1 Aperture antennas constitute a large class of antennas, which emit EM waves through an opening (or aperture).
These antennas have close analogs in acoustics: the megaphone and the parabolic microphone. The pupil of the human eye is a typical aperture receiver for optical radiation. At radio and microwave frequencies, horns, waveguide apertures and reflectors are examples of aperture antennas.
Aperture antennas are commonly used at UHF and above as their gain increases as f2 (approximately). For an efficient and highly directive aperture antenna, its area should be comparable or larger than 2. Hence, these antennas are impractical at low frequencies.
Aperture Antennas
2 Another positive feature of the aperture antennas is their near-real valued input impedance and geometry compatibility with waveguide feeds.
The radiation characteristics of wire antennas can be determined from the current distribution on the wire.
If the current distribution is not known, alternate methods are necessary to compute the radiation characteristics e.g. from the fields on or in the vicinity of the antenna.
One such technique is the Field Equivalence Principle.
3Uniqueness theorem
A solution is said to be unique if it is the only one possible among a given class of solutions.
The EM field in a given region V is uniquely defined if:
a. all sources are given;
b. either the tangential E components or the tangential H components are specified at the boundary S.
4Equivalence principle
The equivalence principle follows from the uniqueness theorem. It allows for building simpler problems.
As long as the equivalent problem preserves the boundary conditions of the original problem for the field at S, it is going to produce only one possible solution for the region outside S.
Through equivalence principle, the fields outside an imaginary closed surface are obtained by placing over the closed surface suitable electric and magnetic current densities which satisfy the boundary conditions.
The current densities are chosen so that the fields inside the closed surface are zero and outside they are equal to the fields produced by the actual sources.
5E1,H1
V1
1,1 1,1
E1,H1V2
S
Original problem
J1 M1
The primary task is to replace the original problem by an equivalent problem which yields the same fields E1 and H1outside S (within V2).
6E,H
V1
1,11,1
E1,H1V2
S
Equivalent problem
n
1sM n E E = r r r
1sJ n H H = r r r
MsJs
The original sources J1 and M1 are removed, and assume there exist fields E and H inside S and fields E1 and H1 outside of S.
7 For this, the fields must satisfy the boundary conditions on tangential electric and magnetic fields components. Hence on imaginary surface there must exist equivalent sources:
These current densities are said to be equivalent only within V2 as they produce same fields outside.
Since the fields E and H within S are not of interest, they can be assumed to be zero:
1
1
s
s
J n H H
M n E E
= =
r r r
r r r
( )( )1 10
1 10
|
|
s H
s E
J n H H n H
M n E E n E=
=
= = = =
r
r
r r r r
r r r r Loves Equivalence Principle
8 We can assume that the boundary S is a perfect electric conductor. This eliminates the surface electric currents, i.e., Js=0, and leaves just surface magnetic currents Ms, which radiate in the presence of a perfect electric surface.
We can apply Loves equivalence principle in three different ways
E,H=0V1
1,1
E1,H1V2
S n
1 0sJ n H= =r r
1sM n E= r r
Electric conductor
9E,H=0V1
1,1
E1,H1V2
S n
1sJ n H= r r
1 0sM n E= =r r
Magnetic conductor
We can assume that the boundary S is a perfect magnetic conductor. This eliminates the surface magnetic currents, i.e. Ms=0, and leaves just surface electric currents Js, which radiate in the presence of a perfect magnetic surface.
10
Make no assumptions about the materials inside S, and define both Js and Ms currents, which are radiating in free space (no fictitious conductors behind them). It can be shown that these equivalent currents create zero fields inside S.
All three approaches lead to the same field solution according to the uniqueness theorem.
The first two approaches are not very accurate in the general case of curvilinear boundary surface S. However, in the case of flat infinite planes (walls), the image theory can be used to reduce the problem to an open one.
Image theory can be successfully applied to curved surfaces provided the curvatures radius is large compared to the wavelength.
