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2EE750, 2003, Dr. Mohamed Bakr
Duality
• Duality means that two differential/integral equationsdescribing the behavior of two different variables have thesame mathematical forms solutions are identical
• Equations describing the case (J≠ 0, M=0) are dual toequations describing the case (J= 0, M ≠ 0)
HE AA jωµ−=×∇ EH FF jωε=×∇
EJH AA jωε+=×∇ HME FF jωµ−−=×∇
JAA µβ −=+∇ 22 MFF εβ −=+∇ 22
VdeR
Rj-
V
′∫∫∫=′
β
πµ J
A4
VdeR
Rj-
V
′∫∫∫=′
β
πε M
F4
3EE750, 2003, Dr. Mohamed Bakr
Duality (Cont’d)
AH ×∇= )/1( µA FE ×∇−= )/1( εF
)).()(/( AAE ∇∇−−= ωµεω jjA )).()(/( FFH ∇∇−−= ωµεω jjF
It follows that the following quantities are identical
E A H F H A E− F J M
A F ε µ εµ
ββ 1/ηη
4EE750, 2003, Dr. Mohamed Bakr
Example
Using Duality, find the fields resulting from an infinitesimalmagnetic dipole Im=azIm
The fields resulting from an electric dipole are
erj
cosθr
lIE rj-e
rβ
βπη
)1
1(2 2
+=
errj
sinθr
lIjE rj-e β
θ ββπηβ
))(
111(
4 2−+=
erj
sinθr
lIjH rj-e β
ϕ βπβ
)1
1(4
+= x
y
z
θ
ArAz
Aθ
θ
Ie=Ieaz
5EE750, 2003, Dr. Mohamed Bakr
Example (Cont’d)
It follows that the fields resulting from the magnetic dipoleare given by
erj
cosθr
lIH rj-m
rβ
βηπ)
11(
2 2+=
errj
sinθr
lIjH rj-m β
θ ββηπβ
))(
111(
4 2−+=
erj
sinθr
lIjE rj-m β
ϕ βπβ
)1
1(4
+−
=
6EE750, 2003, Dr. Mohamed Bakr
Uniqueness Theorem
• This theorem establishes the conditions under which aunique solution exists for a given problem
• Assume that a closed surface S encloses a material withsources Ji, Mi and complex parameters ,
• If there are two possible solutions Ea, Ha and Eb , Hb, theymust satisfy Maxwell’s equations
εεε ′′−′= jµµµ ′′−′= j
HME ai
a jωµ−−=×∇
HME bi
b jωµ−−=×∇
EEJH aai
a jωεσ ++=×∇,
, EEJH bbi
b jωεσ ++=×∇
7EE750, 2003, Dr. Mohamed Bakr
Uniqueness Theorem (Cont’d)
•••• Subtracting the corresponding equations we get
HE ωµδδ j−=×∇ EH δωεσδ )( j+=×∇,
where δE=Ea−Eb and δH=Ha-Hb
•••• Notice that the differential fields satisfy the source-freeMaxwell’s equations
•••• Applying the source-free conservation of energy relation forδE and δH we get
[ ] dvjjV
***
S∫∫∫ ++−=∫∫ × HHEEdsHE δωµδδωεσδδδ ).().().(
dvV
*
S∫∫∫ ′′+′′+−=∫∫ × ))(().(
22HEdsHE δµωδεωσδδ
dv-j-V∫∫∫ ′+′ )(
22HE δµωδεω
8EE750, 2003, Dr. Mohamed Bakr
Uniqueness Theorem (Cont’d)
• Now if we have , this implies that
δE=δH=0 everywhere inside S. Notice that the assumption of
losses existence is important!
• Using the vector identity A•B×C= B•C×A= C•A×B, we have
0).( =∫∫ × dsHE δδ *
S
dsdsdsS
**
S
*
S
EnHHEnnHE δδδδδδ ).().().( ∫∫ ×=∫∫ ×=∫∫ ×
•••• It follows that the condition implies that
one of the following three cases is satisfied:
0).( =∫∫ × dsHE δδ *
S
a) The tangential component of the E field is specified on S,i.e. n ×δE =0 on S
9EE750, 2003, Dr. Mohamed Bakr
Uniqueness Theorem (Cont’d)
b) The tangential component of the H field is specified on S,i.e. n ×δH =0 on S δH*× n =0 on S
c) The tangential component of the E field is specified onpart of S and the tangential component of the H field isspecified on the rest of S, i.e.
n ×δE =0 on S1
n ×δH =0 on S2
S=S1∪ S2
10EE750, 2003, Dr. Mohamed Bakr
Image Theory
• Image theory enables us to evaluate the field generated bysources placed near infinite perfectly conducting boundary
•••• Virtual sources are added to maintain the same tangentialfield boundary conditions for the original problem
εo, µo
σ =∞
direct
reflection
Actual source
Virtual source
11EE750, 2003, Dr. Mohamed Bakr
Image Theory (Cont’d)
εo, µo
Actual source
Virtual source
εo, µo
h
h
θ1
π-θ1
Er2Er1
Eθ1 Eθ2
• Image of a vertical electric dipole is another verticalelectric dipole (same orientation)
• Notice that the tangential electric field componentscancel out
12EE750, 2003, Dr. Mohamed Bakr
Image Theory (Cont’d)
εo, µo
Actual source
Virtual source
εo, µo
h
h
Image for a horizontal electric dipole has the same valuebut opposite orientation (Prove it!)
