Vector Potentials and Antenna Fundamentals

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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