# Antennas and Propagation.L05

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```					                      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

• 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.
2
π       I0l
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

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              = .

6
The maximum effective aperture

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

1
2
2
2π l
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
|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

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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