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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