0
Analysis of Discontinuties in aRectangular Waveguide Using DyadicGreen’s Function Approach inConjunction With Method of MomentsM. D. DeshpandeViGYAN Inc., Hampton, VA. 23681
Contract NAS1-19341
April 1997
National Aeronautics andSpace AdministrationLangley Research CenterHampton, Virginia 23681-0001
NASA CR-201692
1
Contents
List of Figures 2List of Symbols 2Abstract 41. Introduction 4
2. Theory 6Dyadic Green’s function for an electric current sourcein a rectangular waveguide 6
(a) Solution of inhomogeneous Helmholtz equation 6(b) Electromagnetic field due to transverse currents 10(c) Electromagnetic field due to longitudinal current 12(d) Dyadic Green’s function for electric field 15
3. Application 17Analysis of cylindrical post in a rectangular waveguide 17
2. Numerical Results 193. Conclusion 19
References 20
2
List of Figures
Figure 1 Electric current source in a rectangular waveguide 21
Figure 2 Rectangular waveguide with a cylindrical post located at , and
parallel to y-axis 22Figure 3 Reflection coefficient of a y-directed post in a rectangular waveguide as a
function of frequency 23
Figure 4 Transmission coefficient of a y-directed cylindrical post in a rectangular
waveguide as a function of frequency 24
List of Symbolsa,b x-, y-dimensions of rectangular waveguide
magnetic vector potential
x-, y-, and z-components of , respectively
Electric field inside rectangular waveguide
scattered electric field
x-, y-, and z-components of , respectively
incident electric field vector
frequency in cycles per second
dyadic Green’s function for magnetic potential
dyadic Green’s function of electric-type
dyadic Green’s function of electric-type for source free region
xx-, yy-, and zz-components of
functions associated with , respectively
Fourier transforms of , respectively
Magnetic field inside rectangular waveguide
x-, y-, and z- components of magnetic field , respectively
xa2---= z z1–=
A
Ax Ay Az, , A
E
Es
Ex Ey Ez, , E
Ei
f
G
Ge
Ge0
Gxx Gyy Gzz, , G
gxx gyy gzz, , Gxx Gyy Gzz, ,
gxx gyy gzz, , gxx gyy gzz, ,
H
Hx Hy Hz, , H
3
unit impulse current source
amplitude of current
Electric current source
test surface current density
z-component of
=
free-space wave number
=
propagation constants along z-direction inside rectangular waveguide
propagation constant along z-direction
integer associated with waveguide modes
transmission coefficient
reaction of test surface current density with the incident electric field
coordinates of field point in cartesian coordinate system
coordinates of source point in cartesian coordinate system
dummy variables of integration
unit vector along the x-, y-, and z-axis, respectively
self impedance of the y-directed post current
reflection coefficient
gradient operator
delta function
permittivity and permeability of free-space
Neumann’s numbers
free-space impedance
variable of integration
angular freqency equal toFEM finite element methodEM electromagneticEFIE electric field integral equationMoM method of moment
I
I0 J
J
JT
Jz J
j 1–
k0
k k0 εrµr
kz
kI
m n,T
Vy
x y z, ,x' y' z', ,x'' y'' z'', ,x y z, ,
Zyy
Γ
∇δ .( )ε0 µ0,
εm and εn
η0
ϕω 2πf
4
AbstractThe dyadic Green’s function for an electric current source placed in a rectangular
waveguide is derived using a magnetic vector potential approach. A complete solution for the
electric and magnetic fields including the source location is obtained by simple differentiation of
the vector potential around the source location. The simple differentiation approach which gives
electric and magentic fields identical to an earlier derivation is overlooked by the earlier workers
in the derivation of the dyadic Green’s function particularly around the source location. Numeri-
cal results obtained using the Green’s function approach are compared with the results obtained
using the Finite Element Method(FEM).
I. Introduction Analysis and design of dipole, monopole, or aperture radiator to excite high intensity
electromagnetic (EM) fields inside a reverberation chamber can be done using an integral
equation approach. The EM fields inside a reverberation chamber due to a radiator can be
determined by weighting an appropriate dyadic Green’s function with an assumed antenna
current. The Electric Field Integral Equation (EFIE) is then set up by forcing the total tangential
electric field on the antenna surface to be zero. Using the Method of Moments (MoM), EFIE is
then reduced to a matrix equation which can be solved for the antenna current. From the current,
the EM field radiated by the antenna inside a reverberation chamber is determined. Also the
input impedance of the antenna as a function of its location and frequency can be determined.
