ELEC-E4450 Antennas
2017
Exercise 8
18.4.2017
1. Three antenna array configurations with isotropic elements are illustrated below. Design
all arrays (feeds and element spacing) such that 1) their main beam is aligned with the positive z-axis direction and 2) their array patterns have nulls in order to suppress jamming
signals arriving from angles θ1 = 110◦ and θ2 = 150◦ . Plot the array factors for all three
designs.
a)
b)
d
a0
d
a1
a2
a0
a1
c)
jammers
a2
a2
a4
x
a3
dx
a0
a3
a1
dz
This is double problem i.e. worth of 10 exercise points.
1
2
z
Solution a)
Three elements: 5 free parameters – two complex feeding amplitudes (the first element is the
reference for amplitudes and phases) and element spacing d. Design goals are two nulls in the
complex radiation pattern i.e. four conditions. The Schelkunoff polynomial method is used in
this design. The element spacing is optimized to get maximum towards θ = 0◦ .
f (ψ) = e jψ − e jψ1
e jψ − e jψ2 = e j (ψ1 +ψ2 ) − e jψ1 + e jψ2 e jψ + e2 jψ
• ψ = kd cos θ i.e. δ = 0
• Null towards θ = 110◦
=⇒
ψ1 = kd cos 110◦
• Null towards θ = 150◦
=⇒
ψ2 = kd cos 150◦
• Optimize kd such that | f (ψ)| is maximized when θ = 0◦ =⇒ kd = 1.91 i.e.
d = 0.304λ
• The element excitations are: a0 = −(0.67 + j74), a1 = −(0.71 − j1.60) and a2 = 1.
0.6
0.4
0.2
0.2
-0.2
-0.4
-0.6
0.4
0.6
0.8
1.0
Solution b)
The Schelkunoff polynomial method with progressive phase shift is used. Two nulls in the radiation pattern implies, in general, two conjugate pairs of zeros for the polynomial representing
the array factor. Hence, fourth order polynomial with five coefficients are expected. (If a zero is
real, this decreases the order of the polynomial)
• ψ = kd cos θ + δ where δ = −kd to get main lobe towards θ = 0◦
• Null towards θ = 110◦
=⇒
ψ1 = kd cos 110◦
• Null towards θ = 150◦
=⇒
ψ2 = kd cos 150◦
• Optimize kd such that | f (ψ)| is maximized when θ = 0◦ =⇒ kd = 1.91 i.e.
d = 0.304λ
• The element excitations are: a0 = 1.0, a1 = 3.5, a2 = 5.1, a3 = 3.5 and a4 = 1.
0.6
0.4
0.2
0.2
-0.2
-0.4
-0.6
0.4
0.6
0.8
1.0
Solution c)
The design is based on binomial arrays.
Let’s first design a two-element binomial array in z-direction, which is endfire and have a null
towards θ = 150◦ . The array factor is
f 1 (ψ) = 1 + e
jψz
The null must occur at θ = 150◦ =⇒ kdz =
= 1+e
j (kdz cos θ −kdz )
−π
= 1.68 i.e. dz = 0.268
cos 150◦ − 1
The second binomial array is oriented vertically i.e. along x-axis. The array factor is
f 2 (ψ) = 1 + e jψx = 1 + e jkdx cos(θ−π/2)+ jδ
Making the array a broadside type, the maximum radiation directs to θ = 0◦ . Hence δ = 0.
The element spacing is obtained from the condition for the null: kdx cos(110◦ − 90◦ ) = π i.e.
kdx = 3.34 and dx = 0.53
0.4
0.2
=
0.2
-0.2
0.4
0.6
0.8
1.0
X
-0.2
-0.4
a2
a3
dx
a0
a1
dz
dz
dx
a0
a1
a2
a3
= 0.27λ
= 0.53λ
= 1.0 ∠0
= 1.0 ∠ − 1.68
= 1.0 ∠0
= 1.0 ∠ − 1.68