11
=
,
1sM n E= r r
1sM n E= r r
,
sMr
,
12sM n E= r r
, ,
Equivalent problem - Electric Wall
Equivalent problem - images
12
Example: A waveguide aperture is mounted on an infinite ground plane. Assuming that the tangential component of the electric field over the aperture is Ea. Find an equivalent problem that will yield the same fields E, H radiated by the aperture to the right side of the interface.
=
=
aEr
n
S
,
13
S
n, ,
0s
s
J
M =
rr
s
s a
J
M n E=
rr r
0s
s
J
M =
rr
S
n, ,
0
0s
s
J
M
==
rr
0
s
s a
J
M n E
==
rr r
0
0s
s
J
M
==
rr
=
S
n, ,
0
0s
s
J
M
==
rr
0
s
s a
J
M n E
==
rr r
0
0s
s
J
M
==
rr
s aM n E= r r
(Image)
S
n, ,
0
0s
s
J
M
==
rr
0
2s
s a
J
M n E
==
rr r
0
0s
s
J
M
==
rr
Solution:
14
Radiation Equations
'
'
Rr
rr'rr
x,y,z on S
'cosR r rR r
is the angle between r and r
Primed coordinates: sources
Unprimed coordinates: observation point
'4
'4
jkR
VjkR
V
eA J dvR
eF M dvR
=
=
r r
r r
15
'cos
'cos
''4 4
'4 4
'
jkR jkjkr
sS
r
sS
jkjkr
sS
R jkr
sS
N N J e dse eA J dsR r
e eF M dsR r
L L M e ds
==
= =
r r
r r r rr
r r r
The far-zone fields and vector potentials are related as:
( )( )
farA
farF
E j A A
H j F F
= += +
r
r
1far far far farF F A AE H r H E r = = r r r rr r
Since
( ) ( ) farE j A F A F = + + r
16
2 2k
==( )
( )
0
4
4
r rjkr
jkr
E HjkeE L N H
rjkeE L N H
r
+
+
'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
+
+
+
+
= + = + = + = +
17
rrRr
'rr
x
z
y
Rectangular apertures in different planes(Example: y-z plane)
Nonzero components of Js and Ms are:
Jy, Jz, My Mz
( ) ( )'cos '
' ' sin cos sin sin cos
'sin sin 'cos
r
y z x y z
r r a
a y a z a a a
y z
== + + += +
r
' ' 'ds dy dz=
18
Uniform distribution on an infinite ground plane
0 2 ' 2 2 ' 2a yE a E a x a b y b= r
Find the far fields radiated by a rectangular aperture on an infinite ground plane with following field over the opening:
Fields radiated by the aperture:
0
0
0
sin sinsin2
sin sin
sin cos sin
cos c
sin
os
2
2
2
r rjkr
jkr
E H
abkE e X YE j Hr X Y
abkE e X YE j Hr X Y
ka kbX Y
= = = = = =
= = sin cos2
sin sin2
xx ee dx
kaX
kbY
=
=
=
19
The total-field amplitude pattern is, therefore,
2 2 2 2 2sin sin sin sinsin cos cos 1 cos sinX Y X YEX Y X Y
= + = The E-plane (yz-plane: =/2) pattern
0sin sin
22 sin
2
jkrkb
abkE eE j kbr
= The H-plane (xz-plane: =0) pattern
0
sin sin2cos
2 sin2
jkrka
abkE eE j kar
=
Function of dimension b (along y-axis)
Function of dimension a (along x-axis)
20
3 2a b = =
21
For electrically large apertures, the main beam is narrow and the square-root term is negligible, i.e., it is roughly equal to 1 for all observation angles within the main beam.
