Module 6: Antennas 6.0 Introduction

In chapter 2, the fundamental concepts associated with electromagnetic radiation were examined. In this chapter, basic antenna concepts will be reviewed, and several types of antennas will be examined. In particular, the antennas commonly used in making EMC-related measurements will be emphasized.

6.1 The Radiation Mechanism

Antennas produce fields which add in phase at certain points of space. Consider a loop of wire that carries a current.

R1 d I 1

D R 2

d I 2

¡ ¡

Here two elements of current d I 1 and d I 2 are separated by a distance D. The current elements are located at distances R1 and R2, respectively from a distant observation point. If

¢ λ R2-R1 0.1

D ¢ 0.1λ

Then the fields produced by the current elements add out of phase, and the amount of radiation is small. However, if £ λ R2-R1 0.1

D £ 0.1λ

Then the fields produced by the current elements add in phase, and the amount of radiation is large.

-Reception Mechanism

Electromagnetic fields which are incident upon an antenna induce currents on the surface of the antenna which deliver power to the antenna load.

6-1 induced current

incident field

load Z L impedance transmission line

antenna

6.2 Radiated Power

The power radiated by a distribution of sources is that power which passes through a sphere of infinite radius. This, therefore, is the power which leaves the vicinity of the source system, and never returns.

In chapter 2 the time-average Poynting vector was presented

¤ ¤ ¤ 1 P = R e{E × H ∗} 2

At points far from the antenna (the radiation zone)

− jkr

¦ ¦ ¥ µ ¥ e E (r ) ≈ − jω [θ N + φ N ]

4π r θ φ

¨ §

§ × ( ) § r E r ( ) ≈ H r η

where

(θ φ) = ( © ′) jk (r ⋅r ′) ′ N , ∫ J S r e d v v is known as the “radiation vector.” The radiation vector is related to the vector potential by

− jkr © © µ e A (r ,θ , φ) = N (θ , φ) 4π r with

6-2

− jkr

Now in the radiation zone ©

∗ © © 1   r × E   P ≈ R e  E ×   

2   η  

Using the vector identity A × ( B × C ) = B ( A ⋅ C ) − C ( A ⋅ B )

1 P ≈ R e{(E ⋅ E ∗ )r − (E ⋅ r )E ∗}

2η ©

⋅ E E ≈ r ...... because E ⋅ r = 0 2η 2

1  ωµ  ∗ ∗ ≈ r   [N θ N θ + N φ N φ ]

2η  4πr 

∗ ∗ ∗ ∗ ...... because [θ N θ + φ N φ ]⋅[θN θ + φ N φ ] = [N θ N θ + N φ N φ ]

Finally 1 η  2 2  P ≈ r N + N r 2 8 λ2  θ φ 

This represents the average power flow density and lies in the direction of wave propagation.

The power radiated through a sphere of infinite radius is given by W = lim ∫ n ⋅ P ds r →∞ s

Applying the expression for the time average Poynting vector leads to

π 2π η  1 2 2  = ⋅  +  2 θ θ φ W lim ∫ ∫ r r 2 2  N θ N φ  r sin d d r →∞  λ  0 0 r 8

6-3 π 2 π η  2 2  =  N + N  sin θd θd φ ∫ ∫ λ2  θ φ  0 0 8

d W η 2 2 = =  +  Let K 2  N θ N φ  ...... ”radiation intensity” d Ω 8 λ

= power radiated per unit solid angle...... where d Ω = sin θd θd φ .

The total radiated power is then π 2 π W = ∫ ∫ K (θ , φ )d Ω. 0 0

6.3 Antenna Terminology

Antenna Patterns

Radiation pattern - A plot of the radiation characteristics of an antenna. There are two types of radiation patterns:

1. Power pattern - A plot of the radiated power at a constant radius.

2. Field pattern - A plot of the electric or magnetic field magnitude at a constant radius.

An antenna pattern consists of a number of lobes. The largest lobe is usually called the main lobe, while the other smaller lobes are called side lobes. The minima between lobes are called nulls.

main lobe

side lobe

null

Radiation patterns are three-dimensional, but are usually measured and displayed as two-

6-4 dimensional patterns, which are sometimes called cuts. For most antennas, two cuts give a good representation of the three-dimensional pattern. The radiation patterns of linearly polarized antennas are often specified in terms of E- plane and H-plane patterns. The E-plane contains the direction of maximum radiation and the electric field vector. The H-plane contains the direction of maximum radiation and the magnetic field vector.

E-plane

E H-plane H

No antenna has a truly isotropic pattern (one which is the same in all directions). Rather antennas (real ones anyway) tend to radiate more effectively in some directions rather than others.

Directive gain - The ratio of the radiation intensity K(θ,φ) to the uniform radiation intensity for an isotropic radiator with the same total radiation power W.

K (θ , φ ) 4π g (θ , φ ) = = K (θ , φ ) d    W  W  4π   4π  .....where   is the total power radiated by an isotropic radiator per unit solid angle.  W 

Directivity - The maximum value of directive gain.

Gain - Directivity expressed in dB.

G = 10 log10 (directivity) = gain in dB

Beamwidth - The beamwidth of a radiation pattern is the angle between the half-power points of the pattern.

