ECE 6340 Intermediate EM Waves - University of Houstoncourses.egr.uh.edu/ECE/ECE6340/Class...

Click here to load reader

  • date post

    27-Mar-2020
  • Category

    Documents

  • view

    10
  • download

    0

Embed Size (px)

Transcript of ECE 6340 Intermediate EM Waves - University of Houstoncourses.egr.uh.edu/ECE/ECE6340/Class...

  • Prof. David R. Jackson Dept. of ECE

    Fall 2016

    Notes 16

    ECE 6340 Intermediate EM Waves

    1

  • Polarization of Waves Consider a plane wave with both x and y components

    ˆ ˆ( , , ) ( ) jkzx yE x y z x E y E e−= +

    ( = .)y xE Eβ −In general, phase of phase of

    Assume

    Ex

    Ey

    x

    y

    0jx

    jy

    E ae aE be β

    = =

    =

    2

    (polar form of complex numbers)

  • Polarization of Waves (cont.)

    Time Domain:

    ( ) ( )( ) ( )

    Re cos

    Re cos

    j tx

    j j ty

    a e a t

    be e b t

    ω

    β ω

    ω

    ω β

    = =

    = = +

    E

    E

    Depending on b and β, three different cases arise.

    At z = 0

    x

    y

    E (t) |E (t)|

    φ (t)

    3

  • Linear Polarization

    At z = 0:

    ˆ ˆ( ) cosx a y b tω= ±E

    ( )coscos cos

    x

    y

    a tb t b t

    ωω β ω

    =

    = + = ±

    E

    E

    a

    b

    x

    y x´ y´

    E

    β = 0 or π

    (shown for β = 0)

    (+ for 0, - for π)

    4

  • Circular Polarization

    At z = 0:

    2 2 2 2 2 2 2

    2

    cos sinx y a t a t

    a

    ω ω= + = +

    =

    E E E

    ( )coscos cos( / 2) sin

    x

    y

    a tb t a t a t

    ωω β ω π ω

    =

    = + = ± =

    E

    E

    b = a AND β = ± π /2

    5

  • Circular Polarization (cont.)

    ( )1 1 1tan tan ( tan ) tan (tan )yx

    t tφ ω ω− − −

    = = =

    EE

    tφ ω=

    6

    so

    waveddtφω ω== =angular velocity of waveNote:

    a

    x

    y

    ( )tE

    φ(t)

    waveω ω=

    / 2β π= ±

  • Circular Polarization (cont.)

    tφ ω=

    IEEE convention

    ˆ ˆ( cos sin )a x t y tω ω= E

    a

    x

    y

    ( )tE

    φ(t)

    β = +π / 2

    β = −π / 2

    RHCP

    LHCP

    7

    / 2β π= ±

  • Circular Polarization (cont.)

    8

    ( ) ( )( )ˆ ˆcos sina x t kz y t kzω ω= − + −E

    ( ) ( )( )ˆ ˆcos sina x kz y kz= −E

    ( )j t kze ω −Note: The rotation is space is opposite to that in time.

    Time and distance dependence:

    ( ) ( )( )ˆ ˆcos sina x t y tω ω= +E z = 0

    t = 0

    RHCP

    ( )t t kzω ω→ −

  • Circular Polarization (cont.)

    RHCP

    RHCP

    z

    z

    “Snapshot" of wave

    9

  • Circular Polarization (cont.) Animation of LHCP wave

    http://en.wikipedia.org/wiki/Circular_polarization

    (Use pptx version in full-screen mode to see motion.)

    10

  • Unit Rotation Vectors

    1ˆ ˆ ˆ( )2

    1 ˆ ˆ( )2

    r x jy

    l x jy

    = −

    = +

    RHCP (β = - π / 2)

    LHCP (β = + π / 2)

    1 ˆˆ ˆ( )2

    ˆˆ ˆ( )2

    x r l

    jy r l

    = +

    = −

    Add:

    Subtract:

    11

  • General Wave Representation

    ˆ ˆ

    1 1ˆ ˆ( ) ( )2 2

    x y

    x y

    E x E y E

    E r l E j r l

    = +

    = + + −

    1 1( ) ( )2 2RH x y LH x y

    A E j E A E j E= + = −

    Note: Any polarization can be written as combination of RHCP and LHCP waves.

