1

Aperture and Horn AntennasAperture antennas are very common at microwave frequencies. They may take the form of the opening of a waveguide, a horn or a reflector.This chapter will introduce the tools necessary to analyze the radiation characteristics of aperture antennas.Examples of practical configurations (rectangular and circular geometries for aperture and horns) will be developed in detail.

Contents1. Introduction

2. Field equivalence principle

3. Aperture antennasRectangular and circular apertures

4. Babinet’s principle

5. Horn antennasSectoral horns, pyramidal horns, conical horns

2

1. Introduction

• Aperture antennas are most common at microwave frequencies where the transmission media are waveguides.

• Aperture antennas collect radiation through an area (“aperture”)

• They can be flush-mounted on the surface of spacecrafts or aircrafts. The opening can be covered with dielectric material.

• Analysis• Field equivalence principle• Simplifications (far-field, simple shapes) lens corrected conical horns

3

Aperture antennas can take different forms• Aperture in a ground plane• Open end of a waveguide• Horn antennas• Reflector antennas

Pyramidalhorn

Conicalhorn

Rectangularwaveguide

Slot antenna

Reflector antenna

4

2. Field equivalence principle

The field equivalence is a more rigorous (vectorial) way of formulating Huygens’ principle.

propagation

Hyugens’wavelets

secondarysourceHuygens’ principle

Each point on a primary wavefront can be considered to be a new source of a secondary spherical wave.

Plane wave

Diffraction pattern

Aperture edge

The field equivalence principle is based on the uniqueness theorem.

5

Uniqueness theorem

The fields in a region are uniquely specified by1) The sources within the region2) The tangential components of the electric fields over the boundary

or the tangential components of the magnetic fields over the boundary(or a mix of both).

If the tangential electric and/or magnetic fields are completely known over a closed surface S, the field in a source-free region can be determined.

S

,E H

tanE

tanH

JM

6

S

1J1M

Consider sources within a closed surface S :

Equivalence principle - Equivalent sources

ˆ

ˆ

S

S

J n H

M n E

= ´

= - ´

1 1,J M

n : normal unit vector

tanE

tanH

Original problem with actual sources

,E H

Equivalent problem

SSM

SJ

,E H

If the tangential fields are known over the surface S enclosing the sources, an equivalent problem can be defined by equivalent sources over S :

n

0

0

E

H

=

=

7

Remarks:

1) The two problems are only equivalent in the source-free region outside S,i.e., in the region of interest.The fields inside S are set to zero in the equivalent problem.

2) If the surface S is chosen judiciously, so that the tangential E and/or H fields are known a priori, the problem can be greatly simplified.

3) For aperture antennas, the “closed surface” is often chosen as an infinite flat plane.

4) So far, the tangential componentsof both E and H have been used todefine the equivalent problem.It will be shown that an equivalent problem can be found that requiresonly the tangential E or only thetangential H fields.

SsM

SJ

,E H

0

0

E

H

=

=

8

3. Aperture antennas

Aperture antennas can be analyzed using the field equivalence principle.

The imaginary closed surface is chosen as a flat plane extending to infinity.

These current densities will produce the same fields outside of the aperture (z > 0) as the original fields in the aperture.

H

0z =

E n

zIf the tangential fields are known (exactly or approximately) over the entire plane, an equivalent problem can be defined using the current densities

Problem: Fields are usually known only in the aperture.What about the fields outside of the aperture?

ˆ

ˆ

S

S

J n H

M n E

= ´

= - ´

9

Equivalent models near perfect conductors

The chosen closed surface often coincides with a perfectly conducting surface.

Then, using image theory, we find alternative equivalent models:

Example: Equivalent models near a PEC surface

0

ˆ2

S

S

J

M n E

ìï =ïïíï = - ⋅ ´ïïî

Perfect electricconductor (PEC)

ˆ2

0

S

S

J n H

M

ìï = ⋅ ´ïïíï =ïïî

Perfect magneticconductor (PMC)

PEC

ˆSM n E= - ´

s ¥

ˆSJ n H= ´

,e m

Imagetheory

(see Wire Antennas:

p. 20)

,e m,e m ,e m,e m

ˆ2SM n E= - ⋅ ´

0SJ =

10

Equivalent problem for an aperture in a ground plane

Original problemwith known E-field distribution in the aperture.

aE

An infinite flat plane is chosen as imaginary closed surface .

is unknown.S

SJ

ˆS

S

aM n

J

E=- ´

,e mn̂

S

,e m

0S

SJ

M =

0S

SJ

M =

Equivalent problemUsing image theory, the conductor is removed and replaced by imaginary image sources according to the previous page.

