Electromagnetic Wave Theory For Boundary-Value Problems

Fig. 1. 7. Two perfectly conducting radial plates.

2

Plane Wave Propagation

2.1 Uniform Plane Wave Consider electromagnetic wave propagation in a source-free, lossless, unbounded medium. A simple propagating wave, which adequately represents wave propagation often encountered in real situations, is a uniform plane wave. This section introduces the fundamentals of a uniform plane wave and analyzes its propagation characteristics. 2.1.1 Propagation in Lossless Medium

This subsection investigates the characteristics of a uniform plane wave starting from Maxwell's equations. Maxwell's equations (1.33) through (1.36) for a source-free, homogeneous, lossless region (Ji = M i = Pe = Pm = 0 and a = 0) are

-

-

'i1 x E = iWJ1B

(2.1)

\l x H = -iw€E

(2.2)

\l·E = 0

(2.3)

\l·H=O.

(2.4)

Let us derive an equation for E that governs wave propagation. Taking the curl of (2.1) gives

'i1 x 'i1 x E = iWJ.l\l x H .

(2.5)

22

2 Plane Wave Propagation

Substituting (2.2) into (2.5) yields

2 \l x \l x E = w J.L€E .

(2.6)

\l x \l x E - - \l2E + \l (\l . E) = - \l2E .

(2.7)

The vector identity gives

•

'"

.J

o

Substituting (2.7) into (2.6) yields the Helmholtz equation (2.8)

The Helmholtz equation governs the behavior ofthe wave E(r). It would be instructive to consider a one-dimensional (z) wave propagation characteristic by letting

:x =

~

= 0 and E(r) = xE",(z) in (2.8). The one-dimensional

Helmholtz equation is given by (2.9)

where the general solution is written in terms of the wavenumber k = wVJif. as (2.10)

Let us investigate the behavior of Et eikz in the time domain. Transforming ikz Ete into a time-varying form gives

E:(z, t) = E: cos(wt - kz) .

(2.11)

In view of Fig. 2.1, it is seen that Et(z, t) represents the one-dimensional wave that propagates along the +z-direction with a velocity u dz

w

1

dt

k

VJif.

u=-=-=

Substituting E = xEte

ikz

.

(2.12)

into (2.1) and taking its time-varying form yields

+( ) = Et(z, t) H y z, t

(2.13)

'TJ

where TJ =

J.L is the intrinsic impedance of the medium and 1201r (ohms) €

in air. A graphical representation for the propagating wave is illustrated in Fig. 2.2. Since a constant phase front of the wave, Et(z, t) and H;;(z, t), constitutes the x-y plane, the wave is called a uniform plane wave.

2.1 Uniform Plane Wave

23

E ~ (z, t)

at t

at t+dt . ,-~-" ,-, ....

.

'----'

,

,.---+-'• u

,.

'---,

::---J,~~. z Iit:::=~.-~:==~' ' . '----+' ..-, ,,, ::::::'> ••••• .'

~

1------';

dz

..

'

Fig. 2.1. Traveling wave at velocity u along +z-direction.

z

wave propagation

- - - -......- - - y

x

Fig. 2.2. Uniform plane wave propagating along +z direction. Let us consider a three-dimensional case using (2.8). The plane wave solution to (2.8) is assumed to be (2.14)

-

Eo

exp(ik·r)

where the wave vector is k = xk x + yk y + zk z and the position vector is

r = xx + yy + zz. This represents a uniform plane wave that propagates in the k direction. Substituting (2.14) into (2.8) yields the dispersion relation (2.15)

24

2 Plane Wave Propagation

-

211" where the wavenumber is k = Ikj = A and A is the wavelength. Substituting

(2.14) into (2.3) gives (2.16) k· Eo

which results in (2.17)

k·E(r)=O.

Equation (2.17) implies that the wave vector k and the electric field vector E(r) are perpendicular to each other. Furthermore, (2.14) and (2.1) yields -

1 -

H(r) =

WJ.L

k x E(r)

(2.18)

indicating that the three vectors, E(r), H(r), and k, are mutually perpendicular. A uniform plane wave is, therefore, referred to as a TEM (transverse electromagnetic) wave since there are nO electromagnetic field components in the k direction. For a lossless medium, the time-average Poynting vector Sav •

IS

S av = =

~

Re (E x H*)

~IEoI2 2

k wJ.L

(2.19)

which indicates that the real power flows along the k direction. 2.1.2 Propagation in Lossy Medium

Consider a lossy medium where the complex medium permittivity

€e

is (2.20)

Then, the solution to the Helmholtz equation

J2

dz 2

2

+ W J.L€e

Ex (z) = 0

(2.21)

•

IS

(2.22) where k = w.jJ.L€e. Expression (2.22) represents a uniform plane wave that propagates in a lossy medium with complex medium permittivity €e. Consider two different lossy media - good dielectric and good conducting media.

2.1 Uniform Plane Wave

•

25

For the good dielectric medium, a «1 WE

(2.23)

•

For the good conductor,

a WE

»

1

= (1

wJ.La

+ i)

2

(2.24)

•

Then, a wave that propagates within the good conductor is written as

E",(z) = E;t exp (i - 1)

wJ.La

(2.25)

2 z

which shows the wave decays exponentially as it propagates in the lossy medium. As the wave propagates over the distance z =

wJ.La

, the field

amplitude decreases by a factor of e- ~ 0.368. This propagating distance is known as the skin depth J of a good conductor 1

J=

•

2

--. wJ.La

(2.26)

The rate of wave attenuation depends upon the size of the skin depth 7 J. For a good conductor such as copper (a = 5.80 x 10 (siemens/m) ) at 100 (MHz), the skin depth is J = 6.6 X 10- 3 (mm), which is indeed very small compared with the wavelength. The field that exists within the interior of good conductors is, therefore, considered approximately as a null field at 100 (MHz). This means that good conductors like copper can be regarded as PEe (perfect electric conductors) at 100 (MHz) and even higher frequencies. Let us find the magnetic field within good conductors. Substituting E",(z) into (2.27) •

gIves

26

2 Plane Wave Propagation

H (z) =

y =

1 8Ex (z) iW/-l 8z

k E+e ikz

w/-l

.

x

(2.28)

The wave impedance Z in good conductors is given by

=

W/-l

k

.

W/-l

= (1- z)

20'

(2.29)

indicating that the phase difference between electric and magnetic fields 7r in good conductors is 4 (radians).

2.2 Polarization As an electromagnetic wave propagates, the tip of its electric field traces a certain curve on a plane perpendicular to the direction of wave propagation. The trace of the electric field tip is described in terms of the polarization. Let us consider the one-dimensional plane wave propagation of (2.14) with the assumption k x = ky = O. Assume that the electric field phasor is given by

E(z) = Eoe

ikz

(2.30)

where E 1 and E 2 are the magnitudes of Ex and E y , respectively. Then, the time-varying form of E(z) is

E(z, t) = Re [E(z)e-

iwt

]

(2.31 ) where

Ex(z, t) = E 1 cos(wt - kz + Bx )

(2.32)

Ey(z, t) = E 2 cos(wt - kz + By) .

(2.33)

2.2 Polarization

27

Let

= cos(wt - kz

+ Ox)

(2.34)

v = Ey(z, t) -

~

= cos(wt - kz

+ Oy)

.

(2.35)

Then, the relation

(2.36) results in (2.37)

It is possible to transform (2.37) into the equation of an ellipse using coordinates based on the relation

~-1J

Ex(z, t) = Ef. cos'l/J - ETJ sin'l/J

(2.38)

Ey(z, t) = Ef. sin 'l/J + ETJ cos'l/J .

(2.39)

The corresponding polarization ellipse is shown in Fig. 2.3. Substituting Ex(z, t) and Ey(z, t) into (2.37) yields

AEl

+ BE; + CEf.ETJ

2

= sin (Oy - Ox)

(2.40)

(0 _ 0 ) sin 2'l/J cos y x El~

(2.41)

(0 _ 0 ) sin 2'l/J + cos y x E 1 ~

(2.42)

where 2

2

cos 'l/J A - E;

+

2

B _ sin 'l/J - E;

sin 'l/J _

E? 2

+

_ sin 2'l/J CE;

cos 'l/J

E?

+

sin 2'l/J _ 2

E?

(0 _ 0 ) cos 2'l/J cos y x El~ .

(2.43)

28

2 Plane Wave Propagation

Ell

\,

,,

, ""'I'"" "" , "" , "" ""

"" "" ""

~,--'--'--- ' - - - - - - - - +

""

"" ""

E%

Fig. 2.3. Polarization ellipse.

•

Elliptic polarization To obtain the equation of an ellipse, it is necessary to eliminate the cross term E~ E'1 (C = 0) by choosing the rotation angle 'I/J appropriately. The condition (C = 0) gives tan 2'I/J = tan 20 cos( Oy - Ox) where tan 0

(2.44)

E2 • = E . Thus, we obtam 1

(2.45) which represents elliptic polarization where the tip of the electric field traces a tilted ellipse as the wave propagates along the z-direction.

•

Linear polarization nn nn . When Oy - Ox = 0 or n, then 'I/J = 0 + 2 or - 0 + 2 ' where n IS an integer. For instance, when Oy - Ox = 0 and 'I/J = 0, A is shown to be 0 and (2.45) reduces to

E'1 = 0

(2.46)

where the tip of the electric field traces a line on the ECaxis and the wave is said to have linear polarization. Its graphical representation is shown in Fig. 2.4.

•

Circular polarization n When Oy - Ox = ± 2 and E 1 = E 2 , (2.45) becomes

2.2 Polarization

29

E~

Ell

\

\ \ \ \

\ --------..~\-'--=-------.

Ex

\

\ \ \ \

Fig. 2.4. Linear polarization.

,/

,

E~

,/ ,/ ,/

,/

\ \

\

,/ ,/

\

,/

'If ----;---,/-7~\-'--=--i_----. Ex ,/ ,/

\

\ \

,/ ,/ ,/ ,/

,/

Fig. 2.5. Circular polarization.

+

=1

(2.47)

which represents a circle on the ECEf) plane; thus, the wave has circular polarization. Its graphical representation is shown in Fig. 2.5. When Oy - Ox = - ;, the polarization is called right-hand circular polarization, since the electric field behaves as a right-handed screw that advances in the +z-direction. Similarly, when Oy - Ox = ;, (2.47) represents left-hand circular polarization.

30

•

2 Plane Wave Propagation

Unpolarized The intensity of natural sun light is completely random in any direction perpendicular to the direction of wave propagation. Therefore, the trace of the electric field tip is random, and thus irregular. This type of wave is called completely unpolarized.

2.3 Gaussian Beam Consider the problem of laser beam propagation when a laser beam illuminates a large-sized target that is not far away from the source. The amplitude shape of a laser beam is commonly Gaussian due to the Gaussian field distribution across a laser source aperture. This section considers the representations of a Gaussian beam that emanates from line and point sources. The cases of line and point sources correspond to two- and three-dimensional problems, respectively. Note that some relevant discussion is available in [1] and [2, pp. 160-165]. 2.3.1 Line Source

Consider a laser beam E(r) = yE(x, z) that propagates along the +z-direction from the source aperture E(x,O), as shown in Fig. 2.6. The line source is infinitely long in the y-direction and the field is assumed to be uniform with respect to y

~

= 0 . The Gaussian beam E(x, z) for z

> 0 is expressed in

terms of the inverse Fourier transform as x

Gaussian field E(x, z)

z E (x, 0)

source aperture

Fig. 2.6. Gaussian wave propagating along z-direction.

2.3 Gaussian Beam

31

<Xl

1

(2.48)

E(x,z) = 271" -<Xl

The expression E(x, z) should satisfy the Helmholtz equation. Substituting E(x, z) into (2.8) yields

rP 2) ( d 2 +1\, E((,z) = z

.

(2.49)

0

where I\, = Jk 2 - (2 and k = wVfi€. The solution for the beam E((, z) is assumed to propagate along the +z-direction. Therefore -. E((,z) = Eo(()e'''Z . (2.50)

-

Substituting E((,z) into E(x,z) gives <Xl

1 E(x, z) = 271"

(2.51 ) -<Xl

where

-

<Xl

E(x, O)e-i(:Z: dx .

E o(() =

(2.52)

-<Xl

The aim is to express E(x, z) in a mathematically tractable form based on approximations. In practice, the source aperture E(x,O) is modeled as approximately Gaussian in amplitude as 2

E(x, 0) = exp

x - (3'5

+ i¢

(2.53)

where (30 is the beamwidth. To estimate the phase variation ¢ at the source aperture, let us consider the line source placed at z = - Zo, as shown in Fig. 2.7. The circular cylindrical phase front arriving at z = 0 from z = -Zo is

Assuming Zo

•

(2.54)

•

(2.55)

» lxi, then ·k 2 .k z x z Zo + 2 ZQ

Therefore, the source aperture E(x, 0) can be rewritten as 2

E(x,O) = A exp

where A

=

e

ikzo

x - (32

(2.56)

ik .. . and (32 = (3'5 - 2za· SubstItutmg E(x,O) mto (2.52) and 1

1

using the integral formula for Re (P)

>0

32

2 Plane Wave Propagation

x \

\

\ circular phase front \ I

line source

• -z o

, I

I

I

I

Fig. 2.7. Line source placed at z = -zoo

Hence, the Gaussian beam is written as 1 E(x, z) = 21T

00

v:rrA(3exp

(2.59)

-00

The expression (2.59) is still difficult to analytically evaluate unless further approximation is made. Assume that the Gaussian beam is strongly collimated in the z-direction so that its transverse wave vector component «() is much smaller than the longitudinal one (1\;); thus I\;

=

y!k 2 _

(2

~ k _ (2

(2.60)

2k .

Substituting (2.60) into E(x, z) and performing integration yields

A

2

x E(x, z) = ";1 + iZ exp ikz - (32(1 + iZ)

(2.61)

where Z = ; 2zk. Note that E(x, z) represents an approximate form for the Gaussian beam that propagates along the +z-direction. When (30 and ikz E(x, z) reduces to the uniform plane wave Ae as it should.

Zo

---t

00,

2.3 Gaussian Beam

33

2.3.2 Point Source

Gaussian beam propagation from a point source can be similarly analyzed as in the line source case. Consider the laser beam E(r) = yE(x, y, z) that propagates along the +z-direction from the source aperture E(x, y, 0). The Gaussian beam E(x, y, z) for z > 0 is expressed in terms of the inverse Fourier transform 00

1

00

E«(,1J, z)e

E(x, y, z) = (211")2 -00

i

«(Z+71Y)

d( d1J .

(2.62)

-00

The solution for the beam E(x, y, z) propagating along the +z direction is 00

1

00

(2.63)

E(x, y, z) = (211")2 -00

where

K.

=

Jk

2 -

(2 -

1J2

-00

and 00

00

E(x, y, O)e- i «(Z+71Y) dx dy . -00

(2.64)

-00

The source aperture E(x, y, 0) is modeled as approximately Gaussian in amplitude with the beamwidth 130, thereby yielding

E(x, y, 0) = exp -

(x2

+ y2) (35

. + up .

(2.65)

Assume that the point source placed at z = - Zo produces quadratic phase variation ¢ over the source aperture at z = O. The source aperture is, therefore, given by

E(x,y,O) = Aexp where A

=e

ikzo

and ;2

= ;5

(2.66)

- ;:0' Substituting E(x, y, 0) into (2.64) gives (2.67)

Assume that the Gaussian beam is strongly collimated in the z-direction to yield

(2.68)

34

2 Plane Wave Propagation

-

Substituting E o((, TJ) and (2.68) into (2.63) finally gives the approximate Gaussian beam expression for a point source A E(x,y,z) = (l+iZ)exp

(2.69)

2z

where Z = (32 k' When (30 and Zo -+ 00, E(x, y, z) reduces to the uniform plane wave Ae

ikz

.

2.4 Reflection at Plane Boundary When a uniform plane wave impinges on a plane boundary, the reflection and transmission of an incident wave occurs at the boundary. The reflected and transmitted waves naturally take the form of uniform plane waves. This section investigates the reflection and transmission across a plane boundary separating two lossless and isotropic media. TE and TM waves are analyzed to obtain their reflection and transmission coefficients based on the boundary conditions.

2.4.1 TE Wave (Perpendicular Polarization) Consider a TE wave that is incident on a plane boundary between two lossless, dielectric media, as shown in Fig. 2.8. For simplicity, it is assumed that the incident electric field has only a y-component and the incident wave vector lies on the x-z plane (plane of incidence). The incident wave, therefore, is referred to as a TE (transverse electric to the plane of incidence) wave. The electric field vector is perpendicular to the plane of incidence and the TE wave thus

z •

E'y region

x

en

region (II)

at

Fig. 2.8. TE wave incident on plane boundary.

2.4 Reflection at Plane Boundary

35

corresponds to the case of perpendicular polarization. The wavenumbers in regions (I) and (II) are k1 (= W-/f.ll€t) and k 2 (= WVf.l2€2), respectively. The total electric field in region (I) consists of the incident and reflected waves as (2.70) (2.71 ) where RTE is the reflection coefficient for a TE wave. The phase terms are given by k zi = k1 sinOi , k zi = k I COSOi, k zr = k I sinOr, and k zr = k I cos Or. The transmitted wave in region (II) is E~(x, z) = TTE exp(ikztx - ikztz)

(2.72)

where k zt = k 2 sin fh, k zt = k 2 cos Ot, and TT E is the transmission coefficient for a TE wave. Let us enforce the boundary conditions for the tangential field continuity at z = O. The tangential E y continuity at z = 0 (2.73) •

reqmres exp(ikzix)

+ RTE exp(ikzrx)

=

TTE

exp(ikztx) .

(2.74)

Equation (2.74) must be satisfied irrespective of x, thereby yielding the phasematching condition (2.75) which results in sin Oi = sin Or (law of reflection) and k I sin Oi - k 2 sin Ot (Snell's law) simultaneously. Hence, (2.74) becomes 1 + RTE =

TTE·

(2.76)

The accompanying Hz and Hz components can be obtained by substituting (2.70) through (2.72) into Faraday's law \l x E = iWf.lll. Since i BEy h . I H . . 0 Hz = B' t e tangentla z contlllmty at z = Wf.l Z (2.77) •

gIves

k zi (R TE- 1) _ - k zt ,." .lTE· f.lI f.l2

(2.78)

36

2 Plane Wave Propagation

Solving (2.76) and (2.78) for the reflection and transmission coefficients gives

R

where Z2 =

'T/2

()

1

Z1 =

- Z2 - Z1 TE - Z2 + Z1

( 7) 2. 9

2Z2 TTE = Z2 + Z1

(2.80)

1]1

()

J1.2

'T/2 =

1

1

and

1]1

=

J1.1

.

cos t cos i 1:2 1:1 Let us evaluate the time-average power densities entering and leaving the boundary at z = 0 for lossless media, where both Z1 and Z2 are real. The incident time-average power density is given in terms of the Poynting vector as Si =

~Re (~x Jt.)

.(-z)

1

2Z1

(2.81 )

•

Similarly, the reflected and transmitted power densities are

(2.82)

(2.83) A law of power conservation (Sr

+ St

= Si) across the boundary can be easily

proved as Sr

+ St

=

1

2 (2.84)

Let us consider the special case when 1:1 > 1:2 and J1.2 = angle is

-

1-

-

• 2 ()

sm

i.

J1.1.

The transmitted

(2.85)

2.4 Reflection at Plane Boundary

When the incident angle (}i is greater than the critical angle (}e = sin-

37

1

, 1"1

the term cos(h (thus Z2) becomes purely imaginary, thereby yielding IRTEI = 1, Sr = Si, and St = 0. This means that the total internal reflection occurs when a wave impinges on a less dense medium (1"1 > 1"2) at an incident angle greater than the critical angle ((}i > (}e). The condition St = 0 should not be construed as a null field in region (II). The transmitted field in region (II) is evanescent, and it decays exponentially away from the boundary at z = 0.

2.4.2 TM Wave (Parallel Polarization) Consider a TM (transverse magnetic to the plane of incidence) wave that impinges on the plane boundary at z = 0, as shown in Fig. 2.9. A TM wave has no magnetic field components in the plane of incidence. The electric field vector is parallel to the plane of incidence and the TM wave thus corresponds to the case of parallel polarization. In regions (I) (k~ = Wv'J.L1f1) and (II) (k 2 = WVJ.L2f2) the magnetic fields are (2.86) (2.87) H~(x, z) =

TTM

exp(ik:z:tx - ikztz)

(2.88)

where k:z: i = k 1 sin (}i, k zi = k 1 cos (}i, k:z: r = k 1 sin (}r, k:z: t = k 2 sin (}t, k zr = k 1 cos (}r, and k zt = k 2 cos (}t. Since the tangential Hy(x, 0) must be continuous, the continuity condition

z HIy

region (I) = = = = = = x :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::.; .', ...y::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::

region (II)

et

Fig. 2.9. TM wave incident on plane boundary.

38

2 Plane Wave Propagation

(2.89) results in 1 +RTM =

TTM

(2.90)

with the conditions sin Oi = sin Or (law of reflection) and k 1 sin Oi = ~ sin 0t (Snell's law). Ampere's law, \l x H -iWEE, produces the tangential electric field, i 8Hy (). • Ez = 8 ' Hence, E z x,O contmmty

=

WE

Z

(2.91) yields

k zi (RTM

1) -_

-

E1

-

kzt'T' .l.TM·

(2.92)

E2

Solving (2.90) and (2.92) for the reflection and transmission coefficients for a TM wave gives

R

- Y2-Yj

™ - Y2 + Y1

TTM

2Y2 = Y2 + Y1

(2.93) (2.94)

1 , Y1 = 1 , 'T/2 = J.l2, and "II = J.l1 . Note that 'T/2 cos Ot "II cos Oi f2 f1 R TM and TTM are also obtained from (2.79) and (2.80) based on the duality · h . . c h ' 1 d 1 t heorem, wh1C permIts mter angmg "II -+ - an "12 -+ -. "II 'T/2 Let us consider the special case when the reflection coefficient RTM becomes zero for nonmagnetic media (J.l1 = J.l2 = J.lo and E1 f- E2)' The condition Y2 - Y1 = 0 gives

where Y2

=

(2.95) Solving (2.95) and Snell's law

(2.96) for Oi gives II

ui

=

'-1

SIn

= t an

-1

(2.97)

2.5 Infinitely Long Transmission Lines

39

which indicates that the total transmission occurs when a TM wave impinges on the half space at the incident angle ()i. The angle (2.97) is referred to as the Brewster angle. Let us consider the other special case when 101 > 102 and J.L2 = J.Ll = J.Lo. Like the previous TE wave case, the total internal reflection occurs when a TM wave impinges on a less dense medium (101 > 102) at an incident angle greater than the critical angle

()c

= sin-

1 •

2.5 Infinitely Long Transmission Lines Transmission lines are commonly used to transfer electromagnetic energy from one location to another. Parallel wires, parallel plates, and coaxial lines are all typical examples of transmission lines. Transmission lines consist of two conductors that guide TEM (transverse electromagnetic) waves in a low-frequency regime where their field components are all transverse to the wave propagation direction. 2.5.1 Coaxial Line

This section investigates the behavior of wave propagation using an infinitely long coaxial line, as shown in Fig. 2.10. For a practical coaxial line with an inner conductor radius a and outer conductor radius b, the size is usually small (b - a) « A and its annular interior region is filled with dielectric material of permittivity 10 and permeability J.L. For simplicity, an ideal coaxial line is assumed where the dielectric material is lossless and the inner and outer conductors are PEC material. An input voltage source is applied between the inner and outer conductors at z = 0 to send a TEM wave along the z-direction. The TEM wave consists of E p and Hq, components that propagate along the z-direction with the propagation constant k (= wVJif.) p

. . . -- .....H$ I I '..: I P \ I

\,

J

d}/C

b

I

-:a;;+W~~I---------

---

Fig. 2.10. Coaxial line lying along z-axis.

z

40

2 Plane Wave Propagation

(2.98)

H",(p, z) = H",(p)e

ikz

(2.99)

.

Ampere's law \l x

-H = -iw€E - +J

(2.100)

reduces to Ampere's circuital law \lxH=J

(2.101)

for a TEM wave since the electric current density J that flows through the conductors is such that liw€E « Applying Stokes' theorem to Ampere's circuital law gives

JI.

c

H",(p,z)dl = I(p,z).

Substituting the current I(p,z) = Ite

ikz

into (2.102) gives

1+

H (p,z) =

0

'"

(2.102)

27rp

e

ikz

(2.103)

.

Faraday's law \l x E = iWJ.LH

(2.104)

is rewritten as 8Ep (p, z)

8z

=

.

~WJ.L

H (

'" p, z

)

(2.105)

which results in J.L

-H",(p, z) €

-

(2.106)

The time-average power carried by a coaxial line is

J.L -In €

b -

a

•

(2.107)

2.5 Infinitely Long Transmission Lines

41

2.5.2 Voltage and Current As a TEM wave propagates on a coaxial line, the direction of E p and H¢ remains unchanged as if they were scalars. It is thus expedient to represent the wave propagation on transmission lines in terms of scalars such as the voltage and current. Let us reinterpret the transmission line problem by introducing [+(z) and V+(z) where V+(z) and [+(z) are the incident (forward) voltage and current that propagate along the +z-direction, as shown in Fig. 2.11. The incident waves, [+(z) and V+(z), are (2.108) b

V+(z) =

Ep(p, z) dp a J.L

E

,

ln

.

b a

V+ o ikz = V+e 0

(2.109)

.

The time-average power, which is delivered by the voltage and current, is

J.L

-In E

b a

(2.110)

which is identical with (2.107). The characteristic impedance of a transmission line Zo is defined as the ratio ofthe voltage V+(z) to the current [+(z). The characteristic impedance of a coaxial line is, therefore

V+ 0 Z0= + [0

- - t: In ~ 1

21T

a

E

•

(2.111)

Consider the reflected waves, V-(z) and [-(z), as shown in Fig. 2.11. Assuming

r(z) =

[oe-

ikz

(2.112)

and substituting [- (z) into Ampere's circuital law (2.102) gives

r0

H¢(p, z) = 2

1Tp

.kz

e-·

.

(2.113)

42

2 Plane Wave Propagation

---v +(z), [+ (z) y- (Z), [- (Z)

4

z •

Fig. 2.11. Incident [V+(z), I+(z)] and reflected [v-(z), r(z)] waves.

Substituting H¢(p, z) into Faraday's law (2.105) yields

Ep(p,z) = -

/-L

[0

to

21rp

e -ikz

.

(2.114)

Let us define the reflected voltage V - (z) as

V- (z)

b

=

Ep(p, z) dp a

r0 21r '-

/-LIn to

b

-

e

-ikz

"

'V'

11,-

o

- 11,- -ikz

=

0

e

.

(2.115)

Hence, the ratio yields V-(z) _ Vo= -Zo. [-(z) [0

(2.116)

2.6 Terminated Transmission Lines Consider a terminated, lossless transmission line that is excited by the source Vs with internal impedance Zs at z = 0, as shown in Fig. 2.12. Termination with a load Zl at z = l generates reflected waves, thereby resulting in standing waves due to the interference between incident and reflected waves. The total standing wave voltage V (z) consists of the incident V+(z) and reflected V- (z) components V(z) = V+(z)

+ V-(z) (2.117)

2.6 Terminated Transmission Lines

43

I (z)

Zo

+ Zs

• V+(z) V(z)

+

Vs

~

Zl

V- (z)

z

V(z), l(z)

z=O

z =l

Fig. 2.12. Transmission line terminated with load Z/.

Similarly, the current I(z) is

(2.118) Since 10+

=

11,+ 0

Zo

,

=_

and 10-

11,0

Zo '

I(z) = 1 (Vo+e ikz _ Vo-e-ikz) . Zo

(2.119)

It is possible to solve (2.117) and (2.119) for the two unknown voltage amplitudes Vo+ and Vo-' Two boundary conditions must be enforced to determine Vo+ and Vo-' Applying Kirchhoff's voltage law at z = l and 0 results in

V(l)

(2.120)

Zl = I(l) VB = I(O)Zs

+ V(O)

.

(2.121)

Solving (2.120) and (2.121) for Vo+ and Vo- gives

11,+ = o (Zo 11, o =

r.

Ie

Zo v: + ZB) + Il(Zo _ Zs)e2ikl s

2ikl

(2.122)

Zo v: (Zo + Zs) + Il(Zo _ Zs)e2ikl s 'V

11,+ o (2.123) where

rl

Zl - Zo Zl + Zo

is the voltage reflection coefficient at the load z = l.

44

2 Plane Wave Propagation

2.6.1 Reflection Coefficient, Impedance, and Power

The voltage reflection coefficient F(z) at z is

_ V-(z) F(z) = V+(z) _ ZI - Zo 2ik(l-z) _

Z1+ Z 0 " ~

e

-

T' .I.

Ie

2ik(l-z)

.

(2.124)

y

FI

The impedance Z(z) at z is defined as

Z( ) = V(z) = Z 1 + F(z) 01-F(z) . z - fez)

(2.125)

Let us evaluate the time-average power delivery to the load Pav when the load voltage and current are

(2.126)

(2.127)

The result is Pav =

~ Re

[V (l)r (l)] 2

= 1V0+1 Re (1 r,* 2Z . I o

+ r,I -InI 1

2)

(2.128) -

To achieve a maximum power delivery to the load, the reflection coefficient II. must be zero, leading to the relation ZI = Zo, which is called a matched condition. 2.6.2 Voltage Standing-Wave Ratio

Investigating a voltage wave pattern on a transmission line is instructive. The total voltage on a transmission line is

2.6 Terminated Transmission Lines

=

V/ eikz [1 + llei2k(l-z)]

.

45

(2.129)

Assume that the reflection coefficient at the load is (2.130) The magnitude of V(z) then becomes

I

lV(z)1 = iVo+ 1 + 111le i2k (l-z)-ill .

(2.131 )

The maximum of lV(z)1 occurs when 2k(l- zt} - () = 2mr (n: integer) as

lV(z)lmax = iVo+

11 + 11111

(2.132)

while the minimum occurs when 2k(l - Z2) - () = (2n - 1)71" (n: integer) as (2.133) The distance d = Z2 transmission line is

-

Zl

between the minimum and maximum points on a

kd = 71"

2

d=

~ 4

•

(2.134)

The ratio of the maximum to minimum voltage magnitudes is called the voltage standing-wave ratio (VSWR) and is expressed as VSWR = lV(z)lmax

lV(z)lmin

1+ 1111 1-1111·

(2.135)

2.6.3 Cascaded Lines

Consider a two-stage transmission line, as shown in Fig. 2.13 (a). Note that Zl and Z2 are the characteristic impedances of cascaded transmission lines, respectively. At z = Za on the second transmission line (Z2), the voltage reflection coefficient Fa and impedance Za are

_ Zl r.a -

(2.136)

Z - Z 1 +Fa a - 2 1 _ Fa .

