Page 1

0

Analysis of Discontinuties in a

Rectangular Waveguide Using Dyadic

Green’s Function Approach in

Conjunction With Method of Moments

M. D. Deshpande

ViGYAN Inc., Hampton, VA. 23681

Contract NAS1-19341

April 1997

National Aeronautics and

Space Administration

Langley Research Center

Hampton, Virginia 23681-0001

NASA CR-201692

Page 2

1

Contents

List of Figures

List of Symbols

Abstract

1. Introduction

2

2

4

4

2. Theory

6

Dyadic Green’s function for an electric current source

in a rectangular waveguide6

(a) Solution of inhomogeneous Helmholtz equation

(b) Electromagnetic field due to transverse currents

(c) Electromagnetic field due to longitudinal current

(d) Dyadic Green’s function for electric field

3. Application

Analysis of cylindrical post in a rectangular waveguide

2. Numerical Results

3. Conclusion

References

6

10

12

15

17

17

19

19

20

Page 3

2

List of Figures

Figure 1Electric current source in a rectangular waveguide 21

Figure 2Rectangular waveguide with a cylindrical post located at, and

parallel to y-axis

Reflection coefficient of a y-directed post in a rectangular waveguide as a

22

Figure 3

function of frequency 23

Figure 4 Transmission coefficient of a y-directed cylindrical post in a rectangular

waveguide as a function of frequency24

List of Symbols

a,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

x

a

2-- -

=

zz1

–=

A

AxAyAz

,,

A

E

Es

ExEyEz

,,

E

Ei

f

G

Ge

Ge0

GxxGyyGzz

gxxgyygzz

g ˜xxg ˜yyg ˜zz

,,

G

,

,

,

,

GxxGyyGzz

gxxgyygzz

,

,

,

,

H

HxHyHz

,,

H

Page 4

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 to

finite element method

electromagnetic

electric field integral equation

method of moment

FEM

EM

EFIE

MoM

I

I0

J

J

JT

Jz

J

j

k0

1–

kk0εrµr

kz

kI

m n

,

T

Vy

x y z

, ,

x' y' z'

, ,

x'' y'' z''

,

x ˆ y ˆ z ˆ

, ,

,

Zyy

Γ

∇

δ . ( )

ε0µ0

εmand εn

η0

ϕ

ω

,

2πf

Page 5

4

Abstract

The 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

Page 6

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].

Page 7

6

II. Theory

Dyadic Green’s Function for an Electric Current Source in a Rectangular

Waveguide

(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

I

. Equation (5) may be written in compo-

nent form as

J

J

H x y z

, ,()

1

µ0

-----

A

? ? ∇×

=

E x y z

, ,()

j – ω

k0

ejωt

2

-------- - k0

2AA

? ? ∇•

∇

+=

A x y z

, ,()

∇2A x y z

(, ,)

k0

2A x y z

(, ,)

+

µ0

–

Jx' 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'

, , , , ,( ) • Jx' y' z'

, ,() x'

dy'

dz'

d

∫

Source

∫∫

=

∇2G

.

()

k0

2G

.

()

+

µ0

–

Iδ xx'–

( ) δ yy'–

( ) δ zz'–

()

=

Ix ˆx ˆ

y ˆy ˆ

z ˆz ˆ

++=

Page 8

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 yy'–

( ) δ zz'–

()

=

∇2Gyy

.

()

k0

2Gyy

.

()

+

µ0

–

δ x

(

x'–

) δ yy'–

( ) δ zz'–

()

=

∇2Gzz

.

()

k0

2Gzz

.

()

+

µ0

–

δ x

(

x'–

) δ yy'–

( ) δ zz'–

()

=

G

.

()

Gxx

.

()

gxxx' y' z' z

, , ,()

mπx

a

---------- -

cos

nπy

b

-------- -

sin

n

1=

∞

∑

m

0=

∞

∑

=

Gyy

.

()

gyyx' y' z' z

, , ,()

mπx

a

---------- -

sin

nπy

b

-------- -

cos

n

0=

∞

∑

m

1=

∞

∑

=

Gzz

.

()

gzzx' y' z' z

, , ,()

mπx

a

---------- -

sin

nπy

b

-------- -

sin

n

1=

∞

∑

m

1=

∞

∑

=

z2

2

d

dgxx

.

()

kI

2gxx

.

