Post on 10-Dec-2015
description
Vector Potentials 2
∇×E = − jωµ
H ∇ i ε
E = ρ
∇×H = jωε
E +J ∇ i µ
H = 0
Maxwell’s Equations – Time-Harmonic Electromagnetic Fields – Homogeneous Medium
Vector Potentials 3
∇ iB = 0
Since
Now for any vector field
∇ i∇×A = 0
Define
B = µ
H = ∇×
A
or
H = 1
µ∇×A
The Vector Potential
Vector Potentials 4
Substituting
H = 1
µ∇×A
∇×E = − jωµ
H
∇×E = − jωµ 1
µ∇×A
∇×E + jω
A( ) = 0
Note that for any scalar function
∇×∇Φ ≡ 0∇×
E + jω
A( ) = 0 = ∇× −∇Φ( )
E + jω
A = −∇Φ
Φ
Vector Potentials 5
The Wave Equation for A – Time-Harmonic Fields So what?
B = ∇×
A
∇×B = µ∇×
H = µ jωε
E +J( ) = ∇×∇×
A = ∇∇ i
A−∇2
A
µ jωε −∇Φ− jωA( ) + J⎡⎣ ⎤⎦ = ∇∇ i
A−∇2
A
− jωµε∇Φ+ω 2µεA+ µ
J = ∇∇ i
A−∇2
A
∇ ∇ iA( )−∇2
A = µ
J − jωµε∇Φ+ω 2µε
A
Vector Potentials 6
Helmholtz Theorem (the divergence and the curl are independent): Any vector field may be written as: Proof:
C = −∇Φ+∇×
A
Ci = −∇Φ irrotational - zero curl( )Cs = ∇×
A soleniodal - zero divergence( )
∇×C = −∇×∇Φ
≡ 0 +∇×∇×
A
∇ iC = −∇ i∇Φ+∇ i∇×
A
≡ 0
Vector Potentials 7
The Wave Equation for A – Time-Harmonic Fields So what? We specified the curl of as to satisfy We are also free to specify its divergence. (This is known as the Lorentz condition)
∇ ∇ iA( )−∇2
A = µ
J − jωµε∇Φ+ω 2µε
A
⇒∇ ∇ iA( ) = − jωµε∇Φ
⇒−∇2A = µ
J +ω 2µε
A
H = 1
µ∇×A
∇ iA = − jωµεΦ
Vector Potentials 8
The Wave Equation for A – Time-Harmonic Fields Look how nice this is:
E + jω
A = −∇Φ
∇ iE + jω∇ i
A = −∇ i∇Φ
ρε+ jω∇ i
A = −∇2Φ⇒ jω∇ i
A = −∇2Φ− ρ
ε∇ ∇ i
A( ) = − jωµε∇Φ⇒ jω∇ i
A =ω 2µεΦ
⇒ jω∇ iA = −∇2Φ− ρ
ε⇒ω 2µεΦ = −∇2Φ− ρ
ε
⇒∇2Φ+ω 2µεΦ = − ρε
Vector Potentials 9
The Wave Equation for A – Time-Harmonic Fields Look how nice this is:
⇒∇2A+ω 2µε
A = −µ
J
⇒∇2Φ+ω 2µεΦ = − ρε
Vector Potentials 10
Solution of the Inhomogeneous Vector Potential Wave Equation
Once A is known:
∇2A+ β 2
A = −µ
J
H = 1
µ∇×A
E = −∇Φ− jω
A
Φ = jωµε
∇ iA
⎫
⎬⎪
⎭⎪
E = − j
ωµε∇∇ i
A− jω
A
or
∇×H = jωε
E +J ⇒
E = 1
jωε∇×H
Vector Potentials 11
a2d 2 ydt2
+ a1dydt
+ a0 y = f t( ) = δ t( ) = 0, t > 0( )Recall:
Poejωtδ r( ) = 0, r > 0( )
Point Source at origin
Region of Solution (Source Free)
Vector Potentials 12
x
y
z
( ), ,′ ′ ′x y z
ϕ
θ ( ), ,x y z r
Consider as a source the limiting case of a point source current density (z-directed) located at the origin.