That is why, in the theory of large arrays, it is assumed that the amplitude pattern of a rectangular aperture is
2 2 2 2 2sin sin sin sinsin cos cos 1 cos sinX Y X YEX Y X Y
= + =
sin sinX YX Y
22
H-plane pattern for a=20 E-plane pattern for b=10
23
First Null beamwidth for E-plane pattern:
For nulls to occur:
1114.6sin deg 1,2,3,....nn nb = =
n=1 gives FNBW
1 1
sin |2
2sin 57.3sin deg
n
n
kb n
n nkb b
=
= = =
Beamwidth between nulls:
24
Half-power beamwidth (E-plane):
Half-power point:
1
sin sin12 sin | 1.391
22sin2
0.44357.3sin deg
h
h
kbkb
kb
b
=
= = =
Half-power beamwidth:
1 0.443114.6sin degh b =
25
Maximum of first sidelobe:
1 1.43sin | 4.494 57.3sin deg2 s skb
b = = =
Total beamwidth between first sidelobes:
1 1.43114.6sin degs b =
Maximum at the first sidelobe:
( ) ( )sin 4.494 0.217 13.264.494s
E dB = = = =
When evaluating side-lobe levels and beamwidths in the H-plane, one has to include the cos factor, too.
The larger the aperture, the less important this factor is.
26
Directivity:
max0 2
0 2 2
4 4
4 4rad
p em
UD abP
D A A
= =
= =
The physical (Ap) and the effective (Aem) areas of a uniform aperture are equal.
27
The tapered rectangular aperture on a ground plane
The uniform rectangular aperture has the maximum possible effective area (for an aperture-type antenna) equal to its physical area it has the highest possible directivity for all constant-phase excitations of a rectangular aperture.
The uniform distribution excitation produces the highest SLL of all constant-phase excitations of a rectangular aperture.
A reduction of the SLL can be achieved by tapering the equivalent source distribution from a maximum at the apertures center to zero values at its edges.
28
One practical aperture of tapered source distribution is the open rectangular waveguide. The dominant TE10 mode has the following distribution:
0 cos ' 2 ' 2 2 ' 2a yE a E x a x a b y ba = + +
r
29
( )
( )
02
2
02
2
sin cos sin sin2
0
cos sinsin2 2
2cos sincos cos
2
2
22
r rjkr
jkr
E HabkE e X YE j H
r YX
abkE e X YE j Hr
kbY
Y
kX
X
a
= == =
= =
=
=
Fields radiated by the aperture:
30
The E-plane (yz-plane: =/2) pattern
sin sin2
sin2
n
kb
E kb
=
The H-plane (xz-plane: =0) pattern
2 2
sin sin2cos
sin2 2
n
ka
Eka
=
31
H-PLANE PATTERN UNIFORM VS. TAPERED ILLUMINATION (a = 3):
32
Aperture efficiency
Directivity:
0 2 2
4 40.81
0.81
p em
em p
D
A A
A A == =
In general,
0 1em ap p apA A =
33
Example: Consider an aperture in an infinitely conducting ground plane having a length a and width b. The aperture field has a uniform amplitude but a 180 phase difference in the two halves of the aperture as shown below:
Find the far field of the antenna applying the radiation equation of a rectangular aperture.
34
( )( )
2 20'cos 'cos
0 02 2 0
0
0
0
0 'cos 'sin cos 'sin
cos cos 2 ' ' 2 ' '
sin s22 cos cos
2 0.5 0 0.5 0.50 0.5 0.5 0.5
sin
2
b ajkr jkr
b a
z y
s
z y
a E a a x b y bM
x a b y ba E a
N N r x y
L E e dx dy E e dx dy
kb
L E b
=
= = = +
+
+= + +
=
r
20'sin cos 'sin cos
2 0
in sin' '
sin sin2
ajkx jkx
a
e dx e dxkb
+
( )( )
2
0
2
0
sin 2sin2 cos cos2
sin 2sin2 sin2
XYL j E abY X
XYL j E abY X
=
= sin cos
2
sin sin2
xx ee dx
kaX
kbY
=
=
=
35
4
4
jkr
jkr
jkeE Lr
jkeE Lr
=
= +
36
Example (Problem 12.22 Balanis): A rectangular aperture with the dimensions a=/2 and b=/4 and is mounted on an infinite ground plane. What are the corresponding directivities (in dB) when the aperture has:
i) a triangular electric field distribution and an aperture efficiency of 75%, ii) a cosine-square electric field distribution and an aperture efficiency of 66%?
37
Example: A rectangular aperture, of dimensions a and b, is mounted on an infinite ground plane, as shown in the figure. Assuming the tangential field over the aperture given by:
0 2 ' 2 2 ' 2a zE a E a y a b z b= r
Find the far-zone spherical electric and magnetic field components radiated by the aperture.