0

2 ˆS

aS

J

M n E

=

=- ´

,e mn̂

S

0

0S

S

J

M

=

=

0

0S

S

J

M

=

=

,e m

aE

Waveguide

,e mn̂

S

tan 0E =

s ¥

s ¥

To define , is approached by

an imaginary PEC surface (fields are equal to zero on the left of ).

0SJ =

S

S

ˆS

S

aM

J

n E=- ´

,e mn̂

S

0S

S

J

M =

0S

S

J

M =

s ¥

s ¥

s ¥

11

3 Types of equivalent problems for aperture antennas

(1) Aperture in free-space

ˆ

ˆ

S a

S a

J n H

M n E

= ´

= - ´

(2) Aperture in ground plane (3) Aperture in PMC plane

Fields on imaginary screen S are unknownoutside of the aperture.

Approximation: Fields on screenoutside of apertureare set to zero.

and known in the aperture

aE

aH

n̂ n̂ n̂

0

0

S

S

J

M

0

0

S

S

J

M

0

ˆ2

S

S a

J

M n E

=

= - ´

s = ¥

s = ¥

*s = ¥

*s = ¥

0

0

S

S

J

M

=

=

0

0

S

S

J

M

=

=

0

0

S

S

J

M

=

=

0

0

S

S

J

M

=

=

ˆ2

0

S a

S

J n H

M

= ´

=

S S S

known in the apertureaE

known in the apertureaH

(see previous page)

12

Considering linear current densities

,J M Sources

,E H Radiated fields

,A F Vector potentials

Vector potential revisited…

'4

'4

jkR

V

jkR

V

eA J dv

R

eF M dv

R

mp

ep

-

-

=

=

òòò

òòò

,S SJ M

4

4

jkR

S

S

jkR

S

S

eA J ds

R

eF M ds

R

mp

ep

-

-

¢=

¢=

òò

òò

Magnetic

Electric

Radiated fields:

1 1 1 1( )

1 1 1 1( )

A F A

A F F

E E E j A j A F H Fj

H H H A j F j F A Ej

wwme e we e

wm wme m wm

= + = - - ⋅ - ´ = ´ - ´

= + = - ´ - - ⋅ = - ´ - ´

13

In the far-field:Only terms with 1/r are considered; i.e., higher-order terms with are neglected.1/ nr

ˆ

ˆ

A rA

F r F

H j a AE j A

E j a F H j F

ww h

wh w

- ´-

´ -

Geometry considered

( , , )4

( , , )4

jkR

S

S

jkR

S

S

eA J x y z ds

R

eF M x y z ds

R

mp

ep

-

-

¢ ¢ ¢ ¢=

¢ ¢ ¢ ¢=

òò

òò

R: distance source to observation point

x

z

Observation point (x.y,z)( , , )x y z

J

q

f

yq

f

r

R

r

O

14

For phase variations:

For amplitude variations:

Aper

ture

pla

ne

Aper

ture

pla

ne

Far-field approximation

Reference point

cosR r r y¢-

R r

We define y as the angle between and .

( , , )4

( , , )4

jkR

S

S

jkR

S

S

eA J x y z ds

R

eF M x y z ds

R

mp

ep

-

-

¢ ¢ ¢ ¢=

¢ ¢ ¢ ¢=

òò

òò

r

r ¢

r

R R

r ¢

r

r ¢y y

cosr y¢

15

Radiation equations

Under far-fieldconditions

' cos

' cos

' , '4

' , '4

jkR jkrjkr

S S

S S

jkR jkrjkr

S S

S S

e eA J ds N N J e ds

R r

e eF M ds L L M e ds

R r

y

y

m mp p

e ep p

- -

- -

= » =4

= » =4

òò òò

òò òò

Radiated fields are produced by a superposition of the effects causedby both current densities and

Under far-field conditions:

( )

( )

