emlab 1 solution of maxwell’s eqs for simple cases
TRANSCRIPT
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EMLAB
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Solution of Maxwell’s eqs for simple cases
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EMLAB
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,J ,J
Domain : infinite space
Domain : interior of a rectangular cavity
,J,J
Domain and boundary conditionsThe constraints on the behavior of electric and magnetic field near the interface of two media which have different electromagnetic properties. (e.g. PEC, PMC, impedance boundary, …)
Domain : interior of a circular cavity
waveguide
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EMLAB
31-D example : Radiation due to Infinite current sheet
1. Using phasor concept in solving Helmholtz equation,
2
,~~~ 22
ckk JAA
xy
z
2. With an infinitely large surface current on xy-plane, variations of A with coordinates x and y become zero. Then the Laplacian is reduced to derivative with re-spect to z.
JAA ~~~
22
2
kz
)(ˆ 0 zJ sxJ
3. If the current sheet is located at z=0, it can be repre-sented by a delta function with an argument z. If the current flow is in the direction of x-axis, the only non-zero component of A is x-component.
)(~~
~
02
2
2
zJAkz
Asx
x
),(~
),(~
),(~
2
22 rJrArA
c
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EMLAB
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4. With a delta source, it is easier to consider first the re-gion of z≠0.
0~
~2
2
2
xx Ak
z
A
kz
kz
e
CA
jkz
x
sin
cos~
5. Four kinds of candidate solutions can satisfy the differ-ential equation only. Of those, exponential functions can be a propagating wave..
6. Solutions propagating in either direction are
)cos(}Re{
)cos(}Re{
)0(
)0(~)(
)(
2
1
kzte
kzte
zeC
zeCA
kztj
kztj
jkz
jkz
x
)0( z)0( z
7. The condition that A should be continuous at z=0 forces C1=C2
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EMLAB
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ssxx JdzzJdzAkdz
z
A00
22
2 ~)(
~~~
)0(2~~
2
2
jkCjkCejkCez
Adz
z
A jkjkxx
zjkx eCA
8. To find the value of C, integrate both sides of the orig-inal Helmholtz equation.
011~~~
0
0
jk
e
jk
eCdzAdzAdzA
jkjk
xxx
jk
JCJjkC s
s 2
~~
2 00
zjksx e
jk
JzA
2
~)(
~ 0
)0or 0(2
~~ˆ
~ 0
zzeJ
z
A zjksx yAB
)0()space free(377,2
~ˆ
~ 0 zjks e
Jj xAE
t
stjzjks dczJdee
kj
Jt
)/||(
2ˆ
2
)(~
2
1ˆ),( 0
||0 xxrA
2
)/||(ˆ),( 0 czJ
t s
xrB
2
)/||(ˆ),( 0 czJ
t s
xrE
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EMLAB
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)(ˆ 0 zJ sxJ
E
E
H
H
Propagating direction
An infinite current sheet generates uniform plane waves whose amplitude are uniform throughout space.
Plane wave 정의
E
H
Electric field : even symmetry
Magnetic field : odd symmetry
ttJ s cos)(0
Propagating direction
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EMLAB
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,JSource
JAA 22 k
Infinitesimally small current element in free space : 3D
JA
A
2
2
22 1
tc
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EMLAB
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Solution of wave equations in free space
JA
A
2
2
22 1
tc
2
2
22 1
tc
•Boundary condition: Infinite free space solution.
1. As the solutions of two vector potentials are identical, scalar potential is consid-ered first.
2. To decrease the number of independent variables (x, y, z, t), Fourier transform rep-resentation is used.
~~~
),(),(~,),(),(~
2
22
c
dtetdtet tjtj rrrr
3. For convenience, a point source at origin is considered.
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EMLAB
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)(22 r gkg
022 gkg
0)()(
0)(1 2
2
22
2
2
rgkr
rggk
r
rg
r
Green function of free space
ck
where
A suitable solution which is propagating outward from the origin is e-jkr.
kr
kr
e
r
Ag
jkr
cos
sinr
Aeg
jkr
1. The solution of the differential equation with the source function substituted by a delta function is called Green g, and is first sought.
2. With a delta source, consider first the region where delta function has zero value. Then, utilize delta function to find the value of integration constant.
3. With a point source in free space, the solution has a spherical symmetry. That is, g is independent of the variables , , and is a function of r only.
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EMLAB
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1)(22 VVVdgdkgd r
)0(01)1(4
4
sin
0
2
2
0 0 0
222
jk
jkr
jkr
V
ekA
drreAk
drddrr
eAkgdk
)0(4)1(4
sin)1(
0
22
0 2
2
AejkA
ddrr
ejkrA
dggd
jk
jk
SVa r
ergA
jkr
4)(,
4
1
.,4
),( rrrr
RR
eg
jkR
Green function of free space
4. To determine the value of A, apply a volume integral operation to both sides of the differential equations. The volume is a sphere with infinitesimally small radius and its center is at the origin.
