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Simple array of two antennas Consider two identical ... · PDF fileSimple array of two antennas Consider two identical parallel Hertzian dipoles separated by a distance d I2 I1 z x

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  • Antennas

    Amanogawa, 2006 Digital Maestro Series 64

    Simple array of two antennas

    Consider two identical parallel Hertzian dipoles separated by a distance d

    I1I2

    z

    x

    12

    r1r 2 r

    d 2 d 2

    i

    i1

    i2

  • Antennas

    Amanogawa, 2006 Digital Maestro Series 65

    The dipole currents have the same amplitude and total phase difference

    The electric farfield components at the observation point are

    ( )

    ( )

    21 1

    22 2

    cos( 2) I

    cos( 2) I

    jo o

    phasorj

    o ophasor

    I t I t I e

    I t I t I e

    = + =

    = =

    j r jo

    j r jo

    j I z eir

    j I z eir

    1

    1

    2

    2

    2

    1 11

    2

    2 22

    E sin4

    E sin4

    +

  • Antennas

    Amanogawa, 2006 Digital Maestro Series 66

    At long distance, we have

    and the field components can be written as

    1 2

    1 2

    1 2cos cos2 2

    i i i

    d dr r

    d

    r

    r

    r

    +

    >>

    j r d joj I z ei

    r d

    ( ( 2)cos ) 21E 4 ( 2)cos

    +

    ( )j r d j

    oj I z eir d

    ( ( 2)cos ) 22

    sin

    E4 ( 2)cos

    +

    +( ) sin

  • Antennas

    Amanogawa, 2006 Digital Maestro Series 67

    After applying the approximations, the two components can be combined to give the total electric field

    The final result is

    ( ) ( )( )j ro

    j d j d

    j I zi er

    e e

    1 2

    ( 2)cos 2 ( 2)cos 2

    sinE E E4

    + +

    = +

    +

    j roj I z di er

    field of a Hertzian dipole locatedat the center of the arra

    array y

    factor

    sin cosE 2 cos4 2

    +

  • Antennas

    Amanogawa, 2006 Digital Maestro Series 68

    The resultant radiation pattern of the electric field is proportional to

    The unit pattern is proportional to the radiation pattern of the individual antennas, assumed to be identical.

    The group pattern is proportional to the radiation pattern the array would have with isotropic antennas.

    Note: on the xy plane, coincides with the azimuthal angle .

    Following are examples of twoantenna arrays with specific values of dipole distance and current phase difference.

    d

    group patterunit

    patt rn ne

    cossin cos2

    +

  • Antennas

    Amanogawa, 2006 Digital Maestro Series 69

    Broad-side / p2 e0 att rnd = =

    z

    x x

    z

    x

    y

    x

    yUnit Pattern Group Pattern

    x

    yResultant Pattern

    x

    z

  • Antennas

    Amanogawa, 2006 Digital Maestro Series 70

    d Broad-side p/ rn0 t2 at e = =Radiation Pattern for E and H Power Radiation Pattern

  • Antennas

    Amanogawa, 2006 Digital Maestro Series 71

    d End-fir/ 2 1 e pattern80 = =Radiation Pattern for E and H Power Radiation Pattern

  • Antennas

    Amanogawa, 2006 Digital Maestro Series 72

    d Cardioid/ 4 90 pattern = =Radiation Pattern for E and H Power Radiation Pattern

  • Antennas

    Amanogawa, 2006 Digital Maestro Series 73

    d Cardioid patter4 9 n/ 0 = = Radiation Pattern for E and H Power Radiation Pattern

  • Antennas

    Amanogawa, 2006 Digital Maestro Series 74

    d / 2 90 = =Radiation Pattern for E and H Power Radiation Pattern

  • Antennas

    Amanogawa, 2006 Digital Maestro Series 75

    d 0 = =Radiation Pattern for E and H Power Radiation Pattern

  • Antennas

    Amanogawa, 2006 Digital Maestro Series 76

    d 180 = =Radiation Pattern for E and H Power Radiation Pattern

  • Antennas

    Amanogawa, 2006 Digital Maestro Series 77

    CASE STUDY - 1) A radio broadcast transmitter is located 15 km West of the city it needs to serve. The FCC standard is to have 25 mV/m electric field strength in the city. How much radiation power must be provided to a quarter wavelength monopole? We consider =90 for transmission in the plane perpendicular to the antenna

    In a monopole, the lower wire is substituted by the ground. The equivalent radiation resistance is half that of the corresponding dipole. Therefore, the total radiated power is half the power radiated by the half-wavelength dipole, for the same current.

    max

    maxmax

    / 2 dipolecos90E cos2 sin90 2

    E 120 0.025 V/m 6.25 A2 15, 000

    j rj e Iir

    I I

    =

    = =

    =

  • Antennas

    Amanogawa, 2006 Digital Maestro Series 78

    A perfect ground would act like a metal surface, reflecting 100% of the signal. The ground creates an image of the missing wire delivering to a given point above the ground the same signal as a complete dipole. The transmission line connected to the antenna sees only half of the radiation resistance, with total radiated power:

    2 2max

    1 73.070.193978 0.5 6.25 713.622

    W2

    /

    eqR

    totP I

    = = =

    P

    GROUND PLANE

    Monopole

    /4

  • Antennas

    Amanogawa, 2006 Digital Maestro Series 79

    2) Improve the design by using a twoantenna array.

