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Prof. David R. JacksonDept. of ECE
Fall 2013
Notes 26
ECE 6340 Intermediate EM Waves
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Equivalence Principle
Basic idea:We can replace the actual sources in a region by equivalent sources at the boundary of a closed surface.
Keep original fields E , H outside S.
Put zero fields (and no sources) inside S.
E , H
a
b
ANT
S
E , H
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Equivalence Principle (cont.)
Note: (Ea , Ha) and (Eb , Hb) both satisfy Maxwell’s equations.
Eb = 0 Hb = 0
No sources
Ea = E Ha = H
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The B.C.’s on S are violated.
Introduce equivalent sources on the boundary to make B.C.’s valid.
n̂
a
b
esJ
esM
ˆe a bsJ n H H
ˆe a bsM n E E
Equivalence Principle (cont.)
4
Zero fields
Original fields
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Outside S, these sources radiate the same fields as the original antenna,
and produce zero fields inside S.
This is justified by the uniqueness theorem:
Hence ˆ
ˆ
es
es
J n H
M n E
Equivalent sources:
Zerofields
esJ
esMZero
sources
S
Equivalence Principle (cont.)
Maxwell's equations are satisfied along with boundary conditions at the interface.
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Note about materials:
If there are zero fields throughout a region, it doesn’t matter what material is placed there (or removed).
Equivalence Principle (cont.)
This object can be added
Zerofields
esJ
esM
r
(E , H)
S
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Scattering by a PEC
sE
(E , H)
Source
iE
0tE
PEC
S
sJ
i sE E E i sH H H
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Scattering by a PEC (cont.)
ˆ ˆ 0
ˆ
es t
es s
M n E n E
J n H J
S
Source
0 0,
esJesM
(E , H)
(0 ,0)
Equivalent Sources:
n̂
sJ (E , H)
SourcePEC
S
We put zero fields and sources inside of S, and remove the PEC object.
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Conclusion: The conductor can be removed.
Equivalent problem:
Source
0 0, sJ
(E , H)
(0 ,0)
S
(E , H)
Source PEC
S
sJ
Original problem:
Scattering by a PEC (cont.)
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Source
0 0, sJ
(E , H)
(0 ,0)
0,it tsE J E r S
it tsE J E
This integral equation has to be solved numerically.
Integral equation for the unknown current
so
0s it tE E
“Electric Field Integral Equation” (EFIE)
Note: The bracket notation means
“field due to a source.”
Scattering by a PEC (cont.)
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Scattering by Dielectric Body
sE
(E , H)Source
,r r iE
i sE E E
i
s
E
E
incident field
scattered field
(E , H)
Note: The body is assumed to be homogeneous.
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Exterior Equivalence
Replace body by free space(The material doesn’t matter in the zero-field region.)
S
Source
,r r
Ea = E
Ha = H
Eb = 0 Hb = 0
ˆ 0
ˆ 0
e as
e as
J n H
M n E
S
Source
0 0,
esJ
esM
(E , H)(0 ,0)
n̂12
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Summary for Exterior
ˆ
ˆ
es
es
J n H
M n E
,r r
S
Original problem:
Free-space problem:
S
0 0,
esJ
esM
(0 ,0)
n̂
(E , H)
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Interior Equivalence
e es s
e es s
J J
M M
S
No sources or fields outside S
,r r
Ha, Ea
Eb = 0
Hb = 0
Place dielectric material in the "dead region" (region b).
ˆ ˆ ˆ0
ˆ ˆ ˆ0
e a es s
e a es s
J n H n H n H J
M n E n E n E M
ˆ ˆn n
Ea = E
Ha = H
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Interior Equivalence (cont.)
S
e es sJ J
e es sM M
(0 ,0)
E, H
,r r
,r r n̂
When we calculate the fields from these currents, we let them radiate in an infinite dielectric medium.
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Summary for Interior
,r r
S
Original problem:
S
esJ esM
(0 ,0)
E, H
,r r
,r r
Homogeneous-medium problem:
n̂
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Integral Equation
,r r
S
, ,
, ,
e e i e et s t t ss s
e e i e et s t t ss s
E J M E E J M
H J M H H J M
, ,
, ,
e e e e it s t s ts s
e e e e it s t s ts s
E J M E J M E
H J M H J M H
Boundary conditions:
Hence:
The “–” means calculate the fields just inside the surface, assuming an infinite dielectric region.
Note: The “+” means calculate the fields just outside the surface,radiated by the sources in free space.
