-lI-m’-2 (‘2') J
2 2 2 2

H

kb 2 - kb . kb . kb . kb
? J.Ll._-1|+m(-2") J |l-1I-m-2('2_) J |l+lI-Hn'('_2-) J |l+1|-m'-2(7) (7)

Expressions (4) and (5) are used directly in the patch-patch matrix elements relat-
ing radial electric fields to radial surface currents. For patch-patch matrix elements

relating azimuthal electric fields to azimuthal surface currents, the products needed

262
involve 3,3... These can be expressed in terms of products of Ag. already obtained.

3:11,». BQ-lp' i Bit-In Biup' (8)
31:1,». 331+m' i 3:11.»: BQ-m' (9)
From equation l7.( ),
l
31?». = 2' [21413. “Alan-2| -Aljn+2 ] (10)

Using (10), equation (8) is found in terms of Ag... as

1
+ - + - + - + -
Bum B-I-un' + Bz-m B-l+1,1n’ = '4‘ Amp A-l-ln’ + Az-m A—mp' +

1
+73 Az+m+2 A-z—m +2 + Al+l,m+2 143-1,». +2 + An-m-z 144-1.». +2 4' Anna-2 A314,»: «2 +
+—--1 A + A - A + A - A + A - A + A '-
16 454-2 ~—l+1,m’+2 + l-lm-O-Z -l+1,m’+2 + t-uu-z 4+m'+2 + z-m-z -l+l;n’-2 +
—l _ 4. -
+—8 Al+l.m-2 A—t-m' +Az+1n+2 AQ—m' +A1+1 n A-I-l,m’-2 + Ann». A—z-mwz +
‘1 - + - + - + -
+—8 ‘AI-lm-Z A-H-Un' +Az-1,m+2 A-l+l,m' +414,» A-l+l,m'-2 + At-m A—um'n (11)

Using (10), equation (9) is found in terms of Aff, as

1
+ — + - __ _ + - + -
31+”. B-l+l,m' 4‘ BM,» B-z-un' - 4 Am... A-Hun' + 51-1,»; A-z—m' 4'
—1- A* A" +A+ A- +A+ A' A“ A‘ +
+ 16 l+l¢n+2 4+1,»r+2 l+1,m+2 —l+1,m'+2 Hun-2 -l+1,m'+2 + Hun-2 4+1,m'-2
Az-mn A-z-m +2 + 141- -1,m+2 113-1.». +2 + Amp-2 1431-1... +2 + '41- 1,;n-2 AIM.» -2 ] +

-l

4'? 16[A1+1,m-2 Alun' +Az+m+2 AIHM' +Al+l,m A:l+1.m-2 4' Am,” A—m.» +2 ] +
—l _ - .. -

+7 [AM-1.”: A-z-m' +Attm+2 A-t-m' +Altln A-l-l,m’-2 +1411”. A-J-m'n ] (12)

In addition, the patch-feed pin and feed pin-patch involve cross products of A,‘j,,

and 3,3... The terms needed are

263

31:1,». Ai—m' 1'1 31:1.»- A-TI-rm' (13)
3:11,» Al'mn' i 31:1,». A-Tt-ln' (14)
A111,». BII—ln’ i Artur. 33+m' (15)
Alina BIMJn' i Altln BQ-m' , (16)

These are expanded using (15.1.10) to give

1
+ -- + - _ + - + -
BM... A—l-m' - BM... A-mp' - 3 Am... 114-13' - AH... A-mp' +

-1 _ .. - -
+ T [Altman-+2 A-t-m' + Alina-2 A-t-m’ —Al:l.m+2 A-Hm' ‘Attm-z A-l+l.m’] (17)

3111-" A-tmm’ " Bf.“ AII-ln' = % [Alina A3:+m' ‘ A111,»: AQ—m'] +

+ .71' [Azimn A31+1,m' + Alina-2 AIM...’ - A111,”; AIM”: - [4,115-2 143—1.x] (18)
A'TH-M 331-1,": " Alt-1.» BJHW = 'é' [Alt-1,»: AQ-m' " Altlm A::+1.m'] +

4‘ :4'1' [14:11” A3—1.m'+2 + A111,»: AQ—mzz " A111,»; Aland; - Alt”, A3+1fl'.2] (19)
A111,». BIN-1.x ‘ Aztlan BJ-M' = 'é' [14:11... A-Tmn' ‘ Altlun AQ-m'] +

-1 .. _ .. _
+ T [Al-film A-l+1.m’+2 + A111.» A-l+l,m’-2 — A111,». A—z-umz - A111,». A-z-uv—z] (20)

The expansions in (11), (12) and (17) through (20) are all sums of terms of the
forms of (4) and (5).

For the case of l = 0, the expressions (17) through (20) reduce to

.. - 1 - _
...l _ - - .-
+ 4 [ALI-+2 A4”, + Affl‘z A’lfi' -A:1.nl+2 Alan’ ‘14:”...2 Al-M'] (21)

i
2

264

-l _ .. _ ..
+ T [1113": A1,..' + Aim-2 Aw ’A31m2 A-In' ‘Aflm-z A-m'] (22)

- - l - -
Alt»: B-ln’ - 14:1,»: Bln’ = a [Aim A-ln' - Al}, Al»'] +
-l - _ .. _
+ T [Aim A-lm'd-Z + Aim A-ln’—2 " Arlyn A1542 - Ail,“ ADV-2] (23)
_ - 1 _ -
Aim Bun' ’ All»: B-m' = 3 [Aims Alm' - A31.»- A-m'] +
-—l

+ T [Aim Ai-Jll'4-2 + Alta ADV-2 - Aria A:l.ul'+2 ’Arlfl A:1.M'-2] (24)

By inspection of (1) for I = i1, it can be seen that

11:1,,” = A it,” (25)
and that
2.1.0: = A 1:»: = -A it»: (26)

The right hand sides of equations (21) through (24) thus reduce to zero.

The double and quadruple products of spherical Bessel functions needed for equa—
tions 14.1(6) and 14.1(7) will be obtained as a sum of polynomials and polynomials

times simple trigonometric functions.

