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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”
Poynting          Reciprocity



           Interface
Sense of Distributions

   A(r) = {A(r)} + [A(r)]s



 Regular               Singular
Component             Component
Why?


Boundary Conditions
ARE NOT postulation
Why?


Rigorous Treatment
  in Mathematics
Why?


Being able to Describe Interface
    ( Boundary condition is weak to do this )
Review Boundary Trick
               Choose W infinitely close to 0
               to get four B.C.

Gaussian
Surface
           H
   /
Contour         D                       E

      W         B                      H
Three Cases
which will be discussed later
Cases 1


      Dielectric             PEC




interface sustained only electric current.
           no magnetic current
Cases 2


 Dielectric 1       Dielectric 2




Interface sustained no current
Cases 3


       arbitrary          arbitrary
        media              media



            infinite thin film
sustained electric and magnetic current
Review Poynting’s Thm

  pin = pd + pc + pv
  pi = E ⋅ Ji + H ⋅ Mi
      i = d, c, v

dissipated                convection
             conduction
Cases Disscussion
Prerequisite 1

Characteristic function

U(f) = 1 as f > 0
U(f) = 0 as f < 0
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.
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)
Cases 1               Dielectric   PEC




- div Pi = Ei Jdi + Hi Mdi        i = 1, 2



      -n   (P1 - P2) = Es   Js
Cases 2             Dielectric   Dielectric




- div Pi = Ei Jdi + Hi Mdi    i = 1, 2



        -n   (P1 - P2) = 0
Requisite on Surface
                Media   Media




 Ms = - n × Z Js
Cases 3                Media   Media




- div Pi = Ei Jdi + Hi Mdi, i = 1, 2

 -n   (P1 - P2) = Es   Js + Hs   Ms

           = Z(n × Hs)2
Review Reciprocity
           div( Ea × Hb - Eb × Ha )
                      =
( Eb   Jaf - Hb Maf ) - ( Ea Jbf - Ha Mbf )




        <a, b>              <b, a>
Cases Discussion
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
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
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)
Requisite on Surface
                Media   Media




 Ms = - n × Z Js
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
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.
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.
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.
Simple Application 1


Regular Field                Regular Field
Distribution                 Distribution




                Z   [ABCD]
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.
Thank for your attention

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EM Theory Term Presentation Keynotes 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
  • 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>
  • 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