“PMCHWT" Integral Equation*
* Poggio-Miller-Chang-Harrington-Wu-Tsai 17
e es s
e es s
J J
M M
Recall:
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Fields in a Half Space
Equivalent sources:
ˆ
ˆ
es
es
J z H
M z E
(0 ,0)es
es
J
M
(E , H)
Sources, structures,
etc.
z
(E , H)Region of interest
(z > 0)
Equivalence surface S (closed at infinity in the z < 0 region)
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Hence, we have:
Put PEC in zero-field region:
The electric surface current on the PEC does not radiate.
Fields in a Half Space (cont.)
Note: The fields are only correct for z > 0.
19
z
(E , H)es
es
J
MPEC
ˆesM z E
(E , H)
esMPEC z
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Note: The fields are still correct for z > 0.
Now use image theory:
ˆ2sM z E
Fields in a Half Space (cont.)
This is useful whenever the electric field on the z = 0 plane is known.
20
Incorrect fields
(E , H)
sMCorrect fields
0 0( , )
2 es sM M z
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Fields in a Half Space: Summary
Sources + objects
z
(E , H)Region of interest
(z > 0)
21
Incorrect fields
(E , H)
sMCorrect fields
0 0( , )
ˆ2sM z E
z
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Alternative (better when H is known on the interface):
Mse does not radiate on PMC,
and is therefore not included.
Image theory:
Incorrect fields
(E , H)
sJCorrect fields0 0( , )
ˆ2 2es sJ J z H
Fields in a Half Space (cont.)
22
esJPMC
z
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Example: Radiation from Waveguide
zb
y
0ˆ( , ,0) cosx
E x y y Ea
b
a
x
y
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Example (cont.)
Step #1
0ˆ2 coss
xM x E
a
Step #2
0
ˆ
ˆ cos
esM z E
xx E
a
PEC
z
z
24
The feeding waveguide was removed from the dead region that was created, and then a continuous PEC plane was introduced.
Image theory is applied to remove the ground plane and double the
magnetic surface current.
Apply equivalence principle
Apply image theory
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0ˆ2 coss
xM x E
a
x
a
b
y
z
Example (cont.)
25
A three-dimensional view
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0ˆ2 coss
xJ x E
a
a
b
x
y
z
ssJ M
E H
H E
Solve for the far field of this problem first.
(This is the “A” problem in the notation of the duality notes.)
Then use:
Example (cont.)
Use the theory of Notes 22 to find the far field from this rectangular strip of electric surface current.
0 0
0 0
26
Step #3
Apply duality
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A radiating electric current can be replaced by a magnetic current, and vice versa.
Volume Equivalence Principle
27
y
x
z
J
y
x
z
M
1 1( , )E H 2 2( , )E H
P. E. Mayes, “The equivalence of electric and magnetic sources,” IEEE Trans. Antennas Propag., vol. 6, pp. 295–29, 1958.
We wish to have the same set of radiated fields.
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Volume Equivalence Principle (Cont.)
28
1 0 1
0 0 1
E j H
j J j E
1 0 1
1 0 1
E j H
H J j E
21 0 1 0E k E j J
Set 1 (electric current source):
Hence
Therefore, we have
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Volume Equivalence Principle (Cont.)
29
2 0 2
0 0 2
E j H M
j j E M
2 0 2
2 0 2
E j H M
H j E
22 0 2E k E M
Set 2 (magnetic current source):
Hence
Therefore, we have
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Volume Equivalence Principle (Cont.)
30
22 0 2E k E M
21 0 1 0E k E j J
0j J M
0
1J M
j
Set
Compare:
Hence
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Volume Equivalence Principle (Cont.)
31
1 2 0 1 0 1E E j H j H M
1 20
1H H M
j
Hence, the two electric fields are equal everywhere, but the magnetic fields are only the same outside the source region.
Next, examine the difference in the two Faraday laws:
1 0 1
2 0 2
E j H
E j H M
so
This gives us
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Summary
32
y
x
z
J
y
x
z
M
0
1J M
j
0
1H J H M M
j
E J E M
Volume Equivalence Principle (Cont.)
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Apply duality on the two curl equations to get two new equations:
33
22 0 2H k H J
21 0 1 0H k H j M
y
x
z
M
1 1( , )E H
y
x
z
J
2 2( , )E H
Volume Equivalence Principle (Cont.)
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34
22 0 2H k H J
21 0 1 0H k H j M
0
1M J
j
1 0 1
2 0 2
H j E
H j E J
1 2 0 1 2H H j E E J
Similarly, from duality we have
Volume Equivalence Principle (Cont.)
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35
y
x
z
M
y
x
z
J
0
1M J
j
0
1E M E J J
j
H M H J
Summary
Volume Equivalence Principle (Cont.)