38
Horn Antennas Horn antennas are popular in the microwave band (above 1 GHz). Horns provide high gain, low VSWR (with waveguide feeds),
relatively wide bandwidth, and they are not difficult to make. There are three basic types of rectangular horns.
39
The horns can be also flared exponentially. This provides better matching in a broad frequency band, but is technologically more difficult and expensive.
The rectangular horns are ideally suited for rectangular waveguide feeders.
The horn acts as a gradual transition from a waveguide mode to a free-space mode of the EM wave.
When the feeder is a cylindrical waveguide, the antenna is usually a conical horn.
40
Why is it necessary to consider the horns separately instead of applying the theory of waveguide aperture antennas directly to the aperture of the horn?
It is because the so-called phase error occurs due to the difference between the length from the center of the feeder to the center of the horn aperture and the length from the center of the feeder to the horn edge.
This complicates the analysis, and makes the uniform-phase aperture results invalid for the horn apertures.
41
E-Plane sectoral Horn
42
lE=eR0=R=1E=EB=b1
21
21
21
' 21
' 21
' 21
1
' ' ' 0
'( ', ') cos '
'( ', ') sin '
'( ', ') cos '
cos
z x y
j kyy
j kyz
j kyx
e e
E E H
E x y E x ea
H x y jE x eka a
EH x y x ea
= = = =
= =
=
( ')1
( ')1
cos '
cos '
jk yy
jk yy
EJ x ea
J E x ea
= =
43
( ) ( ) ( )
( ) ( ) ( )
21
21
21 11 22 2
21 11 22 2
cos2sin 1 cos ,
82 2
cos2cos 1 cos ,
82 2
y
y
xjkr
j k k
x
xjkr
j k k
x
k aa k E e
E j e F t tr k a
k aa k E e
E j e F t tr k a
= + = +
( ) ( ) ( ) ( ) ( )1 2 2 1 2 11 1
1 1 2 11 1
sin cos sin sin
,
1 12 2
x y
y y
k k k k
F t t C t C t j S t S t
kb kbt k t kk k
= = =
= =
44
E-Plane sectoral Horn
E-Plane ( )2 =
( ) ( ) ( )21 2sin 21 1 ' '1 2' 11 1
1
' 12 1
1
0
2 1 cos ,8
sin2
sin2
r
jkr j k
E E
a k E eE j e F t t
r
bkt
bkt
= =
= + = = +
45
H-Plane ( )0 =
( ) ( )1 1 " "1 22 2" 11
1
" 12
1
0
cos sin21 cos ,
8sin
2 2
2
2
r
jkr
E E
kaa k E e
E j F t tr ka
bkt
bkt
= = = +
= = +
46
47
48
The magnitude of the normalized pattern, excluding the factor , can be written as:( )1 cos+
( ) ( ) ( ) ( ) ( )' ' ' ' ' '1 2 2 1 2 1' 11 1
1
21 1 1
21 1
21 1
1
,
sin2
812 1 sin8 4
1 12 1 sin4 8
nE F t t C t C t j S t S t
bkt
b bb
b bs ss
= = =
= = =
49
' 12 1
1
21 1 1
21 1
1
sin2
812 1 sin8 4
1 12 1 sin4
bkt
b bb
bss
= =
=
For a given value of , the normalized field can be plottedas a function of .
These plots are usually referred to as universal curves.
( )1 sinb s
50
From the universal curves, the E-plane pattern of any E-plane sectoral horn can be obtained.