J = az Jz
Vector Potentials 13
∇2A+ β 2
A = −µ
J
J = az Jz ⇒ Ax = Ay = 0
⇒∇2 Az + β2 Az = −µJz
Since a point source must be independent of angle, Az = Az r( )⇒∇2Az + β 2Az = 0
⇒ 1r2
∂∂rr2
∂Az r( )∂r
⎡
⎣⎢⎢
⎤
⎦⎥⎥+ β 2Az = 0
⇒d 2Az r( )dr2
+ 2rdAz r( )dr
+ β 2Az = 0
For all points except the origin: 2 2 0β∇ + =z zA A
Vector Potentials 14
Solutions are known:
( ) ( )
( )
22
2
1 2
2 0
β β
β
− +
+ + =
= +
z zz
j r j r
z
d A r dA rA
dr r dr
e eA r C Cr r
Since Represents an inward traveling radial wave, the source a the origin requires that
β+ j re
2 0≡C
Vector Potentials 15
( ) 1
β−
=j r
zeA r Cr
For the static case: 0 0ω β= ⇒ =
( ) 1=zCA rr
Which is the solution to Poisson’s Equation (the wave equation with no time variation):
2 µ∇ = −z zA J
Vector Potentials 16
For this static case, it is reminiscent of a charge density at the origin, satisfying (from electrostatics): With solution Thus
∇2Φ = − ρ
ε
Φ = 1
4περrV
∫∫∫ d ′v
2
4µµπ
′∇ = − ⇒ = ∫∫∫ zz z z
V
JA J A dvr
Vector Potentials 17
By analogy to the static case, for
2 2
4
βµβ µπ
−
′∇ + = − ⇒ = ∫∫∫j r
z z z z zV
eA A J A J dvr
0ω ≠
r =r
Vector Potentials 18
If the current density had x and y components as well then:
4
4
4
β
β
β
µπµπµπ
−
−
−
′=
′=
′=
∫∫∫
∫∫∫
∫∫∫
j r
x xV
j r
y yV
j r
z zV
eA J dvr
eA J dvr
eA J dvr
A = µ
4πJ e− jβr
rV∫∫∫ d ′v
Vector Potentials 19
x
y
z
( ), ,′ ′ ′x y z
ϕ′
θ ′ ( ), ,x y z ′r
r
R
ϕ
θ
If the source is moved from the origin and place at coordinates : ( ), ,′ ′ ′x y z
A = µ
4πJ ′x , ′y , ′z( ) e− jβR
RV∫∫∫ d ′v
R =R = r − ′r
Vector Potentials 21
Example: A very short linear current element. (An Infinitesimal Dipole)
Ie ′x , ′y , ′z( ) = az Ie
Ie a constant
λ
x
y
z
ϕ
θ ( ), ,x y zr
Vector Potentials 22
An Infinitesimal Dipole
A = µ
4πIe ′x , ′y , ′z( ) e− jβR
RC∫ d ′
since → 0, R ≈ r ′r = 0( )
= az
µ4π
Ie
− 2
2
∫e− jβR
Rd ′z = az