This work is divided into two parts. In the first part we derive the appropriate dyadic Green’s
function for an electric current source located inside a rectangular waveguide and cavity. Detailed
steps involved in this derivation are reported in this document. The second part of this work,
which will be reported in subsequent documents, consists of an application of the dyadic Green’s
5
function to analyze a dipole antenna placed in a reverberation chamber.
Knowledge of a dyadic Green’s function for cylindrical waveguides and cavities is
essential for analyzing and designing antennas and arbitrarily shaped objects placed inside a
cylindrical waveguide and cavity [1,2]. A detailed derivation of a dyadic Green’s function for the
rectangular waveguide was presented by Tai [3]. In deriving these dyadic Green’s function valid
for both source and source free regions, an additional term must be added to the classical
representation of the field expressions [4]. To include the additional term in the classical
representation, Tai [5] has presented an approach based upon the use of eigenvector functions. In
[6], an electric-type dyadic Green’s function is obtained through a magnetic-type dyadic Green’s
function obtained using the theory of distributions.
The purpose of this communication is to present a simple method using the vector
potential approach to determine the dyadic Green’s function valid in the entire region of a
cylindrical waveguide. For an arbitrarily oriented electric current source in a rectangular
waveguide, expressions for the magnetic vector potential are obtained by solving the
inhomogeneous Helmholtz equation. The electric fields and hence the dyadic Green’s function of
the electric-type is then obtained by taking the derivatives of the magnetic vector potential. In the
process of finding the electric field, if the derivatives of the vector potential are carefully defined,
the additional term discussed in [4-6] automatically follows. Reflection and transmission
coefficients due to a y-directed cylindrical post placed in a rectangular waveguide and excited by
a dominant mode are derived and numerical results are compared with the results obtained by the
Finite Element Method [7].
6
II. TheoryDyadic Green’s Function for an Electric Current Source in a RectangularWaveguide
(a) Solution of Inhomogeneous Helmholtz Equation:
Consider an infinite rectangular waveguide with electric current source as shown in
figure 1. The electromagnetic fields inside the waveguide due to can be determined from
(1)
(2)
where the assumed time variation has been suppressed. The magnetic vector potential
appearing in (1) and (2) statisfies the inhomogeneous wave equation
(3)
If is the dyadic Green’s function for the rectangular waveguide for a unit
impulse current source inside the waveguide, then the magnetic vector potential
can be written in the form
(4)
Substituting (4) in (3) we get
(5)
where is an unit dyadic, defined as . Equation (5) may be written in compo-
nent form as
J
J
H x y z, ,( ) 1µ0----- A∇×=
E x y z, ,( ) j– ω
k02
--------- k02A A∇•
∇+=
ejωt
A x y z, ,( )
∇2A x y z, ,( ) k0
2A x y z, ,( )+ µ0– J x' y' z', ,( )=
G x y z x' y' z', , , , ,( )
I x' y' z', ,( )
A x y z, ,( )
A x y z, ,( ) G x y z x' y' z', , , , ,( ) • J x' y' z', ,( ) x'd y'd z'd∫Source
∫∫=
∇2G .( ) k0
2G .( )+ µ0– Iδ x x'–( ) δ y y'–( ) δ z z'–( )=
I I xx yy zz+ +=
7
(6)
(7)
(8)
Because of the nature of the problem and the boundary conditions, the other components of the
dyadic Green’s function will not be excited and hence are not considered. The solutions
of (6), (7), and (8) may be assumed in the following forms
(9)
(10)
(11)
Substituting (9) in (6), (10) in (7) and (11) in (8) we get
(12)
(13)
(14)
∇2Gxx .( ) k0
