Dipole Antennas

ANTENNAS Vector and Scalar Potentials Maxwell's Equations ∇ × E = − jωB

(M1)

∇ × H = J + jωD

(M2)

∇·D = ρ

(M3)

∇·B=0

(M4)

D = εE B= µH For a linear, homogeneous, isotropic medium µ and ε are contant. Since ∇· B = 0 , there exists a vector A such that B = ∇ × A and ∇· (∇ × A ) = 0. A is called the magnetic vector potential . There are infinitely many vectors A that satisfy B = ∇ × A thus we later need to specify ∇ · A. (M1) can be written as ∇ × E = -jω ∇ × A ⇒ ∇ × (E + jωA) = 0

(1)

∇ × Vector = 0 ⇒ Vector = ∇(some scalar)

(2)

E + jωA = -∇ϕ

ϕ : scalar potential

E = - jωA -∇ϕ

(3) (4)

(M2) can be written as ∇ × B = µJ + jωµεE

(5)

∇ × ∇ × A = µJ + jωεµ (−jωA − ∇ϕ)

(6)

− ∇2 A + ∇(∇ · A) = µJ + ω 2µεA −jωεµ∇ϕ

(7)

2 To simplify the equation, choose ∇ · A as ∇ · A = - jωµεϕ

(8)

This is called the Lorentz Condition . This leads to ∇ 2 A + ω2µεA = - µJ

D'Alembert's Equation (9)

The problem now consists of finding the vector potential A due to a source J. From the knowledge of A, E and H can be determined. Review spherical coordinates, gradient, divergence, curl, and laplacian in spherical coordinates. (Textbook, Appendix A pp. 689-691). In spherical coordinates, the solution for the vector potential A(r) is given by J(r' )e − jβ|r − r'| dv' V' 4 π | r − r' |

A(r) = µ ∫∫∫

(10)

In the above equation, V' is the volume defining the source over which the integration is performed. r is the vector from the origin of the coordinate system, O, to the observer and r' is the vector from O to a source point within V'. observer r -r' r' O

r

V' source Figure 1

From A, use Maxwell's Equations to derive E and H. B= ∇×A

(11)

jωεE = ∇ × H

(12)

3 Hertzian Dipole Assume a source point infinitely small in size located at the origin of the coordinate system with elemental length dl, and driven by a current with strength Io in the +z direction. The equation for the current density of such a system is given by J(r') = iz I odlδ(x')δ(y')δ(z')

(13)

Upon substitution in (10) A(r) = iz

µIodl exp(-jβr) 4πr

(14)

In spherical coordinates: iz = ir cosθ - iθ sinθ A(r) = (ir cosθ - iθ sinθ)

Ar =

µIodl 4πr

exp(-jβr)

(15)

µ I odl cosθ exp(-jβr) 4πr

(16)

µ I odl sinθ exp(-jβr) 4π

(17)

Aθ = Calculate E and H fields. H=

1 ∇×A µ

Hr =

1 ∂ ∂A θ  ( A φ sin θ) − =0  ∂φ  r sin θ  ∂θ

(18)

(19)

 1  1 ∂A r ∂ Hθ =  − ( rA φ ) = 0 r  sin θ ∂φ ∂r 

(20)

1 ∂ ∂A r  H φ =  ( rA θ ) − ∂θ  r  ∂r

(21)

I dl 1 Hφ = o sinθ exp(-jβr) jβ(1+ ) 4πr jβr

(22)

4

Using E =

1 ∇ × H, we can derive the components of the electric field. jωε Er =

1 ∂  1 (H φ sin θ)  r sin θ  ∂θ  jωε

(23)

Er =

2 cosθI o dl − jβr  1  1 e jβ1 +  2 4 πr jβr  jωε 

(24)

{

}

1 ∂  1 E θ = − ( rH φ ) r  ∂r  jωε Eθ = −

(25)

  jβI o dl 1 1  1 sin θ e − jβr − jβ1 + − 2 2  jβr β r   jωε 4 πr  

{

}

(26)

