ECE 6340
Intermediate EM Waves
Fall 2016
Prof. David R. Jackson
Dept. of ECE
Notes 23
1
Duality
Assume a region of space with sources,
and allow for all types of sources and charges:
  E   M  jc H
i
  H  J  j c E
i
  ( E )  v
Maxwell’s equations
now have a completely
symmetric form.
  (  H )  vm
 c    j / 
c    j m / 
Note:
A magnetic conductivity m is introduced for
generality, though it is always zero in
practice.
2
Duality (cont.)
The substitutions shown below leave Maxwell’s equations unaffected.
EH
J M
H  E
M  J
 ,  c   , c
 , c   ,  c
v  
i
i
i
i
m
v
  v
m
v
3
Duality (cont.)
Example:
  E   M i  jc H
  H  J i  j c E
 
    E   M  jc H
  H  J  j c E
 E   M  jc H
  H    J  j c   E 
i
i
i
i
4
Duality (cont.)
Duality allows us to find the radiation from a magnetic current.
Steps:
1)
Start with the given magnetic current source of interest.
2)
Consider the dual problem that has an electric current
source with the same shape or form (Case A).
3)
Find the radiation from the electric current source.
4)
Apply the dual substitutions to find the radiation from the
original magnetic current (Case B).
Note:
When we apply the duality substitution, the numerical values of the permittivity and
permeability switch. However, once we have the formula for radiation from the
magnetic current, we can let the values be replaced by their conventional ones.
5
Radiation From Magnetic Current (cont.)
Problem of interest:
M
i
M  x, y , z   f  x, y , z 
i
A magnetic current is radiating in free space.
Free-space is assumed for simplicity. To be
more general, we can replace
0  c
0 c
6
Radiation From Magnetic Current
The shape of the electric
current is the same as that of
the original magnetic current
that was given.
Case (A): J i only
J
i
J  x, y , z   f  x, y , z 
i
 jk r  r 
e 0
i
A  0  J ( r )
dV 
4 r  r 
V
H
1
0
 A
7
Radiation From Magnetic Current (cont.)
Also
E   j A  
and
  A   j0 0
so
E   j A 
1
j0 0
(  A)
8
Radiation From Magnetic Current (cont.)
Case (B): M i only
M
i
M  x, y , z   f  x, y , z 
i
From duality:
H
1
0
 A
Case A
where
E 
1
0
 F
Case B
where
A  0 
V
 jk r  r 
e 0
J ( r )
dV 
4 r  r 
i
 jk r  r 
e 0
i
F   0  M ( r )
dV 
4 r  r 
V
9
Radiation From Magnetic Current (cont.)
Also,
E   j A 
H   j F 
1
j0 0
1
j 0 0
(  A)
Case A
(   F )
Case B
10
Radiation from Magnetic Current
Summary
M
i
F  0 
V
 jk r  r 
e 0
i
M ( r )
dV 
4 r  r 
E
1
0
H   j F 
F
1
j 0 0
(   F )
11
General Radiation Formula
i
Both J and M
i:
M
i
J
i
J A
i
M F
i
Use superposition:
E
H
1
0
1
0
  F  j A 
  A  j F 
1
j0 0
1
j 0 0
(  A)
(   F )
Note: Duality also holds with the vector potentials, If we include these:
AF
F  A
12
Far-Field
Case (A) (electric current only):
E ~  j A t ( r ,  ,  )
 0 
A ~
  r  a ( ,  )
 4 
1
H ~ ( rˆ  E )
0
where
a ( ,  )   J ( r ) e
i
V
 jk  r 
dV 
e  jk0r
 r 
r
13
Far-Field (cont.)
Case (B) (magnetic current only):
H ~  j Ft (r , ,  )
 0 
F ~
  r  f ( ,  )
 4 
E ~  0 ( rˆ  H )
where

f ( ,  )   M ( r ) e jk r dV 
i
V
14
Example
z
Radiation from a
magnetic dipole
K magnetic Amps
l
y
Case B
x
z
I electric Amps
Dual problem:
l
x
y
Case A
15
Example (cont.)
For the electric dipole:
( Il )  jk0r  jk0 1 
ˆ
H 
e
 2  sin 

4
r 
 r
Duality:
IK
H  E
So, for the magnetic dipole:
( Kl )  jk0r  jk0 1 
ˆ
E  
e
 2  sin 

4
r 
 r
16
Example
A small loop antenna is equivalent to a magnetic dipole antenna,
z
 0k0a 2 I 0   e  jk0r 
Far-field: E  
 sin 

4

 r 
a
y
I
Kl  j0  AI 0 
x
z
A   a2
Kl
y
Far-field:
x
( Kl )  jk0r  jk0 
E  
e

 sin 
4
 r 
This is the correct far field: The loop can be modeled as a magnetic dipole.
17
Boundary Conditions
n̂
1
Electric surface current:
Js
2
ˆn  ( H (1)  H (2) )  J s
Duality:
n̂
1
Ms
2
nˆ  ( E  E )  M s
(1)
(2)
18
Boundary Conditions (cont.)
nˆ  ( E  E )   M s
(1)
(2)
n̂
1
Ms
2
19
Example: Magnetic Current on PEC
A magnetic surface current flows on top of a PEC.
PEC
Ms
n̂
Ms
1
2
PEC


(1)
( 2)
nˆ  E  E   M s
20
Example (cont.)
ˆn  E (1)   M s
or



ˆn  nˆ  E t (1)   nˆ  M s
1
E t  nˆ  M s
Dropping the superscript, we have
Note:
The relation between the direction of the
electric field and the direction of the
magnetic current obeys a “left-hand rule”.
E t  nˆ  M s
Et  0
PEC
n̂
  
Ms
21
Example (cont.)
n̂
Et  0
  
PEC
Ms
E t  nˆ  M s
We then have
nˆ  E t  nˆ   nˆ  M s 
nˆ  E t   M s
Note: The tangential subscript can be dropped.
Hence
M s  nˆ  E
This allows us to find the magnetic
surface current necessary to produce
any desired tangential electric field.
22
Modeling of Slot Antenna Problem
PEC
Et
Et  0
Et  0
Et  0
    
Side view
M s   nˆ  E
Et  0
Et  0
Ms
n̂
Et  0
23