    This could be useful in dealing with CP antennas.

    ˆ RH LHE r A l A= +

    12

  • Circularly Polarized Antennas Method 1: The antenna is fundamentally CP.

    A helical antenna for satellite reception is shown here.

    The helical antenna is shown radiating above a circular

    ground plane.

    13

  • Circularly Polarized Antennas (cont.)

    + - Vy = -j

    z

    x

    y

    Vx = 1

    RHCP

    Method 2: Two perpendicular antennas are used, fed 90o out of phase.

    Two antennas are shown here being fed by a common feed point.

    14

    A more complicated version using four antennas

    (omni CP) Note: LHCP is radiated in the negative z direction.

  • Circularly Polarized Antennas (cont.) Method 3: A single antenna is used, but two perpendicular modes are excited 90o out of phase.

    (a) A square microstrip antenna with two perpendicular modes being excited.

    (b) A circular microstrip antenna using a single feed and slots. The feed is located at 45o from the slot axis.

    (c) A square patch with two corners chopped off, fed at 45o from the edge.

    (c)

    (a) (b)

    15

  • Circularly Polarized Antennas (cont.) Method 4: Sequential rotation of antennas is used (shown for N = 4).

    I = 1 ∠ 0o

    I = 1 ∠ -90o I = 1 ∠ -90o

    I = 1 ∠ -180o

    I = 1 ∠ -270o

    LP elements CP elements

    The elements may be either LP or CP.

    16

    This is an extension of method 2 when used for LP elements.

    Using multiple CP elements instead of one CP element often gives better CP in practice (due to symmetry).

  • Pros and Cons of CP

    Alignment between transmit and receive antennas is not important.

    The reception does not depend on rotation of the electric field vector that might occur†.

    When a RHCP bounces off an object, it mainly changes to a LHCP wave. Therefore, a RHCP receive antenna will not be as sensitive to “multipath” signals that bound off of other objects.

    † Satellite transmission often uses CP because of Faraday rotation in the ionosphere. 17

    A CP system is usually more complicated and expensive.

    Often, simple wire antennas are used for reception of signals in wireless communications, and hence they are already directly compatible with linear polarization. (Using a CP transmitted signal will result in 50% of the transmitted power being wasted, or a 3 dB drop in the received signal).

    Pros

    Cons

  • 18

    Pros and Cons of CP (cont.)

    LP-LP

    Effects of alignment errors

    α = alignment angle error

    cos2α : The received signal can vary from zero dB to -∞ dB

    CP-LP

    TX-RX Gain Loss

    -3 dB, no matter what the alignment

    LP-CP -3 dB, no matter what the alignment

    CP-CP 0 dB (polarizations same) or -∞ dB (polarizations opposite), no matter what the alignment

  • Examples of Polarization

    AM Radio (0.540-1.6 MHz): linearly polarized (vertical) (Note 1)

    FM Radio (88-108 MHz): linearly polarized (horizontal)

    TV (VHF, 54-216 MHz; UHF, 470-698 MHz: linearly polarized (horizontal) (some transmitters are CP)

    Cell phone antenna (about 2 GHz, depending on system): linearly polarized (direction arbitrary)

    Cell phone base-station antennas (about 2 GHz, depending on system): often linearly polarized (dual-linear slant 45o) (Note 2)

    DBS Satellite TV (11.7-12.5 GHz): transmits both LHCP and RHCP (frequency reuse) (Note 3)

    GPS (1.574 GHz): RHCP (Note 3)

    Notes: 1) Low-frequency waves travel better along the earth when they are polarized vertically instead of horizontally. 2) Slant linear is used to switch between whichever polarization is stronger. 3) Satellite transmission often uses CP because of Faraday rotation in the ionosphere.

    19

  • Elliptical Polarization

    Includes all other cases

    β ≠ 0 or π (not linear)

    β ≠ ±π / 2 (not CP)

    β = ± π / 2 and b≠ a (not CP)

    x

    y

    ( )tE

    φ(t)

    Ellipse

    20

  • Elliptical Polarization (cont.)

