Module 6:Antennas

6-1

6.0 Introduction

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

6.1 The Radiation Mechanism

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

D

d I

1

d I

2

R1

R 2

Here two elements of current and are separated by a distance D. The current elementsd I

1 d I

2

are located at distances R1 and R2, respectively from a distant observation point. If

R2-R1 0.1λ

D 0.1λ

Then the fields produced by the current elements add out of phase, and the amount of radiation issmall. However, if

R2-R1 0.1λ

D 0.1λ

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

-Reception Mechanism

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

6-2

an tenna

t ransmiss ionl ine

Z Ll oad

impedance

inc identfield

inducedcurrent

6.2 Radiated Power

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

In chapter 2 the time-average Poynting vector was presented

P E H= × ∗1

2R e

At points far from the antenna (the radiation zone)

( ) [ ] E r j

e

rN N

jkr

≈ − +−

ωµπ

θ φθ φ4

( ) ( )

H rr E r

≈×

ηwhere

( ) ( ) ( ) N J r e d vS

v

jk r rθ φ,

= ′ ′∫ ⋅ ′

is known as the “radiation vector.” The radiation vector is related to the vector potential by

( ) ( ) A r

e

rN

jkr

, , ,θ φµπ

θ φ=−

4with

6-3

( ) ( ) ( ) A r

e

rJ r e d v

jkr

Sv

jk r r≈ ′ ′−

⋅ ′∫µπ4

Now in the radiation zone

P Er E

≈ ××

∗1

2Re

η

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

( ) ( ) P E E r E r E≈ ⋅ − ⋅∗ ∗1

2ηRe

......because ≈⋅ ∗

rE E

2η

E r⋅ =

0

[ ]≈

+∗ ∗r

rN N N N

1

2 4

2

ηωµπ θ θ φ φ

......because [ ] [ ] [ ] θ φ θ φθ φ θ φ θ θ φ φN N N N N N N N+ ⋅ + = +∗ ∗ ∗ ∗

FinallyP r

rN N≈ +

1

82 2

2 2ηλ 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 n P dsr

s

= ⋅→∞ ∫l i m

Applying the expression for the time average Poynting vector leads to

W r rr

N N r d dr

= ⋅ +

→∞ ∫∫l i m

sin

1

82 2

2 22

0

2

0

ηλ

θ θ φθ φ

ππ

6-4

= +

∫∫

ηλ

θ θ φππ

θ φ8 20

2

0

2 2

N N d dsin

Let ......” radiation intensity”Kd W

dN N= = +

Ω

ηλ θ φ8 2

2 2

= power radiated per unit solid angle. ......where d d dΩ = sin .θ θ φ

The total radiated power is then

( )W K d= ∫∫ θ φππ

, .0

2

0

Ω

6.3 Antenna Terminology

Antenna Patterns

Radiation pattern - A plot of the radiation characteristics of an antenna. There are twotypes 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 themain lobe, while the other smaller lobes are called side lobes. The minima between lobes arecalled nulls.

mainlobe

side lobe

null

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

6-5

dimensional patterns, which are sometimes called cuts. For most antennas, two cuts give a goodrepresentation 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 theelectric field vector. The H-plane contains the direction of maximum radiation and the magneticfield vector.

E

H

H-plane

E-plane

No antenna has a truly isotropic pattern (one which is the same in all directions). Ratherantennas (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 foran isotropic radiator with the same total radiation power W.

( ) ( ) ( )gK

W WKd θ φ

θ φ

π

πθ φ,

,,=

=

4

4

.....where is the total power radiated by an isotropic radiator per unit solid angle.4πW

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 ofthe pattern.

6-6

φ

s ide lobe

3 dB (ha l f powe r )po in ts

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

EW

W W L

=+

......where E = radiation efficiencyW = power radiatedWL = power lost

Radiation resistance - The radiation resistance of an antenna is the equivalent resistance throughwhich its input current must flow in order that the power dissipated in the resistance is equal tothe total radiated power.

1

2 0 0I I R Wr∗ =

or ......radiation resistanceRW

I Ir = ∗

2

0 0

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

From the stand point of the source that drives an antenna, radiation resistance isindistinguishable from Ohmic resistance. In both cases, the source must continuously supplyenergy to the antenna in order to keep the current amplitude constant with time. In the case ofOhmic 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.

