Aperture Radiation: Huygen’s Equation
E = x̂E x
Radiating patch, assume
uniform plane wave:
E⊥H, E/H = 377Ω
y
x
ˆx
α
Superposition of contributions
from radiating patches
R
ϕx
θ z
ϕy
r(x, y)
Aperture in X-Y plane “A”
E⎫
⎬ → JS surface current → Eff radiated by integral
H⎭
of current elements
Huygen’s superposition integral
j
− j( 2π λ )r ( x,y )
Eeff ( θ, φ,R ) ≅
dxdy
(1 + cos θ )( −αˆ x ) ∫A E x ( x,y ) e
2Rλ
phase lag ( x,y )
R1
Huygen’s Equation: Geometric approximations
j
j(2λ π )r (x,y )
(1 + cos θ )(− αˆ x )∫A E x (x, y )e−
Eeff (θ, φ,R ) ≅
dxdy
2Rλ
phase lag (x,y )
x
2-D case
Aperture in x-y plane
r(x)
x
0
R
ϕx
x sin ϕx
z
For ϕ x , ϕ y << 1 : r (x, y ) ≅ R − x sin ϕ x − y sin ϕ y ≅ R − xϕ x − yϕ y
Thus E(θ, φ,R ) ≅
2π
-j R
je λ
(1 + cos θ )αˆ x • ∫A E x (x, y )e
R
λ 2
+j
(
2π
xϕ x + yϕ y
λ
)
dxdy
K
R2
Huygen’s Equation: Geometric approximations
2π
R
je λ
-j
Thus E(θ, φ,R ) ≅
(1 + cos θ )αˆ x • ∫A E x (x, y )e
R
λ 2
+j
(
)
2π
xϕ x + yϕ y
λ
dxdy
K
(
E(ϕ x , ϕ y ) vm
(
x̂E x (x, y ) vm
−1
) ≅ K ∫A Ex (x, y )e
−1
)
+j
(
2π
xϕ x + yϕ y
λ
x̂
≅ ∫2π E x (ϕ x , ϕ y )e
K
−j
(
)
dxdy
2π
xϕ x + yϕ y
λ
)
dϕ x dϕ y
R3
Huygen’s Equation: Geometric approximations
(
E(ϕ x , ϕ y ) vm
−1
) ≅ K ∫A Ex (x, y )e
+j
(
2π
xϕ x + yϕ y
λ
)
dxdy
(
x̂
x̂E x (x, y ) vm ≅ ∫ E x (ϕ x , ϕ y )e
K 2π
x
y
∆
Let x λ
, y λ = , 1 + cos θ ≅ 2
λ
λ
(
(
−1
)
−j
2π
xϕ x + yϕ y
λ
(
)
dϕ x dϕ y
)dx
) ≅ λ K ∫A Ex (xλ , yλ )e
λ dy λ
2
− j2π(x ϕ + y ϕ )
λ
x̂
(
)
ϕ
ϕ
x̂E(x λ , y λ )(vm −1) ≅
E
,
e
dϕ x dϕ y
∫2π x x y
K
E(ϕ x , ϕ y ) vm
−1
2
+ j2π x λ ϕ x + y λ ϕ y
λ x
λ y
This is a Fourier transform pair
[Recall X(f ) = ∫ x(t )e− j2πftdt ; x(t ) = ∫ X(f )e+ j2πftdf ]
R4
Fourier Transform Relations
Thus
Aperture
(pulse signal)
(
E( x, y ) Vm
(
−1
↓
)
)[
RE x τλ x , τλ y Vm
~
↔
]
-1 2
~
↔
[
]
E x (ϕ x , ϕ y ) Vm −1 at R
E x (ϕ x , ϕ y )
↓
2
[Vm ] at R
-1 2
7
∝ S(ϕ x , ϕ y ) =
E x (ϕ x , ϕ y )
2ηo
2
[W m-2 ]
R5
Directivity D(θ, φ) of an Aperture Antenna
Let P = radiation intensity and
PTR = total power radiated (W)
-2
(
)
P
θ
,
φ
,
f,
R
[
W
m
]
D(θ, φ ) ∆
PTR 4πR 2 [ W m- 2 ]
(ϕx , ϕy << 1)
E x (x, y )e
∫
A
(1 + cos θ )
D≅
2
2ηo (2Rλ )2
D=
π(1 + cos θ )2
λ2
j
(
2π
xϕ x + yϕ y
λ
)
2
dx dy
1
2
2
(
)
E
x
,
y
dx