(2.137)

Z2 Zl + Z2

46

2 Plane Wave Propagation

Zs

vs

+ -

I-

l1

Z1

, I• I I I I I

-I

l2

original problem (a)

Zs

vs

+ -

equivalent problem (b)

Fig. 2.13. Two-stage transmission line.

At Z = Zb on the second transmission line (Z2), the voltage reflection coefficient and impedance Zb are

n

(2.138)

Z_Zl+ b2 1

n' n

(2.139)

Therefore, the original problem of a two-stage transmission line is simplified to a single-stage transmission line with the equivalent impedance Zb, as shown in Fig. 2.13 (b). The geometry in Fig. 2.13 (b) is identical with that of the single-stage transmission line in Fig. 2.12. The solution to the problem in Fig. 2.12 is thus available from (2.122) and (2.123). It is often desirable to eliminate the reflected wave on Z1 by appropriately choosing Z2. No reflection on Z1 requires the matched condition (2.140) To realize the matched condition, the length l2 = • • mgm

~

is chosen, thereby result-

Problems for Chapter 2

Zl - Z2

-

Zl

47

(2.141)

+ Z2

_ Z? Z b - Zl .

(2.142)

Assume that the load impedance Zl is real. Equating (2.140) with (2.142) • gIVes (2.143) The inserted ~-IOng transmission line with Z2 = ";ZlZI is called the quarterwave transformer.

Problems for Chapter 2 1. Derive (2.40) and (2.44). 2. Derive (2.61) and (2.69).

•

z

HIy

region (I) ·... ........ ....... .. .. .. .. .. .. .......... .. .. "" ..."" ..."" ..."" ..."" ..."" ..."" ...=.."" ..."-"" ..."-"" ..."" ...-. .............................................. .y • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •• ••

•••

•• • • • • • • •

•

• • • • • • ••

• •••••••••••••

x

® H'; d region (II) · ... ................ .. ... .. .. . . . . . .. .... .. ... .. ·. .. ........ ............. ...... ................ .

"......

. .

. .

region (III)

Fig. 2.14. TM wave impinging on dielectric slab of thickness d.

3. Consider a TM wave that impinges on a dielectric slab, as shown in Fig. 2.14. In regions (I), (II), and (III), the total magnetic field representations are given by

48

2 Plane Wave Propagation

H~(x, z) =

Rexp(ik",rx + ikzrz)

H;(x, z) = exp(ik"'dX) [A exp(ikzdZ) H~(x,z) = Texp(ik",tx - ikztz) .

(2.145)

+ B exp( -ikzdz)]

(2.146) (2.147)

To determine the unknown coefficients, A, B, R, and T, the four boundary conditions for the field continuities, Hy(x,O), Hy(x,-d), E",(x,O), and E",(x, -d), must be enforced. Derive A, B, R, and T using the boundary conditions. 4. In circuit analysis, a lossless coaxial line is modeled as a transmission line with the shunt capacitances C and series inductances L. Given a coaxial line of inner and outer radii a and b, filled with dielectric of J.l and i, derive its shunt capacitance and series inductance per unit length. 5. Consider the terminated transmission line, as shown in Fig. 2.12. Derive · ffi . ]-(z) t h e current re fl ectlOn coe Clent ] + (z) . •

3

Waveguides

3.1 Cylindrical Waveguides The understanding of wave propagation along cylindrical waveguides is of practical interest in microwave and optical communications. Typical cylindrical waveguides include rectangular waveguides, optical fibers, and microstrip lines. This section investigates electromagnetic waves that can propagate along an infinitely long waveguide, as shown in Fig. 3.1. Wave propagation is conveniently analyzed in terms of the magnetic and electric vector potentials A and F. The vector potentials in a source-free cylindrical waveguide satisfy the Helmholtz equations

(3.1 ) (3.2) where k (= wVJIf) is the wavenumber. To understand the wave propagation behavior, the Helmholtz equations must be solved for A and F subject to the boundary conditions imposed at the waveguide boundary.

cylindrical waveguide z -'it~T=-I---------­

Ez

p

Fig. 3.1. Infinitely long waveguide that lies along z-direction .

•

50

3 Waveguides

Let us discuss the possible types of waves that can propagate along a waveguide. The propagating waves are conveniently represented in terms of the TE and TM waves. The TE wave refers to the wave whose electric field components are transverse to the wave propagation direction z (Ez = 0). Similarly, the TM wave refers to the wave whose magnetic field components are transverse to the wave propagation direction z (Hz = 0). The description of TM and TE waves is possible using the longitudinal components of A and F, Az(x,y,z) and Fz(x,y,z). •

When F = zFz(x, y, z) and A = 0, then Ez(x, y, z) = 0; this type of wave is called a TE wave (transverse electric to the wave propagation direction z). A TE wave satisfies the Helmholtz equation (3.3)

The explicit field expressions for a TE wave using rectangular and cylindrical coordinates are shown in Tables 3.1 and 3.2, respectively. •

When A = zAz(x,y,z) and F = 0, Hz(x,y,z) = 0 and a TM wave (transverse magnetic to the z-direction) is obtained. A TM wave satisfies the Helmholtz equation (3.4)

A hollow waveguide surrounded with metallic walls can support either TM or TE wave propagation. •

When the wavenumber in the z-direction is k, the propagating term takes the form of e±ikz and the longitudinal field components become zero Ez(x,y,z) = Hz(x,y,z) = 0, as seen in Tables 3.1 and 3.2. This type of wave is referred to as a TEM wave. Transmission lines (coaxial line, paired wire, etc.) consisting of two conducting terminals can support TEM wave propagation.

•

A dielectric waveguide or a partially filled waveguide with metallic walls can support the propagation of a hybrid wave, which is a combination of TM and TE waves. Both Az(x,y,z) and Fz(x,y,z) must be used for a hybrid wave analysis.

3.2 Rectangular Waveguide A rectangular waveguide with conducting walls is commonly used to transmit power in microwave communication and antenna applications. The field analysis for a rectangular waveguide is a canonical boundary-value problem. This section investigates the behavior of TM and TE wave propagation along a rectangular waveguide filled with homogeneous dielectric material (J1., i).

3.2 Rectangular Waveguide

51

Table 3.1. Field representations using rectangular coordinates (x, y, z) Fields

TM wave t

{)2A.

W J.I.€

{)x () Z

•

E",

t

{)2A.

WJ.I.€

{)y {)z

•

Ey

/ {)2

•

t

E.

W J.I.€

\

TE wave 1 {)F•

--€

1 {)F• € {)x

\

() 2 Z

+k

2

A.

0

/

•

t

{)2 F.

WJ.I.€

{)x {)z

•

1 {)A.

-

H",

J.I.

{)y

1

{)A.

t

{)2 F.

J.I.

{)x

WJ.I.€

{)y {)z

--

Hy

•

H.

{)y

t

0

W J.I.€

I

{)2 ()Z2

2'

+k

F.

3.2.1 TM Wave Consider TM wave propagation along a rectangular waveguide with perfect conducting walls, as shown in Fig. 3.2. A TM wave is assumed to propagate along the z-direction with the propagation constant k z . The vector potentials associated with the TM wave are given by Fz(x,y,z) = 0 and y

b

z (a> b)

x

Fig. 3.2. Rectangular waveguide with cross section a x b (a > b).

52

3 Waveguides

Table 3.2. Field representations using cylindrical coordinates (p, c/J, z)

Fields

TM wave

TE wave

•

Ep

z o2A. wltf op oz

1 of• -fp oc/J

•

1 of• f op

E",

02A. wltfP oc/J oz

z

•

2 /0 Z o 2 Z wlt f

~

•

E.

}

D

A.

loA. ItP oc/J

z 02F. wltf op oz

loA. -It op

02F. wltfP oc/J oz

Hp

H",

+k

2

•

•

z

2 / 0 Z o 2 Z wlt f

~

•

D

H.

+k

2

}

F.

Az(x, y, z) = Az(x, y) exp(ikzz). To determine the propagation constant k z , 2 2 it is necessary to solve the Helmholtz equation (\7 + k ) Az(x, y, z) = 0 subject to the boundary conditions. The Helmholtz equation can be rewritten as (3.5) Based on the separation of variables technique, Az(x, y) takes the form of

(3.6)

Az(x,y) = X(x)Y(y) .

Substituting Az(x, y) into the Helmholtz equation (3.5) and dividing by X(x)Y(y) yields 1

J2 X(x)

X(x)

dx2

+

J2y(y) Y(y) dy 2 1

2

+ k - k z = O.

(3.7)

J2y(y) To satisfy (3.7), the terms X(x) dx 2 and Y(y) d 2 must be equal to y constants independent of x and y, respectively. Let the separation constants 2 2 be -kx and _k y as 1

J2 X(x)

2

1

3.2 Rectangular Waveguide

53

J2X(x) X(x) dx 2

(3.8)

J2y(y) Y(y) dy 2

(3.9)

1

1

Then, the dispersion relation is given by

+ k; + k;

k;

= k

2

(3.10)

.

Since E z (hence A z ) must vanish at the boundaries x = 0, a and y = 0, b, the vector potential AAx, y, z) is chosen as

Az(x,y,z) = sink",x sinkyY exp(ikzz) where k", =

":7r, k y =

n;, and

(3.11)

mand n are integers (m, n > 1). The remain-

ing field components are available by substituting Az(x, y, z) into Table 3.1 in Section 3.1. The dispersion relation yields the propagation constant for the TM mn mode (3.12)

while its cutoff wavenumber is given by k c =

m7r)2 (n7r)2 ( a + b . When the

operating frequency is high enough to satisfy the condition k > k c , the TM mn mode can propagate. When k < kc , the TM mn mode cannot propagate and becomes attenuated. Note that the lowest-order TM wave is the TM l1 mode.

3.2.2 TE Wave The analysis of TE wave propagation within a rectangular waveguide is somewhat similar to the analysis of TM waves. Let us start with the assumption Az(x, y, z) = 0 and Fz(x, y, z) = Fz(x, y) exp(ikzz), where the vector potential Fz (x, y) satisfies the Helmholtz equation

2 8

8x 2

2 8 2 2 + 8 +k -k z Fz(x,y) =0. y

2

(3.13)

Solving the Helmholtz equation is possible using the separation of variables technique. The tangential electric fields must vanish on the metallic surfaces, implying that the normal derivatives of Fz(x,y) must be zero,

8FAx,y) 8y of

y=O,b

= 0 and 8Fz (x,y) 8x

= O. Hence, Fz(x,y,z) takes the form "'=O,a

54

3 Waveguides (3.14)

m7r a'

n7rb'

.mtegers

where kz = ky = and m and n are (m > 0, n > 0, m + n :j:. 0). The propagation constant k z and cutoff wavenumber kc for the TE mn mode are (3.15)

7r C: f + (nb7rf .

kc =

When a fields are

(3.16)

> b, the lowest-order TE wave is the TE IO mode, where its nonzero (3.17)

Hz =

-ikza

7r

. H o sm

7r

· 0sm

•

Ey = where H o =

i Wj.Lf

tWj.La H

(7r)2 - and k z =

k2

a

(7rX) (. ) a exp tkzz

(7rX) ('k) a exp t

(7r)2

-

-

(3.18) (3.19)

zZ

.

a

Let us consider a rectangular waveguide of dimension a = 2b, where the lowest propagating wave is TE IO and the next higher wave is TE20 or TEol . If

. . (7r) the operatmg frequency IS chosen as a


27ra

' a rectangular wave-

guide will only support single-mode TE lO propagation. The power density that is delivered by the TE lO mode is

-*)

-Sav = 2Re 1 (Ex H

=

~Re

(-zEyH;

+ xEyH;)

.

(3.20)

The time-average power delivery along the z-direction is a

b

P av =

Sav .

o

0

b

-

zdxdy

o

a

1

-Re (EyH;) dxdy o 2 (3.21 )

3.3 Dielectric Slab Waveguide

55

3.3 Dielectric Slab Waveguide A dielectric slab waveguide is one of the major waveguide components used in optical and microwave communications. For simplicity, the dielectric slab waveguide analyzed in this section is two-dimensional, as illustrated in Fig. 3.3. A dielectric slab waveguide consists of a denser dielectric slab [region (II)] that lies between two less dense dielectric media [regions (I) and (III)]. While a dielectric slab waveguide can generally support either TM or TE wave propagation, the TM case is considered first. y

en

region

d/2

---. ........... ............. .......... ....... .... ..... . .... ......... . ·. .... . ........ . .. . . .. . . . ...... . · -........ . · ........... ..................... .............. .

x

d -

region (II) ----{ .....:::..::::: : • •• • •••

• • • ••

•

•

•

•

• •

•

:

••

•

•

•

••

:: • • • • ••

••

-

.. ..

--------z

: ::::..:. ~d.'2.·.·.·.·.·:.·.·.·:.·.·.·.·.·.·.·.·.·:::::::.·.·.· . • •• • •••••••••••••••••••••••••••••••••••••••••••

• • •• • • •

•

•

•

region (III)

Fig. 3.3. Dielectric slab waveguide.

3.3.1 TM Wave Assume that the TM wave (Hz = 0 and E z :j; 0) is guided by a dielectric slab along the z-direction. The wavenumbers in regions (I), (II), and (III) are k 1 (= WVJ.Llfd, k 2 (= WVJ.L2 f 2) , and k 3 (= WVJ.L3f3), respectively. There is no

:x = 0

field variation in the x-direction

. The vector potentials in each

region also satisfy the respective Helmholtz equations 2

2

2 8 8 A~(y, z) = 0 1 8 y 2 + 8z 2 + k 2

2

8 8 A~I(y,z)=O 8 y 2 + 8z 2 + k~ 2

(3.22)

(3.23)

2

2 8 8 A~II (y, z) = 0 . 3 8 y 2 + 8z 2 + k

(3.24)

56

3 Waveguides

Region (II) is assumed to be denser than regions (I) and (III) (f2 > f}, f3) to guarantee wave guidance within region (II). The field in region (I) must be evanescent in the +y-direction for y > 0 and propagate along the z-direction. Hence, A;(y,z) is written as (3.25)

where

01

= Jk~

-

W2J1.1f1.

d

The wave in region (II) experiences reflection at y = ± 2' thereby exhibiting standing wave characteristics along the y-direction. The vector potential in region (II) is

A;I (y, z) = (A~ cos kyy + A~ sin kyy) exp(ikzz) (3.26)

+ k z2

k 22 = -

h k were W 2 J1.2 f 2· Similarly, since the field in region (III) must be evanescent along the -ydirection for y < 0, 2 y

--

(3.27)

-

where 03 = Jk~ W2J1.3f3' To represent the vector potentials, A; (y, z), A;I (y, z), and A; II (y, z), four unknown coefficients, A 2 , 'IjJ, A 3 , and k z , are introduced, respectively, whereas A} is considered a known coefficient. The aim is to determine the four unknown coefficients utilizing the four boundary conditions for the E z and Hz field continuities at y =

±~.

The E z and Hz field components are obtained by substituting A;(y,z), A;I(y,z), and A;II(y,Z) into Table 3.1 in Section 3.1. The E z continuities at y = ± ~ are (3.28)

d -- z

2'

l

•

Since E z = - WJ1.f

•

(3.29)

2

8 2 8z 2 + k A z , (3.28) and (3.29) can be rewritten as (3.30)

(3.31)

3.3 Dielectric Slab Waveguide

continuities at y =

Similarly, the Hz

±~

57

yield

(3.32)

ky d a3 . - - A 2 sm -ky 2 + 1/J = A 3 exp( -a3 d / 2 ) . M2 M3

(3.33)

Taking the ratio of (3.30) and (3.32) results in

al

d

+ 1/J -

ky

.

(3.34)

•

(3.35)

tan a ± tan{3 tan (a ± {3) = ..,------'---:: 1 =f tan a tan {3

(3.36)

-

tan

€l

kY 2

€2

Similarly, (3.31) and (3.33) gives a3

d

+ 1/J - Y2

--=- tan -k €3

ky

Using the formula

it is possible to eliminate 1/J in (3.34) and (3.35). The result yields the dispersion relation

a a tan(kyd) - €2 l = €2 3 1 + tan(kyd) €} k y €3ky When region (III) is perfectly conducting (PEe, relation reduces to

€3

a €2 l . €l k y

-+

00),

(3.37) the dispersion

(3.38) The transcendental relation (3.37) can be solved for kz numerically. In general more than one solution exist.

3.3.2 TE Wave The analysis of TE wave propagation is quite analogous to that of TM wave propagation. The TE wave (E z = 0 and Hz f:- 0) is assumed to propagate along a slab waveguide with the propagation constant k z . The associated vector potential F z (y, z) takes the form of

Ff (y, z) = B 1 exp( -alY) exp(ikzz)

(3.39) (3.40) (3.41)

58

3 Waveguides

Enforcing the boundary conditions for the Ex and Hz continuities at y = ± ~ results in the dispersion relation for the TE wave •

(3.42)

Note that the dispersion relation (3.42) can also be directly obtained from (3.37) based on the duality theorem (Section 6.5), which permits interchanging 10 1,2,3 ~

J.l1,2,3·

3.4 Circular Waveguide In addition to rectangular waveguides, circular conducting tubes, known as circular waveguides, are also popular guiding structures for microwave applications. This section investigates wave propagation within a circular waveguide that is surrounded with a conducting wall. Both TM and TE waves are considered in the following subsections.

3.4.1 TM Wave

Consider a TM wave (Hz = 0 and E z =j:. 0) that propagates inside a circular waveguide, as shown in Fig. 3.4. Assume that the circular waveguide is filled with dielectric material of (J.l,€). When the wave propagates along the zdirection with the propagation constant k z , the vector potential A(p, ¢, z) takes the form of (3.43)

The corresponding Helmholtz equation drical coordinates is rewritten as

(\7 + k 2

2

)

A z (p, ¢, z) = 0 using cylin-

y

a Z + - - - -W- d - - ' - - - f - - - - - - - - - - -

x

PEC

Fig. 3.4. Circular waveguide with radius a.

3.4 Circular Waveguide

1 8 p 8p

8 p8p

59

(3.44)

or (3.45)

2 where k~ = k - k;. Based on the separation of variables technique, let

Az(p,¢) = R(p)p(¢) .

(3.46)

Substituting Az(p, ¢) into the Helmholtz equation and dividing by R(p)p(¢) yields

d

dR(p) - p dp pR(p) dp 1

1

1

+ (Ji p(¢)

~p(¢)

d¢2

2

+ kp =

0.

(3.47)

Substituting 1

~p(¢)

p(¢)

d¢2

2

(3.48)

=-m

into (3.47) results in the following two ordinary differential equations ~

2

d¢2 +m 1 d

-

pdp

d pdp

p(¢) = 0

(3.49)

R(p) = O.

(3.50)

2

m + k~ - p2

The solution to (3.49) is sinm¢

p(¢) =

(3.51 )

•

cosm¢ Since the field is periodic in the azimuthal direction with a 27T-periodicity, the parameter m should be an integer. Equation (3.50) is known as Bessel's equation (see Appendix B) and its solution is represented as the linear combination of the Bessel functions Jm(kpp), Nm(kpp), Hg),(2)(k pp), .... Among these Bessel functions, the appropriate choice for R(p) is Jm(kpp), since Nm(kpp) and Hg),(2) (kpp) become infinite at p = O. Hence, A z (p, ¢, z) is given by sinm¢

exp(ikzz) . cosm¢

(3.52)

60

3 Waveguides

The explicit field components are obtained by substituting Az(p, ¢, z) into Table 3.2 in Section 3.1. Since E z and E", must vanish at p = a, Jm(kpmna) = 0, where kpmna is the nth root of the mth order Bessel function. The TM wave with the eigenvalue k pmn is called the TM mn mode, where the roots of Jm(kpmna) = 0 are tabulated in Table 3.3. The propagation constant for the TM mn mode is • gIVen as (3.53)

In order for the TM mn mode to propagate, the wavenumber must be higher than k pmn k (= w...fiii.)

> k pmn

(3.54)

where k pmn is the cutoff wavenumber for the TM mn mode. The lowest cutoff . 2.405 ( ) wavenumber IS k po1 ~ TM Ol mode . a Table 3.3. Values of kpmna m=O

m=l

m=2

m=3

n=l

2.405

3.832

5.136

6.380

n=2

5.520

7.016

8.417

9.761

n=3

8.654

10.173

11.620

13.015

3.4.2 Power Delivery by TMm.n Mode

Let us evaluate the power that is delivered by the TM mn mode when k Since

> k pmn . (3.55)

the transverse field components are

•

3.4 Circular Waveguide

E _

61

2

8 A z (p, ¢, z)

i

p - WJ-Li

8p 8z

z = - kpk J:n(kpp)(A sin m¢+ Bcosm¢) exp(ikzz) WJLi 2 E _ i 8 A z (p,¢,z) cf> WJ-Lip 8¢ 8z

(3.56)

z = - mk Jm(kpp)(Acosm¢ - Bsinm¢) exp(ikzz) WJ-Lip

(3.57)

H _ 1 8A z (p, ¢, z) p - J-Lp 8¢ = m Jm(kpp)(Acosm¢ - B sin m¢) exp(ikzz) J-Lp

(3.58)

H __ .!. 8A z (p,¢,z) cf> J-L 8p =

-~J:n(kpp)(Asinm¢ + Bcosm¢)

exp(ikzz)

J-L

where J:r,(kpp) denotes differentiation with respect to the argument

(3.59)

d~(k~;r).

The total time-average power Pa1J is given by a

1

P a1J

2"

= 2 Re

X

0

=

0

a

~Re 2

(E 1l*) .zpd¢dp

2"

(EpH; - Ecf>H;) pd¢dp . 0

(3.60)

0

Let us first consider a case in which m yield

i

O. The ¢ integrations for m

2" ~m2k E H* d¢ = z J2 (k p) o cf> p WJ-L2 ip2 m p

(IAI + IBI 2

2

)

•

i

0

(3.62)

Therefore, P a1J becomes

pdp.

(3.63)

62

3 VVavegtrides

The evaluation of Pav is tedious but straightforward. Let kpp = u, then

o

=- I

(3.64)

.

Using recurrence formulas for the Bessel function

J;"(u) =

~ [-Jm+1(u) + Jm-1(u)]

(3.65) (3.66)

•

gIves

:>:1[2 2 ] 2 Jm+1(u) + Jm_1(u) udu

1=

(3.67)

o where kpa = x. Based on the formula

J~(u)udu

2

=~

[J~(u) - Jm- 1(u)Jm+1(u)]

(3.68)

I is integrated to

x2

4 [J;'+l (x) -

1=

Jm(x) Jm+2 (x)

+J~_l (x) - Jm- 2 (x)Jm(x)] .

(3.69)

Since Jm(x) = 0, 2

1= ;-

[J~+l (x) + J~_l (x)] .

(3.70)

Further simplification is possible using the relations

Jm+1(x)

= -Jm-1(x) = -J;"(x).

(3.71)

The final result for m = 1,2,3,· .. is 2

1= P av

;-J;" 2(X)

(3.72)

2 2 = (IAI + IBI )

~k~ J.L

(k pa)2 J;" 2(kpa) .

(3.73)

f.

Similarly, it can be shown that for m = 0 P av

2 1fkz = IB 1 2 2 WJ.L

(

f.

)2 12( ) kpa J o kpa .

(3.74)

3.5 Circular Dielectric Waveguide

63

3.4.3 TE Wave The analysis of TE wave (Hz t= 0 and E z = 0) propagation inside a metallic circular waveguide is similar to the analysis of TM wave propagation. The solution to the Helmholtz equation 1 0 pop

p

0

(3.75)

op

•

IS

sinm¢ (3.76) cosm¢ The explicit field components are obtained by substituting F z (p, ¢, z) into Table 3.2 in Section 3.1. Since E¢ vanishes at p = a, J:n(k~mna) = 0, where k~mna is the nth root of the derivative of the mth order Bessel function. The roots of J:n(k~mna) = 0 are tabulated in Table 3.4. Note that the propagation constant is given by k z = Jk 2 - (k~mn)2. The lowest cutoff wavenumber of the TEmn mode is

k~ll ~

1.841, indicating that the lowest propagating a mode among TE and TM waves is the TEll mode. Table 3.4. Values of k~mna

m=O

m=1

m=2

m=3

n=1

3.832

1.841

3.054

4.201

n=2

7.016

5.331

6.706

8.015

n=3

10.173

8.536

9.969

11.346

3.5 Circular Dielectric Waveguide Consider electromagnetic wave guidance by an infinitely long circular dielectric waveguide, as shown in Fig. 3.5. A circular waveguide (/-Ll , €}) is embedded in a background medium (/-L, 10), thus constituting a model for an optical fiber with a core and cladding. It is assumed that the circular dielectric waveguide is denser than the background medium (€} > 10). Let us represent the propagating waves in regions (/-L}, €d and (/-L,€) based on the boundary conditions. The

64

3 Waveguides

y (~, e)

a z

--~

"7''---'

,....7--'--+-----------

Fig. 3.5. Circular dielectric waveguide with radius a.

boundary conditions require that the four tangential electric and magnetic field (E,p, E z , H,p, and Hz) continuities must be simultaneously satisfied at p = a. The four boundary conditions are available by assuming that both the E z and Hz components are nonzero. Therefore, the guided wave is a hybrid mode that is a combination of TE (Hz i- 0) and TM (E z i- 0) waves. Inside the waveguide (p < a), both E z (equivalently A z ) and Hz (equivalently F z ) satisfy the Helmholtz equations

= O.

(3.77)

Hence, A~ and F[ take the form of (3.78) (3.79)

where the dispersion relation is

2 k z

k2 + p

2 k 2 = 1 = W ILl £1

.

(3.80)

The choice of the azimuthal field variation, cos m¢ and sin m¢, in A z and Fz facilitates the enforcement of the boundary conditions later on. Outside the waveguide (p > a), the fields are chosen as evanescent types that vanish at infinjty p -+ 00. The vector potentials satisfy the Helmholtz equations

-0 .

(3.81 )

3.5 Circular Dielectric Waveguide

65

The corresponding A~I and pII are assumed to be (3.82) (3.83) where K m is the modified Bessel function of the second kind and the dispersion relation is (3.84) In order for the wave to be properly guided by the dielectric waveguide, the parameters k z , k p , and Cl p must all be real numbers. The boundary conditions and field components must be used to find the propagation constant k z • The field components in the respective media for P < a and P > a are straightforwardly available from Table 3.2 in Section 3.1. The condition of E,p continuity at P = a then gives

(3.85) where the prime denotes differentiation with respect to the argument as J' (kpa) = dJm(kpp) . m d(kpp) p=a Similarly, matching the remaining boundary conditions for the E z , H,p, and Hz continuities at P = a produces the other three equations for A, B, C, and D. A set of the four simultaneous equations can be written in a matrix form p

k J' (k a) £1

m

p

o

o

-~J' (k a) fJ.1 m p

o

o

66

3 Waveguides A

B = 0.

•

(3.86)

c D The dispersion relation is obtained by setting the determinant to zero as p

J

o where kpa

= Ul, Ci.pa -

-K

o mk z U2,

-

_

= p,

J

_

= Ul

W The dispersion relation can be rewritten as p

-

p

--

-K

U~ -

o

o

P

p

to

-J

p

J:"(Ul) _ K:"(U2) J ( )' and K = U2 K ( ). m Ul m U2

K

/-l

-

-J

J

-/-ll

p --

(3.88)

K

or (3.89)

or

(3.90)

3.6 Shielded Stripline

67

In principle, the propagation constant k z can be determined by solving the dispersion relation (3.90). Let us introduce the normalized frequency V as (3.91) An alternative way of determining k z is to solve the simultaneous equations (3.90) and (3.91) graphically. Plotting (3.90) and (3.91) on the UI-U2 plane and identifying intersections of two curves yields the solution k z . Note that (3.91) represents a circle of radius V on the Ui-U2 plane. For each index m, there are a number of solutions k z = k zmn . Once k zmn is determined, the ratios of the field amplitudes A, B, C, and D can subsequently be obtained from (3.86). The hybrid waves are designated as HE mn or EH mn modes. The HE mn mode refers to when the Hz component is dominant, while the EH mn mode refers to when the E z component is dominant. Table 3.5. Mode cutoff frequencies Modes

Jm(Ul) = 0

Ul

HEll

Jl (Ul) = 0

Ul = 0

TEO! I TMO! , HE21

JO(Ul)=0

Ul = 2.4048

Table 3.5 shows that the HEll mode is the lowest order with zero cutoff frequency. Single-mode HEll propagation along a dielectric circular waveguide is possible if 0 < Ul < 2.4048. Considering the constraint (3.91), the radius of circle V on the UI-U2 plane must be V < 2.4048 to guarantee single-mode HEn propagation. Hence, the condition for single-mode propagation is written as

o < Jkr - k a < 2.4048. 2

(3.92)

Consider a practioal single-mode silica fiber that operates at 1.55 (J.lm) , where the refractive indices for the core and cladding are nl = 1.46 and n = 1.46 0.002, respectively. The condition (3.92) shows that the core radius a must be a < 7.77 (J.lm) to support single-mode propagation.

3.6 Shielded Stripline Due to recent advances in integrated-circuit technologies, planar transmission lines such as striplines are now commonly used in microwave circuits. This section investigates a shielded stripline, as shown in Fig. 3.6. Regions (I), (II),

68

3 Waveguides y

PEC/. /.

•

,

I, I, I ,

'I 'I ' !

original problem (a)

,

I

!,

PEC

wJ-//fl'fl'fl'A0'fl'fl'/Lf//fl'fl'/A.mfl'fl'fl'/fl'////fl'/fl'fl'fl'//fl'//L

I ,I , I,

' I' I

, I, I

I '(I) I ' I,

~ ~ ~ ~ ~[~ ~ ~ ~ ~ ~ ~ ~ I'l""'~<.<:~~~~~~~~~~~[~ ~ 5~!~~ ~ ~~~~ ~~~~X~~~ ~

-r----------J.! I ,

T

i,

T

-I

(III)

////)///////////////////////////J//////////////////////////////h

equivalent problem (b)

Fig. 3.6. Cross section of shielded stripline. and (III) are filled with a dielectric material and a conducting strip is embedded in the center of region (II). Regions (I), (II), and (III) are covered with PEe shielding walls. When a voltage source is applied between the conducting strip and the shielding plane, an electromagnetic wave propagates along the z-direction through regions (I), (II), and (III). For the analysis of the shielded transmission lines, it is possible to use the image method and Fourier transform to obtain rigorous series solutions [3]. The image method and Fourier transform are utilized in this section to obtain an analytic series solution for a shielded stripline. Although the derivation procedure is somewhat involved, the final series solution for the dispersion is fast convergent, and is thus useful for practical application.