6-5 3 dB (half power) φ points

side lobe

Radiation efficiency - The radiation efficiency of an antenna is the ratio of the power radiated by the antenna to the total power supplied to the antenna. The total power supplied to the antenna consists of the power radiated and the power given up to resistive losses.

W = E + W W L ...... where E = radiation efficiency W = power radiated

WL = power lost

Radiation resistance - The radiation resistance of an antenna is the equivalent resistance through which its input current must flow in order that the power dissipated in the resistance is equal to the total radiated power. 1 I I ∗R = W 2 0 0 r

2W = or R r ∗ ...... radiation resistance I 0 I 0

...... where I0 is the input current to the antenna.

From the stand point of the source that drives an antenna, radiation resistance is indistinguishable from Ohmic resistance. In both cases, the source must continuously supply energy to the antenna in order to keep the current amplitude constant with time. In the case of Ohmic resistance, this resistance converts energy into propagating electromagnetic waves.

Input impedance - An arbitrary antenna with a pair of input terminals ‘a’ and ‘b’ is shown below.

6-6 Z in I in a V b Antenna

When the antenna is not receiving power from waves generated by other sources the Thevenin equivalent circuit looking into the terminals of the antenna consists only of an impedance V = = + Z in R in jX in I in

where Rin is the input resistance and Xin is the input reactance. The input resistance is the sum of two components = + Rin R ri R L

where Rri is the input radiation resistance and RL is the input loss resistance. RL accounts for that portion of the input power that is dissipated as heat, while the input radiation resistance Rri accounts for power that is radiated by the antenna. Rri is related to Rr by

2  I  = m ax R ri   R r .  I 0  Radiation efficiency can be expressed P R η = ra d = ri . r + Pin R ri R L

6.4 Hertzian Dipole

The simplest radiation source consists of a short segment of current

6-7

d z A , B , E

I r

d z y

− d z 2

A Hertzian dipole consists of a uniform current I flowing in a short wire dz terminated by point charges.

Here

−d z d z  δ ( )δ ( ) Iz x y ...... fo r ≤ z ≤ J (r ) =  2 2  0 ...... e lse w h ere

It is seen that I = ∫ J ⋅ d s = ∫ ∫ Iδ ( x )δ (y )d xd y = I x y

The charge associated with the current is found using the continuity equation

1 d J ∇ ⋅ J = − jωρ ⇒ ρ = − z jω d z

The current density may be expressed   d z   d z   J = Iz δ (x )δ (y )u  z +  − u  z −     2   2   where u(t) represents the unit step function.

6-8 J

z d z d z − 2 2

1 d   d z   d z   ρ = Iδ ( x )δ (y ) u  z +  − u  z −   jω d y   2   2  

I   d z   d z   du(t) = δ ( x )δ (y )−δ  z +  + δ  z −   ...... because = δ(t) jω   2   2   dt

d z d z = (+) point charge at z = , (-) point charge at z = − 2 2

Vector potential

− jkR

µ e ( ) = 0 ( ′) ′ A r π ∫ J r dv 4 v R

where

R = r − r ′ ≈ r = r for r >> d z so,

− jkr

µ e ( ) ≈ 0 ( ′) ′ A r π ∫ J r d v 4 r v dz − jkr 2

µ e ≈ 0 δ ( ′)δ ( ′) ′ ′ ′ π ∫ ∫ x y d x d y ∫ Iz d z 4 r x y dz z ′=− 2 or

− jkr

µ e A(r ) ≈ z 0 Idz 4π r

6-9

now use z = r cosθ − θ sin θ to get

− jkr

µ e A (r ) ≈ (r c osθ − θ sin θ ) 0 Id z 4π r

E-M fields

φ  ∂ ∂A  = ∇ × = − r B A  (rAθ ) 

r ∂r ∂θ  µ φ  ∂ ∂  e − jkr   = 0 Idz  (− sin θe − jkr )−  co sθ   4π r ∂r ∂θ  r   µ φ θ  sin  − = 0 Id z  jk sin θ + e jkr 4π r  r 

so µ 1 + jkr B = φ 0 Id z sin θe − jkr

4π r 2

j = − ∇ × = E ωµ ε B at all points where J 0

0 0 ! j  r! ∂ θ ∂  = −  ( θ ) − ( ) ωµ ε θ ∂θ Bφ sin ∂ rB φ

0 0  r sin r r  ! j Id z  r! ∂  1 + jkr  θ ∂  1 + jkr   = −   sin 2 θe − jkr  −  sin θe − jkr   ωε π θ ∂θ  2  ∂  

0 4  r sin r r r r  #

" j Idz  r 1 + jkr E = − 2 sin θ cosθe − jkr ωε π  θ 2

0 4  r sin r # θ  k jkr − 1 − jkr   − sin θ  − j [1 + jkr ]+  e − jkr  r  r r 2  

6-10 2 2  !    j Id z ! 1 + jkr 1 + jkr − k r = − r 2 co sθ + θ sin θ   e − jkr ωε π 3  3  0 4  r r  now use 1 1 µ η µ = 0 = 0 .....where η = 0 ωε ω µ ε ε 0 ε 0 0 0 0 k 0 Then

− jkr − jkr ! $  η   η  ! Id z e 2 2 Id z e 1 E = r  0 +  co sθ + θ  jωµ + 0 +  sin θ . π ωε 2 π 0 ωε 2 4 r  r j 0 r  4 r  r j 0 r 

Thus the fields may be expressed: ! $ µ Id z  jk 1  − B = φ 0 sin θ  + e jkr 4π  r r 2 

Idz  1 j  − E = 2η co sθ  − e jkr r 4π 0  r 2 kr 3 

  Id z jk 1 j − jkr E θ = η sin θ  + − e 4π 0  r r 2 kr 3 

It is seen that these expressions contain terms having three different rates of decay: 1/r, 1/r2, and 1/r3.

