EM Theory Term Presentation PDF Version
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This document discusses Poynting's theorem and reciprocity relations for discontinuous electromagnetic fields across interfaces using a distributional approach. It examines three cases: 1) a dielectric-PEC interface with only electric surface current, 2) a dielectric-dielectric interface with no surface currents, and 3) an interface between two arbitrary media with both electric and magnetic surface currents. For each case, it applies Poynting's theorem and the reciprocity theorem to derive boundary conditions in terms of a surface impedance tensor Z. The document concludes that interfaces can be rigorously treated as parts of space using a distributional approach and described using surface impedance tensors rather than boundary conditions.
![Sense of Distributions
A(r) = {A(r)} + [A(r)]s
Regular Singular
Component Component](https://image.slidesharecdn.com/finalkeynotes-100115030813-phpapp01/85/EM-Theory-Term-Presentation-PDF-Version-3-320.jpg)
![Cases 1 Dielectric PEC
div ( Eai × Hbi - Ebi × Hai ) =
( Ebi Jai - Hbi Mai ) -
( Eai Jbi - Hai Mbi ) i = 1, 2
n × [(Ea1 × Hb1 - Eb1 × Ha1) -
(Ea2 ×Hb2 - Eb2 ×Ha2)]s
= Ebs Jas - Eas Jbs](https://image.slidesharecdn.com/finalkeynotes-100115030813-phpapp01/85/EM-Theory-Term-Presentation-PDF-Version-23-320.jpg)
![Cases 2 Dielectric Dielectric
div ( Eai × Hbi - Ebi × Hai ) =
( Ebi Jai - Hbi Mai ) -
( Eai Jbi - Hai Mbi ) i = 1, 2
n × [(Ea1 × Hb1 - Eb1 × Ha1) -
(Ea2 ×Hb2 - Eb2 ×Ha2)]s
=0](https://image.slidesharecdn.com/finalkeynotes-100115030813-phpapp01/85/EM-Theory-Term-Presentation-PDF-Version-24-320.jpg)
![Cases 3 Media Media
div ( Ea1 × Hb1 - Eb1 × Ha1 ) =
( Eb1 Ja1 - Hb1 Ma1 ) -
( Ea1 Jb1 - Ha1 Mb1 )
n × [(Ea1 × Hb1 - Eb1 × Ha1)]s
= (Ebs Jas - Ebs Mas)
- (Eas Jbs - Has Mbs)](https://image.slidesharecdn.com/finalkeynotes-100115030813-phpapp01/85/EM-Theory-Term-Presentation-PDF-Version-25-320.jpg)
![Cases 3 application
Ebs Jas - Ebs Mas =
[(n × Has) (Z - ZT) (n × Hbs)]/2
Eas Jbs - Has Mbs =
[(n × Hbs) (Z - ZT) (n × Has)]/2](https://image.slidesharecdn.com/finalkeynotes-100115030813-phpapp01/85/EM-Theory-Term-Presentation-PDF-Version-27-320.jpg)
![Cases 3 application
When Z = ZT
n × [(Ea1 × Hb1 - Eb1 × Ha1)]s = 0.
From the aspect of microwave engineering,
we could say this is a reciprocal element, so
that we say this is a Reciprocal Interface.](https://image.slidesharecdn.com/finalkeynotes-100115030813-phpapp01/85/EM-Theory-Term-Presentation-PDF-Version-28-320.jpg)
![Simple Application 1
Regular Field Regular Field
Distribution Distribution
Z [ABCD]](https://image.slidesharecdn.com/finalkeynotes-100115030813-phpapp01/85/EM-Theory-Term-Presentation-PDF-Version-31-320.jpg)
EM Theory Term Presentation PDF Version
- 1. Poynting and Reciprocity on Discontinuous Field ( J. W. Hsu ) R98941103 ( P. E. Chang ) R97943086 Panel - 15 - Main Reference: B. Polat, “On Poynting’s Theorem and Reciprocity Relations for Discontinuous Fields”
- 2. Poynting Reciprocity Interface
- 3. Sense of Distributions A(r) = {A(r)} + [A(r)]s Regular Singular Component Component
- 4. Why? Boundary Conditions ARE NOT postulation
- 5. Why? Rigorous Treatment in Mathematics
- 6. Why? Being able to Describe Interface ( Boundary condition is weak to do this )
- 7. Review Boundary Trick Choose W infinitely close to 0 to get four B.C. Gaussian Surface H / Contour D E W B H
- 8. Three Cases which will be discussed later
- 9. Cases 1 Dielectric PEC interface sustained only electric current. no magnetic current
- 10. Cases 2 Dielectric 1 Dielectric 2 Interface sustained no current
- 11. Cases 3 arbitrary arbitrary media media infinite thin film sustained electric and magnetic current
- 12. Review Poynting’s Thm pin = pd + pc + pv pi = E ⋅ Ji + H ⋅ Mi i = d, c, v dissipated convection conduction
- 14. Prerequisite 1 Characteristic function U(f) = 1 as f > 0 U(f) = 0 as f < 0
- 15. Prerequisite 2 A = A1U(f) + A2U(-f) A = E, D, H, B on Surface, It converges to Average of Field of both side of interface.
