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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 = j

H i

E =

H = j

E +

J i

H = 0

Maxwells Equations Time-Harmonic Electromagnetic Fields Homogeneous Medium

• Vector Potentials 3

iB = 0

Since

Now for any vector field

iA = 0

Define

B =

H =

A

or

H = 1

A

The Vector Potential

• Vector Potentials 4

Substituting

H = 1

A

E = j

H

E = j 1

A

E + j

A( ) = 0

Note that for any scalar function

0

E + j

A( ) = 0 = ( )

E + j

A =

• Vector Potentials 5

The Wave Equation for A Time-Harmonic Fields So what?

B =

A

B =

H = j

E +

J( ) = A = i A2 A

j jA( ) + J = i

A2

A

j+ 2A+

J = i

A2

A

iA( )2 A = J j+ 2 A

• Vector Potentials 6

Helmholtz Theorem (the divergence and the curl are independent): Any vector field may be written as: Proof:

C = +

A

Ci = irrotational - zero curl( )Cs =

A soleniodal - zero divergence( )

C =

0 +

A

iC = i+ 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 = J j+ 2 A

iA( ) = j

2A =

J + 2

A

H = 1

A

iA = j

• Vector Potentials 8

The Wave Equation for A Time-Harmonic Fields Look how nice this is:

E + j

A =

iE + j i

A = i

+ j i

A = 2 j i

A = 2

i

A( ) = j j i A = 2

j iA = 2

2 = 2

2+ 2 =

• Vector Potentials 9

The Wave Equation for A Time-Harmonic Fields Look how nice this is:

2A+ 2

A =

J

2+ 2 =

• Vector Potentials 10

Solution of the Inhomogeneous Vector Potential Wave Equation

Once A is known:

2A+ 2

A =

J

H = 1

A

E = j

A

= j

iA

E = j

i

A j

A

or

H = j

E +

J

E = 1

j

H

• Vector Potentials 11

a2d 2 ydt2

+ a1dydt

+ a0 y = f t( ) = t( ) = 0, t > 0( )Recall:

Poejt r( ) = 0, r > 0( )

Point Source at origin

Region of Solution (Source Free)

• Vector Potentials 12

x

y

z

( ), , x y z

( ), ,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+ 2

A =

J

J = az Jz Ax = Ay = 0

2 Az + 2 Az = Jz

Since a point source must be independent of angle, Az = Az r( )2Az +

2Az = 0

1r2

rr2

Az r( )r

+ 2Az = 0

d 2Az r( )dr2

+ 2rdAz r( )dr

+ 2Az = 0

For all points except the origin: 2 2 0 + =z zA A

• Vector Potentials 14

Solutions are known:

( ) ( )

( )

22

2

1 2

2 0

+

+ + =

= +

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

+ j re

2 0C

• Vector Potentials 15

( ) 1

=j r

zeA r Cr

For the static case: 0 0 = =

( ) 1=zCA rr

Which is the solution to Poissons Equation (the wave equation with no time variation):

2 = 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

2 =

= 1

4rV

d v

2

4

= = zz z zV

JA J A dvr

• Vector Potentials 17

By analogy to the static case, for

2 2

4

+ = = j r

z z z z zV

eA A J A J dvr

0

r =r

• Vector Potentials 18

If the current density had x and y components as well then:

4

4

4

=

=

=

j r

x xV

j r

y yV

j r

z zV

eA J dvr

eA J dvr

eA J dvr

A =

4J e

jr

rV d v

• Vector Potentials 19

x

y

z

( ), , x y z

( ), ,x y z r

r

R

If the source is moved from the origin and place at coordinates : ( ), , x y z

A =

4J x , y , z( ) e

jR

RV d v

R =

R = r r

• Vector Potentials 20

For filamentary currents:

A =

4Ie x , y , z( ) e

jR

RC d

• Vector Potentials 21

Example: A very short linear current element. (An Infinitesimal Dipole)

Ie x , y , z( ) = az IeIe a constant

x

y

z

( ), ,x y zr

• Vector Potentials 22

An Infinitesimal Dipole

A =

4Ie x , y , z( ) e

jR

RC d

since 0, R r r = 0( )