From [16] (equations (10.1.8) and (10.1.9), Abramowitz and Stegun, page 437),

 

 

in (Z) = 2'1 [P (n+1/2, z) sin(z-n 1V2) + Q(n+l/2, z) cos(z—n 1:12) ] (27)
j_(,.+1)(z) = 2‘1 [P (n +1/2, z) cos(z-mn/Z) - Q(n+l/2, z) sin(z+n1r/2) ] (28)
where
P(n+1/2, z) = [3:2,] (—1)" (" +2“! (22)-2" (29)
0 (2k)! (rt-2k)!
Q(n+1/2. 2) = “"2321 (-l)" (£32;ng (22)-2’"l (30)

From equation (28),

265

j.,.(z) = 2'1 [P (n—1/2, z) cos(z+(n—1)1t/2) - Q(n -1/2, 2) sin(z+(n-1)1t/2) ] (31)
Using

cos(z +(n-l)1d2) = sin(z+n1t/2) (32)
and

sin(z +(n-l)1t/2) = -cos(z+n1t/2) (33)

and letting —n —> n,
5(2) = z'1 [P(|n+1|/2, z) sin(z-n1t/2)+ Q(|n+1|/2, z) cos(z—nun) ] (34)
Equation (34) holds for all integer n.
Using (34) then gives
2 j,(z) 1,,(2) = z-1 [P(|n+1|/2,z)P(|m+1/2|,z)sin(z—n1r/2)sin(z-m1t/2)+
+ Q(|n+1|/2, z) Q(|m+1|/2, z)cos(z-n1U2)cos(z—m1t/2)

+ P(|n+1|/2, z) Q(|m+1l/2, z)sin(z-n1t/2)cos(z—m1t/2)+

+ Q(|n+l|/2,z)P(|m+1|/2, z)cos(z-n1L/2) sin(z—mn/Z) ] (35)
Using
2 cos(z) cos(y) = cos(x —y) + cos(x +y) (36a)
2 sin(x) sin(y) = cos(x—y) - cos(x +y) (36b)
2 sin(z) cos(y) = sin(x—y) + sin(x+y) (36c)

equation (35) becomes
2 1°.(2)J'..(Z) = (21)’1 [P(|n+1|/2. 2)P(|m+1|/2. 2) ( 008((n-M)1r/2)- 608(21-(m+n)n/2))+
+ Q(ln+1|/2. Z) Q(|m+1|/2. Z)(COS((n-m)1t/2)+ 008(22-(M+n)1t/2))
+ P(|n+1|/2, z) Q(|m+l l/2, z) ( sin((n-m)1r/2) + sin(Zz—(m+n)1cl2) ) +

+ Q(|n+1|/2,z)P(|m+l|/2,2) ( —sin((n-m)1:/2) + sin(zz—(m+n)1u2) ) ] (37)

266

Using
cos(x —y) = cos(x) cos(y) + sin(z) sin(y) (38a)
sin(x—y) = sin(z) cos(y) - cos(z) sin(y) (38b)
to obtain
cos(2z -(m +n mm = cos((n +m)1t/2) cos(2z) + sin((n +m)1t/2) mm) (39)
sin(22 -(m +n )1t/2) = cos((n +m )1t/2) sin(22) — sin((n +m)n/2) cos(2z) (40)

equation (37) can be rewritten as

2 j.(z)j...(z) = my1 x

x {[P(|n+1|/2, z)P(|m+1|/2, z) + Q(|n+1|/2, z) Q(|m+1|/2, 2)] cos((n-m)1t/2) +

+ [P(|n+l |/2, z) Q(|m+l |/2, z) - Q(|n+l |/2, z) P(|m+l |/2, 2)] sin((n-m)1t/2) +

- cos(22) [[P(|n+1|/2,z)P(|m+l|/2,z) - Q(|n+l|/2,z) Q(|m+1|/2,z)] cos((n+m)1r/2) +
+ [P(|n+l|/2,z) Q(|m+1|/2,z)+ Q(|n+1|/2,z) P(|m+1 l/2,z)] sin((n+m)1r./2)]

- sin(22) [[P(|n+l|/2,z)P(|m+1|/2,z) — Q(|n+1|/2,z) Q(|m+l|/2,z)] sin((n+m)1t/2) +

- [P(|n+1|/2,z) Q(|m+l|/2,z) + Q(|n+1|/2,z) P(|m+l |/2,z)] cos((n+m)nl2)]}
(41)

Making the definitions
D..*,...(z)=(2z)‘1 [P(ln+llf2.2)P(IM+1|/2.z)iQ(|n+ll/2.z)Q(IM+1|/2.z)] (42)
and

5,3,,(2) = (zzr1 [P(|n+1|/2, z) Q(|m+1|/2, z) :t Q(|n+1|/2, z) P(|m+l |/2, 2)] (43)

267
equation (35) becomes

2 1'. (Z) J'..(1) = 0:...(2) 008((n -M)1t/2) - 5;...(2) sin((n -M)1V2) +
+ cos(22) [-—D,;’_,,,(z) cos((n +m )1t/2) — Effie) sin((n +m )n/2) ]

+ sin(2z) [-D;,,(z) sin((n +m)1t/2) + 5:,(2) cos((n+m)1rl2) ] (44)

Equation (44) is sufficient for A,” and BM, but not for products of them. Using

equation (44), and making the definitions

0 (n) s cos(a—215) (45a)
and

s(n) -=— Sim-2515) (45b)

the product of four spherical Bessel function is found as

268

2’ 1342) 1.42) 1.42) 1.42) =
=D:,.(z)D. .II. 42) c(n—m) c(n —m') + E...(z) 15:. 42) s(n—m) stn an)
- 0.3.(2) E.'I,..I(2) C(n-M) S(n'-m') - E.3.(z) D.”I,..I(2) SDI-In) C(n’-m’)

+ 008(22) [-D.3. (z)D.'I..I I(z) C(n-m) C(n '+M') +E.,..(z) E.*I,..I (z) sou-m) 5(n ’+m')
- 0.3.(2) 5.2541) C(n-M) S(n'+M') + 5;..(1) 0.7.542) SUI-m) C(n'+M')
-D.‘*I,..(z)D. ..I I0) C(n+In) C(n ’—m') +E .342) E5...I(2) S(n+m) 3(n’-m’)
+D.,.(2)E.'I..I I(z) C(n+m) s(n ’-m’)- E. ,...(2)DI .I,..I I(z)S(n+m) 601 '-M’)]

+ sin(zz) [—D.t.(z) D.’I,.4z) C(n-m) S(n’+m’) - 5:..(2) 5.2.42) S(n-m) c<n'+m3
+ 0.3.(2) E.“I,..I(Z) C(n-M) C(n'+m') + 5.3.(2) D.'I,..I(Z) SUI-m) S(N'+m')
-D.’,..(z)D.*II,. I(Z)S(n+m) C(n ’Um’) E. ,..'II(2)E... I(Z) C(n+m) 3(n ’-M')
+ D..(z) 5.7.. (I) s(n+m) so: ’-m') + E. ..(z) D. .II. 42) c<n+m) c(n '—m3]

+ sin(4z)

2 [1).-MR) Dn7’,,,I(z) c(n+m) s(n’+m’) - E,f,,(z) EJ'MIQ) s(n+m) c(n'+m')

—D,,H(z)E I'mI I(z) c(n+m) c(n ’E+m’)+ Hfl(z)D,fl I(z) s(n+m) s(n ’+m')

