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Let's first consider the first goal for our PML, or lossy material is that we don't want to have any reflection
from this interface between material 1 and material 2, so let's consider a case; let's calculate the reflection
coefficient and we'll consider TEz waves, so this is 2-dimensional case that we'll consider, parallel polarization.
So let's say that we have some incoming wave at some angle.
Our H inc is coming out of the screen or out of the board or out of the paper here, slide, any of
the above and this E inc points upward and to the left, and this comes in at an angle theta.
So let's say that this is our normal and, surface normal, and then I was
doing dotted lines, so let's say that this then is our transmitted wave,
H trans and we have E transmitted and then -- put a circle around there, and then let's
say then we can draw out our reflected wave H reflect and this would be E reflect.
All right. And let's say this is the X direction. This is region 1, region number 1 and region number 2.
So region number 1 is lossless.
We just have mu 1, epsilon 1, and region number 2 is lossy and we have mu 2 epsilon 2, sigma and sigma star.
So for this analysis we are going to use phasor analysis, so we have curl of E is equal to minus J Omega mu H.
This is in our lossless material.
Curl of H is equal to J Omega epsilon E, and these bars of course
indicate that these are vectors and so this is lossless case.
And for the lossy case, you have curl of E minus sigma star plus
J omega mu H and curl of H is sigma plus J omega epsilon E.
Be consistent with my arrows. So this is for the lossy side.
So we're using phasor analysis so for our lossy material, or lossless material
alpha here is zero; beta is equal to K is equal to omega squared of mu epsilon.
All right so, now let's try to write out the fields that we have propagating in a different region.
So in region 1, and then we will look at calculating the reflection coefficient.
In region 1 we have H inc is H not E to the minus J.
I'm going to use beta, beta 1 x x plus beta 1 y y and it's oriented in a z direction. So beta 1x here.
This is beta 1 cosign theta.
Beta 1 y is sign beta 1 sign theta and beta 1 is equal to omega mu 1 epsilon 1 is equal to k 1.
Okay. That should be enough and then H region 1 also has reflected
H field and this would be H not and our reflection coefficient,
E to the minus J minus beta 1 x so we have to consider our directions
here and how they would change for a reflective field and z hat.
All right and then one more for -- we have in region 2 we have our transmitted field and
that would be H or transmission coefficient e to the minus J beta 2xx plus beta 2yy.
The 2 is for material 2. 1 is for material 1 and then z hat.
So we can also write this out for beta 1 x but beta 2 x squared plus beta 2 y squared is beta 2 squared.
And this would also equal minus sigma star plus j omega mu 2 times sigma plus j omega epsilon 2.
We can then rearrange and this will be equal to omega squared mu to epsilon
2 and this is equal to k 2 squared times 1+ sigma star j omega mu 2,
1+ sigma over J omega epsilon 2.
All right. So we have written out our fields instant reflected, all of our fields in region 1 and region 2, so now what we're going
to do next is we'll write out the total fields in H and we can use that to solve for our reflection coefficient.