Near-zone fields (induction zone)

λ The near zone fields are those which are strongest near r = 0 , or when r << . Thus

2π

% & these are the terms of B which vary as 1/r2 and the terms of E which vary as 1/r3:

− jkr

( ' Id z e η ( E ≈ − j 0 [2 r co s θ + θ sin θ ] 4π r 3 k

− jkr ( ' µ e B ≈ φ 0 Id z sin θ 4π r 2

6-11 & The near zone E -field looks like the field of an electrostatic dipole. The near zone fields do not contribute to radiated power. Instead they result in reactive power (time changing energy stored in the fields near the antenna). Only the 1/r terms contribute to radiated power.

Far-zone fields (radiation zone fields)

The far-zone fields are those which are strongest as r → ∞ . Thus, these are the terms

& ) of E and B which vary as 1/r:

− jkr ( ' µ Id z e B ≈ φ 0 jk sin θ 4π r

− jkr + * jk Idz e E ≈ θ η sin θ 4π 0 r

The radiation zone fields are those that contribute to radiated power. Note that fields form a transverse wave.

Radiated Power - W = lim ∫ n, ⋅ P ds r →∞

- - - . 1 ∗ . P = R e{E × H } where n = r and d s = r 2 dΩ

/ 0 B has terms that go as 1/r and 1/r2 and E has terms that go as 1/r, 1/r2 and 1/r3.

Since d s ≈ r 2 as r → ∞ only the 1/r terms contribute to W.

- - 1 , W = lim ∫ Re{E r × H r ∗} ⋅ r r 2d Ω

r →∞ 2

1 0 0 1 where E r and H r are the 1/r terms from E and H called the “radiation zone” fields.

π 2 π 1 r r ∗ 2 W = ∫ ∫ E θ H φ r d Ω 0 0 2

6-12 π π 1  Id z  2 2 ωµ  k  = ∫ ∫ 0 sin θ sin θ r 2 sin θd θd φ  π     2 4 0 0 r r

π π 1  Idz  2 2 =   ωµ k ∫ dφ ∫ sin 3θd θ

2  4π  0 354

2 0 0

6 6 7 =2 π =4 3 1 I 2dz2  8π  µ = ωµ k  use: ωµ = ω µ ε 0 = kη π 2 0   0 0 0 ε 0 2 16 3 0

1 I 2dz2 2π = k 2η use: k = 12 π 0 λ

π  d z  2 = η I 2   Watts 0 3  λ 

 d z 2  W = 40π 2 I 2    λ2  Radiation resistance 1 π  d z  2 II ∗ R = W = η I 2   2 r 0 3  λ 

2π  d z  2  d z  2 or R = η   = 80π 2   r 0 3  λ   λ 

Example: Calculate the radiation resistance of a 1 cm length of uniform current if the frequency is 900 MHz and the host medium is air: c 3 × 108 m λ = = s = 33.3cm f 900 × 106 H z

2  1× 10−2 m  R = 80π 2   = 0.711Ω r  33.3× 10−2 m 

It is seen that the radiation resistance of a short current segment is only on the order of a fraction

6-13 of an Ohm, making it a relatively inefficient radiator.

Directive gain

π π K (θ ) 2 g (θ ) = W = K (θ )d Ω d π ∫ ∫ W / 4 0 0

1  Idz  2   [ωµ sin θ ][k sin θ ] 2  4π  0 4π g (θ ) = = sin 2 θ d  2    1 1  Idz   4  ( π ) 4     ωµ k (2π )   2   4π  2  4π  0  3   3

3 g (θ ) = sin 2 θ d 2 Directivity  π  D = m ax[g (θ )] = g   = 1.5 d d  2 

= = ( ) = G gain 10log10 1.5 1.76dB

6.5 Radiation from a cylindrical dipole

The cylindrical dipole antenna is one of the most commonly used antenna in the VHF/UHF frequency range.

The cylindrical dipole may be viewed as an open-ended transmission line which has been flared out. I(z)

R g l +

V g -

z=0 Open circuited transmission line

6-14 z=l I(z)

R g + +

V g V0 - -

z=0

z=-l Dipole antenna

Far zone fields

8 8 8 B (r8 ), E (r )

z=l

8 8

I (z ) R 9

9 r r ′ I0 + V _0 y

z=-l

It is noted that I (z = ± l ) = 0 (current at the tips of the antenna is zero).