2Gxx .( )+ µ0– δ x x'–( ) δ y y'–( ) δ z z'–( )=
∇2Gyy .( ) k0
2Gyy .( )+ µ0– δ x x'–( ) δ y y'–( ) δ z z'–( )=
∇2Gzz .( ) k0
2Gzz .( )+ µ0– δ x x'–( ) δ y y'–( ) δ z z'–( )=
G .( )
Gxx .( ) gxx x' y' z' z, , ,( ) mπxa
----------- cos
nπyb
--------- sin
n 1=
∞
∑m 0=
∞
∑=
Gyy .( ) gyy x' y' z' z, , ,( ) mπxa
----------- sin
nπyb
--------- cos
n 0=
∞
∑m 1=
∞
∑=
Gzz .( ) gzz x' y' z' z, , ,( ) mπxa
----------- sin
nπyb
--------- sin
n 1=
∞
∑m 1=
∞
∑=
z2
2
d
dgxx .( ) kI
2gxx .( )+
mπx
a-----------
cosnπyb
--------- sin µ0– δ x x'–( ) δ y y'–( ) δ z z'–( )=
z2
2
d
dgyy .( ) kI
2gyy .( )+
mπx
a-----------
sinnπyb
--------- cos µ0– δ x x'–( ) δ y y'–( ) δ z z'–( )=
z2
2
d
dgzz .( ) kI
2gzz .( )+
mπx
a-----------
sinnπy
b---------
sin µ0– δ x x'–( ) δ y y'–( ) δ z z'–( )=
8
where . Multiply (12) by and integrate over
the cross section of waveguide we get
(15)
Likewise the equations (13) and (14) yield
(16)
(17)
where and are Neumann’s numbers [7] and equal to for and 2 for . In
order to determine the solution of the inhomogeneous differential equation (15) let us
assume (18)
Substitution of (18) in (15), multiplying by and integrating over z leads to
(19)
Substitution of (19) in (18) yields
(20)
The integrand in equation (20) has poles at , therefore the integral in (20) can be
evaluated using contour integration in the complex domain [8]. Hence
kI2
k02 mπ
a-------
2–
nπb
------ 2
–=m'πx
a------------
cosn'πy
b----------
sin
z2
2
d
dgxx .( ) kI
2gxx .( )+
µ0
εmεn
ab-----------–
mπx'a
------------ cos
nπy'b
---------- sin δ z z'–( )=
z2
2
d
dgyy .( ) kI
2gyy .( )+
µ0
εmεn
ab-----------–
mπx'a
------------ sin
nπy'b
---------- cos δ z z'–( )=
z2
2
d
dgzz .( ) kI
2gzz .( )+
µ0
εmεn
ab-----------–
mπx'a
------------ sin
nπy'b
---------- sin δ z z'–( )=
εm εn 1 m 0= m 0≠
gxx .( ) gxx˜ kz( ) e
jkzz kzd
∞–
∞
∫=
ej– kz'z
gxx˜ kz( ) µ0
εmεn
2πab-------------–
mπx'a
------------ cos
nπy'b
---------- sin
k– z2
kI2
+ ------------------------------------------------------e
j– kzz'=
gxx .( ) µ0
εmεn
2πab------------- mπx'
a------------
cosnπy'
b----------
sin1
kz2
kI2
– -----------------------e
j– kzz'ejkzz kzd
∞–
∞
∫=
kz kI±=
9
(21)
where + sign in the exponential is taken when and - sign in the exponential is taken
when . Likewise, and are obtained as
(22)
(23)
Substituting (21), (22), and (23) in (9), (10), and (11), respectively, the x-, y-, and z-components
of the dyadic Green’s function are obtained as
(24)
(25)
(26)
The x-, y- and z-components of the magnetic vector potential due to the x-,y-, and z-directed
currents are then obtained as
(27)
gxx .( )j– µ0
2kI-----------
εmεn
ab----------- mπx'
a------------
cosnπy'
b----------
sin ej± kI z z'–( )
=
z z'–( ) 0<
z z'–( ) 0> gyy .( ) gzz z( )
gyy .( )j– µ0
2kI-----------
εmεn
ab----------- mπx'
a------------
sinnπy'
b----------
cos ej± kI z z'–( )
=
gzz .( )j– µ0
2kI-----------
εmεn
ab----------- mπx'
a------------
sinnπy'
b----------
sin ej± kI z z'–( )
=
Gxx .( )j– µ0
2kI-----------
εmεn
ab----------- mπx'
a------------
cosnπy'
b----------
sinmπx
a-----------
cosnπy
b---------
sin ej± kI z z'–( )
n 0=
∞
∑m 0=
∞
∑=
Gyy .( )j– µ0
2kI-----------
εmεn
ab----------- mπx'
a------------
sinnπy'
b----------
cosmπx
a-----------
sinnπy
b---------
cos ej± kI z z'–( )
n 0=
∞
∑m 0=
∞
∑=
Gzz .( )j– µ0
2kI-----------
εmεn
ab----------- mπx'
a------------
sinnπy'
b----------
sinmπx
a-----------
sinnπyb
--------- sin e
j± kI z z'–( )
n 0=
∞
∑m 0=
∞
∑=
Ax x y z, ,( )j– µ0
2kI-----------
εmεn
ab----------- mπx
a-----------
cosnπy
b---------
sinn 0=
∞
∑m 0=
∞
∑=
Jx x' y' z', ,( ) mπx'a
------------ cos
nπy'b
---------- sin e
j± kI z z'–( )vd∫∫
Source∫
10
(28)
(29)
The expressions in (27)-(29) are the required solution of inhomogeneous Helmoltz equation given
in (3).