+D.,'II..(z)D... I(Z)S(n+In) C(n ’+m')- E. 3.3..(2)E .II,.. I(2)C(n+m)S(n’+m’)
+ 0.3.(2) E.*I,..I(z) S(n+In) S(n’+M’) - E.3.(2) D.“I,..I(2) COW") C(n’+m’) ]
+ $925131[0,,(z)o,-.,..(z)c(.+m)c(. +m') +5 ..(z) 5.342) s(n+m) s(n’+m')
+D.‘,..(z)E.*I..I (2) com") S(n ’+rn') +E .3.(z)D.’ I...I(2) sown) C(n’+m’)]
+ .1:°_92§(_‘El[ D..3.(2)D.‘I,..I I(z) S(n+m) so: ’+M’) +E .3.(z) E.I I(2) C(n+m) C(n’+m’)
-D.,..(2)E.I ..I I(2) S(n+m) C(n '+m’)- E. ..‘II(z)D... I(2) C(rWn) 3(n ’+m')]

(46)

17.2 Asymptotic Forms for Spectral Integrands.

In this section, the asymptotic forms for evaluating the matrix elements are
obtained. The matrix elements relating electric field over the patch surface to Tcheby-

chef surface current distributions on the patch surface are

21: 1‘2 51,-1' x
m1 (P1 + P2 COMP: (1))

 

.. __1_"
III 2m,-

X k2 _k_2 P1+P2 WNW")
‘ 2 €IP1+P2tanh(Pzd)

] [Bum B-t-m' + BM... B—H-Ln' ] +

 

1:2 p1 +122 tanh(pzd) [ ]

+ B- p B_ B - I 1
2 e, p1+p2| h(Pzd) 3m... (+1... + 1 1... -: 1.»: ) ( )
1 .. 2k 1'2 51,-1'

 

out = dk . X
9! 27! i J (081 (P1+P2 0031(1’2 ‘0)

 

k2 p1+pztanh(pzd) [
2__ A A _ . A- A .]
xflkl 2 £.p1+p2tmh(pzd) ”1’" "'1'"+ '1’" ’M’" +

 

[:2 P1 +P2 WM?!” [ ]
-— A A I A- A _ I 2
2 8. p1+p2 tanh(p2d) (+1.. 4+1... + 1 1,». -4 1,». ( )
“2 51,-4'

 

1 " 2k
z!;:..—_— dk , x
r? I J (081 (P1 +P2 o‘3“‘(P2 4))

 

x _j k2_k_2 p1+p2tanh(pzd)
l 2 e,p1+p2tanh(p2d)

J [Bum A-I-l...’ " 81-1,»: A-mm' ] 4"

iii P1+P2 mmpzd)
+
2 Er P1+P2 tanh(pzd)

 

[Bum A4+1fll' - BM... A-t-m' ] } (3)

1:2 8:..II

...,, 1 " 21:
Z ,m,=— & .— x
" 27! i I met (n+1): comm d»

 

x . 1.2-3 p1+pztanh(pzd)
J 1 2 e,p1+p2tanh(p2d)

 

] [AH-1,»: B-l-m' - 141-1.». B-l+l,m' ] +

+ . k_’ m+pztanh(pzd)
j 2 €rP1+P2tanh(Pzd)

 

[Anna B-m...’ " AH... B—l-ln' ] } (4)

269

270
The products of A and B are expressed as products of four spherical Bessel func-

2
tions of argument 1‘22- times [-%b-] in section 17.1. For kb less than some value p, the

product of four spherical Bessel functions is expressed as a power series. For kb
greater than u, the product of four spherical Bessel functions of a single argument 2
times that argument squared is that obtained in 17.1(39). The product is obtained in
terms of a polynomial in 22 times each of the following: 1, sin(22), cos(zz), sin(4z) or
cos(4z). Since a polynomial is simply a power series, the products of A and B are
expressable as a sum over terms, each of which is a power series in kb times one of 1,
sin(kb), cos(kb), sin(2kb) or cos(2kb). Labeling these various terms as P,-(kb) where each
P,-(kb) is a power of kb times 1 or one of the sin or cos factors, each of (1) through (4)

can be written as

 

 

 

25"" “8"” Idle 1‘ I“: Z P (kb)+
Int: - c. .
7' 10061 ,. (P1+P2¢0th(Pzd)) ,- ’ ’
2 + tanh d
+ k 1 P1 P2 (P2) 201 Pj(kb) +

? (P1+P290th(Pzd)) 8. p1+p2tanh(p2d) ,I

,1}. [...] (5)

where the integration from O to u will be dealt with in later sections. Equation (5) can

be written as

 

 

1:8 . " " 11
1081 i u 1‘ u
where
k3 1 p1+pztanh(pzd)
k =— 7
g() 2 (P1+P2C0‘h(P2d)) 8,p1+p2tanh(p2d) ()
and
Ick’
h(k)= ‘ (8)

 

(P1+ P2 COWP: 4))

271
Each of the integrations of P,(kb) times g(k) and 110:) in Eq. (6) are performed
individually and stored. Once all necessary integrations have been performed, the
matrix elements are assembled using Eq. (6). The asymptotic portions of the integra-
tions are subtracted off, performed analytically, then added to the portions obtained via
numerical integration. The separation of the asymptotic portion of the integrands is

done by finding the bahavior of Mk) and g(k) as 1: becomes large.

 

 

 

Define

glans? gig 3;“? (9)
01‘

_ 2 .5 _2_ .L _ __l.‘_2_

“(H—k 2 e,+l 2k ‘ 2(e,+1) (10)
Then let

gf(k)Eg(k)-gr(k) (11)
01'

_ _ k2 k p1+p2tanh(pzd) _ 1

810‘)- 2 (P1+P200th(Pzd)) er p1+p2tanh(p2d) (Er'l'l) (12)

Now let

82(k)'}i_rp_sf(k) (13)
Taking the limit by using

#122

P13 =k(1- ‘2?) (14)

yields
r 1
(II.2 +1.22)
g m- "2 " 2k - 2" 1 (15)
2 "2' 1 (k2+k2)—(8 +1)
21: -— (1.2 +1.2) -i1__2_ '
2k 1 2 ‘(e,+l)k 2k

 

 

 

272

01’

k2 1 1
82"”- 2 I} (a. +1)

(6. + 1) - 8.7

Putting the right hand side of Eq. (16) over a common denominator gives

II?
2 (e,+1)—(e,+l)+?e,

820‘) = -2— k]:
(8, + 1)((€r + 1) - er?)

 

01'

8r
(a. +1)2

 

820‘):

Now define
820‘) 5 80¢) - 820‘) " 810‘)
01'

g_(k)_ 1 k3 p1+p2tanh(pzd)
2 __

k2

his.