6-15

The current density present on the dipole is given by <

:  ( )δ ( )δ ( ) ≤ ; : z I z x y for z l J (r ) =  

 0  ; elsew h ere sin k (l − z ) ....where I (z ) = I 0 sin kl

The radiation vector is

: :

> = : ( ) N (θ , φ) = ∫ J (r ′)e jk r ⋅r ′ d v ′

A ABA

@ C where r? ′ = z ′z r ⋅ r ′ = z ′(z ⋅ r ) = z ′ co sθ

< l : I z N (θ , φ) = 0 ∫ ∫ δ (x ′)δ (y ′)dx ′d y ′ ∫ sin k (l − z ′ )e jkz ′co sθd z ′ sin kl x y z ′=−l

I z< l = 0 ∫ sin k (l − z ′ )e jkz ′ co sθd z ′

sin kl z ′=−l < I z< l I z 0 = 0 ∫ sin k (l − z ′)e jkz ′co sθ dz ′ + 0 ∫ sin k (l + z ′)e jkz ′co sθ d z ′ sin kl 0 sin kl −l

Using the relationship e ax ∫ e ax sin (b x + c )d x = [a sin (b x + c ) − b co s(b x + c )] a 2 + b 2 gives

: ( ) < zI co s kl co sθ − co s kl N (θ , φ) = z 0 k sin kl sin 2 θ Let co s(kl co sθ ) − co s kl F (θ , kl ) = “radiation function” 0 sin θ

then < : ( ) 2 I F θ , kl <

N (θ , φ ) = 0 0 (r co sθ − θ sin θ )

D E

FIF H k sin kl sin θ FGF z>

6-16

− jkr

< < : µ : e E (r ) = − jω 0 [θ N + φ N ] 4π r θ φ

− jkr J jωµ I e < = 0 F (θ , kl )θ ...... far zone E -field 2πk r co s(kl co sθ ) − co s(kl ) where F (θ , kl ) =

sin (kl ) sin θ

: <

: − jkr < : × ( ) ωµ : r E r j I e H (r ) = = 0 0 F (θ , k l)φ η 2πk η r ωµ = η Use 0 k 0 then

− jkr < : η : j I e E (r ) = θ 0 0 F (θ , kl ) Radiation zone fields 2π r

− jkr

< :

: jI e H (r ) = φ 0 F (θ , kl ) produced by a dipole 2π r

Radiation pattern

η 2 (θ φ ) = 0  2 +  K , 2  N θ N φ  8 λ so 2 η 2 η 4 I K (θ ) = 0 N = 0 0 F 2 (θ , kl ) 8 λ2 θ 8 λ2 k 2

2 I = η 0 F 2 (θ , kl ) 0 8π 2

Often times K (θ , φ ) is plotted as opposed to K (θ , φ ) since K describes the pattern of the far-zone field: co s(kl co sθ ) − co s kl K (θ ) ≈ F (θ , kl ) = sin θ x 2 x 4 co s x = 1 − + + K 2 ! 4 !

6-17 Special case: kl << 1 → l << λ 1 1 1 − (kl co sθ )2 − 1 + (kl )2 F (θ , kl << 1) ≈ 2 2 sin θ 1 (kl )2 [1 − co s 2 θ ] ≈ 2 sin θ

1 2 ≈ (kl ) sin θ → Same pattern as Hertzian dipole. 2

 1  0.707  power point  2 

9 0 L θ = 90 L 1 broadside direction

b eamwidth = 90 L

Other cases:

6-18 90 1 90 2 120 60 120 60

0.5 1 150 30 150 30

180 0 180 0

210 330 210 330

240 300 240 300 270 270 2l= lam bda/2 2l= lam bda 90 1.3985 90 2.3376 120 60 120 60 0.69923 1.1688 150 30 150 30

180 0 180 0

210 330 210 330

240 300 240 300 270 270 2l= (3/2)lam bda 2l= 2lam bda Most often the length l=λ/4 (half-wave dipole) is used, since if it is a nearly resonant structure with its current maximum at the driving point (z=0). For a dipole with a non-zero wire radius, the length must be slightly shorter than λ/4 to produce resonance.

Note if we define the input impedance Z in as V input voltage = 0 Z in I 0 input current then resonance occurs when Z0 is purely real (just as in circuit theory).

Radiated power

2 π 2 π I 2 π π W = K (θ , φ )d Ω = η 0 d φ F 2 (θ , kl )sin θd θ ∫ ∫ 0 π 2 ∫ ∫ θ =0 φ=0 8 0 0 2 I π = η 0 F 2 (θ , kl )sin θd θ 0 π ∫ 4 0 This expression must be integrated numerically.