6-7

a

b

I in

Z in

AntennaV

When the antenna is not receiving power from waves generated by other sources theThevenin equivalent circuit looking into the terminals of the antenna consists only of animpedance

ZV

IR jXin

inin in= = +

where Rin is the input resistance and X in is the input reactance. The input resistance is the sum oftwo components

R R Rin ri L= +

where Rri is the input radiation resistance and RL is the input loss resistance. RL accounts for thatportion 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

.RI

IRri r=

max

0

2

Radiation efficiency can be expressed

.ηrra d

in

ri

ri L

P

P

R

R R= =

+

6.4 Hertzian Dipole

The simplest radiation source consists of a short segment of current

6-8

I

r

y

x

z

A B E, ,

−d z

2

d z

2

d z

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

Here

( ) ( ) ( ) J r

Iz x y fo rd z

zd z

e lse w h ere=

−≤ ≤

. . . . . .

. . . . . .

δ δ0

2 2

It is seen that

( ) ( )I J d s I x y d xd y Iyx

= ⋅ = =∫ ∫∫

δ δ

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

∇ ⋅ = − ⇒ = −J j

j

d J

d zzωρ ρ

ω1

The current density may be expressed

( ) ( )J Iz x y u z

d zu z

d z= +

− −

δ δ

2 2

where u(t) represents the unit step function.

6-9

z

J

−d z

2

d z

2

( ) ( )ρω

δ δ= +

− −

1

2 2jI x y

d

d yu z

d zu z

d z

......because ( ) ( )= − +

+ −

I

jx y z

d zz

d z

ωδ δ δ δ

2 2

( ) ( )d u t

d tt= δ

= (+) point charge at , (-) point charge at zd z

=2

zd z

= −2

Vector potential

( ) ( ) A r J r

e

Rdv

v

jkR

= ′ ′∫−µ

π0

4where

for R r r r r= − ′ ≈ =

r d z>>so,

( ) ( ) A r

e

rJ r d v

jkr

v

≈ ′ ′−

∫µπ0

4

( ) ( )≈ ′ ′ ′ ′ ′−

′=−

∫∫ ∫µπ

δ δ0

2

2

4

e

rx y d x d y Iz d z

jkr

yxz

dz

dz

or

( ) A r z Idz

e

r

jkr

≈− µ

π0

4

6-10

now use to get

co s

sinz r= −θ θ θ

( ) ( ) A r r Id z

e

r

jkr

≈ −−

c os

s inθ θ θµπ0

4

E-M fields

( ) B A

r rrA

A r= ∇ × = −

φ ∂

∂∂∂θθ

( )= − −

−

−µπ

φ ∂∂

θ∂

∂θθ0

4Idz

r re

e

rjkr

jkr

sin co s

= +

−µπ

φθ

θ0

4Id z

rjk

re jkr

s insin

soB Id z

jkr

re jkr=

+ −

s inφµπ

θ024

1

at all points where

Ej

B= − ∇ ×ωµ ε0 0

J = 0

( ) ( )= − −

j r

rB

r rrB

ωµ ε θ∂

∂θθ

θ ∂∂φ φ

0 0

!

sinsin

!

= −+

−

+

− −j Id z r

r

jkr

re

r r

jkr

rejkr jkr

ωε π θ∂

∂θθ

θ ∂∂

θ0

22

4

1 1!

sinsin

!

sin

[ ]

"E

j Idz r

r

jkr

re

rj

k

rjkr

jkr jkr

re

jkr

jkr

= −+

− − + +− −

−

−

ωε π θθ θ

θθ

02

2

4

12

11

#

sinsin cos

#

sin

6-11

= −+

++ −

−j Id z

rjkr

r

jkr k r

re jkr

ωε πθ θ θ

03

2 2

34

12

1!co s

!sin

now use

.....where 1 1

0 0 0

0

0

0

ωε ω µ εµε

η= =

kη

µε0

0

0

=

Then

.$

E rId z e

r r j r

Id z e

rj

r j r

jkr jkr

= +

+ + +

− −!co s

!s in

4

2 2

4

10

02 0

0

02π

ηωε

θ θπ

ωµη

ωεθ

Thus the fields may be expressed:$B

Id z jk

r re jkr= +

−!

