belajar teknik elektro

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Auxiliary material to the lecture

Theorie der Elektrotechnik (engl.)

(437.253)

by

Univ.-Prof. Dipl.-Ing. Dr.techn. Oszkár Bíró

1



. ,

. ,

. 0,

. ,

I curlt

II curlt

III div

IV div ρ

∂= +

∂

∂= −

∂

=

=

DH J

BE

B

D

1. Maxwell‘s equations

generalized Ampere‘s law,

Faraday‘s law,

magnetic flux density is source free,

Gauss‘ law.

Consequence: law of continuity

.t

div∂

∂−= ρ

J

(implied by the divergence of the first equation

+ the fourth equation:

.),0 ρ ==

∂

∂+ D

DJ div

t

div2



Charges are separated in voltage sources due to non-electric

(e.g. chemical) processes. The impressed field intensity Ee is

a fictitious electric field intensity which would give rise to

the same amount of charge separation:

Ideal voltage source:

Non-ideal voltage source:

,2

1

1

2

∫∫ ⋅−=⋅=P

P

e

P

P

eq d d u rErE

Ee + + + + + + + + +

_ _ _ _ _

ui P1

P2

Field quantities in voltage sources

∫ ⋅=2

1

P

P

d u rE

∫∫ ⋅−=⋅ΓΓ

=2

1

1

2

P

P

P

P

q d d iR rJ

rJ

γ γ

∞→−=⇒= γ ,equu EE

γ

JEE +−=⇒−= eqq iRuu

)( eEEJ += γ

J EΓ: Cross section

γ : conductivity

3



Material relationships:

).(,, eEEJHBED +=== γ µ ε

Energy and power density:

BHDE ⋅+⋅=+=2

1

2

1me www

.

2

γ

J= p

,00

⋅+⋅=

∫∫

B D

d d BHDE

., ∫∫ΩΩ

Ω=Ω= pd Pwd W

Energy and power loss in a volume Ω:

4



1.1 Classification of electromagnetics

1. Static fields

=

∂∂ 0t

, , 0, , 0.curl curl div div div ρ = = = = =H J E 0 B D J

electrostatic field:, , .curl div ρ ε = = =E 0 D D E

magnetostatic field: , 0, .curl div µ = = =H J B B H

static current field: , 0, ( ).ecurl div γ = = = +E 0 J J E E

5



2. Quasi-static field

∂

∂>>

t

DJ

,

,

0,

curl

curlt

div

=

∂= −

∂=

H J

BE

B

).(, eEEJHB +== γ µ

6



,

,

0,

,

curlt

curlt

div

div ρ

∂= +∂

∂= −

∂

=

=

DH J

BE

B

D

3. Electromagnetic waves

).(,, eEEJHBED +=== γ µ ε

7



1.2 Fundamentals of circuit theory

Circuit signals:

,2

1

12 ∫ ⋅= rE d u ,∫Γ

Γ Γ⋅= d i nJ

ideal capacitor:

resistor: , Riu =

,CuQ =

ideal inductor: . Li=Φ

Circuit elements:

,

∫∫∫ ΓΩΩΩ Γ⋅=Ω=Ω= d d divd Q nDD ρ .

∫ΓΓ Γ⋅=Φ d nB

8



ii Γ

t ∂∂D

Current-voltage relationships:

C

Capacitor

iu

Ω

Integrating the law of continuity over Ω:

n

dt

duC i =

0d dQ

div d d d i .t dt dt

ρ ρ

Ω Γ Ω

∂ + Ω = ⋅ Γ + Ω = − + = ∂

∫ ∫ ∫J J n

9



Inductor

C Γ

u

i

B

Γ n

Integrating Faraday’s law over Γ:

u

i L

0.C

d d curl d d d ut dt dt

ΓΓ Γ

∂ Φ + ⋅ Γ = ⋅ + ⋅ Γ = − + = ∂ ∫ ∫ ∫

BE n E r B n

dt

di Lu =

10



Kirchhoff’s current law:

Γn

Ω

i1 i2

i3

i4

0 on , since

cuts no capacitor.

t ∂ = Γ Γ∂D

Integrating the law of continuity over Ω:

∑ =ν

ν 0i

0div d div d d d .t t t

ρ

Ω Ω Γ Γ

∂ ∂ ∂ + Ω = + Ω = + ⋅ Γ = ⋅ Γ = ∂ ∂ ∂ ∫ ∫ ∫ ∫D DJ J J n J n

11



Kirchhoff‘s voltage law:

u L1

uC 1

u R1

uq1

u Lµ

uC µ

u Rµ

uqµ

uqm u Rm

uCm

u Lm

Γ

Integrating Faraday’s law over Γ:

0.C

d curl d d d

t dt ΓΓ Γ

∂ + ⋅ Γ = ⋅ + ⋅ Γ = ∂

∫ ∫ ∫B

E n E r B n

C Γ

( ) 01

=+++∑=

m

LC Rq uuuuµ

µ µ µ µ 12



1.3 Energy conversion in electromagnetic field

curlt

curlt

∂− ⋅ = − ⋅ − ⋅∂

∂⋅ = − ⋅

∂

DE H E J E

BH E H

curl curlt t

∂ ∂⋅ − ⋅ = − ⋅ − ⋅ − ⋅∂ ∂

B DH E E H H E J E

I. MGl. (-E)

II. MGl. H +

⋅

⋅

( ) ( )curl curl⋅ − ⋅ = ⋅ ∇ × − ⋅ ∇ × =H E E H H E E H

( ) ( ) ( ) ( )HEHEHEHE ×=×⋅∇=×⋅∇+×⋅∇= divcc

13



Integration over a volume Ω:

( )∫∫∫ΩΩΩ

Ω×+Ω⋅=Ω

∂∂⋅+

∂∂⋅− d divd d

t t HEJE

DE

BH

∂

∂=⋅

∂

∂=

∂

∂=⋅

∂

∂=

∂

∂⋅ ∫

2

0

2

1

2

1HBHBH

BH µ

t t t

wd

t t

m

B

( ) JEJE

J

JEE

J

EEEJ ⋅−=⋅−=⋅⇒−=⇒+= eeee pγ γ γ

2

∂

∂=⋅

∂

∂=

∂

∂=⋅

∂

∂=

∂

∂⋅ ∫

2

02

1

2

1EDEDE

DE ε

t t t

wd

t t

e

D

for

linearmedia

( ) ( )∫∫ΓΩ

Γ⋅×=Ω× d d div nHEHE (Gauss’ theorem)

14



( ) ( )∫∫ ∫∫ΓΩ ΩΩ

Γ⋅×+Ω⋅−Ω=Ω+− d d d d wwdt

d eem nHEJE

J

γ

2

Poynting’s theorem:

right hand side:

reasons for the decrease of energy in a volume

HES ×= Poynting’s vector: power through unit surface

∫Γ

Γ⋅ d nS is the power leaving the volume Ω through

radiation.

15



1.4 Existence of a unique solution to Maxwell’s equations

The solution of Maxwell’s equations in a volume Ω with the boundary Γ is unique for , provided that the

1. initial conditions

,),(),(),(),( 0000 Ω∈∀==== rrErErHrH anf anf t t t t and the

2. boundary conditions for the tangential components

0 0 0( , ) ( , ) or ( , ) ( , ), ,t tan t tant t t t t t = = ∀ ∈ Γ ≥H r H r E r E r r

0t t >

are fulfilled. Both the functions ),(),( 00 rErH anf anf

0 0( , ) and ( , )tan tant t H r E r and the impressed field intensity Ee

must be known. The material properties are assumed to be linear. 16



Proof :

two solutions exist., and ,′ ′ ′′ ′′E H E H Assumption:

Their differences: HHHEEE ′′−′=′′−′= 00 ,satisfy Maxwell’s equations (these are linear). The initial and

boundary conditions for the difference fields are homogenousand Ee0=0. Therefore, Poynting’s theorem :

( )∫∫∫ΓΩΩ

Γ⋅×+Ω=Ω

+− d d d dt

d nHE

JEH 00

2

02

0

2

02

1

2

1

γ ε µ

Since the boundary conditions imply that either E0 or H0

point in the normal direction n, the vector 000 HES ×=has no normal component. Therefore, the surface integral is zero.

17



∫∫ΩΩ

Ω=Ω

+− d d

dt

d

γ

ε µ

2

02

0

2

0

2

1

2

1 JEH

The right hand side is obviously non-negative, therefore

the quantity whose negative time derivative is written in

the left hand side can never increase. The initial

conditions, however, fix it as zero at the time instant t 0 and, since it is obviously non-negative, it can never

decrease either. Therefore it is always zero:

.02

1

2

1 2

0

2

0

=Ω

+

∫Ωd EH ε µ

Therefore, 0 0, , i.e. und .′ ′′ ′ ′′= = = =E 0 H 0 E E H H

q. e. d.

18



2. Static fields

,

,

0,

,

0.

curl

curl

div

div

div

ρ

=

=

=

=

=

H J

E 0

B

D

J

electrostatic

field:

,,

.

curldiv ρ

ε

==

=

E 0D

D E

magnetostatic

field:

,0,

.

curldiv

µ

==

=

H JB

B H

static current

field:

,0,

( ).e

curldiv

γ

==

= +

E 0J

J E E

Maxwell’s equations :0

=∂

∂

t

19



2.1 Boundary value problems for the scalar potential

Electrostatic field and static current field:

,curl gradV = ⇒ = −E 0 E V : electric scalar potential

Magnetostatic field, if J=0:

0 ,curl grad ψ = ⇒ = −H H ψ : magnetic scalar potential

Differential equations:

,)( ρ ε ρ =−⇒= gradV divdivD

,0)(0 =−⇒= gradV divdiv γ J

,0)(0 =−⇒= ψ µ grad divdivBgeneralized Laplace-Poisson

and Laplace equation.

20



The solution of the Laplace-Poisson equation in

unbounded free space (ε =ε 0):

0

1 ( )( ) , ( ) 0.

4

d V V

ρ

πε Ω

′ ′Ω= → ∞ =

′−∫r

r rr r

In regions free of charges, the electrostatic field is also

described by the generalized Laplace equation:

.0)( =− gradV div ε

21



Boundary conditions:

Dirichlet boundary condition: 0

(known) on , D

V V = Γ

0 (known) on . D

ψ ψ = Γ

D

Γ is typically constituted by electrodes in case of

electrostatic and static current field. Indeed:

konstant.=⇒= V 0Et

Magnetostatic field: interface to highly permeable

regions, to magnetic walls ).( ∞→µ

0µ µ = ∞→µ

0→ EisenH0Ht =since Eisen=t tH H

This means the prescription of Et or Ht.

22



Neumann boundary condition: (known) on , N

V

nε σ

∂= Γ

∂

0 on , N V n

γ ∂ = Γ∂

(known) on . N bn

ψ µ

∂= Γ

∂

This means the prescription of Dn , J n or Bn.