(b) Electromagnetic Fields Due to Transverse Currents:
The electric and magnetic fields due to are obtained from (2) as
(30)
(31)
Ay x y z, ,( )j– µ0
2kI-----------
εmεn
ab----------- mπx
a-----------
sinnπy
b---------
cosn 0=
∞
∑m 0=
∞
∑=
Jy x' y' z', ,( ) mπx'a
------------ sin
nπy'b
---------- cos e
j± kI z z'–( )vd∫∫
Source∫
Az x y z, ,( )j– µ0
2kI-----------
εmεn
ab----------- mπx
a-----------
sinnπy
b---------
sinn 0=
∞
∑m 0=
∞
∑=
Jz x' y' z', ,( ) mπx'a
------------ sin
nπy'b
---------- sin e
j± kI z z'–( )vd∫∫
Source∫
Ax x y z, ,( )
Ex Ax( ) j– ω
k02
---------j– µ0
2kI-----------
εmεn
ab----------- k0
2 mπa
------- 2
– mπx
a-----------
cosnπyb
--------- sin
n 0=
∞
∑m 0=
∞
∑=
Jx x' y' z', ,( ) mπx'a
------------ cos
nπy'b
---------- sin e
j± kI z z'–( )vd∫∫
Source∫
Ey Ax( ) j– ω
k02
---------j– µ0
2kI-----------
εmεn
ab----------- mπ
a-------–
nπb
------ mπxa
----------- sin
nπyb
--------- cos
n 0=
∞
∑m 0=
∞
∑=
Jx x' y' z', ,( ) mπx'a
------------ cos
nπy'b
---------- sin e
j± kI z z'–( )vd∫∫
Source∫
Ez Ax( ) j– ω
k02
---------j– µ0
2kI-----------
εmεn
ab----------- jkI±( ) mπ
a-------–
mπxa
----------- sin
nπyb
--------- sin
n 0=
∞
∑m 0=
∞
∑=
11
(32)
(33)
(34)
(35)
Similarly, the electric and magnetic fields due to are obtained as
(36)
(37)
Jx x' y' z', ,( ) mπx'a
------------ cos
nπy'b
---------- sin e
j± kI z z'–( )vd∫∫
Source∫
Hx Ax( ) 0=
Hy Ax( ) j–2kI-------
εmεn
ab----------- jkI±( ) mπx
a-----------
cosnπy
b---------
sinn 0=
∞
∑m 0=
∞
∑=
Jx x' y' z', ,( ) mπx'a
------------ cos
nπy'b
---------- sin e
j± kI z z'–( )vd∫∫
Source∫
Hz Ax( ) j2kI-------
εmεn
ab----------- nπy
b---------
mπxa
----------- cos
nπyb
--------- sin
n 0=
∞
∑m 0=
∞
∑=
Jx x' y' z', ,( ) mπx'a
------------ cos
nπy'b
---------- sin e
j± kI z z'–( )vd∫∫
Source∫
Ay x y z, ,( )
Ex Ay( ) j– ω
k02
---------j– µ0
2kI-----------
εmεn
ab----------- nπ
b------–
mπa
------- mπx
a-----------
cosnπy
b---------
sinn 0=
∞
∑m 0=
∞
∑=
Jy x' y' z', ,( ) mπx'a
------------ sin
nπy'b
---------- cos e
j± kI z z'–( )vd∫∫
Source∫
Ey Ay( ) j– ω
k02
---------j– µ0
2kI-----------
εmεn
ab----------- k0
2 nπb
------ 2
– mπx
a-----------
sinnπy
b---------
cosn 0=
∞
∑m 0=
∞
∑=
Jy x' y' z', ,( ) mπx'a
------------ sin
nπy'b
---------- cos e
j± kI z z'–( )vd∫∫
Source∫
Ez Ay( ) j– ω
k02
---------j– µ0
2kI-----------
εmεn
ab----------- jkI±( ) nπ
b------–
mπxa
----------- sin
nπyb
--------- sin
n 0=
∞
∑m 0=
∞
∑=
12
(38)
(39)
(40)
(41)
(c) Electromagnetic Fields Due to Longitudinal Current:
The transverse electric fields due to are obtained from (2) as
(42)
(43)
In obtaining the longitudinal electric field representation due to , special
attention is required in performing the differentiation with respect to z on . Since
Jy x' y' z', ,( ) mπx'a
------------ sin
nπy'b
---------- cos e
j± kI z z'–( )vd∫∫
Source∫
Hx Ay( ) j2kI-------
εmεn
ab----------- jkI±( ) mπx
a-----------
sinnπy
b---------
cosn 0=
∞
∑m 1=
∞
∑=
Jy x' y' z', ,( ) mπx'a