()

+

mπx

a

---------- -

cos

nπy

b

-------- -

sin

µ0

–

δ x

(

x'–

) δ yy'–

( ) δ zz'–

()

=

z2

2

d

dgyy

.

()

kI

2gyy

.

()

+

mπx

a

---------- -

sin

nπy

b

-------- -

cos

µ0

–

δ x

(

x'–

) δ yy'–

() δ zz'–

()

=

z2

2

d

dgzz

.

()

kI

2gzz

.

()

+

mπx

a

---------- -

sin

nπy

b

-------- -

sin

µ0

–

δ x

(

x'–

) δ yy'–

( ) δ zz'–

()

=

Page 9

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

kI

2

k0

2

mπ

a

-------

2

–

nπ

b

----- -

2

–=

m'πx

a

----------- -

cos

n'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

kz

d

∞

–

∞

∫

=

e

j – kz'z

gxx

˜

kz

()µ0

εmεn

2πab

------------ -

–

mπx'

a

----------- -

cos

----------------------------------------------------- -e

nπy'

b

--------- -

sin

k

–z

2

kI

2

+

j – kzz'

=

gxx

.

()µ0

εmεn

2πab

------------ -

mπx'

a

----------- -

cos

nπy'

b

--------- -

sin

1

kz

2

kI

2

–

---------------------- -e

j – kzz'e

jkzz

kz

d

∞

–

∞

∫

=

kz

kI

±

=

Page 10

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

----------- -

cos

nπy'

b

--------- -

sin

e

j ± kIzz'–

()

=

zz'–

()

0

<

zz'–

()

0

>

gyy

.

()

gzzz ( )

gyy

.

()

j – µ0

2kI

---------- -εmεn

ab

---------- -

mπx'

a

----------- -

sin

nπy'

b

--------- -

cos

e

j ± kIzz'–

()

=

gzz

.

()

j – µ0

2kI

---------- -εmεn

ab

---------- -

mπx'

a

----------- -

sin

nπy'

b

--------- -

sin

e

j ± kIzz'–

()

=

Gxx

.

()

j – µ0

2kI

0

---------- -εmεn

ab

---------- -

mπx'

a

----------- -

cos

nπy'

b

--------- -

sin

mπx

a

---------- -

cos

nπy

b

-------- -

sin

e

j ± kIzz'–

()

n

=

∞

∑

m

0=

∞

∑

=

Gyy

.

()

j – µ0

2kI

0

---------- -εmεn

ab

---------- -

mπx'

a

----------- -

sin

nπy'

b

--------- -

cos

mπx

a

---------- -

sin

nπy

b

-------- -

cos

e

j ± kIzz'–

()

n

=

∞

∑

m

0=

∞

∑

=

Gzz

.

()

j – µ0

2kI

0

---------- -εmεn

ab

---------- -

mπx'

a

----------- -

sin

nπy'

b

--------- -

sin

mπx

a

---------- -

sin

nπy

b

-------- -

sin

e

j ± kIzz'–

()

n

=

∞

∑

m

0=

∞

∑

=

Axx y z

, ,()

j – µ0

2kI

0

---------- -εmεn

ab

---------- -

mπx

a

---------- -

cos

nπy

b

-------- -

sin

n

=

∞

∑

m

0=

∞

∑

=

Jxx' y' z'

, ,()

mπx'

a

----------- -

cos

nπy'

b

--------- -

sin

e

j ± kIzz'–

()

vd

∫ ∫

Source

∫

Page 11

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)

Ayx y z

, ,()

j – µ0

2kI

0

---------- -εmεn

ab

---------- -

mπx

a

---------- -

sin

nπy

b

-------- -

cos

n

=

∞

∑

m

0=

∞

∑

=

Jyx' y' z'

, ,()

mπx'

a

----------- -

sin

nπy'

b

--------- -

cos

e

j ± kIzz'–

()

vd

∫ ∫

Source

∫

Azx y z

, ,()

j – µ0

2kI

0

---------- -εmεn

ab

---------- -

mπx

a

---------- -

sin

nπy

b

-------- -

sin

n

=

∞

∑

m

0=

∞

∑

=

Jzx' y' z'

, ,()

mπx'

a

----------- -

sin

nπy'

b

--------- -

sin

e

j ± kIzz'–

()

vd

∫ ∫

Source

∫

Axx y z

, ,()