First determine the value of . For that value of , the field strength (in dB) as a function of is obtained from the following figure. Finally, the value of , normalized to 0 dB and written as , is added to that number to get the required field strength.
s s( )1 sinb ( )1 cos+( )1020log 1 cos 2 +
51
52
Directivity of E-plane sectoral Horn:
2 2max 1 1 1
1 1 1
4 642 2E rad
U a b bD C SP b
= = +
53
If the values of , which correspond to the maximum directivities are plotted versus their corresponding values of , it can be shown that each optimum directivity occurs when
The corresponding value of is:
1b
1
1 12b b
s
14
s =
54
H-Plane Sectoral Horn
55
Aperture Fields:
( ) ( )
( ) ( )
( )
' '
''2
1
'' 2
1
2
22
0
' cos '
' cos '
1 '' cos2
x y
jk xy
jk xx
h h
E H
E x E x ea
EH x x ea
xx
= = = =
= =
Note: The aperture fields for E-plane sectoral hornwere functions of x and y, whereas for H-planesectoral horn, they are functions of x only
56
Radiated Fields: Over the aperture the equivalent current densities are
( )
( )
'2
1
'2
1
0
cos '
cos '
x z y z
jk xy
jk xx
J J M M
EJ x ea
M E x ea
= = = = = =
Note: The equivalent current densities for E-planesectoral horn and H-plane sectoral horns are sameexcept the complex exponential term and the constant.
57
Radiated Fields
'coscos cos cos sin sin 'jkrx y zS
N J J J e ds + = + From the radiation equations for aperture, we have
So, in the present case this becomes,
'coscos sin 'jkryS
N J e ds + = For aperture in the XY plane,
'cos 'sin cos 'sin sin' ' '
r x yds dx dy
= +=
58
Hence,
( )
21 2
2'sin sin
12
2' 'sin cos
22
cos sin
sin sin sin2'sin sin
2
cos ' '
bjky
b
ajk x x
a
EN I I
kb
I e dy bkb
I x e dxa
++
+
= = =
=
59
( ) ( )
( ) ( ) ( ) ( ) ( )
1 222 1 2 1 2
1 2 2 1 2 1
2 22 2
1 2
1 11 2 2 2
2 2
11 2
2
cos sin sin ', ' ", "2
,
' "sin sin
2 2 21 1' ' ' '
2 2
1" "2
jf jf
x x
x x
x
b YN E e F t t e F t tk Y
F t t C t C t j S t S t
k k kbf f Yk k
ka kat k t kk k
kat kk
= +
= = = =
= = + = 12 2
2
1" "2
' sin cos " sin cos
x
x x
kat kk
k k k ka a
= + = + =
Finally,
60
( ) ( )( ) ( )( ) ( )
1 2
1 2
1 2
22 1 2 1 2
22 1 2 1 2
22 1 2 1 2
22
cos sin sin ', ' ", "2
cos sin ', ' ", "2
cos cos sin ', ' ", "2
sin2
jf jf
jf jf
jf jf
b YN E e F t t e F t tk Y
b YN E e F t t e F t tk Y
b YL E e F t t e F t tk Y
bL Ek
= + = + = + +
= ( ) ( )1 21 2 1 2sin ', ' ", "jf jfY e F t t e F t tY +
Similarly can be found., ,N L L
61
The far-zone electric field components can then be found out:
( ) ( )0 4 4jkr jkr
rke keE E j L N E j L N
r r
= + = +
( ) ( ) ( )
( ) ( ) ( )
1 2
1 2
22
1 2 1 2
22
1 2 1 2
0
8sinsin 1 cos ', ' ", "
8sincos 1 cos ', ' ", "
rjkr
jf jf
jkr
jf jf
E
kb eE jEr
Y e F t t e F t tY
kb eE jEr
Y e F t t e F t tY
==
+ + =
+ +
62
The Electric Field in the E-Plane (y-z plane)2 =
( ) ( ) ( )1 22
2
1 2 1 2
1
1
0
8sin1 cos ', ' ", "
sin2
'
"
rjkr
jf jf
x
x
E E
kb eE jErY e F t t e F t t
YkbY
ka
ka
= =
= + +
=
=
=
63
The Electric Field in the H-Plane (x-z plane)0 =
( ) ( ) ( ){ }1 22
2
1 2 1 2
1
1
0
8
1 cos ', ' ", "
' sin
" sin
rjkr
jf jf
x
x
E E
kb eE jEr
e F t t e F t t
k ka
k ka
= ==
+ + = +
=
64
65
The Directivity of a H-plane sectoral Horn
( ) ( ) ( ) ( ){ }2 221
2 1
1 2
2 1
1 2
4
12
12
HbD C u C v S u S va
aua
ava
= + = + =
The Directivity is optimum when
1 23a =
66
Pyramidal Horn
Most widely used horn.
