prof. david r. jackson dept. of ece fall 2013 notes 22 ece 6340 intermediate em waves 1
TRANSCRIPT
Prof. David R. JacksonDept. of ECE
Fall 2013
Notes 22
ECE 6340 Intermediate EM Waves
1
Radiation
Infinitesimal Dipole
Current model:
iz
I IlJ
x y x y z
1kl
y
x
l
z
I
y
x y
z
x
z l
2
Radiation (cont.)
Define
Then
1,
( ) 2 2
0
x xx
P x x
otherwise
( , , ) ( ) ( ) ( ) ( )izJ x y z Il P x P y P z
As 0, ( ) ( )x P x x
, , 0x y z Letting
( , , ) ( ) ( ) ( ) ( )izJ x y z Il x y z
( , , ) ( ) ( )izJ x y z Il ror
3
Maxwell’s Equations:
From (3):
(1)
(2)
0 (3)
/ (4)
ic
v
E j H
H J j E
H
E
1H A
(5)
Note: c accounts for conductivity.
Radiation (cont.)
Note: The Harrington book uses
H A
4
ˆ( , , ) ( ) ( )iJ x y z z Il r
From (1) and (5):
or
1( )E j A
( ) 0E j A
Hence E j A
E j A (6)
Radiation (cont.)
This is the “mixed potential” form for E.5
or
Substitute (5) and (6) into (2):
Use the vector Laplacian identity:
1( ) [ ]i
cA J j j A
2( ) icA k A J j
2 ( ) ( )A A A
The vector Laplacian has his nice property in rectangular coordinates:
2 2 2 2ˆ ˆ ˆx y zA x A y A z A
icH J j E
Radiation (cont.)
6
Hence
or
Then we have
2 2 ( )icA k A J A j
2 2( ) icA A k A J j
cA j
Lorenz Gauge:
2 2 iA k A J
Choose
Radiation (cont.)
2 2ck
Note: Lorenz, not Lorentz (as in Lorentz force law)!
7
Take the z component:
Hence we have
2 2 iz z zA k A J
2 2 ( ) ( )z zA k A Il r
Note:2 2 0x xA k A 2 2 0y yA k A
0xA 0yA so
Radiation (cont.)
8
Spherical shell model of 3D delta function:
2 2 ( ) ( )z zA k A Il r
Radiation (cont.)
y
x
a
z
Sa
The Az field should be symmetric (a function
of r only) since the source is.
2
1( )
4r r a
a
In spherical coordinates:
24 1a
V a
r dV r r dr
Note :
9
Assume
For
( ) ( )zA r R r
0r 2 2 0z zA k A
so that 2 22
10
d dRr k R
r dr dr
Next, let1
( ) ( )R r h rr
we have that
Radiation (cont.)
2 22
( ) 1 1( ) ( )
( ) ( )
dR rr r h r h r
dr r r
rh r h r
10
22
22
2
10
10
1( ) ( ) 0
d hrh h k
r dr rh
h rh h kr r
h r k h rr
( ) jkr jkrh r Ae Be
We choose exp (-jkr) for r > a to satisfy the radiation condition at infinity.
Solution:
Radiation (cont.)
We choose sin(kr) for r < a to ensure that the potential is finite at the origin.
( ) / .R r h r rRecall that
11
,
sin ,
jkrAe r ah r
B kr r a
Radiation (cont.)Hence
2 22 2
1 1( ) ( ) ( )
4
d dRr k R Il r Il r a
r dr dr a
22
1 1( ) ( ) ( )
4h r k h r Il r a
r a
The R function satisfies
2 1( ) ( ) ( )
4h r k h r Il r a
a
12
Also, from the differential equation we have
Radiation (cont.)
2 1( ) ( ) ( )
4
a a a
a a a
h r dr k h r dr Il r a dra
1( )
4h a h a Il
a
h a h a
We require
13
(BC #1)
(BC #2)
Hence
Radiation (cont.)