 

Doing the same for h(k), define

_. 2 I M")
"I<">="l“3172‘

or

h1(k) = 0
so

h{(1‘) 5 h(k)-h1(k) = Mk)
Continuing,

h20¢) a £132. thC)

2 (p1+pzcoth(pzd» eIIP1+P2t311h(P2d)-(€.I+1)-(fir'l‘l)2

(16)

(17)

(13)

(19)

(20)

(21)

(22)

(23)

(24)

Fri-ii}... a

273
Taking the limit gives

 

k?
’12“) = ‘2- (25)
Lastly,
I: 2,2 k?
h" k = - —
2( ) (P1+ P2 cOIMP2 4)) 2 (25)
The integrals involving 30:) are
jdk g(k)P,-(kb) (26)
p

The integral in (26) is performed in by separating the asymptotic portions of g(k) and

evaluating these pieces analytically. The factor F, can be written as

pl. = m (27)
(kb)
where
cos(mkb)
csp(mkb) represents sin(mlcb) (28)

1

where m is either 1 or 2.

The separation of the asympotic portions depends on n, and on whether csp is a
trigonometric function or simply I. The cases where csp is trigonometric will be

hand— led first, followed by csp = 1. First, for n=2 the integral is broken up as follows.

Idkg(k)-c—SM Idkg2(k)-c—SL(m£b-1+Idk 81(k)£sflk;’fil+
u u u

ch’ = kbz
+jdk g;(k) —L——l°s g?” (29)
p

In Eq. (29), the integral involving 50:) is handled numerically, while the other

two are handled analytically. For n=3 or 4 the integral is broken up as

274
Jdk 8(k) 35%?)- =Idk 34k) 91319-1 +Jdk 81‘(lc)-°s-%,mkfl (30)
1: u u

with the integral involving gf'(k) handled numerically, while the one involving g1(lc) is
handled analytically. For n greater than 4, g(k) is used and the integration is per-

formed numerically.

The asymptotic integrations needed for (29) and (30) are

1: dz M2. (31)
u z

and
Id: ——°°“f") (32)
u z

for n = 0, 1, 2 and for each of m equal to 1 and 2. For n equal to 0, the integration
must be performed with the understanding that the integrand contains an infinitessimal
decay factor, due to small but finite loss in the dielectric. With this understanding,

(31) and (32) become, for n=0,

1 dz sin(mz) =1 dz sin(mz) - I dz sin(mz) (33)
01'

1‘” sin(mz) = -;—; [1- cos(wn)] (34)
and

1 dz cos(mz) =1 dz cos(mz) - Ed: cos(mZ) (35)

id: eos(mz)= 7;- [0+ sin(wn) ]= £13,531 (36)
p.

“‘— vs.-

275
For n = 1 and 2, (31) and (32) are obtained in terms of sin and cos integrals [16].

For n=1, using Eqs. (5.2.26) and (5.2.27) in [ ], (31) and (32) are found to be

jdz EEESZZ‘L). E—Siflun) (37)
u

and
Id: may; s-Citum) (38)
'2

Finally, for n=2, (31) and (32) become, using Eqs. (4.3.120) and (4.3.124) from [16],

[dz-9%?” [El—"gmfll—cuum] (39)
u
and
.. cos(mz) __ cos(a) _ . _1:_
;[dz —-—22 —m I: W" Sr(|.1m)-1I2 ] (40)

For csp = 1, the integration is broken up as follows. For n=3, the integration is

not needed since all terms involving this case cancel exactly. For rid, use

 

 

135-».

- k) .. 810‘) .. 82-05)
dkflik-l: dkgz( + dk + dk— (41)
kb’ ,I, kb‘ 5 kb‘ .1

In Eq. (41), the integral involving g;(k) is handled numerically, while the other

two are handled analytically. For n=5 or 6 the integral is broken up as

19—53

dk ENC) =J'dk 81“) +Jdk gl-(k) (42)
kb' 11 kb' kb'

with the integral involving gf(k) handled numerically, while the one involving g,(k)
handled analytically. For n greater than 6, g(k) is used and the integration is per-

formed numerically.

The integration needed in (41) and (42) for the asymptotic integrations is

_ “in

276

 

 

th-el

1
dz 7— = "_1 “kl (43)

forn=2.3and4.

The integrations involving 110:) are handled in the same manner as g(k), but with

h,(k) equal to zero

The matrix elements relating Tchebychef current distributions on the patch to

tangential electric field along the feed pin are

 

 

g - jaflj-ly'efl’eo "’ 2 Sinhwzd) [ 7 — 7 ' ]

2.. - me] i die k '77.. BI... BI-.. Jz(kro)Jo(ka) (44)
z — _an(_1)l'e Ileo .- 2 Sinh(p2d) [ 7 ' 7 ' ] ’

Zorn. - are, {dk 1‘ ——p2 T... A: +1... +414... Jr(k70)10(k0) (45)

The components of the matrix elements relating current on the feed pin to electric field

tangential to the patch surface are

_- no. " sinh(p d)
= —L%:—— {dk ’52 "E—Ti" [31’1er 1-1,In ] 11000) 100m) (46)
“o " sinh(pzd)
2;“ = 3%]— t[die k2_ pz “[4... + AH. ]Jt(’<To) Joaca) (47)

Equations (44) through (47) may be written in the form

 

a1: e’w° . u
2=——jdkf(k)1,(kro)10(ka)ZC,-,P +£dk [...] (48)
where
P1"2
k = 49
f( ) P2 (8' P1¢°th(Pzd)+P2) ( )
and where

with csp(x) defined in (28). Again, the integration from 0 to p. will be handled in a

277

later section.

Substituting for Pi, the integration from u to infinity in (48) may be expressed as

a summation over integrals of the form

Jdk f(k)11(b'o) Joaca) 9952,11
.. (kb)

Define

f1(k) 5k 11m A?

01'

It
(8. + 1)

 

f1(k)=

Then let
ff(k) Ef(k)-f1(k)
or

P1 1‘2 k

_ k = _
f1 ( ) P2 (8r P100th(Pzd)+P2) (8r + 1)

Now define
l . _
f2(k)-‘k':1_m~kf1(k)

Taking the limit by using (14) yields

2

*1
“172' .I

 

-.l-
f2(k)- k hm”

 

k?-

 

kz’ kz’
k(l-Tz-) k(l--;{)+€,k(l-?
or, working to first order only in we”,
1 . It? 1:22 [:2
f2(k)- k ill-12- (1 k2 + k2) k2

2
(er+l)-_k'T—erF

k}-

k2

(1+ 8,)

k2
(1+ 8,)

 

 

(51)

(52)

(53)

(54)

(55)

(56)

(57)

(53)

278

 

 

 

or
2 .k2 k2 2 2 k2+ k2 2
f2(k)='l-1im k -1-—i+._2 ..L—...,, k 2 €r1_ k
“H“ ‘8'“) *2 k2 (€I+1) (8.+l) k2(e,+1) (e,+1)
or
1 . kzz-k.’ 122+“?
k=—1
f2() kkfl €r+1+(e,+1)2
Using

e, 1:? =10}

and taking the limit,

 

, <a.-1) we} ]

- _1_ _ __
f2(k) — k [k1 (8. + 1) (8, +1)2

01'

klz (8,. - 1) 28,.
k

f2(k) = — (8. +1) — (8, +1)2

(59)

(60)

(61)

(62)

(63)

Again, the asymptotic portions of the integration in (51) is separated and handled

analytically, with the remainder evaluated numerically. For n=l, (51) becomes

Idk f(k)11(kro)Jo(ka) M = I die m.) 1,0...) 10m.) M +
. (kb) ,, (kb)

+ III dk Mk) 1'0”) “a” ’ 9% +1 dk fr“) J:(kro) Jo(ka) MSW":

For n = 2 or 3, (51) becomes

Jdk f(k) 11(kro)Jo(ka) 9143,11 = I die f1(k)11(kro)lo(ka)-c-S'Eg£;l +
11 (kb) u (lib)

+IdJc ff(k)11(k70)10(k0) M
u (kb)

(64)

(65 )

 

 

 

 

279
The integrals in (64) and (65) involving f,(k) and f 2(k) are performed in closed form,
while those involving ff“) and f5“) are handled numerically. For n greater than 3,

(51) is integrated numerically.

