Post on 15-Apr-2018
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