In case of static current field, Γ N is the interface to

the non-conducting region.

If σ =0 in the electrostatic case and b=0 in themagnetostatic case, Γ N is a surface parallel to the

field lines. Otherwise, Dn , Bn is known on Γ N .

23



Boundary value problems:

0

( ) in

on , on . D N

div gradV

V V V

n

ε ρ

ε σ

− = Ω,

∂= Γ = Γ

∂

electrostaticfield:

0

( ) 0 in

on , on . D N

div grad

bn

µ ψ

ψ ψ ψ µ

− = Ω,

∂= Γ = Γ

∂

magnetostaticfield:

0

( ) 0 in

on , 0 on . D N

div gradV

V V V

n

γ

γ

− = Ω,

∂= Γ = Γ∂

static currentfield:

24



2.2 Analytic methods for solving the Laplace equation

Laplace equation: ,02

2

2

2

2

2

=∂

∂

+∂

∂

+∂

∂

=∆= z

V

y

V

x

V

V divgradV

electrostatic field: 00, constant ( ), ρ ε ε = = =

magnetostatic field: 0, constant ( ),µ µ = = =J 0

static current field: constant.γ =

).0( =ψ divgrad

0 is a harmonic function.V V ∆ = ⇒

There exist infinitely many harmonic functions!

25



Dirichlet or Neumann boundary conditions :

V =constant on Γ D

Ω=∆ in0V Ω∉ Ω∉

Analytic methods:

Generation of harmonic functions and selection of the one

satisfying the boundary conditions.

This yields the true solution, since that is unique.

0 on N

V

n

∂= Γ

∂

26



2.2.1 Method of fictitious charges (method of images)

The potential function of a point charge is harmonic in all

points in space except in the point where the charge is

located.

x

y

z ),,( z y x ′′′Q

( ) ( ) ( )[ ] 2

1222

04),,(

−′−+′−+′−= z z y y x x

Q z y xV

πε

, z y x z y x eeer ++=

r

r′ z y x z y x eeer ′+′+′=′

27



( ) ( ) ( ) ( )[ ] =′

−+′

−+′

−′

−−=∂

∂ −2

3222

04 z z y y x x x x

Q

x

V

πε

( )3

04

−′−′−−= rr x xQ

πε

( )[5

23

0

2

2

34

−−

′−′−−′−−=∂∂ rrrr x xQ xV

πε

( ) ( ) ( )=

′−

′−+′−+′−−

′−

−=∂

∂+

∂

∂+

∂

∂5

222

3

0

2

2

2

2

2

2

33

4 rrrr

z z y y x xQ

z

V

y

V

x

V

πε 2

3 5

0

13 0, if .

4

Q

πε

′−′= − − = ≠

′ ′− −

r rr r

r r r r

Proof:

q. e. d.28



In two dimensions (planar problems, ),0=∂

∂

z

the potential function of an infinitely long line charge isharmonic.

τ ),( y x ′′

( ) ( )[ ]21

22

0

0

ln2

),(

y y x x

r y xV

′−+′−

=πε

τ

x

y

r

r′

( ) ( )[ ] ′−+′−−= 220

0

ln21ln

2),( y y x xr y xV

πε τ

, y x y x eer += y x y x eer ′+′=′

29



( ) ( )22

02 y y x x

x x

x

V

′−+′−

′−−=

∂

∂

πε

τ

( ) ( ) ( )

( ) ( )[ ] =

′−+′−

′−−′−+′−−=

∂

∂222

222

0

2

2 2

2 y y x x

x x y y x x

x

V

πε

τ

( ) ( )( ) ( )[ ]222

22

02 y y x x y y x x

′−+′−

′−−

′−= πε

τ

( ) ( ) ( ) ( )

( ) ( )[ ]

0

2

222

2222

0

2

2

2

2

=

′−+′−

′−−′−+′−−′−=

∂

∂+

∂

∂

y y x x

x x y y y y x x

y

V

x

V

πε

τ

q. e. d.

30



Due to its linearity, the Laplace equation is satisfied by the

potential function of an arbitrary charge distribution (in regions

free of charges).

Satisfaction of the boundary conditions: a fictitious chargedistribution within the electrodes should have equipotential

surfaces which coincide with the electrodes.

Examples:

• One point charge: concentric spherical surfaces (sphericalcapacitor)

• One infinitely long line charge: concentric cylindrical

surfaces (cylindrical capacitor)

• Line dipole: non-concentric cylindrical surfaces• Mirror image over a plane: infinite conducting plane

• Multiple mirror images over planes: planar electrodes

forming an angle )/180( n=α

31



2.2.2 Separation of variables

General orthogonal coordinates x1, x2, x3

Cartesian coordinates: .,, 321 z x y x x x ===

Cylindrical coordinates:

.,, 321 z x xr x === φ

Spherical coordinates:

.,, 321 φ θ === x xr x

),,( zr φ

x

y

z

φ r φ φ sin,cos r yr x ==

),,( φ θ r

x

y

z

φ

r

θ

,sinsin,cossin φ θ φ θ r yr x ==

θ cosr z =32



1 2 3, , form a right handed system: x x x

x1

x2

x3

⋅ ⋅⋅

Generally, x1

, x2

and x3

do not

represent lengths.

The differential distance ds1 between the points ( x1, x2, x3)

and ( x1+dx1, x2, x3) is h1dx1.

x1

x2

x3

),,( 321 x x x11 dx x +

22 dx x +

33 dx x +11dxh

22dxh

33dxhIn general:iii dxhds =

h1, h2, h3: metric coefficients

33



Metric coefficients in various coordinate systems:

Cartesian coordinate system: ,,, 321 dzdsdydsdxds ===

.1,1,1 321 === hhh z x y x x x === 321 ,,

Cylindrical coordinate system: z x xr x === 321 ,, φ

,,, 321 dzdsrd dsdr ds === φ .1,,1 321 === hr hh

Spherical coordinate system:

φ θ === 321 ,, x xr x

,sin,, 321 φ θ θ d r dsrd dsdr ds ===

.sin,,1 321 θ r hr hh ===

34



Gradient in a general orthogonal coordinate system:

),,( 321 x x xuLet be a scalar function.

( )111

3213211

01

1),,(),,(lim

1 x

u

hs

x x xu x x x xugradu

s ∂

∂=

∆

−∆+=

→∆

333

222

111

321

111

),,( eee x

u

h x

u

h x

u

h x x xgradu ∂

∂

+∂

∂

+∂

∂

=

z y x z

u

y

u

x

u z y xgradu eee

∂∂

+∂∂

+∂∂

=),,(

zr zuu

r r u zr gradu eee

∂∂+∂∂+∂∂= φ φ

φ 1),,(

φ θ φ θ θ

φ θ eee∂∂

+∂∂

+∂∂

=u

r

u

r r

ur gradu r

sin

11),,(

35



Divergence in a general orthogonal coordinate system:

),,( 321 x x xv

∫Γ

→ΩΓ⋅

Ω= d div nvv

1lim

0

x1

x3

x2 v1

1

1

11 dx

x

vv

∂∂

+

111 dxhds =

333 dxhds =222 dxhds =

1Γ

2ΓΩ

32321

1

dxdxhhvd −=Γ⋅∫Γ

nv

=

∂

∂+

∂

∂+=Γ⋅∫

Γ

321

1

32321

1

11

2

dxdxdx x

hhhhdx

x

vvd nv

( ) )( 3211321

1

32132321 dxdxdxdxOdxdxdx

xhhvdxdxhhv +

∂∂+=

Let be a vector function.

36



( )321

1

321

2121

dxdxdx x

hhvd d d

∂

∂=Γ⋅+Γ⋅=Γ⋅ ∫∫∫

ΓΓΓ+Γ

nvnvnv

( ) ( ) ( ) 321

3

213

2

312

1

321 dxdxdx x

hhv

x

hhv

x

hhvd

∂

∂+∂

∂+∂

∂=Γ⋅∫Γ nv

321321321 dxdxdxhhhdsdsds ==Ω

( ) ( ) ( )

∂∂+∂∂+∂∂= 213

3

312

2

321

1321

321 1),,( hhv x

hhv x

hhv xhhh

x x xdivv

∂

∂+

∂

∂+

∂

∂=

z

v

y

v

x

v z y xdiv z y x),,(v

( ) z

vv

r r

rv

r zr div zr

∂∂+

∂∂+

∂∂=

φ φ

φ 11),,(v

( )

φ θ θ

θ

θ

φ θ φ θ

∂

∂+

∂

∂+

∂

∂=

v

r

v

r r

vr

r

r div r

sin

1sin

sin

11),,(

2

2v

37



Curl in a general orthogonal coordinate system:

( )1

1

1 01

1lim

C

curl d Γ →

= ⋅Γ ∫v v r

x1

x3 x2

333 dxhds =222 dxhds =1Γ

222)1(

1

dxhvd

C

=⋅∫ rv

=

∂

∂+

∂

∂+−=Γ⋅∫ 23

3

223

3

22

)3(1

dxdx x

hhdx

x

vvd

C

nv

( ))(

332323

22

222

dxdxdxOdxdx x

hvdxhv +

∂

∂−−=

)2(

1C

)1(1C

)3(

1C

)4(1C

)4(1

)3(1

)2(1

)1(11 C C C C C +++=

( )32

3

22

)3(1

)1(1

dxdx x

hvd

C C ∂

∂−=⋅∫

+

rv ( )

32

2

33

)4(1

)2(1

dxdx x

hvd

C C ∂

∂=⋅∫

+

rv

),,( 321 x x xvLet be a vector function.

