1 emlab radiation principles. 2 emlab radiation field 계산 식 유도
TRANSCRIPT
1
EMLAB
Radiation principles
2
EMLAB
Radiation field 계산 식 유도
R
l
t
IlI
R
ej
Rt
jkR
/4)(
4)(rE
)(tI
3
EMLAB
Basic laws of EM theory
0
B
D
DJH
BE
t
t1) Maxwell’s equations
0
t
J2) Continuity equation (the relation between current
density and charge density in a space)
)(0
0
MHB
PED
3) Constitutive relation (explains the properties of materials)
4) Boundary conditions ( should be satisfied at the interface of two materials by E, H, D, B.)
4
EMLAB
0)1( B
Potentials of time-varying EM theory
AB 0)( A
tt
tt
AE
AE
AE
AE
0)(
t
BE)2(
t
D
JH)3(
ttt
AJ
EJA
1
AJA
Att
2
22
D)4(
t
t
)(2 A
A
0
0)(
A
Vector identity
5
EMLAB
AB
To find unique value of vector potential A, the divergence and the curl of A should be known.
Only the curl of A is physically observed, divergence of A can be arbitrarily set.
0
t
A
For the above choice, (3), (4) become
Lorentz condition
JA
AJA
A
2
2
22
2
22 1
tct
2
2
22
2
22 1
tct
0
t
A Lorentz condition
Lorentz condition
6
EMLAB
Solution of wave equations in free space
JA
A
2
2
22 1
tc
2
2
22 1
tc
•boundary condition : free space
1. Because the number of variables are as many as four (x, y, z, t), we apply Fourier transform to the above equations.
dedec
detdet
tjtj
tjtj
~
2
1~~
2
1
),(~2
1),(,),(
~
2
1),(
2
22
rrrr
2. For a non-homogeneous differential equation, it is easier to substitute the source term with a delta function located at origin. (effectively it is an impulse response.)
~~~
2
22
c
7
EMLAB
)(22 r gkg
1. The impulse response with the source replaced with a delta function is called a Green function g.
2. For the differential equation with a delta source, a solution is sought first in the region other than origin. Then an integration constant is generated and its value can be found by the delta function.
022 gkg
3. For free space, the green function of the point source is spherically sym-metric. That is, g is a function of r only.
0)()(
0)(1 2
2
22
2
2
rgkr
rggk
r
rg
r
Scalar Green function of free space
ck
where
kr
kr
e
r
Ag
jkr
cos
sinr
Aeg
jkr
8
EMLAB
4. To find the value of A, a volume integral is performed for the sphere at the origin and of radius .
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
5. If the source is located at r’, the solution is
',4
)',( rrrr
RR
eg
jkR
Green function of free space
9
EMLAB
~~~ 22 k ')'()'(~)(~
' d
Vrrrr
)'()',()',( 22 rrrrrr gkg ',4
)',( rrrr
RR
eg
jkR
''
'4
)'(~')',(
)'(~),(
~V
jkR
Vd
R
edg
r
rrr
r
6. The source can be represented by an integral of weighted delta functions, the scalar potential is
'
'
)/(
'
'
'4
)/,'(
'),'(~2
1
4
1
''4
),'(~
2
1
),(~
2
1),(
V
V
cRtj
tj
V
jk
tj
dR
cRt
ddeR
dede
det
r
r
rr
r
rr
rr
10
EMLAB
'
'4
)/,'(),(
Vd
R
cRtt
r
r
Retarded potential
(Retarded potential)
'
'4
)/,'(),(
Vd
R
cRtt
rJ
rA
In the same manner, a vector potential is obtained.
The potentials A and is not independent. They are related by Lorentz condition.
0
t
A
In the electro-dynamic solution, the time variable is retarded by the distance from the source.
11
EMLAB
z4
)/(
R
cRtI
)(tI
Observation point
Source current
R
Due to the finite speed of electromagnetic waves, an observed signal at a distance R is delayed by R/c.
