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transcript
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Auxiliary material to the lecture
Theorie der Elektrotechnik (engl.)
(437.253)
by
Univ.-Prof. Dipl.-Ing. Dr.techn. Oszkár Bíró
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. ,
. ,
. 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
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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
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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:
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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
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2. Quasi-static field
∂
∂>>
t
DJ
,
,
0,
curl
curlt
div
=
∂= −
∂=
H J
BE
B
).(, eEEJHB +== γ µ
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,
,
0,
,
curlt
curlt
div
div ρ
∂= +∂
∂= −
∂
=
=
DH J
BE
B
D
3. Electromagnetic waves
).(,, eEEJHBED +=== γ µ ε
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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
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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
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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 =
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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
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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
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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
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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)
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( ) ( )∫∫ ∫∫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.
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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
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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.
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∫∫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.
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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
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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.
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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
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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.
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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 .
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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:
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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!
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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
∂
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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 ′+′+′=′
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( ) ( ) ( ) ( )[ ] =′
−+′
−+′
−′
−−=∂
∂ −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
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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 ′+′=′
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( ) ( )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.
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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
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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
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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
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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 ===
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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),,(
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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.
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( )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
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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
( )
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( ) ( )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:
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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
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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
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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± ±± ±= +∑
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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
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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
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),(
)(
),(
)(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:
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).()( 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 ± ± ± ±= + + + +∑
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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
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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:
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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
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.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.
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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
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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
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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
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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
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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 − − − − =
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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
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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 +=+−
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+
+=
−
−−
−−−
−
−−−
−
−
−
−
−
−−
−
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.
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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.
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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
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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
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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.
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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)
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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
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Linear interpolation of the potential function within elements
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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
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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
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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:
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,
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
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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
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8-node quadrilateral elements
13
2
4
5
6
7
8
72
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20-node hexahedral elements
73
Shape and basis functions for 20-node hexahedral elements
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Shape and basis functions for 20 node hexahedral elements
74
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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
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2.5.1 Planar 2D problems
.),(),(,),(:0 y y x x z y x B y x B y x J z
eeBeJ +===∂
∂
Magnetostatic field:
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
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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
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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
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Boundary conditions:
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
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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
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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
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2.5.2 Axisymmetric 2D problems
.),(),(,),(:0 z zr r zr B zr B zr J eeBeJ +===∂
∂φ
φ
Magnetostatic field:
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
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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
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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
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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
Boundary conditions:
Prescription of Bn or of H t on Γ.
87
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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
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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-
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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
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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
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lΩ
iΩ
)(t J
Example:
92
Boundary and interface conditions:
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iΩ
non-conducting region
JlΩ
conducting region
liΓ ,: nH × nB ⋅ continuous
: BΓ b−=⋅nB
: HiΓ
=×nH
: E Γ 0nE =×
: HlΓ 0nH =×
n
n
n
Boundary and interface conditions:
93
Summary
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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
boundary conditions:
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
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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
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( )* * * * *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:
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=+×+=×= )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
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=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
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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.
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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
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,)()( 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
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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
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3.1.2 Current flow in an infinite conducting plate
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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:
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).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
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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
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),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:
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( ) 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
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,
)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
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, ,
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
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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:
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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
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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
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T i i liu i∂ ∂
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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
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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
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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
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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
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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
z
t t t + ∆ z
U e α + −
z v t ∆ = ∆
λ
123
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0 1 2 3 4 5 6 7
-1
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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
-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
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
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0
0.2
0.4
0.6
0.8
1
124
Similarly,
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( )( ) 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
+ − −= +
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( ) ,
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
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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:
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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
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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
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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− −
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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:
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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 :
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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.
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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 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 :
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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
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4.2 Planar waves
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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:
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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:
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( , ) ( ), 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
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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
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( , ) ( ) 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 =
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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
∂ ∂ ∂= = ⇒ ∆ =
∂ ∂ ∂
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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
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( ), ( ) : 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 =
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( )( ) 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
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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
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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
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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
∆ =
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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
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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:
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( ), ( ), ( ) 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∂
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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
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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:
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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
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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
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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
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1
Using the Lorenz gauge, ψ can be eliminated:
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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
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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
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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
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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:
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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
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2 2( ) ( )d X x d Y y
The following ordinary differential equations are obtained:
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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
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2 ,k −∆ − =F F 0 .k ω µε =
( , , ) ( , ) : j zx y z F x y e
β −=F e
TE waves:
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( , , ) ( , ) : 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:
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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:
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:
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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
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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≥ ⇒ ≥ +
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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.