ExAx

()

j – ω

k0

2

-------- -

j – µ0

2kI

0

---------- -εmεn

ab

---------- -

k0

2

mπ

a

-------

2

–

mπx

a

---------- -

cos

nπy

b

-------- -

sin

n

=

∞

∑

m

0=

∞

∑

=

Jxx' y' z'

, ,()

mπx'

a

----------- -

cos

nπy'

b

--------- -

sin

e

j ± kIzz'–

()

vd

∫ ∫

Source

∫

EyAx

()

j – ω

k0

2

-------- -

j – µ0

2kI

0

---------- -εmεn

ab

---------- -

mπ

a

-------–

nπ

----- -

b

mπx

a

---------- -

sin

nπy

b

-------- -

cos

n

=

∞

∑

m

0=

∞

∑

=

Jxx' y' z'

, ,()

mπx'

a

----------- -

cos

nπy'

b

--------- -

sin

e

j ± kIzz'–

()

vd

∫ ∫

Source

∫

EzAx

()

j – ω

k0

2

-------- -

j – µ0

2kI

0

---------- -εmεn

ab

---------- -

jkI

±()

mπ

a

-------–

mπx

a

---------- -

sin

nπy

b

-------- -

sin

n

=

∞

∑

m

0=

∞

∑

=

Page 12

11

(32)

(33)

(34)

(35)

Similarly, the electric and magnetic fields due to are obtained as

(36)

(37)

Jxx' y' z'

, ,()

mπx'

a

----------- -

cos

nπy'

b

--------- -

sin

e

j ± kIzz'–

()

vd

∫ ∫

Source

∫

HxAx

()

0=

HyAx

()

j –

2kI

------ -εmεn

---------- -

ab

jkI

±()

mπx

a

---------- -

cos

nπy

b

-------- -

sin

n

0=

∞

∑

m

0=

∞

∑

=

Jxx' y' z'

, ,()

mπx'

a

----------- -

cos

nπy'

b

--------- -

sin

e

j ± kIzz'–

()

vd

∫ ∫

Source

∫

HzAx

()

j

2kI

------ -εmεn

ab

---------- -nπy

b

-------- -

mπx

a

---------- -

cos

nπy

b

-------- -

sin

n

0=

∞

∑

m

0=

∞

∑

=

Jxx' y' z'

, ,()

mπx'

a

----------- -

cos

nπy'

b

--------- -

sin

e

j ± kIzz'–

()

vd

∫ ∫

Source

∫

Ayx y z

, ,()

ExAy

()

j – ω

k0

2

-------- -

j – µ0

2kI

0

---------- -εmεn

ab

---------- -

nπ

b

----- -–

mπ

-------

a

mπx

a

---------- -

cos

nπy

b

-------- -

sin

n

=

∞

∑

m

0=

∞

∑

=

Jyx' y' z'

, ,()

mπx'

a

----------- -

sin

nπy'

b

--------- -

cos

e

j ± kIzz'–

()

vd

∫ ∫

Source

∫

EyAy

()

j – ω

k0

2

-------- -

j – µ0

2kI

0

---------- -εmεn

ab

---------- - k0

2

nπ

b

----- -

2

–

mπx

a

---------- -

sin

nπy

b

-------- -

cos

n

=

∞

∑

m

0=

∞

∑

=

Jyx' y' z'

, ,()

mπx'

a

----------- -

sin

nπy'

b

--------- -

cos

e

j ± kIzz'–

()

vd

∫ ∫

Source

∫

EzAy

()

j – ω

k0

2

-------- -

j – µ0

2kI

0

---------- -εmεn

ab

---------- -

jkI

±()

nπ

b

----- -–

mπx

a

---------- -

sin

nπy

b

-------- -

sin

n

=

∞

∑

m

0=

∞

∑

=

Page 13

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

Jyx' y' z'

, ,()

mπx'

a

----------- -

sin

nπy'

b

--------- -

cos

e

j ± kIzz'–

()

vd

∫ ∫

Source

∫

HxAy

()

j

2kI

------ -εmεn

---------- -

ab

jkI

±()

mπx

a

---------- -

sin

nπy

b

-------- -

cos

n

0=

∞

∑

m

1=

∞

∑

=

Jyx' y' z'