38

( )



( ) ( )32

3

2232

2

33

1

dxdx x

hvdxdx

x

hvd

C ∂

∂−

∂

∂=⋅∫ rv 3232321 dxdxhhdsds ==Γ

( )

( ) ( )

1 2 3 1

3 3 2 2

2 3 2 3

( , , )

1

curl x x x

v h v hh h x x

=

∂ ∂= − ∂ ∂

v 1 2 3

2 3 3 1 1 2

1 2 3

1 1 2 2 3 3

1 1 1

h h h h h h

curl x x x

h v h v h v

∂ ∂ ∂=

∂ ∂ ∂

e e e

v

x y z

x y z

curl x y z

v v v

∂ ∂ ∂

= ∂ ∂ ∂

e e e

v

e. g. Cartesian coordinates:

39



Laplacian in a general orthogonal coordinate system:

( )gradudivu =∆

∂

∂

∂

∂+

∂

∂

∂

∂+

∂

∂

∂

∂=∆

33

21

322

13

211

32

1321

1

x

u

h

hh

x x

u

h

hh

x x

u

h

hh

xhhhu

2

2

2

2

2

2

),,( z

u

y

u

x

u z y xu

∂

∂+

∂

∂+

∂

∂=∆

2

2

2

2

2

11),,(

z

uu

r r

ur

r r zr u

∂

∂+

∂

∂+

∂

∂

∂

∂=∆

φ φ

2

2

222

2

2 sin

1sin

sin

11),,(

φ θ θ θ

θ θ φ θ

∂∂+

∂∂

∂∂+

∂∂

∂∂=∆

u

r

u

r r

ur

r r r u

40



Cartesian coordinates, 2D case:

Assumption: )()(),( yY x X y xV =

Laplace equation: .02

2

2

2

=∂∂+

∂∂

yV

xV

0)(

)()(

)(2

2

2

2

=+dy

yY d x X

dx

x X d yY

Division by X ( x)Y ( y) yields:

This is only possible if

( ) constant, ( ) constant. f x g y= =

0)(

)(

1)(

)(

1

)(

2

2

)(

2

2

=+

yg x f

dy

yY d

yY dx

x X d

x X

42



Hence, the ordinary differential equations are:

),(

)(

),(

)(2

2

2

2

ygY dy

yY d

x fX dx

x X d

==

:, 22 pg p f =−= pxC pxC x X sincos)( 21 +=

pyC pyC eC eC yY py py sinhcosh)( 4343

′+′=+= −

where . f g −=

One solution:

A further solution is obtained by interchanging x and y:

cos sinn n p y p y

n n n n

n

V A e p x B e p x± ±± ±= +∑

cos sinn n p x p x

n n n n

n

V A e p y B e p y± ±± ±= +∑

43



x

Example:

y

a

b

0=V

0=V 0=V

)( x f V =

boundary conditions:

.0,)4(

;0,0)3(

;0,0)2(

);(,)1(

==

==

==

==

V a x

V x

V y

x f V b y

....2,1,

,/,sin

,sinh

=

=⇒

⇒

n

an p x p

y p

nn

n

π

:sinsinh),(1

=

∑

∞

= a

xn

a

yn B y xV

nn

π π

Fourier series of f ( x):

(2), (3), (4)

erfüllt.

.sin)(1

∑∞

=

=

n

na

xnb x f

π

⇒

=

∑

∞

= a

xn

a

bn Bb xV

n

n

π π sinsinh),(

1

nn b

a

bn B =

π sinh

.sin

sinh

sinh

),(1

= ∑∞

= a

xn

a

bn

a

yn

b y xV n

n

π

π

π

44



Cylindrical coordinates, 2D case:

Assumption: )()(),( φ φ Φ= r Rr V

Laplace equation:

Multiplication by r 2 and division by R(r )Φ(φ ) yield:

( ) constant, ( ) constant. f r g φ = =

.0112

2

2 =∂∂+

∂∂

∂∂

φ V

r r V r

r r

.0)(1

)()(1

)(2

2

2 =

Φ+

Φφ

φ φ

d

d

r r R

dr

r dRr

dr

d

r

.0)(

)(

1)(

)(

1

)(

2

2

)(

=Φ

Φ+

φ

φ

φ

φ

gr f

d

d

dr

r dRr

dr

d r

r R

This is only possible if

45



),(

)(

),(

)(2

2

φ φ

φ

Φ=

Φ

=

gd

d

r fRdr

r dR

r dr

d

r

where . f g −=

Since Φ(φ ) must be periodic with the period 2π , the only

possibility is g = -n2 ( n: integer)).0(sincos)( 43 >+=Φ nnC nC φ φ φ

The case n = 0 yields .)( 43 φ φ C C +=Φ

This is only periodic if C 4 = 0. The case ,04 ≠C makes sense only if the problem region is max0 φ φ ≤≤

maxwhere 2 .φ π <Ω

Hence, the ordinary differential equations are:

46



).()( 2

r Rndr

r dRr

dr

d r =

:0a) =n .ln)( ,)(

211 C r C r RC dr

r dRr +==

:0 b) >n Assumption:

,)(

,)(

,)( 121 −− =

== α α α α α α r

dr

r dRr

dr

d r

dr

r dRr r

dr

r dR

,)( α r r R =

⇒= α α α r nr 22 :n±=α nn

r C r C r R −+= 21)(

Solution:

( )( ) ( )1 2 3 4ln cos sinn n

n n

n

V C r C C C A r n B r nφ φ φ ± ± ± ±= + + + +∑

47



Example: cylindrical capacitor

i Ra R

iV

aV

.ln0 21 C r C V +=⇒=∂

∂φ Axisymmetry:

.ln

,ln

21

21

C RC V

C RC V

aa

ii

+=

+= boundary conditions:

.ln

lnln,

ln21

ia

iaai

ia

ia

R R

RV RV C

R R

V V C

−=−=

.

ln

lnln

ia

aiia

R R

Rr V Rr V V

−=

The same solution is obtained by the method of fictitious

charges.

48

E l di l t i li d i h fi ld



Example: dielectric cylinder in a homogeneous field

x E eE 00 =

R

1>r ε homogeneous field: .cos0 φ r E V −=

Proof:

.sin1

,cos 00 φ φ

φ φ E V

r E E

r

V E r −=

∂∂

−==∂∂

−=

r φ

x

y

r E

φ E

,sincossincos 0

2

0

2

0 E E E E E E r x =+=−= φ φ φ φ φ

.0cossinsincoscossin 00 =−=+= φ φ φ φ φ φ φ E E E E E r y

q. e. d.

>

<=

.wenn),,(

,wenn),,(),(

Rr r V

Rr r V r V

a

i

φ

φ φ

⇒∞<→),(lim

0φ r V i

r

⇒−=∞→

φ φ cos),(lim 0r E r V ar

,sincos1∑≥

+=n

ninnini nr Bnr AV φ φ

.sincoscos1

0 ∑≥

−− ++−=n

n

an

n

ana nr Bnr Ar E V φ φ φ

boundary

conditions:

49



Boundary and interface conditions:

⇒=⇒=== ),(),()()( φ φ φ φ RV RV Rr E Rr E aiai

⇒∂

∂=

∂

∂⇒===

r

RV

r

RV Rr D Rr D ai

r rari

),(),()()(

φ φ ε

n

an

n

in

n

an

n

inai R B R Bn R A R A R A R E R A −−− =>=+−= ),1(,1

101

),1(,

112

101 >−=−−=

−−−−

nnR AnR A R A E A

n

an

n

inr air ε ε .11 −−− −= n

an

n

inr nR BnR Bε

Solution: ,1

1,

1

2 2

0101 R E A E Ar

r a

r

i +−

=+

−=

ε

ε

ε

),1(0 >== n A A anin .0== anin B B

50



.cos11cos),(

,cos1

2),(

2

00

0

φ ε ε φ φ

φ ε

φ

r R E r E r V

r E

r V

r

r a

r

i

+−+−=

+−

=

The field within the

cylinder is homogeneous.

51



2.3 Numerical methods for solving the boundary value

problems for the scalar potential

Disadvantages of analytical methods:

• special geometry

• homogeneous materials

Numerical methods:

• geometry discretized

• taking account of non-homogeneous materials

52



Discretization of geometry:

V =constant on Γ D

Ω= in0)( gradV div ε

Ω∉ Ω∉

Numerical methods:

on N

V

nε σ

∂ = Γ∂

The potential is approximated in discrete nodes. The field iscomputed by numerical differentiation. The Dirichlet

boundary conditions in the nodes are satisfied exactly, the

Neumann boundary conditions can only be approximately

fulfilled.53



2.3.1 Method of finite differences

Two-dimensional planar problems are treated only, the

generalization to 3D problems is straightforward.

Homogeneous materials are assumed: Laplace equation.

Generalization for piecewise homogeneous materials is

possible.

Uniform mesh,generalization for non-

uniform mesh is

possible.Ω

Γ h

h 54

432



a

bc

d

A

0=∆V

...!4

1

!3

1

!2

1

!1

1 4

4

43

3

32

2

2

+∂

∂+

∂

∂+

∂

∂+

∂

∂+= h

x

V h

x

V h

x

V h

x

V V V Aa

...!4

1

!3

1

!2

1

!1

1 4

4

43

3

32

2

2

+∂

∂

+∂

∂

+∂

∂

+∂

∂

+= h y

V

h y

V

h y

V

h y

V

V V Ab

...!4

1

!3

1

!2

1

!1

1 4

4

43

3

32

2

2

±∂

∂+

∂

∂−

∂

∂+

∂

∂−= h

x

V h

x

V h

x

V h

x

V V V Ac

...!4

1

!3

1

!2

1

!1

1 4

4

43

3

32

2

2

±∂

∂+∂

∂−∂

∂+∂

∂−= h y

V h y

V h y

V h y

V V V Ad

0

4

0

2

2

2

2

2 )()(

4

≈=

+∂∂+∂∂+

+=+++

hO yV

xV h

V V V V V Ad cba

h

h

04 =−−−− d cba A V V V V V

+

x

y

55



The equation

( )d cba A V V V V V +++= 4

1

corresponds approximately to the mean value

theorem of potential theory:

Sphere

M R

2

1( )

4Sphere

V M Vd Rπ

= Γ∫

a A

dc

b

4 0, i.e. A a b c d V V V V V − − − − =

56



h

h

Γ N

Γ D

e

f g

h

ik

lm

n

o

Taking account of boundary conditions:

Dirichlet boundary condition on Γ D:0 (known).gV V =

⇒=−−−− 04 hg f ei V V V V V 04 V V V V V h f ei =−−−

Neumann bondary condition on Γ N :

0 at the point .V on

ε σ ∂ =∂k : fictitious node outside of Ω

( ) σ ε ε =−≈∂∂ hV V o

nV mk

200

0

2ε

σ hV V mk +=⇒

⇒=−−−− 04 nmlk o V V V V V 0

224

ε

σ hV V V V

nmlo =−−−

57

E l



Example:V =U

V =0

1 2 3

4 5 6 7 84′ 8′

node 1: U V V V 24 521 =−−

10 σ ε =∂∂

n

V 20 σ ε =

∂∂

n

V

node 2: U V V V V =−−+− 6321 4h

node 3: U V V V 24 732 =−+−node 4:

154

02

σ ε =−′

h

V V

0

154

2

ε

σ hV V +=′

⇓

278

02

σ ε =−′

h

V V

0

278

2

ε

σ hV V +=′

⇓

U V V V =−+− ′ 544 4

0

154

224

ε

σ hU V V +=−

node 5: 04 6541 =−+−− V V V V

node 6: 04 7652 =−+−− V V V V

node 7: 04 8763 =−+−− V V V V

node 8: U V V V =−+− ′887 4

0

287

242

ε

σ hU V V +=+−

58



+

+=

−

−−

−−−

−

−−−

−

−

−

−

−

−−

−

02

01

8

7

6

5

4

3

2

1

/2

0

00

/2

2

2

4200

1410

01410014

0000

0100

00101001

0002

0100

0010

0001

4000

0410

0141

0014

ε σ

ε σ

hU

hU

U

U

U

V

V

V V

V

V

V

V

Equations system in matrix form:

Sparse matrix.