Special case #1: kl<<1 (Short dipole)  1  2 F 2 (θ , kl ) ≈  (kl ) sin θ   2 

6-19 2 π I 1 ≈ η 0 ( )2 3 θ θ η ≈ πΩ

W kl ∫ sMONin d 120

P Q 0 π P 0 4 4 0 4 3 2 I 1  2lπ  2 4 W ≈ (120π) 0   4π 4  λ  3 2 2  l  ≈ 40π 2 I   Watts 0  λ

 π  Special case #2:  kl =  (Half-wave dipole)  2   π   π   π  co s co sθ  − co s  co s co sθ   π   2   2   2  F θ ,  = =  2   π  sin θ sin   sin θ  2   π  2 2  θ  π co s  co s  I 0 2 W = η ∫ d θ

0 4π sin θ N

M 0

PIPRP PIPSP Q 1.22 by numerical integration thus = 2 W 36.6 I 0

Directivity

m ax {K (θ )} D = W 4π

Special case #1: kl<<1 (Short dipole)

2 2 2 I 0 I 0  1  K (θ ) = η F 2 (θ , kl ) ≈ η  (kl ) sin θ  0 8π 2 0 8π 2  2 

6-20 π 2   I 0 1 2 ⇒ K (θ ) = K θ =  = η (kl ) m ax  2  0 8π 2 4 so 2 I 1 η 0 (kl )2 0 8π 2 4 3 D = =  2  I 2 1  η 0 1 ( )2 4   0 kl  4π  4π 4 3 

= ( ) = (3 ) = G 10log10 D 10log10 2 1.76d B

π Special case #2: kl = (half-wave dipole) 2   π  π    2 2       π co s co s     I 0  2 2  K (θ ) = K θ =  = η m ax  2  0 8π 2   π    sin      2   2 I 0 = η 0 8π 2 2 I η 0 0 8π 2 2 D = = = 1.64 more directive than short dipole  2  I 1.22 1 η 0 ( )  0 1.22  4π  4π 

= = G 10log10 (1.64) 2.15dB

Radiation resistance 2W 2W R = = r I I ∗ 2 0 0 I 0

6-21 π = 60∫ F 2(θ,kl)sinθdθ 0

Special case #1: kl<<1 (short dipole)

π  1  2 R = 60∫ (kl ) sinθ sinθdθ r   0 2  l  2 = π 2   Ω 80  λ

= π Special case #2: kl 2 (half-wave dipole) 2 2 (36.6 I ) = 0 = Ω R r 2 73.2 I 0

Ω Note that a resonant half-wave dipole (Zin = Rin + jXin = 73.2 + j0) is matched quite well by a 75 coaxial cable transmission line.

Example  l  1 What input currents are needed to a short dipole of length   = and to a half-wave  λ  10 dipole to radiate a power of 1kW?

I 0

R r

 1  2 short dipole: R = 80π 2   = 7.9Ω → I = 15.9 A r  10 0 = Ω → = half-wave dipole: R r 73.2 I 0 5.23A

6.6 Radiation from a small loop antenna

6-22

U U U E ( U r ), B( r )

rT V

I U R

a y

rT ′

The fields associated with a small loop antenna can be shown to be

2 − jkr X W η ( ) W I k a e E (r ) = φ 0 sin θ

4 r

W X

W 2 − jkr X W × ( ) ( ) W r E r I k a e H (r ) = = −θ sin θ η 4 r

Radiated Power η  2 2  η 2 K (θ , φ ) = K (θ ) = 0 N + N = 0 N 8 λ2  θ φ  8 λ2 φ

η 2 = 0 I 2 k 2 (πa 2 ) sin 2 θ 8 λ2

2π using η = 120π λ = 0 k

π 15 4 K (θ , φ ) = I 2(ka ) sin 2 θ 4

6-23 Note that the radiation pattern of the small loop is identical to the radiation pattern of the Hertzian dipole. The small loop is the magnetic dipole analog of the electric (Hertzian) dipole

2π π π 15 4 W = ∫ ∫ K (θ )d Ω = ∫ d φ ∫ I 2(ka ) sin3 θd θ φ=0 θ=0 4

π 15π 2 = I 2(ka)4 ∫ sin3 θdθ

2 θ= Z

Y 0

[ [ \ 4 3

= 10π 2 I 2 (ka)4Watts

Radiation resistance

2W 4 R = = 20π2(ka ) Ω r I 2

a = → = π 2[ π ( )]4 = Ω example: λ 0.5 R r 20 2 0.05 1.92

Directivity

Since the radiation pattern of the small loop is the same as that of the Hertzian dipole, the directive gain, the directivity, and the gain are the same:

D = 3/2 G = 1.76dB

6.7 Currents above a perfectly conducting ground

Antennas are often placed above conducting surfaces, for purposes of measurements in the lab, or in practical situations as when an antenna is placed on a car roof. Often the earth itself is modeled as a perfect conductor (although this not always a good approximation because the conductivity of the earth is fairly low). Fields produced by currents by antennas above a ground plane can be calculated using the method of images.

Method of images

Consider a current carrying element above a perfectly conducting plane:

6-24

z V V I E , B = ?

] ]

p erfect conductor, E = B = 0 ^ The current element will produce an E -field which will induce currents to flow on the surface of the conductor. These currents will produce an additional “scattered field”. The total field must = obey the boundary condition E tangential 0 on the conductor surface at z = 0. We may replace the problem by an equivalent problem. The ground plane is removed and replaced by an “image” current.

Free space

Free space "image" current

Here the fields produced by both the current and its image will be identical to the fields produced by the current above the ground plane as long as the boundary condition on the total = field E tan 0 at z = 0 is obeyed.

Question: What image current will result in the boundary condition being satisfied?

