APERTURE ANTENNAS
Dense phased arrays ~apertures:
Pairs of radiators cancel (e.g., a,b)
Sum of pairs therefore = 0
First null at θ = ~λ/D
Aperture antennas:
Parabolic
reflector
Aperture
screen
Array of 14
elements
D
a cancel
b θ
null
λ/2
ẑ
Phased
array
Derivation of far-field aperture radiation:
Find surface current Js that could produce the same aperture fields
Find the far fields radiated by an infinitesimal element of Js
Integrate the contributions from the entire aperture
Simplify the expressions by using small-angle approximations
Note the Fourier relationship between aperture and far fields
L20-1
APERTURE ANTENNAS
Assume UPW in aperture:
E+ = ẑ Eo e
E- = ẑ Eo e
-jky
+jky
⇒ H+ = xˆ
⇒ H- = -xˆ
Eo
ηo
Eo
ηo
Satisfies E// continuous at y = 0
e
z
-jky
e
+jky
Lz
x
Equivalent surface current:
E
⎯Js
q
θ
H
aperture A
rpq p
r̂
E
αz
αx
y
Lx
Js = ŷ × [H(y=0+ ) - H(y=0- )]
= ŷ × x[E
ˆ o /ηo + Eo /ηo ]
Js = -ẑ 2Eo /ηo [A m-1]
ẑ I dz = Jsdx dz (Hertzian dipole radiator)
Radiated far fields:
-jkr
jηo
E (θ,φ) = θ̂
sinθ ∫∫A Jzs (x,z) e pq dx dz
ff
2λr
-jkr
j
= -θ̂
sinθ ∫∫A Eoz (x,z) e pq dx dz
λr
L20-2
FAR-FIELD RADIATION
Assume UPW in aperture:
-jkr
j
E (θ,φ) = - θ̂
sinθ ∫∫A Eoz (x,z)e pq dx dz
ff
λr
e
− jkr
≅e
− jkr + jkr̂ i r
o
q
Fraunhofer
approximation
(else, Fresnel)
z
z
rpq
q
rq
0
ro
r̂
y
rˆi rq
Lz
x
E
⎯Js
q
θ
H
aperture A
rpq p
r̂ E
αx
αz
y
Lx
j − jkro
+j2π(xα +zα )
ˆ
Eff (αx ,αz ) ≅ −θ e
∫∫ A Eoz (x,z)e λ x z dx dz
λr
Close to the y axis
~Fourier (angle ⇔ space)
Fourier relationship:
∞
S(f ) = ∫−∞
s(t)e− j2πft dt
∞
s( t ) = ∫−∞
S(f )e+ j2πft df (frequency ⇔ time)
L20-3
EXAMPLE: RECTANGULAR APERTURE
Assume UPW in aperture:
(Observer close to the y axis)
+j2πα x x
j − jkro Lz /2 +j2παz λz Lx/2
λ
Eff (αx , αz ) ≅ −θ̂ e
dx dz
∫− Lz / 2 e
∫− Lx /2 Eoz (x,z)e
λr
πα L
L x /2
∫−Lx / 2 Eoz (x,z)e
+j2παx x
λ
dx = Eoz
λ (e
j2πα x
L
+jπα x x
λ
-e
L
-jπαx x
λ
) = EozL x
sin
x
x
λ
πα x L x
λ
πα xL x
*
λ
− jkr
πα L
πα L
j
Eff (αx , αz ) ≅ − θ̂ Eoz e o L xL z sinc x x sinc z z
λr
λ
λ
EozL x sinc
G(αx,αz) ∝ |Eff|2 ∝ sinc2(παxLx/λ)sinc2(παzLz/λ)
αz
aperture A
Null at αz = λ/Lz
First
Antenna Gain:
z
q
Lz
E
θ
x
r̂
rpq
p
sidelobe
αz
α y
G(αx,αz)
x
Lx
αx
*sinc θ ≡ (sin θ)/θ [= 1 for θ = 0]
Null at αx = λ/Lx
L20-4
RECTANGULAR APERTURE (2)
Antenna Gain G(αx,αz):
G(α x , α z ) =
β
I(α x , α z ,r)
Pt [W]
(Pt / 4πr 2 )
|Eoz |2
Pt = A
[W]
2ηo
(A=L xL z )
A [m2]
Go
G(αx,αz)
I(αx,αz,r) [W m-2]
|Eff |2
πα L
πα L
[W/m2 ] = 1 ( 1 )2 |Eoz |2 A 2sinc 2 ( x x )sinc 2 ( z z )
I(α x , α z ,r) =
2ηo
2ηo λr
λ
λ
G(α x ,α z ) = A(4π/λ 2 )sinc 2 (πα xL x /λ)sinc 2 (πα zL z /λ)
2 *
4π
λ
Go =A
⇒ A = A e = Go
2
4π
λ
[ sinθ → 1 as θ → 0]
θ
ηA = “aperture efficiency” ≡ Ae/A ≈ 0.65 typically
(non-uniform aperture amplitude and/or phase ⇒ Ae < A)
Huygen’s approximation:
-jkr
Eff ≅ θˆ (j/2λr) (1+cosβ) ∫∫ A Eoz (x,z)e
dx dz
*True for any uniformly illuminated aperture
pq
L20-5
EXAMPLE: SLIT DIFFRACTION
Pair of rectangular slots:
Uniform illumination
Spatial
domain
δ(x,z) δ(x,z-D)
*
L
D
z=0
=
z
D
Fourier
transforms
λ/D
Angular
domain
=
0
λ/L
L20-6
DIFFRACTION
Aperture tapering: Eff ≅ θˆ (j/2λr) (1+cosβ) ∫∫ A Eoz (x,z)e-jkr dx dz
pq
z
Parabolic
reflector
θ
E(z)
Tapered illumination reduces
sidelobes,broadens beamwidth
G(θ)
⇔
Fourier
blocked for
Fresnel zone plate
Fresnel zone plate:
Blocks rings of negative phase,
maximizing Huygen’s integral
Aperture radius = r1 maximizes
received signal. Additional
transparent rings increase it
further, yielding lense behavior
Adjacent rings cancel if β ≈ 1
r1
r = (r12 + ro2)0.5 = ro + λ/2
ro
receiver
r = ro + λ
Fresnel
zone
transparent rings
L20-7
Negative
contribution
r1
Incident
UPW
r = r0 + λ/2
r0
r = r0 + λ
Fresnel zone Positive contribution
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phone
MIT OpenCourseWare
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6.013 Electromagnetics and Applications
Spring 2009
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