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Antennas and Propagation.L05

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Antennas and Propagation.L05 Powered By Docstoc
					                      Linear Wire Antennas
                                 Ranga Rodrigo
                                 August 24, 2010


    Lecture notes are fully based on Balanis [?]. Some diagrams and text are directly
from the books.



Contents
1 Infinitesimal Dipole                                                              1

2 Small Dipole                                                                     7

3 Finite-Length Dipole                                                             9




    • Wire antennas, linear or curved, are some of the oldest, simplest, cheap-
      est, and in many cases the most versatile for many applications.


1 Infinitesimal Dipole


    • An infinitesimal linear wire (l  λ) is positioned symmetrically at the ori-
      gin of the coordinate system and oriented along the z axis.

    • The spatial variation of the current is assumed to be constant and given
      by
                                              ˆ
                                     I (z ) = a z I 0
      where I 0 is a constant.




                                         1
• The source only carries an electric current I e . I m and the potential func-
  tion F are zero. To find A we write
                                         µ                          e − j kR
                      A(x, y, z) =               I e (x , y , z )            dl .
                                        4π   C                         R
  where
    (x, y, z) : Observation point coordinates.
  (x , y , z ) : Coordinates of the source.
          R : The distance from any point on the source to the observation point.
          C : Path along the length of the source.

                                                       ˆ
                                    I e (x , y , z ) = a z I 0 .
                x = 0, y = 0, z = 0,         for the infinitesimal dipole.

     R=     (x − x )2 + (y − y )2 + (z − z )2 =              x2 + y 2 + z2 = r          (constant).
    dl = dz .
                                    µI 0 − j kr      l /2               µI 0 l − j kr
                             ˆ
                A(x, y, z) = a z        e                         ˆ
                                                            d z = az          e       .
                                    4π              −l /2                4π
                                z




                                             (x, y, z)
                                    θ




                 (x , y , z )

                                φ
                                                                                    y



            x




                                             2
                     1
    Next: H A : H A =  ∇ × A, E A : ∇ × H A = J + j ω E A . Transformation from
                     µ
rectangular to spherical coordinates:

                             sin θ cos φ sin θ sin φ cos θ
                                                           
                    Ar                                           Ax
                   A θ  = cos θ cos φ cos θ sin φ − sin θ   A y 
                    Aφ         − sin φ     cos φ        0        Az

A x = 0, A y = 0, A z = 0.

                                               µI 0 l e − j kr
                             A r = A z cos θ =                 cos θ.
                                                   4πr
                                                   µI 0 l e − j kr
                             A θ = −A z sin θ = −                  sin θ.
                                                        4πr
                             A φ = 0.


          ˆ
         ar     ∂                  ∂      ˆ
                                          aθ  1 ∂         ∂           aφ ∂
                                                                      ˆ                 ∂
∇× A =             (A φ sin θ) −     Aθ +            A r − (r A φ ) +       (r A θ ) −    Ar
       r sin θ ∂θ                 ∂φ      r sin θ ∂φ      ∂r           r ∂r            ∂θ
       1 aφ ∂
          ˆ                  ∂
    H=           (r A θ ) −    Ar
       µ r ∂r               ∂θ

                              Hr = Hθ = 0.
                                      k I 0 l sin θ      1
                             Hφ = j                 1+      e − j kr .
                                          4πr          j kr
                                                   1
                                     E =EA =         ∇× H.
                                                  jω

                              I 0 l cos θ         1
                       Er = η                1+        e − j kr .
                                 2πr 2          j kr
                                k I 0 l sin θ        1          1
                       Eθ = j η               1+         −        e − j kr .
                                     4πr           j kr (kr )2
                      E φ = 0.

The E and H components are valid everywhere except on the source itself.




                                                  3
                                           z


                                                                    ˆ
                                                                    ar , E r , H r
                                                                         ˆ
                                                                         aφ , E φ , H φ


                                                    θ
                                                                       ˆ
                                                                       aθ , E θ , H θ




                                           φ
                                                                                             y



                x


Power Density and Radiation Resistance
   • For a lossless antenna, the real part of the input impedance is designated
     as the radiation resistance, that power is transferred from the guided wave
     to the free space wave.