59



2.3.2 Variational problem of electrostatics

0

( ) in

on , on D N

div gradV

V V V

n

ε ρ

ε σ

− = Ω,

∂= Γ = Γ

∂

The boundary value problem of electrostatic field

is equivalent to the following variational problem:

Find V (V=V 0 on Γ D) , so that the functional

∫∫∫ ΓΩΩΓ−Ω−Ω=

N

Vd Vd Vd grad V W σ ρ ε 2

21)(

attains a minimum.

60



Physical meaning of the functional

:21

21 2

∫∫ ΩΩΩ⋅=Ω d Vd grad DEε energy of electrical field: W e

:∫∫ΓΩ

Γ+Ω N

Vd Vd σ ρ potential energy of the charges: W p

is the action!e pW W W = −

The variational problem corresponds to the principleof least action.

61

2 3 3 Rit ‘s procedure



2.3.3 Ritz s procedure

Since the variational problem and the boundary value problem

are equivalent, an approximate solution of the variational problem is simultaneously an approximate solution of the

boundary value problem.

The variational problem can be solved approximately by means

of the Ritz’s procedure.An approximate solution is sought in the following form:

,1

)( ∑=

+=≈n

j

j j D

n wV V V V 0: arbitrary function with on ,

, 1, 2, ..., : numerical parameters,, 1, 2, ..., : basis functions with

0 on .

D D D

j

j

j D

V V V

V j nw j n

w

= Γ

==

= Γ( ) satisfies the Dirichlet boundary conditions for arbitrary !n

jV V 62



The unknown parameters V j, j = 1, 2, ..., n are determined from

the condition that the approximate solution minimizes the

functional. The necessary conditions are:

....,2,1, ,0)( )(

niV

V W

i

n

==∂

∂

These are n equations (the Ritz equations system), allowing

the determination of the n unknowns V j, j = 1, 2, ..., n.

63



21

2 N

( n )( n ) ( n ) ( n )

i i i i

W (V )grad V d V d V d

V V V V ε ρ σ

Ω Ω Γ

∂ ∂ ∂ ∂= Ω − Ω − Γ =

∂ ∂ ∂ ∂∫ ∫ ∫

.)()(

)()(

∫∫∫ΓΩΩ

Γ∂

∂−Ω

∂

∂−Ω⋅

∂

∂=

N

d V

V d

V

V d gradV

V

V grad

i

n

i

nn

i

n

σ ρ ε

.)(1

)(

i

n

j

j j D

ii

n

wwV V V V

V =+∂∂=∂∂ ∑=

The Ritz equations system:

,1 ∫∫∫∫∑ ΓΩΩΩ=

Γ+Ω+Ω⋅−=Ω⋅ N

d wd wd gradV gradwd gradwgradwV ii Di ji

n

j

j σ ρ ε ε

....,2,1, ni =Symmetric matrix!64

2 3 4 The method of finite elements (Finite Element Method=FEM)



2.3.4 The method of finite elements (Finite Element Method=FEM)

Ω

Discretization of the geometry

Triangular elements: simplest possible approach

Unknowns: potential values in nodesPotential in each element: low order polynomial

finite elements

65



Linear interpolation of the potential function within elements

66



i

k

j

1

iV

i N

V i N i

V

i

k

j

1

jV

j N

V j N j

V

i

k

j

1k V

k N

V k N k

V

i

k

j

k V

V i N i + V j N j +V k N k

iV

jV

+ + =

=

Shape functions

67

V



1 N j

V

j

∑=

=nn

j

j j

n N V V

1

)(nn: number of nodes

V j: nodal potential values

1 in node

0 in all other nodes j

j N

=

68

B i f tiV D V 0



Basis functions:

....,2,1, , n j N w j j == Γ D

1

2

n

n+1n+2 ...

nn ...

D

=+== ∑∑∑=+==

n

j

j j

n

n j

j j

n

j

j j

n N V N V N V V

nn

111

)( .1

∑=

+n

j

j j D N V V

69

Rit eq ations:



,

1

∫∫∫∫∑

ΓΩΩ

Ω=

Γ+Ω+Ω⋅−=

=Ω⋅

N

d N d N d gradV gradN

d gradN gradN V

ii Di

ji

n

j j

σ ρ ε

ε

Ritz equations:

....,2,1, ni =

.i jij bV A =

,∫Ω

Ω⋅= d gradN gradN A jiij ε

.∫∫∫ΓΩΩ

Γ+Ω+Ω⋅−= N

d N d N d gradV gradN b ii Dii σ ρ ε

In matrix form:

70

N N



i j

i N j N

i

j

i N j N

,0≠Ω⋅= ∫Ω

d gradN gradN A jiij ε

.0=Ω⋅= ∫Ω

d gradN gradN A jiij ε

Sparse matrix.

71



8-node quadrilateral elements

13

2

4

5

6

7

8

72



20-node hexahedral elements

73

Shape and basis functions for 20-node hexahedral elements



Shape and basis functions for 20 node hexahedral elements

74



2.5 Boundary value problems for the vector potential

Magnetostatic field:

A: magnetic vector potential

Differential equation:

1( ) ,curl curl curlµ

= ⇒ =H J A J

,div curl= ⇒ =B 0 B A

75



2.5.1 Planar 2D problems

.),(),(,),(:0 y y x x z y x B y x B y x J z

eeBeJ +===∂

∂


J

Field lines of B z y x A eA ),(=

0 .

0 0

x y z

x y

A Acurl

x y y x

A

∂ ∂ ∂ ∂= = = −

∂ ∂ ∂ ∂

e e e

B A e e

76



Differential equation for the single component vector potential in

planar 2D case:

1[ ( )] 0

1 1 0

x y z

zcurl curl A x y

A A y x

µ

µ µ

∂ ∂= =

∂ ∂

∂ ∂−∂ ∂

e e e

e

1 1 1( ). z z

A Adiv gradA

x x y yµ µ µ

∂ ∂ ∂ ∂= − + = − ∂ ∂ ∂ ∂

e e

,)1

( J gradAdiv =−µ

generalized Laplace-Poisson equation.

77



Magnetic flux by means of A:

.C

d curl d d Γ Γ

Φ = ⋅ Γ = ⋅ Γ = ⋅∫ ∫ ∫B n A n A r

C

Γ

n Γd

B

A

rd

z=constant

x

y

PQ zP A e)(

zQ A e)(

C PQ

PQΓ

Planar 2D case:

1

ΓPQ: surface of unit length

through the points P and Q

)()( Q AP Ad PQC

PQ −=⋅=Φ ∫ rA

ze

78




Dirichlet boundary condition: 0 (known) on , D A A= Γ

This means the prescription of Bn:

( ) ( ) ( )n z z z B curl curl A gradA gradA= ⋅ = ⋅ = ⋅ × = ⋅ × =n A n e n e e n

DΓ

n

ze

tne =× z

: tangential derivative of ! A

gradA At

∂= ⋅ =

∂t

In most cases A0=constant:

Then, the section of Γ D with a plane z=constant is

a flux line. The differences in the values of A0

yield the flux per unit length between the lines.

80



Duality between the boundary conditions for the scalar potential

and the single component vector potential in the planar 2D case

Magnetic wall:

constant,ψ =

1 0. Anµ ∂ =∂

Magnetic wall

Flux line:

0, constant. An

ψ µ

∂= =

∂

Flux line

82

Boundary value problem for the single component vector



y p g p

potential function in the planar 2D case:

0

1

( ) in

1 on , on . D N

div gradA J

A A A

n

µ

α µ

− = Ω,

∂= Γ = Γ

∂

magnetostaticfield:

Similar boundary value problem to the ones for the scalar potential

functions.

83



2.5.2 Axisymmetric 2D problems

.),(),(,),(:0 z zr r zr B zr B zr J eeBeJ +===∂

∂φ

φ


φ eA ),( zr A=

1 1

0

0 0

r zr r

curlr z

rA

φ

∂ ∂= = =

∂ ∂

e e e

B A

.)(1

zr r

rA

r z

Aee

∂

∂+

∂

∂−=

84

Differential equation for the single component vector potential in the



1 1

1[ ( )] 0

1 1 ( )

0

r zr r

curl curl Ar z

A rA

z r r

φ

φ µ

µ µ

∂ ∂= =

∂ ∂∂ ∂

− ∂ ∂

e e e

e

Differential equation for the single component vector potential in the

axisymmetric case:

1 1 1( rA ) Adiv( gradA ).

r r r z z ϕ ϕ

µ µ µ

∂ ∂ ∂ ∂= − + ≠ − ∂ ∂ ∂ ∂

e e

1 1 1 1 A Adiv( gradA ) r r r r z zµ µ µ

∂ ∂ ∂ ∂= + ∂ ∂ ∂ ∂

1 1( rA ) A J

r r r z zµ µ

∂ ∂ ∂ ∂− + = ∂ ∂ ∂ ∂

85



Q

P

φ e)(Q A

φ e)(P AC PQ

PQΓ

Flux in axisymmetric 2D case:

ΓPQ: Conical surface throughthe points P and Q

)].()([2 Q Ar P Ar d QP

C

PQ

PQ

−=⋅=Φ ∫ π rA

z

constantφ =

constantφ =

Qr

Pr

Flux lines: Lines of constant rA(r , z).

86



2.5.3 3D problems

The vector potential function is not unique:

( )curl curl gradu= = +B A A u is an arbitrary scalar function,

Differential equation:

1( ) in ,curl curl curlµ

= ⇒ = ΩH J A J


Prescription of Bn or of H t on Γ.

87



Special case: vector potential due to a given current

density in infinite free space 0( and ) :µ µ = Γ → ∞

0 ,

( ) 0 ( ( ) or ( ) ).

curlcurl

curl curl

µ =

∞ = ⇒ ⋅ ∞ = ∞ × =

A J

A n A 0 A n 0

Only B=curlA is defined uniquely, but not A.

A becomes unique if divA is additionally defined: gauging.

The choice divA = 0 is the Coulomb gauge.

88



The gauge makes A unique:

( ) 0, .( ) ( ) ( ) ( ) ,

div div gradu u gradugradu gradu

= + ⇒ ∆ = ⇒ =∞ = ∞ + ∞ ⇒ ∞ = A A 0

A A 0

Boundary value problem for A:

0 ,

0,

( ) .

curlcurl

div

µ =

= ⇒∞ =

A J

A

A 0

0 ,

( ) .

curlcurl graddiv µ − = −∆ =

∞ =

A A A J

A 0

Vector Laplace-Poisson differential equation.

⇓

89

The Coulomb gauge follows from the vector Laplace-



Solution of

0 ( )( ) .4

d µ π Ω

′ ′Ω= ′−∫ J rA rr r

0 ,

( ) .

curlcurl graddiv µ − = −∆ = ⇒∞ =

A A A J

A 0

0

( ) 0

( )

0.