The integrations needed to evaluate the asymptotic integrations in (64) and (65)

are of the form
I dx J, (ax) J... (bx) csp(cx) (cx )"' (66)
u.

where m is zero for (64) and (65), but will be left general for later use.

For csp(x) trigonometric,

NI—

csp(x) = J p(x) [1%] (67)
where p = :1/2. Using (67), (66) can be written as
NE c'“ jdx J,(aJt:)J,,,(bx)J‘,(c1c)x""‘1’2 (68)
u

Letting 7t = -n + 3/2, (68) becomes
- u
\/-§ c‘” [g dx J,(ax)Jo(bx)Jp(cx)x"'1—t[cbc J,(ax) 1,,(bx) Jp(cx) x7H (69)

The integration in (69) from O to u can be done numerically. The integration from O

to infinity is found in [19].

1P1 I an
111(41)Jo(bx)lp(cx)x”'1dx= 2 a b I‘lQ»+l+m+p)/2)

cm“ r(z+1) r(m+1) m-ot +1 + m - p)/2)

l l . a2 b2
x174 -2-(A+l+m—p),E-(l+l+m+p),l+l,m+l;:;,-;; (70)

where

Re(l+l+m+p)>0; Re(k)<-%; c>a+b (71a,b,c)

280
and where [20]

 

F4 [(1. B; 7. Y; x. y] = 2.0 2:30 (1:31:52721 " " (72)
where

(a ). - [{ffil (73)

For csp(x) = l, (66) becomes

:[dx J,(ax) J,.(bx) (cx)‘“ = c'“ [1 dx J;(ax) J,,(bx) 1:" +1 (1: J,(ax) J..(bx) x'“ (74)

The integral from O to u may be performed analytically, while the one from O to
infinity is found in [21],

 

 

 

 

 

a,r[l+m—n+l]
" 2
£dx1,(a.x).l,.(bx)x"= : , x
2,,b,_,+1r[-l+m+n+l]“
2
XF [1+m2—n+l’l-m2n+l’l+l;a:] (75)
where[16]
'- (a).(B). ,
F[a.mx]=§o (7).»: x (76)

The asymptotic portions of the components of the matrix elements relating the
singular current distribution to tangential electric field at the patch surface will now be

separated. Those matrix components are

 

 

2;“: [an/W I dk k P1+P2tanh(Pzd) -k12 x
wet p1+pzcoth(p2d) e,p1+p2tanh(p2d)

2
x [mm 4.11... ]Jz(k70)a 2[f.~(kR)J.-(kR)-f.-(ka)J.-(ka)] (77)

ill

281

Z’“=;afl 1"" k P1+P2mh(Pzd) -k2
P1+P2C0m(Pzd) e,p1+p2tanh(p2d) ‘

2
x [1411... MA... ]J.<kro)a §[fz(kR)J.-(kR)-f;(ka)J.-(ka)] (78)

Equations (77) and (78) may be written in the form

21'»- ammo-d" k kr 2
: -T‘,[ S( WK 0” §[f;(kR)J;(kR)—fg(ka)J,-(Ica) ] 2C,- Pj(kb)+
:8 I

 

u
+ {d}: [~- ] (79)
where
1: P1 4’ pztanh(p2d)
k E - 2
S() P1+P200th(Pzd) [8,p1+p2tanh(p2d) kl] (80)

Make the definition

 

 

 

310:) a k2 an. I]? s(k) (81)

or
k2

310:) E 8' +1 (82)
then let

Sf(k)ES(k)-si(k) (83)

sf“) 5 [’1 + chkotmpzd) ;;1++p;:a;hnh(p(;:3) - 12 — 3% (84)

Now define

520‘) = at: £06) (85)

Using Equation (14) and (85) in (84),

282

 

 

 

 

 

 

 

k(l klz)+k(1 [‘22)
520:) = lim k *2 k2 - 1:12 - "2 (86)
it... 1:} k} k} k,2 e, +1
k(l—-k7)+k(l-F _e'k(l—-k—2)+k(l--k_2)
Working to order UP, and using his, = k}, (86) becomes
i k? [:22
13(2 — —— - —)
. 1 k? kl’ [:2 k2 2 k2
32(k)'i"..’l4[2+k2 k2] 1,22 'k‘ 'e,+1 (87)
(8, +1)- Z'k—z

 

 

01'

k2 k2 2 k2 k2 k2 2
82(k)=ii-I+n-l[2+-kL 2][—L‘—(2-—1-—-i+——4—-—2 —k12]—e'k—+(88)

 

 

 

 

 

 

4 2+7 e,+1 [:2 [:2 (e,+1) k2) l
_ - "Z 1 2 2 __ 2 _ 2 4 2 _ if, 1‘
32(k)-:l_£n_ e,+1 + 2(e,+1) [k1+k2 k1 k2 + ta,+lk2 2 -c,+l (89)
s(k)=lim 1 4 k2 -E (90)
2 z—.~2(e,+l) e,+l2 2
or, taking the limit
21:22 1:?
k = -—
32( ) (8, + 1)2 2 (91)
Finally, define
s5“) = sac) — mic) - 5,0.) (92)

Since the functions in (79), f,(x), are polynomials, when csp(kb) is substituted for

P,(kb), the result is expressable in terms of integrals of the form
Jdk sac) 1:070) J.-(kt) csp<kb> cab)" (93)
11

Substituting for 30:) using either (83) or (92), the integrations involving 31(k) and sz(k)
are of the form of (66). The integrations involving sf(k) and s;(k) are performed

numerically.

283
The final matrix element relates the electric field on the feed pin due to the feed

pin and singular patch current distributions. This matrix element is, from 16.4(13),

 

8r p22 p27."