                             1                1
                        W = E × H ∗ = (a r E r + a θ E θ ) × a φ Hφ .
                                                  ˆ       ˆ           ˆ   ∗
                             2                2
                             1              ∗            ∗
                                 ˆ               ˆ
                           = a r E θ Hφ − a θ E r Hφ .
                             2
                             η I 0 l 2 sin2 θ               1
                        Wr =                        1− j          .
                             8 λ            r 2          (kr )3
                                 k|I 0 l |2 cos θ sin θ            1
                        Wθ = j η              2r 3
                                                          1+ j          .
                                        16π                     (kr )2

   The complex power moving int eh radial direction
                                               2π       π
          P=             W ·ds =                            (a r Wr + a θ Wθ ) · a r r 2 sin θd θd φ
                                                             ˆ
                        S                  0        0
                        2π       π
            =                        Wr r 2 sin θd θd φ
                    0        0
                                     2
               π I0l                              1
            =η                           1− j          .
               3 λ                              (kr )3

                                                                4
The transverse component Wθ does not contribute to the integrals. Thus P
does not represent the total complex power radiated by the antenna. Wθ is
purely imaginary, and does not contribute to any real radiated power. It con-
tributes to the imaginary (reactive) power.
    The reactive power density, which is most dominant for small values of kr ,
has both radial and transverse components. It merely changes between out-
ward and inward directions to form a standing wave at a rate twice per cycle. It
also moves in the transverse direction.
    Time average power radiated is
                                                           2
                                             π       I0l
                                p rad = η                      .
                                             3        λ
For large values of kr (kr    1), the reactive power diminishes. For free space
η 120π,
                                                2                      2
                                   2π       l                      l
                       R rad = η                    = 80π2 .               .
                                    3       λ                      λ


Near-Field Region kr     1


                                       I 0 l e − j kr
                             Er    −jη                cos θ.
                                         2πkr 3
                                       I 0 l e − j kr
                             Eθ − j η                 sin θ.
                                         4πkr 3
                             E φ = Hr = Hθ = 0.
                                    I 0 l e − j kr
                             Hφ                    sin θ.
                                        4πr
E r and E θ are in time-phase. E r and E θ are in time-phase quadrature with Hφ .
Therefore, there is no time-average power flow associated with them.

Intermediate-Field Region kr > 1


                                    I 0 l e − j kr
                             Er η                   cos θ.
                                      2πkr 2
                                      k I 0 l e − j kr
                             Eθ j η                    sin θ.
                                           4πkr
                             E φ = Hr = Hθ = 0.
                                      k I 0 l e − j kr
                             Hφ     j                  sin θ.
                                           4πr

                                            5
E r and E θ approach time-phase quadrature. They form a rotating vector whose
tip traces and ellipse in a plane parallel to the direction of propagation: cross
field.

Far Field kr      1


                                         k I 0 l e − j kr
                               Eθ     jη                  sin θ.
                                            4πkr
                               Er     E φ = Hr = Hθ = 0.
                                          k I 0 l e − j kr
                               Hφ     j                    sin θ.
                                               4πr
The ratio of E θ to Hφ is equal to

                                                  Eθ
                                     Zw =               η.
                                                  Hφ

where Z w is the wave impedance and η is the intrinsic impedance (377 120πΩ
for free-space.) E - and H -field components are perpendicular to each other,
transverse to the direction of propagation, and r variations are separable from
those of θ and φ. This relationship is applicable in the far-field region of all
antennas of finite dimensions.

Directivity
   The average power density
                                                                          2
                     1                  1               η k I0l               sin2 θ
               W av = Re E × H ∗ = a r
                                   ˆ      |E θ |2 = a r
                                                    ˆ                                .
                     2                 2η               2 4π                    r2

The radiation intensity
                                              2
                                 η k I0l                     r2              2
                  U = r 2Wav =                    sin2 θ =      E θ (r, θ, φ) .
                                 2 4π                        2η

The maximum value occurs at θ = π/2:
                                                             2
                                                  η k I0l
                                    Umax =                       .
                                                  2 4π

                                                  Umax 3
                                  D 0 = 4π              = .
                                                  P rad  2


                                                  6
The maximum effective aperture

                                              λ2      3λ2
                                  A em =         D0 =     .
                                              4π      8π

                                               1
                                     P rad =     |I 0 |2 R rad .
                                               2
                                                              2
                                                 2π l
                                     R rad = η                    .
                                                  3 λ


2 Small Dipole
The current distribution of the infinitesimal dipole (l < λ/50) is I 0 , a constant.
For a small dipole (λ/50 ≤ l ≤ λ/10) the triangular current distribution approx-
imation must be used.
                                          z


                                                I0
                                                       |I |




The current distribution is
                                                                  l
                                         a z I0 1 − 2 z , 0 ≤ z ≤ 2 ,
                                         ˆ          l
                    I e (x , y , z ) =                       l
                                         a z I 0 1 + 2 z , − 2 ≤ z ≤ 0.
                                         ˆ           l

                 µ       0          2  e − j kR                           l /2         2  e − j kR
 A(x, y, z) =      ˆ
                   az         I0 1 + z                ˆ
                                                d z + az                         I0 1 − z          dz .
                4π      −l /2       l     R                           0                l     R
Because the overall length of the dipole is small, the value of R for different
values of z along the length of the wire are not much different from r .