Retarded potential
12
EMLAB
Electric field in a phasor form
'
' 4
2
2
2
''
'4
)]ˆ(ˆ[
')()(33)(1
4
1
)(1
'
V
jkR
V
jkR
VV
dR
ej
dR
kRjkR
R
kRjkR
R
ej
dgj
gdj
JRRJ
JRRJ
rJJE
jj
t
)( AA
AE
''
)(1
)]()([1
VVdg
jdRg
jj
rJrJ
A
R
eRgdRg
jkR
V
4)(,')()(
'
rJA
JJRR jkRjkR e
R
jkRe
R
kRjkRg
35
2
4
1)(
4
)(33
13
EMLAB
Example – 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
14
EMLAB
Green’s function
dtftf )()()(
2
)()|(2 rrrr G
V
dG )|(
)()( rr
rr
V
Vd
V rrP
rrrrP
),(
,0)()( V)(rP
V
V
V
d
dG
dG
)()(
)|()(
)|()(
2
22
rrr
rrr
rrr
15
EMLAB
)()|(2 rrrr G
2
1. G 는 원래 미분 방정식을 만족 .
2. G 는 경계조건도 만족시켜야 한다 .
)()|()( 22 rrrr Gk )( 22 k
Poisson’s equation
Helmholtz’ equation
JEE jk 2
)(2 rrIGG kVector wave equation
rrrr
4
1)|(G
Free space scalar green function
rrrr
rr
4)|(
jkeG
rrIrrG
rr
41
)|(2
jke
k
Green’s function for physical laws
16
EMLAB
Domain
,J
,J
경계가 infinite space 인 경우
17
EMLAB
V
jkRe
e
V
jkRe
0
0
e
dR
eρ
dR
e
μ
j
4
4
1
JA
AH
AE
A
A
Example - Free space
,J
Boundary condition : radiation condition
Source
JAA 22 k
18
EMLAB
Example – Parallel plate
JA
JAA
22
2
2
2
2
2
22
kzyx
k
x
yz
V ee
V e0
0
e
gdρ
gd
μ
j
JA
AH
AE
A
A
1
J0z
dz
nyx
ykxkjyxx
x
dkdkekkAd
znA
JAkzyx
yx )(
02
2
2
2
2
2
2
),(sin
19
EMLAB
Derivation of dyadic Green’s function for vector wave eq.
''
jji,
'
jji
2
2ji,
2
'''1
1
VVV
djdJGjdJxx
g
kgj
kj
jjj
JG
AAA
AAE
222 )'()'()'(,4
zzyyxxRR
eg
jkR
R
xx
R
ejkR
x
R
R
g
x
g jkr )(
4
)1( jj
2jj
R
xx
x
R )( jj
j
jiijjijiji
2
x
R
R
g
xx
R
x
R
R
g
Rx
R
R
g
xx
g
xxx
g
ji3
jjii
2
jjii
2
2
jiijj2
2
))(())((
x
R
R
g
xR
xxxx
R
g
R
xxxx
R
g
x
R
R
g
xx
R
x
R
RR
g
x
R
R
g
JJRR jkRjkR
jkRjkR
eR
jkRe
R
kRjkR
eR
jkRxxxxe
R
kRjkR
RR
g
R
xxxx
R
g
R
gR
35
2
ji,3jjii5
2
ji,3
jjii2
2
4
1)(
4
)(33
4
1))((
4
)(33
))((
'
)'(
'
2
2
dj
k
jk
V
JGE
rrIGG
JEE
안테나의 radiated E field 를 구하기 위한 Green function.