,
sin ,
jkrAe r ah r
B kr r a
sin
1cos ( )
4
jka
jka
Ae B ka
A jk e Bk ka Ila
Recall
14
(BC #2)
(BC #1)
Substituting Ae-jka from the first equation into the second, we have
Radiation (cont.)
1sin cos ( )
4jk B ka Bk ka Il
a
Hence we have
1 1
( )4 sin cos
B Ila k j ka ka
sin1( )
4 sin cos
jkae kaA Il
a k j ka ka
From the first equation we then have
15
Letting a 0, we have
Radiation (cont.)
sin1( )
4 sin cos
1( )
4
jkae kaA Il
a k j ka ka
Il
1( )
4A Il
Hence
1 1 1( )
4jkr jkr
zA r R r h r Ae Il e r ar r r
16
so
Final result:
ˆ ( )4
ˆˆ cos sin ( )4
jkr
jkr
eA z Il
r
er Il
r
Radiation (cont.)
yl
z
I
( , , ) ( ) ( )izJ x y z Il rx
Note: It is the moment Il that is important. 17
Extensions
(a) Dipole not at the origin
ˆ ( )4
jk r reA z Il
r r
y
x
r
z
r r
r
18
Extensions (cont.)
(b) Dipole not in the z direction
x
y
r
z
r r
rp̂
ˆ ( )4
jk r reA p Il
r r
19
(c) Volume current
y
x
( )iJ r
z
( )4
jk r ri
V
eA J r dV
r r
Extensions (cont.)
dl
ˆ
ˆ( )( )
i izJ dV z J dS dl
z I dl
dS
20
ˆ ( )4 4
jk r r jk r rie e
d A z Idl J dVr r r r
Hence
Consider the z component of the current:
The same result is obtained for the current components in the x and y directions, so this is a general result.
(d) Volume, surface, and filamental currents
( ) ( )4
( ) ( )4
ˆ( ) ( ) ( )4
jk r ri
V
jk r ris
S
jk r ri
C
eJ r dV
r r
eA r J r dS
r r
er I r d
r r
volume current density
surface current density
filament of current
( )
( )
ˆ ( )
i
i is
J dV
J dV J dS
I d
volume current density
surface current density
filamentary (wire) current
Extensions (cont.)
21Note: The current Ii is the current that flows in the direction of the unit vector.
ˆ ( ) ( )4
jk r ri
C
eA r r I r d
r r
Extensions (cont.)
22
Note: The current Ii is the current that flows in the direction of the contour.
Filament of current:
CA
BI
Fields
To find the fields:
1
1( 0).
c
H A
E H rj
valid for
1
c
E j A
j A Aj
Note: Ampere’s law (the equation above) for finding the electric field is an alternative to using the following equation:
23
cA j Recall :
Fields (cont.)
Results for infinitesimal, z-directed dipole at the origin:
2
2
1 11 cos
2
1 1 1( ) 1 sin
4 ( )
1 1( ) 1 sin
4
jkrr
jkr
jkr
IlE e
r jkr
IlE j e
r jkr jkr
IlH jk e
r jkr
c
with
24
Note on Dissipated PowerNote: If the medium is lossy, there will be an infinite amount of power dissipated by the infinitesimal dipole.
2
22 2
0 0 0
22 2 2
0 0 0
2 2 2
0
2 22 2
30
1
2
1sin
2
1sin
2
1 4 22
2 3 3
1 4 2 12 1 2
2 3 3 4
d
V
r
N Nr
c
P E dV
E r d d dr
E E r d d dr
E E r dr
Ilr dr
r
2
0
2
0
4sin sin
3
2cos sin
3
d
d
Note:
The N superscript denotes that the dependence is suppressed in the fields.
(power dissipation inside of a small spherical region V of radius )
25
Far Field
Far-field: ( 1)kr
2
10
~ ( ) sin4
~ ( ) sin4
r
jkr
jkr
E Or
jE Il e
r
jkH Il e
r
behaves as
E
H k
Note that The far-zone field acts like a homogeneous plane wave.