5. With a source at r’, the solution is translated such that
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EMLAB
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~~~ 22 k
dV
)()'(~)(~ rrrr
)(),(),( 22 rrrrrr gkg rrrr
RR
eg
jkR
,4
),(
V
jkR
Vd
R
edg
4
)(~),(
)(~),(
~ rrr
rr
V
V
cRtj
tj
V
jk
tj
dR
cRt
ddeR
dede
det
4
)/,(
),(~2
1
4
1
4
),(~
2
1
),(~
2
1),(
)/(
r
rr
r
6. As the original source function can be represented by an integral of a weighted delta function, the solution to the scalar potential is also an integral of a weighted Green function.
7. Taking the inverse Fourier transform, the time domain solution is obtained as follows.
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EMLAB
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V
dR
cRtt
4
)/,(),(
rr
Retarded potential
(Retarded potential)
V
dR
cRtt
4
)/,(),(
rJrA
0
t
A
The distinct point from a static solution is that a time is retarded by R/c. This newly derived potential is called a retarded potential.
The vector potential also contains a retarded time variable.
Those A and are related to each other by Lorentz condition.
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EMLAB
13Solution in time & freq. domain
' 4
)/,(),(
Vd
R
cRtt
r
r
V
dR
cRtt
4
)/,(),(
rJrA
V
jkR
dR
e
4
)(),(
rr
V
jkR
dR
e
4
)(),(
rJrA
V
jkR
V
jkR
V
jkR
deR
jkRd
R
e
dR
e
2
1ˆ4
1)(
4
4
)(),(
1),(
JRrJ
rJrArH
V
jkR
dR
kRjkR
R
kRjkR
R
e
k
)(
)(33)(1
4
1),(
4
2
2
2
2JRRJrE
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EMLAB
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V
jkR
dR
ej
4)ˆ(ˆ)( JRRJrE
V
jkR
dR
ejk
)ˆ(
4)( JRrH
Far field approximation
Electrostatic solution
V R
d2
ˆ
4
11),(
RJArH
Vd
R
24
ˆ)(),(
RrrE
Biot-Savart’s law
0k
Coulomb’s law
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EMLAB
15Electric field in a phasor form
V
V
jkR
VV
dtR
dR
ej
dgj
gdj
)]ˆ(ˆ[4
1
4)]ˆ(ˆ[
)(1
'
JRRJ
JRRJ
rJJE
jj
t
)( AA
AE
''
)(1
)]()([1
VVdg
jdRg
jj
rJrJ
A
R
eRgdRg
jkR
V
4)(,)()(
'
rJA
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EMLAB
16Radiation pattern of an infinitesimally small current
R
ezIjd
R
ej
jkR
V
jkR
4sinˆ
4)]ˆ(ˆ[
JRRJE
sinˆ)cosˆˆ()ˆ(ˆ
ˆ
00
0
JzJ
Jz
rJrrJ
J
ˆ
ˆˆ
0sincos
cossincossinsin
sincoscoscossin
ˆ
ˆ
ˆ r
z
y
x
z
r
sinz
I
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EMLAB
17Example – wire antenna
coscos222 zrzrrrR rr
z
o
r
C
zjkjkr
V
jkR
zdezJr
ej
dR
ej
cos)(4
sin
4)ˆ(ˆ)( JRRJrE
2/l
2/l
02/)2/(sin
2/0)2/(sin)(
0
0
zlzlkI
lzzlkIzJ
2coscos
2cos
2sin 0 klkl
r
eIj
jkr
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EMLAB
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r
z
z
o
r
R coscos222 zrzrzrR
cosz
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EMLAB
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C
zjkjkr
zdezJr
ej
cos)(
4sin)(rE
N
n
dnjkneIAF
1
cosArray factor :
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EMLAB
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Typical array configurations
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EMLAB
21Array antenna
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EMLAB
22Poynting’s theorem and wave power
)()()( ttt HES
Electromagnetic wave power per unit area(Poynting vector)
}Re{2
1 *HES
Average wave power per unit area
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EMLAB
23Derivation of Poynting’s theorem
t
t
DJH
BE
)2(
)1(
t
t
DEJEHE
BHEH
)(2
1
2
1)(
)2()1(
BHDEJEHE
BH
DEJEHEEH
tt
tt
VVS d
dt
ddd HHEEJEaHE
2
1
2
1)(
,J
HEL WWP ,,
VS
Thermal loss
EW
LPHW
Electric energy
Magnetic energy
Electromagnetic wave power per unit area(Poynting vector)
)()()( ttt HES