Since any current distribution can be viewed as a superposition of Hertzian dipoles, we only need to identify the image of a Hertzian dipole.

The field due to a Hertzian dipole on the z-axis is given by

6-25

− jkr − jkr X W  η   η  X Id z e 2 2 Id z e 1 E = r  0 +  co sθ + θ  jωµ + 0 +  sin θ π ωε 2 π 0 ωε 2 4 r  r j 0 r  4 r  r j 0 r 

Case I: Horizontal dipole

I I θ θ

d r r

_ _ _ _

E , E i E , E i

E ri ri E r ri d i E θ E θ i θ θ i i

I i I i ` = = ` = − choose: d i d , I i I , I i I

= θ = θ thus: ri r , i

and: (E ) = −(E ) , (E θ ) = −(E θ )

ri tan r tan i tan tan

a a So, (E + E ) = 0 → then boundary condition is satisfied. i tan

6-26 Case II: Vertical dipole

θ θ

I I

d r r

b b b b

E , E i E , E i

E r E ri ri ri d i θ θ E θ E θ i i i

I i

I i c = = c = choose: d i d , I i I , I i I

= θ = π − θ → θ = − θ thus: ri r , i cos i cos

and: (E ) = −(E ) , (E θ ) = −(E θ )

ri tan r tan i tan tan

d d So, (E + E ) = 0 → then boundary condition is satisfied. i tan

Summary: 1. Horizontal currents image in opposite directions. 2. Vertical currents image in the same direction.

6-27

6.8 Monopole above a ground plane

e e free E , B = ? I space l

I 0

coaxial cable

The equivalent problem is a dipole.

l I

l = I i I

Thus the fields radiated by a monopole above a ground plane are identical to those of a dipole in free space. However, the radiated power, and thus the radiation resistance are half that of the dipole, since the monopole only radiates into half the space that the dipole does.

2 π 2 1 I 0 W = W = η F 2 (θ , kl )sin θd θ m o nopole dipole 0 π ∫ 2 4 0

π 1 2 R = R = 60 ∫ F 2 (θ,kl )sinθd θ rm on op loe rdip ole 2 0

6.9 Broadband antennas

Although dipole antennas possess many attractive characteristics for measurement of

6-28 radiated emissions, they are not ideal for gathering data over a wide range of frequencies. The radiated emissions range typically extends from 30 MHz to 16 GHz, and the length of a dipole must be physically adjusted to provide a length of ½ λ at each measurement frequency. A more practical technique is to employ broadband measurement antennas. A broadband antenna has the following characteristics: 1. The input/output impedance is fairly constant over the frequency band. 2. The antenna pattern is fairly constant over the frequency band.

Two types of broadband antennas will be examined: The biconical antenna, and the log-periodic antenna.

The biconical antenna is typically used in the frequency range of 30 MHz to 200 MHz.

The log-periodic antenna is typically used in the band from 200 MHz to 1 GHz.

Biconical Antennas θ An infinite biconical antenna consists of two cones of half angle h with a small gap at the feed point.

E θ

free θ h space

H φ

θ r

voltage source

θ h

6-29

f g

In the space surrounding the cones J = 0 . Symmetry suggests that the fields are H = φ H φ and

j i E = θ E θ . Maxwell’s equations can be solved to give the form of the field as

− H e jk 0 r H = 0 φ sin θ r and − k H e jk 0 r E = 0 0 = η H θ ωε θ 0 φ 0 sin r

Where H0 is constant. It is noted that these fields form transverse electromagnetic (TEM) waves (the electric and magnetic fields are orthogonal and transverse to the direction of propagation). Therefore, a unique voltage between two points on the cones may be defined. The voltage produced between two points on opposite cones that are both a distance ‘r’ from the feed point is θ

k k V (r ) = − ∫ E ⋅ d l θ =π −θ h

−  1  = 2η H e jk 0 r ln  c ot θ  0  2 h 

The current on the surface of the cones is given by

2 π ( ) I r = ∫ H φr sin θd φ φ=0

− = π jk 0 r 2 H 0 e and the input impedance is then V (r ) η  1  Z = = 0 ln  co t θ  in ( ) π  h  I r r =0 2

 1  = 120ln cot θ   2 h which is purely resistive.

6-30 Usually, the cone half-angle is chosen to provide a match to the feed line characteristic impedance. The total time average radiated power is given by

W = ∫ P ⋅ d s s

π π −θ 2 2 h E = θ 2 θ θ φ ∫ ∫ η r sin d d φ= θ =θ 2 0 h 0

θ h d θ = πη H 2 0 0 ∫ θ θ =0 sin

2  1  = 2πη H ln  c ot θ  0 0  2 n 

Radiation resistance is given by 2  1  4πη H ln  c ot θ  2W 0 0  2 h  R = = r ( ) 2 π 2 2 I 0 4 H 0 or  1  R = 120ln cot θ  r  2 h 

which is the same as the input impedance Z in.

i l

It is noted that the radiated fields are spherical waves with E in theθ direction and H in m the direction. For linearly polarized waves incident on the antenna from the broadside direction (θ = 90o ), the antenna responds to the field component that is parallel to its axis. Also the input impedance and pattern are theoretically constant over an infinite range of frequencies. Infinite length cones are obviously impossible to construct, therefore real biconical antennas consist of truncated cones. The finite length of the cones causes reflections as the waves travel outward along the cones. This produces standing waves that result in the input impedance having an imaginary (reactive) component, rather than being purely real.