13EE750, 2003, Dr. Mohamed Bakr
Example
εo, µo
σ =∞
h
Ie=Ieaz
•••• Obtain an expression for the far fields generated by avertical electric dipole of length l placed near an infiniteconducting wall
14EE750, 2003, Dr. Mohamed Bakr
Example (Cont’d)
εo, µo
εo, µo
h
h
θ1
θ2
Pr1
r2
r
θ
For the far fields we have
eθsinr
lIjE rj-e 1
11
1
4β
θ πηβ= eθsin
r
lIjE rj-e 2
22
2
4β
θ πηβ=and
θcoshrhrr 22221 −+=
θcoshrhrr 22222 ++=
r>>h θcos1 hrr −=θcos2 hrr +=
15EE750, 2003, Dr. Mohamed Bakr
Example (Cont’d)
• Further, we can use for the amplitude r=r1=r2
• The total field in the top half space is the sum of thefield generated by both the actual and virtual sources
)(4
21 eeesin θr
lIjEEE hcosj-hcosjrj-e θβθββ
θθθ πηβ +=+=
z ≥ 0
Eθ=0, z<0
•••• Combining terms we get
)cos(4
2 θβπ
ηβ βθ hcosesin θ
r
lIjE rj-e=
array factorelement factor
16EE750, 2003, Dr. Mohamed Bakr
Reciprocity Theorem
• Reciprocity theorem in circuit theory states that we canchange the location of the source and observation pointswithout affecting the measured values
• A similar theorem can be derived for electromagnetics
• Assume that two sets of sources J1, M1 and J2, M2 radiatewithin a linear isotropic medium
• Using Maxwell’s equations, we have
HME 111 ωµj−−=×∇ HME 222 ωµj−−=×∇
EJH 111 ωεj+=×∇ EJH 222 ωεj+=×∇
17EE750, 2003, Dr. Mohamed Bakr
Reciprocity Theorem (Cont’d)
•••• Dot multiplying the H2 curl equation by E1 and dotmultiplying the E1 curl equation by H2 and subtracting weget
HHEEMHJEEHHE 212112211221 ...... ωµωε jj +++=×∇−×∇
HHEEMHJEHE 2121122121 ....).( ωµωε jj +++=×∇−
•••• Similarly, we obtain
HHEEMHJEHE 1212211212 ....).( ωµωε jj +++=×∇−
•••• Subtracting these two expressions, we obtain
MHJEMHJEHEHE 211212211221 ....).( −−+=×−×∇−
18EE750, 2003, Dr. Mohamed Bakr
Reciprocity Theorem (Cont’d)
•••• Applying divergence theorem, we obtain the Lorentzreciprocity theorem
=∫∫ ×−×−S
dsHEHE ).( 1221 ∫∫∫ −−+V
dV)....( 21121221 MHJEMHJE
•••• For a source-free region we have0)( 1221 =∫∫ ×−×− •
S
dsHEHE
•••• If S is taken as a sphere of infinite radius, we have
0)....( 21121221 =∫∫∫ −−+V
dVMHJEMHJE
∫∫∫ −=∫∫∫ −VV
dVdV )..()..( 12122121 MHJEMHJE
19EE750, 2003, Dr. Mohamed Bakr
Surface Equivalence Theorem
• This theorem is based on the uniqueness theorem
• It obtains the fields outside an imaginary surface byplacing electric and magnetic sources on the boundary sothat the same boundary conditions are satisfied
• Assume that sources J1 and M1 radiate in an unboundedmedium
• We place a virtual surface S that encloses these sources
20EE750, 2003, Dr. Mohamed Bakr
Surface Equivalence Theorem (Cont’d)
V1
V2
E1, H1
E1, H1
S J1 M1
Remove the sources and assumearbitrary fields E, H inside S. Forthe boundary conditions to besatisfied we add the boundarysources )( 1 HHnJ −×=s
)( 1 EEnM −×−=s
ε1, µ1ε1, µ1
V1
V2
E1, H1
E, H
S ε1, µ1ε1, µ1
Js
Ms
As E and H are chosenarbitrarily we may chooseE=0, H=0
21EE750, 2003, Dr. Mohamed Bakr
Surface Equivalence Theorem (Cont’d)
V1
V2
E1, H1
0, 0
S ε1, µ1ε1, µ1
Js=n×H1
Ms=-n ×E1
As Js and Ms represent thetangential components of theH1 and E1 fields, only one ofthem is needed according touniqueness theorem
V2
E1, H1
S ε1, µ1
Js=0
Ms=-n ×E1
Electric Conductor Notice that Js is shorted out
22EE750, 2003, Dr. Mohamed Bakr
V2
E1, H1
S ε1, µ1
Ms=0
Js=n ×H1
Magnetic Conductor Notice that Ms is shorted out
Surface Equivalence Theorem (Cont’d)