------------ sin
nπy'b
---------- cos e
j± kI z z'–( )vd∫∫
Source∫
Hy Ay( ) 0=
Hz Ay( ) j–2kI-------
εmεn
ab----------- nπy
b---------–
mπxa
----------- sin
nπyb
--------- sin
n 0=
∞
∑m 0=
∞
∑=
Jy x' y' z', ,( ) mπx'a
------------ sin
nπy'b
---------- cos e
j± kI z z'–( )vd∫∫
Source∫
Az x y z, ,( )
Ex Az( ) j– ω
k02
---------j– µ0
2kI-----------
εmεn
ab----------- jkI±( ) mπ
a-------
mπxa
----------- cos
nπyb
--------- sin
n 0=
∞
∑m 0=
∞
∑=
Jz x' y' z', ,( ) mπx'a
------------ sin
nπy'b
---------- sin e
j± kI z z'–( )vd∫∫
Source∫
Ey Az( ) j– ω
k02
---------j– µ0
2kI-----------
εmεn
ab----------- jkI±( ) nπ
b------
mπxa
----------- sin
nπyb
--------- cos
n 0=
∞
∑m 0=
∞
∑=
Jz x' y' z', ,( ) mπx'a
------------ sin
nπy'b
---------- sin e
j± kI z z'–( )vd∫∫
Source∫
Az x y z, ,( )
Az x y z, ,( )
13
is continuous as a function of z, the first derivative of is straightforward,
and therefore causes no difficulty. Hence
(44)
where double prime quantities are the dummy variables of integration. Clearly is
discontinuous at , so in performing the derivative of with respect to z
around , care must be exercised to account for the jump in as one crosses
the point. The behavior at is properly accounted for by an impulse function at
the point whereas the differentiation throughout the rest of region poses no problem, therefore,
(45)
Integrating on in the second term of equation (45) yields
Az x y z, ,( ) Az x y z, ,( )
z∂∂
Az x y z, ,( )j– µ0
2kI-----------
εmεn
ab----------- jkI±( ) mπx
a-----------
sinnπy
b---------
sinn 0=
∞
∑m 0=
∞
∑=
Jz x'' y'' z'', ,( ) mπx''a
------------- sin
nπy''b
----------- sin e
j± kI z z''–( )v''d∫∫
Source∫
z∂∂
Az x y z, ,( )
z z''=z∂
∂Az x y z, ,( )
z z''=z∂
∂Az x y z, ,( )
z z''= z z''=
z2
2
∂
∂Az x y z, ,( )
j– µ0
2kI-----------
εmεn
ab----------- kI
2–
mπxa
----------- sin
nπyb
--------- sin
n 0=
∞
∑m 0=
∞
∑=
Jz x'' y'' z'', ,( ) mπx''a
------------- sin
nπy''b
----------- sin e
j± kI z z''–( )v''d∫∫
Source∫
j– µ0
2-----------
εmεn
ab----------- 2j–( ) mπx
a-----------
sinnπyb
--------- sin
n 0=
∞
∑m 0=
∞
∑+
Jz x'' y'' z'', ,( ) δ z z''–( ) mπx''a
------------- sin
nπy''b
----------- sin e
j± kI z z''–( )v''d∫∫
Source∫
z''
14
(46)
Expanding in the Fourier sine series over the domains and
where and [9], it can be shown that
(47)
Using (47), (46) can be written as
(48)
The longitudinal component of the electric field is then obtained using (2) as
(49)
z2
2
∂
∂Az x y z, ,( )
j– µ0
2kI-----------
εmεn
ab----------- kI
2–
mπxa
----------- sin
nπyb
--------- sin
n 0=
∞
∑m 0=
∞
∑=
Jz x'' y'' z'', ,( ) mπx''a
------------- sin
nπy''b
----------- sin e
j± kI z z''–( )v''d∫∫
Source∫
µ0– Jz x'' y'' z, ,( ) 4ab------ mπx
a-----------
sinnπyb
--------- sin
mπx''a
------------- sin
nπy''b
----------- sin
n 0=
∞
∑m 0=
∞
∑source
∫∫
dx''dy''
δ x x''–( ) δ y y''–( ) 0 x a≤ ≤
0 y b≤ ≤ 0 x'' a≤ ≤ 0 y'' b≤ ≤
δ x x''–( ) δ y y''–( ) 4ab------ mπx
a-----------
sinnπy