3.6 Shielded Stripline

69

3.6.1 Field Representations

Consider the original problem of Fig. 3.6 (a), where regions (I), (II), and (III) are filled with a dielectric material of permittivity E. Due to the electric walls placed at x = ~ and ~ - T, it is possible to apply the image method to transform the problem in Fig. 3.6 (a) into the equivalent problem in Fig. 3.6 (b). Figure 3.6 (b) illustrates the problem geometry of an infinite number of striplines that are placed between two PEC parallel plates. Since regions (I) through (III) are homogeneous, either a TM or TE wave, transverse to the z-direction, propagates in regions (I) through (III) in Fig. 3.6 (b). This section investigates TE wave propagation (E z = 0 and Hz ::f 0). The electric vector potential is necessary for field representation as F z (x, y, z) = Fz(x, y) exp(ikzz). The vector potentials should satisfy the Helmholtz equations (3.93)

where k =w..jJif.. In each region, the guided waves are written as

F!(x,y) =

2~

00

(Pt eil1Y +Pi e -il1Y ) e-i(zd(

(3.94)

-00 00

F!/(x,y) =

L

[S::'cos(~my)+Q~sin(~mY)]

m=O,2,4···

. cosam(x - nT)

•

(3.95)

2 - k 2 T} = Jk 2 - 1"2 - k 2 and n = integers where a m = m7r ) ~m c = Jk 2 - a m z, ~ z' . a ote that the constraint (m = 0,2; 4, ... ) is required to make the tangential electric field in region (II) vanish at the vertical PEC walls as

(3.97)

In view of Table 3.1 in Section 3.1, the fields in the respective regions can be obtained in terms of the vector potentials. In region (I) 00

iT} (Pt eil1Y -

p/- e- il1Y ) e-i(z d(

(3.98)

-00

H / (x,y) Z

kz 1

= WJ.LE 27r •

H;(x,y)=

t WJ.LE

00 -00

(

_

+. l1Y

i( F e'

_

.l1Y)

+ F-e-'

/

(-k~+k2)F!(x,y).

?

e-"z

d(

(3.99)

/

(3.100)

70

3 Waveguides In region (II) 1

EII(x,y)=-x .

E

00

L m=O,2,4·· .

. cosam(x - nT)

(3.101)

00

H~I (x, y) = :;E L

am [S::-' cos(emY)

+ Q~ sin(emY)]

m=O,2,4·· .

. sin am (x - nT) •

H;I (x, y) =

~

WJ.lE

(-k~

+k

2

(3.102) F;I (x, y) .

)

(3.103)

In region (III) 00 •

~1)

(3.104)

HIII(X,y) = k z x

W J.lE

1 211'

00

i(

-00

. [Pi! Iei1)(y+d) + PII Ie -i1)(Y+d)] e -i(x d( •

H; II (x, y) =

t (_ WJ.lE

2 k~ + k )

F; II (x, y) .

(3.105) (3.106)

Since the number of striplines is infinite, the modal coefficients S::-, and Q~ should be independent of the stripline location n. The shielded stripline in the original problem can have either an even or odd excitation. When even (or odd), the field configuration becomes even (or odd) with respect to the a T center x = - - of the shielded stripline. The even and odd fields are

2

2 illustrated in Fig. 3.7. Therefore, it is possible to let S::-, -+ (±l)n sm and Q~ -+ (±l)n qm for the even [(+l)n] and odd [(_1)n] excitations. 3.6.2 Boundary Conditions

To determine the propagation constant k z , matching the boundary conditions is necessary. The vector potential expressions (3.94) through (3.96) employ I' This means that six six unknown modal coefficients-P!, qm, Sm, and simultaneous equations must be otained by matching the boundary conditions. The boundary conditions require that the tangential electric and magnetic

PlI

3.6 Shielded Stripline

71

PEC

..

.............. __ .

.

T

Fig. 3.7. Even and odd field configurations.

fields must be continuous at the boundaries y = b1 , 0, -d, and (-d - ba). The boundary conditions to be enforced at each boundary are given as

o

otherwise

E~II (x, -d - ba) = 0

(3.110)

< x < nT + a

EiI(x, -d)

for nT

o

otherwise

H;I (x, -d) = H; II (x, -d) .

(3.112)

Note that the boundary conditions, Hi(x,O) = H;I(x,O) and H;l(x,-d) = HiII (x, -d), are automatically satisfied due to (3.109) and (3.112). It is necessary to obtain a set of simultaneous equations for the modal coefficients qm and Sm . •

The condition Ei(x,b 1 ) = 0 gives (3.113)

•

Next consider the condition

Eil(x,O) E~(x,O) =

o

for nT

< x < nT+a (3.114)

otherwise

72

3 Waveguides which can be written as 00

1

27r

iTJ

(Ft - Fi) e-i(z d(

-00 00

L

~m(±l)nqm

m=O,2,4···

-

< x < nT + a

. cos am(x - nT)

for nT

o

otherwise.

Applying the inverse Fourier transform

1

27r

(3.115)

00

(. )ei(z

dx

to (3.115)

-00

yields

=

00

00

L

L

00

00

nT+a

cosam(x - nT)ei(z dx nT

=2: n=-oo m=O,2,4· ..

(3.116)

where (3.117)

Fi

Solving (3.113) and (3.116) for 00

Ft = 2:

00

2:

iTJ (1_le i2 1/ bl)

n=-oo m=O,2,4,'" 00

gives

(±l)nqm~mG~()

(3.118)

00

It is possible to show that the tangential electric field in region (I) van-

ishes at the vertical PEC walls as follows:

[E; (~ + pT, y) = 0]

using

Ft

and

Fl-

3.6 Shielded Stripline

EI(~

2

Y

T

+ p ,y

73

)=~aFl(x,y)

ax

f.

•

00

271'f.

-00

Z

-

( (Pt ei'1Y + PI e- i'1Y) e- i«( ~+pT) d(

00 •

-00

(3.120) Since the integrand [ . ] is an odd function of (, the condition E~

°

•

(~ + pT, y)

= is satisfied irrespective of y. Consider the condition for pT < x

< pT + a ll;(x,O) = ll;I(X, 0)

(3.121)

which can be written as 1

00

L

(±l)Psmcosam(x - pT) .

(3.122)

m=O,2,4 .. ·

Substituting 00

00

into (3.122), multiplying (3.122) by cosal(X - pT), and integrating over pT < x < pT + a yields

n=-oo m=O,2,4,.··

. 1 271'

00 -00

1 G~(() 7/ tan 7/b 1

pT+a

cos al (x - pT)e-i(x dx

d(

pT

00

L

(±l)Ps m

m=O,2,4 .. ·

pT+a

cos al (x - pT) cos am (x - pT) dx

•

pT

.

(3.124)

74

3 Waveguides

Note pT+a

COSal(X - pT)e-i(x dx = Gj(-()

(3.125)

pT pT+a

cosal(X - pT) cos am (x - pT) dx pT

= c5m1 e m

a

2

(3.126)

where c5ml is the Kronecker delta, eo = 2, and em = 1 (m = 2,4,6'" ). Therefore, (3.124) becomes

where l = 0,2,4, ... and p = integers. Therefore, (3.127) is rewritten as

a (±1)PSI2 el

(3.128)

m=O,2 ,4···

where

n=-oo

00 -00

Jt

•

1}

1( b ) G~(()Gj( -() d( .

tan

1} 1

can be transformed into a fast convergent series based on residue calis given in the next subsection. culus. The detailed derivation of The condition E;II(x, -d - b3) = yields

Jt

°

F+ -i21)b3 F III = IIT e .

•

(3.129)

(3.130)

Consider the boundary condition (3.111). Taking the inverse Fourier transform of (3.111) results in

=

f:

f: ~m

n=-oo m=0,2,4°o' ~1}

(3.131) •

-d) = HIII(x -d) which is Consider the boundary condition HII(X z' z" rewritten as

3.6 Shielded Stripline

75

00

L

[(±l)P sm cos((m d ) - (±l)P qm sin((md )]

m=O,2,4···

. cos am(x - pT)

(3.132) Substituting

FiJI + FJI I = -

00

00

L

L

n

-00

m=O,2,4,.··

1 b (m [(±l)n sm sin((md) 1J tan 1J 3

+ (±l)n qm cos((md)]

·G~(()

(3.133)

into (3.132), multiplying (3.132) by cos al (x - pT), and integrating over pT < x < pT + a yields

00

+

L

(m [sm sin((m d ) + qm cos((m d )]

Ji = 0

(3.134)

m=O,2,4···

where 00

Ji= L

00

-00

n=-oo

1( b )G~(()Gf(-()d(. 1J tan 1J 3

(3.135)

To determine the propagation constant k z , the dispersion relation must be obtained by setting the determinant of the simultaneous equations (3.128) and (3.134) to zero. In particular, when the strip width d --+ 0, the dispersion relation is given by (3.136) where the sign ± corresponds to the even and odd excitations, respectively. Although the preceding derivation of the dispersion relation is rather lengthy, its final expression (3.136) is remarkably simple and thus efficient for numerical evaluation.

3.6.3 Residue Calculus for

J;:

Let us transform the integral J;(q = 1,3) into a fast convergent series based on residue calculus. The integral J q is given by

76

3 Waveguides t

00 - 00'"

n=-oo

where anm «() =

i(

an ",

(1 - ei(a) . (2

e,(nT , m a =

2 -am

Let

l(b)a~«()af(-()d(

m1l'

a

(3.137)

q

and 11 = Jk 2 "/

2 (2 - k z'

-

(3.138) n=-oo

where Iq

00

1 =-

21l'

00

f«() d( .

(3.139)

-00

Let us evaluate I q for three different cases, namely, n - p = 0, n - p n - p < o. •

> 0, and

Consider I q when n = p. It is more convenient to transform I q into 1

I q =21l' 00

-

f«() d( .

(3.140)

-00

Note that f«() contains simple poles at ( = simple poles at ( = ±(v = ±

k2

V1l'

-

bq

±am when

m = l and other

2

- k;,

which are the solutions

to the characteristic equation", tan(",b q ) = O. Consider the contour integral along the closed path r, which consists of r 1 , r 2 , r 3 , and r 4 in the complex (-plane with R --7 00, as shown in Fig. 3.8. The paths r2 , r3 , and n denote semicircles. Assume that an imaginary part of the wavenumber k is slightly positive. The imaginary part of +(v is therefore always positive. The residue theorem gives 00

r

f«() d( = 21l'i

L v=o

Res f«()

(3.141)

(=(.

where 00

L

21l'i

v=o •

00

L v=o

= Z

Res

(=(.

f (()

«( -

2(2(1 _ ei(a) (v) «(2 _ a2 )«(2 _ a 2 ) ",tan(",b ) m

I

q

•

(3.142)

3.6 Shielded Stripline

77

1m (~)

••• x x

-~

v

••• :::..... ......:.-JL-_

x -am

Re

(~)

x •• •

Fig. 3.8. Complex (-plane with simple poles.

L'Hopital's rule gives 1

-

(3.143)

Hence, the residue is written as (3.144) In addition, the contour integrations along the path sider

r, r4

r

are necessary. Con-

f(() d( = I q

(3.145)

f(()d(=O.

(3.146)

For m = 2,4,6,···, two different simple poles at ( = ±am exist. Their residue contributions are

78

3 Waveguides

r2

f(() d( = -1ri<5ml Res

(=-a~

f(()

(---t-a~

(3.147)

r3

f (() d( =

-1riOm l Res

(=a~

f (() (3.148)

For m = 0, a simple pole at ( = 0 exists. Its residue contribution is

~

f (() d(

= -1ri<501 Res (=0

f (0

i 1 2(2(1 - ei(a) = - - <501 -------;c---;-, _:........:...--;-;o----=2 TJtan(TJb q ) (3

a = -<501 TJ tan( TJb q )

(---to

(3.149)

. (---to

Substituting (3.144) through (3.149) into (3.141) yields • 00 2" (1 i( a) a I - 0 em _ ..::. '>11 - e • q ml 2TJ tan (TJb q ) (---ta~ bq 11=0 ell ((~ - a~) ((~ -

L

•

Consider I q when n

(3.150) af) .

> p. The complex (-plane is shown in Fig.

3.9. Note 2

that the function

f (() contains simple poles at ±(v = ±

k2 -

V1r bq

-

k~,

which are the solutions to the equation TJ tan(TJb q ) = O. Thus 00

f(O d( +

f(() d( = 21ri r2

L Res f(() . v=O (-(.

(3.151)

Since

r2

f(() d( = I q

(3.152)

f(() d( = 0

(3.153)

3.6 Shielded Stripline

79

1m (~)

~v X

X X

x'"

rl x, ,

,

Fig. 3.9. Complex (-plane with simple poles.

I q is given by 00

I q = 21Ti

L

v=o

Res f(C)

(=(v

(3.154) •

Consider I q when n < p. Let us choose a semicircular path in the lower half-plane and perform residue calculus. The result is 00

I q = -21Ti

L

v=o

Res f(C)

(--(v

(3.155) The expressions (3.150), (3.154), and (3.155) can be combined into

(3.156)

80

3 Waveguides

where (3.157) Substituting (3.156) into (3.138) gives ()()

(3.158) Summing series over n in

Ji yields

()()

n=-oo

(3.159)

(3.160)

(3.161)

•

Therefore 2 [1 ±

()()

L

(±l)nAn = (±l)P

ei(v T - ei(v a

1 =f

=f ei(v(T-a)]

ei(v T

•

(3.162)

n=-oo

Substituting (3.162) into (3.158) gives the final series expression for

•

Ji as

(3.163)

Problems for Chapter 3 1. The TE mn mode propagates along a rectangular waveguide with dimensions (a x b), as shown in Fig. 3.2. Compute the power delivered by the TE mn mode.

Problems for Chapter 3

81

2. Consider the dielectric slab waveguide shown in Fig. 3.3. The dielectric medium in region (III) is replaced with a perfect magnetic conductor (PMC). Determine the dispersion relation of a TE wave. 3. Construct (3.86) using the boundary conditions for the field continuities. 4. Derive (3.134) from (3.132) and (3.133). 5. Derive (3.155) using residue calculus. 6. Consider the shielded stripline shown in Fig. 3.6. Assume TM wave propagation in the z-direction (Hz = 0 and E z i 0). Following the TE wave analysis in Section 3.6, derive the dispersion relation for a TM wave.

4

Cavity Resonators and Coupler

4.1 Rectangular Cavity Resonator Resonant cavities are basic microwave components that store electromagnetic energy. Microwave resonant cavities are known to have large quality factors. A rectangular cavity resonator is relatively easy to analyze, yet it provides physical insight into the resonance mechanism. This section investigates wave resonance in a rectangular resonant cavity that is surrounded by electric walls. 4.1.1 TE Mode

Consider a rectangular cavity resonator that is surrounded by perfectly conducting walls, as shown in Fig. 4.1. A resonant cavity is filled with a dielectric medium with permeability J..t and permittivity E. A rectangular cavity resonator of dimensions (a x b x d) is viewed as a rectangular waveguide of dimensions (a x b) with perfectly conducting walls placed at z = 0 and d. z

y

d

b

a

, Fig. 4.1. Rectangular cavity resonator surrounded by perfectly conducting walls.

84

4 Cavity Resonators and Coupler

Assume that the TE mn mode propagates along the ±z-direction with A(x,y,z) = 0 and F(x,y,z) = zFz(x,y,z). The Helmholtz equation for Fz(x, y, z) is given by

(4.1)

, Let

n1r (Ok Fz(x, y, z) = cos a x cos b y Ae' zZ m1r

ok

+ Be-'

zZ

)

(4.2)

where

1r

C:

f

+ (nb1rf + k;

= k

2

(4.3)

and A and B are unknown amplitudes. Since the tangential electric fields at z = 0 and d must vanish,

Exl z=O,d

= _~ 8Fz (x,y,z) to 8y

=0

(4.4)

-0 .

(4.5)

z=O,d

E y z=O,d

= ~ 8Fz (x,y,z) to 8x z=O,d

Hence, the unknown coefficients A and B are chosen accordingly to yield m1r

n1r

(4.6)

Fz(x,y,z) = Gcos a x cos b y sinkzz

rnr d'

where k z (m,n) = 0,1,2,"" (m > 0, n ~ 0, m+n f:. 0), and p = 1,2,3, .... The wave associated with F z (x, y, z) is called the rectangular cavity TE mnp mode, whereas its field components can be obtained by substituting Fz(x,y,z) into Table 3.1 in Section 3.1. 4.1.2 TM Mode

Similarly, it is possible to obtain the vector potentials for the rectangular cavity TM mnp mode. Assume

n1r (Ok Az(x,y,z) = sin a x sin b yA'e' zZ m1r

+ B'e-'

zZ

)

(4.7) (4.8)

F z (x, y, z) = 0 .

The tangential electric fields at z boundary conditions

ok

o and

d must vanish. Therefore, the.

4.1 Rectangular Cavity Resonator

Exl

=

z=O,d =

E Y

z=O,d

iFAz(x,y,z) WJ-LE 8x 8z i

85

=0

(4.9)

=0

(4.10)

z=O,d

2

8 A z (x,y,z) WJ-LE 8y 8z i

z=O,d

yield

"

m1l"

.

n1l"

rm

A z (X, y, z ) = C sm a x sm b y cos d z

(4.11)

where m, n = 1,2,3, ... and p = 0,1,2,· ... The resonant wavenumber k mnp for the TEmnp and TM mnp modes is given by

=W# where

W

(4.12)

is the resonant angular frequency.

4.1.3 Quality Factor

When conducting walls that surround a cavity are imperfect with finite conductivity (J, ohmic energy loss due to the finite (J occurs. To measure the energy storage and loss for a resonant cavity, the quality factor Q of a resonant cavity is used as (4.13) where U is the energy storage and WI is the time-average power loss. Let us evaluate Q for a rectangular cavity that is surrounded by imperfect but good conducting walls. The field configuration near the surface is shown in Fig. 4.2. Assume that current flow through good conductors exists almost near the conducting surface due to the small skin depth. For a good conductor like 3 copper, the skin depth is 6 = 6.6 X 10- (mm) at 100 (MHz). The surface current density J s on a good conductor is approximated by the surface current on a PEC surface as

•

-J

8

:::::!

nx

-H

(4.14)

where H is the magnetic field on an idealized PEC surface. The average power loss on the imperfect conductor surface S is then 1

WI =-Re

2

(4.15)

86

4 Cavity Resonators and Coupler

n'"

-

II

cavity interior (Il, €) '/j

goo con uctor (Ill' cr)

Fig. 4.2. Fields in good conducting surface. I

According to (2.29), the relation between the electric and magnetic fields inside a good conductor is given in terms of the wave impedance Z as E~ZnxH

(4.16)

where WJ.LJ

Z = (1 - i)

2(J

•

(4.17)

ds

(4.18)

Hence 1

-

R J8 WI =2 s where R =

WJ.Ll

2

is the surface resistivity of a good conductor.

2(J For illustration, let us consider a case when the TE 101 mode is excited within a rectangular resonant cavity. Substituting the electric vector potential for the TE 101 mode

FzC

x, y, z) = C cos ex) sin Gz)

(4.19)

into Table 3.1 of Section 3.1 results in the field components. They are given by Ex = E z = 0

(4.20)

(4.21)

87

4.1 Rectangular Cavity Resonator

(1f) (1f) C ad sin ;;x cos d

i

Hx = -

1f2

WJ-L€

(4.22)

Z

(4.23)

H y =0

(1f)2 cos (1f). (1f a x sm d

i Hz = WJ-L€ C a

Z

)

(4.24)

•

First, consider the current flow on the top surface (z = d) as

Js

= z=d

nx H

z=d

+ zHz )

= -z x (xHx = -fjHx

z=d

.

= -y"

z=d

Z

WJ-L€

2

C1fd sm . (1f-x ) a

a

(4.25)

.

The power loss from the top surface is -

R Js

2

z=d

ds

S

-

1

2

a

b

o

0

1

R

2

WJ-L€

(1f2 ) 2 (1fX ) IC/ ad sin a dx dy 2

2

(4.26)

where w=

1f

Ja 2 +d2

(4.27)

----~-.

#

ad

Similarly, it is possible, to evaluate the power loss associated with the bottom and side surfaces. The total power loss is shown to be

ICI1f

2

WJ-L€

2

2b(a3

+ d3) + ad(a + ~) 2

a 4 d2

•

(4.28)

The energy storage is the sum of the electric and magnetic energy as

u=~4

(4.29)

88

4 Cavity Resonators and Coupler

where d

1 -2 1 4 V fiE I dv = 4

0

b

o o (4.30)

(4.31) Note that the stored time-average electric energy is equal to the stored timeaverage magnetic energy. Substituting U and WI into the quality factor Q finally yields Q = wU WI

1f 2R

/-L

f

b(a 2 + ~)1.5 2b(a 3 + d 3 ) + ad(a 2

+ d2)

.

(4.32)

For a cube cavity (a = b = d),

Q = .j21f 6R

/-L

-

•

(4.33)

4.2 Circular Cavity Resonator A hollow circular cylinder is another simple cavity structure often used for confining electromagnetic wave energy. This section analyzes electromagnetic resonance within a circular cavity resonator (/-L and f) covered with perfect conducting walls, as shown in Fig. 4.3. 4.2.1 TM Mode Consider a TM mn mode that propagates along the ±z-directions in a circular cavity resonator. The vector potentials associated with a TM mode are F(p,
4.2 Circular Cavity Resonator

89

z

d ~--p

Fig. 4.3. Circular cavity resonator surrounded by perfect conducting walls.

cosm¢ (4.35)

sinm¢ where m = 0,1,2, .. " A and B are unknown coefficients, and k~ + k~ = k • Since the tangential electric fields, Etf> and E p , must vanish at z = 0 and d, the boundary conditions 2

=

E P z=O,d

2

8 A z (p,¢,z) WJ.Lf. 8p 8z i

=0

(4.36)

z=O,d

=0

(4.37)

z=O,d

yield cosm¢ (4.38)

sinm¢ p7r

where k z = d and p = 0,1,2, .... Since the tangential electric fields, Etf> and E z , must vanish at p = a, the boundary conditions require •

Ez

I

p=a.

t

=0

=-WJ.Lf.

(4.39)

p=a

= 0. p=a

The resonance condition is thus determined by

(4.40)

90

4 Cavity Resonators and Coupler

(4.41) There exist a number of roots that satisfy the resonance condition. Let the nth root be denoted by k p = k pmn , where Jm(kpmna) = 0, m = 0,1,2"", and n = 1,2,3, .... The resonant wavenumber k mnp for the TM mnp mode is

k~mn + (~f·

(4.42)

The field expressions for the TM mnp mode can subsequently be obtained by substituting A z (p, ¢>, z) into Table 3.2 in Section 3.1. 4.2.2 Quality Factor for TM olO Mode

When a conducting surface is imperfect, power loss due to the conduction current occurs. Let us evaluate the quality factor Q that is associated with the simple TM olO mode (m = 0, n = 1, and p = 0), where kp = k and ka = 2.405. The magnetic vector potential for the TM olO mode is

Az(p, ¢>, z) = HoJo(kp)

(4.43)

and its respective field components are ik 2 Ez = H oJo(kp) W J.Lf.

(4.44)

Hq, = - k HoJ~(kp) J.L

k

= -HOJI (kp) .

(4.45)

J.L

The time-average electric and magnetic energy storage U is given by

U=~

4

a

o

[JJ(kp)

+ J[(kp)] pdp.

(4.46)

Since 2

J~(u)udu = ~ [J~(u) - Jm-l(U)Jm+I(u)]

(4.47)

4.2 Circular Cavity Resonator

91

U can be rewritten as (4.48)

The power loss from the top surface with surface resistivity R is

2

~ 2

a 0

R IHoik J 1 (kp) J.L

2-rrpdp

2

o = -rr R IH l (ka)2 J2(ka) . 2 J.L2 1

(4.49)

The power loss from the side is thus

(4.50)

The total power loss is

= -rrRIHoI2 k 2a(a + d)J~(ka) . J.L2

(4.51 )

Hence, the quality factor Q is shown to be

_

J.L 1.202d fR(a+d) .

(4.52)

4.2.3 TE Mode

The wave analysis for the circular cavity TEmnp mode is similar to the previous TM mnp case. It is assumed that the TE mn mode propagates along the ±z-directions with the vector potentials A(p, ¢, z) = 0 and F(p, ¢, z) = zFz (p, ¢, z). The electric vector potential F z (p, ¢, z) is given by cosm¢ (4.53)

sinm¢

92

4 Cavity Resonators and Coupler

where k~ + k; = k . The boundary conditions 2

I

E

p z=O,d

= _~ 8Fz (p, ¢, z) fp 8¢

=0

(4.54)

z=O,d

EI q,

=!8Fz (p,¢,z) z=O,d

=0

8p

f

(4.55)

z=O,d

yield cosm¢ (4.56)

sinm¢ p1r

where k z = d and p = 1,2,3, .... Another boundary condition requires E q,

I

=

p=a

! 8Fz (p,¢,z) 8p

f

= 0.

(4.57)

=0

(4.58)

p=a

The parameter k p =

k~mn

is chosen such that

p=a

where m = 0,1,2, ... and n = 1,2,3,···. Hence, the field expressions for the TEmnp mode are made available by substitution of the electric vector potential cosm¢ sin kzz

(4.59)

sinm¢ into Table 3.2 in Section 3.1. The resonant wavenumber

k:n

np

for the TEmnp mode is

(4.60)

4.3 Spherical Cavity Resonator

93

4.3 Spherical Cavity Resonator Let us consider electromagnetic resonance in a perfectly conducting spherical cavity resonator, as shown in Fig. 4.4. A cavity is filled with a dielectric material of permeability J.L and permittivity €. A boundary-value problem of a spherical cavity resonator can be solved using the separation of variables technique. It is convenient to decompose a wave into TE and TM components to the r-direction where the TE (or TM) components refer to the waves whose electric (or magnetic) field is transverse to the r-direction, respectively.

4.3.1 TM Mode First, consider the TM case by choosing A(r, 0, ¢) = f Ar(r, 0, ¢) and F(r, 0, ¢) = O. Consider source-free Maxwell's equations

'V x E = iWJ.LH 'V x H =

-iw€E .

(4.61) (4.62)

Substituting -

1

-

H=-'VxA

(4.63)

J.L

into (4.61) yields 'V x (E - iwA) = 0 .

(4.64)

z

PEe

x

Fig. 4.4. Spherical cavity resonator surrounded by perfect conducting wall.

94

4 Cavity Resonators and Coupler

Let

-

-

E-iwA = -\1¢e.

Eliminating E and

H from

(4.65)

(4.62), (4.63), and (4.65) gives

- \1 x \1 x A + w WA 2

+ iWJ.Lf.\1 ¢e

= 0.

(4.66)

The f component of \12 A using spherical coordinates is not equal to the 2 Laplacian of A r, namely, (\1 A)r i- \12 A r . Therefore, the Lorentz condition \1 . A = iWJ.Lf.¢e will not reduce (4.66) into the simple scalar equation that we wish to obtain. Another gauge condition is, therefore, needed to make (4.66) a single scalar equation. To that end, let us rewrite (4.66) explicitly using spherical coordinates A

r r2

8

.

8() sin ()

if;

8A r

8 Ar r sin () 8r 8¢

sm () 8() A8¢e

r

2

+ 8r

()A 1 8¢e

1 8¢e -r 8() +'1' r sin () 8¢

J..

A

-0 .

(4.67)

A

Consider the condition that makes the () and ¢ components of (4.67) zero as 1 8

r 8()

8

1

r sin () 8¢

=0

(4.68)

-0 .

(4.69)

If the gauge condition

8A r - 8r

.

A..

+ tWJ.Lf.'I'e

=

0

(4.70)

is chosen, (4.68) and (4.69) can be simultaneously satisfied and (4.67) becomes the scalar equation

8 2 Ar

8r 2

1

8

+ r 2 sin () 8()

. () 8A r

sm

8()

(4.71)

or equivalently

(4.72) where k (= w#) is the wavenumber. Based on the separation of variables technique, it is possible to assume (4.73)

4.3 Spherical Cavity Resonator

95

Substituting (4.73) into (4.71), dividing (4.71) by 1h(r)'l/J2(())'l/J3(¢), and multiplying by r 2 results in 1

+ sm .2 ()'l/J3


Let us introduce the separation constants m and n as (4.75) 1 d 'l/J2 sin () dB

. d'l/J2 sm () dB

m

2

. 2

sm ()

= -n(n

+ 1) .

(4.76)

Then, (4.74) results in the following three ordinary differential equations (4.77) 1

-

d

sin () dB

. () d'l/J2 sm dB

(4.78)

(4.79) •

•

The solution to (4.77) is the circular harmonic functions, cos m¢ and sin m¢. Since A r is periodic with a 21T-periodicity, the parameter m must be an integer (m = 0,1,2, ... ). Two independent solutions to (4.78) are the associated Legendre functions of the first and second kinds, P;:' (cos ()) and Q~ (cos ()), respectively. A discussion on the associated Legendre functions is available in Appendix D. When n =j:. an integer, P;:'(cos()) becomes singular (unbounded) at () = 1T and O. For P;:' (cos ()) to be a solution to the spherical cavity problem considered in this section, n should be an integer. Since P;:'(cos()) = 0 when m > n, the integer m should be 0, 1, 2, ... ,n. The function Pg (cos ()) (= 1) is an unacceptable solution since the resulting electric and magnetic fields all vanish. Furthermore, since PO" (cos ()) = 0, the integer n should be n = 1,2,3,···. Since Q~(cos()) becomes singular at () = 1T and 0, Q~(cos()) cannot be solutions. Hence, 'l/J2(()) is written as (4.80)

•

The solution to (4.79) is 'l/Jl(r) = Bn(kr) [In(kr), Nn(kr), iI~l)(kr), ... ], which are related to the spherical Bessel functions bn(kr) as (4.81) The spherical Bessel functions bn (kr) satisfy

96

4 Cavity Resonators and Coupler

dPbn(kr) dr 2 r 2

dbn(kr) + 2r dr

+ [k

2 2

r

n(n + 1)] bn(kr) = O.

-

(4.82)

A discussion on the spherical Bessel functions is provided in Appendix B. Since "pI (r) must remain finite at r = 0, the spherical Bessel function of the first kind is chosen as the solution A

"pI (r) = In(kr) .

(4.83)

The solution A r , therefore, is written as cosm¢ •

(4.84)

sinm¢ The resulting TM wave components can be derived from •

E=

z

V'xV'xA

(4.85)

WIJ-f..

-

1

-

H=-V'xA.

(4.86)

IJ-

The explicit relations between the TM wave components and A r are shown in Table 4.1. Since the tangential electric fields, Ef) and E"" must vanish at r = a, the resonant condition is given by

dJn(kr) d(kr)

= 0.

(4.87)

r=a

A number of solutions to (4.87) exist. The pth root of (4.87) k = k mnp is the resonant wavenumber that is associated with the TM mnp mode of a spherical cavity resonator. 4.3.2 Quality Factor for TMoll Mode

The lowest-order mode corresponds to the cases (m = 0, n = 1, P = 1) and (m = 1, n = 1, p = 1). Consider the lowest TM oll mode

Ar

0

= JI(kr)PI (cos B) A

(4.88)

where

1rkr 2 J1.5(kr) sin kr

kr

pf(cos B) =

k

- cos r

PI (cos B) = cos B .