µIe
4πe− jβr
rd ′z
− 2
2
∫
Ie ′x , ′y , ′z( ) = az Ie
Ie a constant
Vector Potentials 23
An Infinitesimal Dipole
A = az
µIe
4πe− jβr
rd ′z
− 2
2
∫
= az
µIe
4πe− jβr
rd ′z
− 2
2
∫
= az
µIe4πr
e− jβr
Vector Potentials 24
Rectangular to Spherical Conversion
sin cos sin sin coscos cos cos sin sinsin cos 0
sin cos cos cos sinsin sin cos sin coscos sin 0
θ
ϕ
θ
ϕ
θ ϕ θ ϕ θθ ϕ θ ϕ θ
ϕ ϕ
θ ϕ θ ϕ ϕθ ϕ θ ϕ ϕ
θ θ
⎛ ⎞ ⎛ ⎞⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟= −⎜ ⎟ ⎜ ⎟⎜ ⎟
⎜ ⎟⎜ ⎟⎜ ⎟ −⎝ ⎠⎝ ⎠⎝ ⎠
⎛ ⎞−⎛ ⎞ ⎛ ⎞⎜ ⎟⎜ ⎟ ⎜ ⎟= ⎜ ⎟⎜ ⎟ ⎜ ⎟
⎜ ⎟ ⎜ ⎟⎜ ⎟−⎝ ⎠ ⎝ ⎠⎝ ⎠
r x
y
z
x r
y
z
A AA AA A
A AA AA A
Vector Potentials 25
Ar
Aθ
Aϕ
⎛
⎝
⎜⎜⎜⎜
⎞
⎠
⎟⎟⎟⎟
=sinθ cosϕ sinθ sinϕ cosθcosθ cosϕ cosθ sinϕ −sinθ−sinϕ cosϕ 0
⎛
⎝
⎜⎜⎜
⎞
⎠
⎟⎟⎟
00Az
⎛
⎝
⎜⎜⎜
⎞
⎠
⎟⎟⎟
Ar = cosθ Az = cosθµIe4πr
e− jβr
Aθ = −sinθ Az = −sinθµIe4πr
e− jβr
Aϕ = 0
An Infinitesimal Dipole
Vector Potentials 26
Ar = cosθ Az = cosθ
µIe4πr
e− jβr , Aθ = −sinθ Az = −sinθµIe4πr
e− jβr
An Infinitesimal Dipole
H = 1
µ∇×A
E = 1
jωε∇×H
∇×A =
ar
r sinθ∂ Aϕ sinθ( )
∂θ−∂Aθ
∂ϕ
⎡
⎣⎢⎢
⎤
⎦⎥⎥+
aθ
r1
sinθ∂Ar
∂ϕ−∂ rAϕ( )∂r
⎡
⎣⎢⎢
⎤
⎦⎥⎥
+aϕ
r∂ rAθ( )∂r
−∂Ar
∂θ⎡
⎣⎢⎢
⎤
⎦⎥⎥
Vector Potentials 27
An Infinitesimal Dipole
H = 1
µ∇×A =
aϕ
r1µ
∂ rAθ( )∂r
−∂Ar
∂θ⎡
⎣⎢⎢
⎤
⎦⎥⎥
=aϕ
r1µ
∂ −r sinθµIe4πr
e− jβr⎛⎝⎜
⎞⎠⎟
∂r−∂ cosθ
µIe4πr
e− jβr⎛⎝⎜
⎞⎠⎟
∂θ
⎡
⎣
⎢⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥⎥
= −aϕ
rsinθ
Ie4π
∂ e− jβr( )∂r
+Ie
4πr∂ cosθ( )
∂θe− jβr
⎡
⎣⎢⎢
⎤
⎦⎥⎥
= −aϕ
r− jβ sinθ
Ie4π
e− jβr −Ie
4πrsinθe− jβr⎡
⎣⎢
⎤
⎦⎥
=Ie4π
sinθaϕ
re− jβr 1
r+ jβ⎡
⎣⎢
⎤
⎦⎥
Vector Potentials 28
An Infinitesimal Dipole
E = 1
jωε∇×H = 1
jωεar
r sinθ∂ Hϕ sinθ( )
∂θ−
aθ
r
∂ rHϕ( )∂r
⎡
⎣⎢⎢
⎤
⎦⎥⎥
= 1jωε
ar
r∂Hϕ
∂θ+ ar
cosθr sinθ
Hϕ − aθ
∂Hϕ
∂r− aθ
Hϕ
r⎡
⎣⎢⎢
⎤
⎦⎥⎥
= 1jωε
ar
r∂Hϕ
∂θ+ ar
cosθr sinθ
Hϕ − aθ
∂Hϕ
∂r− aθ
Hϕ
r⎡
⎣⎢⎢
⎤
⎦⎥⎥
= 1jωε
ar
r∂Hϕ
∂θ+ ar
cosθr sinθ
Hϕ − aθ
∂Hϕ
∂r− aθ
Hϕ
r⎡
⎣⎢⎢
⎤
⎦⎥⎥
Hϕ = jβ Ie4πr