( ) 0,

div

div curlcurl graddiv div

div

div

µ

∆

− =

⇒ =∞ =

A

A A J

A

A

0 , ( ) :µ −∆ = ∞ =A J A 0

The Coulomb gauge follows from the vector Laplace

Poisson differential equation:

90

3 Quasi static fields



3. Quasi-static fields

Maxwell’s equations :

∂

∂>>

t

DJ

,

,

0,

curl

curlt

div

=

∂= −∂

=

H J

BE

B

).(, 0EEJHB === eγ µ

In conducting media (Ωl): In non-conducting media (Ωi):

,

0,

curl

div

==

H J

B

.HB µ =

J is unknown:

time dependent

quasi-static field

J is known:

time dependent

magnetostatic field 91

E l



lΩ

iΩ

)(t J

Example:

92




iΩ

non-conducting region

JlΩ

conducting region

liΓ ,: nH × nB ⋅ continuous

: BΓ b−=⋅nB

: HiΓ

=×nH

: E Γ 0nE =×

: HlΓ 0nH =×

n

n

n


93

Summary



Differential equations in Ωi

(eddy current free region):

0

,

i i

i

i i i i

curldiv

µ ν

==

= =

H JB

B H H B


0nH =×l on HlΓ ,

0nE =×l on E Γ ,KnH =×i

on HiΓ ,

bi −=⋅ nB on BΓ .

interface conditions on Γli:

0=⋅+⋅=×+×

iill

iill

nBnB0nHnH

initial conditions at t =0:iiilll

Ω=Ω= in,in00

BBBB

Differential equations in

Ωl (eddy current region):

0

, ,

l l

ll

l

l l l l l l

curl

curlt

div

∂

∂

µ ν γ

=

= −

=

= = =

H JB

E

B

B H H B J E

94

C l t ti f ti h i titi



Complex notation for time harmonic quantities:

Complex amplitude:

)(

)(ˆ)(

r

rBrB

ϕ j

e=Time derivative:

Maxwell’s equations for complex amplitudes(quasi-static case):

0curl , curl j , div .ω = = − =H J E B B

))(cos()(ˆ),( rrBrB ϕ ω += t t

Time function :

specially for linear polarization:

ˆ( , ) ( )cos( ( )) x x x

B t B t ω ϕ = +r r r

ˆ( , ) ( )cos( ( )) y y y B t B t ω ϕ = +r r rˆ( , ) ( )cos( ( ))

z z z B t B t ω ϕ = +r r r

in time domain multiplication by in frequency domain jt

ω ∂

→∂

95

Poynting’s theorem for complex amplitudes in quasi-static



( )* * * * *1 1 1 1 1

2 2 2 2 2curl curl div jω ⋅ − ⋅ = − × = ⋅ + ⋅E H H E E H E J B H

y g p p q

case:

( ) S d d jd =Γ⋅×−=Ω⋅+Ω⋅ ∫∫∫ ΓΩΩnHEHBJE

***

2

1

2

1

2

1

ω

S : complex power flowing into the region Ω through the

boundary Γ.

96

Proof:



=+×+=×= )cos(ˆ)cos(ˆ)()()( H E t t t t t ϕ ω ϕ ω HEHES

[ ] =+++−×= )2cos()cos(ˆˆ2

1 H E H E t ϕ ϕ ω ϕ ϕ HE

[ ]

Effective part

1 ˆ ˆ cos( ) 1 cos 2( )2

E H E t ϕ ϕ ω ϕ = × − + + +E H

[ ]

Reactive part

1 ˆ ˆ sin( ) sin 2( )2

E H E t ϕ ϕ ω ϕ + × − +E H

∫Γ

Γ⋅−×−= d P H E nHE )cos(ˆˆ21 ϕ ϕ

∫Γ

Γ⋅−×−= d Q H E nHE )sin(ˆˆ2

1ϕ ϕ

Effective power:

Reactive power:

Time function of Poynting’s vector:

97

∫ ˆˆ1



=Γ⋅−+−×−=+= ∫Γ

d j jQPS H E H E nHE )]sin()[cos(ˆˆ2

1ϕ ϕ ϕ ϕ

=Γ⋅×−=Γ⋅×−= ∫∫Γ

−

Γ

−d eed e H E H E j j j

nHEnHE ϕ ϕ ϕ ϕ ˆˆ

2

1ˆˆ2

1 )(

( ) .

2

1 * Γ⋅×−= ∫Γ

d nHE q. e. d.

Complex Poynting’s vector:*

2

1HES ×=

),(,2

1

2

12

*

0E

J

JE =Ω=Ω⋅= ∫∫ ΩΩ

ed d P γ

.2

1

2

1 2*

∫∫ΩΩ

Ω=Ω⋅= d d Q HHB µ ω ω

Effective power:

Reactive power:

98

3.1 Some analytical solutions of the boundary value problem for



In conducting region (Ωl): 0 ,div curl= ⇒ =B B A

( ) 0 .curl

curl curl curl gradV t t t t

∂ ∂ ∂ ∂+ = + = + = ⇒ = − −

∂ ∂ ∂ ∂

B A A AE E E E

Differential equations:

1( ) ,curl curl curl gradV

t γ γ

µ

∂− = ⇒ + + =

∂

AH J 0 A 0

=

∂∂−−⇒= .0)()(0

t divgradV divdiv

AJ γ γ

boundary conditions: ( ) 0, ( ) 0.V ∞ = ∞ =A

3.1 Some analytical solutions of the boundary value problem for

the magnetic vector potential

99

Special case: µ = constant, γ = constant, time harmonic case.



Coulomb gauge: divA=0

( ) ( ) 0 0

0 0

( ) 0

div gradV j div V j div

V V

V

γ ω γ ω

⇓

− − = ⇒ −∆ = =

∆ = ⇒ =

∞ =

A A

Differential equation in Ωl:

0AA =+∆− ωµγ j vector diffusion equation.

Planar 2D problems:

0),(),( =+∆− y x A j y x A ωµγ scalar diffusion equation.

100



,)()( 1

py

e A j y A j y E

−−=−= ω ω

Field quantities:

,)()( 1

pye A j y A j y J

−−=−= ωγ ωγ

,

)(

)( 1

py

e pAdy

ydA

y B

−

−==

Determination of the constant A1: the current through a

conductor of width b is assumed to be given.

.)(1

)( 1

pye A p

dy

ydA y H −−==

µ µ

103

z IbyHdd ===⋅=Γ⋅ ∫∫ )0(rHnJ



b x

y

I l

0Γ0C

I b y H d d C

Γ ∫∫Γ

)0(

00

rHnJ

pb I A I A pb µ

µ −=⇒=− 11

,)( δ δ

γ

ωµ y y

j jpy ee I

b

pe I

pb

j y E

−−− ==

,)()( δ δ γ y y

j

ee I b

p y E y J

−−

==

.)( ,)( δ δ δ δ µ

y y j

y y j

ee

b

I y H ee I

b

y B−−−−

==

δ

δ

y

eb

I y J

−

= 2)(

yδ

)( y J

Magnitude of the

current density

decays exponentially

Skin effect:

104



3.1.2 Current flow in an infinite conducting plate



f f g p

x

y

z

2h

2b2b2h

I

),()( 2

2

2

y A pdy

y Ad = .

1

2

1

δ ωµγ

ωµγ j j

j p +

=+

==

Diffusion equation:

,)()( z z y E y A j eeE =−= ω ,)()( z z y J y A j eeEJ =−== ωγ γ

,)()(

x x y Bdy

ydAeeB == .)(

)(1 x x y H

dy

ydAeeH ==

µ

106

Solution of the diffusion equation:



).sinh()cosh()( 2121 pyC pyC e Ae A y A py py +=+= −

)()( y A j y J ωγ −=The current density has to be an even function

( J ( y) = J (-y) ): .02 =C

).cosh()( 1 pyC y A =

)cosh()()( 1 pyC j y A j y E ω ω −=−=Field quantities:

)cosh()()( 1 pyC j y A j y J ωγ ωγ −=−=

)sinh()(

)( 1 py pC dy

ydA y B ==

)sinh()(1

)( 1 pyC p

dy

ydA y H

µ µ ==

107

D i i f h C C h h



Determination of the constant C 1: Current through a

conductor of width b is assumed to be given.

x z

y0Γ

I

0C

2h y =

2h y −=

=−=+=−=⋅=Γ⋅ ∫∫Γ

bh y H bh y H d d C

)2()2(

00

rHnJ

b

,)2sinh(2

)2(2 1 I ph

C pb

bh y H =−==−= µ

.

)2

sinh(21 ph

pb

I C

µ −=

108

IIj



),cosh(

)

2

sinh(2

)cosh(

)

2

sinh(2

)( py ph

b

I p py

ph pb

I j y E

γ

ωµ ==

),cosh(

)2

sinh(2

)()( py ph

b

I p y E y J == γ

),sinh(

)2

sinh(2

)( py ph

b

I y B

µ −=

).sinh(

)2

sinh(2

)( py ph

b

I y H −=

109

Impedance of a conductor of width b and length l:



( ) Γ⋅×−==

∫Γd Z I S nHE

*2

2

1

2

1

y x z y H y E y H y E eeeHE )()()()( *** =×=×

for 2, for 2, otherwise . y y y y h y h= = = − = −n e n e n e

=−=−=+==−= blh

y H h

y E blh

y H h

y E Z I )2

()2

(2

1)

2()

2(

2

1

2

1 **2

bl

h I

x

y

z

Γ

.2

)2

sinh(2

)2

cosh()

2()

2(

*

*bl

b

I

phb

ph I p

blh

y H h

y E

γ

===−=

Impedance of a conductor of width b and length l:

110

ph



,

)2sinh(2

)2

cosh(

phb

ph pl

Z

γ

= .

2

1

2

1

2 2

+=

+=

ωµγ

δ

h jh j ph

Low frequency:

High frequency:

.2

)2

sinh(,1)2

cosh(12

ph ph ph ph≈≈⇒<<

:bhl Z

γ ≈ like D. C. in the entire height h.

).2

sinh()2

cosh(12

ph ph ph≈⇒>>

:2

)1(2 b

l j

b

pl Z

γδ γ +=≈ D. C. resistance of two layers each of

thickness δ , reactance same as

resistance.111

4. Electromagnetic waves



, ,

0, ,

curl curlt t

div div ρ

∂ ∂= + = −∂ ∂

= =

D BH J E

B D , , ( )eε µ γ = = = +D E B H J E E

The full set of Maxwell’s equations:

describes electromagnetic waves.

t

∂

∂

D

( )t H

t

∂−

∂

B

( )t E

The time varying electric and

magnetic fields mutually sustaineach other:

electromagnetic field

in vacuum:

112

4.1 Transmission lines



Transmission line: a very (infinitely) long arrangement of

two parallel conductors of constant cross section:

e. g.:

General transmission line

z

d

0r 0r

Two parallel wires of circular cross section

Coaxial cable

z

z

ir

ar

Material properties:

, ,µ γ ε around the conductors,

lγ in the conductors.