b---------
sinmπx''
a-------------
sinnπy''
b-----------
sinn 0=
∞
∑m 0=
∞
∑=
z2
2
∂
∂Az x y z, ,( ) µ0– Jz x y z, ,( )
j– µ0
2kI-----------
εmεn
ab----------- kI
2–
mπxa
----------- sin
nπyb
--------- sin
n 0=
∞
∑m 0=
∞
∑+=
Jz x'' y'' z'', ,( ) mπx''a
------------- sin
nπy''b
----------- sin e
j± kI z z''–( )v''d∫∫
Source∫
Ez Az( ) jω
k02
------µ0Jz x y z, ,( ) j– ω
k02
---------j– µ0
2kI-----------
εmεn
ab----------- k0
2kI
2–
mπxa
----------- sin
nπyb
--------- sin
n 1=
∞
∑m 1=
∞
∑+=
Jz x' y' z', ,( ) mπx'a
------------ sin
nπy'b
---------- sin e
j± kI z z'–( )vd∫∫
Source∫
15
The magnetic field components due to are obtained as
(50)
(51)
(52)
The total electric and magnetic fields inside the waveguide due to is then obtained by
superpostion of the electromagnetic fields due to , , and .
(d) Dyadic Green’s Function for Electric Field:
It is instructive at this point to defined the dyadic Green’s function for the electri field
formulation. To this end, we write the vector wave equation for the electric field as
(53)
If the electric field in terms of the dyadic Green’s function is given as
(54)
Substituting (54) in (53) the wave equation for the dyadic Green’s function of electric-type is
obtained as
Az
Hx Az( ) j–2kI-------
εmεn
ab-----------nπ
b------ mπx
a-----------
sinnπy
b---------
cosn 0=
∞
∑m 0=
∞
∑=
Jz x' y' z', ,( ) mπx'a
------------ sin
nπy'b
---------- sin e
j± kI z z'–( )vd∫∫
Source∫
Hy Az( ) j2kI-------
εmεn
ab-----------mπ
a------- mπx
a-----------
cosnπyb
--------- sin
n 0=
∞
∑m 0=
∞
∑=
Jz x' y' z', ,( ) mπx'a
------------ sin
nπy'b
---------- sin e
j± kI z z'–( )vd∫∫
Source∫
Hz Az( ) 0=
J
Ax Ay Az
E∇×∇× k02E– j– ωµ0J=
Ge x' y' z' x y z, ,⁄, ,( )
E x y z, ,( ) jωµ0 Ge x' y' z' x y z, ,⁄, ,( ) J x' y' z', ,( )• v'd∫∫∫–=
16
(55)
From equations (30)-(32), (36)-(38), (42), (43), and (49), the dyadic Green’s function can be
written as
(56)
where is given by
(57)
Ge .( )∇×∇× k02Ge .( )– Iδ x x'–( ) δ y y'–( ) δ z z'–( )=
Ge .( ) Ge0 .( ) δ x x'–( ) δ y y'–( ) δ z z'–( )
k02
-------------------------------------------------------------------– zz=
Ge0 .( )
Ge0 .( ) 1
k02
----- j–2kI-------
εmεn
ab-----------e
j± kI z z'–( )
n 0=
∞
∑m 0=
∞
∑=
k02 mπ
a-------
2–
mπxa
----------- cos
nπyb
--------- sin
mπx'a
------------ cos
nπy'b
---------- xxsin
mπa
-------– nπ
b------ mπx
a-----------
sinnπyb
--------- mπx'
a------------
cosnπy'
b----------
yxsincos+
jkI±( ) mπa
-------– mπx
a-----------
sinnπy
b---------
mπx'a
------------ cos
nπy'b
---------- zxsinsin+
nπb
------– mπ
a-------
mπxa
----------- cos
nπyb
--------- sin
mπx'a
------------ sin
nπy'b
---------- cos xy+
k02 nπ
b------
2–+
mπxa
----------- sin
nπyb
--------- cos
mπx'a
------------ sin
nπy'b
---------- yycos
jkI±( ) nπb
------– mπx
a-----------
sinnπyb
--------- sin
mπx'a
------------ sin
nπy'b
---------- zysin+
jkI±( ) mπa
------- mπx
a-----------
cosnπyb
--------- sin
mπx'a
------------ sin
nπy'b
---------- sin+ xz