s inφµ

πθ0

24

1

EIdz

r

j

krer

jkr= −

−

42

10 2 3π

η θco s

EId z jk

r r

j

kre jkr

θ πη θ= + −

−

4

10 2 3sin

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)

The near zone fields are those which are strongest near , or when . Thusr = 0 r <<λπ2

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

&E

[ ]'E j

Id z e

r kr

jkr

≈ − +−

423

0

πη

θ θ θ(

co s(

s in

'B Id z

e

r

jkr

≈−(

s inφµπ

θ024

6-12

The near zone -field looks like the field of an electrostatic dipole.&

EThe 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 contributeto radiated power.

Far-zone fields (radiation zone fields)

The far-zone fields are those which are strongest as . Thus, these are the termsr → ∞of and which vary as 1/r:&

E)B

'B

Id zjk

e

r

jkr

≈−(

s inφµ

πθ0

4

*E

jk Id z e

r

jkr

≈−+

sinθπ

η θ4 0

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

Radiated Power

W n P dsr

s

= ⋅→∞ ∫lim

, -

where and - - -P E H= × ∗1

2Re

. .n r= d s r d= 2 Ω

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

0E

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

W E H r r dr

r r= × ⋅→∞

∗∫l i m Re,1

22- -

Ω

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

E r0

H r0

E1

H

W E H r dr r= ∫∫ ∗1

20

2

0

2

ππ

θ φ Ω

6-13

=

∫∫

1

2 4

20

0

2

0

2Id z

r

k

rr d d

πωµ

θ θ θ θ φππ

sin sin sin

2=

∫ ∫

= =

1

2 4

2

00

2

2

3

0

43

Id zk d d

πωµ φ θ θ

π

π

π

sin3546 76

use: =

1

2 1 6

8

3

2 2

2 0

I d zk

πωµ

πωµ ω µ ε

µε

η0 0 00

00= = k

use: =1

1 2

2 22

0

I d zk

πη k =

2πλ

Watts=

η

πλ0

2

2

3I

d z

W Id z

=

4 0 2 2

2

2πλ

Radiation resistance

1

2 302

2

I I R W Id z

r∗ = =

η

πλ

or Rd z d z

r =

=

η

πλ

πλ0

2

2

22

38 0

Example: Calculate the radiation resistance of a 1 cm length of uniform current if the frequency is900 MHz and the host medium is air:

λ = =×

×=

c

f

ms

H zcm

3 1 0

9 0 0 1 03 3 3

8

6 .

Rm

mr =×

×

=

−

−8 01 1 0

3 3 3 1 00 7 1 12

2

2

2

π.

. Ω

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

6-14

of an Ohm, making it a relatively inefficient radiator.

Directive gain

( ) ( )g

K

Wd θθ

π=

/ 4( )W K d= ∫∫ θ

ππ

0

2

0

Ω

( )[ ][ ]

( ) ( )g

Idzk

Idzk

d θ πωµ θ θ

π π ωµ π

π

πθ=

=

12 4

14

12 4

243

4

243

2

0

2

0

2

sin sinsin

( )g d θ θ=3

22sin

Directivity

( )[ ]D g gd d= =

=m ax .θ

π2

1 5

( )G g a in d B= = =1 0 1 5 1 7 61 0lo g . .

6.5 Radiation from a cylindrical dipole

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

The cylindrical dipole may be viewed as an open-ended transmission line which has been flaredout.

V g

+

-

R g

z=0

I(z)

l

Open circuited transmission line

6-15

z = - l

V g

+

-

R g

z = 0

I(z)

z = l

V0

+

-

Dipole antenna

Far zone fields

z = - l

z = l

y

I 0

z

+

_V 0

( )8I z

8R

9r9

′r

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

x

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

6-16

The current density present on the dipole is given by

( ) ( ) ( ) ( ): : ;;

J rz I z x y for z l

e lsew h ere=

≤<

δ δ0

....where ( ) ( )I z I

k l z

k l=

−0

si n

si nThe radiation vector is

( ) ( ) ( ): : : =N J r e d v

v

jk r rθ φ,>

= ′ ′∫ ⋅ ′

where ?