113

Distributed circuit parameters:



Resistance, inductance, conductance and capacitance per unit

length: R, L, G, C . Units: Ω/m, H/m, S/m, F/m.

Series parameters:

z z + dz

1du

2du( , )i z t

( , )i z t

1 2 ,du du du iRdz= + =

R depends on thegeometry and on γ l.

,d iLdzψ = L depends on the

geometry and on µ .

Current field in the

conductorslγ

z z + dz

( , )i z t

( , )i z t

d ψ Magnetic fieldaround the

conductors

µ

114



Parallel parameters:

z z + dz

( , )u z t Electric field

around the

conductors

ε

,di uGdz=

G depends on the

geometry and on γ . z z + dz

( , )u z t di Current fieldaround the

conductors

γ

dQ

dQ−,dQ uCdz=

C depends on thegeometry and on ε .

115



T i i liu i∂ ∂



Transmission line

equations:,

u i Ri L

z t

∂ ∂− = +

∂ ∂

.i u

Gu C z t

∂ ∂− = +

∂ ∂

2 2

2 ,

u i i i i

R L R L z z z t z t z

∂ ∂ ∂ ∂ ∂ ∂

− = + = +∂ ∂ ∂ ∂ ∂ ∂ ∂2 2

2 2( ) .

u u u LC RC LG RGu

z t t

∂ ∂ ∂= + + +

∂ ∂ ∂

Similarly:

2 2

2 2( ) .

i i i LC RC LG RGi

z t t

∂ ∂ ∂= + + +

∂ ∂ ∂118

4.1.1 Solution of the transmission line equations in time harmonic



case

Assumption: ˆ( , ) ( )cos( ( )),uu z t U z t zω ϕ = +

ˆ( , ) ( )cos( ( )).ii z t I z t zω ϕ = +

Complex amplitudes: ( )ˆ( ) ( ) ,u j zU z U z e

ϕ =

( )ˆ( ) ( ) .i j z I z I z e

ϕ =

Transmission line equations in the frequency domain:

( ) ( ) ( ),dU z R j L I zdz

ω − = +

( )( ) ( ).

dI zG j C U z

dzω − = +

119

2 ( ) ( )d d



2

2

( ) ( )( ) ( )( ) ( ).

d U z dI z R j L R j L G j C U z

dz dzω ω ω = − + = + +

2 ( )( ), p R j L G j C ω ω = + +

( )( ) : p R j L G j C ω ω = + + propagation coefficient .

22

2

( )( ).

d U z p U z

dz=

( ) . pz pzU z U e U e+ − −= +

Re( ) 0, Im( ) 0. p pα β = ≥ = ≥, p jα β = +

120



1 ( )( ) , pz pzdU z p p

I z U e U e R j L dz R j L R j Lω ω ω

+ − −= − = −+ + +

0 : R j L R j L

Z p G j C

ω ω

ω

+ += =

+wave impedance.

0 0

( ) . pz pzU U I z e e

Z Z

+ −−= −

121



0 1 2 3 4 5 6 7

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

z

t t t + ∆ z

U e α + −

z v t ∆ = ∆

λ

123



0 1 2 3 4 5 6 7

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6 7

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6 7

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6 7

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6 7

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6 7

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6 7

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6 7

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6 7

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6 7

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6 7

-1

-0.8

-0.6

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-0.2

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6 7

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6 7

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6 7

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6 7

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6 7

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6 7

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6 7

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6 7

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6 7

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6 7

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6 7

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6 7

-1

-0.8

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-0.2

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6 7

-1

-0.8

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0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6 7

-1

-0.8

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0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6 7

-1

-0.8

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0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6 7

-1

-0.8

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0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6 7

-1

-0.8

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0

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0.6

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1

0 1 2 3 4 5 6 7

-1

-0.8

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0

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1

0 1 2 3 4 5 6 7

-1

-0.8

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0

0.2

0.4

0.6

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1

0 1 2 3 4 5 6 7

-1

-0.8

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0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6 7

-1

-0.8

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-0.2

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6 7

-1

-0.8

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0

0.2

0.4

0.6

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1

0 1 2 3 4 5 6 7

-1

-0.8

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0

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1

0 1 2 3 4 5 6 7

-1

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0

0.2

0.4

0.6

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1

0 1 2 3 4 5 6 7

-1

-0.8

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0

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0.6

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1

0 1 2 3 4 5 6 7

-1

-0.8

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0

0.2

0.4

0.6

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1

0 1 2 3 4 5 6 7

-1

-0.8

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0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6 7

-1

-0.8

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0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6 7

-1

-0.8

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-0.4

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0

0.2

0.4

0.6

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1

0 1 2 3 4 5 6 7

-1

-0.8

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0

0.2

0.4

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1

0 1 2 3 4 5 6 7

-1

-0.8

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0

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0.6

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1

0 1 2 3 4 5 6 7

-1

-0.8

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0

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0.4

0.6

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1

0 1 2 3 4 5 6 7

-1

-0.8

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0

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0.4

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1

0 1 2 3 4 5 6 7

-1

-0.8

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0

0.2

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1

0 1 2 3 4 5 6 7

-1

-0.8

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0

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0.4

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1

0 1 2 3 4 5 6 7

-1

-0.8

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0

0.2

0.4

0.6

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1

0 1 2 3 4 5 6 7

-1

-0.8

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0

0.2

0.4

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1

0 1 2 3 4 5 6 7

-1

-0.8

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-0.2

0

0.2

0.4

0.6

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1

0 1 2 3 4 5 6 7

-1

-0.8

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-0.2

0

0.2

0.4

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0 1 2 3 4 5 6 7

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6 7

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6 7

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6 7

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6 7

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6 7

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6 7

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6 7

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6 7

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6 7

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6 7

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6 7

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6 7

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6 7

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6 7

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6 7

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6 7

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6 7

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6 7

-1

-0.8

-0.6

-0.4

-0.2

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0.4

0.6

0.8

1

0 1 2 3 4 5 6 7

-1

-0.8

-0.6

-0.4

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0

0.2

0.4

0.6

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0 1 2 3 4 5 6 7

-1

-0.8

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-0.4

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0

0.2

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0.6

0.8

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0 1 2 3 4 5 6 7

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6 7

-1

-0.8

-0.6

-0.4

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0.4

0.6

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0 1 2 3 4 5 6 7

-1

-0.8

-0.6

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0 1 2 3 4 5 6 7

-1

-0.8

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0 1 2 3 4 5 6 7

-1

-0.8

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0.6

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0 1 2 3 4 5 6 7

-1

-0.8

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0 1 2 3 4 5 6 7

-1

-0.8

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0 1 2 3 4 5 6 7

-1

-0.8

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0 1 2 3 4 5 6 7

-1

-0.8

-0.6

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0

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0.4

0.6

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0 1 2 3 4 5 6 7

-1

-0.8

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0

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0.6

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0 1 2 3 4 5 6 7

-1

-0.8

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0 1 2 3 4 5 6 7

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-0.8

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0 1 2 3 4 5 6 7

-1

-0.8

-0.6

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0 1 2 3 4 5 6 7

-1

-0.8

-0.6

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0 1 2 3 4 5 6 7

-1

-0.8

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0

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0 1 2 3 4 5 6 7

-1

-0.8

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0

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0.6

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0 1 2 3 4 5 6 7

-1

-0.8

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-0.4

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0

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0.6

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0 1 2 3 4 5 6 7

-1

-0.8

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-0.2

0

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0.6

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1

0 1 2 3 4 5 6 7

-1

-0.8

-0.6

-0.4

-0.2

0

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0.6

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1

0 1 2 3 4 5 6 7

-1

-0.8

-0.6

-0.4

-0.2

0

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0.4

0.6

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1

0 1 2 3 4 5 6 7

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6 7

-1

-0.8

-0.6

-0.4

-0.2

0

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0.6

0.8

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0 1 2 3 4 5 6 7

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6 7

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6 7

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6 7

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6 7

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6 7

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6 7

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

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1

0 1 2 3 4 5 6 7

-1

-0.8

-0.6

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0

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0 1 2 3 4 5 6 7

-1

-0.8

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0

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0 1 2 3 4 5 6 7

-1

-0.8

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0 1 2 3 4 5 6 7

-1

-0.8

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0 1 2 3 4 5 6 7

-1

-0.8

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0 1 2 3 4 5 6 7

-1

-0.8

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0 1 2 3 4 5 6 7

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-0.8

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0 1 2 3 4 5 6 7

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-0.8

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0 1 2 3 4 5 6 7

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-0.8

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0 1 2 3 4 5 6 7

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-0.8

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0 1 2 3 4 5 6 7

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0 1 2 3 4 5 6 7

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-0.8

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0 1 2 3 4 5 6 7

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-0.8

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0 1 2 3 4 5 6 7

-1

-0.8

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0 1 2 3 4 5 6 7

-1

-0.8

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0 1 2 3 4 5 6 7

-1

-0.8

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0 1 2 3 4 5 6 7

-1

-0.8

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0 1 2 3 4 5 6 7

-1

-0.8

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0 1 2 3 4 5 6 7

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-0.8

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0 1 2 3 4 5 6 7

-1

-0.8

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0 1 2 3 4 5 6 7

-1

-0.8

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0 1 2 3 4 5 6 7

-1

-0.8

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0 1 2 3 4 5 6 7

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0 1 2 3 4 5 6 7

-1

-0.8

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0 1 2 3 4 5 6 7

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-0.8

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0 1 2 3 4 5 6 7

-1

-0.8

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0 1 2 3 4 5 6 7

-1

-0.8

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1

124

Similarly,



( )( ) pz j zU z U e U e

α β − − − += =

describes an attenuated wave propagating in the negative z-direction:

( , ) cos( ). zu z t U e t z

α ω β − −= +

Condition for the same phase at the time instant t and location z as well as at the time instant t +∆t and at location z+∆ z:

( ) ( )t z t t z zω β ω β + = + ∆ + + ∆ ,t zω β ⇒ ∆ = − ∆ . z

vt

ω

β

∆= = −

∆

: ,attenuation constant α : . phase constant β

125

General solution: ( ) , pz pzU z U e U e

+ − −= +



( ) ,

0 0( ) .

pz pzU U

I z e e Z Z

+ −−

= −

The wave impedance Z 0 is the ratio between the complex

amplitudes of the voltage and current waves propagating in

the positive direction.

The same ratio is - Z 0 for the waves propagating in the

negative direction.

126

4.1.2 Ideal transmission line



0, 0. R G= =

( )( ) 0, . p j L j C j LC LC ω ω ω α β ω = = ⇒ = =

No attenuation!

0 . j L L

Z j C C

ω

ω = = The wave impedance is real!