, ,()

mπx'

a

----------- -

sin

nπy'

b

--------- -

cos

e

j ± kIzz'–

()

vd

∫ ∫

Source

∫

HyAy

()

0=

HzAy

()

j –

2kI

------ -εmεn

---------- -

ab

nπy

b

-------- -–

mπx

a

---------- -

sin

nπy

b

-------- -

sin

n

0=

∞

∑

m

0=

∞

∑

=

Jyx' y' z'

, ,()

mπx'

a

----------- -

sin

nπy'

b

--------- -

cos

e

j ± kIzz'–

()

vd

∫ ∫

Source

∫

Azx y z

, ,()

ExAz

()

j – ω

k0

2

-------- -

j – µ0

2kI

0

---------- -εmεn

ab

---------- -

jkI

±()

mπ

a

-------

mπx

a

---------- -

cos

nπy

b

-------- -

sin

n

=

∞

∑

m

0=

∞

∑

=

Jzx' y' z'

, ,()

mπx'

a

----------- -

sin

nπy'

b

--------- -

sin

e

j ± kIzz'–

()

vd

∫ ∫

Source

∫

EyAz

()

j – ω

k0

2

-------- -

j – µ0

2kI

0

---------- -εmεn

ab

---------- -

jkI

±()

nπ

b

----- -

mπx

a

---------- -

sin

nπy

b

-------- -

cos

n

=

∞

∑

m

0=

∞

∑

=

Jzx' y' z'

, ,()

mπx'

a

----------- -

sin

nπy'

b

--------- -

sin

e

j ± kIzz'–

()

vd

∫ ∫

Source

∫

Azx y z

, ,()

Azx y z

, ,()

Page 14

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

Azx y z

, ,()

Azx y z

, ,()

z

∂

∂Azx y z

, ,()

j – µ0

2kI

0

---------- -εmεn

ab

---------- -

jkI

±()

mπx

a

---------- -

sin

nπy

b

-------- -

sin

n

=

∞

∑

m

0=

∞

∑

=

Jzx'' y'' z''

,,()

mπx''

a

------------ -

sin

nπy''

b

-----------

sin

e

j ± kIzz''–

()

v''

d

∫ ∫

Source

∫

z

∂

∂Azx y z

, ,()

zz''=

z

∂

∂Azx y z

, ,()

zz''=

z

∂

∂Azx y z

, ,()

zz''=

zz''=

z2

2

∂

∂Azx y z

, ,()

j – µ0

2kI

0

---------- -εmεn

ab

---------- -

kI

2

–

mπx

a

---------- -

sin

nπy

b

-------- -

sin

n

=

∞

∑

m

0=

∞

∑

=

Jzx'' y'' z''

,,()

mπx''

a

------------ -

sin

nπy''

b

-----------

sin

e

j ± kIzz''–

()

v''

d

∫ ∫

Source

∫

j – µ0

2

0

---------- -εmεn

ab

---------- -

2j

–

()

mπx

a

---------- -

sin

nπy

b

-------- -

sin

n

=

∞

∑

m

0=

∞

∑

+

Jzx'' y'' z''

,,( ) δ zz''–

()

mπx''

a

------------ -

sin

nπy''

b

-----------

sin

e

j ± kIzz''–

()

v''

d

∫ ∫

Source

∫

z''

Page 15

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

∂

∂Azx y z

, ,()

j – µ0

2kI

0

---------- -εmεn

ab

---------- -

kI

2

–

mπx

a

---------- -

sin

nπy

b

-------- -

sin

n

=

∞

∑

m

0=

∞

∑

=

Jzx'' y'' z''

,,()

mπx''

a

------------ -

sin

nπy''

b

-----------

sin

e

j ± kIzz''–

()

v''

d

∫ ∫

Source

∫

µ0

–

Jzx'' y'' z

,,()4

ab

----- -

mπx

a

---------- -

sin

nπy

b

-------- -

sin

mπx''

a

------------ -

sin

nπy''

b

-----------

sin

n

0=

∞

∑

m

0=

∞

∑

source

∫∫

dx''dy''

δ x

(

x''–

) δ yy''–

()

0

xa

≤≤

0

yb

≤≤

0

x''

a

≤≤

0

y''

b

≤≤

δ x

(

x''–

) δ yy''–

()