sinθ 1+ 1jβr
⎡
⎣⎢
⎤
⎦⎥e− jβr
Hr = Hθ = 0
Vector Potentials 29
An Infinitesimal Dipole
E = 1
jωεar
r∂Hϕ
∂θ+ ar
cosθr sinθ
Hϕ − aθ
∂Hϕ
∂r− aθ
Hϕ
r⎡
⎣⎢⎢
⎤
⎦⎥⎥
Hϕ = jβ Ie4πr
sinθ 1+ 1jβr
⎡
⎣⎢
⎤
⎦⎥e− jβr
Vector Potentials 30
An Infinitesimal Dipole
E = 1
jωε
ar
r∂∂θ
jβ Ie4πr
sinθ 1+ 1jβr
⎡
⎣⎢
⎤
⎦⎥e− jβr⎛
⎝⎜⎞⎠⎟
+arcosθr sinθ
jβ Ie4πr
sinθ 1+ 1jβr
⎡
⎣⎢
⎤
⎦⎥e− jβr
−aθ∂∂r
jβ Ie4πr
sinθ 1+ 1jβr
⎡
⎣⎢
⎤
⎦⎥e− jβr⎛
⎝⎜⎞⎠⎟
−aθ1r
jβ Ie4πr
sinθ 1+ 1jβr
⎡
⎣⎢
⎤
⎦⎥e− jβr + 0 i aϕ
⎡
⎣
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥
Vector Potentials 31
An Infinitesimal Dipole
Er =cosθωε
1+ 1jβr
⎛⎝⎜
⎞⎠⎟β Ie4πr 2 e− jβr
= cosθωε
1+ 1jβr
⎛⎝⎜
⎞⎠⎟ω µε Ie
4πr 2 e− jβr
= µε
cosθ 1+ 1jβr
⎛⎝⎜
⎞⎠⎟
Ie4πr 2 e− jβr
=ηcosθ 1+ 1jβr
⎛⎝⎜
⎞⎠⎟
Ie4πr 2 e− jβr
Vector Potentials 32
An Infinitesimal Dipole
Eθ = − 1ωε
β Ie4π
sinθ ∂∂r
1r+ 1
jβr 2
⎡
⎣⎢
⎤
⎦⎥e− jβr⎛
⎝⎜⎞
⎠⎟+ 1
r 2 1+ 1jβr
⎡
⎣⎢
⎤
⎦⎥e− jβr
⎡
⎣⎢⎢
⎤
⎦⎥⎥
= − 1ωε
β Ie4π
sinθ
1r+ 1
jβr 2
⎛⎝⎜
⎞⎠⎟∂∂r
e− jβr( ) + e− jβr ∂∂r
1r+ 1
jβr 2
⎛⎝⎜
⎞⎠⎟
+ 1r 2 1+ 1
jβr⎛⎝⎜
⎞⎠⎟
e− jβr
⎡
⎣
⎢⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥⎥
= −ηIe4π
sinθ− jβ 1
r+ 1
jβr 2
⎛⎝⎜
⎞⎠⎟− 1
r 2 +2
jβr3
⎛⎝⎜
⎞⎠⎟
+ 1r 2 1+ 1
jβr⎛⎝⎜
⎞⎠⎟
⎡
⎣
⎢⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥⎥
e− jβr
Vector Potentials 33
An Infinitesimal Dipole
Eθ = −ηIe4π
sinθ− jβ 1
r+ 1
jβr 2
⎛⎝⎜
⎞⎠⎟− 1
r 2 +2
jβr3
⎛⎝⎜
⎞⎠⎟
+ 1r 2 1+ 1
jβr⎛⎝⎜
⎞⎠⎟
⎡
⎣
⎢⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥⎥
e− jβr
= −ηIe4π
sinθ − jβ 1r+ 1
jβr 2
⎛⎝⎜
⎞⎠⎟− 1
jβr3
⎡
⎣⎢
⎤
⎦⎥e− jβr
= jβηIe4π
sinθ 1r+ 1
jβr 2 −1
β 2r3
⎡
⎣⎢
⎤
⎦⎥e− jβr
= jβηIe4πr
sinθ 1+ 1jβr
− 1β 2r 2
⎡
⎣⎢
⎤
⎦⎥e− jβr
Vector Potentials 34
An Infinitesimal Dipole
Er =ηcosθ 1+ 1jβr
⎛⎝⎜
⎞⎠⎟
Ie4πr 2 e− jβr
Eθ = jβηIe4πr
sinθ 1+ 1jβr
− 1β 2r 2
⎡
⎣⎢
⎤
⎦⎥e− jβr
Eϕ = 0
Hϕ = jβ Ie4πr
sinθ 1+ 1jβr
⎡
⎣⎢
⎤
⎦⎥e− jβr
Hr = Hθ = 0
Valid everywhere except on the current element.
