# transfer matrix method in solving em problem

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Transfer Matrix Method In Solving EM Problem. Produced by. Yaoxuan Li , Weijia Wang , Shaojie Ma. Presented by Y.X.Li. 4. Theoretical Analyzing. Introducing Transfer Matrix in Solving Laplace Equation. 1. 2. General Properties for TMM in Multi-layer Shell. 3. - PowerPoint PPT PresentationTRANSCRIPT

Consider a series of co-central spherical shells with n, 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

Then we would easily manifest An and Bn in terms of An-1 and Bn-1 as

and further in matrix form where

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.

We now start some general solution for general conditions, solutions to be we apply the same B.C and trick in calculation

the TMM notation will be and elements when l=1, the results would automatically turn to the same results in the previous story.

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.

Though tough, but physics For 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.

Consider a multi-layer withn 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

Then got the solution Simply we can change the expression into matrix

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

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.

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 as1=1000, 2= -2000, d1=d2= 2mm, then we get = 2*0.850 GHz, there is a perfect transmission. From a COMSOL simulation we can see the S21 is near to 1.

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