(4.89) (4.90)

4.3 Spherical Cavity Resonator

97

Table 4.1. Field representations using spherical coordinates

Fields

TM wave 2 /8 t 8 2 WJ-lf , T •

Er

+k

2

TE wave ,

Ar

0

/

2

•

8 Ar WJ-lfT 8T 8(} t

Eo

1 8Fr fT sin () 8

2

•

8 Ar WJ-lfT sin () 8T 8 t

E",

1 aFr fT 8(} •

Hr

t

0

WJ-lf

•

1 8A r J-lT sin () 8

Ho

, +k

2

Fr I

2

a Fr WJ-lfT 8T 8(} t

•

1 8A r -J-lT 8(}

H",

/ a2 a 2 T

2

8 Fr WJ-lfT sin () 8T 8 t

Substituting A r into the TM wave components of Table 4.1 yields three nonzero components, E r , Eo, and H",. For instance, the H", component is given by

H __ 1 oAr '" J.LT 00 1 . 0 = sm J.LT

sin kr k kr - cos r

•

(4.91)

The resonant frequency is determined from the boundary condition

Eol r=a =0.

(4.92)

The resonant condition

=0

(4.93)

r=a

gives the characteristic equation

ka -----:---:-7 = tan ka 1 - (ka)2

(4.94)

98

4 Cavity Resonators and Coupler

where the solution ka = 2.744 determines the cavity resonance frequency. The time-average electric and magnetic energy storage (or equivalently peak magnetic energy storage) U is given by

a

1

-

2

o

0

a [

471"

3J.L

1

J.Lr

o

]1 (kr)

] 2

• sin OJI (kr)

dr

0

a

--

kr [J1.5(kr)]2 dr.

3J.L

(4.95)

0

Since (4.96)

U can be rewritten as 2

U=

=

7I"

ka 3J.L

2

71"2 ka 2

3J.L 2k 71"

a

2

{ 2 [J1.5(ka)] [0.512

2

-

} Jo. (ka)J2. (ka) 5

5

0.187 x 0.373)

x 0.192 .

(4.97)

3J.L The power loss from a cavity inner surface with surface resistivity R is 1 WI =2

-

1 2 1 2

s

R

IJ81~=a •

s

R -f x ¢Htf>

211' 0

ds

11' 0

2

ds r=a

0 R sin ]1 (ka) J.La

2 2

a sin 0 dO d¢

(4.98)

4.3 Spherical Cavity Resonator

99

Hence, the quality factor Q is shown to be

Q= wU WI

1.005

- ----==-R

(4.99)

•

4.3.3 TE Mode

Similarly, a TE wave in a spherical cavity resonator can be obtained from the assumption A(r,(),¢J) = 0 and F(r,(),¢J) = fFr(r,(),¢J). Its analysis is somewhat analogous to that of TM waves. The governing equation for the electric vector potential F is

F +w

- V' x V' x

2

/-Lf.F

+ iW/-Lf.V' ¢Jm =

0.

(4.100)

Using the gauge condition A,

__

'I'm -

i 8Fr W/-Lf. 8r

(4.101)

•

gIves . () 8Fr sm 8()

(4.102)

where its solution is given by cosm¢J •

(4.103)

sin m¢J A TE wave is represented in terms of F as 1 E=--V'xF

(4.104)

f. •

-

H=

t

-

V'xV'xF.

(4.105)

W/-Lf.

For convenience, the explicit field components for a TE wave are listed in Table 4.1. Since the tangential electric fields, E(J and E¢, must vanish at r = a, the resonance condition is given by (4.106) where the pth root k = k mnp gives the resonant wavenumber for the TEmnp mode of a spherical cavity resonator.

100

4 Cavity Resonators and Coupler

4.4 Groove Guide Coupler An open groove guide is known to have a low-loss capacity at millimeter wavelengths. An open groove guide consists of rectangular grooves that are covered with PEC walls, as shown in Fig. 4.5. The groove guide interior is filled with a homogeneous dielectric medium, thereby supporting either TE (E z = 0 and Hz f:. 0) or TM (Hz = 0 and E z f:. 0) wave propagation. TE waves are known to be dominant since TE waves have a lower cutoff frequency than TM waves. A groove guide coupler, which consists of a double groove guide, is potentially a high-power coupling structure. A cross-sectional view of a groove guide coupler is shown in Fig. 4.6, where a wave is assumed to propagate along the z-direction. This section investigates the guiding and coupling characteristics of a groove guide coupler using the Fourier transform and mode matching [4). The Fourier transform and mode matching approach provides a rigorous way of analyzing a groove guide coupler. 4.4.1 Field Analysis

Consider a double rectangular groove guide that is filled with a homogeneous dielectric material with permittivity E and permeability JJ-. A TE wave propagates along the z-direction with the electric vector potential ~(x,y,z) = z~z(x,y,z)

(4.107)

The electric vector potential ~z(x,y,z) should satisfy the Helmholtz equation (\7 2 + k 2 ) ~z(x,y,z) = O. To determine the propagation constant k z , the boundary conditions for the field continuities must be utilized.

Fig. 4.5. Single groove guide.

4.4 Groove Guide Coupler

101

y

(III)

b

region (II)

~Z~0_ _---=2a region (l)

T

T+2a

...------+x

(l)

Fig. 4.6. Groove guide coupler: double groove guide.

For analysis, the guide interior is divided into regions (I) (-d ~ y < 0), (II) (0 ~ y < b), and (III) (b < Y < b + d), where their respective Fz components • are gIVen as 00

F[ (x,y) =

L

Q~ cosam(x - nT) cos~m(Y + d)

(4.108)

m=O 00

(4.109) -00 00

F[II (x, y) =

L

S~ cosam(x - nT) cos~m(Y - b - d)

(4.110)

m=O

where (4.111) (4.112) am =

m1l"

2a

, k = wVJif. =

211"

,and n = 0,1. In (4.108) through (4.110),

A the modal coefficients-Q~, S:;:', F+, and F--are introduced to represent the

fields. The aim is to constitute a set of simultaneous equations for the discrete modal coefficients Q~ and S:;:'. Note that (4.108) and (4.110) are constructed

102

4 Cavity Resonators and Coupler

so as to satisfy the boundary conditions at the PEC walls where the tangential electric field is required to vanish. 1BFz -Since Ex(x,y,z) = Ex(x,y) exp(ikzz) , the continuity of Ex to

By

at y = 0 is written as

< x < nT + 2a (n = 0,1)

E~(x,O)

for nT

o

otherwise.

Applying the inverse Fourier transform

2~

00

(')ei(X dx to (4.113) yields -00

(4.114) where

em

n (()

= i([l- (_1)m ei(2a] i(nT (2 2 e . -a m

(4.115)

The Ex continuity at y = b is written as E~ll(X,b)

E~l (x, b) =

for nT

< x < nT+2a (4.116)

o

otherwise.

Applying the inverse Fourier transform to (4.116) yields

Solving (4.114) and (4.117) for 1

00

F+=-LL

F+ and F- results in

e-il
+ snm

(4.118)

n=Om=O 1

00

F-=-LL n=Om=O

47l'1\: sin I\:b

(4.119)

Note that Hz is given by l

•

Hz(x, y, z) = - -

(4.120)

WJ-LtO

where Hz(x,y,z) = Hz(x,y)exp(ikzz). Hence, the Hz continuity at y = 0 and pT < x < pT + 2a (p = 0,1)

4.4 Groove Guide Coupler

103

H;I(X, 0) = H;(x,O)

(4.121)

Q::;" cosam(x - pT) cos(~md) .

(4.122)

can be rewritten as 00

-00 00

L

=

m=O

H ( ) · Smce z x, y, z =

2

i

a Faz (x,y,z) h H a 't e z

. . 0 d (pT contmwty at y = an

<

x Z X < pT + 2a) is seen to be automatically satisfied from differentiating (4.122) with respect to x. Substituting F+ and F- into (4.122) gives 1

WJ.Lf.

00

-LL

n=Om=O

00

-00

1

27rK

(Q~

.

cot Kb + S~ csc Kb)~m sm(~md)

00

=

L

Q::;" cosam(x - pT) cos(~md) .

(4.123)

m=O

It is possible to simplify (4.123) by using the orthogonality of the sinusoidal function cosam(x - pT). Multiplying (4.123) by cosal(X - pT), (p = 0,1) and integrating over (pT < x < pT + 2a) with respect to x yields 1

00

L L

n=Om=O

{Q~ [~m sin(€md)h + acos(~md)OmlOnpcm] (4.124)

where co = 2,

C1

=

C2

= ... = 1, Oml is the Kronecker delta, and

h=

(4.125) 00 -00

csc(Kb)G~(()Gf(_()d(.

(4.126)

27rK

Based on residue calculus, the integrals h and 12 can be transformed into a rapidly convergent series. An evaluation of h and 12 is given in the next subsection. It is necessary to derive another set of simultaneous equations based on the boundary condition at y = b. The analytic procedure for enforcing the boundary condition at y = b is similar to that for enforcing the boundary

104

4 Cavity Resonators and Coupler

condition at y = O. Applying the orthogonality property to the Hz continuity at y = b and pT < x < pT + 2a (4.127) yields 1

00

L L n=Om=O

{Q~~m sin(~md)h (4.128)

Equations (4.124) and (4.128) constitute a set of simultaneous equations for the discrete modal coefficients Q~ and S:;'. The dispersion relation is formed by setting the determinant of the simultaneous equations to zero

=0

(4.129)

where the elements of 1Ji1 and 1Ji2 are (4.130) (4.131) Solving the dispersion relation (4.129) finally yields the propagation constant kz . 4.4.2 Residue Calculus for 1 1 and 1 2

Let us rewrite h as

h= (4.132) where A = [(_1)m+1

+ 1] ei«n-plT

_( _1)mei<[(n-plT+2a] _ (_1)le i<[(n- pl T-2a J

•

(4.133)

4.4 Groove Guide Coupler

•

105

First, consider the case n = p. When m + l= odd, II = O. When m + l= even, h becomes

h=

00

(2

-00

1r'" tan (",b)

[1- (_1)m ei2(a] ((2 - a;") ((2 - an d(

00

f(() d( .

(4.134)

-00

In the complex (-plane, as shown in Fig. 4.7, the term '" sin( "'b) in h

yields an infinite number of poles at (

= ±(" = ±

k2 -

c;)

k~

2 -

for

v = 0,1,2,···. When m = l, the condition ( = ±am yields two simple poles for m = 1,2,3,··· and one simple pole for m = O. Performing contour integration in the upper half-plane along

r 1 , r 2 , r 3 , and r 4

gives

f(() d(

n

f(() d(

+

r2

f(() d(

+

f(() d( ra

+

r.

f(() d(

00

= 21ri

L

,,=0

Res f(() .

(=(y

x •• •

Fig. 4.7. Complex (-plane with simple poles.

(4.135)

106

4 Cavity Resonators and Coupler

Note

f(() d( = h

~

r3

r4

(4.136)

f(() d( = -7fi Res f(() C=-a m

= 13

(4.137)

f(() d( = -7fi Res f(() - 14

(4.138)

f(() d( = 0 .

(4.139)

C=a m

Hence 00

h = 27fi L Res f(() v=o C=Cu

Is

+ 14

for m = 1,2,3, ...

(4.140) for m = 0 .

13

Consider

13 (m - 1 , 2, 3 , ... ) = -7fi Res

C=-a m

=

(2

[1 _ (_1)mei2ca]

7f"- tan("-b) ((2 - a:,J((2 - an iSm1iS

np

• -7f~

. (4.141) C=-a m

Since the term [1 - (_1)m ei2 ca] becomes zero at ( = -am, let us use l'Hopital's rule

[1- (_1)m ei2 ca]

(( + am)

= -i2a.

(4.142)

C=-a m

Therefore a

iSm1iS np

Is(m = 1,2,3,' .. ) = - 2 ~m tan(~mb) .

Similarly, I 4 (m = 1,2,3",,) and Is(m = 0) are

(4.143)

4.4 Groove Guide Coupler

[1- (_1)m ei2(a]

(2 =

-1ri

Res

m1 np 1rK tan (Kb) ((2 - a;,)((2 - an c5 c5

(=a m

=

(2

•

-1rt

a

1rK tan(Kb)

[1- (_1)m ei2(a] m1 np (( + a m )2(( - am) c5 c5

c5m1 c5np

--

107

(4.144)

2 ~m tan(~mb)

I 3 (m = 0) = -1ri Res (=0

(4.145) Therefore (4.146) Consider Res f(() (=(v

[1 _ (_1)m ei2(a]

(2 = Res (=(v

1rK tan(Kb) ((2 - a;,)((2 - an c5

(; [1- (_1)m ei2(v a] =

-; ((; - a;,)((; - an c5

np

1

np

•

(4.147)

When ( = (0 (K = 0), Res (=(0

1 Ktan(Kb)

1 2(ob .

(4.148)

1 K tan (Kb)

1 (vb'

(4.149)

When ( = (1 = (2, ... , Res (=(v

Therefore, II can be rewritten as

-

108

4 Cavity Resonators and Coupler

1m (~)

x •• •

Fig. 4.8. Complex (-plane with simple poles.

•

Second, consider the case n f; p. Note that It contains an infinite number of poles at ( = ±(v, which are the solutions to I\: sin (I\:b) = 0. When n-p > 0, let us use the contour shown in Fig. 4.8 to evaluate It. 00

It = 21l"i 2:= Res

(=(u

v=o

(2A = 21l"i'"' L..J 21l"((2 _ a2 )((2 _ a2) v=o m I 00

i

= -

(=(u

(=("

(A

00

b2:= c

v=o v

Similarly, when n - p

Res

1 I\: tan (I\:b)

((2 - a 2 ) ((2 - a 2 ) m

I

(4.151)

. (=("

< 0,

i II = b

(A C ((2 _ a2 )((2 _ v=o v m 00

2:=

a 2) I

.

(4.152)

(=-(u

It is possible to combine (4.151) and (4.152) into i

(vA It .= - b v=o Cv ((2v _ am2 )((2v _ aI2 )

A=

00

2:=

[(_1)m+1

(4.153)

+ 1J ei("ln-pIT

_(_1)mei(uICn-p)T+2al _ (_1)lei("ICn-p)T-2al .

(4.154)

Problems for Chapter 4

109

Rewriting (4.150) and (4.153) in a compact form results in (4.155) Similarly, it can be shown that (4.156)

Problems for Chapter 4 1. Consider the TE mnp mode in the circular cavity resonator in Fig. 4.3.

Determine the lowest resonant frequency and energy stored in the cavity resonator. 2. Consider the TE mnp mode in the spherical cavity resonator in Fig. 4.4. Determine the lowest resonant frequency. 3. Consider the integral 12 given by (4.126). Using residue calculus, show that (4.126) is represented by the series given by (4.156). 4. Consider the groove guide coupler shown in Fig. 4.6. Assume TM wave propagation in the z-direction (Hz = 0 and E z ::p 0). Following the TE wave analysis in Section 4.4, derive the dispersion relation for a TM wave.

5

Propagation

•

In

Anisotropic Media

5.1 Propagation in Anisotropic Media Materials such as crystals, magnetoplasma, and ferrites possess anisotropic properties at optical or microwave frequencies. In anisotropic materials, the direction of E (or H) usually does not coincide with that of D (or Ii). For instance, the relation between D and E using rectangular coordinates is

-

•

(5.1)

Equation (5.1) can be rewritten as

15 =

f.E

(5.2)

where f is a tensor permittivity represented by a (3 x 3) matrix. Similarly, it is possible to relate B to H in terms of the tensor permeability J..Las

Ii = /iH .

(5.3)

Wave propagation in anisotropic materials depends on the wave propagation direction with respect to the tensor permittivity or permeability. 5.1.1 Dispersion Relation

This section investigates the wave propagation characteristics in unbounded anisotropic materials and derives a dispersion relation using the tensor permittivity. Assume that a plane wave

112

5 Propagation in Anisotropic Media

E =

Eo

exp (ik.

r) = Eo exp(ik",x + ikyY + ikzz)

(5.4)

propagates within an unbounded, source-free, and anisotropic medium with tensor permittivity f. Note that for a constant vector Eo \l x E

= \l x [Eo

exp (ik . r)]

= ik x E

(5.5)

.

Therefore, Maxwell's equations \l x

-E

\l x

H = -iwEE

(5.7)

(5.8) (5.9)

=

iWJ.1oH

(5.6)

become

Eliminating

ik

x

E = iWfJ,H

ik

x

H = -iw l E .

H from (5.8) and (5.9) gives k x (k x IE) + w J.LE E = 0 2

(5.10)

where the vector identity

k x (k x

IE) = - (k· k) E + k (k . IE)

(5.11)

can be used for further simplification. It is possible to express f in a diagonal form using newly rotated coordinates. The axes of the newly rotated coordinates, which make the tensor permittivity l diagonal, are called the principal axes. Assume that the tensor permittivity f using the principal coordinates (x, Y, z) is represented by

o -E =

o

0

o

o (5.12)

•

o

Substituting l into (5.10) results in

E", Ey

=0.(5.13)

5.1 Propagation in Anisotropic Media

113

Let (5.14) (5.15) k 2

where k = k;

+ k~ + k~.

2

2

- W JU'.z

=

(5.16)

S2

Consequently, (5.13) can be rewritten as

Ey

= O.

(5.17)

The dispersion relation is obtained by setting its determinant to zero

(5.18)

which can be further simplified to

(pqk z )2

+ (pSk y)2 + (qsk:zY

= (pqS)2 .

(5.19) 2

The dispersion relation (5.19) is a quadratic equation of k . Therefore, there are two solutions for k 2 . Once the wavenumbers k are determined from the dispersion relation, the field components E"" E y , and E z can be obtained from (5.17). 5.1.2 Uniaxial Medium

For certain anisotropic crystals known as a uniaxial medium, € takes the following special form using newly rotated coordinates (x, y, z)

o -

€

-

o

o o

o (5.20)

o

where the z-direction is called the optical axis. Quartz and lithium niobate are uniaxial crystals. For lithium niobate, €l = (2.300? €() and €2 = (2.208)2 €Q. Assume that a plane wave

114

5 Propagation in Anisotropic Media

E = Eo

exp (ik·

r) = xE., + yEy + zEz

(5.21)

propagates within an unbounded uniaxial medium, where the incident wave vector k is

k = k (x sin 0 cos ¢ + y sin 0 sin ¢ + zcos 0) = xk.,

+ yky + zk z

(5.22)

.

The dispersion relation (5.19) becomes

p2 [_p2 k; _ s2(k~

+ k~) + S2 p 2]

= O.

(5.23)

Let us solve (5.23) to investigate the wave propagation characteristics. •

The first solution p2 = 0 yields the wavenumber k = WV/lf1. The relation

E., Ey

yields E z = 0 and

~.,

= -

y

~y

= 0

(5.24)

= - tan ¢, which represents an ordinary

.,

wave that propagates isotropically through a uniaxial medium. The electric field vector of the ordinary wave is perpendicular to the wave vector k and optical axis z since (5.25)

-E·z=O.

(5.26)

The time-average Poynting vector Sav

for the ordinary wave is

-*)

-Sav = 2 1 Re (Ex H

= 2

1

w/l

2

1

w/l

Re

Re

IEI -',---''- k

( E . E*) k - E* ,(E . k),

.

o

2

2w/l

.

(5.27)

The expression (5.27) indicates that the power flows in the k-direction. A graphical representation of the ordinary wave is shown in Fig. 5.1.

5.1 Propagation in Anisotropic Media

115

-E •• ••••••••••••••••••••••••••••••••••••••••••••••• •••........• -• • • . . ........ . . .. . . . ... . . .. . . ......................... . • • • • • • • • • • • •• • ••••••••••••••••••••••••••••••••••••••••••••• · • , .... ....... ........ . . • • • • • •• • •••••••••••••••••••••••••••••••••••••••••••• ··... ......... . . .. .,.....,........... ,. .. , , .. ··...., . . ...,,. . ..... ...,....... ................. ..... ... . .., , , , , .. ·.. , . , .. , , ·.... ,..... ... , .. , .. , .. , , . • •• •• • • •• • • • •

•

•

'

• •

• • • • •• ••

_

•

•

•• ••

• •• •

•

• •

•

• •• • • • • ••

'

"

H··:····,·:··:·······'::::··:··:· '.'::: ··.·:.':::. ,. . , , ... . , ,. ., , , .. ,... ·.. , .. , . . · ... ..... ..... ...... .,... .. ..., . . ....... .... .

-

........ ........... ,

:, :

,

::.'.'.'.' :.•..::, : '

, ,

,,

,

,

.,

,

,

: ..'

. . ,.

..

-

,............' . , .. ............... ... ............ . . ..... .. .... , .·.. ...·············k···· ....... ... .. , .....···S··························· .. . . , .. _ ... .:::::::. ·av·::··::::··:::::::::··:··:··· ···Z·:::::::: ...::::::::.'::::::::. ... , , .. , .. .. .. , , .... , ' .. , , , .. , . .. .. , , ,, , , .. , , . . . .... . . .. . .. .. . . .. . .. . ... . . ,

Fig. 5.1. Ordinary wave in uniaxial medium.

•

The second solution _p2k~ -

S2

(k; + k~) + S2 p 2 =

0

(5.28)

yields the wavenumber

k=w

(5.29)

and field components s2k x Ex Ez p 2k z

Ey Ez

-

s2k y p 2k z

(5.30)

•

(5.31 )

The second solution represents an extraordinary wave that propagates anisotropically in a uniaxial medium. The time-average Poynting vector Sav for the extraordinary wave is -Sav = 1 2 Re

(-E x -H*) (5.32)

Since (5.33)

the electric field vector of the extraordinary wave is, in general, not perpendicular to the wave vector k. The direction of the power flow does not

116

5 Propagation in Anisotropic Media

coincide with the k-direction. The electric field vectors of ordinary (Eord ) and extraordinary (E ezt ) waves are perpendicular to each other since

-E

ord

.E

ezt

= 0.

(5.34)

A graphical representation of the extraordinary wave is shown in Fig. 5.2.

-H • ••• • ••• • •• • • • • • • • •• ....................... ................. ......

• ••••••••••••••••••••••••••••••••••••••••••••••• .

.. , , . ··....-.......................... . .. .. . ............ .. . ·· .. ................. . -..... . . • • • • • • • • • • • • • • • • • • • • • • • • • •• • ••••••••••••••••••••••••••••••••••••••••••• · . . . . ......... . .. .. . . ... . .. . . ·.. .... ............ . .... .. ......... .. . ·..... . ..... . ......... . . ........ ............. . ·• ·. . .. . . .. ..... . ............... . . . .... .. . • • • • • • ••.. • •.. • •.. • •... • • •.•.. • •.•• • •.•.•... ••• .• ••••••••••••••••••••••••••••••••••••••• ·......... ...•.... . . . .. . . . . . ... . .. . . . .. .. . .. .. . . · • • • • • • • • • • • • • • • • • • • • • • • • • •• • •••••••••••••••••••••••••••••••••••••• ··.•...• ... .•.. . .... ........................ .. . . . .... ................. . ..................................... .............. . . . ... ..." .. . . ...... . • • • • • • • • • • • • • •• • • • • • • • • ••••••••••••••••••••••••••••••• • •.. • •.•• • • • •.•.•.•.•.. • •.• .. • • •. . •••••••••••••••••••••••••••••• ·••..•••••• • • ••. .••.••.. ..•.•.... ... ...... . . .. . . • •

•

•

_. •

•

•

•

•

•

........... ......... . .... ...... .......... ....... ..... .. ···..·............. ............ ...... . . .... . ... .. ...... . ....... ...... . .... . . ........... ........... ....... .... .. ............. .. .. .... . . ....... . .. .. . .. .. .............. .. ........ ................ .............. .. . . ...... . . ....... ....................... ....... .

. :..;.............. ;;.';;.':;.'. k'"...... :: :.....;....... ::..:;:.....:..: ..;:......;...... :::.':.:.'. :.:.' ..':.""'..:: ... .. ..

... ... ........ .. .................. .•.•..•••.•.••.•..

.

. :

: . ········································z········ . ..

.................................... .................................. .......

.

..

... .. .

Fig. 5.2. Extraordinary wave in uniaxial medium.

A uniaxial medium permits two types of waves, namely, ordinary and extraordinary waves. A uniaxial medium is, therefore, said to possess a birefringence or double refraction property.

5.2 Propagation in Ferrites Certain magnetic materials, called ferrites, find wide applications as phaseshifting elements in microwave circuits and antennas. Commonly used ferrite devices include the ferrite gyrators, ferrite isolators, and ferrite circulators. This section investigates the wave propagation characteristics in ferrite materials in terms of the dispersion relation.

5.2.1 Magnetized Ferrite When ferrites are subject to a DC magnetic field, i.e. when ferrites are magnetized, they exhibit anisotropy and their permeability takes a tensor form Jj.. The permeability of ferrite materials depends on the direction and state of the internal DC magnetic field. If the applied magnetic field is, for example, in the z-direction, the ferrite permeability J.L is represented as

5.2 Propagation in Ferrites

iK

-

-

-tK

o

0

o

•

117

(5.35)

0

where the explicit expressions for the ferrite parameters, J1 and K, are given in [2, page 235]. Assume that a plane wave 1l = H 0 exp (ik . r) propagates in an unbounded ferrite medium, where the governing equation is written as (5.36) First, consider wave propagation in any arbitrary direction with the wavenumber k = xkz + fjk y + ikz . Substituting the vector identity k x (k x

1l) = k (k .H) - (k . k) H

(5.37)

into (5.36) yields Hz Hy

= O. (5.38)

This can be rewritten as Hz Hy

2 = k2 where p2 -- k 2 _ W 2 r-' /1f. q2 -k 2 _ W 2 r-u /1nf. , and k x Setting the determinant to zero

=0

(5.39)

+ k y2 + k z·2

(5.40)

yields the dispersion relation (5.41)

118

5 Propagation in Anisotropic Media

When the wavenumber

= k (x sin Bcos ¢

+ f) sin Bsin ¢ + z cos B)

(5.42)

is assumed, the dispersion relation becomes (5.43) where (5.44) (5.45) (5.46) 5.2.2 Transversely Magnetized Ferrite

As a special case, consider a wave that propagates in the direction transverse to the z-axis (k = xk z + f)k y ). Since k z = 0 and k~ + k; = k2 , (5.39) becomes

o

o

Hz

o

Hy

=

o.

(5.47)

o

The dispersion relation (5.41) reduces to (5.48) ",2

which yields the two solutions k = w.,jf..Lo€ or w •

€

f..L-f..L

.

When k = w.,jf..Lo€, a TEM wave (Hz =I 0 and Hz = Hy = 0) propagates and the ferrite material behaves like an isotropic medium. ",2

•

When k = w

€

f..L - f..L

, the field components are related by

Hz JIl,f..L + (f..L2 - ",2) cos ¢ sin ¢ . 2 2 Hy f..L2 cos ¢ + ",2 sin ¢

(5.49)

5.2 Propagation in Ferrites

119

5.2.3 Longitudinally Magnetized Ferrite Let us consider another special case in which kx - ky = 0 and k = zk z , where the ~irection of wave pr~a~ion coi~cides with ~he direction of the DC magnetic field. Then, the dIspersIOn relatIOn (5.41) gIves (5.50) Substituting k+ into (5.39) results in Z/'i,

o

-

o

•

-/'i,

•

- Z/'i,

o

/'i,

Hy

=0

(5.51)

o

which gives (5.52) Substituting (5.52) into Ampere's law "V x H = -iWfE

(5.53)

•

gIves (5.54) Similarly, the condition k z = k- gives (5.55) Ex = -iEy

(5.56)

.

Therefore, the electric field is represented as (5.57)

(E

(E-)

which indicates left-hand ~) and right-hand circular polarizations. Let us consider a linear combination of two circularly polarized waves

(5.58)

120

5 Propagation in Anisotropic Media

where their time-varying form is

Ex (z, t) = cos(k+ z - wt)

+ cos(k- z -

wt)

k+ - k2 z cos

= 2 cos

(5.59)

E y (z, t) = sin(k+ z - wt) - sin(k- z - wt)

k+ +kcos - - - z - w t 2

= 2 sin

•

(5.60)

Taking the ratio of

Ey(z, t) -=:=-:----:- = tan Ex (z, t)

(5.61)

shows that the polarization state of (5.58) is not only linear but also rotates as the wave propagates along the z-direction. This behavior is known as a Faraday rotation. The principle of a ferrite gyrator, ferrite isolator, and ferrite circulator is based on the Faraday rotation of a wave that propagates in ferrites at microwave frequencies. Further discussion on the microwave gyrator, microwave isolator, and microwave circulator can be found, for instance, in [5].

5.3 Propagation Along Ferrite-Filled Parallel-Plate Waveguide Waveguides containing magnetized ferrites find practical applications in microwave circuits and systems. For instance, a ferrite-loaded rectangular waveguide is used as a nonreciprocal phase shifter in microwave frequencies. This section considers a conducting parallel-plate waveguide filled with transversely magnetized ferrite. The problem geometry is shown in Fig. 5.3. Note that some relevant discussion is available in [6]. Assume that a DC magnetic field is applied along the z-direction and the ferrite tensor permeability Ii is given by i It

0

o

-

•

(5.62)

o A wave is assumed to propagate in the y-direction. Assume no field variation in the z-direction

8 = 0 .

8z

5.3 Propagation Along Ferrite-Filled Parallel-Plate Waveguide

121

x

PEe

a

ferrite medium

'\f'v-

wave propagation

Fig. 5.3. Conducting parallel-plate waveguide filled with ferrite.

Faraday's law \l x E = iw"jiH

(5.63)

can be rewritten as 8Ez

•

(H 8y = zw JL z

z 8E - 8x =

.

8Ey 8Ez 8x - 8y =

ZW

(

(5.64)

H ) + JL y

(5.65)

+ ZK

H -ZK z

. ZWJ.Lo

. H )

.