Vector Potentials 35
What have we learned? The fields produced by finite antennas are spherical waves, i.e., they contain the radial factor Also, the fields decay by a factor of The general form for the potential in spherical coordinates is
β− j re
1r n , n = 1, 2, 3,…
A = ar Ar r,θ ,ϕ( ) + aθ Aθ r,θ ,ϕ( ) + aϕ Aϕ r,θ ,ϕ( )
Vector Potentials 36
What have we learned? The potential may be written as
A = ar Ar r,θ ,ϕ( ) + aθ Aθ r,θ ,ϕ( ) + aϕ Aϕ r,θ ,ϕ( )= 1
re− jβr ar Ar
′ θ ,ϕ( ) + aθ Aθ′ θ ,ϕ( ) + aϕ Aϕ
′ θ ,ϕ( )⎡⎣⎢
⎤⎦⎥
+ 1r 2 e− jβr ar Ar
′′ θ ,ϕ( ) + aθ Aθ′′ θ ,ϕ( ) + aϕ Aϕ
′′ θ ,ϕ( )⎡⎣⎢
⎤⎦⎥
+ 1r3 e− jβr ar Ar
′′′ θ ,ϕ( ) + aθ Aθ′′′ θ ,ϕ( ) + aϕ Aϕ
′′′ θ ,ϕ( )⎡⎣⎢
⎤⎦⎥
+This type of r-separation will occur in all we do.
Vector Potentials 37
The Far Field
As r →∞A 1
re− jβr ar Ar
′ θ ,ϕ( ) + aθ Aθ′ θ ,ϕ( ) + aϕ Aϕ
′ θ ,ϕ( )⎡⎣⎢
⎤⎦⎥
E = − jω
A− j 1
ωµε∇ ∇ i
A( )
∇ iA = 1
r 2
∂∂r
r 2 Ar( ) + 1r sinθ
∂∂θ
sinθ Aθ( ) + 1r sinθ
∂Aϕ
∂ϕ
∇ψ = ar∂ψ∂r
+ aθ1r∂ψ∂θ
+ aϕ1
r sinθ∂ψ∂ϕ
Vector Potentials 38
∇ iA = 1
r 2
∂∂r
r 2 Ar( ) + 1r sinθ
∂∂θ
sinθ Aθ( ) + 1r sinθ
∂Aϕ
∂ϕA 1
re− jβr ar Ar
′ θ ,ϕ( ) + aθ Aθ′ θ ,ϕ( ) + aϕ Aϕ
′ θ ,ϕ( )⎡⎣⎢
⎤⎦⎥
∇ iA = 1
r 2
∂∂r
r 2 1r
e− jβr Ar′ θ ,ϕ( )⎡
⎣⎢
⎤
⎦⎥ +
1r sinθ
∂∂θ
sinθ 1r
e− jβr Aθ′ θ ,ϕ( )⎡
⎣⎢
⎤
⎦⎥
+ 1r sinθ
∂∂ϕ
1r
e− jβr Aϕ′ θ ,ϕ( )⎡
⎣⎢
⎤
⎦⎥
= 1r 2 Ar
′ θ ,ϕ( ) ∂∂rre− jβr⎡⎣ ⎤⎦ +
e− jβr
r 2 sinθ∂∂θ
sinθ Aθ′ θ ,ϕ( )⎡
⎣⎢⎤⎦⎥
+ e− jβr
r 2 sinθ∂∂ϕ
Aϕ′ θ ,ϕ( )⎡
⎣⎢⎤⎦⎥
Vector Potentials 39
∇ iA = 1
r 2 Ar′ θ ,ϕ( ) ∂∂r
re− jβr⎡⎣ ⎤⎦ +e− jβr
r 2 sinθ∂∂θ
sinθ Aθ′ θ ,ϕ( )⎡
⎣⎢⎤⎦⎥
+ e− jβr
