prof. david r. jackson ece dept. spring 2014 notes 24 ece 6341 1
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Prof. David R. JacksonECE Dept.
Spring 2014
Notes 24
ECE 6341
2
Dipole Radiation
Using the Lorenz gauge (the solution from ECE 6340):
2 2 ( ) ( ) ( )LG LGz z zA k A J x y z
y
x
z
1I
4
jkrLGz
eA
r
Note: This is not the same
magnetic vector potential that we get using the Debye potential!
ˆ zA z A
2 2 iA k A J
3
Dipole Radiation (cont.)
Note:
2 3
2 3
2
1 1cos
2
1 1sin
4
1 1sin
4
jkrr
jkr
jkr
E er j r
jE e
r r j r
jkH e
r r
0rH
The fields are (from ECE 6340):
(TMr)
4
Dipole Radiation (cont.)
(2)ˆcosr n nA AP H kr
0 ( )
( )
( )n
m
Q z
n z
No
Note : no variation
axis included
axis included
0
1
2
cos 1
cos cos
1cos 3cos 2 1
4
P
P
P
TMr
Examine some
values of n:
5
Dipole Radiation (cont.)
22
2
1r rE k A
j r
From the handout,
cosrE From the ECE 6340 solution,
Hence cosrA
Hence, choose 1n
6
Dipole Radiation (cont.)
(2)1
ˆcosrA A H kr
(2) (2)1 3/2 3/2 3/2
ˆ ( )2 2
xH x x H x J x jY x
x
Therefore we have
Next, simplify the Schelkunoff Hankel function:
3/ 2 3/ 2Y x J x cos
sin
J x J xY x
7
Dipole Radiation (cont.)
(2)1
2 sin 2 cosˆ cos sin2
x x xH x x j x
x x x x
3/ 2
3/ 2
2 sincos
2 cossin
xJ x x
x x
xJ x x
x x
We have (from the Schaum’s Math Handbook):
Hence
8
Dipole Radiation (cont.)
(2)1
sin cosˆ cos sin
cos sin cos sin
cos sin cos sin
1 jx
x xH x x j x
x x
jx j x x j x
xj
x j x x j xx
je
x
Hence
(2)1
2 sin 2 cosˆ cos sin2
x x xH x x j x
x x x x
(2)1
ˆ 1 jxjH x e
x
so
9
Dipole Radiation (cont.)
Hence
cos 1 jkr
r
jA A e
kr
The final step is to determine the coefficient A.
10
Dipole Radiation (cont.)
The final result is
2
22
2 3
1cos 1 1
1 1cos
2
jkrr
jkr
jE k A e
j r kr
er j r
4
j k
A
(from ECE 6340 solution)
(from TMr table)
Compare the Er field:2
22
1r rE k A
j r
11
Dipole Radiation (cont.)
Hence we have
cos 14
jkrr
j k jA e
kr
12
Dipole Radiation (cont.)
Debye potentials (using Debye Gauge):
ˆ cos 14
jkrj k jA r e
kr
Lorenz Gauge:
ˆ4
jkrA z er
Comparison