Poynting and Reciprocity on Discontinuous Field

許家瑋 ( J. W. Hsu ) R98941103張沛恩 ( 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

RegularComponent

SingularComponent

Why?

Boundary Conditions ARE NOT postulation

Why?

Rigorous Treatmentin Mathematics

Why?

Being able to Describe Interface( Boundary condition is weak to do this )

Review Boundary Trick

H

W

Choose W infinitely close to 0 to get four B.C.

D E

B H

GaussianSurface

/Contour

Three Caseswhich will be discussed later

Cases 1

PECDielectric

interface sustained only electric current. no magnetic current

Cases 2

Dielectric 2Dielectric 1

Interface sustained no current

Cases 3

arbitrarymedia

arbitrarymedia

infinite thin filmsustained electric and magnetic current

Review Poynting’s Thm

pi = E ⋅ Ji + H ⋅ Mi

i = d, c, v

dissipatedconduction

convection

pin = pd + pc + pv

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 1pin = - 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)

PECDielectric

Cases 1 PECDielectric

- div Pi = Ei．Jdi + Hi．Mdi i = 1, 2

- n．(P1 - P2) = Es．Js

Cases 2

- div Pi = Ei．Jdi + Hi．Mdi i = 1, 2

- n．(P1 - P2) = 0

Dielectric Dielectric

Requisite on Surface

Ms = - n × Z．Js

Media Media

Cases 3

- div Pi = Ei．Jdi + Hi．Mdi, i = 1, 2

- n．(P1 - P2) = Es．Js + Hs．Ms

= Z(n × Hs)2

Media Media

Review Reciprocity

div( Ea × Hb - Eb × Ha ) =

( Eb．Jaf - Hb．Maf ) - ( Ea．Jbf - Ha．Mbf )

<a, b> <b, a>

Cases Discussion

Cases 1div ( 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

PECDielectric

Cases 2div ( 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

Dielectric Dielectric

Cases 3div ( 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)

Media Media

Requisite on Surface

Ms = - n × Z．Js

Media Media

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

Z ⇔ [ABCD]

Regular Field Distribution

Regular Field Distribution

Simple Application 2

Z ⇔ εeμeσe

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 avoid this status.

Thank for your attention