r 2 sinθ∂∂ϕ
Aϕ′ θ ,ϕ( )⎡
⎣⎢⎤⎦⎥
= 1− jβrr 2 Ar
′ θ ,ϕ( )e− jβr + e− jβr
r 2 sinθsinθ
∂Aθ′ θ ,ϕ( )∂θ
+ cosθ Aθ′ θ ,ϕ( )
⎡
⎣
⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥
+ e− jβr
r 2 sinθ∂Aϕ
′ θ ,ϕ( )∂ϕ
Vector Potentials 40
∇ iA = 1− jβr
r 2 Ar′ θ ,ϕ( )e− jβr + e− jβr
r 2 sinθsinθ
∂Aθ′ θ ,ϕ( )∂θ
+ cosθ Aθ′ θ ,ϕ( )
⎡
⎣⎢⎢
⎤
⎦⎥⎥
+ e− jβr
r 2 sinθ∂Aϕ
′ θ ,ϕ( )∂ϕ
∇ ∇ iA( ) = ar
∂∂r
+ aθ1r∂∂θ
+ aϕ1
r sinθ∂∂ϕ
⎛⎝⎜
⎞⎠⎟∇ iA
= ar∂∂r
1− jβrr 2 Ar
′ θ ,ϕ( )e− jβr + e− jβr
r 2 sinθsinθ
∂Aθ′ θ ,ϕ( )∂θ
+ cosθ Aθ′ θ ,ϕ( )
⎡
⎣⎢⎢
⎤
⎦⎥⎥
+ e− jβr
r 2 sinθ∂Aϕ
′ θ ,ϕ( )∂ϕ
⎧
⎨
⎪⎪⎪
⎩
⎪⎪⎪
⎫
⎬
⎪⎪⎪
⎭
⎪⎪⎪
+aθ1r∂∂θ
1− jβrr 2 Ar
′ θ ,ϕ( )e− jβr + e− jβr
r 2 sinθsinθ
∂Aθ′ θ ,ϕ( )∂θ
+ cosθ Aθ′ θ ,ϕ( )
⎡
⎣⎢⎢
⎤
⎦⎥⎥
+ e− jβr
r 2 sinθ∂Aϕ
′ θ ,ϕ( )∂ϕ
⎧
⎨
⎪⎪⎪
⎩
⎪⎪⎪
⎫
⎬
⎪⎪⎪
⎭
⎪⎪⎪
+aϕ1
r sinθ∂∂ϕ
1− jβrr 2 Ar
′ θ ,ϕ( )e− jβr + e− jβr
r 2 sinθsinθ
∂Aθ′ θ ,ϕ( )∂θ
+ cosθ Aθ′ θ ,ϕ( )
⎡
⎣⎢⎢
⎤
⎦⎥⎥
+ e− jβr
r 2 sinθ∂Aϕ
′ θ ,ϕ( )∂ϕ
⎧
⎨
⎪⎪⎪
⎩
⎪⎪⎪
⎫
⎬
⎪⎪⎪
⎭
⎪⎪⎪
Vector Potentials 41
∇ ∇ iA( ) = ar Ar
′ ∂∂r
1− jβrr 2 e− jβr⎡
⎣⎢
⎤
⎦⎥ + sinθ
∂Aθ′
∂θ+ cosθ Aθ
′ +∂Aϕ
′
∂ϕ
⎡
⎣⎢⎢
⎤
⎦⎥⎥∂∂r
e− jβr
r 2
⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪
+aθ1r
1− jβrr 2 e− jβr ∂Ar
′
∂θ+ e− jβr
r 2
∂∂θ
1sinθ
sinθ∂Aθ
′
∂θ+ cosθ Aθ
′ +∂Aϕ
′
∂ϕ
⎡
⎣⎢⎢
⎤
⎦⎥⎥
⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪
+aϕ1
r sinθ1− jβr
r 2
∂Ar′
∂ϕe− jβr + e− jβr
r 2 sinθsinθ
∂2 Aθ′
∂ϕ ∂θ+ cosθ
∂Aθ′
∂ϕ+∂2 Aϕ
′
∂ϕ 2
⎡
⎣⎢⎢
⎤
⎦⎥⎥
⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪
Vector Potentials 42
∇ ∇ iA( ) = are
− jβr Ar′ − jβ 1− jβr
r 2 + − jβr 2 −1+ j2βr 2