′ = ′r z z@ ( )A ABA

cosr r z z r z⋅ ′ = ′ ⋅ = ′C

θ

( ) ( ) ( ) ( ):

NI z

k lx y dx d y k l z e d z

yx

jkz

z l

l

θ φ δ δ θ,

<

si nsi n cos= ′ ′ ′ ′ − ′ ′∫∫ ∫ ′

′=−

0

( )= − ′ ′′=−

′∫I z

k lk l z e d z

z l

l

jkz0

<

sinsin cosθ

( ) ( )= − ′ ′ + + ′ ′∫ ∫′

−

′I z

k lk l z e dz

I z

k lk l z e d z

l

jkz

l

jkz0

0

00<

sinsin

<

sinsincos cosθ θ

Using the relationship

( ) ( ) ( )[ ]e b x c d xe

a ba b x c b b x cax

ax

∫ + =+

+ − +sin sin cos2 2

gives

( ) ( ):N z

zI

k k l

k l k lθ φ

θθ

,<

si n

cos cos cos

si n=

−02

Let

“ radiation function”( ) ( )F kl

k l k l0 θ

θθ

,cos cos cos

sin=

−

then

( ) ( )( ):D EFGF HFIFN

I

k kl

F klr

z

θ φθ

θθ θ θ,

si n

,

si n

<cos

<si n

>= −

2 0 0

6-17

( ) [ ]: :E r j

e

rN N

jkr

= − +−

ωµπ

θ φθ φ0

4

< <

......far zone -field( )=−j I

k

e

rF kl

jkrωµπ

θ θ0

2,< J

E

where ( ) ( ) ( )( )F kl

k l k l

k lθ

θθ

,cos cos cos

sin sin=

−

( ) ( ) ( ): :

: :H r

r E r j I

k

e

rF k l

jkr

=×

=−<

,<

ηωµπ η

θ φ0 0

2Use thenωµ η0 0= k

Radiation zone fields( ) ( ): :

E rj I e

rF kl

jkr

=−<

,θη

πθ0 0

2

produced by a dipole( ) ( ): :

H rjI e

rF kl

jkr

=−<

,φπ

θ0

2

Radiation pattern

( )K N Nθ φηλ θ φ, = +

02

2 2

8so

( ) ( )K NI

kF klθ

ηλ

ηλ

θθ= =02

2 02

0

2

22

8 8

4,

( )= ηπ

θ0

0

2

22

8

IF kl,

Often times is plotted as opposed to since describes the pattern of the( )K θ φ, ( )K θ φ, K

far-zone field:

( ) ( ) ( )K F kl

k l k lθ θ

θθ

≈ =−

,co s co s co s

sin

co s! !

xx x

= − + +12 4

2 4

K

6-18

Special case: k l l<< → <<1 λ

( )( ) ( )

F klk l k l

θθ

θ,

co s

sin<< ≈

− − +1

112

112

2 2

( ) [ ]≈

−12

12 2k l co s

sin

θ

θ

Same pattern as Hertzian dipole.( )≈ →1

22k l sin θ

9 0 L

0 .707 1

2 pow er poin t

z

1

θ = 90 Lbroadsidedirection

b eam w id th = 90 L

Other cases:

6-19

2l= lambda/2

0.5

1

30

210

60

240

90

270

120

300

150

330

180 0

2l= lambda

1

2

30

210

60

240

90

270

120

300

150

330

180 0

2l= (3/2)lam bda

0.69923

1.3985

30

210

60

240

90

270

120

300

150

330

180 0

2l= 2lam bda

1.1688

2.3376

30

210

60

240

90

270

120

300

150

330

180 0

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

Note if we define the input impedance asZ in

ZV

Iin = 0

0

in p u t vo lta g e

in p u t cu rren tthen resonance occurs when Z0 is purely real (just as in circuit theory).

Radiated power

( ) ( )W K dI

d F kl d= ===∫∫ ∫∫θ φ η

πφ θ θ θ

φ

π

θ

π ππ

, , sin0

2

00

0

2

22

00

2

8Ω

( )= ∫ηπ

θ θ θπ

00

2

2

04

IF kl d, sin

This expression must be integrated numerically.

