Chapter Chapter 12 12 : Aperture Antenna: Aperture Antenna

• Huygen’s Principle

• Application of Huygen’s principle

• Rectangular apertures

• Circular apertures

1

• Circular apertures

Aperture Antenna Example

Huygen’s principleHuygen’s principle

3

Equivalent surface currents Js, Ms radiate fields E1,H1 in V2

(same as actual problem)

However, Js, Ms do NOT radiate fields E1,H1 in V1.

Thus, actual problem and equivalent problem are equivalent

only in V2 (outside V1)

Actual problem Equivalent problem

Huygen’s principle (Huygen’s principle (22))

Since the fields E,H radiated by Js, Ms in V1 are arbitrary, we

can assume they are zero.

4

Love’s

equivalent

Electric Conductor

equivalent

Magnetic Conductor

equivalent

Equivalence principle models

Equivalent model exampleEquivalent model example

Equivalent models for magnetic source radiation near

a perfect electric conductor

via image

theory

Waveguide aperture exampleWaveguide aperture example

6

via Huygen’s principle via image theory

Aperture fields (Ea,Ha) are known.

FarFar--zone fieldszone fields

Far-zone fields due to electric and magnetic surface currents

are given by:

1

')]'(ˆˆˆ)'([4

)( 'ˆ

ejk

dserrrrrr

ejkr

jkr

rrjk

S

s

jkrr

rrr

rrr

−

⋅−

∫∫ ××−×−≈ JME s ηπ

7

')]'(1

ˆˆˆ)'([4

)( 'ˆdserrrrr

r

ejkr

rrjk

S

s

jkrrrrr ⋅

−

∫∫ ××+×≈ MJH s ηπr

rˆ;

ˆ×=

×= HE

EH η

ηRecall that:

and

]'coscos'sinsin)'[cos('

cos'sin)'cos('

cos'sin)sin'cos'(cos''ˆ

θθθθφφθθφφρ

θθφφς

+−=

+−=

++==⋅

r

z

zyxrrrr

Rectangular AperturesRectangular Apertures

Rectangular aperture positions

for

antenna system analysis

8

In general, the non-zero components of Js and Ms are:

(a) Figure

,,, zyzy MMJJ

(b) Figure

,,, zxzx MMJJ

(c) Figure

,,, yxyx MMJJ

antenna system analysis

Rectangular Apertures (Rectangular Apertures (22))

Rectangular aperture with uniform distribution on an infinite

ground plane: find far-zone fields for z > 0.

Aperture field: 2/'2/;2/'2/ˆ0 bybaxaEya ≤≤−≤≤−=E

Equivalent problem for z > 0

9

apertureon 2ˆ

ˆˆ2

ˆ2

0

0

Ex

Eyz

n as

=

×−=

×−= EM

aperture outside 0JM == ss

Rectangular Apertures (Rectangular Apertures (33))The far-zone electric field can then be given by

''ˆ)'(4

)( 'ˆ2/

2/

2/

2/dxdyerr

r

ejkr

rrjka

a

b

b

jkrrrr ⋅

− −

−

∫ ∫ ×−≈ sMEπ

Since

)sinˆcoscosˆ(2

ˆ2ˆˆ0

φθφθφ −−=

×=×

E

rExrsM

10

)0'for (sin)sin'cos'('ˆ =+=⋅ zyxrr θφφr

)sinˆcoscosˆ(2 0 φθφθφ −−= E

∫ ∫− −

⋅−

+≈2/

2/

2/

2/

'ˆ

0 '')sinˆcoscosˆ(24

)(a

a

b

b

rrjkjkr

dxdyeEr

ejkr

rrφθφθφ

πE

and

Rectangular Apertures (Rectangular Apertures (44))

The far-zone fields are given by

Since

− jkr

=

= ∫∫∫ ∫ −−− −

⋅

φθφθ

φθφθ

sinsin2

sinccossin2

sinc

''''2/

2/

sinsin'2/

2/

cossin'2/

2/

2/

2/

'ˆ

kbkaab

dyedxedxdyeb

b

jkya

a

jkxa

a

b

b

rrjkr

11

The far-zone fields are given by

)sinˆcoscosˆ(

sinsin2

sinccossin2

sinc24

)( 0

φθφθφ

φθφθπ

+×

=−

kbkaabE

r

ejkr

jkrr

E

)sinˆcoscosˆ(

sinsin2

sinccossin2

sinc24

)( 0

φφφθθ

φθφθηπ

+−×

=−

kbkaabE

r

ejkr

jkrr

H

Radiation PatternRadiation Pattern

12

Normalized field pattern: a = 3λ, b = 2λ

Radiation PatternRadiation Pattern

13

Normalized field pattern: a = 3λ, b = 3λ

Radiation Pattern (Radiation Pattern (33))

In many applications, only principle plane patterns are usually

sufficient.