    00

    β ππ β

    < <− < <

    LHEPRHEP

    x

    LHEP

    RHEP

    y

    φ(t)

    Ellipse

    21

    Note: The rotation is not at a constant speed.

  • Proof of Ellipse Property

    cos( )

    cos cos sin siny b t

    b t b tω β

    ω β ω β

    = +

    = −

    E

    so 2

    cos sin 1x xy b ba aβ β

    = − −

    E EE

    22 2cos siny x x

    b bba a

    β β − = − − E E E

    2 22 2 2 2 2cos 2 cos siny x x y x

    b b bba a a

    β β β + − = −

    E E E E E

    cosx a tω=E

    22

  • Consider the following quadratic form:

    ( )2

    2 2 2 2 2 2cos sin 2 cos sinx y x yb b ba a

    β β β β + + − =

    E E E E

    22 2 2 22 cos sinx x y y

    b b ba a

    β β + − + = E E E E

    2 2x x y yA B C D+ + =E E E E

    2 22 2 2 2 2cos 2 cos siny x x y x

    b b bba a a

    β β β + − = −

    E E E E E

    Proof of Ellipse Property (cont.)

    23

  • 2

    2 22

    22

    4

    4 cos 4

    4 cos 1

    B AC

    b ba a

    ba

    β

    β

    ∆ = −

    = −

    = −

    224 sin 0b

    aβ ∆ = − ∆ =∆ <

    hyperbolalineellipse

    From analytic geometry:

    If β = 0 or π we have ∆ = 0, and this is linear polarization. 24

  • Rotation Property

    coscos( )

    x

    y

    a tb t

    ωω β

    == +

    EE

    00

    β ππ β< <

    − < <LHEPRHEP

    x

    LHEP

    RHEP

    y

    φ(t)

    ellipse

    25

    We now prove the rotation property:

    cos cos sin siny b t b tω β ω β= −E

    Recall that

  • Rotation Property (cont.)

    Take the derivative:

    [ ]tan cos tan sinyx

    b ta

    φ β ω β= = −EE

    ( )( )2 2sec sec sind b tdt aφφ ω ω β = −

    so

    ( )( )2 2sin cos secd b tdt aφ β φ ω ω = −

    ( ) 0 0

    ( ) 0 0

    dadtdbdt

    φβ π

    φπ β

    < < ⇒ <

    − < < ⇒ >

    LHEP

    RHEP (proof complete)

    Hence

    26

  • Phasor Picture

    ( )tE

    x

    y ( ) 0a β π< < LHEP

    Ey leads Ex

    Re

    β

    Ey

    Ex

    Im

    LHEP

    Rule:

    The electric field vector rotates in time from the leading axis to the lagging axis.

    27

    A phase angle 0 < φ < π is a leading phase angle (leading with respect to zero degrees). A phase angle -π < φ < 0 is a lagging phase angle (lagging with respect to zero degrees).

  • Phasor Picture (cont.)

    x

    y

    ( )tE( ) 0b π β− < < RHEP

    Rule:

    The electric field vector rotates in time from the leading axis to the lagging axis.

    Ex

    Ey

    Im

    Re β

    Ey lags Ex

    RHEP

    28

    A phase angle 0 < φ < π is a leading phase angle (leading with respect to zero degrees). A phase angle -π < φ < 0 is a lagging phase angle (lagging with respect to zero degrees).

  • Example

    [ ]ˆˆ (1 ) (2 ) jkyE z j x j e= + + −

    y

    Ez

    z

    x Ex

    What is this wave’s polarization?

    29

  • Example (cont.)

    [ ]ˆˆ (1 ) (2 ) jkyE z j x j e= + + −

    Therefore, in time, the wave rotates from the z axis to the x axis.

    Ez leads Ex Ez

    Ex

    Im

    Re

    30

  • [ ]ˆˆ(1 ) (2 ) jkyE z j x j e= + + −

    y

    Ez

    z

    x Ex

    LHEP or LHCP

    Note: 2x z

    E E πβ≠ ≠ ± (so this is not LHCP)

    Example (cont.)

    and

    31

  • [ ]ˆˆ (1 ) (2 ) jkyE z j x j e= + + −

    LHEP

    Example (cont.)