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

( ) ( )F kl k l2

21

2θ θ, sin≈

6-20

( )WI

kl d≈ ∫ηπ

θ θπ

00

2

2 3

43

04

1

4sinMONP QP η π0 1 2 0≈ Ω

( )WI l

≈

1 2 0

4

1

4

2 4

30

2 2

ππ

πλ

Watts≈

4 0 2

0

22

πλ

Il

Special case #2: (Half-wave dipole)k l =

π2

F θπ

πθ

π

πθ

πθ

θ,

co s co s co s

sin sin

co s co s

sin22 2

2

2

=

−

=

WI

d=

∫ηπ

πθ

θθ

π

00

2 2

0

1 2 2

42

co s co s

sin. b y nu m er ica l integ ra tio n

M NPIPRP QPIPSP

thus

W I= 3 6 6 0

2.

Directivity

( )

DK

W=m ax θ

π4

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

( ) ( ) ( )KI

F klI

k lθ ηπ

θ ηπ

θ= ≈

0

0

2

22

0

0

2

2

2

8 8

1

2, sin

6-21

( ) ( )⇒ = =

=K K

Iklθ θ

πη

πm ax 2 8

1

40

0

2

22

so

( )

( )D

Ikl

Ik l

=

=η

π

πη

π

00

2

22

00

2

2

814

14 4

14

43

3

2

( ) ( )G D d B= = =1 0 1 0 32 1 7 61 0 1 0lo g lo g .

Special case #2: (half-wave dipole)k l =π2

( )K KI

θ θπ

ηπ

π π

πm ax

co s co s

sin= =

=

2 8

2 2

2

00

2

2

2

= ηπ0

0

2

28

I

more directive than short dipole

( )D

I

I=

= =η

π

πη

π

00

2

2

00

28

14 4

1 2 2

2

1 2 21 6 4

..

.

G d B= =1 0 1 6 4 2 1 51 0lo g ( . ) .

Radiation resistance

RW

I I

W

Ir = =∗

2 2

0 0 0

2

6-22

( )= ∫6 0 2

0

F kl dθ θ θπ

, sin

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

( )R kl dr =

∫6 0

1

2

2

0

sin sinθ θ θπ

=

8 0 2

2

πλl

Ω

Special case #2: (half-wave dipole)k l = π2

( )R

I

Ir = =

2 3 6 67 3 2

0

2

0

2

.. Ω

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

Example

What input currents are needed to a short dipole of length and to a half-wavel

λ

=

1

1 0dipole to radiate a power of 1kW?

R r

I 0

short dipole: R I Ar =

= → =8 0

1

1 07 9 1 5 92

2

0π . .Ω

half-wave dipole: R I Ar = → =7 3 2 5 2 30. .Ω

6.6 Radiation from a small loop antenna

6-23

x

y

z

a

T′r

Tr

UR

( ) ( )U U U UE r , B r

VI

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

( ) ( )W WE r

I k a e

r

jkr

=−X

si nφη

θ02

4

( ) ( ) ( )W WW W

H rr E r I k a e

r

jkr

=×

= −−X X

si nη

θ θ2

4

Radiated Power

( ) ( )K K N N Nθ φ θηλ

ηλθ φ φ, = = +

=0

2

2 202

2

8 8

( )=ηλ

π θ02

2 2 2 2 2

8I k a sin

using η π λπ

0 1202

= =k

( ) ( )K I kaθ φπ

θ, sin=15

42 4 2

6-24

Note that the radiation pattern of the small l oop is identical to the radiation pattern of theHertzian dipole. The small loop is the magnetic dipole analog of the electric (Hertzian) dipole

( ) ( )W K d d I ka d= ===∫∫∫∫ θ φ

πθ θ

θ

π

φ

π

Ω15

400

2

2 4 3sin

( )==∫

15

2

22 4 3

0

43

πθ θ

θ

π

I ka dsinY Z[ \[

( )= 10 2 2 4π I ka W a tts

Radiation resistance

( )RW

Ikar = =

2202

2 4π Ω

example: ( )[ ]a R rλ π π= → = =0 5 20 2 0 05 1 922 4. . . Ω

Directivity

Since the radiation pattern of the small loop is the same as that of the Hertzian dipole, thedirective 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 thelab, or in practical situations as when an antenna is placed on a car roof. Often the earth itself ismodeled as a perfect conductor (although this not always a good approximation because theconductivity of the earth is fairly low).