dy
4
π
R
x
2ηo ∫A
∫A
(
)
2π
2
xϕ x + yϕ y
E x (x, y )e λ
dx dy
j
∫A
E x (x, y ) dx dy
2
R6
Directive Gain D(θ,φ) of an Aperture Antenna
D=
∫A E x (x, y )e
π(1 + cos θ )2
λ2
∫A
j
(
2π
xϕ x + yϕ y
λ
)
2
dx dy
E x (x, y ) dx dy
2
Bounds on D(ϕx,ϕy), A(ϕx,ϕy)
Recall “Schwartz Inequality”
(∫ f dx )(∫ g dx )
[ ] dx dy ≤ ( E dx dy ) ( 1
E
e
∫
∫
∫
2
∫ f g dx ≤
Therefore:
4π
2
λ
2
2
j
A x
D(ϕ x , ϕ y ) ≤
2
Ao(m2) is physical
area of aperture
x
∫
• A
Ex
∫A
2
2
dx dy • A o
Ex
2
dx dy
2
A
=
dx dy
)
4πA o
λ2
R7
Directive Gain D(θ, φ) of an Aperture Antenna
D(ϕ x , ϕ y ) ≤
4π
2
λ
∫
• A
Ex
∫A
2
dx dy • A o
Ex
2
dx dy
=
4πA o
λ2
4π A e (ϕ x , ϕ y )
•
⇒ A e (ϕ x , ϕ y ) (effective area) ≤ η A o
But D =
2
R
ηR
λ
where radiation efficiency ηR ≤ 1.0
Define “aperture efficiency” ηA
ηA
A e (max )
∆
≅ 0.65 in practice; = 1 for uniform illumination
ηR Ao
Therefore A e = η A • ηR A o
R8
Uniformly Illuminated Circular Aperture Antennas
x
φx
θ
φ’
y
ϕy
z
D=
π(1 + cos θ )2
λ2
r
∫A 1• e
Aperture coordinates = r, φ′
Source coordinates = ϕx, ϕy for θ << 1
D(f, θ, φ ) =
(
2π
xϕ x + yϕ y
λ
∫A
0
–3 dB
( )
πD
where
λ
Λ1( q) = J1(q) q
r dr dφ'
2
E r dr dφ'
–17.6 dB
–28.8 dB
0 1.63 4 7 10
side lobes
(
πD
(1 + cos θ ) ⎤ Λ12
sin θ
⎢⎣ λ
⎥⎦
λ
=
)
P=0
2
⎡ πD
j
πD
sin θ
λ
)
“Lambda function”
2
= 4 πA o λ
2
at θ = 0
“Bessel function of first kind”
T1
Non-Uniformly Illuminated Circular Apertures
( )
⎡
2r ⎤
Assume E x ( r ) = ⎢1 −
⎥
D
⎣
⎦
Ex(r)
2
P
P=0
P=1
P=2
0
G(θ)
D/2
θ
More
typical
P
θB1/2
θNULL #1
First Side - Lobe
ηA
0
1
3
1.02 λ/D
1.27 λ/D
1.47 λ/D
1.22 λ/D
1.63 λ/D
2.93 λ/D
17.6 dB
24.6 dB
30.4 dB
1.00
0.75
0.56
T2
Sidelobes and Backlobes of Aperture Antennas
Spillover
Main lobe
Feed
Sidelobes
Backlobes
Diffraction
Backlobes
Reflector
T3
Waveguide Horn Feeds
Pyramidal Horn
y
φy
D
Ey(y)
⇒
φx
Ey(x)
Dominant
Waveguide ⇒
Mode TE10
0
Null at ϕ = λ/D
G(ϕy)
High Sidelobes
(∼17.6 dB)
G(ϕx)
Lower Sidelobes
(∼25 dB)
T4
“Scalar” Feed
y
Aperture
λ/4 minimizes
return echo
Ey(y)
∼λ/4 ⇒ open circuit
at wall
Yields very
low sidelobes
Side view
λ/4 grooves cut into wall
T5
Examples of Parabolic Reflector Antennas
Focus
No aperture blockage
Circularly Symmetric
Parabolic Reflector
“Off-Axis Paraboloid”
Lateral scan via