= az

4Ie

2

2

e jR

Rd z = az

Ie4

e jr

rd z

2

2

Ie x , y , z( ) = az IeIe a constant

• Vector Potentials 23

An Infinitesimal Dipole

A = az

Ie4

e jr

rd z

2

2

= azIe4

e jr

rd z

2

2

= azIe4r

e jr

• 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

=

=

r x

y

z

x r

y

z

A AA AA A

A AA AA A

• Vector Potentials 25

ArAA

=sin cos sin sin coscos cos cos sin sinsin cos 0

00Az

Ar = cos Az = cosIe4r

e jr

A = sin Az = sinIe4r

e jr

A = 0

An Infinitesimal Dipole

• Vector Potentials 26

Ar = cos Az = cos

Ie4r

e jr , A = sin Az = sinIe4r

e jr

An Infinitesimal Dipole

H = 1

A

E = 1

j

H

A =

arr sin

A sin( )

A

+

ar

1sin

Ar

rA( )r

+ar

rA( )r

Ar

• Vector Potentials 27

An Infinitesimal Dipole

H = 1

A =

ar

1

rA( )r

Ar

=ar

1

r sinIe4r

e jr

r cos

Ie4r

e jr

= ar

sinIe4

e jr( )r

+Ie

4r cos( )

e jr

= ar

j sinIe4

e jr Ie

4rsine jr

=Ie4

sinar

e jr 1r+ j

• Vector Potentials 28

An Infinitesimal Dipole

E = 1

j

H = 1

jar

r sin H sin( )

ar rH( )r

= 1j

arrH

+ arcosr sin

H aHr

aHr

= 1j

arrH

+ arcosr sin

H aHr

aHr

= 1j

arrH

+ arcosr sin

H aHr

aHr

H = j Ie4r

sin 1+ 1jr

e

jr

Hr = H = 0

• Vector Potentials 29

An Infinitesimal Dipole

E = 1

jarrH

+ arcosr sin

H aHr

aHr

H = j Ie4r

sin 1+ 1jr

e

jr

• Vector Potentials 30

An Infinitesimal Dipole

E = 1

j

arr

j Ie4r

sin 1+ 1jr

e

jr

+arcosr sin

j Ie4r

sin 1+ 1jr

e

jr

ar

j Ie4r

sin 1+ 1jr

e

jr

a1r

j Ie4r

sin 1+ 1jr

e

jr + 0 i a

• Vector Potentials 31

An Infinitesimal Dipole

Er =cos

1+ 1jr

Ie4r 2

e jr

= cos

1+ 1jr

Ie

4r 2e jr

=

cos 1+ 1jr

Ie4r 2

e jr

=cos 1+ 1jr

Ie4r 2

e jr

• Vector Potentials 32

An Infinitesimal Dipole

E = 1

Ie4

sin r

1r+ 1

jr 2

e

jr

+ 1

r 21+ 1

jr

e

jr

= 1

Ie4

sin

1r+ 1

jr 2

r

e jr( ) + e jr r1r+ 1

jr 2

+ 1r 2

1+ 1jr

e jr

= Ie4

sin j 1

r+ 1

jr 2

1

r 2+ 2

jr3

+ 1r 2

1+ 1jr

e jr

• Vector Potentials 33

An Infinitesimal Dipole

E = Ie4

sin j 1

r+ 1

jr 2

1

r 2+ 2

jr3

+ 1r 2

1+ 1jr

e jr

= Ie4

sin j 1r+ 1

jr 2

1

jr3

e jr

= jIe4

sin 1r+ 1

jr 2 1 2r3

e

jr

= jIe4r

sin 1+ 1jr

1 2r 2

e

jr

• Vector Potentials 34

An Infinitesimal Dipole

Er =cos 1+1

jr

Ie4r 2

e jr

E = jIe4r

sin 1+ 1jr

1 2r 2

e

jr

E = 0

H = j Ie4r

sin 1+ 1jr

e

jr

Hr = H = 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

j re

1r n

, n = 1, 2, 3,

A = ar Ar r, ,( ) + a A r, ,( ) + a A r, ,( )

• Vector Potentials 36

What have we learned? The potential may be written as

A = ar