26
Radiated Power (lossless case)
*
2
22 2
0 0
2 2
0
1ˆP Re ( )
2
1Re
2
1Re
2
1sin
2
2sin
2
rad
S
S
S
E H r dS
E H dS
EdS
E r d d
E r d
The radius of the sphere is chosen as infinite in order to simplify the fields.
27
We assume lossless media here.
Radiated Power (cont.)
or2
2 2 2
0
22 3
0
P ( ) (sin ) sin4
( ) sin4
rad Il r dr
Il d
3
0
4sin
3d
Hence2
2 4P ( )
4 3rad Il
Integral result:
28
Use
k
22P ( )12rad
kIl
2k
Simplify using
22 4
P ( )4 3rad
kIl
22
2
4P ( )
12rad Il
Radiated Power (cont.)
Then
or
29
2
2P W3rad
lI
Note: The dipole radiation becomes significant when the dipole length is significant relative to a wavelength.
Radiated Power (cont.)
30
Far-Field Radiation From Arbitrary Current
( , , )4
jk r ri
V
eA J x y z dV
r r
y
x
( )iJ r
z
r r
r
r
r
r
31
12 2 2 2
12 2 2 2 2 2 2
2 2 2 2
2
2
2 2
ˆ1 2
r r x x y y z z
x y z x y z xx yy zz
r r r r r r r r
r r rr
r r
211 ~ 1 ...
2 8
xx x
Far- Field (cont.)
Use Taylor series expansion:
x
32
2 2
3
ˆ ˆ1 1 2 11
2 8
r r r r rr r r O
r r r r
2 2
2
1 1 1 1ˆ ˆ
2 2r r r r r r r r O
r r
As r
ˆr r r r r
Far- Field (cont.)
33
ˆr r r r r
Physical interpretation:
Far- Field (cont.)
Far-field point (at infinity)
y
x
z
r r
r
r
r
34
r̂
Hence
( )jkre
rr
Define:
( , ) ( )i jk r
V
a J r e dV
Far- Field (cont.)
ˆ
( ) ( , , )4
~ ( )4
jk r ri
V
jkri jkr r
V
eA r J x y z dV
r r
eJ r e dV
r
ˆ ˆ ˆ ˆsin cos sin sin cosk kr x k y k z k
(This is the k vector for a plane wave propagating towards the observation point.)
and
35
ˆr r r r r
The we have
( )jkre
rr
where
( , ) ( )i jk r
V
a J r e dV
Far- Field (cont.)
( ) ~ ,4
A r r a
sin cos sin sin cosk r x k y k z k
and
36
with
( , ) ( )i jk r
V
a J r e dV
Fourier Transform Interpretation
Then
sin cos
sin sin
cos
x
y
z
k k
k k
k k
Denote
( , ) ( , , ) x y zj k x k y k zi
V
a J x y z e dV
The array factor is the 3D Fourier transform of the current.
37
Magnetic Field
1H A
We next calculate the fields far away from the source, starting with the magnetic field.
38
In the far field the spherical wave acts as a plane wave, and hence
ˆjk jkr
1ˆH jk r A
~ ,4
A r a
1ˆ
4H jk r r a
Recall Hence
We then have
Note: Please see the Appendix for an alternative calculation of the far field.
ˆ~ ( , ) ( )4
jkH r a r
The far-zone electric field is then given by
1
1ˆ
1ˆ ˆ
4
ˆ ˆ4
c
c
c
E Hj
jkr Hj
jkjkr r a
j
j r r a
Electric FieldWe start with
39
ˆ ˆ ,4
,4
t
E j r r a
j a
Electric Field (cont.)We next simplify the expression for the far-zone electric field.
40
( ) ~ ,4
A r r a
tE j A
Recall that
Hence
Hence, we have
~ , ,tE j A r
Electric Field (cont.)