Often wires are used to approximate the cone surfaces:

6-31 truncated biconical antenna composed of wire elements

Other variations:

cone

ground plane

Discone Bowtie

The fields of the discone antenna above the ground plane are the same as those of the biconical antenna by the method of images. The radiation resistance of the discone is ½ that of the biconical antenna. The bow-tie antenna consists of flat triangular plates or a wire which outlines the same area as the plates. The bow-tie antenna is frequently used for reception of the UHF television signals. Using wires instead of solid metal triangles tends to reduce the bandwidth of the bow-tie antenna.

Log-periodic antennas

The log-periodic antenna achieves a large operational bandwidth through repetitive dimensioning of structures. The structural dimensions increase in proportion to the distance from the origin of the structure. As a result, the input impedance and radiation properties repeat periodically and are functions of the logarithm of frequency. The log-periodic dipole array is a common log-periodic measurement antenna. This antenna shares the properties of all log-periodic antennas in that element distances, lengths, and separations are related by a constant such that l d R τ = n = n = n ln −1 d n −1 R n −1

6-32 ln −1 l l + n α n 1

R n d n

log-periodic array

The most efficient way of operating a log-periodic array is such that the currents on adjacent elements are reversed in phase.

criss-cross feed method

In this way the shorter elements will not interfere with elements to the right. The bandwidth of the log-periodic antenna is approximated by determining the frequency at which the shortest element is, and the frequency at which the longest element is ½ λ. At a particular frequency only the elements which are at or near a resonance are active. Thus, the active region of the antenna adjusts depending on frequency.

6-33 high frequency low frequency measurement measurement

emitter emitter

active region

6.10 Aperture antennas

Aperture antennas are characterized by an aperture or opening from which radiated fields are emitted. These include horns, slots and microstrip patch antennas. The operation of such antennas is best explained by Huygen’s principle:

Each point in an advancing wavefront acts as a source of spherical, secondary wavelets, that propagate outward.

Incident field

The secondary wavelets cause the wave to spread as it travels away from the aperture.

6-34 This is known as diffraction. The antenna pattern of an aperture antenna is actually a diffraction pattern. In the far zone of a simple aperture the magnitude of the electric field is given by

− jk r −r ′ n n jk e

( ) ≈ ( ) n E r E r ′ n d s′ π ∫ a − ′ 2 S r r where S is the aperture surface, E (r ′) is the magnitude of the electric field in the aperture and a o r ′ is a position vector that sweeps over points in the aperture.

At points far from the aperture p rp − r ′ ≈ r − z ′ co sθ Assuming that the field in the aperture is uniform and that the aperture width is small

   π a θ ∆ − jkr a / 2 − jkr sin  co s  jk xe ′ θ e λ E (r ,θ , φ) ≈ E e jkz co s d z ′ = jE ∆ × π ∫ a a π θ 2 r −a / 2 r co s

The size of the main lobe in the resulting pattern is inversely proportional to the aperture width ‘a’. Thus in order for the main lobes to be narrow, the aperture dimensions must be on the order of a wavelength or greater.

S θ x r No rm alized far-zone E-field vs. Observation angle 1

0. 9

r − r ′ 0. 8

0. 7 ) 0. 6 ed

a z i l ds ′ r − z ′ c os θ a

m 0. 5 r no (

| 0. 4 E |

0. 3

0. 2

∆x 0. 1

0 Radiation from a narrow aperture 0 20 40 60 80 100 120 140 160 180 Theta (degrees)

Horn antennas

6-35 Horn antennas are flared waveguides. There are three basic types of horn antennas.

E-plane horn - flare is in the plane that contains the Eo -field vector. q H-plane horn - flare is in the plane that contains the H -field vector. Pyramidal horn - flare is in both planes.

The aperture distribution of a horn is typically the same as the mode of the feeding waveguide, with a phase taper across the aperture.

E 13 r

2.57λ z

H 0.5λ Geometry of a typical horn antenna

The antenna as a receiving element

Any transmitting antenna can also be used for the purpose of “receiving” – intercepting a portion of the power radiated by some source. Instead of being driven by transmission line, a receiving antenna delivers power to a load connected at its terminals. Consider a spherical wave radiated by some distant source, and incident on a receiving antenna. Over the local region of the antenna, the spherical wave can be approximated as a plane wave. The plane wave induces a current in the antenna, which in turn produces an additional scattered field. But the induced current also causes a voltage to appear across the load impedance. This voltage then acts like a driving voltage causing additional currents to flow, just as in a transmitting antenna, which produce still another scattered field. Thus, by superposition, the total current flowing on the antenna may be viewed as that due to a scatterer interacting with a plane wave plus that of a transmitting antenna.