Fields produced by currents by antennas above a ground plane can be calculated using themethod of images.

Method of images

Consider a current carrying element above a perfectly conducting plane:

6-25

x

z

V VE, B = ?

I

perf ect conductor , E = B = 0] ]

The current element will produce an -field which will i nduce currents to flow on the surface of^

Ethe conductor. These currents will produce an additional “ scattered field”. The total field mustobey the boundary condition on the conductor surface at z = 0.E tangenti al = 0

We may replace the problem by an equivalent problem. The ground plane is removed andreplaced by an “image” current.

z

x

Free space

Free space" image" current

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

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 needto identify the image of a Hertzian dipole.

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

6-26

W

E rId z e

r r j r

Id z e

rj

r j r

jkr jkr

= +

+ + +

− −Xcos

Xsi n

4

2 2

4

10

02 0

0

02π

ηωε

θ θπ

ωµη

ωεθ

Case I: Horizontal dipole

θ

I

θi

I i

r

r i

_ _E , E i

E θ Eiθ

θ

I

θi

I i

r

ri

_ _E , E i

E riE rd i

d

choose: , ,d di = I Ii =` `I Ii = −

thus: ,r ri = θ θi =

and: ,( ) ( )E Eri rtan tan= − ( ) ( )E Eiθ θtan tan

= −

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

tan0

6-27

Case II: Vertical dipole

I

I i

r

ri

b bE , E i

E θ Eiθ

θ

I

θi

I i

r

ri

b bE , E i

E r i

E r

d i

d

θi

θ

choose: , ,d di = I Ii =c cI Ii =

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

and: ,( ) ( )E Eri rtan tan= − ( ) ( )E Eiθ θtan tan

= −

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

tan0

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

6-28

6.8 Monopole above a ground plane

coaxia lcable

e eE, B = ?

I 0

I

l

f reespace

The equivalent problem is a dipole.

l

l I Ii =

I

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

( )W WI

F kl dm o nop ole dip ole= = ∫1

2 400

2

2

0

2

ηπ

θ θ θπ

, si n

( )R R F kl dr rm on op loe dip ole= = ∫

1

260 2

0

2

θ θ θπ

, si n

6.9 Broadband antennas

Although dipole antennas possess many attractive characteristics for measurement of

6-29

radiated emissions, they are not ideal for gathering data over a wide range of frequencies. Theradiated emissions range typically extends from 30 MHz to 16 GHz, and the length of a dipolemust 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 broadbandantenna 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-periodicantenna.

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

An infinite biconical antenna consists of two cones of half angle with a small gap at theθh

feed point.

z

E θ

r

f rees p a c e

θh

θh

H φ

θ

vo l tagesou rce

6-30

In the space surrounding the cones . Symmetry suggests that the fields are andfJ = 0

gH H=

hφ φ

. Maxwell’s equations can be solved to give the form of the field asi

E E=j

θ θ

HH e

r

jk r

φ θ=

−0

0

si nand

Ek H e

rH

jk r

θ φωε θη= =

−0

0

00

0

si n

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, aunique 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

( )V r E dlh

h

= − ⋅= −∫k k

θ π θ

θ

=

−21

200η θH e jk r

hl n cot

The current on the surface of the cones is given by

( )I r H r d==∫ φ

φ

π

θ φsi n0

2

= −2 00πH e jk r

and the input impedance is then

( )( )Z

V r

I rin

r

h= =

=0

0 1

2

ηπ

θl n cot

=

120

1

2l n cot θh

which is purely resistive.

6-31

Usually, the cone half-angle is chosen to provide a match to the feed line characteristicimpedance.

The total time average radiated power is given by

W P d ss

= ⋅∫

==

−

=∫∫

Er d d

h

h

θ

θ θ

π θ

φ

π

ηθ θ φ

2

00

22

2si n

==∫πη

θθθ

θ

0 02

0

Hdh

si n

=

2

1

20 0

2πη θH nl n cot

Radiation resistance is given by

( )

RW

I

H

Hr

h

= =

2

0

412

42

0 0

2

20

2

πη θ

π

l n cot

or

R r h=

120

1

2l n cot θ

which is the same as the input impedance Z in.