In summary, the exact solutions for the fields of an infinitesimal antenna are given by Er =

 1 jβI o dl − jβr 1  µ e 2 cos θ  +  4 πr  jβr ( jβr )2  ε

(27)

Eθ =

 jβI o dl − jβr 1 1  µ + e sin θ 1 +  jβr ( jβr )2  ε 4 πr 

(28)

Eφ = 0 Hφ =

 jβI o dl − jβr 1  e sin θ 1 +  jβr  4 πr 

(29)

H r = Hθ = 0 In most practical cases, the observer is located several wavelengths away from the source. This defines a far field region which is the region where the distance from the source to the 2π observer is much larger than the wavelength λ = . In this case r >> λ so that βr >>1; β consequently, the terms varying as 1/r 2 and 1/r 3 can be neglected. The far-field solutions for the infinitely small antenna thus become Er = 0 Eθ =

(30) µ jβI o dl − jβr e sinθ ε 4 πr

(31)

5

Hφ =

jβI o dl − jβr e sinθ 4 πr

(32)

Note that the ratio Eθ /H φ is the characteristic impedance η of the propagation medium. Over a small region, the far field solution is a plane-wave solution since the electric and magnetic fields are in phase, perpendicular to each other and their ratio is the intrinsic impedance η, and are perpendicular to the direction of propagation. However, unlike plane waves, the far field solution is a function of the elevation angle θ, and does not have constant magnitude (1/r dependence). Radiation Patterns The graph that describes the far-field strength versus the elevation angle at a fixed distance is called the radiation pattern of the antenna. In general, radiation patterns vary with θ and φ. The distance from the dipole to a point on the radiation pattern is proportional to the field intensity or power density observed in that direction. Figure 2 shows the E-field and power density radiation patterns of a Hertzian dipole. As can be verified these patterns are based on the sinθ and sin 2θ dependence of the E-field and power density respectively. z

z θ

θ

1

(a)

1

(b)

Figure 2. (a) Radiation pattern for E field (b) Radiation pattern for power density Time Average Power in Radiation Zone In order to calculate the power radiated in the far field, we need to determine the timeaverage Poynting vector or power density

.

6

=

 µ  i 1 Real E × H* = r Real  | H φ |2  2 2  ε 

[

]

iη  βI dl  < P > = r Real  o  sin 2 θ 2  4 πr 

(33)

2

(34)

The total power radiated PT at a distance r is by definition obtained by integrating the Poynting vector over a sphere of radius r. 2π π

Total Power = PT = ∫ ∫ < P > ⋅ dS 0

(35)

0

dS is the elemental surface of radius r and is given by dS = i r r 2 sin θ dθ dφ

(36)

so that 2π π

2

η  βI o dl  2 3 r sin θ dθ dφ   r π 2 4 0

PT = ∫ ∫ 0

PT =

PT =

η βI o dl 2 4π

2

π

2 π ∫ sin 3 θ dθ

4 πη βI o dl 4π 3

(37)

(38)

0 2

=π

η dl 2 2 Io 3 λ

(39)

The directive gain is a figure of merit defined as Poynting power density Directive Gain = AveragePoynting power density over area of sphere of radius r or Directive Gain =

For an infinitesimal antenna, we get

PT / 4 πr 2

(40)

7 η βI o dl sin 2 θ 3 2 4 πr Directive Gain = 4 πr 2 = sin 2 θ 2 2 4 πη βI o dl 3 4π 2

(41)

The directivity is the directive gain in the direction of its maximum value. For an infinitesimal antenna, the direction of maximum value is for θ = π/2 and the directivity is 1.5. Radiation Resistance The radiation resistance of an antenna is defined as the value of the resistor that would dissipate an equal amount of power than the power radiated for the same value of current. Using PT =

1 2 2 Rrad I o

(42)

We get Rrad =

2P T Io2

(43)

For an infinitesimal antenna, we get R rad

Using β =

2 4 π βI dl = 2  η o Io  3  4π

2

(44)

2π , and η = 120π ohms, we obtain λ dl R rad = 80 π 2    λ

2

Ω

(45)