    32

    y

    Ez

    z

    x Ex

  • Axial Ratio (AR) and Tilt Angle (τ )

    x

    y ( )tE

    τ

    A

    B C

    D

    AR 1ABCD

    ≡ = ≥major axisminor axis

    ( )dB 10AR 20log AR=33

  • Axial Ratio (AR) and Tilt Angle (τ )

    ( )

    ( )

    o o

    0

    0

    45 45

    sin 2 sin 2 sin

    AR cot

    ξ

    ξ

    ξ

    ξ γ β

    ξ

    >

    <

    − ≤ ≤ +

    =

    =

    implies LHEP

    implies RHEP

    Axial Ratio

    tan 2 tan 2 cosτ γ β=Tilt Angle

    Note: The tilt angle τ is ambiguous by the addition of ± 90o.

    (We cannot tell the difference between the major and minor axes.)

    1tan ba

    γ − =

    We first calculate γ :

    34

    0jx

    jy

    E ae aE be β

    = =

    =

    (-90o < τ < 90o)

  • ( )AR cot ξ=

    x

    y ( )tE

    τ

    A

    B C

    D |ξ |

    Axial Ratio (AR) and Tilt Angle (τ) (cont.)

    Physical interpretation of the angle |ξ|

    35

    o

    LP CP

    0 45ξ<

  • Special Case

    tan 2 tan 2 cos 00 / 2

    τ γ βτ π

    = =⇒ = or

    Tilt Angle:

    The tilt angle is zero or 90o if:

    / 2β π= ±

    x

    y

    ( )tE

    coscos( / 2)

    sin

    x

    y

    a tb t

    b t

    ωω π

    ω

    =

    = ±

    =

    EE

    22

    2 2 1yx

    a b+ =

    EE

    36

  • 37

    Sometimes it is useful to know the ratio of the LHCP and RHCP wave amplitudes.

    LHCP/RHCP Ratio

    1 1ˆ ˆ ˆ ˆ ˆ( ) ( )2 2

    r x jy l x jy= − = +

    ˆ ˆ RH LHj jRH LH RH LHE r A l A r A e l A eφ φ= + = +

    : ( ) RH LHA t A A= +Point E

    : ( ) RH LHB t A A= −Point Ex

    y ( )tE

    A

    B

    AR RH LH RH LHRH LH RH LH

    A A A AA A A A

    + += =

    − −

    Hence

  • 38

    LHCP/RHCP Ratio (Cont.) Hence

    1 /AR

    1 /LH RH

    LH RH

    A AA A

    +=

    AR 1 AR 1AR 1 AR 1

    LH

    RH

    AA

    − +=

    + −or

    From this we can also solve for the ratio of the CP components, if we know the axial ratio:

    1 /AR

    1 /LH RH

    LH RH

    A AA A

    += ± −

    or

    (We can’t tell which one is correct from knowing only the AR.)

    (Use whichever sign gives AR > 0.)

  • 39

    z

    y

    Ez

    x Ex

    LHEP

    Re-label the coordinate system:

    [ ]ˆˆ (1 ) (2 ) jkyE z j x j e= + + −

    x xz yy z

    →→→ −

    Find the axial ratio and tilt angle.

    Example

    Find the ratio of the CP amplitudes.

  • 40

    y

    z

    Ez

    x Ex

    LHEP

    ˆ ˆ(2 ) (1 ) jkzE x j y j e− = − + +

    1.249 o1 0.2 0.6 0.6324 0.6324 71.5652

    y j

    x

    E j j eE j

    += = + = = ∠

    ( )1 1tan tan 0.632 0.564ba

    γ − − = = =

    radians

    Example (cont.)

    o/ 0.6324, 71.565b a β⇒ = =

  • 41

    tan 2 tan 2 cosτ γ β=

    o o45 45

    LHEP

    RHEP

    sin 2 sin 2 sin

    AR cot

    0 :0 :

    ξ

    ξ γ β

    ξ

    ξξ

    − ≤ ≤ +

    =

    =

    ><

    o

    o o

    16.84516.845 90

    τ =

    −or

    o29.499ξ =

    LHEP

    Results

    AR 1.768=

    Example (cont.) Formulas

  • 42

    o16.845τ =

    AR 1.768=

    x

    y

    ( )tE

    τ

    Example (cont.)