H

Y

(5.66)

z·

Similarly, Ampere's law (5.67)

\l x H = -iWEE

can be rewritten as (5.68) (5.69) 8Hy

8x

8Hz

8y =

. -ZWE

E

z·

(5.70)

Equations (5.64), (5.65), and (5.70) govern the field components (E z , Hz, and H y ), whereas (5.66), (5.68), and (5.69) govern the field components (Hz, E z , and E y ), respectively. The field components (E z , Hz, and H y ) and (Hz,

122

5 Propagation in Anisotropic Media

Ex. and E y ) constitute TE (E y = 0) and TM (H y = 0) waves, respectively. Thus, a parallel-plate waveguide filled with transversely magnetized ferrite admits TE and TM waves. •

Let us first consider the TE case. Solving (5.64) and (5.65) for H x and H y yields

Hx

Hy

= =

. 8Ez -ZJ-L 8y

8Ez + I', 8x (5.71)

W (J-L2 - 1',2)

. 8Ez ZJ-L 8x

8Ez + I', 8y (5.72)

•

W (J-L2 - 1',2)

Substituting H x and H y into (5.70) gives the governing equation (5.73) •

Based on the separation of variables technique, E z takes the form of (5.74) where n = 1,2,3· . '. The expression E z is chosen to satisfy the boundary

= O.

condition E z x=O,a

Substituting (5.74) into (5.73) produces the propagation constant k y (5.75) Subsequently, substituting (5.74) into (5.71) and (5.72) results in the magnetic field components

_ Anexp(ikyY) [k . Hx (2 2) J-L Y sm WJ-L-K,

_ Anexp(ikyY) [. Hy (2 2) ZJ-L WJ-L-K,

•

(n1l') n1l ' (n1l')] x + cos x I',

a

a

a

n1l' cos (n1l') . k . (n1l' )] x + ZK, Y sm x a

a

(5.76)

.

-·a

.

(5.77)

Next, consider the TM wave case. Substituting Ex and E y from (5.68) and (5.69) into (5.66) gives

Hz = O.

(5.78)

Problems for Chapter 5

123

Since E y must vanish at x = 0 and a, Hz takes the form of

Hz = B n cos

(na x) exp(ikyY) '7r

(5.79)

where n = 0,1,2,' . '. Substituting (5.79) into (5.78) finally gives the solution (5.80) The TM wave propagates as if the waveguide were filled with an isotropic medium with permittivity € and permeability J..Lo.

Problems for Chapter 5 1. Prove the vector identity k x (k x E) = - (k .k) E+ k (k· E). 2. Show that (5.41) reduces to (5.43). 3. Consider a TE wave propagating in a ferrite-filled parallel-plate waveguide, as shown in Fig. 5.3. Compute the time-average power delivered by the TE wave.

6

Electromagnetic Theorems j

When an electromagnetic wave impinges on scattering objects, the incident wave not only scatters off but also penetrates the scattering objects. If the scattering objects are complex-shaped, the scattering analysis becomes complicated; hence, electromagnetic theorems can be used to facilitate the problem formulation. This chapter introduces some useful theorems and principles that are often used in electromagnetic scattering analyses.

6.1 Uniqueness Theorem When solving electromagnetic boundary-value problems, it is important to ensure the uniqueness of the solutions. The uniqueness theorem provides the boundary conditions that guarantee the uniqueness of solutions to boundaryvalue problems. Consider the electromagnetic fields E and H within volume V that contains the electric and magnetic sources Ji and Mi. The problem geometry is shown in Fig. 6.1. The uniqueness theorem can be stated as follows: The fields E and H can be uniquely determined if one of the following

A

n

-E,H -

- -

::::::::\ V ...... •• • • • • • ••••••• • ••••••• ••

s

.•.•..• •...... .. ... .... . . . . • • • • •• • • • • • • • • •• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • ... • • •.. • •• •.•. •.... • • • •.. • •.. • •.•.•.•. •.. • •. •... • • •.•.•.•.•.•.•.•.•.•.•.•.. • •.•.•.. • •.. ••• . . ••••••••••••••••••••••••••••••••••••••••••• • • • • • • • • • • • •• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •• ••• •• • • • • • • • • • • • • • • •• • • • • • • •• • • • • • • • • • • • • • • • • •

Fig. 6.1. Bounded volume V

•

containing sources Ji

and Mi.

126

6 Electromagnetic Theorems

boundary conditions is specified on the surface S. 1. The tangential component n x E is given on S. 2. The tangential component n x H is given on S. 3. The tangential component n x E is given on a portion of S and the tangential component n x H is given on the rest of S. Proof Let us assume that two different fields (E a, H a) and (Eb' ll b) exist within V. Each field must obey Maxwell's equations as

\7 x E a = iWj1.H a \7 x \7

X

\7

X

Mi

(6.1)

-H a = -iwf.E -+ Ji a

(6.2)

-

-

-

E b = iWj1.H b - M i

Hb = -iWf.Eb + J i

(6.3) (6.4)

.

Subtracting (6.3) and (6.4) from (6.1) and (6.2), respectively, yields (6.5)

\7 x H d = -iwf.Ed - - -

---

where Ed = E a - Eb and Hd = H a Let us consider

\7. (Ed x

(6.6)

-

H b·

1l:) = 1l:. (\7 =

X

Ed) - Ed· (\7 x H:)

1l:. (iwj1.H d) -

Ed· (iw{·E:)

(6.7) where the symbol 0* denotes the complex conjugate of (.). Integrating (6.7) over V and utilizing the divergence theorem gives

v

iwj1.

2 Hd 1

*' dv- v iwf./ Ed

2

dv.

(6.8)

6.2 Image Method

127

For analytic convenience, the medium is considered to be lossy with a complex permeability J1, J.tr + iJ.ti and complex permittivity E = Er + iEi, where J.ti and Ei account for medium loss. Substituting J1, and E into (6.8) finally produces

f

s

(Ed

X

H~)

. dS =

-

+i

v

W (J.ti

v

w

IH d

(J.tr

2

Rd l

+ Ei Edn 2 - Er

dv

E dJ2) dv.

(6.9)

Consider the vector identities

If one of the conditions 1, 2, or 3 is satisfied, then, either n x Ed = 0 or

n x H d = 0 on the surface S. Therefore

indicating that the left-hand side of (6.9) becomes zero. The integrands of the right-hand side of (6.9) are all real numbers. The only way to make the right-hand side of (6.9) zero is to let (6.12) (6.13) Therefore, as long as either condition 1, 2, or 3 is specified on the boundary S, the field must be unique, such as (Ea = Eb and R a = R b). This completes the proof. A lossless mediu]Il is considered to be a limiting case of a lossy medium with an infinitesimally small Ei and J.ti (Ei -+ 0 and J.ti -+ 0), making the uniqueness theorem also applicable to a lossless medium.

6.2 Image Method The image method is a useful concept for solving problems associated with infinitely extended PEC or PMC plane boundaries. This section introduces the image method and its application to radiation from a small current above a perfect electric conducting plane. 6.2.1 Image Method Using Infinite Planes

Let us consider the original problem in Fig. 6.2, illustrating the real sources J and M above a PEC plane boundary of infinite extent. The real source -J represents a small electric current density, which is a small electric dipole

128

6 Electromagnetic Theorems

-J

-J

+

M

M

+

real sources

-

+

-

region CD

original problem

-J

-J

+

M

M

+

real sources region CD

------------------------------•

Image sources

region (II)

+

+

-

+

equivalent problem

Fig. 6.2. Image method with PEC (perfect electric conductor) plane. Ej3: positive electric charge, e: negative electric charge, -+: electric current density J, EE: positive magnetic charge, 8: negative magnetic charge, -*: magnetic current density M . •

consisting of two opposite electric point charges. Similarly, the real source M represents a small magnetic current density, which is a small magnetic dipole consisting of two opposite magnetic point charges. Since the outside field produced by a small magnetic dipole is identical to the outside field produced by a small electric current loop, a small magnetic dipole is considered a small electric current loop. The total field in region (I) is the sum of the real source contribution in the absence of a boundary and the reflection from the boundary. The reflection is regarded as the response of an image source that is placed in free space without a PEe boundary. The image sources must be constructed so as to satisfy the tangential electric field continuity condition

/

6.2 Image Method

129

requiring a zero tangential electric field at the boundary. Figure 6.2 illustrates that the equivalent problem gives the same field in region (I) as the original problem. The image method indicates that the total field in region (I) is the response due to the real and imaginary sources. As a dual problem, consider the real sources J and M that are placed above the PMC boundary, as shown in Fig. 6.3. The image sources must also be constructed so as to satisfy the tangential magnetic field continuity requiring a zero tangential magnetic field at the boundary.

-J

-J

+

M

M

+

real sources

-

+

region

en

original problem

J +

real sources

-

J

M

M

+

-

region (I)

------------------------------•

Image sources

+

region (II)

+

equivalent problem

Fig. 6.3. Image theorem with PMC (perfect magnetic conductor) plane. EB: positive electric charge, e: negative electric charge, -+: electric current density J, EE: positive magnetic charge, 8: negative magnetic charge, -*: magnetic current density M.

130

6 Electromagnetic Theorems

6.2.2 Current Above Perfect Electric Conducting Plane

z

- E (r r/

-M(r') h

h

---y

--------

--------

/

/ /

/

/

/

/

r2

/

-___

<1>/ /

-------- /" /

/ /

e r-....

---

)

'777:~~rY I PEe I

____

----0

x

Fig. 6.4. Small current above perfect electric conducting plane.

Consider an infinitesimally small magnetic current source M(r')

M(r') = iM J(x')J(y')J(z' - h)

(6.14)

located above a perfect electric conductor, as shown in Fig. 6.4. Let us evaluate the fields E(1') and 1/(1') radiating from M(1") when r » h. Based on the image method, the electric vector potential F(1') is the sum of the contributions from the real and image sources that are placed in· free space. Substituting M = iM J(x')J(y') [J(z' - h) - J(z' + h)] into a free-space solution (1.102) • gIVes -

iklr-r'l

€

') e , (l' F(1') = 47T M _'I dv v' r- r

A€M = z-,-47T

r

ikr1 e

(6.15)

rl

where € and k are the medium permittivity and wavenumber, respectively. Note that rl and r2 are the distances from the real and image sources to the observing point r, respectively. F( r) is expediently expressed using spherical coordinates (r,(},
r»h

6.3 Equivalence Principle rl~r-hcos()

(6.16)

+ hcos()

(6.17)

r2 ~ r

into F(r) with

131

z = f cos () - 0sin () yields

_

F(r) = -

if.M e

ikr

27rr

(

.)

f cos () - () sin () sin (kh cos ()) .

(6.18)

Substituting F(r) into -

1

-

E = --\7 x F(r)

(6.19)

f.

and collecting the

~ r

terms gives _ • Mke ikr E = ¢ 2 sin () sin(kh cos ()) . ,

.

7rr

(6.20)

J

E¢ Subsequently, substituting F(r) into •

H = iwF(r) +

t

Wllf.

\7 [\7 . F(r)]

(6.21)

1

and collecting the - terms gives r •

H = -()

f.

-E¢.

(6.22)

Il

6.3 Equivalence Principle The equivalence principle is a useful tool for solving boundary-value problems in electromagnetic scattering and radiation. The equivalence principle allows the actual sources to be replaced by equivalent, fictitious sources that produce the same field within a region of interest. Some relevant discussion on the equivalence principle is available in [7]. This section introduces Love's equivalence principle and its application to the problem of transmission through a circular aperture. 6.3.1 Love's Equivalence Principle

Let us consider problem (1) in Fig. 6.5, where the current sources J and M radiate the fields (E, H) throughout regions (I) and (II). In problem (2), the

132

6 Electromagnetic Theorems

-J

s

~M_

region (I)

region (II)

H

-E

-H

-E

problem (1)

- " -

fL).. . . . . . . . . ._---.....; . . ,Js = n X H region (II) null field

-H

-E

region (l)

,,Ms= - nxE

problem (2)

Fig. 6.5. Equivalent problems.

sources J and M are removed and the fields are assumed to be null in region (I) and (E, H) in region (II), respectively. To retain a null field in region (I) and (E, H) in region (II), the surface currents J s and Ms must be placed on the surface boundary

-J

s

=

Ms =

nx H -

-n x E.

(6.23) (6.24)

Then, the surface current densities ]. and M. radiate (E, H) in region (II), while maintaining a null field in region (I). Problem (2) is thus equivalent to problem (1) as far as region (II) is concerned. 6.3.2 Transmission Through Circular Aperture

Consider electromagnetic transmission through a circular aperture in a perfectly conducting plane, as shown in Fig. 6.6. The incident field impinges on the aperture from below a perfectly conducting plane. Let us evaluate the far-zone transmitted field for z > O. For simplicity, the aperture electric field

6.3 Equivalence Principle

133

z

-F(r)

--y

x

i wave incidence Fig. 6.6. Circular aperture in conducting plane.

is assumed to be E ap = xEo. It is convenient to transform the original problem into an equivalent problem (b), as shown in Fig. 6.7, where the surface magnetic current is M 8 (7;<') = -ft x lEap. The equivalent problem (b) in Fig. 6.7 can then be transformed into the equivalent problem (c) in Fig. 6.7, where a PEC sheet is placed beneath the boundary S. The surface electric current 1 8 (r') is short-circuited by the PEC sheet beneath the boundary S, and the field (E, 1l) in region (II) is solely due to the surface magnetic current M8(1"). According to the image method, the surface magnetic current M 8(r') on an infinitely extended PEC can be considered as 2M 8(1") in free space, and the electric vector potential for the radiated field for z > 0 is _

eik!r-r'1

t

F(r) = 41T

s'

2M s (1")

1_r - -'I r

ds'

iklr-r'l

41T S' A

(-2ft x E ap )

tEo

~-r - _'I r

ds'

eik1r-r' 1

=-y-21T S' I1'-1''I ds' .

Let us evaluate the electric vector potential F(r) when r

Ir -

r'

(6.25)

» r'. Note

I~ r -

r' cos 'l/J = r - r' sin 0 cos(¢' - ¢) .

(6.26)

134

6 Electromagnetic Theorems

z

-E y E ox x

2a

-E

-

.. x

t . 'd wave mCl ence

original problem (a) z n'"

-E

====== '-+x 2a null field

equivalent problem (b) z

-E

Ms x

~

2a

PEe equivalent problem (c) z

-E .. x

~-----

2a equivalent problem (d)

Fig. 6.7. Equivalent problems for circular aperture in conducting plane.

6.3 Equivalence Principle

135

The electric vector potential F(r) is -F(-) r

o .f.E ~-y21r

ikr e exp [-ikr' sin/:lcos(¢' - ¢)] ds' r

S'

a ikr • f.E0 e =-y 21r r o

211'

exp [-ikr' sin 0 cos(¢' - ¢)] r' d¢' dr' .(6.27) 0

Substituting the identity 00

L

exp[-ikr'sinBcos(¢'-¢)] =

inJn(-kr'sinB)e in (¢'-¢)

(6.28)

n=-oo

into F(r) results in

_ eikr F(r) = -fjf.Eo r

a

Jo(-kr'sinO)r'dr'.

(6.29)

0

Since fj =

•

•

rsin Bsin ¢ + Bcos Bsin ¢ + ¢ cos ¢

(6.30) (6.31)

F(r) can be written as F(r) = - (rsinBsin¢+OcosBsin¢+¢cos¢) ikr 2 e a J 1 (ka sin B) ·f.Eo r k ' B asm •

= f Fr + BFo + ¢F¢ . •

(6.32)

1 Let us evaluate E and H in the far zone. Substituting F( r) into E = - - \7 x I'.

F(r) and collecting the 1 terms gives r

•

-E=-~ ik (-BF¢+¢Fo . .) .

(6.33) •

Similarly, substituting F(r) into H = iwF(r) +

t

WJlf.

\7 [\7. F(r)] and col-

1

lecting the - terms gives r (6.34)

136

6 Electromagnetic Theorems

6.4 Induction Theorem The induction theorem is often used to solve electromagnetic scattering problems. This section first introduces the induction theorem, then considers its application to scattering from a conducting rectangular plate. 6.4.1 Equivalence Based on Induction Theorem

•

i (E ,Hi)

Consider an incident wave

that impinges on a dielectric scat-

terer, as shown in the original problem (1) in Fig. 6.8. The sources

(E, Hi)

produce the incident field

(J, M)

in the absence of the scatterer. The

total field in region (II) consists of the incident and scattered components

(E, Hi)

and t is (Et,H ).

(E H S

,

),

respectively. The transmitted field in region (I)

-. -.

sourees(E " H') (IfS,BS ) :::::::-::.:::::::-::.:-:::-:::.:.:-:-:-:.::: . ..

_ J /

S

M . :.::-:-::::-::-:-::::-:-::':-:':::: ...

~

. .:.:

·.... ·....

·:::.::-::::-:-:::.seatterer

..... - t -t ( E , H ).,:::::: region (II) • ••••

• ••••• ••••••

• •

.-

region (I)

••

.

• • • • • • • •• • • • • • • • • • • • • • • • •• • • • • •• •• • • • • •

:.:.:

.

•• ••••••••••••••

. .

.

problem (1)

-

-

n"

............... -

....::::::-:-:::::::

.... '::::-:-::':'::... J-

.. ::::::.':'"

"::::::. • •••••

• • •• • •• • • • • •• • • • • •• • • •• ••• • •• • • • • • • ••••

·..... .... . ... .

• ••••• • •••• • ••••

. ...... ..

.::::::::::.:.. seatterer ...............

.::::'.:-'-::::-:~::::::::.~.:: : ...

...... ... .. ..

.::.:::::::;:7,... ·

. • •

•

• • • • • •• • • • • • • • • • • • • • • •• • • • • • •• ••• • •••

~

-

S

1'\

=-nx

H- i

/\_.

Ms= nxE'

..

problem (2)

Fig. 6.8. Equivalent problems.

6.4 Induction Theorem

•

137

Based on the induction theorem, the original problem (1) can be transformed into the equivalent problem (2), where the field It) is as-

(Itt,

sumed in region (1) and the field

(Its,Jr)

is assumed in region (II).

1s

= -=i fI. x E must

At the boundary between regions (1) and (II), the surface currents

-S -t) fI. x (H - H = -fI. x H

•

(-S -t) E - E =

and M S = -fI. x be impressed in order to satisfy the boundary conditions. It is of interest to consider what happens when region (1) becomes a PEC scatterer. Note that

Itt

i

= H

t

= 0 in region (1) and fI. x

(F + E

S

between regions (1) and (II). The impressed surface electric current -fI.

S

X

= 0

)

Hi is short-circuited on the PEC surface, and the field (E , H

1s = S )

in

region (II) is due solely to the impressed surface magnetic current M S = -='E nx . •

A

6.4.2 Scattering from Conducting Rectangular Plate

Let us consider the problem of scattering from a perfectly conducting rectangular plate based on the induction theorem. A uniform plane wave

It(x, y, z) =

xEo exp [i(ky sin Bi

-

kz cos Bi )]

(6.35)

is incident on an (a x b) perfectly conducting rectangular plate, as shown in Fig. 6.9. Let us evaluate the far-zone scattered field for z > O. If the size of the plate is relatively large compared to the wavelength (a, b » >'), it is possible to assume that the plate is infinitely extended. According to the image method, the surface magnetic current M S on an infinitely extended PEC can be considered as 2M S in free space. The electric vector potential for z > 0 is given by -

.

€

F(r) = 411"

~ -

S'

€

411" S'

eik1r-r' 1

2f1. x

It (x', y', 0+) 1_r - _'I ds' r

2f1. x

F

€eikr

(x', y', 0+) ~exp [ik(r - r' . f)] ds'

b/2

a/2

-b/2

-a/2

r

--,-----

411"r

•

2f1. x

It (x', y' ,0+) exp (-ikr' . f) dx' dy'

.

(6.36)

Let us evaluate the electric vector potential using spherical coordinates (r, B, ¢J). Substituting

138

6 Electromagnetic Theorems

z

,.,

n

:....---.... y

a

b x

Fig. 6.9. Perfectly conducting rectangular plate of size (a x b) . •

n x If' (x',y',O+)

= f)Eoexp(iky'sin(}i)

(6.37)

r' . f = (Xx' + f)y') . (x sin (} cos 4> + Ysin (} sin 4> + Z cos (}) = x' sin (} cos 4> + y' sin (} sin 4>

(6.38)

into F(r) and performing integration gives

F(r) =

(f sin (} sin 4> + {; cos(} sin 4> + ¢ cos 4» .

f.e

ikr

21rr

Eoab A

sin Q Q

sin f3

f3

A

+ (}Fe + 4>FcfJ

(6.39)

Q

ka . (} ..I. = 2 sm cos 'I'

(6.40)

f3

=

= fFr where

k; (sin (} sin 4> -

sin (}i) .

(6.41)

6.5 Duality Theorem

-

139

-

The far-zone scattered fields E and H are given by

E =

ik ( . -~ -()F

.) + ¢Fe

H=iw(OFe+¢F
(6.42) (6.43)

.

The far-zone fields (6.42) and (6.43) are good approximations of the exact solution as long as (()i and ()) « ~ and (a and b) » A.

6.5 Duality Theorem The duality theorem utilizes the symmetric property of Maxwell's equations. Maxwell's equations for the electric source (J ¥- 0 and M = 0) are given as

\7 x E e = iWJ.LHe

(6.44)

\7 x He = -iwfEe + J

(6.45)

\7 . fEe = Pe

(6.46)

-

(6.47)

Similarly, Maxwell's equations for the magnetic source are

\7 x H m = -iwfEm -

-

-\7 x Em = -iwJ.LHm + M

-

(M ¥-

0 and J = 0)

(6.48) (6.49)

\7 . J.LH m = Pm

(6.50)

\7 . fE m = 0 .

(6.51)

A dual set for the electric and magnetic sources is shown in Fig. 6.10. Maxwell's equations (6.48) through (6.51) can be constructed from (6.44) through (6.47) by interchanging the quantities, as shown in Table 6.1 (E e -+ H m , He -+ -Em,"')' Equations (6.44) through (6.47) and (6.48) through (6.51) are termed dual to each other since they take a symmetric mathematical form. If the solution (Ee and He) to (6.44) through (6.47) is known,

140

6 Electromagnetic Theorems

,...-- ......... /

~

,I

, I

"-

I m \ Em' H / \\ \. /

\(-)1 \ E., He / \

'-

.... _-

/

,....-- ....... "M\\

I( -

'-

-)

.... _-

./

/ ./

Fig. 6.10. Dual set.

Table 6.1. Dual quantities Electric source

-E

Magnetic source -

e

Hm

He

-Em

-

J

-M

pe

pm

J.L

f

f

J.L

A

F

the solution to (6.48) through (6.51) can immediately be constructed by interchanging the quantities, as shown in Table 6.1 (E e -+ H m , He -+ -Em, ... ).

Consider radiation from a small electric current] above a perfect magnetic conductor plane, as shown in Fig. 6.11. The real current J above the PMC surface is equivalent to the sum of the real J and its image source in a medium of infinite extent. The problem considered in Fig. 6.11 is dual to the problem considered in Fig. 6.4 in Section 6.2, and the radiation field in the far zone

6.6 Reciprocity Theorem

141

z

-E (r-

)

I

PMC

-

e

h

/"---

n, / 't' /

h -L,t

---- -..-...

/

------ -- __ d

x

Fig. 6.11. Small electric current above perfect magnetic conductor plane.

is immediately given from the duality theorem. The solution to the problem in Fig. 6.4 is given by (6.20) and (6.22). The solution to the dual problem in Fig. 6.11 is thus ikr "ke

_ H = -<jJJ 2

1I'T

sin osin(khcos 0)

"ke ikr E = -OT]J sin 0 sin(kh cosO)

_

(6.52) (6.53)

211'T

where T] =

•

6.6 Reciprocity Theorem In circuit theory, the reciprocity theorem holds for any linear network. In electromagnetic theory, the Lorentz reciprocity theorem holds for any linear medium. The reciprocity theorem in circuit theory is a special form of the Lorentz reciprocity theorem in electromagnetic theory. This section derives the Lorentz reciprocity theorem, starting from Maxwell's equations.

142

6 Electromagnetic Theorems

v

s -(J , M a

a

(E , H a

a

)

)

set (a)

v

s (E b' H

b )

set (b)

Fig. 6.12. Two different sources and fields.

6.6.1 Lorentz Reciprocity Theorem Consider two sets of Maxwell's equations in the same linear medium with different sources. The problem geometry is shown in Fig. 6.12. Set (a) is given by \1 x

-E

a=

iWJ.LH a - M a

(6.54)

\1 x H a = -iwfEa + J a .

(6.55)

Set (b) is (6.56) \1

X

-H

b = -iwfEb + J b .

The vector operations \1. (E a x H b ) and \1. (E b \1. (E a x H b) = Hb' \1 x E a - Ea' \1

X

X

(6.57)

H a ) give

ll b

6.6 Reciprocity Theorem

143

Subtracting (6.59) from (6.58) leads to

\l.

(Ea x Hb - E b

X

H a) = (6.60)

The expression (6.60) is called the Lorentz reciprocity theorem in differential form. Taking volume integration of (6.60) and utilizing the divergence theorem on the left-hand side yields the Lorentz reciprocity theorem in integral form as

s v

(E a

x Hb

-

Eb

X

H a ) . dS =

[- (Ea' J b - H a · M b) + (E b · J a - H b · Ma)J dv.

(6.61)

Two special cases of (6.61) are of practical interest. If region V is free of sources (Ja = M a = Jb = Mb = 0), (6.61) reduces to

s

(E a x H b - E b

X

H a ) . dS =

0.

(6.62)

If region V contains sources and V extends to infinity, the fields (E a , Eb' H a , _ eikr and H b ) take the asymptotic form at r -+ 00. The fields have the B and r ¢ components that satisfy the relations, Eo = TJH¢ and E¢ = -TJHo. Thus, the left-hand side terms of (6.61) all vanish at r -+ 00, and (6.61) reduces to

v

(E a . J b- H a . M b) dv =

(Eb·Ja-Hb·Ma) dv.

(6.63)

V

6.6.2 Reciprocity for Antennas Let us apply (6.63) to the problem of two small dipole antennas in free space, as shown in Fig. 6.13. Assume that the electric current density J a at antenna 1 produces E a , which is the electric field at antenna 2. Similarly, the electric current density J b at antenna 2 produces E b , which is the electric field at antenna 1. Therefore, two antennas satisfy the Lorentz reciprocity theorem as

v

Ea' Jb dv =

(6.64)

Let us reinterpret (6.64) in terms of the voltages and currents. The voltages and currents are given as

144

6 Electromagnetic Theorems

VI

+ •

11

-E -J

-E

b

-J

a

antenna 1

a

12

+

V2 b

antenna 2

Fig. 6.13. Two small dipole antennas.

(6.65) (6.66) (6.67)

12 =

Jb'

as .

(6.68)

Therefore, the Lorentz reciprocity theorem (6.64) can be rewritten as (6.69) which amounts to the reciprocity theorem in circuit theory.

Problems for Chapter 6 1. Consider a small dipole electric current placed within a conducting parallel-plate waveguide, as shown in Fig. 6.14. Construct the equivalent problem based on the image method. 2. A uniform plane wave is normally incident on a thin PMC circular plate of radius a. Assume a A. Evaluate the hackscattered far-zone field using the induction theorem.

»

Problems for Chapter 6

145

x

PEe

t

a

Q8(x-x )o(y-y I

I

)8(z)

z

Fig. 6.14. Small dipole within conducting parallel-plate waveguide.

3. Consider radiation from a small electric current J above a perfect magnetic conductor plane, as shown in Fig. 6.11. Derive the solution (6.52) and (6.53) without recourse to the duality theorem.

7

Wave Scattering

7.1 Dielectric Circular Cylinder The problem of scattering from a dielectric circular cylinder is a canonical boundary-value problem that can be solved with the separation of variables technique. This section investigates TM and TE scattering from a long dielectric circular cylinder under the illumination of a uniform plane wave.

7.1.1 TM Scattering Assume that an incident TM wave (Hz = 0 and E z =I- 0) impinges on a cylinder that is infinitely long in the z-direction, as shown in Fig. 7.1. The incident magikp ikx netic vector potential A~ = e= e- cos ¢ is chosen to yield the incident electric field E~ = Eoe- ikx . The incident TM wave has nonzero E z , H p , and

H¢ components. No field variation is assumed in the z-direction

:z = 0

so that the geometry of the scattering problem becomes two-dimensional. The incident electric field is a periodic function of if; with period of 21l'. Therefore, the incident magnetic vector potential A~(p, if;) can be expanded in a Fourier series as 00

A~ (p, if;) =

L

Bne

in

¢

(7.1)

n=-oo

where 1 Bn = -

21l'

211"

e-ikpcos¢-in¢ dif; 0

(7.2)

148

7 Wave Scattering y ........... ... .....- --. .. . . • •• •

J.L,€

• • • •• • • •

•• ••

p= a

'-:,:-:

----l4.~ep!-.~,d,.,:..--

t:::,?-,:

z

.".

x

....-tV'v incident wave

••••

..•... . .. . . • • • •• • •• • •••••••• .. .... ...... -

Fig. 7.1. Cross section of dielectric circular cylinder.

Hence, A~ (p, 4» is given by 00

A~(p,4» =

L

i-nJn(kp)ein.

(7.3)

n=-oo

Scattering from a circular cylinder generates the scattered magnetic vector potential A: for p > a (k = w..jji€) and the transmitted magnetic vector potential for p < a (k 1 = Wy'/Llfl), respectively. The scattered magnetic vector potential A:(p,4» satisfies the Helmholtz equation using cylindrical coordinates

A;

a

1 pop

(7.4)

Based on the separation of variables technique, it is possible to solve (7.4), where (p, 4» takes the form of a Hankel function of the first kind

A:

00

A~(p, 4» =

L

i-nanH~l)(kp)ein

(7.5)

n=-ex>

representing an outward-traveling wave that vanishes at infinity. Similarly, the transmitted magnetic vector potential can be written as 00

A;(p,4» =

L

i-nbnJn(klP)ein.

(7.6)

n=-oo

To determine the unknown coefficients an and bn , the boundary conditions for the E z and H field continuities must be enforced at p = a. The field components are obtained by substituting A~(p, 4», A:(p, 4», and A;(p, 4» into Table 3.2 in Section 3.1. The continuity of the tangential electric field at p = a

7.1 Dielectric Circular Cylinder

149

•

reqUIres (7.8) . The continuity of the tangential magnetic field at p = a (7.9) yields

(7.10) where the symbol

I

denotes differentiation with respect to the argument

J' (k ) = dJn(kp) n a d(kp)

•

p=a

Solving (7.8) and (7.10) for an and bn yields

~Jn(k1a)J~(ka) lO1

J.ll

lO1

1

J~(k1a)Jn(ka)

J.l

J.l1

J~ (k 1a)H~l) (ka) -

~Jn (k 1 a)H~l)1 (ka) J.l

(7.11)

--(7.12)

where the Bessel function Wronskian

(7.13) is used to simplify bn . The scattered field in the far zone (kp asymptotic expression for H~l) (kp) as

»

1) is written in terms of an

00

E;

kp:oo

Eo

2: n=-ex:>

2 a

~kp

eikp+in(tj>-1f) -i1f / 4 n

.