0

4

4

0

4

4

r

jkr

jkr

r

jkr

jkr

E

jkeE L N

rjke

E L Nr

H

jke LH N

r

LjkeH N

r

q f q

f q f

qq f

ff q

hp

hp

p h

p h

-

-

-

-

- +

+ -

æ ö÷ç+ - ÷ç ÷ç ÷çè ø

æ ö÷ç- + ÷ç ÷ç ÷çè ø

SJ

SM

ˆ

ˆ

A rA

F r F

H j a AE j A

E j a F H j F

ww h

wh w

- ´-

´ -

16' cosr y = ? (difference in paths from source elements to observation point)

cos

cos

cos

cos

[ cos cos cos sin sin ]

[ sin cos ]

[ cos cos cos sin sin ]

[ sin cos ]

jkrx y z

S

jkrx y

S

jkrx y z

S

jkrx y

S

J J J e ds

J J e ds

M M M e ds

M M e ds

y

y

y

y

q f q f q

f f

q f q f q

f f

¢+

¢+

¢+

¢+

¢= + -

¢= - +

¢= + -

¢= - +

òò

òò

òò

òò

Rectangular-to-spherical component transformation

Computation of N and L

' cos

' cos

'

'

jkrS

S

jkrS

S

N J e ds

L M e ds

y

y

=

=

òò

òò

typically known in rectangular coordinates.,S SJ M

cos

cos

cos

cos

jkr

S

jkr

S

jkr

S

jkr

S

N J e ds

N J e ds

L M e ds

L M e ds

yq q

yf f

yq q

yf f

¢+

¢+

¢+

¢+

¢=

¢=

¢=

¢=

òò

òò

òò

òò

sin cos sin sin cos

cos cos cos sin sin

sin cos 0

r x

y

z

J J

J J

J J

q

f

q f q f q

q f q f q

f f

é ù é ù é ùê ú ê ú ê úê ú ê ú ê ú

= -ê ú ê ú ê úê ú ê ú ê úê ú ê ú ê ú-ê ú ê ú ê úë û ë û ë û

17

z

x

ya

b

dy¢

dx ¢

r

R

r ¢

y

Rectangular aperturex

y

za

b

dz ¢

dy¢

r

R

r ¢

y

y

z

xa

b

dx ¢

dz ¢

r

R

r ¢

y

(a)

(b) (c)

3 different configurations

in a rectangular coordinate system

Different analytical

description of the same physical

problem

•The aperture lies in one of the fundamental planes

•Different definitions of the angles q and f

18

cos ?r y¢ =Differential paths:

( ) ( )

( ) ( )

( )

ˆ ˆ ˆ ˆ ˆ ˆcos sin cos sin sin cos

sin sin cos

ˆ ˆ ˆ ˆ ˆ ˆcos sin cos sin sin cos

sin cos cos

ˆ ˆ ˆ ˆcos sin cos

r y z x y z

r x z x y z

r x y x

r y z

y z

r x z

x z

r x y

y q f q f q

q f q

y q f q f q

q f q

y q f

¢ ¢ ¢ ¢= ⋅ = + ⋅ + +

¢ ¢= +

¢ ¢ ¢ ¢= ⋅ = + ⋅ + +

¢ ¢= +

¢ ¢ ¢ ¢= ⋅ = + ⋅ +

(a)

(b)

(c)

r a a a a a a

r a a a a a a

r a a a a( )ˆ ˆsin sin cos

sin cos sin sin

y z

x y

q f q

q f q f

+

¢ ¢= +

a a

Aperturein yz plane

Aperturein xz plane

Aperturein xy plane

ds dy dz

ds dx dz

ds dx dy

¢ ¢ ¢=

¢ ¢ ¢=

¢ ¢ ¢=

(a)

(b)

(c)

Differential areas: r

R

r ¢

y

ds¢

( )ˆ ˆ ˆcos r rrr ry ¢¢ ¢ ¢= ⋅ = ⋅

a a r a

19

Example: Aperture with constant field distribution.

Consider a rectangular aperture mounted in an infinite ground plane (xy-plane). The field distribution is constant over the aperture:

0ˆa yE a E=

/2 /2( sin cos sin sin )

/2 /2

/2 /2( sin cos sin sin )

/2 /2

0

cos cos

sin

b ajk x y

x

b a

b ajk x y

x

b a

N N

L M e dx dy

L M e dx dy

q f

q f q fq

q f q ff

q f

f

¢ ¢+

- -

¢ ¢+

- -

= =

é ùê ú¢ ¢= ê úê úë û

é ùê ú¢ ¢= - ê úê úë û

ò ò

ò ò

2) solve the radiation integrals (p. 16 + 18)

