ECE 1100 Introduction to Electrical and Computer Engineeringcourses.egr.uh.edu/ECE/ECE6341/Class...
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Prof. David R. Jackson ECE Dept. Spring 2016 Notes 15 ECE 6341 1
Transcript of ECE 1100 Introduction to Electrical and Computer Engineeringcourses.egr.uh.edu/ECE/ECE6341/Class...
Prof. David R. Jackson ECE Dept.
Spring 2016
Notes 15
ECE 6341
1
Arbitrary Line Current
TMz: ( , )ρzA z
Introduce Fourier Transform:
1( ) ( )2
zjk zz zI z I k e dk
π
+∞−
−∞
= ∫
( )( ) zjk zzI k I z e dz
+∞+
−∞
= ∫
2
y
z
x
( )I z
Arbitrary Line Current (cont.) View this as a collection of phased line currents::
Then
0( ) zjk zdI z I e−=
01 ( )
2π=
z zI I k dk
( ) ( )
( ) ( )
20 00
200
41 ( )
4 2
z
z
jk zz
jk zz z
IdA H k ej
I k dk H k ej
ρ
ρ
µ ρ
µ ρπ
−
−
=
=
(from Notes 11)
3
1( ) ( )2
zjk zz zI z I k e dk
π
+∞−
−∞
= ∫
Hence, from superposition, the total magnetic vector potential is
( ) ( )200( )
8zjk z
z z zA I k H k e dkj ρ
µ ρπ
+∞−
−∞
= ∫
Arbitrary Line Current (cont.)
4
( )1/22 2zk k kρ = −
2 2
2 2
,
,z z
z z
k k k kk
j k k k kρ
− ≤= − − ≥
Example
Uniform phased line current
00( ) zjk zI z I e−=
0
0
0
( )0
0 0
( )
2 ( )
z z
z z
jk z jk zz
j k k z
z z
I k I e e dz
I e dz
I k kπδ
+∞− +
−∞
+∞−
−∞
=
=
= −
∫
∫
( ) 12
12
x
x
jk xx
jk xx
x e dk
e dk
δπ
π
∞−
−∞
∞+
−∞
=
=
∫
∫
5
y
z
x
( )I z
Note:
Example (cont.) Hence
( ) ( )
[ ] ( ) ( )
( ) ( )( ) ( )
0
0
200
200 0 0
200 0 0
20 00 0
( )8
2 ( )8
28
4
z
z
z
z
jk zz z z
jk zz z z
jk z
jk z
A I k H k e dkj
I k k H k e dkj
I H k ej
I H k ej
ρ
ρ
ρ
ρ
µ ρπ
µ πδ ρπµ π ρπµ ρ
+∞−
−∞
+∞−
−∞
−
−
=
= −
=
=
∫
∫
( )1/22 20 0zk k kρ = −where
6
Example (cont.) Hence
( )1/22 20 0zk k kρ = −
where
0(2)00 0( )
4zjk z
zIA H k ej ρ
µ ρ − =
7
If kz0 is real, then
2 20 0
0 2 20 0
,
,z z
z z
k k k kk
j k k k kρ
− ≤= − − ≥
This is the correct choice of the wavenumber.
y
z
x
( )I z
Example (cont.)
( )1/22 20 0zk k kρ = −
8
If kz0 is complex:
( )( )
0 0
0 0
0 : Im 0 ( )
: Im 0 ( )
z
z
k k
k kρ
ρ
β
β
< < >
> <
improper
proper
0 0 0z z zk jβ α= −
This is the “physical” choice of the wavenumber.
Note that the radiation condition at infinity is violated (z → - ∞), so we lose uniqueness. However, we can still talk about what choice of the square root is “physical.”
y
z
x
( )I z
Example (cont.)
9
( )0 00 : Im 0 ( )z k kρβ< < > improper
A semi-infinite leaky-wave line source produces a cone of radiation in the near field.
y
z
x
( )I z
Cone of leakage Region of exponential growth
Example
Dipole
( ) ( )I z Il zδ=
( )zI k Il=
10
y
z
x
Il
Example (cont.)
Also,
( ) ( )200 ( )
8zjk z
z zA I H k e dkj ρ
µ ρπ
+∞ −
−∞= ∫
( ) 0
4
jkr
zeA I
rµπ
−
=
(from ECE 6340)
( ) ( )200( )
8zjk z
z z zA I k H k e dkj ρ
µ ρπ
+∞−
−∞
= ∫
Hence
11
2 2
2 2
,
,z z
z z
k k k kk
j k k k kρ
− ≤= − − ≥
Example (cont.) Hence,
A spherical wave is thus expressed as a collection of cylindrical waves.
( )20
1 ( )2
z
jkrjk z
ze H k e dk
r j ρ ρ− +∞ −
−∞= ∫
12
Example (cont.) Use a change of variables:
( ) ( )
( )
2 20 0
20
1 1( ) ( )2 2
1 ( )2
z z
CCW
z
CW
jk z jk zz
zC
jk z
zC
kH k e dk H k e dk
j j k
kH k e dk
j k
ρρ ρ ρ
ρρ ρ
ρ ρ
ρ
+∞ − −
−∞
−
−=
=
∫ ∫
∫
13
( )
( )
1/22 20
1/22 20
z
zz
k k k
k kdk dk dk
kk k
ρ
ρ ρρ ρ
ρ
= −
= − = −−
( )20
1 ( )2
z
jkrjk z
z
ke H k e dkr j k
ρρ ρρ
− +∞ −
−∞
=
∫
We then have
or
(The path has been deformed back to the real axis, please see the next two slides)
Example (cont.)