“ kzd ‘11 k2
Z}=21ra£dk _ka 13m) 2 sm (pzd) ac +241) _192p +__2_ +
[0381 kaz k0 p22

 

 

 

sinh J Ica 2 4 6

+ (P’d)°()10(kk)[-[A+B £2-+C-R—‘+D£;-]+
PzTn a a a
8 R2 R‘ 1920 R2
c — +30 — — — -
+ 2 [ a2 a‘ ] M4 a2 ]

Sinh(P2d)Jo(k0) ' 1 [ ] 16 3341)

— [)sz 11(ka) ha ZB+4C+GD —ka3(C+GD)+ [ms +

 

R

inh Jka 3 s
+8 (”1)“) [235+4c-—3+60-R—5]—
a a a

P 2T».

 

11(kR)

 

'31—,

 

3
-..L6. [05+w57]+ 3843 5]} (94)
a a ha a

The dominant term in the integrand is the term involving J 0(ka) Jo(kR ). Using

smh(p2d) sinh(pzd) 1
= — 95
P 2T»: ( )

P P
P2 [gwsmpzd Hfsinh(P2d)] szcom(P2d)+-;1]
1 l

 

this term becomes

-Jo(ka )

 

2 4 a
Jo(kR)[A+B-R—2+C-R—‘-+DR—6] (96)
] a a a

P2 [c,coth(P2d) + _I;_2
1

Taking the limit as k approaches infinity in p1, p2 and coth(p2d), the integrand becomes

‘Jo(ka)~’o(kR) R2 R‘ R‘5
(e,+l) [A+B?+C a“ +0 06] (97)

With the asymptotic portion separated out, (94) becomes

lea 2 kzzd sinh(pzd) 8C + 240 1920 kz’
100“!) 2 2 4 + _2 -
1 8r P2 P27}. ka ka p2

Z:I=21tal[dk',—- - -
10x:

 

_VC;
Clear
- g" \_

2,.

284

 

sinh(pzd) 1 R2 R‘ R“
_ p27.“ —k(8,+1) ]Jo(ka)lo(kR) [144-3 y+C7+DF]+

 

 

 

 

+52%kaawom) [7:7 [c%+3o%]-%%]—
_ “”35“” Ada) [11: [23 +4C+GD ]—-;%(C+60 )+ 3:? ]+
+ shh(p;f%nlo(ka) 11(kR) [:15- [ZB §+4C fig-+60 2%]-
—11—1} .-
‘).
$22-34" Mfgjfgm [_C_D_] (98) ‘*

The first integral in (98) is handled numerically and the second is performed using
(75).

17.3 Products of Bessel Functions for Argument Approaching Zero.

In this section, the necessary products of Aff, and Bf... are obtained in a form use-

ful for small arguments. From 17.1(3)

“-1122.- $.52}! 52- H»
m- 2 2 “Ham-2 2 Jllzl-m 2 + 2 Jllz|+m 2 I'll-2M4 ‘2' x
(:1)' 120
x (-:1)‘ (<0 (I)

As the argument kb/2 approaches zero, it is efficient to express the spherical

Bessel functions above in a power series [16].

 

 

 

 

 

. _ (z/2)” _ 22/8 (22/8)2 _, . .
"("2)“ 1.3 - s (2!: +1) {1 (2n+3) + 21 (2n+3)(2n+5) } " 2° (2)
and
j (2,2) = (_1), (—1)- 1 - 3 (—2n - 3) 1- 22/8 + (2%)2 _ . .. n < 0(3)
" (2/2)“ (Zn-+3) 2! (2n+3)(2n+5)

Equations (2) and (3) may be approximated, for 2/2 small, as

 

j, (z /2) = (z/2)” c,I [l - (2 /2)2 d, + (2/2)4 c. ] (4)
where
l
4l'3'$"°(2n+l) n20»
C. = (5)
(_1)l(_1).l.3.5...(.2’!_3) "<0

 

 

_ l
— 2(3+2n) (6)

 

4.

_ 1
’ 8(3+2n )(5+2n) (7)

 

en

285

286
Using equations (4) through (7), (1) can be written as

b2 (il)' :20
Alta = T (kb/Z)”I ' {(_ :1)! (<0 }

 

<ka
x Cmm—2 CHI-u. l-T (dlllm—z +dlll-n )+
2 2 2 2

4
+ (16 (C|z|+m-2+C|z|—n+dlrlm-zdltl-m)]+
2 2 2 2

E

+ C 111+». Cm-m-z
2 2

ka
[l-fi-L(d.z....+d,:.....z)+
2 2

(kb)‘ 8
+ 16 (3mm +CHI-m—2 + 4411+». djll-m-Z) ()
2 2 2 2

 

 

Using equation (8),

+ - b4 lll+|l'l 1 12 0 (”up 12 0
All” A13"! = T (kb/Z) . (‘1), (<0 . 1 (<0 x

 

 

 

 

kb 2 kb 4
x al—L—Layr-S—La, (9)
4 16
where
015 Cyrus—2 ciH-m + C 121+». C 111-m-2 ' C [I'l-Hn-Z Cm-m + C mm C [mm-2 (10)
L 2 2 2 2 2 2 2 2

r
“2" Cmm—z CUL-m(d|1|+n-2 +d|l|-m )+C1u+m cfli-m-Z (4mm +d|l|-m-2 ) ] x
2 2 2 2 2 2 2 2

,
x cfl’im-Z C ll'lm + C mm C ll’l-m-Z +
2 2 2 2

L

+ iCmm-z cLli—m +C|1|+m Cpl-«P2 ]X
2 2 2 2

P

X C [Hm-2 Cum-m (dlrlm-z + d|l'|-—m )+ Cm“. cil'l-m-Z (dll’lm + d|l'|-m-2 ) (11)
2 2 2 2 2 2 2 2

 

 

035

X

287

Cums-2 Cpl-... (dlllm-z +dIII-nn )+C|1|+.n Clzl-m-z (4mm +dIII-m-2 ) 1X
2 2 2 2 2 2 2 2

 

2

 

cil’fim-Z cll’l-m (dlz'lm—z + dll'I-m )+ C mm C Il'l-m-Z (dll'lm + dlz'I-m-z ) ] +
2 2 2 2 2 2 2

Cmm-z CJIL-m (Cmm-z +€|z|—m +dlllm-2 dJu-n. )+
2 2 2 2 2 2

 

b

+ C mm c|I|m-2 (9mm +e|l|—m—2 +d|l|+m dlll-m-Z ) J X
2 2 2 2 2 2

 

C trim—2 cil’l-m + C m». C |l’|-4n—2 +
2 2 2 2

h

C [gm-2 ch’L—m (eJl'J-m-Z + C [VI-m + dfl’lm—z 4423—». )+
2 2 2 2 2 2

 

L

+ C 112+». CLI'J-m-z (C I"l+"' + C |l'l-fll-2 + dmm dll'I-m-z ) +
2 2 2 2 2 2

C II|+m-2 lel-m + C m». ch-m-Z . (12)
2 2 2 2

 

Equation (9) along with equations (10) through (12) is used to obtain the necessary

products of A,” from equations 17.1(4) and 17.1(5). The products involving 8,, are

then obtained from the products of A, ... folowing the development of section 17.1.