It is noted that the radiated fields are spherical waves with in the direction and ini

E θl

Hthe m

direction. For linearly polarized waves incident on the antenna from the broadside

direction ( ), the antenna responds to the field component that is parallel to its axis. Alsoθ = 90o

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 wavestravel outward along the cones. This produces standing waves that result in the input impedancehaving an imaginary (reactive) component, rather than being purely real.

Often wires are used to approximate the cone surfaces:

6-32

truncated biconical antenna composed of wire elements

Other variations:

groundplane

cone

Discone Bowt ie

The fields of the discone antenna above the ground plane are the same as those of thebiconical antenna by the method of images. The radiation resistance of the discone is ½ that ofthe biconical antenna.

The bow-tie antenna consists of f lat triangular plates or a wire which outlines the samearea as the plates. The bow-tie antenna is frequently used for reception of the UHF televisionsignals. Using wires instead of solid metal triangles tends to reduce the bandwidth of the bow-tieantenna.

Log-periodic antennas

The log-periodic antenna achieves a large operational bandwidth through repetitivedimensioning of structures. The structural dimensions increase in proportion to the distance fromthe origin of the structure. As a result, the input impedance and radiation properties repeatperiodically and are functions of the logarithm of frequency.

The log-periodic dipole array is a common log-periodic measurement antenna. Thisantenna shares the properties of all log-periodic antennas in that element distances, lengths, andseparations are related by a constant such that

τ = = =− − −

l

l

d

d

R

Rn

n

n

n

n

n1 1 1

6-33

ln 1+

ln 1−ln

α

R nd n

log-per iodic array

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

cr iss-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.

6-34

emit ter emit ter

h igh f requencymeasurement

low f requencymeasurement

act ivereg ion

act ivereg ion

6.10 Aperture antennas

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

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

Inc identfield

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

6-35

This is known as diffraction. The antenna pattern of an aperture antenna is actually a diff ractionpattern.

In the far zone of a simple aperture the magnitude of the electric field is given by

( ) ( )E rjk

E re

r rd sa

jk r r

S

n nn n≈ ′

− ′′

− − ′

∫2π

where S is the aperture surface, is the magnitude of the electric field in the aperture and( )E ra o ′ is a position vector that sweeps over points in the aperture.′r

At points far from the aperture

p pr r r z− ′ ≈ − ′ cosθ

Assuming that the field in the aperture is uniform and that the aperture width is small

( )E rjk xe

rE e d z jE

e

r

ajkr

ajk z

a

a

a

jkr

, ,si n cos

coscos

/

/

θ φπ

πλ

θ

π θθ≈ ′ = ×

−

′

−

−

∫∆

∆2 2

2

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

ds ′

z

∆x

a

xSθ

r r− ′

r

r z− ′ cosθ

Rad ia t i on f rom a na r row aper tu re 0 20 40 60 80 100 120 140 160 1800

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

No rm aliz ed far-z on e E -field vs . O bs ervat io n a ngle

The ta (d egre es )

|E|

(no

rma

lize

d)

Horn antennas

6-36

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

E-plane horn - flare is in the plane that contains the -field vector.oEH-plane horn - flare is in the plane that contains the -field vector.

qH

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.

z

0 5. λ

2 57. λ

13 r E

H

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 aportion of the power radiated by some source. Instead of being driven by transmission line, areceiving antenna delivers power to a load connected at its terminals.

Consider a spherical wave radiated by some distant source, and incident on a receivingantenna. Over the local region of the antenna, the spherical wave can be approximated as a planewave. The plane wave induces a current in the antenna, which in turn produces an additionalscattered field. But the induced current also causes a voltage to appear across the loadimpedance. This voltage then acts like a driving voltage causing additional currents to flow, justas 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 dueto a scatterer interacting with a plane wave plus that of a transmitting antenna.

