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School of Electrical Engineering
Lecture 4: Aperture antennas
Aperture antennas
KTH School of Electrical Engineering • www.ee.kth.se 2
• They are antennas in which we know the distribution of the fields in asurface (typically flat).
• They are typically employed for high power applications, and spaceapplications.
Guided waves
Radiation
Aperture
Horn antennas
KTH School of Electrical Engineering • www.ee.kth.se 3
• To increase the directivity of the radiated antenna,normally, a horn is employed.
Horn antennas
KTH School of Electrical Engineering • www.ee.kth.se 4
• Electric field in a horn antenna:
- Transition between waveguide and horn.
Side view Top view
TE10 mode
Equivalence theory
KTH School of Electrical Engineering • www.ee.kth.se 5
Our Problem Equivalent problem
Equivalence principle models
KTH School of Electrical Engineering • www.ee.kth.se 6
Love’s equivalent Electric conductor equivalent
Magnetic conductor
equivalent
][ˆ
][ˆ
1
1
EEnM
HHnJ
S
S
Theory
KTH School of Electrical Engineering • www.ee.kth.se 7
• The radiation can be derived from:
1. Magnetic potential vector:
2. Electric potential vector:
'
' '4
S
jkR
S dSR
eJA
'4
'
'
S
jkR
S dSR
eMF
Theory
KTH School of Electrical Engineering • www.ee.kth.se 8
• In far-field: cos'rrR Phase
rR Amplitude
Nr
edSeJ
r
edS
R
eJA
jkr
S
jkr
S
jkr
S
jkR
S
4'
4'
4'
cos'
'
Lr
edSeM
r
edS
R
eMF
jkr
S
jkr
S
jkr
S
jkR
S
4'
4'
4'
cos'
'
''
cos'
S
jkr
S dSeML
'
cos' 'S
jkr
S dSeJN
Theory
KTH School of Electrical Engineering • www.ee.kth.se 9
• Again, since we are in the far field, we will have onlycomponents θ and ϕ:
FjE
FjE
AjE
AjE
F
F
A
A
FjH
FjH
AjH
AjH
F
F
A
A
FHj
FAjAjEEE AFA
1111
FFA Ej
AFjFjAHHH
1111
Fields
KTH School of Electrical Engineering • www.ee.kth.se 10
• The fields will be:
jkr
jkr
r
eL
Nr
kjH
eL
Nr
kjH
H
4
4
0
jkr
jkr
r
eNLr
kjE
eNLr
kjE
E
4
4
0Nr
eA
jkr
4
Lr
eF
jkr
4
FjE
FjE
AjE
AjE
F
F
A
A
FjH
FjH
AjH
AjH
F
F
A
A
Fields
KTH School of Electrical Engineering • www.ee.kth.se 11
• Where our N and L components will be obtained as:
'cossin
'sinsincoscoscos
'
cos'
'
cos'
S
jkr
yx
S
jkr
zyx
dSeMML
dSeMMML
'cossin
'sinsincoscoscos
'
cos'
'
cos'
