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Vector Potentials 1 Vector Potentials and Antennas

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Vector Potentials and Antenna Fundamentals

### Transcript of Vector Potentials and Antenna Fundamentals

Vector Potentials 1

Vector Potentials and Antennas

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 20

For filamentary currents:

A = µ

4πIe ′x , ′y , ′z( ) e− jβR

RC∫ d ′

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

⎜⎜⎜⎜

⎟⎟⎟⎟

=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

⎣⎢⎢

⎦⎥⎥+

r1

sinθ∂Ar

∂ϕ−∂ rAϕ( )∂r

⎣⎢⎢

⎦⎥⎥

+aϕ

r∂ rAθ( )∂r

−∂Ar

∂θ⎡

⎣⎢⎢

⎦⎥⎥

Vector Potentials 27

An Infinitesimal Dipole

H = 1

µ∇×A =

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−

r

∂ rHϕ( )∂r

⎣⎢⎢

⎦⎥⎥

= 1jωε

ar

r∂Hϕ

∂θ+ ar

cosθr sinθ

Hϕ − aθ

∂Hϕ

∂r− aθ

r⎡

⎣⎢⎢

⎦⎥⎥

= 1jωε

ar

r∂Hϕ

∂θ+ ar

cosθr sinθ

Hϕ − aθ

∂Hϕ

∂r− aθ

r⎡

⎣⎢⎢

⎦⎥⎥

= 1jωε

ar

r∂Hϕ

∂θ+ ar

cosθr sinθ

Hϕ − aθ

∂Hϕ

∂r− aθ

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

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

⎣⎢⎢

⎦⎥⎥+

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

⎢⎢⎢

⎥⎥⎥+

r1

sinθe− jβr

r∂Ar

∂ϕ− Aϕ

′∂ e− jβr( )

∂r

⎣⎢⎢

⎦⎥⎥

+aϕ

rAθ′ ∂ e− jβr( )∂r

− e− jβr

r∂Ar

∂θ

⎢⎢

⎥⎥

r →∞

∇×A = −

rAϕ′∂ e− jβr( )

∂r+

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

Vector Potentials 49

For the Far Field

E = − jω

A

H = 1

ηar ×E = − jω

ηar ×A