(7.14)

150

7 Wave Scattering

When a circular cylinder is a perfect electric conductor (PEC, an incident wave cannot penetrate the cylinder. Thus, bn=O and

£)

-+ 00),

(7.15) The electric current density that is induced on the cylindrical surface at p = a • )s

-J=nxH -

p=a p=a

= i [H~(a, ¢) 2' -

-

" Z

+ H;(a, ¢)] in¢

00

t

""'

L....J

J.L1ra n=-oo

'-n e t --;::-;---

H~))(ka)'

(7.16)

7.1.2 TE Scattering

Assume that an incident TE wave impinges on a cylinder that is infinitely long in the z-direction. The incident TE wave has nonzero Hz, E p , and E¢ components. The analysis for TE scattering is somewhat similar to that of the previous TM scattering case. The incident electric vector potential is written as

F;(p, ¢) = e- ikx = e-ikpcos¢ 00

=

L

in en In(kp)e ¢ .

(7.17)

n=-oo

Similarly, the scattered and transmitted electric vector potentials are, respectively 00

F:(p, ¢) =

L

encnH~l)(kp)ein¢

(7.18)

i-ndnJn(klP)ein¢.

(7.19)

n=-oo 00

F;(p, ¢) =

L

n=-(X)

The continuity of the tangential magnetic field at p = a H~(a, ¢)

+ H:(a, ¢)

= H;(a, ¢)

(7.20)

7.1 Dielectric Circular Cylinder

151

yields (7.21) The continuity of the tangential electric field at p = a E~(a, ¢)

+ E;'(a, ¢) =

E~(a, ¢)

(7.22)

yields (7.23) where the symbol I denotes differentiation with respect to the argument. Solving (7.21) and (7.23) for C n and d n yields j), Cn

to

= j),l tOl

In(kla)J~(ka) -

1

J~(kla)HAl)(ka) _

j),l tOl

J~(kla)Jn(ka) (7.24)

j), In(kla)H~l)'(ka)

to

Note that the modal coefficients en and dn are a dual set of an and bn since en and dn can be obtained from (7.11) and (7.12) (an and bn ) by interchanging j), and to. 7.1.3 Electrostatic Fields Let us consider the field distribution due to a circular cylinder when the applied field is static (k = 0 and E~ = Eo). It is convenient to work with the electrostatic potential ¢e using Laplace's equation

\72¢e = 0

(7.26)

which is rewritten as

+

[P¢e 8¢2 = O.

(7.27)

152

7 Wave Scattering

y • • •• • • •• . .... . . ... • • • • • •

• • ••

....

• •

•

•

•

.

• ••• • ••••••

.

p==a

•

••

•

••

•

••

_+..~.!.. -';;':':':.1-X 1':.:. z ::.:.: r:.:.;-:. •

••

•

static E-field .. ....... . • • • • ••

• ••••••

.

Fig. 7.2. Cross section of circular cylindrical dielectric cylinder.

Note that there is no field variation in the z-direction

O¢e

oz

= 0 . Laplace's

equation is a special form of Poisson's equation in Section 1.6, where the electric charge is assumed to be zero (Pe = 0). A solution is available from the separation of variables technique, as was discussed in Section 3.4. Substituting the product form for ¢e (p, ¢)

¢e(P, ¢) = R(p) p(¢)

(7.28)

into Laplace's equation and dividing by R(p) p(¢) yields

2cPR (P) dR(p) R(p) P dp2 +p dp 1

1 cPp(¢) _ 0 + p(¢) d¢2 - .

(7.29)

'-

Thus

cP R(p) p dp2 2

dR(p) _ 2R() - 0 + P dp m p-

cP;~¢)

2 + m p(¢) = 0 .

(7.30)

(7.31)

Equation (7.30) is a type of Euler-Cauchy equation. Substituting the assumption

R(p) = pP

(7.32)

into (7.30) yields (7.33)

7.1 Dielectric Circular Cylinder

153

Therefore, p = ±m and

R(p)

= Apm + Bp-m .

(7.34)

Since p( ¢) is periodic in ¢ with 27r periodicity, m = 1, 2,3, .. " and the solution to (7.31) is

p(¢) = Ccosm¢+Dsinm¢

(7.35)

for every integer m > 1. The total potential for the region p > a consists of the applied and scattered potentials ¢~ (p, ¢) and ¢: (p, ¢), respectively, ¢~(p, ¢)

= -Boy = -Eopsin¢

(7.36)

00

L

¢:(p,¢) =

BmP-msinm¢.

(7.37)

m=l

The transmitted potential for p

~

a is

00

¢~(p, ¢) =

L

Ampm sin m¢ .

(7.38)

m=l

To determine the unknown modal coefficients Am and B m , the boundary conditions must be enforced. The continuity boundary conditions (7.39) (7.40) p=a

p=a

yield (7.41) (7.42) and all other Am = B m = 0 for m

~

2. Hence

Al = _ 2€Eo €l

BI =

€l €l

+€ €

+€

2

a Eo .

Therefore, the electric field can be obtained from

(7.43) (7.44)

154

7 Wave Scattering S

E = -\hP: = t

E = =

(psin
A) 1:1
2

a 2"

I:

+ I: P

(7.45)

-"V
(p sin cos
21: 1:1

+ I:

AE 21: = Yo· 1:1

(7.46)

+ I:

Note that the direction of the transmitted field E direction of the applied field Ea.

t

,

y, coincides with the

7.2 Dielectric Sphere The boundary-value problem of electromagnetic wave scattering from a dielectric sphere was first solved by Mie; thus, its solution is known as Mie scattering. The Mie scattering solution has been widely used in practical applications for electromagnetic wave propagation in a medium embedded with spherical particles. This section derives the Mie scattering solution using magnetic and electric vector potentials. 7.2.1 Electromagnetic Case

Let us consider scattering when the uniform plane wave E' = xEoeikrcos() is incident on a dielectric spMere of radius a, as shown in Fig. 7.3. The wavenumbers for r > a and r < a are k = w# and k1 = WVJ.ll 1:1, respectively. In view of the field analysis in Section 4.3, the vector potentials, A = rA r and F = rFr , can be separated into the components TM and TE to the rAr Fr direction. When A = rA r and F = rFr are chosen, - and satisfy the r r scalar Helmholtz equations

r

-0 .

(7.47)

r It is possible to solve (7.47) using the separation of variables technique, as in Section 4.3. Since the scattering geometry is a sphere, it is expedient to rewrite the incident plane wave using spherical coordinates (r, 0,
is represented in terms of the r components of the vector potentials F', where the f component of the incident wave is •

•

A'

and

7.2 Dielectric Sphere

155

z (J.I., €)

--- -- a .::.:.::. :::::::

x

::::::::: ..

y

.

.s,:.:::::::::.

:/..

____-":·2:-:::::::::ff.:if::::::.:.:: :::::.7:.23::::: . :.:...:.:.. :..:..:..:..:..:..:..:...:.:...:.: ..:.:.:.:.:.:.:.:.:.:.:.:.:.:..::.' • • • • • • • • • • • • •• • •••••••••••• • • • • • • • • • • •• • •••••••••••• ••••••••••• •••••••••• • • • • • • • •• • •••••••• • • • • • •• • •••••

Fig. 1.3. Dielectric sphere with radius a. E~

= cos ¢sin OEoeikr cos 0 = cos ¢ Eo B (eikr cos 0) -ikr BO

.

(7.48)

Since the generating function for the spherical Bessel function (see Appendix B) gives ..--00

n 2~r ~)2n + l)i In+~(kr)Pn(cOSO) n=O

eikrcosO = 00

=

L

•

n i (2n

+ 1) Jn~~r) Pn(cosO)

(7.49)

n=O

E; is written as 00

E; = cos¢ -i~r)2 ~ i (2n n

+ l)Jn (kr) ~ [Pn(cosO)]

00

= cos ¢ -if:r)2

~i

n

(2n

+ l)Jn (kr)

[-P~(cos 0)] (7.50)

where pJ (cos 0) = O. The magnetic vector potential A~ generates E~, while the electric vector potential does not. Substituting

F;

156

7 Wave Scattering

(7.51)

into •

Ei = r

=

t W J.Lf.

i wJ.Lf.

fn(n;l)EocoS¢anjn(kr)P~(coSB) n=l

r

w

(7.52)

yields

+ 1) n(n + 1)

n

i (2n an = -

Similarly, from

.

(7.53)

H;, the incident electric potential F: takes the form of (7.54)

In view of A~ and written as

F:, the scattered potentials A: and F: (r ~ a) can be (7.55)

F: =

00

~Eosin¢ LenH~l)(kr)P~(cos()).

(7.56)

n=l

The transmitted potentials A~ and F; (r

< a) take the form of (7.57)

00

F; =

i-EoSin¢ Lenjn(klr)P~(cos()). 1 n=l

(7.58)

Four unknown modal coefficients-bn , en, d n , and en-are introduced to represent the scattered and transmitted vector potentials. To determine the unknown coefficients-bn , en, dn , and en, the boundary conditions for the field continuities

7.2 Dielectric Sphere

(E~ + Eo)

r=a

t -- E 9

157

(7.59)

r=a

(E~ + E;) Ir=a = E~ r=a

(7.60)

(H~ + He) Ir=a = H~ r=a

(7.61)

t

r=a

-- H

(7.62)

r=a.

must be enforced, thereby resulting in i

2.(A + AB) J.L r r 1 8(A~ + A:) 8r J.Lf. i

2.(F + F f. r r

B

= r=a

r=a

)

1 8(F; + F:) 8r J.Lf.

r=a

r=a

~At J.LI

(7.63)

r

r=a

1 8Atr J.LI f.1 8r

t = ..!.F f.1 r

(7.64) r=a

(7.65) r=a

t 1 8Fr J.Llf.1 8r

•

(7.66)

r=a

Rewriting (7.63) through (7.66) explicitly gives (7.67) (7.68) 1 [~ k anJn(U)

2. J.L

~ (I)

+ en H n

] _

1

~

(u) - k;enJn(U d

[anj~(U) + CnfI~I)I(u)]

=

1 J.LI

enj~(UI)

(7.69) (7.70)

where ka = u and k1a = UI. Here, the prime denotes differentiation with respect to the argument. Solving (7.67) and (7.68) for bn and dn gives b _ n -

-JJ.Lf.lj~(U)jn(UI)

+ JJ.Llf.jn(u)j~(ud JJ.Lf.lfI~I)'(U)jn(ud _ JJ.Llf.fI~I)(u)j~(ud an J.LI

d

n

=

%[fI~I)I(U)jn(U) -

(7.71)

fI~I)(u)j~(u)]

JJ.Lf.lfI~I)'(U)jn(UI) _ JJ.Llf.fI~I)(u)j~(ul) an'

(7.72)

158

7 Wave Scattering

Since

=

1r2U [H~l)'(U)JlI(U) -

=

1r; i [N~(u)JlI(U) -

H~l)(U)J~(U)]

NlI(u)J~(U)]

•

(7.73)

=l

where

1I

= n

1

+ 2'

the modal coefficient dn is simplified to

J.Ll

fl .

-l

n d = VJ.Lf.JI$.l)'(u)Jn(ud _

J.L

VJ.LlfiI$.l)(u)J~(Ul) an·

(7.74)

Similarly, (7.69) and (7.70) produces

-VJ.LflJn(U)J~(Ul) + VJ.Lld~(u)Jn(Ul) en = VJ.LfliI$.l)(U)J~(Ul) _ VJ.LlfiI$.l)'(U)Jn(ud an

-J.Ll en =

%[iI$.l)'(U)Jn(u) -

iI~l)(U)J~(u)]

VJ.LfliI$.l)(U)J~(ud _ VJ.LlfiI$.l)'(U)Jn(ud an -J.Ll

=

(7.75)

'(1).

fl .

-l

J.L

·(1)·

VJ.LflHn (U)J~(Ul) - VJ.LlfHn '(u)Jn(ud

an,

(7.76)

A:

The scattered TM and TE fields can be evaluated by substituting and into Table 4.1 in Section 4.3. Within the low-frequency limit k l a « 1, the Mie scattering solution reduces to a much simpler form of Rayleigh scattering. • (1) i . (kr)2 Using the low-frequency approximation HI (kr) >::J - kr and J l (kr) >::J 3

F:

yields the Rayleigh scattering solution bn = en = 0 (n

> 2) and (7.77)

J.Ll - J.L J.Ll + 2J.L

•

(7.78)

When sunlight impinges upon small particles that are present in the sky, blue light (higher k) tends to scatter more than red light (lower k), thus causing the sky to appear blue. Within the static limit k -? 0

7.2 Dielectric Sphere

159

where Pl(cos(}) = sin(}. Substituting the vector potentials into Table 4.1 in Section 4.3 yields 1':1 - I': E = (f2 sin () cos ¢ - () cos () cos ¢ + ¢ sin ¢)Eo 1':1 + 21':

---8

~

A

(7.80)

31': E = ifsin(}cos¢+(}c~s(}cos¢-¢sin¢2EoI':1 +21':

-t

.

•

•

x

=xEo

(7.81)

•

The direction of the transmitted field the incident field If'.

Il,

X, coincides with the direction of

7.2.2 Electrostatic Case

Let us consider the field distribution due to a dielectric sphere of radius r = a when the applied field E~ = Eo is static (k = 0). It is convenient to work with the electrostatic potential ¢e using Laplace's equation (7.82)

where a solution is available from the separation of variables technique, as discussed in Section 4.3. Substituting (7.83)

into Laplace's equation yields the following three ordinary differential equations (7.84)

1

d

-

sin () dB

+

n(n + 1) -

m

2

. 2 sm ()

1/J2

= 0

(7.85)

(7.86)

where the solutions for every integer n

~

0 and m

> 0 are

160

7 Wave Scattering

(7.87) (7.88)

'l/J3 (¢) = D cos m¢ + E sin m¢ .

(7.89)

The total potential for the region r > a consists of the applied and scattered potentials ¢~(r,(),¢) and ¢:(r,(),¢), respectively, whereas the potential for the region r < a is given by the transmitted potential ¢~(r,(),¢). The continuity boundary conditions require ¢~ (a, (), ¢)

+ ¢: (a, (), ¢)

= ¢~ (a, (), ¢)

(7.90)

•

(7.91)

r=a

Since the applied potential is in the form of ¢~(r,(),¢)

= -Eox = -Eorsin()cos¢

(7.92)

the scattered potential takes the form of 00

¢:(r, (), ¢) =

L

n [Anr + Bnr-(nH)] P~(cos()) cos ¢ .

(7.93)

n=O

In view of the associated Legendre polynomials P~(cos()) = 0

(7.94)

pl (cos ()) =

sin ()

(7.95)

pi (cos ()) =

3 sin () cos ()

(7.96) (7.97)

the scattered potential can be written as

(7.98) Furthermore, since the scattered potential vanishes at infinity (r -t 00), Al = o and

7.3 Step in Parallel-Plate Waveguide

Similarly, the transmitted potential for r 00

¢;(r, (), ¢) =

L

161

:s a can be written as

n Cnr P~ (cos ()) cos ¢

n=O

(7.100) Substituting ¢~(r,(),¢), ¢:(r,(),¢), and ~(r,(),
C1 = -

310 101

+ 210

Eo .

(7.102)

Hence

a

3

(7.103)

2" sin () cos r

310

¢;(r, (), ¢) = -Eo 101

+ 210

•

()

r sm cos

A.. 'I' .

(7.104)

Note that the electric fields from (7.105) -t

t

E = -\le(r,(),¢)

(7.106)

are shown to be identical with the expressions (7.80) and (7.81).

7.3 Step in Parallel-Plate Waveguide Consider a step between two perfectly conducting parallel-plate waveguides, as shown in Fig. 7.4. Assume that a TE wave E~(x, z) impinges on a step between perfectly conducting parallel-plate waveguides. When an incident wave E~(x, z) impinges on the step at z = 0, scattering takes place due to a step discontinuity. No field variation is assumed in the y-direction

a=0 ay

and

nonzero field components are E y , Hz, and Hz. The electric field Ey(x, z) should satisfy the Helmholtz equation

162

7 Wave Scattering

x

PEe ( II )

(1)

a

b

Ei'\J\ry

Fig. 1.4. Step between two perfect conducting parallel-plate waveguides. (7.107)

where k (= w.,jjIf.) is the wavenumber. This section uses a mode-matching technique to analyze scattering. The field in region (1) (z S 0) consists of the incident and reflected components (7.108) 00

E;(x,z) =

L

Rmsin(amx)exp(-ikz1mz)

(7.109)

m=l

where k z1m = Jk2 - a~ and am = m1r. Note that ~(x,z) and E;(x,z) are a chosen to satisfy the boundary conditions (7.110)

Similarly, the transmitted field Et(x, z) in region (II) (z

> 0) is set as

00

E~(x, z) =

L

T m sin(bmx) exp(ik z2m z)

(7.111)

m=l

where k z2m = Jk 2 - b~ and bm = ~1r. To determine the reflection and transmission coefficients Rm and T m , respectively, the boundary conditions for the tangential field continuities must be enforced. The tangential electric field continuity at z = 0 is

7.3 Step in Parallel-Plate Waveguide

163

00

:L Tm sin(bmx) m=l 00

sin(anx)

+ :L Rm sin(amx)

for 0

<x
m=l

(7.112)

o

for a

< x < b.

The orthogonality property of the sinusoidal function sin(bmx) can be used for further simplification. Multiplying (7.112) by sin(bpx) and integrating with respect to x for 0 < x < b yields 2 T p =b

00

Gnp

+ :L RmGmp

(7.113)

m=l

where (7.114)

The tangential magnetic field is given by

_ H x (x, z ) -

i 8Ey(x,z) WJ-L

8

Z

.

(7.115)

Similarly, multiplying the tangential magnetic field continuity at z = 0 and

O<x
k z1n sin(anx) -

:L kz1mRm sin(amx) m=l

00

=

:L k z2m T m sin(bmx)

(7.116)

m=l by sin(apx) and integrating with respect to x for 0

< x < a yields (7.117)

where c5np is the Kronecker delta. Equations (7.113) and (7.117) constitute a set of simultaneous systems for the unknown coefficients T m and R m . Once T m and Rm are determined, it is possible to evaluate the reflected and transmitted powers. The reflected magnetic field is given by

164

7 Wave Scattering

H

r (

) _ :z: X,Z -

i 8E;(x, z) ~ uZ

WJ.L 00

= 1 WJ.L

2: kz1mRm sin(amx) exp( -ikz1mz) .

(7.118)

m=l

A Poynting vector gives the reflected time-average power a

1

Pr = 2"Re

E;(x, z)H;*(x, z) dx o M}

a '"" = 4 L..J WJ.L m=l

IRmI

2

(7.119)

k z1m

and the transmitted time-average power (7.120)

where M 1 and M 2 denote the numbers of propagation modes as M 1 and M 2

ka

<-

< kb, respectively. 1r

7.4 Slit in Conducting Plane Electromagnetic scattering from a slit in a conducting plane finds many applications in antenna and electromagnetic interference problems. One of the basic radiating elements for aperture antennas is a slit in a conducting plane. An analysis of scattering from a slit in a thick conducting plane is available in [8) based on the mode matching and Fourier transform. Meanwhile, this section investigates TE wave scattering from a slit in a conducting plane using the mode matching and Fourier transform.

7.4.1 Field Analysis Consider a TE wave E~(x, y) impinging on a thick slit in a perfectly conducting plane. The problem geometry is shown in Fig. 7.5 where the wavenumbers in regions (I), (II), and (III) are ko (= wJJ.Lof.o) , kl (= WVJ.Llf.l), and k 2 (= WVJ.L2f.2) , respectively. No field variation is assumed in the z-direction

8 8z = 0 and field components are Ez(x,y), H:z:(x,y), and Hy(x,y). The field in region (I) (x < 0) consists of the incident, reflected, and scattered components, although the inclusion of the reflected component in region (I) is arbitrary. The incident and reflected fields are assumed to be

7.4 Slit in Conducting Plane

incident wave

165

scattered wave (I) ..--.y

--

x=d

(III)

x

transmitted wave

Fig. 7.5. Thick slit in conducting plane.

E~(x, y) = exp(ik",x

+ ikyY)

(7.121) (7.122)

where k", = ko cos ()i and k y = ko sin ()i' It is possible to represent the scattered field in region (I) in terms of a continuous spectrum. Let the scattered field E; (x, y) be represented in terms of the Fourier transform 00

E; (x, ()e-i(y d(

E; (x, y) =

(7.123)

-00

where its inverse is -

E; (x, () =

00

1

E;(x, y)ei(Y dy .

271"

(7.124)

-00

Substituting

E; (x, y)

into the Helmholtz equation (7.125)

•

gIVes
2

+ ~o

E;(x, () = 0

(7.126)

Jk3 -

where KO = (2. The solution that satisfies the radiation condition E;( -00, () --+ 0 is given by (7.127)

166

7 Wave Scattering

-

Substituting E;(x, () into (7.123) finally gives the representation for E;(x, y) 00

E;(x,y) =

E;(()e-i(y-iItOZ d(.

(7.128)

-00

The fields, E~(x,y) and E;(x,y), in regions (II) and (III) respectively, are 00

E~(x, y) =

L

(b m cos~mx

+ Cm sin~mx) sin am(y + a)

(7.129)

m=l 00

t()

E z x, y =

E-zt (()e iIt2 (z-d)e- i(y d.,r

(7.130)

-00

where am =

~:' ~m = Jk~ - a:n, and 1',2 = Jk~

-

(2.

Note that

E~(x,y) is

constructed so as to satisfy the boundary condition E~(x, ±a) = O. Equations (7.128), (7.129), and (7.130) represent the scattered fields in regions (I), (II), and (III), respectively, in terms of the four unknown modal coefficients-namely, E;(O, bm , Cm , and E;(O. The boundary conditions must be used to obtain simultaneous equations for the modal coefficients. Let us enforce the four boundary conditions for the continuities of the tangential fields-Ez(O,y), Hy(O,y), Ez(d,y), and Hy(d,y). Applying the inverse Fourier transform to the tangential electric field continuity at x = 0 E~(O,y)

E:(O, y) =

for Iyl
o

otherwise

•

glVes

b

00

E;(() =

L

2

m~;a Fm((a)

(7.132)

m=l

[e iU ( -l)m _ e- iU ]

where Fm(u) = -=---------,,--=7r u2 -

(n; r

.

The continuity of the tangential magnetic field at x - 0 and • reqmres H~(O,y) + H;(O,y)

+ H;(O,y)

= H~(O,y) .

Iyl <

a

(7.133)

Since H ( ) _ y x, y -

(7.133) can be rewritten as

i 8E z (x,y) l:l

WJ.L

uX

(7.134)

7.4 Slit in Conducting Plane

2k", ik Y - e 1/ J.lo •

_

t

1 +J.Lo

167

-00

00

L~mcmsinam(y+a).

(7.135)

J.ll m=l

The orthogonality property ofthe sinusoidal function sin am(y+a) can be used for further simplification. Multiplying (7.135) by sin an(y + a) and integrating with respect to y for Iyl < a yields (7.136)

where 00

I(k o) =

(7.137) -00

An evaluation of I(ko) is given in Section 7.4.3 using residue calculus. Let us enforce the boundary conditions at x = d. Applying the inverse Fourier transform to the tangential electric field continuity at x = d E~(d,y)

E;(d,y) =

o

for Iyl
otherwise

•

glVes (7.139)

IYI

Similarly, the continuity of the tangential magnetic field at x = d and
results in

(7.141)

where 00

(7.142) -00

168

7 Wave Scattering

Equations (7.136) and (7.141) constitute a set of simultaneous equations for the discrete modal coefficients bm and em' Once (7.136) and (7.141) are numerically solved, it is possible to evaluate the scattered and transmitted fields, E;(x,y) and E~(x,y), respectively. While the scattering solution requires the evaluation of (7.137) and (7.142), an approximate simple solution is available within the high-frequency limit [8]. When leoa -t 00,

[(leo) -t 27T~m <5mn aam

(7.143) (7.144)

where Pm = Jk'tJ - a~ and qm = Jk~ - a~. Substituting (7.143) and (7.144) into (7.136) and (7.141) yields an approximate high-frequency solution for bn and en in closed form .J.Lo

Pn

-2

C <,n

J.Ll

en

-

•

(7.145)

o 7.4.2 Far Field and Transmission Coefficient It is possible to obtain the far-zone transmitted field (k 2 P » 1) based on the saddle-point method [9]. Substituting the relations

x-d=pcos() y = psin ()

(7.146) (7.147)

( = k2 sin 1/J "'2 =

Vki -

(7.148) (2 = k2 cos 1/J

(7.149)

into the transmitted field (7.130) gives

E~

=

E~ (k 2 sin 1/J) exp [ik 2 P~os( 1/J + ()~] k 2 cos 1/J d1/J .

.

(7.150)

f(1/J) The condition k 2 P» 1,

f' (1/Jo)

= 0 yields the saddle point 1/Jo - -(). Hence, when

•

7.4 Slit in Conducting Plane

E;

::::;-E;(-~sinO) ~2: exp [i (k 2P- ~)] k 2 cosO.

169

(7.151)

Substituting (7.139) into (7.151) finally yields the far-zone transmitted field

k2 cosO 27rp 00

L

(b m cos ~md + em sin ~md)ama2Fm (-k2a sin 0) .

(7.152)

m=]

Pi =

a

-a a

-

~ Re [E(O,y) 1

2Re

X

H\O,y)*] . xdy

[.

.

]

-E~(O,y)H;(O,y)*

dy

-a

-

(7.153)

•

WJ.-Lo

Similarly, the power transmitted through a slit at x = d is

Pt =

a -a

~ Re [-E:(d,y)H;(d,y)*]

dy.

(7.154)

Since 00

E:(d, y) =

L

(b m cos ~md + em sin ~md) sin am(y

+ a)

(7.155)

m=]

x=d •

=

z

00

L

Will ,.. m=]

. sin am(y

~m (-b m sin ~md + Cm cos ~md)

+ a)

(7.156)

the transmitted power Pt becomes 00

Pt =

1m - 2: ] L ~~ [- Ibm J.L

2

1

COS

~md(sin ~md)*

m=l

+bmc~1 COS ~mdl2 - b~eml sin ~mdl2

2

+lcml

sin ~md(cos ~md)*]

•

(7.157)

170

7 Wave Scattering

The transmission coefficient

can be readily obtained from (7.153)

and (7.157). 7.4.3 Residue Calculus for I(k o )

Consider the integral

•

00

2

I(k o) = -00

a F «(a)F (-(a)Ko d(. m n ,

(7.158)

;

....

f«() When m When m

+n +n

is odd, f«() is an odd function of (. Therefore, I(ko) = is even, I(ko) can be rewritten as

o.

00

I(k o) =

(7.159)

Ko d( .

- ----"

-00

~""----.......

f«()

It is expedient to evaluate I(ko) in the complex (-plane using residue calculus and contour integration. The integrand contains branch points at ( = ±ko and simple poles at ( = ±am when m = n, as shown in Fig. 7.6. Assume R -+ 00. For analytic convenience, the medium wavenumber ko = kr + ik i is assumed to have an infinitesimally small positive imaginary part (k i > 0). Considering the contour which consists of r 1 , 2 , ... , 7 in the upper half-plane, the residue theorem gives

r,

r

f«()d(=

r

r

(7.160)

f«()d(=O.

The paths n and r3 denote infinitesimal semicircles and the path r6 denotes an infinitesimal circle. The path r 4 denotes an infinite semicircle. Note

f«() d( = I(k o)

(7.161)

(7.162) •

(7.163)

r.

f«()d( = 0

(7.164)

7.4 Slit in Conducting Plane

171

1m (~) branch cut

14

17

G

12

-k 0

16

-a m 11 am k o

Re

(~)

branch cut

Fig. 7.6. Complex (-plane.

rs

ra

f(() d( = K j

(7.165)

f(() d( = 0

(7.166)

(7.167)

Hence (7.168)

where K 1 and K 2 are integrations along the branch cut r 5 and n associated with the branch point ko. In order to evaluate K j and K 2 , consider KO = Jk~ - (2. Introducing the polar forms

(- leo

(7.169)

ei02 r2

(7.170)

= rje

( + ko = leads to

i1h

172

7 Wave Scattering

(7.171)

The corresponding phase diagram is illustrated in Fig. 7.7. The phase variations along the branch cut path AI, A 2 , A 3 , and A 4 are in Table 7.1, showing that the difference in 1/J at Al and A 4 (or at A 2 and A 3 ) is 1r. Therefore (7.172) (7.173)

1m (~) (infinity)A 4

Al (infinity)

k·-O I

Re k·-O I

Fig. 1.7. Phase diagram.

(~)

7.4 Slit in Conducting Plane

173

Table 7.1. Phase variations along branch cut

Al

A2 A3 A4

(h

(12

1r

1r

-2

1r

-2

31r

-2

0

31r

1r

2

1r

31r

0

-2 --

1/J

4 1r

--4

2

0

7.4.4 Thin Slit Within High-Frequency Limit The previous subsections performed a rigorous analysis of the scattering from a slit in a conducting plane. While a rigorous scattering analysis provides an exact solution, it requires further tedious computations to obtain the scattering characteristics. This subsection presents an alternative approximate solution to the problem based on Love's equivalence principle and image method. Consider an incident wave E~(x, y) that impinges on an infinitesimally thin slit (d = 0), as shown in Fig. 7.8 (a). Assume that the medium wavenumber is k( = wJJii.) throughout. Let the transmitted (x > 0) wave be denoted by -t

E (x,y). Based on Love's equivalence principle, the original problem (a) can be transformed into an equivalent problem (b). In problem (b), the surface im-t -t pressed currents J. = fi x H and M. = -fi x E are placed at x = 0, thereby -t -t radiating E and H for x > 0 and the null field for x < O. When a perfect electric conductor is placed at x = 0 beneath the impressed current in problem (c), the impressed electric current ]. becomes short-circuited on the PEC. The impressed current M. on the infinite PEC plane in problem (c) is equivalent to the impressed current -t

(7.174)

2M. = -2x x E (O,y) t

t

in a medium of infinite extent in problem (d), as far as the field E and H for x > 0 is concer ed. t To determine - 2x x E (0, y)], the tangential field continuities at x = 0 must be used. The continuity of the tangential electric field for E~(O, y)

+ E~(O, y) + E;(O, y)

= E;(O, y) .

When the operating frequency is high enough that ka

•

»

Iyl < a requires (7.175)

1, the condition

174

7 Wave Scattering

e· I

:zz:-- y

2a

e x

original problem (a)

-

null field

~===="-+Y 1+--+--+1 "-

n

2a

•

x

equivalent problem (b)

PEe /

2a x

equivalent problem (c)

?JJs ':------+y

1+--+--+1

2a

x

equivalent problem (d)

Fig. 1.8. Thin slit in conducting plane.

7.4 Slit in Conducting Plane

IE~(O,y)+E;(O,y)l«

applies for -a

E~(O,y)1

175

(7.176)

< y < a. Therefore E;(O,y) ~ E~(O,y)

(7.177)

•

2M s (0,y') = ii'2E(0,y') .

(7.178)

The electric vector potential F(p) due to 2 M s (0, y') in a medium of infinite extent is given in terms of the free-space Green's function in Section 1.5.2. The electric vector potential F(p) is thus written as a

F(p) =

E

.