- 16. Cases 1 Dielectric PEC pin = - div P1U(f) - div P2U(-f) - n (P1 - P2) δ(S) pd = (E1 Jd1 + H1 Md1)U(f) + (E2 Jd2 + H2 Md2)U(-f) pc = Es Js δ(S)
- 17. Cases 1 Dielectric PEC - div Pi = Ei Jdi + Hi Mdi i = 1, 2 -n (P1 - P2) = Es Js
- 18. Cases 2 Dielectric Dielectric - div Pi = Ei Jdi + Hi Mdi i = 1, 2 -n (P1 - P2) = 0
- 19. Requisite on Surface Media Media Ms = - n × Z Js
- 20. Cases 3 Media Media - div Pi = Ei Jdi + Hi Mdi, i = 1, 2 -n (P1 - P2) = Es Js + Hs Ms = Z(n × Hs)2
- 21. Review Reciprocity div( Ea × Hb - Eb × Ha ) = ( Eb Jaf - Hb Maf ) - ( Ea Jbf - Ha Mbf ) <a, b> <b, a>
- 22. Cases Discussion
- 23. Cases 1 Dielectric PEC div ( Eai × Hbi - Ebi × Hai ) = ( Ebi Jai - Hbi Mai ) - ( Eai Jbi - Hai Mbi ) i = 1, 2 n × [(Ea1 × Hb1 - Eb1 × Ha1) - (Ea2 ×Hb2 - Eb2 ×Ha2)]s = Ebs Jas - Eas Jbs
- 24. Cases 2 Dielectric Dielectric div ( Eai × Hbi - Ebi × Hai ) = ( Ebi Jai - Hbi Mai ) - ( Eai Jbi - Hai Mbi ) i = 1, 2 n × [(Ea1 × Hb1 - Eb1 × Ha1) - (Ea2 ×Hb2 - Eb2 ×Ha2)]s =0
- 25. Cases 3 Media Media div ( Ea1 × Hb1 - Eb1 × Ha1 ) = ( Eb1 Ja1 - Hb1 Ma1 ) - ( Ea1 Jb1 - Ha1 Mb1 ) n × [(Ea1 × Hb1 - Eb1 × Ha1)]s = (Ebs Jas - Ebs Mas) - (Eas Jbs - Has Mbs)
- 26. Requisite on Surface Media Media Ms = - n × Z Js
- 27. Cases 3 application Ebs Jas - Ebs Mas = [(n × Has) (Z - ZT) (n × Hbs)]/2 Eas Jbs - Has Mbs = [(n × Hbs) (Z - ZT) (n × Has)]/2
- 28. Cases 3 application When Z = ZT n × [(Ea1 × Hb1 - Eb1 × Ha1)]s = 0. From the aspect of microwave engineering, we could say this is a reciprocal element, so that we say this is a Reciprocal Interface.
- 29. Conclusion With sense of distribution, there is no Boundary, only Fields Distribution on Volume and Interface. You can derive Z tensor from material parameters of both side and replace boundary conditions with Z tensor.
- 30. Conclusion In the past, Fields react on interface COULD NOT be described with boundary conditions. With sense of distribution, interface could be described and treated as part of space or as an element.
- 31. Simple Application 1 Regular Field Regular Field Distribution Distribution Z [ABCD]
- 32. Simple Application 2 In FDTD simulation programs, it will cause field broken if grid point just locate on boundary. Using Z tensor to calculate effective parameters help us to Z εeμeσe avoid this status.
- 33. Thank for your attention