14
( )1/22 20zk k kρ= −
Branch cut
( )Re kρ
CCCW
( )Im kρ
0k
Mapping equation for C:
Note: The value of kz is opposite across the branch cut.
CCCW: counterclockwise around branch cut
0k−
( ),CCW
z
k C
kρ ∈
∈ −∞ ∞
The square root is defined so that when kρ is on the real axis we have:
2 20 0,zk j k k k kρ ρ= − − >
Example (cont.)
15
( )Re kρ
CCW
( )Im kρ
0k
CCW: clockwise around branch cut
Deform path
0k−
( )Re kρ
( )Im kρ
0k
Branch cut
0k−
“Sommerfeld path”
Example (cont.)
16
Alternative form:
( )
( ) ( )
( ) ( )
20
0 2 20 00
0 2 20 00
0
( )2
( ) ( )2 2
( ) ( )2 2
2
z
z z
z z
jkrjk z
z
jk z jk z
z z
jk z jk z
z z
z
ke j H k e dkr k
k kj jH k e dk H k e dkk k
k kj jH k e dk H k e dkk k
kj Hk
ρρ ρ
ρ ρρ ρ ρ ρ
ρ ρρ ρ ρ ρ
ρ
ρ
ρ ρ
ρ ρ
− +∞ −
−∞
+∞− −
−∞
+∞− −
∞
= −
= − − ′
′ ′= − − −
′ = +
∫
∫ ∫
∫ ∫
( ) ( )
( ) ( )
( )
0 1 200
1 20 00 0
00
( ) ( )2
( ) ( )2 2
z z
z z
z
jk z jk z
z
jk z jk z
z z
jk z
z
kjk e dk H k e dkk
k kj jH k e dk H k e dkk k
kj J k e dk
k
ρρ ρ ρ ρ
ρ ρρ ρ ρ ρ
ρρ ρ
ρ ρ
ρ ρ
ρ
+∞− −
∞
∞ +∞− −
∞ −
′ ′ −
′
′ ′= − −
= −
∫ ∫
∫ ∫
∫
k kρ ρ′ = −
( ) ( ) ( ) ( )2 10 0H z H z− = − +
( ) ( ) ( ) ( ) ( )( )1 20 0 0
12
J z H z H z= +
Example (cont.)
17
00
1 ( ) z
jkrjk z
z
ke J k e dkr j k
ρρ ρρ
− +∞ − =
∫
Sommerfeld Identities
Hence we have:
( )20
1 ( )2
z
jkrjk z
z
ke H k e dkr j k
ρρ ρρ
− +∞ −
−∞
=
∫
The integrals are along a “Sommerfeld path” that stays slightly above or below the branch cuts.
Far-Field Identity This identity is useful for calculating the far-field of finite (3-D) sources in cylindrical coordinates.
Note: We assume that the current decays at z = ± ∞ fast enough so that a 3-D far field exists.
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y
z
x
( )I zr θ
Far-Field Identity (cont.)
( )200( ) ( )
8zjk z
z z zA I k H k e dkj ρ
µ ρπ
+∞ −
−∞= ∫
Exact solution:
19
y
z
x
( )I zr θ
Far-Field Identity (cont.) From ECE 6340, as →∞r
0~ ( , )4
jkreA ar
µ θ φπ
−
( ) ( )sin cos sin sin cos
cos0
( , )
ˆ ( )4
jk x y z
V
jkrj kz
a J r e dx dy dz
ez I z e dzr
θ φ θ φ θ
θ
θ φ
µπ
′ ′ ′+ +
− +∞ ′
−∞
′ ′ ′ ′=
′ ′=
∫
∫
Hence
0~ ( cos )4
jkr
zeA I k
rµ θπ
−
20
Far-Field Identity (cont.)
Hence, comparing these two,
( )20 00( ) ( ) ~ ( cos )
8 4z
jkrjk z
z zeI k H k e dk I k
j rρµ µρ θπ π
−+∞ −
−∞
∫
r →∞as
( )20( ) ( ) ~ 2 ( cos )z
jkrjk z
z zeI k H k e dk j I k
rρ ρ θ−+∞ −
−∞
∫
or
21
Far-Field Identity (cont.)
To generalize this identity, use
( ) ( )2 2 42( ) ~nj x
nH x ex
π π
π− − −
Hence
( ) ( )2 20( ) ~ ( )n
nH x j H x
22
Far-Field Identity (cont.)
Therefore, we have
( )2 1( ) ( ) ~ 2 ( cos )z
jkrjk z n
z n zeI k H k e dk j I k
rρ ρ θ−+∞ − +
−∞
∫
Note: This is valid for ρ →∞
, 0, 180r θ→∞ ≠ Hence, this is valid for
23
Far-Field Identity (cont.)
Since the current function is arbitrary, we can write
( )2 1( ) ( ) ~ 2 ( cos )z
jkrjk z n
z n zef k H k e dk j f k
rρ ρ θ−+∞ − +
−∞
∫
for , 0, 180r θ→∞ ≠
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