4

ab

----- -

mπx

a

---------- -

sin

nπy

b

-------- -

sin

mπx''

a

------------ -

sin

nπy''

b

-----------

sin

n

0=

∞

∑

m

0=

∞

∑

=

z2

2

∂

∂Azx y z

, ,()µ0

–

Jzx y z

, ,()

j – µ0

2kI

0

---------- -εmεn

ab

---------- -

kI

2

–

mπx

a

---------- -

sin

nπy

b

-------- -

sin

n

=

∞

∑

m

0=

∞

∑

+=

Jzx'' y'' z''

,,()

mπx''

a

------------ -

sin

nπy''

b

-----------

sin

e

j ± kIzz''–

()

v''

d

∫ ∫

Source

∫

EzAz

()

jω

k0

2

----- -µ0Jzx y z

, ,()

j – ω

k0

2

-------- -

j – µ0

2kI

1

---------- -εmεn

ab

---------- - k0

2

kI

2

–

mπx

a

---------- -

sin

nπy

b

-------- -

sin

n

=

∞

∑

m

1=

∞

∑

+=

Jzx' y' z'

, ,()

mπx'

a

----------- -

sin

nπy'

b

--------- -

sin

e

j ± kIzz'–

()

vd

∫ ∫

Source

∫

Page 16

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

HxAz

()

j –

2kI

------ -εmεn

ab

---------- -nπ

b

----- -

mπx

a

---------- -

sin

nπy

b

-------- -

cos

n

0=

∞

∑

m

0=

∞

∑

=

Jzx' y' z'

, ,()

mπx'

a

----------- -

sin

nπy'

b

--------- -

sin

e

j ± kIzz'–

()

vd

∫ ∫

Source

∫

HyAz

()

j

2kI

------ -εmεn

ab

---------- -mπ

a

-------

mπx

a

---------- -

cos

nπy

b

-------- -

sin

n

0=

∞

∑

m

0=

∞

∑

=

Jzx' y' z'

, ,()

mπx'

a

----------- -

sin

nπy'

b

--------- -

sin

e

j ± kIzz'–

()

vd

∫ ∫

Source

∫

HzAz

()

0=

J

AxAy

Az

E

? ? ∇× ? ? ∇×

k0

2E

–

j – ωµ0J

=

Gex' y' z' x y z

, ,⁄ , ,()

E x y z

, ,()

jωµ0

Gex' y' z' x y z

, ,⁄ , ,()

Jx' y' z'

, ,()•

v'

d

∫ ∫ ∫

–=

Page 17

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

.

() ? ? ∇× ? ? ∇×

k0

2Ge

.

()

–

Iδ xx'–

( ) δ yy'–

( ) δ zz'–

()

=

Ge

.

()

Ge0

.

()

δ x

(

-------------------------------------------------------------------

x'–

) δ yy'–

2

( ) δ zz'–

()

k0

–

z ˆz ˆ

=

Ge0

.

()

Ge0

.

()

1

---- -

k0

2

j –

2kI

------ -εmεn

---------- -e

ab

j ± kIzz'–

()

n

0=

∞

∑

m

0=

∞

∑

=

k0

2

mπ

a

-------

2

–

mπx

a

---------- -

cos

nπy

b

-------- -

sin

mπx'

a

----------- -

cos

nπy'

b

--------- -x ˆx ˆ

sin

mπ

a

-------–

nπ

----- -

b

mπx

a

---------- -

sin

nπy

b

-------- -

mπx'

a

----------- -

cos

nπy'

b

--------- -

y ˆx ˆ

sincos+

jkI

±()

mπ

a

-------–

mπx

a

---------- -

sin

nπy

b

-------- -

mπx'

a

----------- -

cos

nπy'

b

--------- -

z ˆx ˆ

sinsin+

nπ

b

----- -–

mπ

-------

a

mπx

a

---------- -

cos

nπy

b

-------- -

sin

mπx'

a

----------- -

sin

nπy'

b

--------- -

cos

x ˆy ˆ

+

k0

2

nπ

b

----- -

2

–+

mπx

a

---------- -

sin

nπy

b

-------- -

cos

mπx'

a

----------- -

sin

nπy'

b

--------- -

y ˆy ˆ

cos

jkI

±()

nπ

b

----- -–

mπx

a

---------- -

sin

nπy

b

-------- -

sin

mπx'

a

----------- -

sin

nπy'

b

--------- -

z ˆy ˆ

sin+

jkI

±()