                                                 7
                                    z




                                                                   P (r, θ, φ)
                                         θ
                                                    R
                   l /2   dz
                                        θ           r
                                z



                   l /2
                                    φ=φ
                                                                                             y



               x

                                                                                  kl     π
Maximum phase error due to the assumption R                                r is   2
                                                                                       = 10 = 18◦ for λ/10.

                                                        1 µI 0 e − j kr
                                   ˆ        ˆ
                               A = a z Az = a z                         ,
                                                        2    4πr
which is one half of that obtained in the previous section for the infinitesimal
dipole.

Far-Zone Fields, kr       1


                                                k I 0 l e − j kr
                                    Eθ       jη                  sin θ.
                                                   8πkr
                                    Er       E φ = Hr = Hθ = 0.
                                                 k I 0 l e − j kr
                                    Hφ       j                    sin θ.
                                                      8πr
Directivity and the maximum effective area are the same as for the infinitesimal
dipole.
                                                    2
                                  2P rad        2 l
                          R rad =         = 20π
                                  |I 0 |2         λ
which is 1/4 of the value for the infinitesimal dipole.



                                                        8
3 Finite-Length Dipole
We can analyze the radiation characteristics of a dipole with any length us-
ing magnetic vector potential A. For a thin, center-fed finite-length dipole
(l ≥ λ/10, d      λ), the approximate current distribution can be written as
                                      
                                      a z I 0 sin k l − z , 0 ≤ z ≤ l ,
                                        ˆ            2               2
             I e (x = 0, y = 0, z ) =
                                      a z I 0 sin k l + z , − l ≤ z ≤ 0.
                                        ˆ            2         2



    In the far field, we have, z                        r, θ       θ . For amplitude: R                  r . For phase:
R    r − z cos θ.

                           z                                                                        z


                                                         P (r, θ, φ)                                                     P (r, θ, φ)
                               θ                                                                           R
                                          R                                                         θ
       l /2       dz                                                                       dz
                               θ          r
                       z                                                                        z   θ          r


       l /2                                                       y                                                             y
                           φ=φ                                                                      φ=φ
                                                                                       x
              x

                                                                                       kl     π
Maximum phase error due to the assumption R                                     r is   2
                                                                                            = 10 = 18◦ for λ/10.

                                                   µ                           e − j kR
                               A(x, y, z) =                 I e (x , y , z )            dl .
                                                  4π    C                         R
                           µe − j kR
                       A(x, y, z) =    I e (x , y , z )e j kz cos θ d z .
                             4πr     C
The finite dipole antenna is subdivided into a number of infinitesimal dipoles
of length δz . For an infinitesimal dipole of length d z positioned along the
z-axis at z
                                              j ηk I e (x , y , z )e − j kR
                                   d Eθ                                     sin θd z .
                                                         4πR
                                   d Er       d E φ = d Hr = d Hθ = 0
                                              j k I e (x , y , z )e − j kR
                               d Hφ                                        sin θd z .
                                                         4πR

                                                              9
Using the far field approximation

                             j ηk I e (x , y , z )e − j kr               cos θ
                  d Eθ                                     sin θe j kz           dz .
                                        4πr
Summing the contribution from all the infinitesimal elements

               −l /2                ke − j kr          −l /2
                                                                                        cos θ
       Eθ =            d Eθ = j η             sin θ            I e (x , y , z )e j kz           dz
              −l /2                  4πr              −l /2

Simplifying

                                                  kl            kl
                                                                   
                               I 0 e − j kr  cos 2 cos θ − cos 2 
                       Eθ   jη                                        .
                                  2πr                 sin θ

                                            cos kl cos θ − cos kl
                                                                  
                                  − j kr
                              I0e                2             2
                      Hφ    j                                     .
                                2πr                  sin θ



Power Density, Radiation Intensity, and Radiation Resistance


                            1                 1             ∗
                      W av = Re E × H ∗ = Re a θ E θ × a φ Hφ
                                                     ˆ ˆ
                            2                 2
                            1
                                ˆ         ˆ ∗
                           = Re a θ E θ × a φ E θ /η
                            2




                                                 10

				
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