lim ∂Ezs
ρ→∞ ∂ρ
= − jkEzs
lim
∂H
s z
ρ→∞ ∂ρ
=
−
r jkH
s z
(2.10) (2.11)
For the TM and TE polarizations, respectively, where ρ is
the radial variable in cylindrical coordinates.
r E
s
r H
s
= =
− ∇
jωµ0
r J
×
r J
−
r + ∇∇ ⋅ J
jωε 0
jωε
0
r K
+
−
∇
×
r K
∇∇
⋅
r K
jωµ0
*G *G
(2.25) (2.26)
A differentiation followed by an integration.
12
Source-field relationships in homogeneous space
Splitting the fields into two parts
One associated with the primary source located somewhere outside the scatterer
Incident fields
r E
inc
r H
inc
Another associated with the equivalent induced source
( ) G
=
1 4j
H
(2
0
)
k
ρr
(2.20)
To summarize, the above procedure is the integration-followed-bydifferentiation procedure, which is NOT well suited for numerical implementation.
field, symbolically denoted
( ) G ej =
1 j ωε 0
∇∇
+ k2I G
( ) G ek = −∇ × I G
(2.29) (2.30)
and and are the dyadic Green’s functions for the magnetic
field,
( ) G mj = ∇ × I G
rr
A = J *G
r F
=
r K
*G
(2.16) (2.17)
where the scalar function G is the well-known three-dimensional
Green’s function
G
=
e − jk rr
4π rr
(2.18)
And the asterisk (*)denote three-dimensional convolution, that is,
(2.5) (2.6)
5
General description of a scattering problem (III)
Some additional explanations:
( ) (1) The vector Laplacian:
∇2
r E
=
∇
∇
⋅
r E
−
∇
×
∇
×
r E
(2.7)
(2)our interests is the case of an excitation produced by some source in the far zone. Often, we will consider the incident field to be a uniform plane wave.
those in the right.
___r________________________r_______
Er
Hr Jr
K
_
Hr E
r
_KJr
ρe
ρm
ρm
_ ρe
εµr
µ
εr
Ar
_Fr
__F________________________A_________
14
Conclusions
•General description of a scattering problem •Source-field relationships in homogeneous space ---- four forms •Principle of Duality
The description of a scattering problem
Part 2
1
Quick summary of last class (I)
ε0
µ0
r E
inc
kˆ
r H
inc
ε 0ε r
µ0µr
r Er
H
s
s
2
Quick summary of last class (II)
How to describe the electromagnetic problems? Maxwell’s equations and boundary conditions
inc
+
r E
s
(2.1)
r H
=
r H
inc
+
r H
s
(2.2)
where
∇2
r E
inc
+
k
2
r E
inc
=
0
(2.3)
∇2
r H
inc
+
k
2
r H
inc
=
0
(2.4)
∇2 ∇2
r E
s
r H
s
+
k
2
r E
s
+
k
2
r H
= s=
j−ω∇µ0×JrJr−+∇jj∇ωωεε⋅ 0J0rKr+−∇∇×j∇ωKrµ⋅ K0r
---- a fourth form (I)
A simple interchange of integration and differentiation results in
r E
s
=
r J * Gej
+
r K
* Gek
r H
s
=
r J * Gmj
+
r K
* Gmk
(2.27) (2.28)
Where and are the dyadic Green’s functions for the electric
∫∫∫ Ar(rr) =
( ) r
J
rr′
e− jk rr−rr′
4π rr′
drr′
(2.19)
9
Source-field relationships in homogeneous space ---- a first form (III)
The two-dimensional Green’s function is
of the magnetic .according to
(2.12)
jωε 0
r H
s
=
∇×
r A+
∇∇
⋅
r F
+
k
2
r F
(2.13)
jωµ0
And the vector potential satisfy
∇2
r A
+
k
2
r A
=
−
r J
∇2
r F
+
k
2
r F
=
−Kr
(2.14） （2.15）
8
(3) Radiation conditions in a three-dimensional problem
lim
rˆ
×
∇
×
r E
s
=
r jkE
s
r→∞
lim
rˆ
×
∇
×
r H
s
=
r jkH
s
r→∞
where r is the conventional spherical coordinate variable.
r E
s
r H
s
= =
−
jωµr0
r A
∇× A−
j−ω∇εΦ0Fre
− −
∇
×
r F
∇Φ m
(2.21) (2.22)
where Φe and Φm are scalar potential functions and given by
Φe
=
ρe ε0
*G
(2.23)
Φm
=
ρm µ0
*G
(2.24)
15
7
Source-field relationships in homogeneous space
---- a first form (I)
How to solve equations (2.5) and (2.6)?
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