EMLAB

1

Solution of Maxwell’s eqs for simple cases

EMLAB

2

,J ,J

Domain : infinite space

Domain : interior of a rectangular cavity

,J,J

Domain and boundary conditionsThe constraints on the behavior of electric and magnetic field near the interface of two media which have different electromagnetic properties. (e.g. PEC, PMC, impedance boundary, …)

Domain : interior of a circular cavity

waveguide

EMLAB

31-D example : Radiation due to Infinite current sheet

1. Using phasor concept in solving Helmholtz equation,

2

,~~~ 22

ckk JAA

xy

z

2. With an infinitely large surface current on xy-plane, variations of A with coordinates x and y become zero. Then the Laplacian is reduced to derivative with re-spect to z.

JAA ~~~

22

2

kz

)(ˆ 0 zJ sxJ

3. If the current sheet is located at z=0, it can be repre-sented by a delta function with an argument z. If the current flow is in the direction of x-axis, the only non-zero component of A is x-component.

)(~~

~

02

2

2

zJAkz

Asx

x

),(~

),(~

),(~

2

22 rJrArA

c

EMLAB

4

4. With a delta source, it is easier to consider first the re-gion of z≠0.

0~

~2

2

2

xx Ak

z

A

kz

kz

e

CA

jkz

x

sin

cos~

5. Four kinds of candidate solutions can satisfy the differ-ential equation only. Of those, exponential functions can be a propagating wave..

6. Solutions propagating in either direction are

)cos(}Re{

)cos(}Re{

)0(

)0(~)(

)(

2

1

kzte

kzte

zeC

zeCA

kztj

kztj

jkz

jkz

x

)0( z)0( z

7. The condition that A should be continuous at z=0 forces C1=C2

EMLAB

5

ssxx JdzzJdzAkdz

z

A00

22

2 ~)(

~~~

)0(2~~

2

2

jkCjkCejkCez

Adz

z

A jkjkxx

zjkx eCA

8. To find the value of C, integrate both sides of the orig-inal Helmholtz equation.

011~~~

0

0

jk

e

jk

eCdzAdzAdzA

jkjk

xxx

jk

JCJjkC s

s 2

~~

2 00

zjksx e

jk

JzA

2

~)(

~ 0

)0or 0(2

~~ˆ

~ 0

zzeJ

z

A zjksx yAB

)0()space free(377,2

~ˆ

~ 0 zjks e

Jj xAE

t

stjzjks dczJdee

kj

Jt

)/||(

2ˆ

2

)(~

2

1ˆ),( 0

||0 xxrA

2

)/||(ˆ),( 0 czJ

t s

xrB

2

)/||(ˆ),( 0 czJ

t s

xrE

EMLAB

6

)(ˆ 0 zJ sxJ

E

E

H

H

Propagating direction

An infinite current sheet generates uniform plane waves whose amplitude are uniform throughout space.

Plane wave 정의

E

H

Electric field : even symmetry

Magnetic field : odd symmetry

ttJ s cos)(0

Propagating direction

EMLAB

7

,JSource

JAA 22 k

Infinitesimally small current element in free space : 3D

JA

A

2

2

22 1

tc

EMLAB

8

Solution of wave equations in free space

JA

A

2

2

22 1

tc

2

2

22 1

tc

•Boundary condition: Infinite free space solution.

1. As the solutions of two vector potentials are identical, scalar potential is consid-ered first.

2. To decrease the number of independent variables (x, y, z, t), Fourier transform rep-resentation is used.

~~~

),(),(~,),(),(~

2

22

c

dtetdtet tjtj rrrr

3. For convenience, a point source at origin is considered.

EMLAB

9

)(22 r gkg

022 gkg

0)()(

0)(1 2

2

22

2

2

rgkr

rggk

r

rg

r

Green function of free space

ck

where

A suitable solution which is propagating outward from the origin is e-jkr.

kr

kr

e

r

Ag

jkr

cos

sinr

Aeg

jkr

1. The solution of the differential equation with the source function substituted by a delta function is called Green g, and is first sought.

2. With a delta source, consider first the region where delta function has zero value. Then, utilize delta function to find the value of integration constant.

3. With a point source in free space, the solution has a spherical symmetry. That is, g is independent of the variables , , and is a function of r only.

EMLAB

10

1)(22 VVVdgdkgd r

)0(01)1(4

4

sin

0

2

2

0 0 0

222

jk

jkr

jkr

V

ekA

drreAk

drddrr

eAkgdk

)0(4)1(4

sin)1(

0

22

0 2

2

AejkA

ddrr

ejkrA

dggd

jk

jk

SVa r

ergA

jkr

4)(,

4

1

.,4

),( rrrr

RR

eg

jkR

Green function of free space

4. To determine the value of A, apply a volume integral operation to both sides of the differential equations. The volume is a sphere with infinitesimally small radius and its center is at the origin.