1.v

LC

ω

β

= = in general: . LC µε =

In vacuum: 8

0 0

12.9979 10 :

mv c

sµ ε = ≈ ⋅ = velocity of light!

127

Ideal transmission line at general time dependence:



Transmission line equations: ,u i

L

z t

∂ ∂− =

∂ ∂

.i u

C

z t

∂ ∂− =

∂ ∂2 2

2 2,

u u LC

z t

∂ ∂=

∂ ∂

2 2

2 2:

i i LC

z t

∂ ∂=

∂ ∂scalar 1D wave equation.

All functions of the form ( , ) ( ) z f z t f t v

= are solutions

of the 1D wave equation1

( ) :v LC

=

2

2 21 ( ), f z f t z v v

∂′′=∂

2

2 21( ) ( ). f z z LC LCf t f t t v v v∂ ′′ ′′= =∂

q. e. d.

128

( ) z

f t v

− describes a wave propagating with the velocity v



vin the positive z-direction:

e.g. ( ) f x− x

( , ) f z t 0t = t t = ∆

z

v t ∆

129

4.1.3 Termination of transmission lines



0U 2 Z

z0 l

( ) I z

( )U z

0(0) ,U U = 2( ) ( ).U l Z I l=

( ) , pl plU l U e U e

+ − −= +0 0

( ) . pl plU U I l e e

Z Z

+ −−= −

2 0 0

1

.1

pl

pl pl pl

pl pl pl

pl

U e

U e U e U e Z Z Z U eU e U e

U e

−

+ − − + −

−+ − −

+ −

++

= =− −

130

2 pl

lU e U− −



2

2 : pl

pl

U e U r e

U e U + − +

= = Reflection coefficient .

22 0

2

1,

1

r Z Z

r

+=

−2 0

2

2 0

. Z Z

r Z Z

−=

+

( ) ( )

2( ) ( ), pl p l z p l zU z U e e r e+ − − − −= +

( ) ( )

2

0

( ) ( ). pl p l z p l zU I z e e r e

Z

+− − − −= −

incident waves reflected waves

131

At the beginning of the transmission line:



At the beginning of the transmission line:2

0 2(0) (1 ), plU U U r e+ −= = +

0

2

2

,1 pl

U U

r e

+−

=+

2

2 . plU U r e

− + −=

If 2 0 2: 0. Z Z r = = No reflection:

( ) ,

pz

U z U e

+ −=0

( ) . pzU I z e

Z

+−= 0

( ).

( )

U z Z

I z=

Matching:

2 02

2 0

. Z Z

r Z Z

−=

+

132

Short circuit :



If 2 20 : 1. Z r = = −

( ) ( ), pl px pxU x U e e e

+ − −= −

Using the notation : x l z= −

0

( ) ( ). pl px pxU I x e e e

Z

+− −= +

Special case: ideal transmission line ( 0, ) p jα β = =

( ) ( ) 2 sin , j l j x j x j lU x U e e e jU e x

β β β β β + − − + −= − =

0 0

( ) ( ) 2 cos . j l j x j x j lU U I x e e e e x Z Z

β β β β β + +

− − −= + =

Standing waves.133

( , )u x t 1t

t t>

Voltage and current in case of short circuit.



0 1 2 3 4 5 6 7-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 6.2832-1

0

1

x

1

2 1t t >

0

3 2t t >

4 3t t >

5 4t t >

0 6.2832-1

0

1

x 1t 2 1t t >

03 2t t >

4 3t t >

( , )i x t

5 4t t >

134

Open circuit :



If 2 2: 1. Z r → ∞ =

For ideal transmission lines:

( ) ( ) 2 cos , j l j x j x j lU x U e e e U e x

β β β β β + − − + −= + =

0 0

( ) ( ) 2 sin . j l j x j x j lU U I x e e e j e x

Z Z

β β β β β + +

− − −= − =

135



4.2 Planar waves



Maxwell’s equations in vacuum (µ = µ 0, ε = ε 0), in

absence of charges and currents ( ρ = 0, J = 0):

0 ,curlt

ε ∂

=∂E

H 0 ,curlt

µ ∂

= −∂H

E

0,div =H 0.div =E2

0 0 0 2,curlcurl graddiv curl

t t ε ε µ

=

∂ ∂= − ∆ = = −

∂ ∂0

HH H H E

2

0 0 2,

t ε µ ∂∆ =

∂HH

Similarly:2

0 0 2.

t ε µ

∂∆ =

∂

EE

vector 3D wave equation

137

Assumption: 0, 0. x y

∂ ∂= =

∂ ∂The electromagnetic field is

constant in any plane z = constant:



y constant in any plane z constant:

Planar waves:2 2

0 02 2,

z t ε µ

∂ ∂=

∂ ∂

H H2 2

0 02 2:

z t ε µ

∂ ∂=

∂ ∂

E E

Vector 1D wave equation.

All components E x, E y, E z, H x, H y, H z of the electromagnetic field

satisfy the scalar 1D wave equation:

2 2

0 02 2. f f

z t ε µ ∂ ∂=

∂ ∂Solution:

0 0

1( ), . z f t v cv µ ε

= =

138

Waves propagating with light velocity:



( , ) ( ), x x

z E z t E t

c

= ( , ) ( ), y y

z E z t E t

c

= ( , ) ( ), z z

z E z t E t

c

=

( , ) ( ), x x

z H z t H t

c= ( , ) ( ), y y

z H z t H t

c= ( , ) ( ).

z z

z H z t H t

c=

Relationship between E and H:

0 ,

x y z

x y z

curl x y z t

H H H

ε ∂ ∂ ∂ ∂

= =∂ ∂ ∂ ∂

e e e

EH

10, 0, .

x y z c t

∂ ∂ ∂ ∂= = =

∂ ∂ ∂ ∂Planar waves:

139

x y ze e e

e e e



1 1 10 0 0 0 1 ( ).

x y z

z

x y z x y z

curl

c t c t c t H H H H H H

∂ ∂ ∂= = = ×

∂ ∂ ∂

e e e

H e H

0

0 0

0

1 1

( ) ( ) ; ( ) . z z zc t t c

ε

ε ε µ

∂ ∂

× = ⇒ × = × =∂ ∂

E

e H e H E e H E

Similarly, from Faraday’s law: 0

0

( ) . z

µ

ε × = ±e E H

Sign above: propagation in the positive z-direction,

Sign below: propagation in the negative z-direction.

140

( , ) ( ) z

z t t c

= −E E ( , ) ( ) z

z t t c

= +H H

direction of propagation direction of propagationS S



( , ) ( ) z z t t c

= −H H

ze

( , ) ( ) z

z t t c= +E E

zep p g p p g

The direction of Poynting’s vector coincides with the

direction of propagation.

2 20 00

0 0 0 0

1( ) , z z z

µ µ µ

ε ε µ ε = × = × × = ± = ±S E H e H H e H e H

2 20 00

0 0 0 0

1[ ( )] , z z z

ε ε ε µ µ µ ε = × = × ± × = ± = ±S E H E e E e E e E

2 2

0 0

1( ).

2 zc ε µ = ± +S e E H

141

More general material properties: , , konstant.µ ε γ =



Maxwell’s equations in absence of charges ( ρ = 0):

2

2,curlcurl graddiv curl curl

t t t γ ε γµ εµ

=

∂ ∂ ∂= − ∆ = + = − −∂ ∂ ∂

0

H HH H H E E

2

2

,t t

γµ εµ ∂ ∂

∆ − − =∂ ∂

H HH 0

Similarly:2

2.

t t γµ εµ

∂ ∂∆ − − =

∂ ∂

E EE 0

, ,

0, 0.

curl curlt t

div div

γ ε ∂ ∂

= + = −∂ ∂

= =

E BH E E

H E

142

For planar waves:2

20, 0 .

x y z

∂ ∂ ∂= = ⇒ ∆ =

∂ ∂ ∂



y

2

2 2,

z t t µε µγ ∂ ∂ ∂= +

∂ ∂ ∂

2

E E E

2

2 2.

z t t µε µγ ∂ ∂ ∂= +

∂ ∂ ∂

2

H H H

Total analogy with the transmission line equations:

2 2

2 2( ) ,

u u u LC RC LG RGu

z t t

∂ ∂ ∂= + + +

∂ ∂ ∂2 2

2 2( ) .

i i i LC RC LG RGi

z t t

∂ ∂ ∂= + + +

∂ ∂ ∂, , 0, , , .u i R L G C µ γ ε ⇔ ⇔ ⇔ ⇔ ⇔ ⇔E H

143

Time harmonic case, complex notation



( ), ( ) : complex amplitudes. z zE H

Solution (due to analogy):

( ) , pz pz z e e

+ − −= +E E E0 0

( ) . pz pz z e e Z Z

+ −−= −

E EH

Propagation coefficient: ( ) , p j j jωµ γ ωε α β = + = +

Wave impedance:0 .

j Z

j

ωµ

γ ωε =

+

Attenuated waves propagating in the positive and

negative z-direction.

144

Lossless medium: 0.γ =



( )( ) 0, . p j j jωµ ωε ω µε α β ω µε = = ⇒ = =

0 . j

Z j

ωµ µ

ωε ε = =

0 0

1 1 1 .r r r r

cv cω β µε µ ε µ ε µ ε = = = = ≤

Optics: ,c

vn

= n: refraction index

Maxwell’s relationship:2; ,r r n nε ε = =

r since for optically transparent media 1.µ =145

4.3 Electromagnetic waves in homogeneous, infinite space



Assumptions:

• current density J and charge density ρ are known everywhere at

any time instant:

• material properties µ and ε are constant everywhere, e. g. µ = µ 0,

ε = ε 0,• lossless medium: γ = 0.

( , ) and ( ,t) are given,t ρ J r r

e. g. antenna:

( , )t J r

homogeneous medium

0 0(e. g. vacuum or air: , )µ ε

electromagnetic field: E(r,t ), H(r,t )

146

4.3.1. Solution of Maxwell’s equations with retarded potentials



Maxwell’s equations:

,curlt

∂= +

∂D

H J ,curlt

∂= −

∂B

E

0,div =B ,div ρ =D

Potentials:0

1, ,curl curl

µ = =B A H A

0 0, .gradV gradV t t

ε ε ∂ ∂= − − = − −∂ ∂A A

E D

0 0, .µ ε = =B H D E

147

2

0 0 0 0 02.

V curlcurl graddiv grad

t tµ µ ε µ ε

∂ ∂= − ∆ = − −

∂ ∂

AA A A J



t t ∂ ∂

The divergence of A can be freely chosen:

0 0 :V

divt

µ ε ∂

= −∂

A Lorenz gauge.