17.4 Asymptotic Forms for Spectral Integrands as Argument Approaches
Zero.

From the form of A, ... from the last section, the components of the integrand of
the matrix elements of 17.2(1-4), and 17.2(44-47) are all well behaved as k approaches
zero. The numerical integrations from O to u in 172(5) and 17.2(48) may thus be car-
ried out easily. From equations 17.2(77), 17.208) and 17.2(80), the integration from O
to u in 17.209) may be written as

["90 ll 2
2.7"": 345—— !dk sac) [Art]... +A.:1...]J.(kro)a )3 f.-(kR)J.-(kR)-f.-(ka)J.-(ka)](1)

1 i=1
As I: approaches 0, s(k) is proportional to k, and
[A1113 +Az:1.m] = (kb)'*‘ (2)

so as k approaches zero, the integrand in (1) becomes proportional to

2
k (kb)"*“ Jl(kr0)Z[fi(kR)Ji(kR)-fi(ka)1i(ka)] (3)

i=1
In the worst case for integration to the limit k = 0, l = 0 or ii. For these cases,
(kb)”*“ 11070) = k (4)

so (3) becomes

2
k’ 2 [mm J.-(kR) -f.-(ka) mm] (5)

i=1

For (5) to be integrable to O,
2
2 [f1(kR) J.-(kR) -f.-(ka) mm] (6)

i=1

must diverge to infinity no quicker than 1 Ik’ as k approaches 0. Letting r be either a

288

 

289
orR, look at the term

f1(k¢)11(k3)+f2(kt)12(k1) (7)

From 14.2(55) and 14.2(56)

 

 

_- A 1.2L 3 s 3
mm- *0 (21122215 (211+ am), (40“) -48(k:) ) (8)
_A+ 3 2+ 4_ 2 6_ 4 2
f2(kt)- Ira 273' (21)+—,kac (0“) 80d) )+ “(1)-7(40“) Mk!) +1920“) ) (9)
Since
1
mm ik—‘L (10)

2'1!

for k approaching zero, by inspection the terms involving coefficients A , B and C are
sufficiently well-behaved, as are portions of the terms involving D . The portions of

flan) and f2(k1) that must be examined more carefully are

f1(kt)~ - £7 481/“? (11)
f2(’“) = (TE—)7 1920“)2 (12)

Using (10) through (12), (7) becomes, for k approaching zero

D —48(Iu)3 fl+ 19201:)2 3—L- 4=s—

gk12‘_ glc ‘

Thus the nonconverging term cancels, so the integral in (1) exists.

The final matrix element integration to check is that shown in equation 17.2(98).

Rewriting this here,

" 2d 2
21,, = 2M dk .1221 130‘“) It: _ sinh(pzd) BC + 24D_19ZD k_2 +
[(081 8

 

_ [Sinh(P2d) 1

R2 R: R‘
FZTM — k(€, + l) — +

]Jo(ka)Jo(kR)[A A+B—-+C-- +D
a

 

290

 

 

 

sinh(pzd)Jo(lca) 3 R2 R‘ 1920 R2
+ m... ’°‘”’[m[67+”—]'77 ‘
sinh(pzd) Jo(ka) ' 1 3_8_4D
- p27”... 11(ka) E[ZB+4C+GD]-—(C+ +GD)+— ka ]+
. 5
+ smhwfgjfla) mm) “[23 —+4c —+GD— R as]—
2 m

 

16 R R3 3840 R
'm [C a +w7]+7:]} (14>

Separating the terms in (14) which pose no difficulty at the lower integration limit

from those which do, (14) becomes

" kzd sinh [:2
Z:'=2M1dk'_k'a_ 130m) _2_2.__£P;2."_)_2? 1,
mar mo; 12sz p2

 

 

 

 

 

 

 

 

_ [5???) - k(£,1+1) ]Jo(ka).lo(kR) [A +3 f—z +C §—:+ D ‘Z—Z ]+
_ Sinhwpf;m10(ka)11(ka) [11: [28 +4C +6D ] ]+
+ Sinhw’d”°(ka)1,(kk)-l- [280 5 +4C — +60 15; J}
p21,. [ca
,2“ Ed, 7%, {1m [5135") [8:540 - 133?] ]
+ smh@:;mj°(ka)10(m [i2 [c72- R +31) 5;}‘13425'; _
_ smhwpzj;mlo(ka) 110m) [—%(C +60 )+ 32:? ]+

 

 

. 3
+ 8mh(Pzd)Jo("a) MR) [__li [c gm) L ]+ 33“” g ] }+ (15)

P 2T»:

 

291
The second integration in (15) will be manipulated to make explicit the conver-

gence at the lower limit. Rearrange the second integration to

 

 

" ka sinh(Pzd)-’o(ka) 8C+240 1920
2"“ {‘0‘ 7627 p21... {'M‘a) [ (kc)2 ' (ka)‘ ]

2 4 2
+Jo(kR)[ 8 [c%+3ok—] 1920i}

(“)2 a4 — (“)4 a2
+Jl(lca) -k%(C+3D+3D)—£k‘ai?]-

 

16 R R3 R3 3840 R
_Jl(kR) (“)3 [C a +30 a3 +30 03 ]- am), a }+ (16)
A Bessel identity gives [16]
21
J:+1(Z) = ‘2' 11(2) - 114(2) (17)

Using (17) with (=1, (16) becomes

480
] ”10”) (kc? —

 

" ka sinh(pzd) Jo(ka) BC + 240 1921)
2 dk J ka -
Ml! 10381 p27”... { 2( ) [ Ica2 (ka)4

 

8 R2 R‘ 1920 R2 480 R3
4,“, [W [C a, ,4 ]- ,2 1-18, a, } (18)

Using (17) with (=2, (18) becomes

 

" ka sinh(pzd) Jo(ka) 8C + 241) 48D
211'“ It!" jme, p27." {120(0) _kaz - 13050) —(kal)3

8 R2 R“ 480 R3
-Jz(kR)-— [C7+3D-‘17]+J3(kR)-(-ka—)3-?} (19)
By inspections using (10), equation (19) may be integrated to the limit 0 without
difficulty.

Using (19) in place of the second integral in (15) and combining with the first

integral gives

 

1292

no 2 - 2

 

1 8r P22 P21... p22

 

sinh(pzd) 1 R2 R‘ R‘
- [W-m]10(ka)10(kk) [A +3 :2-+C 7+0 F ]+
smh(p2d)Jo(ka) 1
- p27... J1(ka) [T [23 +4C +60 ] ]+

+ Sinthd) 100“?)
psz

 

l R R3 R5
J1(kR)E [237+4C-a—3+GD-a—5]}

 

sinh(pzd) Jo(ka) 8C + 241) 480
P21." {12(ka) kaz -J3(ka) (“)3

8 R2 R4 48D R3
_J2(kR) (ha), [C a2 +31) 0, ]+J,(kR) (M3 as } (20)

Equation (20) is in the form used for integration to the lower limit of 0, replacing the

form in equation 17.2(98).