( )/2

/2

sin2

2

cj z

c

ce dz c

ca

a

a

+

-

é ùê úê ú= ê úê úê úë û

ò

Use integral form

z

x

ya

b

r

q

f

aE

ds¢

1) Equivalent problem (p. 11)

0 0

0

ˆ ˆ ˆ2 2

S

S z y x

J

M a a E a E

ìï =ïïíï = - ´ =ïïî

xMyEb

aIn the aperture:

20

RESULTSUniform distribution

aperture on ground planeUniform distribution

aperture in free-spaceTE10-mode distribution

aperture on ground plane

Aperture distribution

Geometry

Equivalent problem

Far-zone fields

0

/ 2 /2ˆ

/2 /2a y

a x aE a E

b y b

ì ¢- £ £ïïï= íï ¢- £ £ïïî

0

0

ˆ /2 /2

/2 /2ˆ /

a y

a x

E a E a x a

b y bH a E h

ì ¢= - £ £ïïïíï ¢- £ £=- ïïî

0

/ 2 /2ˆ cos

/2 /2a y

a x aE a E x

a b y b

p ì ¢- £ £ïæ ö ïï÷¢ç= í÷ç ÷è ø ï ¢- £ £ïïî

z

x

ya

b

ˆ2

0

0

a

S

S

n EM

J

ìï- ´ïï= íïïïî=

in aperture

elsewhere

everywhere

ˆ

ˆ

0

S a

S a

S S

M n E

J n H

M J

üï= - ´ ïïýï= ´ ïïþ

in aperture

elsewhere

z

x

ya

bz

x

ya

b

0sin cos sin sin2 2 2

jkrka kb abkE eX Y C j

rq f q f

p

-

= = =with

0

sin sinsin

sin sincos cos

/

/

r rE H

X YE C

X YX Y

E CX Y

H E

H E

q

f

q f

f q

f

q f

h

h

= =

=

=

=-

=

( )

( )

0

sin sin1 cos sin

2sin sin

1 cos cos2

/

/

r rE H

C X YE

X YC X Y

EX Y

H E

H E

q

f

q f

f q

q f

q f

h

h

= =

= +

= +

=-

=

( )

( )

22

22

0

cos sinsin

2 /2

cos sincos cos

2 /2

/

/

r rE H

X YE C

YX

X YE

YX

H E

H E

q

f

q f

f q

pf

pp

q fp

h

h

= =

=--

=--

=-

=

ˆ2

0

0

a

S

S

n EM

J

ìï- ´ïï= íïïïî=

in aperture

elsewhere

everywhere

21

RESULTS (2)Uniform distribution

aperture on ground planeUniform distribution

aperture in free-spaceTE10-mode distribution

aperture on ground plane

Half-power beamwidth(degrees)

E-planeb>>l

H-planea>>l

First null beamwidth(degrees)

E-planeb>>l

H-planea>>l

First side lobe level (relative to

main maximum)

E-plane -13.26 dB ≈ -13.26 dB -13.26 dB

H-planea>>l

-13.26 dB ≈ -13.26 dB -23 dB

Directivity D0

(dimensionless)

50.6/b l

50.6/b l

»50.6/b l

50.6/a l

50.6/a l

»68.8/a l

114.6/b l

114.6/b l

»114.6

/b l

114.6/a l

114.6/a l

»171.9

/a l

2 2

44

abpp

l læ ö÷ç⋅ = ÷ç ÷çè ø

area 2 2

44

abpp

l læ ö÷ç⋅ = ÷ç ÷çè ø

area2 2 2

84 4

ab abp p

p l lé ùæ ö æ ö÷ ÷ç çê ú = ⋅÷ ÷ç ç÷ ÷ç çè ø è øê úë û

0.81

22

Remarks on rectangular apertures:

1) Use different equivalent problems depending on the presence or absence of ground plane.

notice the term in far fields of aperture in free space.1 cos2

q+

3) The third case considers the open-end of a rectangular waveguide.

a

bx

y

zTE10

TE01

TE20

2) The beamwidth and directivity are dependent on the aperture dimensions

The non-constant field distribution over the aperture has an effect on beamwidth, sidelobe level and directivity.