Free space
rrR
20
EMLAB
)()(33)(1
4
1
)(1)(
)(33
4
1
4
)(1)(
4
)(33
1
4
)(1))((
4
)(33
4
2
2
2
2
2
2
4
2
2
3
2
5
2
2
jji,ji
2
2jji,i
ji,3
2
jjii5
2
ji,2
JRRJ
JJRR
JJRR
JG
R
kRjkR
R
kRjkR
R
e
k
R
kRjkR
R
kRjkR
R
e
k
R
kRjkR
R
kRjkR
k
e
Jgxx
g
kJG
eR
kRjkRxxxxe
R
kRjkRGk
jkR
jkR
jkR
jkRjkR
21
EMLAB
Equivalence theorem for Electric field E
)'(, 22 rrIGGJEE kjk
')'()ˆ()ˆ(')|'()'(
')'()ˆ()'(ˆ')|'()'()(
')ˆ()'()'(ˆ
')''()''(
dajdj
dadj
da
d
SV
SV
S
V
GEnGHnrrGrJ
GEnGEnrrGrJrE
BnABAn
BABA
S
JiE
n
V
')|'()'()(
')'()ˆ()ˆ()(
dj
daj
V
i
S
i
rrGrJrE
GEnGHnErE
)(rE
Enˆ
Hnˆ Scatterer 의 표면에서 E 와 H 의 접선 성분만 알면 공간 내의 모든 점에서 E, H 를 구할 수 있다 .
22
EMLAB
'
)'(
'
2
2
dj
k
jk
V
JGE
rrIGG
JEE
Derivation of Equivalence theorem for E-field
arrIaGaG
JEE
)'(2
2
k
jk
aErraGEaGE
JaGaGEEaG
)'()()(
)()()(2
2
k
jk
arEJaGaGEEaG )'()()()(VV
djd
S
V
d
d
aQPPQ
PQQP
arEJaGSEaGaGE )'()()()(VS
djd
SV
ddj SEaGaGEJaGarE )()()()'(
SV
dSdj )ˆ()()()ˆ()()'( EnaGaGEnJaGarE
SV
dSjdj )()ˆ()()ˆ()()'( aGHnaGEnaGJarE
SV
dSjdj ')'|()'ˆ()'|(')'ˆ(')'|()( rrGHnrrGEnrrGJrE
SJiE
'n
V
)(rE
En'ˆ
Hn'ˆ
'n
(Vector Green’s identity)
gxx
g
kG ji,
ji
2
2ji,
1)|'( rrG
SV
dSjdj )|'()ˆ()|'()ˆ()|'()()'( rrGHnrrGEnrrGrJrE
g
xx
g
kG ji,
ji
2
2ji,
1)',( rrG
(a 는 임의의 상수 벡터 .)
23
EMLAB
Equivalence theorem
)'(, 22 rrIGGJEE kjk
')'()'ˆ()'ˆ(')'|()'(
')'()'ˆ()'('ˆ')'|()'()(
')ˆ()'()'(ˆ
')''()''(
dajdj
dSdj
dS
d
SV
SV
S
V
GEnGHnrrGrJ
GEnGEnrrGrJrE
BnABAn
BABA
S
JiE
'n
V
')'|()'()(
')'()ˆ()ˆ()(
dj
daj
V
i
S
i
rrGrJrE
GEnGHnErE
)(rE
Enˆ
Hnˆ
Scatterer 의 표면에서 E 와 H 의 접선 성분만 알면 공간 내의 모든 점에서 E, H 를 구할 수 있다 .
면적 요소 n 의 방향이 바뀜 .
