em theory term presentation pdf version
TRANSCRIPT
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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”
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Poynting Reciprocity
Interface
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Sense of Distributions
A(r) = {A(r)} + [A(r)]s
RegularComponent
SingularComponent
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Why?
Boundary Conditions ARE NOT postulation
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Why?
Rigorous Treatmentin Mathematics
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Why?
Being able to Describe Interface( Boundary condition is weak to do this )
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Review Boundary Trick
H
W
Choose W infinitely close to 0 to get four B.C.
D E
B H
GaussianSurface
/Contour
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Three Caseswhich will be discussed later
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Cases 1
PECDielectric
interface sustained only electric current. no magnetic current
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Cases 2
Dielectric 2Dielectric 1
Interface sustained no current
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Cases 3
arbitrarymedia
arbitrarymedia
infinite thin filmsustained electric and magnetic current
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Review Poynting’s Thm
pi = E ⋅ Ji + H ⋅ Mi
i = d, c, v
dissipatedconduction
convection
pin = pd + pc + pv
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Cases Disscussion
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Prerequisite 1
Characteristic function
U(f) = 1 as f > 0 U(f) = 0 as f < 0
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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.
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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
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Cases 1 PECDielectric
- div Pi = Ei.Jdi + Hi.Mdi i = 1, 2
- n.(P1 - P2) = Es.Js
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Cases 2
- div Pi = Ei.Jdi + Hi.Mdi i = 1, 2
- n.(P1 - P2) = 0
Dielectric Dielectric
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Requisite on Surface
Ms = - n × Z.Js
Media Media
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Cases 3
- div Pi = Ei.Jdi + Hi.Mdi, i = 1, 2
- n.(P1 - P2) = Es.Js + Hs.Ms
= Z(n × Hs)2
Media Media
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Review Reciprocity
div( Ea × Hb - Eb × Ha ) =
( Eb.Jaf - Hb.Maf ) - ( Ea.Jbf - Ha.Mbf )
<a, b> <b, a>
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Cases Discussion
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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
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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
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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
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Requisite on Surface
Ms = - n × Z.Js
Media Media
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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
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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.
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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.
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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.
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Simple Application 1
Z ⇔ [ABCD]
Regular Field Distribution
Regular Field Distribution
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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.
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Thank for your attention