ch10_2005

Scalar potential Green’s function
Starting from Maxwell’s equation and defining a vector potential function A and scalar potential Φ with Lorentz condition ∇ ⋅ A = − jωµεΦ , we have the inhomogeneous scalar and vector Helmholtz equation： ( ∇2 + k 2 ) Φ = − ρ / ε ( ∇2 + k 2 ) A = − µ J The scalar Green’s function is then defined by
Integrate over kz first
2 0
⎛ 1 − jk ( x − x ' )− jk y ( y − y ' ) − jk z ( z − z ' ) ⎞ Re s ⎜ 2 e x ⎟ 2 ⎝ k z − k0 z ⎠ 1 − jk ( x − x ' )− jk y ( y − y ' ) − jk z ( z − z ' ) = lim ( k z + k0 z ) 2 e x 2 k z →− k0 z k z − k0 z = 1 − jk ( x − x ' ) − jk y ( y − y ' ) − jk0 z ( z − z ' ) e x −2 k 0 z
Poles in the kz-plane occur at k0 z = ( k − k − k kz-plane is a complex plane. − j k ' + jk " z e ( z z ) = e − jk z ' z e k z " z
2 x
2 1/ 2 y
)
, where
If z > 0 , we have to choose k z " < 0 , which means that we need to deform the contour into the lower plane. If z < 0 , we have to choose k z " > 0, which means that we need to deform the contour into the upper plane.
∇∇ ⋅ Ge ( r , r0 ) − ∇2Ge ( r , r0 ) − k 2Ge ( r , r0 ) = I δ ( r , r0 ) 1 ⎛ ⎞ ∇2 + k 2 ) Ge ( r , r0 ) = − ⎜ I + 2 ∇∇ ⎟ δ ( r , r0 ) ( k ⎝ ⎠ 1 ⎛ ⎞ ∇2 + k 2 ) Ge ( r , r0 ) = ⎜ I + 2 ∇∇ ⎟ ( ∇2 + k 2 ) g ( r , r0 ) ( k ⎝ ⎠
Guidance properties of strips in an inhomogeneous layered medium -second report
報告者：陳登凱 王裕閔
Outline
Formulation Results and discussion
Green’s function for Helmholtz differential equation
k z = k z '+ jk z "
For
z − z' > 0 +∞ 1 1 − jk ( x − x ' ) − jk y ( y − y ' ) − jk z ( z − z ' ) − jk ( x − x ' ) − jk y ( y − y ' ) − jk0 z ( z − z ' ) dk z 2 e x = 2π j e x ∫−∞ k z − k02z −2k0 z −j 1 − jkx ( x − x ')− jk y ( y − y ')− jk0 z ( z − z ') g ( r , r ') = dk x dk y e 2 ∫∫ 2k0 z ( 2π )
The Fourier transform of scalar Green’s and delta functions：
g ( r , r ') =
3 ( 2π ) ∫ ∫ ∫−∞
1
+∞Leabharlann dke− jk ⋅( r − r ')
% g (k )
δ ( r , r ') =
3 ( 2π ) ∫ ∫ ∫−∞
e 0 2 0
Hence, the electric field dyadic Green’s function is related to the scalar Green’s function as
1 ⎛ ⎞ Ge ( r , r0 ) = ⎜ I + 2 ∇∇ ⎟ g ( r , r0 ) k ⎝ ⎠
Upon electromagnetic textbook, we have the solution to the inhomogeneous vector Helmholtz equation
µ A(r ) = 4π
0 e ∫∫∫ dr0 J ( r0 ) r − r0
− jk r − r
ˆ ˆ ˆ If µ J is replaced by a unit vector source ( a x + a y + az ) δ ( r − r0 ) then the vector Green’s function is defined by
2 ∇2φn ( r ) + kn φn ( r ) = 0
The Green’s function can be written ∞ φ ( r1 ) φ ( r ) G ( r1 , r2 ) = ∑ n 2 n 2 2 The general solution is
n =0
k − kn
ψ ( r1 ) = ∫∫∫ G ( r1 , r2 ) ρ ( r2 ) dr2
For an arbitrary current distribution the solution for the vector potential is
A ( r ) = µ ∫∫∫ dr0G ( r , r0 ) ⋅J ( r0 )
0 e G ( r , r0 ) = I 4π r − r0
Electric field dyadic Green’s function
0 e ˆ ˆ ˆ ( ax + a y + az ) 4π r − r 0
− jk r − r
For a general current distribution J, the solution would be obtained by a superposition integral provided we make the rule that the s component of the current Jx is to be associated with the unit vector ax in the Green’s function.