2M s (0, yl)~H61) (kip - p' I) dy'

-a

(7.179) •

where ~ H6 ) (kip - P'I) is the two-dimensional free-space Green's function. Substituting the asymptotic form 1

'Ir~P exp [i (kp -

1

H6 )(klp - p'l) -+ for the far zone p » p' and kp

F(p)=yiE

»

ky' sinO -

~)]

(7.180)

1 into F(p) gives

2 eXP[i(kp_'lr)]Sin[k(sinO-sinOi)a]. 'lrkp 4 k(sin 0 - sin Oi)

(7.181)

.

Fy

Since y = f; sin () + 0cos (), the electric vector potential F(p) using cylindrical coordinates (p, (), z) is

F(p) = f;Fy sin() + OFy cosO. In the far zone kp -

1

,;p

»

1, the radiation field is of the

(7.182) 1

,;p order. Taking the

order terms from -t

1

-

E =--\7xF

(7.183)

E

gives the far-zone field t

EZ

~

1.

--zkFo E

= cosO

2 exp 'lrkp

[i (kp- ~)] sin[k~sinO -.sinO

i)

4

(sm 0 - sm Oi)

a] . (7.184)

176

7 Wave Scattering

7.5 Circular Aperture: Electrostatic Case The subject of wave penetration into a circular aperture in a thick conducting plane is important in electromagnetic interference problems. When the wave frequency is relatively low, low-frequency fields can approximately be described in terms of the static potentials. This section investigates electrostatic potential penetration into a circular aperture in a thick perfectly conducting plane, as shown in Fig. 7.9. A similar discussion is available in [10]. Regions (I) (z ? 0), (II) (-d ~ z ~ 0 and p ~ a), and (III) (z < -d) represent the upper half-space, circular aperture, and lower half-space, respectively. In region (I) (z > 0) the primary potential ~(p, z) is applied to a circular aperture with radius a and depth d in a thick perfectly conducting plane at a zero potential. The electrostatic potential
Our aim is to solve Laplace's equation subject to the boundary conditions. Assume that the primary potential is ~(p,z)

= z.

(7.186)

•

pnmary potential

region (I) scattered potential

z=O z=-d

region (III) transmitted potential

Fig. 1.9. Thick circular aperture in conducting plane.

7.5 Circular Aperture: Electrostatic Case

177

Since the primary potential is assumed to be azimuthally symmetrical, Laplace's equation is independent of
o o
(7.187) Let us first consider the scattered potential in region (I). Since the scattering domain in region (I) is open (p --+ 00) in the radial direction p, it is expedient to use the Hankel transform representation 00

-f(() =

o 00

f(p) =

o

-

f(p)Jo((p)pdp

(7.188)

-f(()Jo((p)( d( .

(7.189)

Substituting the scattered potential based on the Hankel transform 00

(7.190) o

into (7.187) gives (7.191) where the solution (7.192)

is chosen to satisfy the condition

J:(()Jo((p)e-(Z( d( .

(7.193)

o

The total electrostatic potential in region (I) is the sum of the primary and scattered components, namely, tM (p, z) +

L

[bnsinhkn(z

+ d) + cncoshkn(z + d)]Jo(knP)

(7.194)

n=l

where the boundary condition
(7.195)

178

7 Wave Scattering

yields the characteristic equation Jo(kna) = 0, which determines the eigenvalues k n . The representation of the transmitted potential in region (III) is somewhat similar to that in region (I). The transmitted electrostatic potential in region (III) is 00

¢;(p, z) =

¢;(()Jo((p)e((z+d)( d( .

(7.196)

o

To represent the potentials, the four unknown modal coefficients-¢~((), ¢~((), bn , and en-are introduced, thus requiring four boundary conditions at z = 0 and -d. The boundary condition for the continuity of the potential at z = 0 requires ~(p, 0)

+ ¢:(p, 0)

¢~(p, 0)

-

o

for p < a (7.197) otherwise .

For arbitrary Bessel functions Zp and B p [11, page 634],

f3pZp(ap)B p- 1 (f3p) - apZp_l (ap)B p(f3p) a 2 - 13 2

.

(7.198)

00

Applying the Hankel transform o

(·)Jo((p)pdp to (7.197) yields

00

¢:(()

=

L

(b n sinh knd + en cosh knd)

n=l •

Let us substitute ¢~(() into the other boundary condition at

(7.199) z = 0 for

p
(7.201)

o

o

for n =j:. p .

Problems for Chapter 7

179

Multiplying (7.200) by Jo(kpp)p and integrating with respect to P from 0 to a yields 00

f- - L p

(b n sinh knd + Cn cosh k nd)a knkpJ1 (kna )Inp 2

n=l

(7.202) where 00

I np =

(7.203)

o

Similarly, the boundary conditions at z = -d ¢~(p, -d) ¢~ (p,

-d) =

for p


o

otherwise (7.205)

finally give b

00

L

enknJ1 (kna)Inp =

fJ (kpa) .

(7.206)

1

n=l

To evaluate the scattered and transmitted potentials, the simultaneous equations (7.202) and (7.206) must be solved for the unknown modal coefficients bn and en' The behavior of the aperture field is often discussed in terms of the electric polarizability x(z). The electric polarizability is represented in terms of bn and Cn as a

x(z) = 21l' = 21l'a

o

¢~(p, z)p dp

t

[b n sinh kn(z + d) + Cn cosh kn(z + d)] J

n=l

1

~na)

. (7.207)

n

Problems for Chapter 7 1. Derive (7.120). 2. Show that I(k o) is given by (7.143) within the high-frequency limit.

180

7 Wave Scattering

3. Consider electromagnetic scattering from the thick PEC slit, as shown in Fig. 7.5. Assume that the TM uniform plane wave H~(x, y) = exp(ikzx + ikyY) is incident on the slit. Using the Fourier transform and mode matching, obtain a set of simultaneous equations for the modal coefficients. Using the high-frequency approximation and equivalence principle, evaluate approximately the transmitted (x > 0) field for the thin slit (d = 0). 4. Derive (7.206).

8

Green's Functions: Fundamentals

8.1 Delta Function and Sturm-Liouville Equation The delta function is an important concept in dealing with point charges and currents. The Sturm-Liouville equation also plays an important role in the study of wave propagation and radiation. For instance, Bessel's equation is of the Sturm-Liouville equation type. This section introduces the fundamentals of the delta function and Sturm-Liouville equation, which are useful in radiation and scattering formulation. 8.1.1 Delta Function

The delta function c5(r - r') is defined as

J(r - r') = 0

v

J(r - r') dv = 1

when r

=I r'

when V contains r'.

(8.1) (8.2)

The delta function has the sifting property

v

f(r)J(r - r') dv = fer') .

(8.3)

Let us represent the delta function using rectangular, cylindrical, and spherical coordinates, respectively. •

Consider the delta function c5(r - r') where the source is located at the position x = x'. Then

J(r - r') = J(x - x') .

(8.4)

182

8 Green's Functions: Fundamentals Similarly, in two- and three-dimensional rectangular coordinates

•

8(1' - 1") = 8(x - x')8(y - y')

(8.5)

8(r - r') = 8(x - x')8(y - y')8(z - z') .

(8.6)

Consider the delta function 8(1'-r') where the source is located at the position (p',
8(T - r') = A I 8(p - p')8(
(8.7)

Substituting (8.7) into (8.2) yields 211"

00

A 1 8(p - p')8(
o

(8.8)

0

1

Therefore, Al = - and p

1

8(r - r') = -8(p - p')8(¢ - ¢') . p

(8.9)

Consider the delta source 8(1' - r') = A 2 8(p - p')8(¢ -
8(r - r') = -8(p - P')8(¢ - ¢')8(z - z') . p

•

(8.10)

Consider the delta source 8(r - r') = A 3 8(r - r')8((} - (}')8(
11"

00

2

o

A 3 8(r - r')8(0 - O')8(
0

Therefore, A 3 =

1 2 .

r sm

()

8(1' - 1") =

and

r

2

~sm i(r -

r')8((} - O')8(¢ - ¢') .

(8.12)

It is possible to express the delta function in the spectral domain based on transform pairs. Consider the Fourier transform pair

-

00

f(x)e-i(x dx

f(C) =

(8.13)

-00

1 f(x) = 211'

00

(8.14) -00

8.1 Delta Function and Sturm-Liouville Equation

183

-

Substituting f(() into f(x) gives 1

00

00

f(x')e-i(x'dx' ei(x d( .

f(x) = 27r -00

(8.15)

-00

Since 00

J(x - x')f(x') dx'

f(x) =

(8.16)

-00

the delta function can be written as 1

00

J(x - x') = 27r

ei((x-x') d( .

(8.17)

-00

8.1.2 Sturm-Liouville Equation

Consider the differential equation

d

d

dx p(x) dx on the interval a

+ q(x) + Anr(X)

'l/Jn(x) = 0

(8.18)

< x < b subject to the boundary conditions

where the functions p(x), q(x), and r(x) as well as the parameters a1, a2, {31, and {32 are real. Equation (8.18) is called a Sturm-Liouville equation with the eigenfunction 'l/Jn(x) and eigenvalue An· •

When p(x) = 1, q(x) = 0, An = k;, and r(x) = 1, (8.18) becomes the differential equation of simple harmonic motion (8.21)

•

m2

When p(x) = x, q(x) = ,An = k;, and r(x) = x, (8.18) becomes x Bessel's equation •

184

8 Green's Functions: Fundamentals 2

•

m When p(x) = 1- x , q(x) = -1 2' An -x

= n(n + 1), and r(x) = 1, (8.18)

2

becomes the associated Legendre equation

'l/Jn(x) (8.23) Let us show that the eigenfunctions 'l/Jn(x) are orthogonal to each other. Multiplying (8.18) by 'l/J;"'(x) and integrating from a to b gives

a

b .1,. ( ) 'I'm X

d () d'l/Jn (x) dX P x dX

dx

b

+

b

q(x)'l/J;"(X)'l/Jn(x) dx + An

r(x)'l/J;"(x)'l/Jn(x) dx = O.

(8.24)

a

a

Integration by parts gives

a

b. I,. ( ) 'I'm X

d

d

X

P

()d'l/Jn(x) X

d

dx

X

(8.25) Substituting (8.25) into (8.24) gives Ql

Q2

•

p(a) f31 'l/J;" (a)'l/Jn (a) - p(b) f32 'l/Jm(b)'l/Jn(b) _

b

d'l/J;;(X) p(x) d'l/J;(x) dx +

a

X

b q(x)'l/J;" (x)'l/Jn (x)

dx

a

X

b

+A n

(8.26)

r(x)'l/J;"(x)'l/Jn(x) dx = 0 . a

Interchanging the indices m and n and taking the complex conjugate of (8.26) • gIVes Ql

Q2

p(a) f31 'l/Jn(a)'l/J;"(a) - p(b) f32 'l/Jn(b)'l/J;"(b) _

b

d'l/J; (x) p(x) d'l/J;; (x) dx +

a

X

X

b q(x)'l/Jn (x

W;" (x) dx

a

b

+A;"

r(x)'l/Jn(x)'l/J;"(x) dx = 0 . a

(8.27)

8.1 Delta Function and Sturm-Liouville Equation

185

Subtracting (8.27) from (8.26) finally yields b

(An - A;")

r(x)1/In(x)1/I;"(x)dx = O.

(8.28)

a

Assume that r(x) does not change sign on the interval (a,b). For m :/: n (An :/: A;") b

r(x)1/In(x)1/I;"(x) dx = 0 .

(8.29)

a

For m = n b

r(x)1/In(x)1/I~(x) dx

:/: 0

(8.30)

= O.

(8.31)

= 1.

(8.32)

a

An The eigenfunctions

A~

1/In (x) are normalized as b

r(x)1/In(x)1/I~(x) dx a

The condition An = A~ shows that the eigenvalue An is real. The property (8.29) and (8.32) is referred to as the orthogonality of the eigenfunctions 1/In(x). The eigenfunctions 1/In(x) constitute a complete set. This completeness implies that an arbitrary function f(x) can be represented to any desired accuracy as a linear combination of 1/In(x) (8.33) n

Multiplying f(x) by r(x)1/I;"(x) and integrating from a to b yields the expansion coefficient b

r(x)f(x)1/I~(x) dx .

fn =

(8.34)

a

Consider the Sturm-Liouville equation with a delta source 6(x - x')

~

p(x) d~

on the interval a S (x, x')

+ q(x) + Ar(x)

g(x; x') = -6(x - x')

(8.35)

< b, where the boundary conditions are given by

186

8 Green's Functions: Fundamentals

The solution g(x; x'), which is the response to the delta source, is usually known as a Green's function. Let us represent the Green's function and delta function by a series of eigenfunctions

g(x; x') =

L

gn(X ' )1Pn(x)

(8.38)

n

8(x - x') " * ( ') r(x ) = LJ 1Pn x 1Pn(x). ' n

(8.39)

Substituting (8.38) and (8.39) into (8.35) finally gives

g(x; x') = -

L n

1P~~X~~n(X)

(8.40)

n

which is a Green's function representation based on eigenfunction expansions.

8.2 One-Dimensional Green's Function This section discusses one-dimensional Green's functions in free space, half space, and closed space, respectively. The one-dimensional Green's functions are often used in scattering and radiation problem formulations. Let us first consider a free-space case. 8.2.1 Free Space

Consider the Sturm-Liouville equation in one-dimensional free space (-00 < 2 x < 00) with the parameters p(x) = r(x) = 1, q(x) = 0, and >. = k . The Sturm-Liouville equation is given by

::2 + k

2

g(x; x') = -8(x - x')

(8.41)

subject to the radiation condition g(±oo; x') = o. The response g(x; x') at x is due to the delta source at x', as shown in Fig. 8.1. In the following, two different approaches are introduced to derive g(x; x').

s (x-x') --I

g(x;x')

I

•x

x = x' .

Fig. 8.1. Free-space Green's function.

8.2 One-Dimensional Green's Function

187

Approach 1

It is possible to represent g(Xj x') in terms of eigenfunctions. Since the domain is open (-00 < x < 00), g(x; x') is represented in the inverse Fourier transform 00

1 g(XjX ' ) = 21T

g«(j x')ei(z d( .

(8.42)

-00

Substituting (8.42) and the identity 00

J(X _ x') = 1

(8.43)

21T

-00

into (8.41) yields 1 ei«z-z')

00

g(X; x') =

-00

21T «(2 _ k 2 ) d( . , .,

(8.44)

.

f«() The physical condition requires that the delta source response g(Xj x') be an outgoing wave in the form of e±ikz that vanishes as x ---+ ±oo. This condition can be met if the medium is assumed to be slightly lossy, where the medium wavenumber k = k r + iki has an infinitesimally small positive imaginary part (k i > 0). For x-x' > 0, let us use the residue theorem in the complex (-plane, as shown in Fig. 8.2 (a). Performing contour integration along the path r 1 and r 2 yields

n Since

n

f «() d(

f«() d( +

---+ 0 and

r}

n

f«() d( = 21Ti Res f«()

(8.45)

(=k

f( () d( = g(Xj x') as R ---+ 00, •

Z eik(z-z') g( x', x') = 2k .

When x - x' so as to make

n

(8.46)

< 0, the semicircle in the lower half-plane is chosen for

n

f«() d( ---+ 0 as R ---+ 00, as shown in Fig. 8.2 (b). Contour

integration gives

•

I

'k(

Z

g(x; x) = 2k e-'

z-z

')

.

(8.47)

The one-dimensional free-space Green's function is, therefore, given by •

Z g(x; x') = 2k eiklz-z'l

.

(8.48)

188

8 Green's Functions: Fundamentals

k x

x

Re

(~)

-k

contour (a)

1m (~)

k x x

-k

R

contour (b)

Fig. 8.2. Complex (-plane.

Approach 2

The solution g(x; Xl) can also be obtained by another approach as follows: Consider the homogeneous differential equation

cP 2 dx 2 + k

I) 9 X; X = 0 (

(8.49)

when X f:. Xl. The Green's function that satisfies the radiation condition at X = ±oo is

8.2 One-Dimensional Green's Function ikz Ae

for x

189

> x'

g(x; x') =

(8.50) ilcx Be-

for x

< x'

.

The unknown constants A and B can be determined by the boundary conx' ± .11, where .11 denotes an infinitesimally small interval. ditions at x Integrating (8.41) from x = x' - .11 to x = x' + .11 yields

=

dg(x; x') dx

x'+A

x'+A 2

+

x'-A

k g(x; x') dx = -1 .

(8.51 )

x'-A

· dg(x; x') . di t' d ( ') . . I h b d SIDce dx IS scon IDUOUS an 9 x; x IS contlDuous at x ,t e oun ary conditions

dg(x; x') dx

x'+A

= -1

(8.52)

= 0

(8.53)

x'-A x'+A

g(x; x')

x'-A

•

glVe

ik

ikz (Ae '

+

ikz Be')

= -1

(8.54) (8.55)

Solving (8.54) and (8.55) for A and B finally gives •

~ eilc(x-x')

2k

g(x; x') =

for x> x'

•

~ e -ilc(x-x')

for x

< x'

2k •

- 2~k eilclx-x'i .

(8.56)

8.2.2 Half Space

Let us investigate the Sturm-Liouville equation in one-dimensional half space 2 with the parameters p(x) = r(x) = 1, q(x) = 0, and A = k . These types of problems are often encountered in scattering and radiation problems dealing with perfectly conducting boundaries. Consider the half-space Sturm-Liouville equation

:::2 + k

2

g(x; x') = -8(x - x')

(8.57)

for x > 0 subject to the boundary conditions g(O; x') = 0 and g(oo; x') = 0, as shown in Fig. 8.3. Let us introduce four different approaches to obtain g(x; x').

190

8 Green's Functions: Fundamentals

o(X-X')

g (x; X ')

X =X '

X=o

Fig. 8.3. Half-space Green's function.

Approach 1

Due to reflection from the boundary at x = 0, g(X; x') takes a standing wave sin kx for 0 < x < x' and a traveling wave e ikx for x > x'. Assume

Ae

ikx

for x

> x'

g(x; x') =

(8.58)

B sin kx

for 0 -< x

< x'

which satisfies the boundary conditions g( 00; x') = 0 and g(O; x') = O. Integrating (8.57) gives

x'+.a

= -1

(8.59)

x'-.a (8.60) Solving (8.59) and (8.60) for A and B yields

A = sin kx ' k

(8.61 )

eikx ' B = --:-k .

(8.62)

Hence, the half-space Green's function is

. k I sm x

k

ikx

e

for

x> x' (8.63)

e

ikx'

k

sin kx

for 0

< x < x'

.

8.2 One-Dimensional Green's Function

191

Approach 2

It is also possible to obtain g(x; x') using eigenfunction expansions. Since the domain is open for x > 0 and g(O; x') = 0, it is convenient to represent g(x; x') in terms of the Fourier sine transform. Substituting 00

2 g(x; x') = 11'

J(x - x') =

(8.64)

sin (x sin (x' d( .

(8.65)

0 00

~ 11'

g((; x') sin (xd(

0

into (8.57) yields

_(c. ') = sin (x' 9 ,x (2 _ k2 .

(8.66)

Therefore

g(x; x') = 2

o

11'

1

00

-:;

-

sin (x sin (x' d( (2 - k 2

00

-00

sin (x sin (x' (2 _ k2 d(

1

00

411'

-00

ei(x+x') _ ei(x-x') _ ei( -x+x')

+ ei(

(2 _ k2

-x-x')

d(. (8.67)

In view of the free-space Green's function representations (8.44) and (8.48), g(x; x') is written as •

t

g(x; x') = - 4k

(eiklx+x'i _ eiklx-x'i _ eikl-x+x'l

. k x ' ikx sm k e

+ eikl-x-x/l)

for x> x' (8.68)

ikx' e

k sin kx

for 0

< x < x' .

Approach 3

Let us show an approach based on the image method, as shown in Fig. 8.4, where the image source -J(x + x') is added to satisfy the boundary condition g(O; x') = O. Since both J(x - x') and -J(x + x') are assumed to be in free

192

8 Green's Functions: Fundamentals I I

I

-0 (x+x')

0 (x-x')

g (x; x')

-Ir---I;.....---I------I I

.

x

x =x'

I

I I I

X

=0

Fig. 8.4. Half-space Green's function based on image theorem.

space, the response is the sum of two free-space Green's functions •

z g(x; x') = 2k (eiklx-xll - eiklx+xll)

. k I sm x ikx k e

for x> x'

eikx' ----:-- sin kx k

(8.69)

for 0

< x < x' .

Approach 4

A solution to (8.57) consists ofthe incident [gi(X; x')] and scattered [g8 (x; x')] Green's functions. The incident Green's function is a response to a delta source in free space and the scattered one results from reflection from the boundary at x = O. The incident term is the one-dimensional free-space Green's function and the scattered one is given in terms of the reflected waves as •

'kl 'I g'(x; x') = 2k e' x-x .

Z

(8.70) (8.71)

Since

(8.72) the coefficient A is given by (8.73)

8.2 One-Dimensional Green's Function

193

Therefore

g(x; x') = gi(X; x')

+ g8(X; x')

. k I sm x ikz k e

for x> x' (8.74)

eikz ' k sin kx

for 0

< x < x' .

8.2.3 Closed Space

This subsection investigates the Sturm-Liouville equation in one-dimensional closed space. Consider cf2

dx 2

+k

2

g(x; x') = -<5(x - x')

(8.75)

on the interval 0 < x < a subject to the boundary conditions g(O; x') = g(a; x') = 0, as shown in Fig. 8.5. The following are three different approaches to derive the solution g(x; x'). Approach 1

Due to the boundaries at x = 0 and a, the solution g(x; x') can be represented in terms of standing waves, sin kx and cos kx. The choice

Asink(x - a)

for x' < x < a

g(x; x') =

(8.76)

B sin kx

8 (x-x')

for 0 < x < x'

g (x; x')

x = x'

x= 0

x=a

Fig. 8.5. Closed-space Green's function. •

194

8 Green's Functions: Fundamentals

automatically satisfies the boundary conditions g(O; x') = g(a; x') = O. Consider the boundary condition near x = x'. Integrating (8.75) near x = x' • gIves

dg(x;x') dx

z'+L1

=-1

(8.77)

z' - L1

z' + L1

g(x; x')

z' - L1

= 0.

(8.78)

Solving (8.77) and (8.78) for A and B yields

A = _ sinkx' ksinka B=-

(8.79)

sin k(x' - a)

(8.80)

•

ksinka

Hence sin kx' sin k(x - a)

for x'

ksinka

g(x; x') =

-

<x
sin k(x' - a) sin kx

for 0< x

ksinka

< x' .

Approach 2

It is also possible to obtain g(x; x') using eigenfunction expansions, where the eigenfunction 'l/Jm (x) is given by 2 . m1r

(8.82)

-sm x. a a

'l/Jm(x) = Substituting

00

g(x; x') =

L

Am(x')'l/Jm(x)

(8.83)

'l/Jm (x )'l/Jm (x')

(8.84)

'l/Jm(x')

(8.85)

m=l 00

c5(x - x') =

L

m=l

into (8.75) yields

A (x') = _ m

•

k2 -

(m1r)2 a

.

8.2 One-Dimensional Green's Function

195

Hence

f: VJm (Xl):,;(xl

g(x; Xl) = -

m=l

k2 - ( a ) 2 -

00

=-2:

. (m1l' ) . (m1l' I) sm a x sm a x k2 _

m=l

a =

cos

00

m1l'

C:1I')

(Xl - x) - cos

a

'""

11'2 ~ m=l

2

2

k

m -

m1l'

(Xl

a

a

2

+ x) •

(8.86)

11'

Using the series formula for 0 < u <

211'

[11, page 40]

~ cosmu _ 1 _ ~ cos[a(1I' - u)] ~ m2 - a 2 2a 2 2 asina1l'

(8.87)

m=l

in (8.86) yields (8.81). Approach 3

A solution to (8.75) consists of the incident [gi(x; Xl)] and scattered [g8(X; Xl)] Green's functions, where the incident term is a response to a delta source in free space and the scattered one results from multiple reflection from the boundaries at x = 0 and a. Hence •

z gi(x; Xl) = 2k eiklz-z'l

(8.88) (8.89)

Since

[gi(X;X

I )

l

+ g8(x;x )Jz=o,a =

0

(8.90)

A and B are given by A = _ i sin k(a - Xl) 2ksin(ka)

(8.91)

i sin(kxl)e ika B = - ----=-:-'-:----;-;,----,-- • 2ksin(ka)

(8.92)

196

8 Green's Functions: Fundamentals

Therefore

•

z 2k

eiklx-x'i _

. k( ') ikx sm a - x e sin(ka)

sin kx' sin k(x - a) ksinka

-

for x'

sin( kx')e ik ( a-x)

sin(ka)

< x -< a (8.93)

sin k(x' - a) sin kx k sin ka

for 0< x -

< x' .

8.3 Two-Dimensional Green's Function The Helmholtz equation using cylindrical coordinates (p, ¢» with a delta source (line source at p = pi) is

(\7 + k ) g(p; p') 2

2

= -J(p - p') .

(8.94)

Its explicit expression is

1 8 p 8p

8 P8p

1 8

2

+ f}i 8¢>2 + k

2

g(p, ¢>; p', ¢>')

= _ J(p - p') J(¢> _ ¢>') .

(8.95)

p

y

-p - -p'

-

~---J'--

-

p

..

X

Fig. 8.6. Two-dimensional free-space Green's function.

8.3 Two-Dimensional Green's Function

197

The problem geometry is shown in Fig. 8.6. Let us find a response at p due to a delta source at (l using three different approaches. 8.3.1 Approach 1

Based on the separation of variables technique, the Green's function and delta function are given by 00

9(P, ¢; p', ¢') =

L m=-oo 00

o(¢-¢') =

L

m=-oo

im 9m (p; p', ¢')e ¢

(8.96)

1 e im (¢-¢') . 21T

(8.97)

Substituting 9(P, ¢; p', ¢') and o( ¢ - ¢') into (8.95) gives 1 d

d

pdp

dp

-- P

2

_ m 2 p

k2

+

i:(

')

-im¢'

-1,') __ u p - p e () 9m P,P,'I' - 2 - ' 8.98 p 1T (.,

Let us use an approach based on the cylindrical wave propagation characteristics, as shown in Section 3.4. When p:j:. p', (8.98) reduces to Bessel's equation where its solutions 9m(P; p', ¢') are the Bessel and Hankel functions for p

., ,

< p' (8.99)

9m(P; P , ¢ ) = BmHg) (kp)

for p> p' .

Multiplying (8.98) by p and integrating from p = p' - Ll to p = p'

d9m (p; p' , ¢') p dp e

. ..., -,mY"

-----,--

21T

p' +L1

p'+L1

+ p'-L1

p'-L1

2

m -;;"'2 P

+

k2

+ Ll yields

9m(P; p', ¢')p dp

.

(8.100)

Since d9m (Pi p', ¢') and 9m (p; p', ¢') are discontinuous and continuous at p = dp p', respectively, the boundary conditions are written as .

...'

-1m",

p' - AmkJ:n(kp')

+ BmkHgl' (kp')

= - e 21T

(8.101) (8.102)

, ( ') dJm(kp) where Jm kp = d(kp)

. Solving (8.101) and (8.102) for Am yields p=p'

198

8 Green's Functions: Fundamentals

Since

-J:n (kp')H~) (kp')

+ J m (kp')H~)' (kp')

. 2 = ~ 'Trkp'

(8.104)

the coefficients Am and B m as well as the Green's function gm (Pi p', ¢') are given by •

im Am = ~ e- 4>' H~) (kp')

(8.105)

•

im B m = :"e- 4>' Jm(kp')

(8.106)

4

for p

•

gm (p; P" ,¢) = 4"~

< p' (8.107)

for p> p' . Therefore

g(p, ¢; p', ¢') 00

L

=

im gm (p; p', ¢')e 4>

m=-oo 00

L •

-

~

H~)(kp')Jm(kp)eim(4>-4>')

for p

< p'

m=-oo

-

4

(8.108) 00

L

Jm(kp')H~) (kp)e im (4>-4>')

for p

> p' .

m=-oo

By the summation theorem for the Hankel function [11, page 979], the twodimensional free-space Green's function is written as •

g(p,¢;p',¢') = ~H~l) (kip-pI) .

(8.109)

Let us rewrite g(p, ¢; p', ¢') in terms of the superposition of plane waves in rectangular coordinates. Rewriting (8.94) in rectangular coordinates (x, y) • glVes

8.3 Two-Dimensional Green's Function

8

2

8

2

8x 2 + 8 2 + k y

2

l g(x, y; Xl, yl) = -<5(x - x )<5(y - yl) .

199

(8.110)

Substituting 00

1

g( ()ei(z-:z:') d(

g(x, y; Xl, yl) = 211"

(8.111)

-00

into (8.110) results in (8.112) The solution g(() is the one-dimensional free-space Green's function •

g(() = ~ eil
Jk

2 -

(8.113)

(2. Substituting g(() into (8.111) finally gives 00

i H(l) (kl- _ _II) = 1 4 0 P P 211"

-00

•

~ ei(z-:Z:')+il
2K:

d(

(8.114)

which represents the cylindrical wave H~l) (kip - P'I) as the superposition of a continuous spectrum of plane waves. 8.3.2 Approach 2

It is also possible to derive (8.107) based on the residue calculus and Hankel transform approach. Substituting the Hankel transform representations 00

gm (p; p',
(8.115)

o 00

o(p - pI) _

(8.116)

o

P

into (8.98) gives 1 d

gm(() - pdp

2

d pdp

-

m 2 P

+

k2

(8.117) In view of Bessel's equation 1 d --

pdp

P

d

dp

-

m

2

p2

(2 +

(8.118)

200

8 Green's Functions: Fundamentals

(8.117) can be rewritten as (8.119) .

Substituting 9m(() of (8.119) into (8.115) gives • oJ. , -ImY'

I) e I 9m ( p; P , ¢ = 21T

(8.120)

Let us compute the integral (8.120) analytically in the complex (-plane based on a contour integral technique. Consider two different cases, pi > P and / < p. When / > p, substituting (8.121) into (8.120) yields

Since HgJ(-u) = -(-l)mH~J(u), 9m (p;

/, ¢/) =

00

-00

e- im ¢' Jm((p)HgJ((pl) 21T 2((2 - P) ( d( .

(8.123)

.

f(() Note that the function f(() contains simple poles at ( = ±k. The Hankel function Hg J((/) has the branch point at ( = 0 and the branch cut is chosen along the negative real (-axis, as shown in Fig. 8.7. Performing contour integration in the (-plane gives

f(() d( F1

+

f(() d(

+

f(() d( = 21Ti Res f(() . F3

r2

The asymptotic expressions for (-+

00

(on

(=k

r 2)

give

2 cos ((p _ ~ _ m _ _1T) 1T(p 4 2 2 exp 1T(pl

(8.124)

[i ((/ _ 1T4 _

m1T)] . 2

(8.125)

(8.126)

8.3 Two-Dimensional Green's Function

201

k X

branch cut

X

-k

Fig. 8.7. Complex (-plane.