g
gg
ggkgk
gkgk
')'ˆ()',(')'ˆ(
]')'ˆ[()'()'ˆ()',(')'ˆ(
')](')('''[1
)',('
])(''[1
)',(
22
22
EnrrGEn
aEnaEnarrGEn
aaaarrG
aaarrG
24
EMLAB
SV
dajdj ')'|(')'ˆ()'|()'ˆ(')'|()( rrGEnrrGHnrrGJrE
SV
dajdj ')'ˆ()'|(')'ˆ()'|(')'|()( TTT EnrrGHnrrGJrrGrE
g
xx
g
kG ji,
ji
2
2ji,
1)',( rrG
R
eg
jkR
40
; for free space
)'(')(33)(1
4
14
2
2
2
20 JRRJJGR
kRjkR
R
kRjkR
R
e
k
jkR
Equivalence theorem with free space kernel : G0
]'ˆ)'ˆ[(44
)1(]'ˆ)'ˆ[(')'ˆ()',(')'ˆ(
20 REnREnEnrrGEn
R
ejke
R
jkRg
jkRjkR
R
e jkR
4)'ˆ('ˆ
0
JRRJJG
S
jkR
V
jkR
daR
ejd
R
ej '
4
)'ˆ(ˆ)]'ˆ(ˆ[ˆ)'ˆ('
4)'ˆ('ˆ)(
EnR
HnRRHnJRRJrE
Free space Green function 을 이용하여 만든 Green’s theorem 식 .
S
JiE
'n
V
)(rE
Enˆ
Hnˆ
Infinite free space
25
EMLAB
PEC
'S)(rE
,JiE
n
'V
. )(
같다는변하지만
함께도따라선택에의
rE
EiG
G 를 선택할 때 기준은 이 물체의 경계면에서 경계 조건을 만족하는 G 를 쉽게 구할 수 있느냐 이다 . 계산하기 편하게 G 를 정해 놓아도 실제 解와의 오차는 물체의 표면 전류로 보정할 수 있다 .
sm , JJ
air
'S
,JiE
n
'V
sm ',' JJ
두 경우의 G 와 Jm, Je 는 모두 다르다 . 아래 그림의 경우 매질이 균일하므로 free space 의 G 를 선택한다 .
Equivalence theorem
)(rE
26
EMLAB
Dyadic Green’s function for half space – PEC
AE j
2222
2221 )'()'()'(,)'()'()'( zzyyxxRzzyyxxR
tniV iV
V tnV
V
JJdgdg
dJJgdg
dgg
tnJJJ
tnJ
nnttIJttnnnnttA
ˆˆ,''
')ˆˆ('
ˆˆˆˆ,')]ˆˆˆˆ()ˆˆˆˆ([
' 2' 1
' 2' 1
' 21
])()()[(1
)()(11
ttntntntt
ntnt
AAA
AAAH
0H PEC
0nH
0E
0tE
22
112121 4
,4
,,,)ˆˆˆˆ(21
R
eg
R
egggggggdgg
jkRjkR
ntV nt
JnnttA
; image current
전기장의 접선 성분을 0 으로 만들기 위한 Green function 들 .
t 는 지면과 평행한 성분 , n 은 지면에 수직인 성분 .
27
EMLAB
mJ
iJ iJ
Image source for magnetic current mJ
nEJ ˆm
J
iJ
J
iJ
Image source for electric current
HnJ ˆe
Image current due to PEC plane
0][ tan eJG
0][ tan mJG
28
EMLAB
• 실제 상황에서 안테나는 도체판 위에 설치된다 .
• 대부분의 경우 도체 평판은 완전한 평면도 아니고 , 유한한 크기를 가지며 완전도체도 아니다 . • 그러나 문제를 비교적 단순화함으로써 현상에 대한 이해를 분명하게 할 수
있다 . 무한 (Infinite), 평판 (flat), 완전도체 (Perfect conductor) 를 가정 • Image Theory 적용
Antenna
Ground plane
Antenna
Ground plane
Infinite ground plane
Antenna
P.E.C
무한 평판 완전 도체에서의 선형 소자 (1)
29
EMLAB
Image Theory
무한 균일 매질가정 가상의 Imaginary Source 전반사의 법칙 적용 : θ i = θ r
도체속 또는 아래는 관심없음 경계 조건 (Boundary Condition) 무한 도체상에서의 모노폴은 무한균일 매질에서의 다이폴과 동일
무한도체평면
monopole antenna
P.E.C. Equivalent
dipole
30
EMLAB
Love’s equivalence theorem
HE,
Source #1
Source #2 (surface current)
,J
nEJ
HnJ
ˆ
ˆ
m
e
SS HE ,
공간상에서 E, H 를 구하기 위해 공간을 S 를 기준으로 분할한다 .