For a cosine distribution: ( p. 21)- larger beamwidth- Lower sidelobe level- Smaller aperture efficiency (≈ 0.81)

1 EqInstead of for

cos EfqInstead of for

23

Example 1: Different representations

Rectangular aperture in an infinite ground plane, with constant field distribution

3a l=

x

y2b l=

aE

z

x yaE

H-Plane (xz) E-Plane (yz)

z

x y

Amplitude patternLinear representation

dB representation

H-PlaneE-Plane

24

Example 2: Effect of aperture size(dB representations, 40 dB range)

z

x y

a l=

z

x y

2a l=z

x y

5a l=z

x y

10a l=

Square apertureUniform field along ya

aaE

z

x y

/2a l=z

x y

a l

Equivalent to an infinitesimal magnetic dipole along x

25

z

xy

z

xy

Example 3: Effect of aperture distribution

Aperture in a ground plane

4

2

a

b

l

l

=

=a

b

TE10

a

b

H-plane (xz) E-plane (yz)

Uniform

UniformTE10

UniformTE10

(dB representation)

26

z

xy

z

xy

Example 4: Effect of ground plane

3

2

a

b

l

l

=

=

H-plane (xz)

Without ground plane (approx.)

With ground plane

E-plane (yz)

a

b

(dB representation)

with GPwithout GP

1 c s2

1o q

«+1 o

ss

2co

c qq

+«

27

total radiated powpower within cone angl

ee r

12

0 02

0 0

( , )sin

( , )sin

BE

U d d

U d d

p q

p p

q f q q f

q f

q

q q f

1=

=ò òò ò

Beam efficiency

The beam efficiency is frequently used to judge the quality of a transmitting or receiving antenna.

It describes the ability of the antenna to discriminate between signals received through the main lobe and those through minor lobes.

The beam efficiency depends on the sidelobe distribution

z

xy

z

xy

q1

28

(b) using upper abscissa scale

BE 28%

Example: Beam efficiency of a square aperture

Consider a square aperture with a cos3 aperture field distribution. Determine the beam efficiency for q1 = 10° and

(a) a = b = 20(b) a = b = 3

Solution

(a) Use lower abscissa scale for 10 BE 100%(aperture to large !)

cos distribution

cos2 distribution

cos3 distribution

2 3sin10 1.64u

p ll

= =2

A uniform distribution would have(a’) BE 94%(b’) BE 58%

Aperture distributions:

Tapered distributions have lower sidelobes but larger beamwidths

Beam

effi

cien

cy (%

)

Beam angle q1 in degree (a = b = 20l)

uniform

cos( / ), cos( / )x a y bp p¢ ¢

2 2cos ( / ), cos ( / )x a y bp p¢ ¢

3 3cos ( / ), cos ( / )x a y bp p¢ ¢

1sin2ka

u q=

0 1 2 3 4 5 6 7 8 9 10

100

90

80

70

60

50

40

30

20

102 3 41

(a)

(b)

(a’)

(b’)

29

Tangential fields 0»

Reflection

Radiation from the open end of a rectangular waveguide

Full field simulation

Without ground planeWR28: 26 GHz to 40 GHzCut-off TE10: 21.08 GHz

Standing waves in the waveguide

Ey

Frequency: 30 GHz

30

Tangential componentsEy Hx

xx

z 0» 0» 0» 0»ˆ

ˆ

0

S a

S a

S S

M n E

J n H

M J

üï= - ´ ïïýï= ´ ïïþ

in aperture

elsewhere

Equivalent problem:

a

b

x

y

z

x y

z

x y

z

x y

22 GHz 32 GHz 42 GHz

Aperture:a = 0.52lb = 0.26l

Aperture:a = 0.76lb = 0.38l

Aperture:a = 1.0lb = 0.5l(li

near

repr

esen

tatio

n)

Patterns:

31

Circular Apertures z

x

y

r

R

r ¢ y

90

r r

q

¢ ¢=¢ =

f¢ f

q

adr¢ dr f¢ ¢

Circular aperture in an infinite ground plane

- Simple construction- Closed-form expression for the modes

Analysis:Modes are given in cylindrical coordinates

cos

sin

x

y

z z

r f

r f

¢ ¢ ¢=

¢ ¢ ¢=

¢ ¢=

, , , , ,x y z x y zJ J J M M M , , ,N N L Lq f q f, , , , ,z zJ J J M M Mr f r f

cos sin 0

sin cos 0

0 0 1

x

y

z z

J J

J J

J J

r

f

f f

f f

é ù é ùé ù¢ ¢-ê ú ê úê úê ú ê úê ú¢ ¢=ê ú ê úê úê ú ê úê úê ú ê úê úê úê ú ê úë ûë û ë û p. 16