Transmitting/receiving equivalent circuits

6-36 V = Z I + Z I  1 1 1 1 1 2 2  = + network description of transmitting/receiving system V 2 Z 2 1 I 1 Z 2 2 I 2 

<< → Assume: Z 1 2 Z 1 1 for r large (little coupling from receiving to transmitting antenna)

≈

t 0 = + ≈ V 1 Z 11 I 1 Z 12 I 2 Z 11 I 1

V (Z ) = 1 ≈ Z 1 in 11 I 1

I linear + 1 s V1 medium E t - I2

I 2 Z antenna 1 L (transmitting) antenna 2 (receiving) r

I 1

Z 11 + Equivalent circuit for

V1 transmitting antenna -

= − = + V2 I2 Z L Z21I1 Z22 I2

− Z I I = 21 1 2 + Z 22 Z L

6-37 I 2

Z 22 + Equivalent circuit for

Z 21 I1 Z L receiving antenna -

≈ Note: When antenna 2 is in receiving, we can not assume Z 2 1 0 since this term describes the coupling effect between transmitter and receiver.

≈ Note: When antenna 2 is transmitting, then Z 21 0 so that

≈

u 0 = + ≈ V 2 Z 21 I 1 Z 22 I 2 Z 22 I 2

V (Z ) = 2 ≈ Z 2 in 22 I 2

Z ≈ (Z ) 22 2 in

Receiving/transmitting reciprocity

There are three basic reciprocity relations between receiving and transmitting antennas: 1) The antenna pattern for reception is identical to that for transmission. 2) The equivalent impedance in the receiving antenna equivalent circuit is identical to the input impedance of the antenna when it is transmitting. 3) The effective receiving cross-section area of an antenna is proportianal to its diective gain as a transmitting element.

We have already considered (2) above. The others require the use of the Lorentz reciprocity = theorem. From this theorem we can show that Z 12 Z 21 .

Relationship between gain and effective receiving cross-sectional area

W = r 2 Definition: Aer effective receiving cross-section area (m ). Pa v

where: W r = received power (power delivered to the load)

Pa v = average Poynting vector (power density) maintained by transmitting

6-38 antenna (Watts/m2)

Aer is a function of: 1) Load impedance 2) Aspect of antenna to oncoming wave 3) Polarization of oncoming wave

= Note: W r A er Pa v = total power intercepted by receiving antenna.

d W t d Ω Recall: g (θ , φ ) = directive gain of transmitting antenna, W = transmitted power. dt  W  t  t   4π 

4πr 2 P W g = av ⇒ P = t dt av π 2 W t 4 r

A g So: W = W er dt received power in terms of transmitted power. r t 4πr 2

Maximum power transfer relation: Assume load impedance is conjugate matched ⇒ maximum power transferred to load

I 2

Z 2 2 + = ∗ Z 21 I 1 Z L Z 2 2 -

receiving antenna

= ∗ = − Z L Z 22 R2 jX 2

2 2 1 2 1 − Z I Z I = = 21 1 = 21 1 W r I 2 R 2 R 2 2 2 2 R 2 8 R 2

6-39 I 1

Z 11 +

V1 -

transmitting antenna

= + Z11 R1 jX 1

1 2 W = I R t 2 1 1

2 Z 21 I 1 2 W 8 R Z r = 2 = 21 W 1 2 4 R R t I R 1 2 2 1 1 but: 2 W A g Z r = er dt = 21 π 2 W t 4 r 4 R1 R 2 so:

2 R1 R 2 A er g dt Z = 2 1 antenna 1 transmitting, antenna 2 receiving 21 πr 2

2 R 2 R 1 A er g dt Z = 1 2 antenna 1 receiving, antenna 2 transmitting 12 πr 2

= now use Z 1 2 Z 2 1 (Reciprocity of network)

g A = g A dt1 er2 dt 2 er1 or

Aer Aer 1 = 2 g g dt1 dt 2

6-40 Now since antennas 1 and 2 were arbitrary,

A er = constant g d t

For polarization matched conditions (receiving antenna oriented to intercept maximum amount of 4π power), the constant can be shown to be (Derivation is kind of messy). λ2 A λ2 e = Thus, π Universal relationship between gain of an antenna acting as a transmitter g d 4 and effective area of same antenna acting as a receiver.

 λ  2 So, W = W   D D “Friis” equation r t  4πr  r t

Reciprocity between transmission and reception patterns

Case 1) Antenna 1 transmitting, antenna 2 receiving

(2b)

(2a) (1)

Measure transmitting pattern of antenna 1 by varying θ.

W 2  1  r2 = Z   W 21  4 R R  t1 1 2

6-41 2 W Z 21 r2 b = b 2 W r Z 2 a 21 a

Case 2) Antenna 2 transmitting, antenna 1 receiving

(1b) θ (2) (1a)

Measure receiving pattern of antenna 1 by varying θ.

W 2  1  r1 = Z   12   W t 2 4 R1 R 2

2 W Z 1 2 r1 b = b 2 W r Z 1 a 1 2 a

W r W r By reciprocity: Z = Z ⇒ 2 b = 1 b power patterns are equal. 12 21 W W r2 a r1 a

Thus, the reception pattern of an antenna is identical to its transmission pattern.

References 1. Demarast, K. Engineering Electromagnetics, Prentice Hall Inc., 1998 2. Paul, C. Introduction to Electromagnetic Compatibility, John Wiley & Sons, 1992

6-42