r 4
⎡
⎣⎢
⎤
⎦⎥ − sinθ
∂Aθ′
∂θ+ cosθ Aθ
′ +∂Aϕ
′
∂ϕ
⎡
⎣⎢⎢
⎤
⎦⎥⎥
jβr 2 + 2rr 4
⎡
⎣⎢
⎤
⎦⎥
⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪
+aθ1r
1− jβrr 2 e− jβr ∂Ar
′
∂θ+ e− jβr
r 2
∂∂θ
1sinθ
sinθ∂Aθ
′
∂θ+ cosθ Aθ
′ +∂Aϕ
′
∂ϕ
⎡
⎣⎢⎢
⎤
⎦⎥⎥
⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪
+aϕ1
r sinθ1− jβr
r 2
∂Ar′
∂ϕe− jβr + e− jβr
r 2 sinθsinθ
∂2 Aθ′
∂ϕ ∂θ+ cosθ
∂Aθ′
∂ϕ+∂2 Aϕ
′
∂ϕ 2
⎡
⎣⎢⎢
⎤
⎦⎥⎥
⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪
∇ ∇ iA( ) = are
− jβr Ar′ − jβ 1− jβr
r 2 − 1− jβr 2
r 4
⎡
⎣⎢
⎤
⎦⎥ − sinθ
∂Aθ′
∂θ+ cosθ Aθ
′ +∂Aϕ
′
∂ϕ
⎡
⎣⎢⎢
⎤
⎦⎥⎥
jβr 2 + 2rr 4
⎡
⎣⎢
⎤
⎦⎥
⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪
+aθ1r
1− jβrr 2 e− jβr ∂Ar
′
∂θ+ e− jβr
r 2
∂∂θ
1sinθ
sinθ∂Aθ
′
∂θ+ cosθ Aθ
′ +∂Aϕ
′
∂ϕ
⎡
⎣⎢⎢
⎤
⎦⎥⎥
⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪
+aϕ1
r sinθ1− jβr
r 2
∂Ar′
∂ϕe− jβr + e− jβr
r 2 sinθsinθ
∂2 Aθ′
∂ϕ ∂θ+ cosθ
∂Aθ′
∂ϕ+∂2 Aϕ
′
∂ϕ 2
⎡
⎣⎢⎢
⎤
⎦⎥⎥
⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪
Vector Potentials 43
∇ ∇ iA( ) = are
− jβr Ar′ − jβ 1− jβr
r 2 − 1− jβr 2
r 4
⎡
⎣⎢
⎤
⎦⎥ − sinθ
∂Aθ′
∂θ+ cosθ Aθ
′ +∂Aϕ
′
∂ϕ
⎡
⎣⎢⎢
⎤
⎦⎥⎥
jβr 2 + 2rr 4
⎡
⎣⎢
⎤
⎦⎥
⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪
+aθ1r
1− jβrr 2 e− jβr ∂Ar
′
∂θ+ e− jβr
r 2
∂∂θ
1sinθ
sinθ∂Aθ
′
∂θ+ cosθ Aθ
′ +∂Aϕ
′
∂ϕ
⎡
⎣⎢⎢
⎤
⎦⎥⎥
⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪
+aϕ1
r sinθ1− jβr
r 2
∂Ar′
∂ϕe− jβr + e− jβr
r 2 sinθsinθ
∂2 Aθ′
∂ϕ ∂θ+ cosθ
∂Aθ′
∂ϕ+∂2 Aϕ
′
∂ϕ 2
⎡
⎣⎢⎢
⎤
⎦⎥⎥
⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪
As r →∞
∇ ∇ iA( ) = −ar
β 2
re− jβr Ar
′
Vector Potentials 44
The Far Field
As r →∞A 1
re− jβr ar Ar
′ θ ,ϕ( ) + aθ Aθ′ θ ,ϕ( ) + aϕ Aϕ
′ θ ,ϕ( )⎡⎣⎢
⎤⎦⎥
E = − jω
A− j 1
ωµε∇ ∇ i
A( ), ∇ ∇ i
A( ) = −ar
β 2
re− jβr Ar
′
E = − jω
re− jβr ar Ar
′ + aθ Aθ′ + aϕ Aϕ
′⎡⎣⎢
⎤⎦⎥ + j 1
ωµεar
β 2
re− jβr Ar
′
E = − jω
re− jβr ar Ar
′ + aθ Aθ′ + aϕ Aϕ
′⎡⎣⎢
⎤⎦⎥ + jar
ωr
e− jβr Ar′
E = − jω
re− jβr ar Ar
′ − Ar′( ) + aθ Aθ
′ + aϕ Aϕ′⎡
⎣⎢⎤⎦⎥
E = − jω
re− jβr aθ Aθ
′ + aϕ Aϕ′⎡
⎣⎢⎤⎦⎥
Vector Potentials 45
The Far Field
H = 1
µ∇×A
A 1
re− jβr ar Ar
′ θ ,ϕ( ) + aθ Aθ′ θ ,ϕ( ) + aϕ Aϕ
′ θ ,ϕ( )⎡⎣⎢
⎤⎦⎥
∇ ×A =
ar
r sinθ∂ Aϕ sinθ( )
∂θ−∂Aθ
∂ϕ
⎡
⎣⎢⎢
⎤
⎦⎥⎥+
aθ
r1
sinθ∂Ar
∂ϕ−∂ rAϕ( )∂r
⎡
⎣⎢⎢
⎤
⎦⎥⎥
+aϕ
r∂ rAθ( )∂r
−∂Ar
∂θ⎡
⎣⎢⎢
⎤
⎦⎥⎥
Vector Potentials 46
The Far Field
∇×A =
are− jβr
r 2 sinθ
∂ sinθ Aϕ′( )
∂θ−∂Aθ
′
∂ϕ
⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥+
aθ
r1
sinθe− jβr
r∂Ar
′
∂ϕ− Aϕ
′∂ e− jβr( )
∂r
⎡
⎣⎢⎢
⎤
⎦⎥⎥
+aϕ
rAθ′ ∂ e− jβr( )∂r
− e− jβr
r∂Ar
′
∂θ
⎡
⎣
⎢⎢
⎤
⎦
⎥⎥
r →∞
∇×A = −
aθ
rAϕ′∂ e− jβr( )
∂r+
aϕ
rAθ′ ∂ e− jβr( )∂r
= − jβ e− jβr
raθ Aϕ
′ − aϕ Aθ′( )
Vector Potentials 47
The Far Field
r →∞H = 1
µ∇×A
= − j βµ
e− jβr
raθ Aϕ
′ − aϕ Aθ′( )
= − jωη
e− jβr
raθ Aϕ
′ − aϕ Aθ′( )
β =ω µε , η = µε
Vector Potentials 48
The Far Field
r →∞A e− jβr
rar Ar
′ θ ,ϕ( ) + aθ Aθ′ θ ,ϕ( ) + aϕ Aϕ
′ θ ,ϕ( )⎡⎣⎢
⎤⎦⎥
E = − jω e− jβr
raθ Aθ
′ + aϕ Aϕ′( )
H = − jω
ηe− jβr
raθ Aϕ
′ − aϕ Aθ′( )
Er 0 Hr 0
Eθ − jω Aθ Hθ − jωη
Aϕ = −Eϕ
η
Eϕ − jω Aϕ Hϕ + jωη
Aθ = +Eθ
η
⎫
⎬
⎪⎪⎪
⎭
⎪⎪⎪
A pair of plane wavestraveling in the r direction