S
jkr
yx
S
jkr
zyx
dSeJJN
dSeJJJN
Summary
KTH School of Electrical Engineering • www.ee.kth.se 12
1. To select a closed (imaginary) surface withknown and
2. To obtain L and N.
3. To obtain the fields: E and H.
4. To determine the radiated fields (far fields)
SM
SJ
Uniform distribution
KTH School of Electrical Engineering • www.ee.kth.se 13
• Lets assume a uniform distribution in the aperture over a ground plane:
yEEaˆ
0
2'
2
2'
2
by
b
ax
a
Elsewhere
by
b
ax
a
ExEyzEnM aperture
S
02
'2
2'
22ˆˆˆ2ˆ2 00
0SJ
1) M and S
Uniform distribution
KTH School of Electrical Engineering • www.ee.kth.se 14
Elsewhere
by
b
ax
a
ExEyzEnM aperture
S
02
'2
2'
22ˆˆˆ2ˆ2 00
0SJ
2) L and N0 NN
2/
2/
2/
2/
cos'
0
2/
2/
2/
2/
cos'
0
''sin2
''coscos2
b
b
a
a
jkr
b
b
a
a
jkr
dydxeEL
dydxeEL
'cossin
'sinsincoscoscos
'
cos'
'
cos'
S
jkr
yx
S
jkr
zyx
dSeMML
dSeMMML
Uniform distribution
KTH School of Electrical Engineering • www.ee.kth.se 15
2) L and N
2/
2/
2/
2/
cos'
0
2/
2/
2/
2/
cos'
0
''sin2
''coscos2
b
b
a
a
jkr
b
b
a
a
jkr
dydxeEL
dydxeEL
2/
2/
2/
2/
)sinsin'cossin'(
0
2/
2/
2/
2/
)sinsin'cossin'(
0
''sin2
''coscos2
b
b
a
a
yxjk
b
b
a
a
yxjk
dydxeEL
dydxeEL
2
2sin
'
2/
2/
de j
sinsinsin2
sinsincoscos2
0
0
abEL
abEL
sinsin2
cossin2
kb
ka
sinsin'cossin'
)ˆcosˆsinsinˆcos(sin)ˆ'ˆ'(ˆ'cos'
yx
zyxyyxxrrr
Uniform distribution
KTH School of Electrical Engineering • www.ee.kth.se 16
3) E and H
jkrjkr
jkrjkr
r
er
abkEje
L
r
kjH
er
abkEje
L
r
kjH
H
sinsin
2
sin
4
sinsin
2
coscos
4
0
0
0
jkrjkr
jkrjkr
r
er
abkEjeL
r
kjE
er
abkEjeL
r
kjE
E
sinsin
2
coscos
4
sinsin
2
sin
4
0
0
0
0 NN
Uniform distribution
KTH School of Electrical Engineering • www.ee.kth.se 17
4) Far field (E-plane)
jkr
r
ekb
kb
Er
abkjH
H
H
sin2
sin2
sin
2
0
0
0
0
sin2
sin2
sin
2
0
0
E
ekb
kb
Er
abkjE
E
jkr
r
2/
4) Far field (H-plane) 0
jkr
r
eka
ka
Er
abkjE
E
E
sin2
sin2
sin
cos2
0
0
0
0
sin2
sin2
sin
cos2
0
0
H
eka
ka
Er
abkjH
H
jkr
r
TEM Mode:
transversal electric
and magnetic
H
EZwave
H
EZwave
H
ErZwave
ˆ
Uniform distribution: Radiation pattern
KTH School of Electrical Engineering • www.ee.kth.se 18
abAD aperture 22max
44
Other distributions
KTH School of Electrical Engineering • www.ee.kth.se 19
Other shapes: Circular
KTH School of Electrical Engineering • www.ee.kth.se 20
Horns
KTH School of Electrical Engineering • www.ee.kth.se 21
• The are normally used to increase the directivity of an aperture.
• The are fed with a waveguide.
• Normally employed as feeding of reflectors.
waveguide horn
A
L
tan α = A/(2L)α
Horns
KTH School of Electrical Engineering • www.ee.kth.se 22
• Lets assume a TE10 mode in the waveguide.
• The amplitude in the horn will be the same, but wewill have a phase error between the center and thecorners.
ljk
waveguidehorn eEE
Error in the phase-Constant in E-plane
-Cosine in H-plane
yxa
EEwaveguideˆcos1
yx
Horns: Phase error
KTH School of Electrical Engineering • www.ee.kth.se 23
waveguide horn
A
Ltan α = A/(2L)
α
∆l
222][ yLlL
The phase error:
L
Al
8
2
max
Normally,
given in rad: ][max radlk
The maximum error:
L
yLyLl
2
222
y
yexa
EE ljk
hornˆcos1
Horns: Types
KTH School of Electrical Engineering • www.ee.kth.se 24
E-planeH-plane
Pyramidal Conical
yexa
EEL
yjk
hornˆcos
2
1
2
yexa
EEL
xjk
hornˆcos
2
1
2
yeexa
EEL
yjk
L
xjk
hornˆcos
22
1
22
Corrugated horns
KTH School of Electrical Engineering • www.ee.kth.se 25
• The waveguide presentscorrugations.
• These corrugations areemployed to obtain a moresymmetric radiation pattern,to reduce the diffraction inthe borders and x-polarization.
Homework
KTH School of Electrical Engineering • www.ee.kth.se 26
• We have a X-band waveguide, and we are operatingat 10GHz.
• We would like to design a E-plane sectoral horn inwhich our maximum phase error is 50º. The increaseof the size will be to the double in E-plane.
• Calculate the required flare angle α of our horn.