32

cos

cos

cos

cos

[ cos cos cos sin sin ]

[ sin cos ]

[ cos cos cos sin sin ]

[ sin cos ]

jkrx y z

S

jkrx y

S

jkrx y z

S

jkrx y

S

N J J J e ds

N J J e ds

L M M M e ds

L M M e ds

yq

yf

yq

yf

q f q f q

f f

q f q f q

f f

¢+

¢+

¢+

¢+

¢= + -

¢= - +

¢= + -

¢= - +

òò

òò

òò

òò

page 16

, , , , ,x y z x y zJ J J M M M

, , , , ,z zJ J J M M M

( ) ( )

( ) ( )

( ) ( )

( ) ( )

cos

cos

cos

cos

[ cos cos cos sin sin ]

[ sin cos ]

[ cos cos cos sin sin ]

[ sin cos ]

jkrz

S

jkr

S

jkrz

S

jkr

S

N J J J e ds

N J J e ds

L M M M e ds

L M M e ds

yq r f

yf r f

yq r f

yf r f

q f f q f f q

f f f f

q f f q f f q

f f f f

¢+

¢+

¢+

¢+

¢ ¢ ¢= - + - -

¢ ¢ ¢= - - + -

¢ ¢ ¢= - + - -

¢ ¢ ¢= - - + -

òò

òò

òò

òò

cos sin cos sin sin sin cos( )r x y

ds d d

y q f q f r q f f

r r f

¢ ¢ ¢ ¢ ¢= + = -

¢ ¢ ¢ ¢=with

cossin

xyz z

r fr f

ìï ¢ ¢ ¢=ïïï ¢ ¢ ¢¬ =íïï ¢ ¢=ïïî

33

RESULTSUniform distribution

aperture on ground planeTE11-mode distribution

aperture on ground plane

Aperture distribution

Geometry

Equivalent problem

Far-zone fields

0ˆa yE a E ar¢= £

( )

( )

0 1 11 11

0 1 11

ˆ ˆ

/ sin / 1.841

/ cos

aE a E a E a

E E J a

E E J a

r r f f

r

f

r

c r f r c

c r fr

ìïïï= + ¢ £ïïïï¢ ¢ ¢ ¢= ¢ =íïïï ¶¢ ¢ ¢ ¢ ï= ¢=ïï ¢¶ïî

elsewhere

everywhere

ˆ2

0

0

a

S

S

n E aM

J

rìï ¢- ´ £ïï= íïïïî=

11

11

0

( )sin

( )cos cos

/

/

r rE H

J ZE jC

ZJ Z

E jCZ

H E

H E

q

f

q f

f q

f

q f

h

h

= =

=

=

=-

=

( )

12

122

11

0

( )sin

( )cos cos

1 /

/

/

r rE H

J ZE C

ZJ Z

E CZ

H E

H E

q

f

q f

f q

f

q fc

h

h

= =

=

¢=

¢-

= -

=

elsewhere

everywhere

ˆ2

0

0

a

S

S

n E aM

J

rìï ¢- ´ £ïï= íïïïî=

z

x

y

a

z

x

y

a

20

1

0 1 112

11

sin

( )

1.841

jkr

jkr

Z ka

ka E eC j

rkaE J e

C jr

q

c

c

-

-

=

=

¢=

¢ =

Jn: Bessel functions1 0 1( ) ( ) ( )/J Z J Z J Z Z¢ = -

with

34

RESULTS (2)Uniform distribution

aperture on ground planeTE11-mode distribution

aperture on ground plane

Half-power beamwidth(degrees)

E-planea>>l

H-planea>>l

First null beamwidth(degrees)

E-planea>>l

H-planea>>l

First side lobe level (relative to

main maximum)

E-plane -17.6 dB -17.6 dB

H-plane ≈ -17.6 dB -26.2 dB

Directivity D0

(dimensionless)