It can be shown that Jm((p)Hg)((p') -t 0 on n for p' > p, implying that the integral on path r 2 vanishes. It is possible to show that the integral on the infinitesimal semicircular path r 3 also vanishes. Let us consider

fee) d( = 9m(P; p', ¢') -im¢' J (( )H(l)(( ') Resf(() = e m P m P (((-k) 2 (=k 21T 2( (2 - k ) -im¢' = e 81T Jm(kp)Hg)(kP').

(8.127)

(=k

(8.128)

Substituting (8.127) and (8.128) into (8.124) yields the Green's function 9m(Pi p', ¢') as •

9m(PiP',¢') = ~e-im¢'Hg)(kP')Jm(kp). Similarly, when p' as

(8.129)

< p, it is possible to obtain the Green's function 9m(P; p', ¢') •

9m(Pi p', ¢') = ~e-im¢' Hg)(kp)Jm(kp') . Note that (8.129) and (8.130) are identical with (8.107).

(8.130)

202

8 Green's Functions: Fundamentals

8.3.3 Approach 3

The Helmholtz equation (8.94) is rewritten in rectangular coordinates as

2 2 8 8 2 8x 2 + 8 2 + k g(X, y; X', y') = -8(x - x' )8(y - y') . y

(8.131)

Substituting the two-dimensional Fourier transform representations I

')

g(X, Yi x , Y

=

1

00

00

(271")2

-00

-00

g((, 1])e i «z-z')+i"l(V-V') d( d1] 00

00

-00

-00

(8.132)

8(x - x ' )8(y - y') = (2:)2 into (8.131) yields (8.134) Hence 00 I

g(x, y; x , y

')

1

(8.135)

= (271")2 -00

Based on the relations

R = ((x - x') + f}(y -

K

- y')

. = (( + f}1]

(8.136) (8.137)

-

K

R

Fig. 8.8. Polar coordinates.

8.3 Two-Dimensional Green's Function

203

as shown in Fig. 8.8, the Green's function g(x, y; x', y') is rewritten as exp (iK. 17.) g(x, y; x', y') = (271')2 s K2 - k2 ds 1

00

1

(271')2

211" eiK R casC

K2 _ k2

o

0

"'-"'0) K d¢ dK .

(8.138)

Substituting 00

L

=

eiKRcasC-"'o)

in In(KR)einC-o)

(8.139)

n=-oo

into g(x, y; x', y') gives I

( gx,y;x,y

')

00

1 = (271')2

0

1

o

00

.L

inJn(KR)einC-o)Kd¢dK

n=-(X)

_

1

00

271'

0

KJo(KR) dK K2 - k 2

•

(8.140)

Since

~ [H~l)(KR) + H~2)(KR)]

Jo(KR) =

(8.141)

the Green's function becomes I

9 x,y;x ,y (

')

1 = 471'

00

-00

KHCl)(KR) (K - ~)(K + k) dK .

(8.142)

It is possible to evaluate g(x, y; x', y') using the residue theorem. Let us integrate along the path in the complex K-plane, as shown in Fig. 8.9. The integrals along the infinite semicircular path F2 and along the infinitesimal semicircular path F3 both vanish. The result is

1 g(X,y;xl,y') = 4 271'i Res 71'

K=k

KH~l)(KR) (K - k)(K + k)

.KH~l)(KR) = 471' 271'z (K + k) 1

K--+k

•

= ~H~l) (kip - pi!) 4

which is identical with (8.109).

(8.143)

204

8 Green's Functions: Fundamentals 1m (K)

k X

Re (K)

X

branch cut

-k

Fig. 8.9. Complex K-plane.

8.4 Three-Dimensional Green's Function Consider a problem of a three-dimensional Green's function, where the problem geometry is shown in Fig. 8.10. The Helmholtz equation using spherical coordinates (r, B,
-r-r'-r

x

Fig. 8.10. Three-dimensional free-space Green's function.

8.4 Three-Dimensional Green's Function

205

This section shows three different approaches to derive the solution g(r; r').

8.4.1 Approach 1

-II th 8g(f]) 8g(f]) 0 d ~(-I) 8(f]) . h ' al Le t f] = r-r, en 8B = 8 = , an u r-r = 47rf]2 usmg sp enc 1

coordinates (f],B, ¢). Substituting g(f]) = u(f]) into (8.144) gives

f]

cf2

d

f]

+k

2

2

8(f]) U(f]) = - 4 .

(8.145)

7rf]

A solution for f] :j:. 0 is chosen to be the outgoing wave u(f]) =

g({}) = A

e

ikg Ae

and

ikg

. The boundary condition for g(f]) at f] = 0 must be enforced to

f] determine A. Integrating

( \72

+ k 2 ) g(f])

= _ 8(f]) 47r f]2

(8.146)

over an infinitesimal spherical volume centered at f] = 0 results in

v

\72 g ({}) dv

+

v

2 k g(f]) dv = -1 .

(8.147)

Since \72 g(f]) is more singular than g(f]) at f] = 0, it is possible to assume

g(f]) dv = 0

(8.148)

\72 g (f])dv = -1.

(8.149)

v v

Applying the divergence theorem to (8.149) gives

\72 g(f]) dv = V

"

.

",

s

\7g(f]) . as

\7 . \7 g(f])

=A g-+O

=

-47rA .

(8.150)

Hence, A = 1 , and the three-dimensional free-space Green's function g(r; r ' ) •

47r

IS

eik1r-r' 1

g(r;r') = 47rr-r r ='1'

(8.151)

It is useful to express g(r; r ' ) in terms of the superposition of plane waves and cylindrical waves as follows:

206

•

8 Green's Functions: Fundamentals

Let us rewrite (8.144) using rectangular coordinates as

[J2

82

82

8x 2 + 8 y 2 + 8z 2 +k

2

g(x,y,z;x',y',z')

= -8(x - x')8(y - y')8(z - z') .

(8.152)

Substituting

g(x, y, z; x', y', z') <Xl

<Xl

g( (,1] )ei«(Z-Z')+i'l(Y-Y') d( d1] -00

(8.153)

-00

into (8.152) yields

(::2

2 +k -

(2 _1]2)g((,1]) = -8(z -

Z')

(8.154)

where the solution is the one-dimensional free-space Green's function •

g((, 1]) = with II: = relation

jk 2 -

~

211:

. Iz-z 'I e't<

(8.155)

(2 - 1]2. Substituting g( (,1]) into (8.153) finally gives the

eiklr-r'l .

41l'Ir - r'l (8.156)

•

which represents the superposition of a continuous spectrum of plane waves. In view of Fig. 8.8, let R = ((x - x')

+ i}(y -

y')

(8.157)

A

K = (( + i}1] .

(8.158)

Hence eik1r-r' 1

41l'1r-r'l

-

<Xl

1

(21l')2

(8.159) 0

8.4 Three-Dimensional Green's Function

where R = J(x - X')2 Since

+ (y -

y')2 and", =

Vk 2 -

207

K2.

00

L

eiKRcos(¢-¢o) =

inJn(KR)ein(¢-¢o)

(8.160)

n=-CX)

integration with respect to ¢J gives

eik1r-r' 1 47l'Ir -

r'l

1 -

00'

-

27l'

t

2",

0

eil
(8.161)

which represents the three-dimensional free-space Green's function in terms of the cylindrical wave Jo (KR). 8.4.2 Approach 2

The Helmholtz equation (8.144) is explicitly rewritten using spherical coordinates as 1 a

r2

ar

2ag r -=a=-r

a

1

. ()ag

sm

+ r 2 sin () a()

o(r - r')o(() - ()')o( ¢J - ¢J') r 2 sin ()

ag 2

1

2_

a() + r 2SID ' 2 () a¢J2 + k g -

(8.162)

•

It is expedient to represent the Green's function and delta function using spherical harmonics P:;'(cos())e im¢, as shown in Appendix E. The Green's

function g and the delta function

o(() - ()')o(¢J - ¢J')

00

g(r,(),¢J;r',()',¢J') =

. () sm

n

L L

are written as

im gnm(r;r',()',¢J')P:;'(cos())e ¢ (8.163)

n=Om=-n and

o(() - ()')o(¢J - ¢J') _ sin () 00

n

L L

im AnmP::' (cos ()')P::' (cos ())e (¢-¢')

(8.164)

= 2n + 1 (n - m)! . nm 47l' (n + m)!

(8.165)

n=Om=-n where

A .

.

Substltutmg g(r, (), ¢J; r', ()', ¢J') and

o(() - ()')o(¢J - ¢J') . . ()

sm

.

mto (8.162) gives

208

8 Green's Functions: Fundamentals

cPgnm

2 dg nm

+r

-d-::-r-'-;;:2~

+

dr

gnm =

. ..' ~(r - r') m -A nm pm n (cos ()')e-· ", 2 r

Introducing gnm

•

(8.166)

= krg nm gives cPgnm dr 2

+

k2

-

n(n + 1) gnm = 2 r A

.m..' ~(r - r') -AnmP::'(cosB')e-· ", k--'------'- . r

(8.167)

The solution to (8.167) for r f:. r' is the spherical Bessel functions and spherical Hankel functions. Let us assume A

BJn(kr) gnm =

for r

< r' (8.168)

CfI~I)(kr)

for r

> r'

where In(kr) and fI~I)(kr) are the spherical Bessel function of the first kind and the spherical Hankel function of the first kind, respectively. Multiplying (8.167) by r and integrating from r' - L\ to r' + L\ gives

dg nm r dr

r ' +.6

= -AnmP::'(cosB')e-imtfJ' k

(8.169)

= 0

(8.170)

r ' -.6

r ' +.6

gnm

r ' -.6

which are then rewritten as

r' [cfI~I)' (kr') - BJ~(kr')] = -AnmP::'(Cos()')e-imtfJ' CfI~I)(kr') - BJn(kr') = 0 (I) dHn (kr)

(8.171) (8.172)

A

A

(1)'

,

_

where H n (kr) -

d(kr)

. Solving (8.171) and (8.172) for B and C r=r'

•

gIVes

B =

imtfJ' -A nm iI(1)(kr')pm(cos()')en n

r' [In(kr')fI~I)' (kr') - J~(kr')fI~I)(krl)] •

imtfJ fI(1) (kr')pm(cos B')e- '

(8.173)

C = ';AnmJn(kr')P::,(cosB')e-imtfJ' . r

(8.174)

= ~A I nm

r

n

n

•

8.4 Three Dimensional Green's Function

209

Substituting B and C into (8.168) gives •

g(r,B,¢;r',B',¢') = k:r'

00

n

l:: l::

n=Qm=-n AnmP;:' (cos B)P;:' (cos B')e im (t/>-t/>')

.

(8.175)

The Bessel function summation theorem in [11, page 980] shows that (8.175) reduces to (8.151). 8.4.3 Approach 3 The Helmholtz equation (8.144) is rewritten using rectangular coordinates as 2

2

2

8 8 8 2 8x 2 + 8 y2 + 8z 2 + k

9

(

"')

x, y, z; x , y , z =

-o(x - x')o(y - y')o(z - z') .

(8.176)

Substituting the three-dimensional Fourier transform representations

g(x,y,z;x',y',z') = 00

-IX)

00

00

-<Xl

-00

g((, TJ, ~)ei(("'-"")+i1J(Y-Y')+ie(z-z') d( dry ~

(8.177)

o(x - x')o(y - y')o(z - z') = 00

00

-00

-00

00

-(X)

(8.178) into (8.176) results in (8.179)

210

8 Green's Functions: Fundamentals

g(X,y,z;xl,yl,Z') = 00

1

00

00

e i«(x-x')+i1)(y-y')+i{ (%_%') 1"2

(21r )3

-00

-00

-00

..

t2

2

+ TJ + '" -

d( dTJ df.

k2

. (8.180)

Let

R = ((x - x') _

A

+ fJ(y -

y')

+ t(Z -

Zl)

(8.181)

A

K = (( + fJTJ

+ ~~ .

(8.182)

Then

exp(iK.R) 9 ( x,y,z;x ,y ,z = (21r)3 V K2 _ k 2 dv. I

1

')

I

Let us introduce rectangular coordinates ((', TJ ' , ~') whose with the vector R, as shown in Fig. 8.11. Therefore

() 8.183

e-axis coincides

(' = K sin Bcos

(8.184)

TJ' = K sin Bsin

(8.185)

(=KcosB.

(8.186)

Using spherical coordinates (K, B,
(8.187)

K·R=KRcosB 2

dv = K sin Bd dB dK .

(8.188)

Hence I I ') = gx,y,z;x,y,z (

00

1

(21r)3 1

0

o (eiKR _ e- iKR ) K

00

(K - k)(K

o

+ k)

dK.

(8.189)

Since 00

o

o -00

_e-iKRK

(K - k)(K + k) dK = eiKRK

(K - k)(K + k) dK

(8.190)

8.4 Three-Dimensional Green's Function

""

-R

""

" ""-

-

211

"

/

/

TI'

/ 1

I / 1/

/

Fig. 8.11. Spherical coordinates.

the Green's function g(x, y, Zj x', y', ZI) is 1

1

9 ( x,y,ZjX,y,z

')

1

= (27r)2iR

00 -00

eiKRK

(K - k)(K + k) dK .

(8.191)

Let us utilize the residue theorem and contour integration to evaluate (8.191). Integrating along the path in the complex K-plane of Fig. 8.12 gives Im(K)

k X

x ----l'--------JL.-+ Re (K)

- - - L - - -...

-k

r1

Fig. 8.12. Complex K-plane.

212

8 Green's Functions: Fundamentals I

I

9 x, y, z; X , Y ,z (

')

1

.

= (2 )2.R27rZ Res

7r

-

1

27rR

(K - k)(K + k)

K=k

Z

eiKRK (K

+ k)

K--+k

e ikR

47rR eik1r-r' I

(8.192)

47rlr - r ' I . Problems for Chapter 8 1. Consider the half-space Sturm-Liouville equation

d~2 + k £or

X

2

g(x; x') = -15(x - x')

dg(x; x') · h b d d· . >0 su b Ject to t e oun ary con ItlOns dx

(8.193)

0 d = an

z=o g(oo; x') = O. The problem geometry is shown in Fig. 8.13. Derive g(x; x').

o(x-x') X

g (x;

X I)

=x '

x=O

Fig. 8.13. Half-space Green's function.

2. Consider the Sturm-Liouville problem described by (8.75). Derive the solution (8.81) using the image method. 3. Show that a series (8.86) yields (8.81).

9

Green's Functions: Applications

9.1 Currents in Free Space The free-space Green's function is used to evaluate electromagnetic wave scattering and radiation from currents in free space. This section investigates two examples using the free-space Green's function, namely, radiation from a sheet current and radiation from a shell current. 9.1.1 Radiation from Sheet Current

Consider electromagnetic radiation from a constant electric current sheet of infinite extent J = iJc5(x). The geometry of the radiating structure is shown in Fig. 9.1, where the medium wavenumber is k = w.jiii. The field has no vari-

z

7i-----y x

Fig. 9.1. Current sheet in free space.

214

9 Green's FUnctions: Applications

8

8

8y = 8z = 0 . The corresponding Helmholtz

ation with respect to y and z

equation for if = zAz(x) with a delta source is

C~2 + k )A z (x) = -J1,J~(x) . 2

(9.1)

The solution is

A (x) = ifJ-Jeiklzl 2k

z

.

(9.2)

Therefore

j

= -z2

(9.3)

It is also possible to derive E based on the boundary condition at x = 0, which requires

(9.4) Letting

(9.5) (9.6) results in

J

H o ="2'

(9.7)

For the region x > 0, the current sheet produces a uniform plane wave that propagates in the +x-direction with the wavenumber k as

J 'k Hy(x) = _eo Z 2

(9.8) (9.9)

Similarly for the region x

<0 J 'k Hy(x) = --e-' Z 2

Ez(x) =

J

-"2

Equations (9.9) and (9.11) are identical with (9.3).

(9.10) (9.11)

9.1 Currents in Free Space

215

9.1.2 Radiation from Shell Current

Consider radiation from an infinitely long current shell source located at p = p' in Fig. 9.2. The shell current is given by J = ZID(P - pI). The derivation 21rp

for the problem solution is available in [12]. The corresponding Helmholtz equation with a delta source using cylindrical coordinates (p, ¢) is

= -J.l ID(p - p') .

(9.12)

21rp

The following presents three different approaches to derive the solution Az(p, ¢; p', ¢'). Approach 1

Let us rewrite the governing equation for the two-dimensional free-space Green's function

1 8 p8p

1 8

2 2

+ pi 8¢2 + k g(p, ¢; p', ¢/) (9.13)

z

-----y

p= P x

I

~

Fig. 9.2. Infinitely long current shell source located at p = p'.

216

9 Green's Functions: Applications

Equation (9.12) is similar to (9.13), which describes the response to an infinitesimally thin line current at p = pi and ¢ = ¢'. The current shell is considered as a collection of infinitesimally thin line currents at p = pi and ¢ =
o

I

g(p, ¢i pi, ¢') ~ d¢' 11"

2".

o

I

00

L

im gm (Pi pi, ¢')e ¢ ~ d¢'

m=-oo

(9.14)

11"

where H~) (kp')Jm (kp )e- im ¢'

•

I

gm PiP, ¢ (

')

< pi

for p

(9.15)

2

= 4

Jm(kpl)H~)(kp)e- im ¢'

for p > pi .

Since 2".

o

e- im ¢' d¢' = 2m5mo

(9.16)

A z (p; pi) is written as ') A z ( p; p

=

ip,I 4

H~l) (kp')JO(kp)

for p

< pi (9.17)

JO(kpl)H~l) (kp)

for p> pi .

Hence

E = zEz = ziwA z (Pi pi) .

(9.18)

Approach 2 Here, an approach based on the cylindrical wave propagation characteristics is used. Since Az(p;p') has no azimuthal variation in ¢

8

8¢ = 0 , (9.12)

becomes 1 d dAz(Pi pi) - d p d

P P

P

+

k 2A ( . ') _ z

Ic5(p - pi) p, P - -J.L 2 . 1I"p

(9.19)

When pi-pi, (9.19) reduces to a Bessel's equation whose solutions are in the form of Jo(kp), No(kp), H~l)(kp), H~2)(kp), .... The Bessel function of

9.1 Currents in Free Space

217

the first kind, Jo(kp), is not singular at p = O. The Hankel function of the first kind, Hd1)(kp), represents the outgoing wave that vanishes at infinity. Therefore, Az(p; pi) is chosen as for p < pi

AJo(kp) Az(p; pi) =

(9.20)

BHd1) (kp)

for p> p'.

Multiplying (9.19) by p and integrating from p = pi - ..1 to P = pi

dA ( . ') p zP,P dp

pi +L1

pi +L1

+

p' -L1

pi -L1

+ ..1 yields

I 2 k A z (p;pl)pdp=-J-L.

(9.21)

21l"

· (p; pi) an d A z (p; P') are discant'muous and ' . I SIDce dAz dp contmuous, respective y, at p = pi,

pdAz(p; pi) dp

p'+L1 pi _ L1

[-AkJ~(kpl) + BkHd1)1 (kpl)]

= pi

p'+L1

Az(p;p')

= _1.L1

(9.22)

21l"

.

=-AJo(kpl)+BHdl)(kp') =0.

(9.23)

p'-L1

Solving (9.22) and (9.23) for A and B, and substituting A and B into Az(p; pi) • gIVes

1 Hd )(kp')Jo(kp)

')

A z (p; P = iJ-LI 4

1 JO(kp')Hd )(kp)

for p

< pi (9.24)

for p> pi .

Approach 3

In this approach, A z (p; pi) is derived starting from a transform approach. It is expedient to express Az(p; pi) based on the Hankel transform representation 00

(9.25)

o 00

Az «(; pi)

Az(p; p')Jn«(p)pdp .

=

(9.26)

o

Since A z (p; pi) has no azimuthal variation in ¢, the parameter n = 0 is chosen in view of Bessel's equation. Hence 00

Az(p; p') = J(p - pi) _ P

-

(9.27)

o 00

o

Jo«(p) Jo «(p')( d( .

(9.28)

218

9 Green's Functions: Applications

Substituting Az(Pi p') and 6(p - p') into (9.19) and noting P 1 d

dJo((p) P dp p dp

(9.29)

= _(2 Jo((p)

yields (9.30)

Let us first evaluate Az(Pi pi) when p < p'. When Az((j p') and

JO((p') =

~ [H~I)((P') + H~2)((pl)]

(9.31)

are substituted into (9.27), Az(PiP') is written as

( z A PiP

')

J-LI = 27r +J-LI 27r

Since H~I)(-u)

= -H~2)(u) A z (Pi p') =

00

Jo((p)H~I)((pl)(d( 2((2 - P)

0 00

Jo((p)H~2)((pl)(d(

0

and Jo(-u)

2((2 - k 2 )

•

(9.32)

= Jo(u),

~~

(9.33)

Contour integration is performed in the complex (-plane, as shown in Fig. 8.7. Note that the integral evaluation along the path r 2 vanishes since P < p'. The result is

') iJ-LI (1)( ') ( ) ( A z Pi P = 4 H o kp J o kp . Similarly, when p

(9.34)

> pi, A z (Pi P') = iJ-LI 4 H o(1)( kp ) J o( kp') .

(9.35)

Note that the solution Az(PiP') is identical with (9.17).

9.2 Line Current in Rectangular Waveguide This section investigates three cases dealing with radiation from line currents in a rectangular waveguide. The first is radiation in a parallel-plate waveguide, the second is radiation in a shorted parallel-plate waveguide, and the third is radiation in a rectangular waveguide.

9.2 Line Current in Rectangular Waveguide

219

9.2.1 Radiation in Parallel-Plate Waveguide

Let us consider electromagnetic radiation from an infinitely long electric line current J(x', y') within a conducting parallel-plate waveguide. The geometry of the radiating problem is shown in Fig. 9.3. Two different approaches are presented to analyze the problem. Approach 1

The Helmholtz equation that governs radiation from J(x', y') - iJJ(x x')J(y - y') is (9.36)

Using rectangular coordinates (x, y), it can be rewritten as 2

8 8x 2

+

2

8 8 2 y

+k

2

Az(x, y; x', y') = -j.JJJ(x - x')J(y - y').

(9.37)

The boundary condition requires

Ez(x,y;x',y')

x=O,a

=iwAz(x,y;x',y')

x=O,a

=0.

(9.38)

Let us choose eigenfunctions

2 . m7r -sm x a a

(9.39)

and expand A z (x, y; x' ,y') in terms of "pm (x) as 00

Az(x,y;x',y') =

L

am(y;x',y')"pm(x) .

m=l

x

PEe

a

®

J (x', y')

z

Fig. 9.3. Line current J in PEC parallel-plate waveguide.

(9.40)

220

9 Green's Functions: Applications

The delta function o(x - x') is also written as 00

o(x - x') =

L

1/Jm(x)1/Jm(X' ) .

(9.41)

m=l

Substituting A z (x, y; x', y') and o(x - x') into (9.37) yields

dcf22 y

2 (m1l")2 k a + am (y;x,y I

-

')

I = -j.LJ1/Jm ( x ') o(y - y) .

(9.42)

The solution to (9.42) is the following one-dimensional free-space Green's function

ij.LJ1/Jm (x') (. I 'I) am y; x , y = 2(m exp ~(m Y - Y I

(

where

em =

k

2

-

')

(9.43)

m1l") 2 (a .

The vector potential and electric field inside the waveguide are (9.44)

E = zEz = ziwAz(x, y; x', y') .

(9.45)

Approach 2

It is also possible to obtain A z (x, y; x', y') using different expressions for the Green's function. The second approach uses the continuous mode representations for Az(x,y;x',y') and o(y - y') based on the Fourier transform along the y-direction. Let 00

(

I

A z x, y; x , y

')

1 = 21l"

Azei(y d(

(9.46)

ei«(y-y') d( .

(9.47)

-00

1 o(y - y') = 21l"

00

-00

Substituting A z (x, y; x', y') and o(y - y') into (9.37) gives

cf2

dx 2

Jk

+K

2

-

"( ,

(9.48)

A z = -j.LJo(x - x')e-' y

2 - (2. Since A vanishes at x = 0 and a, where K = z one-dimensional closed-space Green's function

Az

is given by the

9.2 Line Current in Rectangular Waveguide l sin K,(x - a) .

. K, sm K,a

sm K,X

for 0

221

< x < Xl (9.49)

I

sin K,X . sin K,(x - a) K,sm Ka Let us rewrite A z (x, y; Xl, yl) for x - Xl

>0

for Xl

<X
as I

00 _

Az(x, y; Xl, yl) =

-00 '-

jJ-J sin K,(x - a) sin K,X i«(y-y') d( . e . 21rK, sm K,a ,

.

(9.50)

f«() It would be expedient to use residue calculus and contour integration to evaluate A z (x, y; Xl, yl). The function f(() has singularities when sin K,a = O. The equation sin K,a = 0 contains simple poles at K, = K,m = (m) where K,m = m1r a and (m = K,~ for m = 1,2, .... First consider when y - yl > O. In view of Fig. 9.4, contour integration is performed to obtain

«(

Jp -

00

AZ(X,y;xl,yl)

+

r2

f(() d( = 21ri

L

> 0 and x -

Note that the function f«() is bounded when y - yl

f(() d( ---+ 0 as R ---+

(9.51)

Res f«() m=l (-('" Xl

> O. Since

AAx, y; Xl, yl) is given as follows:

00,

r2 00

Az(x,y;xl,yl) = 21ri

L

Res f(()

m=l

(=(",

00

= 21ri

I

L

Res

m=l

(=(",

jJ-J sin K,(x - a) sin K,X i«(y-y') e 21rK, sin K,a I

sin K,(x - a) sin K,X i«(y-y') d(sin Ka) e K, d( 00

= -ijJ-J

L

m=l

sin K,(x - a) sin K,X

I

----''--:-:---e

(=(",

i«(y-y')

-a(cosKa

(=(m

(9.52) Similarly, it is possible to show that Az(x, y; Xl, yl) for X - Xl (9.52) . Equation (9.52) agrees with (9.44) for y - yl > O.

< 0 reduces

to

222

9 Green's Functions: Applications

1m

(~)

~m

x··· -.:"':."':"'.vx-;x~x:""""':::x~I--~~r"':l~:""--+l..--+Re X

-

X X

(~)

-~ m x • • •

Fig. 9.4. Complex (-plane with simple poles.

For y - y' < 0, A z (x, Yi x', y') is also shown to be

9.2.2 Radiation in Shorted Parallel-Plate Waveguide

Let us consider radiation when a perfect conducting plane is placed at y = 0 in a parallel-plate waveguide. The problem geometry is shown in Fig. 9.5. The Helmholtz equation that governs radiation from J(x', y') = zJo(x-x')o(y-y') •

IS

x

®J(x',y')

Fig. 9.5.

Line current J

a

in shorted parallel-plate waveguide.

9.2 Line Current in Rectangular Waveguide

(\7 + k 2

2

/ A(x, Yi x', y') = - p.J(x , y') .

)

223

(9.54)

The boundary conditions require

AZ(X,y;x/,y')

= Az(X,YiX/,y') :z:=O,a

=0.

(9.55)

y=O

Approach 1

Following the previous procedure, the solution can be written as 00

Az(x,y;x/,y') =

L

am(y;x/,Y')'l/Jm(x)

(9.56)

m=l

. m7r x. Th en, am ( . fy am (0i x I ,Y') = 0 -2 sm Yi " x ,) y must sabs

where 'l/Jm(x) = and for y

a

>0

a

~2 - (r: f 7r

+k

2

am(Yjx',y') = -f..LJ'l/Jm(xl)J(y_y').

(9.57)

The solution to (9.57) is the one-dimensional half-space Green's function that satisfies the radiation condition at y = 00 I

')

am ( Yi x, Y =

k2

where (m =

f..LJ'l/Jm(x ' )

sin (mY I ei(m y

for y'


(m

7r

-

(r: f. Hence, the vector potential and electric field are 00

A z (x, Yi x', y') =

~ ~~ 'l/Jm (x )'l/Jm (x') sin (mY I ei(m Y

for y'


.

(9.59)

.. A z (x, y; x I , y ') . E = zAEz = zzw

(9.60)

Approach 2 It is possible to transform the original problem in Fig. 9.5 into an equivalent one, as shown in Fig. 9.6, based on the image method. The total field for y > 0 is the sum of two contributions, the real current J(x', y') and the image current - ](x' , _y'). The total vector potential due to the real and image currents is

•

224

9 Green's Functions: Applications

(9.61) which reduces to (9.59). x

-J (x "

® J(x',y')

-y') @

a

z

Fig. 9.6. Equivalent problem based on image method.

9.2.3 Radiation in Rectangular Waveguide

Let us consider radiation from a line current in a rectangular waveguide, as shown in Fig. 9.7. The Helmholtz equation {)2

{)2

+ 8y2

()X 2

+k

2

Az(X,YiX',y') = -j.lJc5(x-x')c5(y-y')

(9.62)

is subject to the boundary conditions

Az(X,YiX',y')

z=O,a

= Az(X,YiX',y')

y=O,b

=0.

(9.63)

Following the previous procedure, the solution is given by 00

Az(x,y;x',y') =

L

am(YiX',y')'l/Jm(x)

(9.64)

m=l

where

'l/Jm (x) =

~ sin m7l" x a

a

and am(Yi x', y') satisfies

The solution am(y; x', y') is known as the one-dimensional closed-space Green's function

9.3 Line Current in Circular Waveguide

225

x b

a

Fig. 9.7. Line current] in rectangular waveguide.

am ( Y;

I

X

j.LJ1/Jm (x') ,Y = - I" • I" b ')

.,m Sln.,m

sin (m(Y' - b) sin (mY

for 0

< y < y'

.

(9.66) sin (mY I sin (m(Y - b)

where (m =

k2

-

for y' < y < b

(":1rf. Hence

sin (m(Y' - b) sin (mY

for 0

< y < y'

.

(9.67) sin (mY I sin (m(Y - b)

for y'


The electric field is •. A z (X, y; X I ,y') . E = z"Ez = z~w

(9.68)

9.3 Line Current in Circular Waveguide Consider radiation from an electric line current at (p', ¢/) inside a perfectly conducting circular waveguide. The geometry of the radiating problem is

:z = 0

.

P')J5( ¢ - ¢/)

i~

shown in Fig. 9.8. No field variation is assumed with respect to z The Helmholtz equation with a delta source of] = written as

zJp J5(p -

226

9 Green's Functions: Applications

(\7 where k as

2

2 + k ) A z (p,
(9.69)

(= w#) is the wavenumber. Equation (9.69) is explicitly rewritten

(9.70) This section illustrates two different approaches to obtain A z (p,
The solution for the problem is derived in [13] using eigenfunction expansions. Let us represent o(