% g (k ) =
g ( r , r ') =
− jk ⋅( r − r ')
dke
− jk ⋅( r − r ')
1 1 = 2 2 k x2 + k y + k z2 − k02 k z − k02z
3 ( 2π ) ∫ ∫ ∫−∞
2 k02z = k02 − k x2 − k y
1
+∞
dk
1 e − jk ⋅( r − r ') 2 2 k z − k0 z
Electric field due to current
The electric field due to current distribution in an unbounded homogeneous medium can be expressed as
E ( r ) = − jωµ ∫∫∫ dr0Ge ( r , r0 ) ⋅ J ( r0 )
Definition of dyadic Green’s function
Then we introduce the dyadic Green’s function, which is defined as a solution of the equation
∇2G ( r , r0 ) + k 2G ( r , r0 ) = − I δ ( r − r0 )
Upon Maxwell’s equation, we have the wave equation for electric field
∇ × ∇ × E − k 2 E = − jωµ J
The electric field dyadic Green’s function is defined as
The inhomogeneous Helmoholtz differential equation is ∇2ψ ( r ) + k 2ψ ( r ) = ρ ( r ) The Green’s function is then defined by ( ∇2 + k 2 ) G ( r1, r2 ) = δ ( r1 − r2 ) Define the basis functions as the solutions to the homogeneous Helmoholtz differential equation
∇ × ∇ × Ge ( r , r0 ) − k 2Ge ( r , r0 ) = I δ ( r , r0 )
Solution for electric field dyadic Green’s function
Start from the scalar Green’s function ( ∇2 + k 2 ) g ( r , r0 ) = −δ ( r − r0 ) The divergence of ∇ × ∇ × Ge ( r , r0 ) − k 2Ge ( r , r0 ) = I δ ( r , r0 ) gives − k 2∇ ⋅ Ge ( r , r0 ) = ∇ ⋅ I δ ( r , r0 ) = ∇δ ( r , r0 ) 1 ∇ ⋅ G ( r , r ) = − ∇δ ( r , r ) k Expand the ∇ × ∇ × operator
1
+∞
dke− jk ⋅( r −r ')
dk = dk x dk y dk z
Substituting into
(∇
2
+k
2 0
) g ( r , r ') = −δ ( r , r ') , we have
% g ( k ) = −∫ ∫ ∫
+∞ −∞
∫∫∫
+∞
−∞
dk ( ∇ 2 + k02 ) e
1 ⎛ ⎞ Ge ( r , r0 ) = ⎜ I + 2 ∇∇ ⎟ g ( r , r0 ) where k ⎝ ⎠
is the solution of ∇ × ∇ × Ge ( r , r0 ) − k 2Ge ( r , r0 ) = I δ ( r , r0 )
Fourier transform of Green’s function
ˆ ˆ ˆ ˆ ˆ ˆ where I is the unit dyadic I = a x a x + a y a y + a z a z The free-space dyadic Green’s function for the vector Helmholtz equation is − jk r − r
(∇
2
+k
2
) g ( r , r ) = −δ ( r − r )
0 0
ρ ( r0 ) Φ ( r ) = ∫∫∫ dr0 g ( r , r0 ) ε
0 e g ( r , r0 ) = 4π r − r0
− jk r − r
Vector potential Green’s function
−j
For z − z ' < 0 g ( r , r ') =
( 2π )
2
∫∫ dk x dk y
1 − jkx ( x − x ')− jk y ( y − y ')+ jk0 z ( z − z ') e 2k0 z