Transmitting/receiving equivalent circuits

6-37

+-

V1

I 1

antenna 1(transmit t ing)

I 2

I 2 Z L

r

antenna 2(receiv ing)

sE t

l inearm e d i u m

network description of transmitting/receiving systemV Z I Z I

V Z I Z I1 11 1 12 2

2 21 1 22 2

= += +

Assume: for large (little coupling from receiving to transmitting antenna)Z Z12 11<< r →

tV Z I Z I Z I1 11 1 12 2

0

11 1= + ≈≈

( )ZV

IZ

in11

111= ≈

Z 1 1

+

-V1

I 1

Equivalent c i rcui t fort ransmit t ing antenna

V I Z Z I Z IL2 2 21 1 22 2= − = +

IZ I

Z Z L2

21 1

22

=−

+

6-38

Z 2 2

+

-Z I2 1 1

I 2

Equivalent c i rcui t forreceiv ing antennaZ L

Note: When antenna 2 is in receiving, we can not assume since this term describes theZ 21 0≈coupling effect between transmitter and receiver.

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

uV Z I Z I Z I2 21 1

0

22 2 22 2= + ≈≈

( )ZV

IZ

in22

222= ≈

( )Z Zin22 2≈

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 reciprocitytheorem. From this theorem we can show that .Z Z12 21=

Relationship between gain and effective receiving cross-sectional area

Definition: effective receiving cross-section area (m2).AW

Perr

a v

=

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

= average Poynting vector (power density) maintained by transmittingPa v

6-39

antenna (Watts/m2)

is a function of: 1) Load impedanceAer

2) Aspect of antenna to oncoming wave3) Polarization of oncoming wave

Note: = total power intercepted by receiving antenna.W A Pr er av=

Recall: directive gain of transmitting antenna, = transmitted power.( )g

d W

dWdt

t

t

θ φ

π

, =

Ω

4

W t

= ⇒ =4

4

2

2

ππ

r P

WP

W g

rav

tav

t d t

So: received power in terms of transmitted power. W WA g

rr ter dt=

4 2π

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

Z 22

+

-Z I2 1 1

I 2

Z ZL = ∗2 2

receiv ing antenna

Z Z R jXL = = −∗22 2 2

W I RZ I

RR

Z I

Rr = =−

=1

2

1

2 2 82

2

221 1

2

2

221 1

2

2

6-40

Z 1 1

+

-V1

I 1

t ransmit t ing antenna

Z R jX11 1 1= +

W I Rt =1

2 1

2

1

W

W

Z I

R

I R

Z

R Rr

t

= =

21 1

2

2

1

2

1

21

2

1 2

812

4

but:

W

W

A g

r

Z

R Rr

t

er dt= =4 42

21

2

1 2πso:

antenna 1 transmitting, antenna 2 receivingZR R A g

rer dt

21

2 1 2

22 1=

π

antenna 1 receiving, antenna 2 transmittingZR R A g

rer dt

12

2 2 1

21 2=

π

now use (Reciprocity of network)Z Z12 21=

g A g Adt er dt er1 2 2 1=

or

A

g

A

ger

dt

er

dt

1

1

2

2

=

6-41

Now since antennas 1 and 2 were arbitrary,

constantA

ger

d t

=

For polarization matched conditions (receiving antenna oriented to intercept maximum amount of

power), the constant can be shown to be (Derivation is kind of messy).4

2

πλ

Thus, Universal relationship between gain of an antenna acting as a transmitterA

ge

d

=λπ

2

4and effective area of same antenna acting as a receiver.

So, “Friis” equationW Wr

D Dr t r t=

λπ4

2

Reciprocity between transmission and reception patterns

Case 1) Antenna 1 transmitting, antenna 2 receiving

θ

(1)

(2a)

(2b)

Measure transmitting pattern of antenna 1 by varying θ.

W

WZ

R Rr

t

2

1

21

2

1 2

1

4=

6-42

W

W

Z

Z

r

r

b

a

b

a

2

2

21

2

21

2=

Case 2) Antenna 2 transmitting, antenna 1 receiving

θ

(1a)

(2)(1b)

Measure receiving pattern of antenna 1 by varying θ.

W

WZ

R Rr

t

1

212

2

1 2

1

4=

W

W

Z

Z

r

r

b

a

b

a

1

1

12

2

12

2=

By reciprocity: power patterns are equal.Z ZW

W

W

Wr

r

r

r

b

a

b

a

12 212

2

1

1

= ⇒ =

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

References1. Demarast, K. Engineering Electromagnetics, Prentice Hall Inc., 19982. Paul, C. Introduction to Electromagnetic Compatibili ty, John Wiley & Sons, 1992