jkI±( ) nπb
------ mπx
a-----------
sinnπy
b---------
cosmπx'
a------------
sinnπy'
b----------
yzsin+
k02
kz2
–mπx
a-----------
sinnπy
b---------
sinmπx'
a------------
sinnπy'
b----------
zzsin+
17
The expression in (57) is identical to the Green’s function reported in reference [1].
III. Application
Analysis of Cylindrical Post in a Rectangular Waveguide:
Consider a rectangular waveguide with a cylindrical post as shown in figure 2. It is
assumed that the waveguide is excited by the dominant mode from the right. For simplicity it is
assumed that the surface current density on the post as
(58)
Let be the scattered electric field due to the current and be the incident electric
field due to TE10 mode. The total electric field inside the waveguide is then given by
. Subjecting the total tangential electric field on the surface of the post to zero, we
get following electric field integral equation:
(59)
where the subscript is for the tangential component. Selecting a testing surface current density
as which resides on the cylindrical surface, Galerkin’s procedure reduces equation (59) to
(60)
Equation (60) can be written in a algebric form as
(61)
J yI0δ x' a2---–
δ z z'–( )=
Es J
J Ei
Es J
Ei+
Es J
Ei+
t 0=
t
JT
Es J
JT•⟨ ⟩ Ei JT•⟨ ⟩+ 0=
ZyyI0 Vy+ 0=
18
where , , with the indicated integration performed in
cylindrical coordinates. Using (54) and (56), the expression for is obtained as
(62)
Assuming an unit amplitude dominant mode can be written as
(63)
Using (63) the quantity can be written as
(64)
The algebric equation (61) can be solved for . The reflected amplitude of the dominant mode
field at a reference plane is then determined from
(65)
The transmitted amplitude of the dominant mode at the reference plane is obatined as
(66)
ZyyI0 Es J
JT•⟨ ⟩= Vy Ei JT•⟨ ⟩=
Zyy
Zyy ωµ02b a⁄( ) 1kIπ-------- e
jkIr0 ϕ( )sin– mπr0
a------------- ϕ( )cos
cos
0
π
∫m 1 3 5 .., , ,=
∑–= dϕ
Ei
Ei y2
ab------
πxa------
ej k0
2 πa---
2–
z–
sin=
Vy
Vy2ba
------2π---
e
j k02 π
a---
2–
z1–
ej k0
2 πa---
2–
r0 ϕ( )sin– πr0
a-------- ϕ( )cos
cos dϕ
0
π
∫=
I0
z 0=
Γk0η0I0–
k02 π
a---
2–
----------------------------------- b
2a------e
j k02 π
a---
2–
2z1–
=
z 2z1=
T 1k0η0I0–
k02 π
a---
2–
----------------------------------- b
2a------+
ej k0
2 πa---
2–
2z1–
=
19
IV. Numerical Results
To validate the Green’s function derived in this report, a y-directed cylindrical post of
radius placed at in a rectangular waveguide with ,
and excited by an unit amplitude dominant mode field is considered. The reflection
coefficient at the plane and the transmission coefficient at the plane due
to the presence of the probe are calculated using expressions (65) and (66) and presented in
figures 3 and 4 along with the numerical results obtained using the FEM method [7,8]. The close
agreement between the results obtained from two different numerical methods confirms the
validity of the Green’s functions derived here.