E-Plane

(φ=π/2,3π/2)

(y-z plane)0;

sinsin2

sinc24

0

===

=−

θ

θ φθπ

HEE

H

kbabE

r

ejkE

jkr

14

(y-z plane)0; === θφ

θφ η

HEE

H

H-Plane

(φ=0,π)

(x-z plane) 0;

coscossin2

sinc24

0

==−=

=−

φθφ

θ

φ

η

φθθπ

HEE

H

kaabE

r

ejkE

jkr

Radiation Pattern (Radiation Pattern (44))

15

E-plane and H-plane

amplitude pattern for

uniform distribution

aperture mounted on an

infinite ground plane

(a=3λ, b=2λ)

BeamwidthBeamwidthFor the E-plane, maximum at θ=0.

Nulls occur when L,2,1sin2

== nnkb

n πθthus

b

nn

λθ 1sin−=

K,2,1,sin2 1 ==Θ −n

b

nN

λBeamwidth between nulls

λ

16

b

bFNBW

λ1sin2 −=ΘFNBW:

Half-power angles:

391.12/1)(sinc2 =⇒= xxb

h

λθ

443.0sin 1−=

HPBW:b

HPBW

λ443.0sin2 1−=Θ

DirectivityDirectivityThe radiated power can be obtained from the power into the

aperture. For uniform distribution, the Poynting vector is given

by

2/'2/;2/'2/2

||ˆ)Re(

2

1 2

0*bybaxa

Ezav ≤≤−≤≤−=×=

ηHEW

thus abE

dxdyE

dPa b ||

''|| 2

02/ 2/

2

0∫ ∫∫ ==⋅= sW

17

thus abE

dxdyE

dPa b

S

avrad ηη 2

||''

2

|| 0

2/ 2/

0∫ ∫∫ − −==⋅= sW

Recall that

)(2

),(

and),(4

),(

222

φθηφθ

φθπφθ

EEr

U

P

UD

rad

+=

=

Directivity (Directivity (22))The maximum radiation intensity is given by

thus

pAababU

D22

2

maxmax

4414

4

λπ

λπ

λπ

π==

==

ηλπηφθ

θ 242

)(),(

2

0

2

2

2

0

22

0max

EabEkabUU

====

18

p

rad

AababP

D22max 4λλλ

π ==

==

Recall that the maximum effective aperture is given by

where Ap is the physical aperture (=ab).

pem ADA == max2

4

λπ

The aperture efficiency 1==p

emap

A

Aε

Quiz

A uniform plane wave traveling in the

direction as shown. The field is given

by

0ˆ :case TM

0ˆ :case TE

0

0

≤=

≤=⋅−

⋅−

zeHy

zeEy

rkji

rkji

H

E

(a) Find the propagation direction.

(b) Find the aperture field.

(c) Find the equivalent current.

Assume that the aperture dimension in the y direction is b.

0ˆ :case TM 0 ≤= zeHyH

Circular AperturesCircular Apertures

Circular aperture with uniform distribution on an infinite

ground plane: find far-zone fields for z > 0.

Aperture field: aEya ≤= 'ˆ0 ρE

Equivalent problem for z > 0

22

aρ'Ex

Eyz

n as

≤=

×−=

×−=

2ˆ

ˆˆ2

ˆ2

0

0

EM

aperture outside 0JM == ss

Circular Apertures (Circular Apertures (22))The far-zone electric field can then be given by

'''ˆ)'(4

)( 'ˆ

0

2

0ρφρ

π

πdderr

r

ejkr

rrjka

jkrrrr ⋅

−

∫ ∫ ×−≈ sME

Since

)sinˆcoscosˆ(2

ˆ2ˆˆ0

φθφθφ −−=

×=×

E

rExrsM

23

)'cos(sin'

)0'(sin)sin'cos'('ˆ

φφθρθφφ

−=

=+=⋅ zyxrrr

)sinˆcoscosˆ(2 0 φθφθφ −−= E

∫ ∫ ⋅−

+≈a

rrjkjkr

ddeEr

ejkr

0

2

0

'ˆ

0 ''')sinˆcoscosˆ(24

)(π

ρφρφθφθφπ

rrE

and

Circular Apertures (Circular Apertures (33))Using

)sin'(2' 0

2

0

)'cos(sin' θρπφπ φφθρ

kJdejk =∫ −

and

)()()( 1010

0 ββββ

JzzJdzzzJ ==∫

24

the far-zone fields can be found to be

)sinˆcoscosˆ(sin

)sin()( 10

2

φθφθφθθ

+=−

ka

kaJ

r

eEjkar

jkrr

E

)sinˆcoscosˆ(sin

)sin()( 10

2

φφφθθθθ

η+−=

−

ka

kaJ

r

eEjkar

jkrr

H

Radiation PatternRadiation Pattern

25

Normalized field pattern: a = 1.5λ