VIII CONCLUSION.

A potentially exact solution was obtained for each structure, allowing off reso-
nance and higher order resonance behavior to be obtained accurately. In the case of
the patch antenna and patch array, a set of current distributions capable of modelling
an arbitrary surface current on a circular patch has been developed. Also, two addi—
tional current distributions were developed, each of which models the surface current
diverging from the feed pin onto the patch surface. One of these was used in the

infinite array and the other was used for the single patch.

For the case of the single patch, the Sommerfeld integrations were performed via
real line integration. For all but one of the matrix elements, the integrand involved a
sum of many terms which were products of two or four spherical Bessel functions.
These products can be expressed most simply in terms of a power series plus power
series multiplied by sin and cos. Rather than integrating each matix element
separately-~which would involve summing many products of Bessel functions, each of
which is a summation of power series--all the terms needed for the power series for
some matrix element are integrated. The matrix elements are then pieced together

from the various integrations.

In addition, the asymptotic forms of the integrations for the matrix elements are

identified, separated and performed analytically.

For the infinite arrays, the Sommerfeld integrals are converted into a doubly
infinite summation, each term of which represents a Flouquet mode. A sufficient
number of Flouquet modes are summed to ensure convergence. The range of terms
necessary for convergence in the patch array is greater for the matrix elements involv-

ing the feed pin.

The current distributions on the patch surcace are obtained at the primary reso-
nance, and at three higher resonances. They match well qualitatively with the

expected resonance currents. The resonances come in the expected order also, but at

293

294
lower frequencies than the simplest models predict. The third and fourth resonances

occur very close together.

The received power is obtained for a constant load impedance, and for a matched
load impedance. For the load impedance matched, the power received is flat until the
frequency is high enough to excite surface waves with wavelength equal to the patch
spacing. At this frequency, the received power varies wildly, and it is believed the

solution is inaccurate here.

For the array of patches, the solution was obtained for a frequency range of over
3 to 1, from below the primary resonance to above the fourth resonance. The develop-
ment is for an arbitrary plane wave incidence angle, but only normal incidence is
investigated. Difficulty was encountered in the vicinity of the frequency where the
lowest order surface wave has a spatial period that matches the spacing of the patches.
Approaching this frequency, the central term in the Flouquet summation approaches

the location of the surface wave pole and diverges.

For the dipole array, the solution is again over a 3 to 1 frequency band for near
normal incidence. The the lowest order surface wave is encountered, but doesn’t seem
to damage the solution. The component of the Green function dyad used contains a
term which cancels the pole for normal incidence. The effect of the surface wave pole

is thus smaller at nearly normal incidence.

Solutions for the dipole array as a function of incidence angle, varying from nor-
mal to nearly grazing, are also obtained at several frequencies. At the frequencies
where the TM 0 surface wave pole is implicated, this occurs away from normal

incidence, so the effect is greater on the array properties.

LIST OF REFERENCES

[l] Pozar, D. M. and D. H. Schaubert, "Comparison of architectures for monolithic
phased array antennas," Microwave journal, pp. 93-104, March 1986.

[2] Pozar, D. M. and D. H. Schaubert, "Analysis of an infinite array of rectangular
microstrip patches with idealized probe feeds," IEEE Transactions on Antennas and
Propagation, vol. AP-32, no. 10, pp. 1101-1107, October 1984.

[3] Pozar, D. M., "Finite phased arrays of rectangular microstrip patches," IEEE Tran-
sactions on Antennas and Propagation, vol. AP-34, no. 5, pp. 658-665, May 1986.

[4] 1.0, Y. T., D. Solomon, and W. F. Richards, "Theory and experiment on micros-
trip antennas," IEEE Transactions on Antennas and Propagation, vol. AP-27, March
1979.

[5] Araki, K. and T. Itoh, "Hankel transform domain analysis of open circular micros-
trip radiating structures," IEEE Transactions on Antennas and Propagation, vol. AP-29,
no. 1, pp. 84-89, January 1981.

[6] Yano, S. and A. Ishimaru, "A theoretical study of the input impedance of a circu-
lar microstrip dish antenna," IEEE Transactions on Antennas and Propagation, vol.
AP-29, no. 1, pp. 77-83, January 1981.

[7] Chew, W. C. and J. A. Kong, "Resonances of nonaxial symmetric modes in circu-
lar microstrip disk antenna," Journal of Mathematical Physics, vol. 21, no. 10, pp.
2590-2598, October 1980.

295

296

[8] Liu, C-C, et. al., "Plane wave reflection from microstrip-patch arrays -- theory and
experiment," IEEE Transactions on Antennas and Propagation, vol. AP-33, no. 4, pp.
426-435, April 1985.

[9] Chen, C-C. "Scattering by a tWO-dimensional periodic array of conducting plates,"
IEEE Transactions on Antennas and Propagation, vol. AP-18, no. 5, pp. 660-665, Sep-
tember 1970.

[10] Chew, W. C. and J. A. Kong, "Analysis of a circular microstrip disk antenna
with a thick dielectric substrate," IEEE Transactions on Antennas and Propagation, vol.
AP-29, no. 1, pp. 68-76, January 1981.

[11] Uzunoglu, N. K., N. G. Alexopoulos, and J. G. Fikioris, "Radiation properties of
microstrip dipoles," IEEE Transactions on Antennas and Propagation, vol. AP, no. 6,
pp. 853-858, November 1979.

[12] Alexopoulos, N. G. and I. B. Rana, "Mutual impedance computation between
printed dipoles," IEEE Transactions on Antennas and Propagation, vol. AP-29, no. 1,
pp. 106-111, January 1981

[13] Long, S. A., L. C. Shen, and P. B. Morel, "Theory of the circular-disk printed-
circuit antenna," Proceedings in the IBB, vol. 125, no. 10., October 1978.

[14] Pozar, D. M., and D. H. Schaubert, "Scan blindness in infinite phased arrays of
printed dipoles," IEEE Transactions on Antennas and Propagation, vol. AP-32, no. 6,
pp. 602-610, June 1984.

[15] Papoulis, A., The Fourier Integral and its Applications, Mcgraw-Hill, New York,
1962,p.47.

[16] Abramowitz, M. A., and Stegun, I. A., Handbook of Mathematical Functions,
Dover Publications, New York, 1970.

[17] Bateman, H., Tables of Integral Transforms, Vol 2., Mcgraw-Hill, New York,
1954, p. 42.

297

[19] Bailey, W. N., "Some infinite integrals involving Bessel Functions," Proceedings
London Mathematical Society (2), Vol. 40, 1936, p. 45.

[20] Bailey, W. N., "A reducible case of the fourth type of Appell’s hypergeometric
functions of two variables," Quarterly Journal of Mathematics (Oxford), 4, (1933), p.
305.

[21] Gradshteyn, I. S., Ryzhik, I. M., Table of Integrals, Seriesgand Products,
Academic Press, Florida, 1980.

 

 

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