mπ

a

-------

mπx

a

---------- -

cos

nπy

b

-------- -

sin

mπx'

a

----------- -

sin

nπy'

b

--------- -

sin+

x ˆz ˆ

jkI

±()

nπ

b

----- -

mπx

a

---------- -

sin

nπy

b

-------- -

cos

mπx'

a

----------- -

sin

nπy'

b

--------- -

y ˆz ˆ

sin+

k0

2

kz

2

–

mπx

a

---------- -

sin

nπy

b

-------- -

sin

mπx'

a

----------- -

sin

nπy'

b

--------- -

z ˆz ˆ

sin+

Page 18

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)

Jy ˆI0δ x'

a

2-- -

–

δ zz'–

()

=

EsJ

JEi

EsJ

Ei

+

EsJ

Ei

+

t

0=

t

JT

EsJ

JT

•〈〉

Ei

JT

•〈〉

+0=

ZyyI0

Vy

+0=

Page 19

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

EsJ

JT

•〈〉

=

Vy

Ei

JT

•〈〉

=

Zyy

Zyy

ωµ02b a

⁄()

1

kIπ

------- - e

jkIr0

ϕ( )

sin–

mπr0

a

------------ -

ϕ ( )

cos

cos

0

π

∫

m

1 3 5 ..

, , ,

=∑

–=

dϕ

Ei

Ei

y ˆ

2

ab

----- -

πx

a

----- -

e

jk0

2

π

a-- -

2

–

z

–

sin=

Vy

Vy

2b

a

----- -2

π-- -

e

jk0

2

π

a-- -

2

–

z1

–

e

jk0

2

π

a-- -

2

–

r0

ϕ ( )

sin–

πr0

a

------- -

ϕ ( )

cos

cos

dϕ

0

π

∫

=

I0

z

0=

Γ

k0η0I0

2

–

–

k0

2

π

a-- -

---------------------------------- -

b

2a

----- -e

jk0

2

π

a-- -

2

–

2z1

–

=

z

2z1

=

T

1

k0η0I0

2

–

–

k0

2

π

a-- -

---------------------------------- -

b

2a

----- -+

e

jk0

2

π

a-- -

2

–

2z1

–

=

Page 20

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. Conclusion

The 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

=

x

a

2-- - z ,

0.5==

a

2.25cm

=

b

1.02cm

=

z

0.0cm

=

z

1.0cm

=

Page 21

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: Intext

Educational 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 the

source region in waveguides and cavities, ” IEEE Trans. Microwave Theory and

Techniques, Vol. MTT-23, pp. 762-765, September 1975.

[7] M. D. Deshpande and C. J. Reddy, “Application of FEM to estimate complex permittivity of

dielectric material at microwave frequency using waveguide measurements, ”, NASA

Contractor Report 198203, August 1995.

[8] M. D. Deshpande, et al, “ A new approach to estimate complex permittivity of dielectric

material at microwave frequencies using waveguide measurements, ” To appear in IEEE

Trans. on Microwave Theory and Techniques, March 1997.

Page 22

21

X

Y

Z

J

Electric Current

Source Inside

Rectangular Waveguide

a

b

Figure 1 Electric current source inside a rectangular waveguide

Page 23

22

X

Y

TE10

Mode

Incident

a

b

a/2

Cylindrical

Probe Radius

r0

Matched

Load

Figure 2 Rectangular waveguide with a cylindrical post parallel to y-axis placed at

x = a/2, z = z1.

z1

Z

Page 24

23

8.5 9.09.510.0 10.511.011.5 12.0

-1.0

-0.5

0.0

0.5

1.0

Γ

Reflection Coeff.

Real

Imaginary

Frequency (GHz)

Real Part

Imaginary Part}Present Method

Real Part

Imaginary Part}FEM Method [ ]

Figure 4 Reflection coefficient of a y-directed post in a rectangular waveguide as a function

of frequency.

8

Page 25

24

8.59.09.5 10.010.5 11.0 11.512.0

-1.0

-0.5

0.0

0.5

1.0

Real Part

Imaginary Part}Present Method

Real Part

}FEM Method [8]

Imaginary Part

Transmission Coeff.

T

Frequency (GHz)

Imaginary

Real

Figure 5 Transmission coefficient of a y-directed post in a rectangular waveguide as

a function of frequency.