ECE 1100 Introduction to Electrical and Computer Engineeringcourses.egr.uh.edu/ECE/ECE6341/Class...

of 24/24
Prof. David R. Jackson ECE Dept. Spring 2016 Notes 15 ECE 6341 1
  • date post

    21-Nov-2020
  • Category

    Documents

  • view

    0
  • download

    0

Embed Size (px)

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 dkπ

    +∞−

    −∞

    = ∫

    ( )( ) 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 dkπ

    +∞−

    −∞

    = ∫

  • Hence, from superposition, the total magnetic vector potential is

    ( ) ( )20 0( )8zjk z

    z z zA I k H k e dkj ρµ 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( ) z

    jk 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( )4

    zjk zz

    IA 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,

    ( ) ( )20 0 ( )8zjk z

    z zA I H k e dkj ρµ ρπ

    +∞ −

    −∞= ∫

    ( ) 04

    jkr

    zeA I

    rµπ

    = (from ECE 6340)

    ( ) ( )20 0( )8zjk z

    z z zA I k H k e dkj ρµ 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 ρρ

    − +∞ −

    −∞= ∫

    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 kρCCCW

    ( )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 kρCCW

    ( )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 012J z H z H z= +

  • Example (cont.)

    17

    00

    1 ( ) zjkr

    jk 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.

    18

    y

    z

    x

    ( )I zr θ

  • Far-Field Identity (cont.)

    ( )200( ) ( )8

    zjk zz 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 4

    z

    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

    rρρ 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( ) ~ ( )

    nnH x j H x

    22

  • Far-Field Identity (cont.)

    Therefore, we have

    ( )2 1( ) ( ) ~ 2 ( cos )zjkr

    jk z nz n z

    eI k H k e dk j I krρ

    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 )zjkr

    jk z nz n z

    ef k H k e dk j f krρ

    ρ θ−+∞ − +

    −∞

    for , 0, 180r θ→∞ ≠

    24

    Slide Number 1Slide Number 2Slide Number 3Slide Number 4Slide Number 5Slide Number 6Slide Number 7Slide Number 8Slide Number 9Slide Number 10Slide Number 11Slide Number 12Slide Number 13Slide Number 14Slide Number 15Slide Number 16Slide Number 17Slide Number 18Slide Number 19Slide Number 20Slide Number 21Slide Number 22Slide Number 23Slide Number 24