Flared in both directions.
Radiation characteristics are combination of the E-and H-plane sectoral horns.
67
The tangential components of the E- and H-fields over the aperture of the horn are approximated by:
( ) ( )
( ) ( )
2 22 1
2 22 1
' ' 20
1
' ' 20
1
' ', ' cos '
' ', ' cos '
j k x yy
j k x yx
E x y E x ea
EH x y x ea
+
+
= =
The equivalent current densities (approximated):
( ) ( )
( ) ( )
2 22 1
2 22 1
' ' 20
1
' ' 20
1
', ' cos '
', ' cos '
j k x yy
j k x yx
EJ x y x ea
M x y E x ea
+
+
= =
We have cosinusoidal amplitude distribution in the x direction and quadratic phase variations in both x and y directions.
68
One can derive the following expressions from the current densities:
0 01 2 1 2
0 1 2 0 1 2
cos sin cos
cos cos sin
E EN I I N I I
L E I I L E I I
= = = = ( )
( ) ( ) ( ) ( ) ( ){ }( ) ( ) ( ) ( ) ( ){ }
22
22
22
' 2 'sin cos21 2
' 222 1 2 1
" 22 1 2 1
cos ' '
1 ' ' ' '2
" " " "
x
x
jk x xa
a
j k k
j k k
I x e dxa
e C t C t j S t S tk
e C t C t j S t S t
+
= =
+
( )
( ) ( ) ( ) ( ) ( ){ }2
1
21
' 2 'sin sin22 2
212 1 2 1
'
y
jk y yb
b
j k k
I e dy
e C t C t j S t S tk
+ =
=
69
Far-zone E-field components are:
( )( )
01 2
01 2
0
sin 1 cos4 4
cos 1 cos4 4
rjkrjkr
jkrjkr
E
kE ekeE j L N j I Ir r
kE ekeE j L N j I Ir r
= = + = + = + + = +
The principal E-plane pattern of a pyramidal horn, aside from a normalization factor, is identical to the E-plane pattern of an E-plane sectoral horn.
The principal H-plane pattern of a pyramidal horn, aside from a normalization factor, is identical to the H-plane pattern of an H-plane sectoral horn.
Hence, the pattern is very narrow in both the principal planes.
( )2 =
( )0 =
70
3-D Field Pattern of a Pyramidal Horn
The maximum radiation isnot necessarily directedalong its axis, becausethe phase error taper atthe aperture is such thatthe rays coming fromdifferent parts of theaperture towards the axisare not in phase and donot add constructively.
71
To physically construct a pyramidal horn, the dimension and should be equal:ep hp
( )
( )
1 22
11
1 22
11
14
14
ee
hh
p b bb
p a aa
= =
The method outlined above gives accurate patterns forangular regions near the main lobe and its closest minorlobes. The accurately predict the field intensity in the minorlobes, diffraction techniques (accounting for diffraction nearthe aperture edges) are utilized. Diffraction becomesdominant in regions where the radiation is of very lowintensity.
72
Co- and Cross-Polarization
The patterns that are plotted represent the main polarization of the field radiated by the antenna. This is called co-polarized pattern.
Ideally the radiated field has no orthogonal (to the main polarization) component (cross-polarized), provided the antenna is symmetrical and is excited in the dominant mode.
In practice, because of nonsymmetries, defects in construction or excitation of higher order modes, all antennas have cross-polarized components (usually of very low intensity).
For good designs, cross-polarized components should be at least 30 dB below the co-polarized.
73
Directivity of a Pyramidal Horn
( ) ( ) ( ) ( ){ }
2
2 21 1 1
1 1 1
2 22
1
2 1
1 2
2 1
1 2
32
642 2
4
12
12
p E H
E
H
D D Dab
a b bD C Sb
bD C u C v S u S va
aua
ava
= = +
= + = + =
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