V. ConclusionThe complete dyadic Green’s function for a electric current source located inside a
rectangular waveguide is derived using the magnetic vector potential approach. The magnetic
vector potential for an electric current source in a rectangular waveguide is obtained by solving
the inhomogeneous Helmholtz’s equation. The electric and magnetic fields are obtained from the
magnetic vector potential through spatial differentiation. The fields which are valid over the
source region are obtained by carefully differentiating the vector potential around the source
location. The electric and magentic field expressions obtained by the present method are found to
be identical with the expressions reported in the literature. Numerical results on the reflection and
transmission coefficients using the Green’s function approach are in a good agreement with the
numerical results obtained using the FEM techniques.
r0 0.1cm= xa2--- z, 0.5= =
a 2.25cm=
b 1.02cm=
z 0.0cm= z 1.0cm=
20
References[1] J. J. H. Wang, “ Analysis of a three-dimensional arbitrarily shaped dielectric or biological
body inside a rectangular waveguide, ”IEEE Trans. on Microwave Theory and Techniques,Vol. MTT-26, No. 7, pp 457-462, July 1978.
[2] M. S. Leong, et al, “Input impedance of a coaxial probe located inside a rectangular cavity:theory and experiment, ”IEEE Trans. on Microwave Theory and Techniques, MTT Vol. 44,No. 7, pp. 1161-1164, July 1996.
[3] C. T. Tai,Dyadic Green’s function in electromagnetic theory, Scranton, PA: IntextEducational Publishers, 1972.
[4] R. E. Collins, “On the incompleteness of E and H modes in waveguides, ”Can. J. Phys.,Vol. 51, pp. 1135-1140, 1973.
[5] C. T. Tai, “On the eigen-function expansion of dyadic Green’s functions, ”Proc. IEEE, Vol.61, pp. 480-481, March 1974.
[6] Y. Rahmat-Samii, “On the question of computation of the dyadic Green’s function at thesource region in waveguides and cavities, ”IEEE Trans. Microwave Theory andTechniques, Vol. MTT-23, pp. 762-765, September 1975.
[7] M. D. Deshpande and C. J. Reddy, “Application of FEM to estimate complex permittivity ofdielectric material at microwave frequency using waveguide measurements, ”,NASAContractor Report 198203, August 1995.
[8] M. D. Deshpande, et al, “ A new approach to estimate complex permittivity of dielectricmaterial at microwave frequencies using waveguide measurements, ” To appear in IEEETrans. on Microwave Theory and Techniques, March 1997.
21
X
Y
Z
J
Electric CurrentSource InsideRectangular Waveguide
a
b
Figure 1 Electric current source inside a rectangular waveguide
22
XY
TE10Mode
Incident
a
b a/2
CylindricalProbe Radiusr0
MatchedLoad
Figure 2 Rectangular waveguide with a cylindrical post parallel to y-axis placed at x = a/2, z = z1.
z1
Z
23
8.5 9.0 9.5 10.0 10.5 11.0 11.5 12.0-1.0
-0.5
0.0
0.5
1.0
ΓR
eflec
tion
Coe
ff.
Real
Imaginary
Frequency (GHz)
Real Part
Imaginary Part} Present Method
Real PartImaginary Part} FEM Method [ ]
Figure 4 Reflection coefficient of a y-directed post in a rectangular waveguide as a function of frequency.
8
24
8.5 9.0 9.5 10.0 10.5 11.0 11.5 12.0-1.0
-0.5
0.0
0.5
1.0
Real Part
Imaginary Part} Present Method
Real Part } FEM Method [8]Imaginary Part
Tra
nsm
issi
on C
oeff.
T
Frequency (GHz)
Imaginary
Real
Figure 5 Transmission coefficient of a y-directed post in a rectangular waveguide as a function of frequency.
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