0 0 02 .t µ ε µ

2∂

−∆ + =∂

A

A J

0

( ) .div

div gradV V t t

ρ

ε

∂ ∂− − = −∆ − =

∂ ∂A A

Using the Lorenz gauge:

0 0 2

0

.V

V t

ρ µ ε

ε

2∂−∆ + =

∂

Non-homogeneous 3D

wave equations

148

In the static case ( 0)t

∂=

∂these equations reduce to the

Laplace Poisson eq ations ∆A J V ρ

∆ =



Laplace-Poisson equations 0 ,µ −∆ =A J0

V ε

−∆ =

whose solutions are known to be

0

1 ( )( ) .

4

d V

ρ

πε Ω

′ ′Ω=

′−∫r

rr r

0 ( )( ) ,

4

d µ

π Ω

′ ′Ω=

′−∫J r

A rr r

In regions with vanishing current density and charge density,

one obtains the wave equations

0 0 2,

t µ ε

2

∂−∆ + =∂

AA 0 0 0 20.V V

t µ ε

2

∂−∆ + =∂

149

The solutions of the non-homogeneous wave equations in

infinite free space are



infinite free space are

0

( , )( , ) ,

4

t d ct

µ

π Ω

′−′ ′− Ω=

′−∫

r rJ r

A rr r

0

( , )1

( , ) ,4

t d cV t

ρ

πε Ω

′−′ ′− Ω=

′−∫r rr

rr r

( , ) and ( , ) are the potentials.t V t retarded A r r

0 0

1( ).c

µ ε =

150

Time harmonic case:



( ), ( ), ( ) and ( ) are complex amplitudes.V ρ ′ ′A r r J r r

ˆ( , ) ( ) cos[ ( ) ( )].t t c c

ω ϕ ′ ′− −

′ ′ ′− = − +r r r r

J r J r r

In frequency domain: 0( )ˆ ( ) ( ) , j

jk j ce e eω

ϕ

′−− ′− −′′ ′=

r r

r rrJ r J r

0 0 0 : wave number (phase factor).k cω ω µ ε = =

0

0 ( )( ) ,

4

jk e d µ

π

′− −

Ω

′ ′Ω=

′−∫

r rJ r

A r

r r

0

0

1 ( )( ) .

4

jk e d

V ρ

πε

′− −

Ω

′ ′Ω=

′−∫

r rr

r

r r

In time domain:

151

Using the Lorenz gauge, V can be eliminated:

V∂



0 0

V div

t

µ ε ∂

= −

∂

A is in the frequency domain 0 0 .div j V ωµ ε = −A

0 0

1.V div

jωµ ε = − A

,curl=B A

0 0

1 j gradV j graddiv j

ω ω ωµ ε

= − − = − + =E A A A

2

0 0

0 0

1( ).graddiv

jω µ ε

ωµ ε = +A A

2

0

0 0

1( ).k graddiv

jωµ ε = +E A A

152

4.4 Guided waves

Transmission lines: transversal (x y) dimensions are much



Transmission lines: transversal ( x, y) dimensions are much

smaller than the wave length.

D

2 2 1.

v D

k f f

π π λ

ω µε µε = = = =

If this condition is not fulfilled: waveguide.

e. g.: cylindrical

waveguideMetallic tube( )γ → ∞

z

a

b ,µ ε

153

Assumptions:



Assumptions:

• sinusoidal time dependence: all quantities are

complex amplitudes,

• material properties are constant:

µ , ε = constant,

• Lossless medium: γ = 0,

• No free charges: ρ = 0.

154

4.4.1 TM and TE waves



Wave equations for the potentials A and V :

2,

t µε

2∂−∆ + =

∂

AA 0

20.

V V

t µε

2∂−∆ + =

∂

In time domain:

In frequency domain:

2 ,ω µε −∆ − =A A 02 ,V V ω µε −∆ − = 0

2 .k −∆ − =A A 0:k ω µε =

155

Electromagnetic field in case ( , , ) ( , , ) : z

x y z A x y z=A e

e e e



1 1 1 1( ) ,

0 0

x y z

z x y A Acurl A x y z y x

A

µ µ µ µ ∂ ∂ ∂ ∂ ∂= = = −∂ ∂ ∂ ∂ ∂

e e e

H e e e

The magnetic field is transversal: TM waves.

21[ ( )] z zk A graddiv A

jωµε

= + =E e e

2 2 22

2

1[ ( ) ]. x y z

A A Ak A

j x z y z zωµε

∂ ∂ ∂= + + +

∂ ∂ ∂ ∂ ∂e e e

0 : the longitudinal component of the magnetic field is zero. z H =

156



1

Using the Lorenz gauge, ψ can be eliminated:



1.div j div

j

ωµεψ ψ

ωµε

= ⇒ =F F

,curl=D F21 ( ).k graddiv

jωµε = − +H F F

1 j grad j graddiv

jω ψ ω

ωµε = − = − =H F F F

21( ).graddiv

jω µε

ωµε = − +F F

158

( , , ) ( , , ) : z

x y z F x y z=F e

e e e

Electromagnetic field in case



1 1 1 1( ) ,

0 0

x y z

z x yF F curl F x y z y x

F

ε ε ε ε ∂ ∂ ∂ ∂ ∂= = = −∂ ∂ ∂ ∂ ∂

e e e

E e e e

21[ ( )] z zk F graddiv F

jωµε

= − + =H e e

2 2 22

2

1[ ( ) ]. x y z

F F F k F

j x z y z zωµε

∂ ∂ ∂= − + + +

∂ ∂ ∂ ∂ ∂e e e

0 : the longitudinal component of the electric field is zero. z E =

The electric field is transversal: TE waves.

159



The general solution of Maxwell’s equations inhomogeneous media can be written as the superposition

of TM and TE waves.

Hence, the single component vector potentials allow the

description of the electromagnetic field by means of twoscalar functions.

160

4.4.2 Waves in rectangular waveguides

z



z

x

y

0,0 x a=

y b=

Assumption:Wave propagation in the

positive z-direction.

TM waves: ( , , ) ( , , ) ( , ) , j z

z z x y z A x y z A x y e β −= =A e e

TE waves: ( , , ) ( , , ) ( , ) . j z

z z x y z F x y z F x y e β −= =F e e

Boundary conditions: on the walls of the waveguide.× =E n 0

0, 0 at 0 and , y z E E x x a= = = =

0, 0 at 0 and . x z E E y y b= = = =161

22

2, . j

z z β β

∂ ∂= − = −

∂ ∂Boundary conditions for the potentials:



2 2 22

2

1

[ ( ) ] x y z

A A A

k A j x z y z zωµε

∂ ∂ ∂

= + + + =∂ ∂ ∂ ∂ ∂E e e e

2 21[ ( ) ]. x y z

A A j j k A

j x y β β β

ωµε

∂ ∂= − + + −

∂ ∂e e e

TM waves:

0 at 0, , 0 and , A x x a y y b= = = = =

TE waves:1 1

. x y

F F

y xε ε

∂ ∂= −

∂ ∂E e e

0 at 0 and ,

F

x x a x

∂

= = =∂

0 at 0 and ,F

y y b y

∂= = =

∂

Dirichlet B.C.

Neumann B.C.

162



2 2( ) ( )d X x d Y y

The following ordinary differential equations are obtained:



2 2

( ) ( )( ), ( ).

d X x d Y y fX x gY y

dx dy

= =

2 2, : x y f k g k = − = − 1 2( ) cos sin , x x x x X x C k x C k x= +

1 2( ) cos sin . y y y yY y C k y C k y= +

Solution:

boundary conditions: 1(0) ( ) 0 0, , 1,2,... , x x X X a C k m ma

π = = ⇒ = = =

1

(0) ( ) 0 0, , 1,2,... . y y

Y Y b C k n nb

π = = ⇒ = = =

164



2 ,k −∆ − =F F 0 .k ω µε =

( , , ) ( , ) : j zx y z F x y e

β −=F e

TE waves:



( , , ) ( , ) : z x y z F x y eF e

2 22 2

2 20.

F F F k F

x y β

∂ ∂− − + − =

∂ ∂

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

2 22 2

2 2

( ) ( )( ) ( ) ( ) ( ) ( ) 0,

d X x d Y yY y X x k X x Y y

dx dy β − − + − =

2 22 2

2 2

( ) ( )

1 ( ) 1 ( )0.

( ) ( ) f x g y

d X x d Y yk

X x dx Y y dy

β − − + − =

Solution by separation:

This is only possible if ( ) constant, ( ) constant. f x g y= =166

2 2( ) ( )d X x d Y y

The following ordinary differential equations are obtained:



2 2

( ) ( )( ), ( ).

d X x d Y y fX x gY y

dx dy

= =

2 2, : x y f k g k = − = − 1 2( ) cos sin , x x x x X x C k x C k x= +

1 2( ) cos sin . y y y yY y C k y C k y= +

Solution:


2

(0) ( )0 0, , 0,1,2,... , x x

dX dX aC k m m

dx dx a

π = = ⇒ = = =

2

(0) ( )0 0, , 0,1,2,... .

y y

dY dY bC k n n

dy dy b

π = = ⇒ = = =

167

1 1 : x yC C C = ( , , ) cos( )cos( ) . j zm nF x y z C x y e

a b

β π π −=

1 1∂ ∂ C

TEmn waves:



1 1:

x y

F F

y xε ε

∂ ∂= −

∂ ∂E e e cos( )sin( ) , j z

x

C n m n E x y e

b a b

β π π π

ε

−= −

sin( ) cos( ) , j z

y

C m m n E x y e

a a b

β π π π

ε

−=

0. z E =

sin( )cos( ) , j z

x

C m m n H x y e

a a b

β β π π π

ωµε

−= −

cos( )sin( ) , j z

y

C n m n H x y eb a b

β β π π π

ωµε

−= −

2 2( )cos( )cos( ) . j z

z

C k m n H x y e

j a b

β β π π

ωµε

−−=

2 21[ ( ) ] : x y z

F F j j k F

j x y β β β

ωµε

∂ ∂= + + −∂ ∂

H e e e

168

Satisfaction of the separation equation:2 2 0. f g k β − − + − =

2 2 2 2 2 2m nπ π



2 2 2 2 0, x yk k k β + + − = 2 2, , x y

m nk k k

a b

π π ω µε = = = .

2 2 2 2( ) ( ) 0,m n

a b

π π β ω µε + + − = m, n = (0), 1, 2, ... .

At given values of n and m (wave modes), the angular

frequency is not arbitrary: cannot be negative!2 β

If 2 0, β < one had , : j z z j j e e

β α β α β α −= ± − = ⇒ =

attenuation only, no wave propagation.

169

2 2 2 2 2 2( ) ( ) 0 ( ) ( )m n m nπ π π

β ω µε ω≥ ⇒ ≥ +



2

( ) ( ) 0 ( ) ( ) .

g f

a b a b

π

β ω µε ω µε

= − − ≥ ⇒ ≥ +

2 21( ) ( ) .

2g

m n f f

a bµε ≥ = +

f g: cut-off frequency.

Any particular mode is only propagable above the cut-off

frequency.

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