1. S 의 외부에서 E, H : 실제 source #1 을 적분한 값 + 등가 면에서 source #2 를 면 적분 한 값 .
2. S 의 내부에서 E, H: source #2 를 면적분한 값 .
31
EMLAB
Equivalence theorem
11 HE ,
22 HE ,
D
11 HE ,
0, 22 HE
D
Free space PEC
nEEJ
HHnJ
ˆ)(
)(ˆ
21m
21S
nEJ
HnJ
ˆ'
ˆ'
1m
1S
(1) 무한 공간에 있는 유전체
(2) 도체로 둘러싸여 있는 유전체
•(1) 과 (2) 의 주변 환경은 다르지만 관심 영역인 유전체 안에 존재하는 E, H 는 같게 만들 수 있다 . 이 경우 Js, Jm 을 적절히 영역의 경계에 포함시켜 주면 된다 .
11,
22 ,
11,
32
EMLAB
Schelkunoff’s field equivalence principles
image current
33
EMLAB
iE
1E
2E
1ˆ EnJ m
도체
Electromagnetic 문제
iE
2ˆ EnJ m
Reflected wave 문제 Transmitted wave 문제
2ˆ2 EnJ m
두 문제는 등가이다 . 오른쪽은 도체를 없애고 전류와 전하의 크기를 2 배로 한 것 . :image current 이용함 .
Image current 이용
34
EMLAB
Fourier transform, Fresnel diffraction
'2
)]'ˆ('ˆ[)( daR
ejk
S
jkR
EnnrE
tan)'ˆ'ˆ()'ˆ('ˆ)'ˆ('ˆ EnnEEnnEnn
'2
)( tandaR
ejk
S
jkR
ErE
)(rE
'')','(2
'')','(2
'')','(2
')',','(2
),,(
'
])'()'('2'2[2
0
2
'
])'()'('2'2[2
0
'
])'()'[(2
0
'
22
22
2222
22
1
dydxeyxueR
jk
dydxeyxuR
jke
dydxeyxuR
jke
dazyxuR
ejkzyxu
S
yxyyxxz
kjz
yxzjk
S
yxyyxxyxz
kjjkz
S
yyxxz
kjjkz
S
jkR
z
yyxxz
z
yyxxz
z
yyxxzzyyxxR
zzzyyxxR
22
2
22
2
22222
222
)'()'(
2
1)'()'(
2
11
)'()'(1)'()'(
0',)'()'()'(
: Fresnel diffraction
35
EMLAB
Far field, near field
2
3
22
2
sincos2
'1sin
2
'1cos'
z
r
z
rzrR
36
EMLAB
Far field, near field
37
EMLAB
z
ky
z
kxUe
R
jk
yxuFeR
jk
dydxeyxueR
jkzyxu
z
yxzjk
z
yxzjk
S
yyxxz
kjz
yxzjk
,2
)}','({2
'')','(2
),,(
0
2
0
2
'
)''(
0
2
22
22
22
Fraunhofer diffraction
: Fourier transform
sin
)sin(2
2
'''2
'')','(2
),,(
120
2
0
2
0
)'')('cos(sin'0
2
'
)''(
0
2
22
22
22
ka
kaJaAe
R
jk
ddeAeR
jk
dydxeyxueR
jkzyxu
z
yxzjk
ayyxxjkz
yxzjk
S
yyxxz
kjz
yxzjk
sinsin,cossin
'sin'','cos''
z
y
z
x
yx
)()(,2
1)( 10
2
0
cos0 xxJxdxxJdexJ jx