Note: The exact field is
E j A
41
The exact electric field has all three components (r, , ) in general, but in the far field there are only (, ) components.
Relation Between E and H
Hence
,EE
H H
ˆ4
ˆ4
ˆ4 / (4 )
t
jkH r a
jkr a
jk Er
j
or
1ˆH r E
Start with
~
4
t
t
E j A
j a
42
Recall:
1c ck
Also
~ , ,tE j A r
Summary of Far Field
( )jkre
rr
( , ) ( )i jk r
V
a J r e dV
( , , ) ~ ,4
A r r a
1ˆH r E
43
(keep only and components of A)
sin cos sin sin cosx y zk r k x k y k z k x k y k z
Poynting Vector
* *
* *
*
* *
1
21
~21 ˆ ˆ ˆ ˆ21
ˆ2
1ˆ
2
t t
S E H
E H
E E H H
r E H E H
E EE Er
44
Assuming a lossless media,
22 2
ˆ ˆ2 2 2
EE ES r r
22 2
* * *ˆ ˆ
2 2 2
EE ES r r
Poynting Vector (cont.)
In a lossless medium, the Poynting vector is purely real (no imaginary power flow in the far field).
45
Far-Field Criterion
Antenna(in free-space)
D : Maximum diameter
D
r r
How large does r have to be to ensure an accurate far-field approximation?z
The antenna is enclosed by a circumscribing sphere of diameter D.
/ 2D
y
x
rr
/ 2r D46
Far-Field Criterion (cont.)
Let Df = maximum phase error in the exponential term inside the integrand.
22
2
ˆ1 1ˆ
2 2
r rrr r r r r O
r r r
22
max
ˆ1
2 2
r rrk
r r
2 2
max
( / 2)
2 2
kr k D
r r
Hence
NeglectError
47
Set
Then
2
max 8
kD
r
omax rad (22.5 )
8
2
8 8
kD
r
Far-Field Criterion (cont.)
48
Hence, we have the restriction
2
0
2Dr
2 2 2
0 0
2 2kD D Dr
Far-Field Criterion (cont.)
max rad8
for
49
“Fraunhofer criterion”
Note on choice of origin:
2
0
2Dr
Far-Field Criterion (cont.)
The diameter D can be minimized by a judicious choice of the origin. This corresponds to selecting the best possible mounting point when mounting an antenna on a rotating measurement platform.
50
aD aD
aD D 2 aD D
Mounted at center Mounted at edge
D = diameter of circumscribing sphereDa = diameter of antenna
Far-Field Criterion (cont.)
51
It is best to mount the antenna at the center of the rotating platform.
Platform (ground plane)Antenna
Rotating pedestal
Wire Antenna
Assume
y
+h
z
I (z')
x-h
0( ) sinI z I k h z
+h-h
Wire antenna in free space
52
Wire Antenna (cont.)
Hence
( , ) ( )i jk r
V
a J r e dV
ˆ ( )iJ dV z I z dz
sin cos sin sin cos
cos
ˆ( , ) ( )
ˆ ( )
ˆ ( )
hjk r
h
hj x k y k z k
h
hjz k
h
a z I z e dz
z I z e dz
z I z e dz
53
cos0ˆ( , ) sin
hjz k
h
a z I k h z e dz
ˆ ˆ ˆ ˆˆ ˆ( , )
ˆ sin
t z z
z
a z a za
a
( ) ( , )4 ttE j A j r a
Wire Antenna (cont.)
Recall that
For the wire antenna we have
54
0 2
cos( cos ) cos( )ˆ( , ) (2 )
sin
kh kha z I
k
The result is
ˆ~ sin ( , )4
jkr
z
eE j a
r
Hence
or
0 2
ˆ~ sin ( , )4
cos( cos ) cos( )ˆ sin (2 )4 sin
jkr
z
jkr
eE j a
r
e kh khj I
r k
k
Simplify using
Wire Antenna (cont.)