5. With a source at r’, the solution is translated such that

EMLAB

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~~~ 22 k

dV

)()'(~)(~ rrrr

)(),(),( 22 rrrrrr gkg rrrr

RR

eg

jkR

,4

),(

V

jkR

Vd

R

edg

4

)(~),(

)(~),(

~ rrr

rr

V

V

cRtj

tj

V

jk

tj

dR

cRt

ddeR

dede

det

4

)/,(

),(~2

1

4

1

4

),(~

2

1

),(~

2

1),(

)/(

r

rr

r

6. As the original source function can be represented by an integral of a weighted delta function, the solution to the scalar potential is also an integral of a weighted Green function.

7. Taking the inverse Fourier transform, the time domain solution is obtained as follows.

EMLAB

12

V

dR

cRtt

4

)/,(),(

rr

Retarded potential

(Retarded potential)

V

dR

cRtt

4

)/,(),(

rJrA

0

t

A

The distinct point from a static solution is that a time is retarded by R/c. This newly derived potential is called a retarded potential.

The vector potential also contains a retarded time variable.

Those A and are related to each other by Lorentz condition.

EMLAB

13Solution in time & freq. domain

' 4

)/,(),(

Vd

R

cRtt

r

r

V

dR

cRtt

4

)/,(),(

rJrA

V

jkR

dR

e

4

)(),(

rr

V

jkR

dR

e

4

)(),(

rJrA

V

jkR

V

jkR

V

jkR

deR

jkRd

R

e

dR

e

2

1ˆ4

1)(

4

4

)(),(

1),(

JRrJ

rJrArH

V

jkR

dR

kRjkR

R

kRjkR

R

e

k

)(

)(33)(1

4

1),(

4

2

2

2

2JRRJrE

EMLAB

14

V

jkR

dR

ej

4)ˆ(ˆ)( JRRJrE

V

jkR

dR

ejk

)ˆ(

4)( JRrH

Far field approximation

Electrostatic solution

V R

d2

ˆ

4

11),(

RJArH

Vd

R

24

ˆ)(),(

RrrE

Biot-Savart’s law

0k

Coulomb’s law

EMLAB

15Electric field in a phasor form

V

V

jkR

VV

dtR

dR

ej

dgj

gdj

)]ˆ(ˆ[4

1

4)]ˆ(ˆ[

)(1

'

JRRJ

JRRJ

rJJE

jj

t

)( AA

AE

''

)(1

)]()([1

VVdg

jdRg

jj

rJrJ

A

R

eRgdRg

jkR

V

4)(,)()(

'

rJA

EMLAB

16Radiation pattern of an infinitesimally small current

R

ezIjd

R

ej

jkR

V

jkR

4sinˆ

4)]ˆ(ˆ[

JRRJE

sinˆ)cosˆˆ()ˆ(ˆ

ˆ

00

0

JzJ

Jz

rJrrJ

J

ˆ

ˆˆ

0sincos

cossincossinsin

sincoscoscossin

ˆ

ˆ

ˆ r

z

y

x

z

r

sinz

I

EMLAB

17Example – wire antenna

coscos222 zrzrrrR rr

z

o

r

C

zjkjkr

V

jkR

zdezJr

ej

dR

ej

cos)(4

sin

4)ˆ(ˆ)( JRRJrE

2/l

2/l

02/)2/(sin

2/0)2/(sin)(

0

0

zlzlkI

lzzlkIzJ

2coscos

2cos

2sin 0 klkl

r

eIj

jkr

EMLAB

18

r

z

z

o

r

R coscos222 zrzrzrR

cosz

EMLAB

19

C

zjkjkr

zdezJr

ej

cos)(

4sin)(rE

N

n

dnjkneIAF

1

cosArray factor :

EMLAB

20

Typical array configurations

EMLAB

21Array antenna

EMLAB

22Poynting’s theorem and wave power

)()()( ttt HES

Electromagnetic wave power per unit area(Poynting vector)

}Re{2

1 *HES

Average wave power per unit area

EMLAB

23Derivation of Poynting’s theorem

t

t

DJH

BE

)2(

)1(

t

t

DEJEHE

BHEH

)(2

1

2

1)(

)2()1(

BHDEJEHE

BH

DEJEHEEH

tt

tt

VVS d

dt

ddd HHEEJEaHE

2

1

2

1)(

,J

HEL WWP ,,

VS

Thermal loss

EW

LPHW

Electric energy

Magnetic energy

Electromagnetic wave power per unit area(Poynting vector)

)()()( ttt HES