29.2/a l

29.2/a l

29.2/a l

37.0/a l

69.9/a l

69.9/a l

69.9/a l

98.0/a l

( )area2

22

4 24

aa

p pp p

l læ ö÷ç⋅ = = ÷ç ÷è ø

22 apl

æ ö÷ç⋅ ÷ç ÷çè ø0.836

35

z

xy

z

xy

Example: Circular aperture with radius a = 1.5mounted in a ground plane

TE11Uniform

H-planeE-plane

H-planeE-plane

36

x

y

z x

y

z

Square aperture vs. circular aperture

3D patterns in dB representation seen from top

Apertures in a ground plane with constant field distribution along y

2

3

A a

a l

=

=

Area:

0 20.5

13.26

D =-

dBSLL: dB

0 20.5

17.6

D =-

dBSLL: dB

a

yE

2 ar

yE

Area: 2

3a

a

A r

r

p

lp

=

=

37

Beam efficiency of circular apertureExample: Determine the beam efficiency of a circular

aperture with radius a for q = 10°. The aperturehas the following characteristics

Aperture distributions:

( )1/22

1 '/aré ù-ê úë û

( )12

1 '/aré ù-ê úë û

( )22

1 '/aré ù-ê úë û

(b) Using upper abscissa scale

BE 92%

Solution

(a) Use lower abscissa scale for 10 BE 100%

1

2 2sin 3 sin(10 ) 3.27u a

p pq l

l l= = =

( )220

1 '/3

aa

a

lr

l

ü= ïïï -ýï= ïïþ

(a) and a taper

(b)

Beam

effi

cien

cy (%

)

Beam angle in degree (a = 20l)

uniform1/221 ( / )aré ù¢-ë û

1sinu ka q=

0 1 2 3 4 5 6 7 8 9 10

100

90

80

70

60

50

40

30

20

102 3 41

121 ( / )aré ù¢-ë û221 ( / )aré ù¢-ë û

(a)(b)

38

4. Babinet’s principle

Sum of diffracted fields= original fields

Absorbing screens

Incident wave

Incident wave

When the field behind a screen with an opening is added to the field of a complementary structure, the sum is equal to the field without screen.

In Optics

39

Brooker’s extension

Brooker’s extension of Babinet’s principle includes

- the effects of polarization

- conducting screens instead of absorbing screens

PEC

It is therefore better suited to RF!

complementary structure ?

PEC: Perfect Electric ConductorPMC: Perfect Magnetic Conductor

40

Radiated fields in the presence of an electric conducting screen with an opening

Radiated fields in the presence of the complementary magneticconducting screen

0

0

e m

e m

E E E

H H H

= +

= +

PMC

Consider the radiated fields of an electric source J in an unbounded medium. The source produces the fields E0, H0 at a point P.

Babinet’s principle equivalents: The same fields can also be produced by combining the following fields

J ,e m

0 0,

/

E H

h m e

ìïïïíï =ïïî

P

J ,e m

,

/e eE H

h m e

ìïïïíï =ïïî

PaS

J ,e m

,

/m mE H

h m e

ìïïïíï =ïïî

P

aS

PEC

(a)

(b)

(c)

41

/

, ,

m d

m d

J M

E H

H E

h

he m m e

ì ïïïï ïïïíï -ïïï ïïïî

0

0

e d

e d

E E H

H H E

h

h

= +

= -

In practice, the dual of (c) is more easily realized:

The electric conducting screen of (b) and the electric conductor of (d) are usually referred as complementary structures.

PEC

,m e,

/d d

d

E H

h e m

ìïïïíï =ïïîaS

M P(d)

42

2) Relationship between the radiated far-zone fields for complementary structures

20

4s cZ Zh

=

2 20 0

,

,

s c s c

ccs s

E H E H

EEH H

q q f f

fqq fh h

- -

Application to slot antennas

s: screenc: complementary structure

1) Relationship between the terminal impedances of complementary structures

Screen with opening Complementary dipole

a

b

sZ

PEC Current line

Transmission feed line

cZ

PECTransmission feed line

43

Example: Find the radiation resistance of a thin half-wavelength slot in an infinite ground plane.

Solution:

( )

,

2

,,

73

120487

4 4

rad dipole c

s c rad slotrad dipole

R

Z Z RR

h p2

» W = Z

= = » W

PEC

Transmission feed line/2l

w

l/2 thin slot (w << l)

/2l

w

Transmission feed line

l/2 flat dipole (w << l)

Dipole vs. slot: identical pattern in shape E and H field are interchanged. Orthogonal polarization to each other

Dipole: Slot:

E

H E

H

44

Microstrip coupledslot antenna

Coplanar waveguide fedslot antenna

Examples of feeding for slot antennas