55
We then have
0
cos( cos ) cos( )ˆ~2 sin
jkre kh khE j I
r
Wire Antenna (cont.)
Note: The pattern goes to zero at = 0, . (You can verify this by using L’Hôpital’s rule.)
56
Wire Antenna: Radiated Power
22 2
0 0
22
0
22
0
2 2 22 2 2
0
0
1sin
2
1(2 ) sin
2
sin
1 1 cos( cos ) cos( )sin
2 sin
radP E r d d
r E d
r E d
kh khr I d
r
57
Radiated Power (cont.)
or
2
20
0
cos( cos ) cos( )
4 sinrad
kh khP I d
58
Input Resistance
2
2
1(0)
22
(0)
rad in
radin
P R I
PR
I
2
20
0
cos( cos ) cos( )
4 sinrad
kh khP I d
59
Zin
Rin
jXin
I (0)
0
0
(0) sin
sin( )
I I k h z
I kh
2 2
0
cos( cos ) cos( )1
2 sin( ) sinin
kh khR d
kh
/2 Dipole: ,4 2
h kh
73inR
Input Resistance (cont.)
(resonant)
60
61
ExampleA small loop antenna
x
y
z
I
a0a
The current is approximately uniform)
0I I
0
2sin cos
0
0
cos jk aya a I e ad
Assume = 0: yE E
ˆˆ( , ) ya ya a
0 sinxk k
cosx a
ˆ ˆˆ ˆ cosl y y
By symmetry:
0yk
The x component of the array factor cancels for points and - .
62
Example (cont.)
2
cos1
0
cos 2jxe d j J x
0
0
0
00 1 0
4
2 sin4
jk r
jk r
eE j a
r
ej j aI J k a
r
0 1 02 sina j aI J k a
Hence
0 sinx k a
Identity:
We then have
63
020 0 0ˆ sin
4
jk rk a I eE
r
1 , 12
xJ x x
Approximation of Bessel function:
The result is
0
0 01 0
ˆ sin2
jk raI eE J k a
r
Hence
Example (cont.)
Appendix
64
In this appendix we derive the far field expression for the magnetic and electric field from a radiating current source by using the curl operations in spherical coordinates, instead of using the plane-wave approximation for the del operator.
Hence
1ˆ, sin
sin
1 1 1 1
sinr r
aa r a
r
ra raa a
r r r r r r
1,a O
r
2
1 1a O O
r r
Therefore
Magnetic Field (cont.)
Note that a is not a function of r, so these derivative terms are O(1).
65
Therefore
2
ˆ ˆ
1ˆ
ˆ~
ˆ
jkr
jkr
jkr
er r
r r r
jkrr e
r
jkr e
r
jkr
Magnetic Field (cont.)
ˆ~4
jkH a r
Hence
1~
4H a
ˆjkr 66
We then have
ˆ~ ( , ) ( )4
jkH r a r
Magnetic Field (cont.)
ˆ~ ( , ) ( )4 t
jkH r a r
or
Note: The t subscript can be added without changing the result.
The t subscript means transverse to r. That is, keep only
the and components.
67
ˆ~ ( , ) ( )4
jkH r a r
The far-zone electric field is then given by
1
1ˆ
4
c
c
E Hj
jkr a
j
Electric FieldWe start with
68
(This follows from the same reasoning as before.)
ˆ ˆ ˆ
ˆ~
r a r a r a
r a
Hence
ˆ ˆ ˆ~
ˆ ˆ
ˆ ˆ
t
r a r a jkr
jk r a r
jk r r a
jk a
Electric Field (cont.)
69
Electric Field (cont.)We next simplify the expression for the far-zone electric field.
70
( , , ) ~ ,4
A r r a
tE j A
Recall that Hence
1ˆ
4
1
4
4
c
t
c
t
jkE r a
j
jkjk a
j
j a