    A good choice of array parameters is

    which gives a cardioid pattern

    dphase(antenna B) phase(antenna A) 90

    / 4

    = =

    =

    15 km

    BA

    d

  • Antennas

    Amanogawa, 2006 Digital Maestro Series 80

    The Poynting vector is given by

    ( )

    ( )

    r

    r

    I dP t i

    r

    Ii

    r

    2

    2

    2 dipole Poynting vecto array factor

    array fa

    22 2max

    2 2 2

    22 2max

    ct

    2

    r

    r

    o

    2 2

    cos cos( ) cos 4 cos

    2 28 sin

    coscos 4 cos cos

    2 48 sin 4

    +=

    =

    R

    sinR sin cosR

    cosR

    Nocos sin cos

    te =

    sin sinR

  • Antennas

    Amanogawa, 2006 Digital Maestro Series 81

    The total radiated power is

    2 220 0

    sin ( )totP r d P t d

    =

    Only /2 because it is a monopole

    ( )2

    22max

    2 2 2

    array facto

    2

    r

    2 cos2 cos sin0 28 sin2

    4cos0 4sin cos

    4

    I dtotr

    P r

    d

    =

  • Antennas

    Amanogawa, 2006 Digital Maestro Series 82

    The integral over the azimuthal angle gives

    For a monopole, we only have the integral

    2 20

    20

    20

    4 cos sin cos4 4

    1 14 cos sin cos2 2 2 21 14 sin sin cos 42 2 2

    d

    d

    d

    = + = + =

    20 2d =

  • Antennas

    Amanogawa, 2006 Digital Maestro Series 83

    In the direction of maximum

    The same field strength (25 mV/m) is obtained by applying half the current of the original monopole to the array elements (also monopoles)

    The total radiated power is proportional to the square of the current, and the integral over gives a factor 4 instead of 2 for the array. Overall, the total radiated power needed by the array, to produce the same electric field, is half that of the individual monopole

    90 & 90 array factor 2 = = =

    max6.25 3.125 A

    2I = =

    714 357 W2tot

    P = =

  • Antennas

    Amanogawa, 2006 Digital Maestro Series 84

    For a monopole Radiated Power = 0.51428.3127 = 714.155635 [ W ] Rad. Resistance = 0.573.129616 = 36.564808 [ ]

    SAME FEEDING CURRENT FOR BOTH DIPOLE AND MONOPOLE

  • Antennas

    Amanogawa, 2006 Digital Maestro Series 85

    For monopoles Radiated Power = 0.5714.1532 = 357.0766 [ W ]

    SAME FEEDING CURRENTS FOR BOTH DIPOLES AND MONOPOLES

  • Antennas

    Amanogawa, 2006 Digital Maestro Series 86

    N-Element Antenna Array

    Assume a uniform array of N identical antennas. The elements are fed by currents with constant amplitude and with phase increasing by an amount from one to the other. The spacing d between the antennas is uniform.

    r ( 1) cosr N d

    d

    1 2 3 N

    ( 1)(1) ; (2) ; ; ( )j j No o oI I I I e I N I e = = =

  • Antennas

    Amanogawa, 2006 Digital Maestro Series 87

    The electric field at the observation point (r,) is of the form

    j r j r d jo o

    j r N d j Noj r j d

    o

    j N d

    jN dj r

    o j d

    r E e E e e

    E e e

    E e e

    e

    eE ee

    ( cos )

    ( ( 1) cos ) ( 1)

    ( cos )

    ( 1)( cos )

    ( cos )

    ( cos )

    E( , )

    1

    11

    +

    +

    +

    +

    = + +

    +

    = + ++

    =

    We have used( cos )1

    ( cos )( cos )

    0

    1 1

    jN dNjn d

    j dn

    eee

    ++

    +=

    =

  • Antennas

    Amanogawa, 2006 Digital Maestro Series 88

    The magnitude of the electric field is given by We have used

    ar

    ( cos )

    ( cos

    ray facto

    )

    r

    1E( , )1sin[ ( cos ) / 2]sin[( cos ) / 2]

    jN do j d

    o

    er EeN dE

    d

    +

    +

    =

    +=

    +

    21 2 sin 2 sin2 2

    jx j xx xe j e = =

  • Antennas

    Amanogawa, 2006 Digital Maestro Series 89

    We can rewrite The group pattern has

    group pattern

    1 sin[ ( cos ) / 2]sin[( cos

    E( ,) / 2]

    ) oN d

    N dr N E

    +

    =+

    Maxima when

    Nulls when

    for

    ( cos ) 2

    cos

    integer

    0,

    0,

    2

    , 2 ,

    , 4

    NN

    d

    d mm N

    + =