    Note: Plotting the ellipse as a function of time will help determine which value is correct for the tilt angle τ.

    ( ) ocos

    0.6324 cos( 71.565 )x

    y

    tt

    ω

    ω

    =

    = +

    E

    E

    Normalized field (a = 1):

  • 43

    AR 1.768=

    x

    y

    ( )tE

    τ

    Example (cont.)

    We know that the polarization is LHEP, so the LHCP amplitude must dominate. Hence, we have

    AR 1 AR 1AR 1 AR 1

    LH

    RH

    AA

    − +=

    + −or

    0.2775 3.604LHRH

    AA

    = or

    Hence

    3.604LHRH

    AA

    =

  • Poincaré Sphere

    ( )AR cot ξ=0

    0

    ξ

    ξ

    >

    <

    implies LHEP

    implies RHEP

    / 2 2θ π ξ= −

    x

    y ( )tE

    τ

    A

    B C

    D |ξ |

    44

    2φ τ=

    x y

    z

    LHCP

    RHCP

    Unit sphere

    longitude = 2τ

    latitude = 2ξ

  • Poincaré Sphere (cont.)

    linear (equator)

    LHCP (north pole)

    RHCP (south pole) φ = 0 (no tilt)

    φ = 90o (45o tilt)

    latitude = 2ξ longitude = 2τ

    LHEP

    RHEP

    ξ = 45o

    ξ = 0o

    ξ = -45o

    x

    y

    45

  • Poincaré Sphere (cont.)

    View in full-screen mode to watch the “world spinning”.

    46

  • Poincaré Sphere (cont.) Two diametrically opposite points on the Poincaré sphere have the property that the electric field vectors (in the phasor domain) are orthogonal in the complex sense. (A proof is given in the Appendix.)

    x

    y

    z

    A

    B

    * 0A BE E⋅ =

    Examples

    -

    ˆ :ˆ :

    A

    B

    x

    x

    E x AE y B

    =

    =

    equator on axis

    equator on axis

    ( )( )

    ˆ ˆ (LHCP) :

    ˆ ˆ (RHCP) :

    A

    B

    E x y j A

    E x y j B

    = + +

    = + −

    north pole

    south pole

    47

    ( )* ˆ 0A BE H z× ⋅ =( ) ( )( )

    ( ) ( )( )

    * *

    0

    * *

    0

    1ˆ ˆ ˆ

    1 ˆ ˆ ˆ

    A B A B

    A B B A

    E H z E z E z

    z E E E E z z

    η

    η

    × ⋅ = × × ⋅

    = ⋅ − ⋅ ⋅

    ( ) ( ) ( )A B C B A C C A B× × = ⋅ − ⋅Note :(orthogonal in the complex power sense)

  • Appendix Proof of orthogonal property

    x

    y

    z

    A

    B

    * 0A BE E⋅ =

    ( ) ( )ˆ ˆ ˆ ˆj jxa y be xb y aeβ β++ → −Hence

    48

    ( ) ( )( ) ( )( ) ( )

    :, , ;

    , , / 2 ;

    , , / 2 ;/ /

    A B

    b a a b

    θ φ π θ φ π

    ξ τ ξ τ π

    β γ π β π γ

    → − +

    → − +

    → + −

    From examining the polarization equations:

    ( )( ) ( )( )*ˆ ˆ ˆ ˆ 0j jxa y be xb y ae ab abβ β++ ⋅ − = − =

    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 24Slide Number 25Slide Number 26Slide Number 27Slide Number 28Slide Number 29Slide Number 30Slide Number 31Slide Number 32Slide Number 33Slide Number 34Slide Number 35Slide Number 36Slide Number 37Slide Number 38Slide Number 39Slide Number 40Slide Number 41Slide Number 42Slide Number 43Slide Number 44Slide Number 45Slide Number 46Slide Number 47Slide Number 48