phased array line feed
Cylindrical Parabola
T6
Spherical Reflector Antennas
Focus at ≥ R/2 R/2
Center of curvature
Focal plane
Feed
Support
Variable-pitch
linear phased
array, positioned
at line focus
Beam
Line Feed
Illuminated Portion
e.g. Aricebo (1000’ = D, 600’ is illuminated)
T7
Toroidal Parabolic Reflector Antenna
Spinning Feeds
Axis of Revolution
Direction
of View
z
Spin
r(z)
ω
Antenna Feeds
Feed Surface
(No Aperture Blockage)
Toroidal
Reflector
z
Toroidal Reflector
Toroidal Reflector
Advantage: many rapidly scanning spinning feeds
T8
Multifeed Arrays
θ
C
B
Parabola
GA(θ)
D
Focus “A”
x
y
GB(θ)
GC(θ)
z
∆
Focal length = f
For f D = 0.5, n ≅ 3 - 5 beams with useable Go and sidelobes
(say ~1 dB gain loss)
η ∝ (f D ) in x - direction
2
e.g. if f D = 7, n x ≅ (7 0.5 ) • 5 ≅ 1000
2
Can do much better with good lens systems
U1
“Scalar” Feed
y
Aperture
λ/4 minimizes
return echo
Ey(y)
∼λ/4 ⇒ open circuit
at wall
Yields very
low sidelobes
Side view
λ/4 grooves cut into wall
U2
Multiple-Horn Feeds
Parabolic
Reflector
Adjacent Feeds
B
A
Cross-over Point
below 3 dB, Far-Field
A
B
Fourier Transform
Focal Plane
Thus
Cross-over ≤ 6 – 10 dB
for Low Sidelobes
Ey(y)
Scalar Feeds
U3
Multiple-Horn Feeds
Patterns
1
1
2
3
4
Feeds
A
2
3
Three-Array Solution
4
B
C
A
C
A
B
A
B
C
C
A
B
Poor Coverage
Feed A′ is assembly of
excited adjacent feeds
U4
“Near-Field” Antenna Coupling
Near-field of aperture >> near field of
Hertzian dipole (r << λ/2π)
λ
•r ≅ D
D
r
D2
r≅
λ
Uniform Phase Front
z
Spherical
Phase Front
2
2D
" Far field" ⇒ r ~
>
λ
U5
“Near-Field” Antenna Coupling
Consider near-field link:
Say: uniformly illuminated apertures
Ao
PT1
Ao/3
PT2
Eo(vm-1)
2
D
r <<
λ
PT =
1
Pr
2
=
Eo
2
2ηo
Eo
• A o watts
2
2ηo
•
Claim Pr =
1
Ao
3
PT
2
3
=
PT
1,
3
=
PT
2
Eo
2
•
2ηo
Ao
3
Pr = ?
(reciprocity) i.e.
1
Pr
2
PT
1
=
1
3
=
Pr
1
PT
2
U6
“Near-Field” Antenna Coupling, Mode Orthogonality
y
y
Ey(y)
Ey(y) received
Illuminate only half =
0
Eo
Eo
2
2ηo
•
Ao
2
[W ]
Half
gets in
(Eo 2)2 A o
2ηo
Eo
Eo
2
2
+
0
Half reflected,
orthogonal to
dominant
(coupled) mode
Only half the power is accepted here!
Waves are not a sum of independent “bullets;” they have
phase, modal structure (classic wave/particle issue).
U7