Download - Transfer Matrix Method In Solving EM Problem
Introducing Transfer Matrix in Solving Laplace Equation1
General Properties for TMM in Multi-layer Shell
2
General use in EM Wave Propagating in Multi-layer3
Introducing Transfer Matrix in Solving Laplace Equation1
General Properties for TMM in Multi-layer Shell
2
General use in EM Wave Propagating in Multi-layer3
Consider a series of co-central spherical shells with εn, at the nth shell, and the radius between the nth
and n+1th level is Rn,n+1.
We see a simple example first.
We apply a uniform field E=E0ex , and then solve the Laplace equation in the spherical coordinate, we got solutions for the 1st order inducing field
and boundary conditions , at r = Rn-1,n
2( ) cosn
n n
BA r
r
1n n 11
n nn nr r
Then we would easily manifest An and Bn in terms of An-1 and Bn-1 as
1 1
11
1,
2 2 2
3 3
n n
n n nn n
n n
BA A
R
1 1
1 1, 1
1 1 2
3 3
n n
n nn n n n nB A R B
and further in matrix form
where
111 12
21 22 1
n n
n n
A AQ Q
Q QB B
1
22
1 2
3
n
nQ
1
21 1,
1
3
n
nn nQ R
1
121,
2 21
3
n
n
n n
QR
1
11
2
3
n
nQ
The matrix is called the transfer matrix for the
n-1,n th level. If at the 1st level there is A1 and B1=0( to ensure converge) and at the infinite space there is An= -E0 and Bn, multiply the transfer matrix again and again we will get
And surly we got the solution of A1 and Bn, then whichever Ak
and Bk you want could be solved by using transfer matrix.
11 121,
21 22n n
Q QQ
Q Q
0 1, 1 1, 2 2,1...
0n n n nn
E AQ Q Q
B
Introducing Transfer Matrix in Solving Laplace Equation1
General Properties for TMM in Multi-layer Shell
2
General use in EM Wave Propagating in Multi-layer3
We now start some general solution for general conditions, solutions to be
we apply the same B.C and trick in calculation
1( ) (cos )
ll l n
n n lll
BA r P
r
, 1 , 11 1, 1 , 1 1
1 1 !, 1 1 , 1 12 2
, 1 , 1
1 1
1 1( 1) ( 1)
l ln n n nl ll l
n n n nn n
l ll ln n
n n n n n n n nl ln n n n
R RR RA A
B BlR l lR lR R
the TMM notation will be
and elements
when l=1, the results would automatically turn to the same results in the previous story.
1, 1
1
l ln nl
n nl ln n
A AQ
B B
122
( 1)
2 1
n
n
l l
Ql
2 11
21 1,
(1 )
2 1
n
lnn n
l
Q Rl
112 2 1
1,
( 1)(1 )1
2 1
n
nln n
l
Ql R
1
11
( 1)
2 1
n
n
l l
Ql
Furthermore, when adding up free boundary charge, the boundary condition will turn to be
we will soon get a solution no more complex than before
where
But the additional term, called charge term, is not that neat.
1, 1 , 1
1, 1 2
, 1 , 1
1
/
2( 1)
2( 1)
l ln n n n
nln n l l
n n n n
n
R
lC
R
l
1, 1 , 1
1
l ln nl l
n n n nl ln n
A AQ C
B B
1n n 11 , 1
n nn n n nr r
0 1, 1 1, 2 2,1 , 1 , 1 , 1... ...
0
ll
n n n n n n k k k kln
E AQ Q Q Q Q C
B
Though tough, but physicsFor a metal layer, the potential is constant, therefore only , otherwise is 0. If in the boundary for n,n+1th layer there is a l order charge ,we would directly got
It means, the surface charge density is just like other external conditions such as E field, would only induce the same order term, as a uniform E inducing only a term.
1, 1 , 1
1 1, 1 2
1 , 1 , 1
1
/
2( 1)
2( 1)
l ln n n n
ln nl
n nl l ln n n n n
n
R
A lC
B R
l
, 1ln n
0lnA
1(cos )P
Introducing Transfer Matrix in Solving Laplace Equation1
General Properties for TMM in Multi-layer Shell
2
General use in EM Wave Propagating in Multi-layer3
Consider a multi-layer withεn in the nth layer, and position of the boundary between nth and n+1th is dn,n+1 along z direction. Assume the wave propagates along z direction, perpendicular to the layer, with the expression of field, where
Solve , we got
As we have assumed there is not any surface current, so the boundary conditions are 1xn xnH H 1yn ynE E
n nik z ik zyn n nE A e B e
0 ( ,0, )x zH H H0 (0, ,0)yE E
n nik z ik zn n
xnn
A e B eH
Z
k H E 55555555555555555555555555 55
k E H 55555555555555555555555555 55
Then got the solution
Simply we can change the expression into matrix
1
1
1 1
1
1 1 1
1 10 01
20 01 1
n n
n n
n nik d ik d
n n n n
ik d ik dn n n n
n n
Z Z
A Z Z Ae e
B Z Z Be eZ Z
1 1 11
1[(1 ) (1 ) ]2
n n nik d ik d ik dn nn n n
n n
Z ZB e A e B e
Z Z
1 1 11
1[(1 ) (1 ) ]2
n n nik d ik d ik dn nn n n
n n
Z ZA e A e B e
Z Z
and in it, we define
If there are n layer (noticing that outside the layers are air, so the 0th and n+1th layer are absolutely air terms), the final solution can be written as
1 0 00 0, , 1 1 1, 2 2 1,0 0 , 1
1 0 0
...nn n n n n n n n n n
n
A A AP M PM P M P M P QB B B
1 1
, 11 1
1 11
21 1
n n
n nn n
n n
n n
Z Z
Z ZM
Z Z
Z Z
0
0
n
n
ik d
n ik d
eP
e
Further if we define the starting terms as ,
the solution could be simplified to the transfer matrix
By using this method we could quickly get the answer of t and r
And from this we could use transfer method to get the propagating properties in any layers.
1
1 0n
n
A t
B
0
0
1A
B r
1
0
tQr
12 2111
22
Q Qt Q
Q 21
22
Qr
Q
Interestingly, from the result above, one could think that, if Q21= 0, the reflective terms would be 0, and further if Q11= 1, meaning no absorption, t=1 , the transmittance behavior would be perfect.
Specifically, we could input some data asε1=1000, ε2= -2000, d1=d2= 2mm, then we get ω= 2π*0.850 GHz, there is a perfect transmission. From a COMSOL simulation we can see the S21 is near to 1.
12 2111
22
Q Qt Q
Q 21
22
Qr
Q