Course: BEng (Hons) in Electrical and Electronic Engineering

1. EEEC6430310 ELECTROMAGNETIC FIELDS AND WAVES
Dipole Arrays
FACULTY OF ENGINEERING AND COMPUTER TECHNOLOGY
BENG (HONS) IN ELECTRICALAND ELECTRONIC ENGINEERING
Ravandran Muttiah BEng (Hons) MSc MIET

2. 1
Two Element Array
Two Element Array in 𝜃 =
π
2
plane (𝑥 − 𝑦 plane)
𝑎 𝑎
𝑦
𝑥
𝜙
𝜙
𝑟1
𝑟2
𝑟
𝐼1
𝐼2
Figure 1

3. 2
Far field 𝑘𝑟 ≫ 1, 𝑟 ≫ 𝑎
𝐸𝜃 𝑟, 𝜃 =
π
2
, 𝜙 =
𝐸1
j𝑘𝑟1
e−j𝑘𝑟1 +
𝐸2
j𝑘𝑟2
e−j𝑘𝑟2 = 𝜂𝐻𝜙 𝑟, 𝜃 =
π
2
, 𝜙
𝐸1 = −
𝐼1𝑑𝑙𝑘2
𝜂
4π
𝐸2 = −
𝐼2𝑑𝑙𝑘2
𝜂
4π
𝑟2 ≈ 𝑟 + 𝑎 cos 𝜙 , 𝑟1 ≈ 𝑟 − 𝑎 cos 𝜙
𝐸𝜃 𝑟, 𝜃 =
π
2
, 𝜙 = 𝜂𝐻𝜙 𝑟, 𝜃 =
π
2
, 𝜙
≈ −
𝑘2
𝜂𝑑𝑙
4πj𝑘𝑟
e−j𝑘𝑟
𝐼1e+j𝑘𝑎 cos 𝜙
+ 𝐼2e−j𝑘𝑎 cos 𝜙
Assume: 𝐼1 = 𝐼, 𝐼2 = 𝐼ej𝜒
⇒ 𝐸1 = 𝐸0, 𝐸2 = 𝐸0 ej𝜒
Element factor Array factor

4. 3
𝐸𝜃 𝑟, 𝜃 =
π
2
, 𝜙 = 𝜂𝐻𝜙 𝑟, 𝜃 =
π
2
, 𝜙
=
𝐸0
j𝑘𝑟
e−j𝑘𝑟
e+j𝑘𝑎 cos 𝜙
+ ej𝜒
e−j𝑘𝑎 cos 𝜙
=
𝐸0
j𝑘𝑟
e−j𝑘𝑟
ej
𝜒
2 e
−j
𝜒
2
−𝑘𝑎 cos 𝜙
+ e
j
𝜒
2
−𝑘𝑎 cos 𝜙
=
2𝐸0
j𝑘𝑟
e−j𝑘𝑟
ej
𝜒
2 cos −
𝜒
2
+ 𝑘𝑎 cos 𝜙
𝑆𝑟 𝑡, 𝜃 =
π
2
, 𝜙 =
1
2
𝐸𝜃
2
𝜂
=
2 𝐸𝜃
2
𝜂 𝑘𝑟 2
cos2
𝑘𝑎 cos 𝜙 −
𝜒
2

5. 4
𝑥 𝑥 𝑥
𝑦 𝑦 𝑦
𝜙
𝑎
−𝑎 𝑎
−𝑎 𝑎
−𝑎
𝑆𝑟 ∝ cos2 π
2
cos 𝜙 , 𝜒 = 0 𝑆𝑟 ∝ cos2 π
2
cos 𝜙 − π
8
, 𝜒 = π
4
𝑆𝑟 ∝ cos2 π
2
cos 𝜙 − π
4
, 𝜒 = π
2
(a) (b) (c)
Broadside
2𝑎 =
𝜆
2

6. 5
𝑥 𝑥
𝑦 𝑦
𝑎 −𝑎 𝑎
−𝑎
(d) (e)
Endfire
Figure 2: The power radiation pattern due to two-point dipoles depends strongly on the
dipole spacing and current phases. With a half wavelength dipole spacing 2𝑎 = 𝜆
2
, the
radiation pattern is drawn for various values of current phase difference in the 𝜃 = π
2
plane.
The broadside array in (a) with the currents in phase 𝜒 = 0 has the power lobe in the
direction perpendicular to the array while the end-fire array in (e) has out-of-phase currents
𝜒 = π with the power lobe in the direction along the array.
𝑆𝑟 ∝ cos2 π
2
cos 𝜙 − 3π
8
, 𝜒 = 3π
4
𝑆𝑟 ∝ cos2 π
2
cos 𝜙 − π
2
, 𝜒 = π
−𝑎

7. 6
Broadside:
2𝑎 =
𝜆
2
, 𝜒 = 0, 2𝑎 =
π
2
𝑆𝑟 =
2 𝐸0
2
𝜂 𝑘𝑟 2
cos2
π
2
cos 𝜙
Endfire:
2𝑎 =
𝜆
2
, 𝜒 = π, 𝑘𝑎 =
π
2
𝑆𝑟 =
2 𝐸0
2
𝜂 𝑘𝑟 2
cos2
π
2
cos 𝜙 − 1

8. 7
Maxima:
𝑘𝑎 cos 𝜙 −
𝜒
2
= ±𝑚π, 𝑚 = 0, 1, 2, …
Minima:
𝑘𝑎 cos 𝜙 −
𝜒
2
= ± 2𝑚 + 1
π
2
, 𝑚 = 0, 1, 2, …
Case Studies:
2𝑎 =
𝜆
2
⇒ 𝑘𝑎 =
2π𝑎
𝜆
=
2π𝑎
4𝑎
=
π
2
2𝑎 = 𝜆 ⇒ 𝑘𝑎 = π
2𝑎 =
𝜆
2
⇒
π
2
cos 𝜙 −
𝜒
2
= ±𝑚π maxima ⇒ cos 𝜙 =
𝜒
π
± 2𝑚
π
2
cos 𝜙 −
𝜒
2
= ± 2𝑚 + 1
π
2
minima ⇒ cos 𝜙 =
𝜒
π
± 2𝑚 + 1

9. 8
𝜆 = 4𝑎
𝜒 cos 𝜙max cos 𝜙min 𝜙max 𝜙min
0 0 1 ±
π
2
0, π Broadside
π
4
1
4
−
3
4
±75.5°
±138.6°
π
2
1
2
−
1
2
±60°
±120°
3π
4
3
4
−
1
4
±41.4°
±104.5°
π 1 0 0, π ±90°
Endfire
𝜆 = 2𝑎 ⇒ cos 𝜙max =
𝜒
2π
± 𝑚, cos 𝜙min =
𝜒
2π
±
1
2
2𝑚 + 1
𝜒 cos 𝜙max cos 𝜙min 𝜙max 𝜙min
0 0, 1
1
2
, −
1
2
0, ±90°
, 180°
±60°
π
4
1
8
, −
7
8
−
3
8
,
5
8
82.8°
, 151°
51°
, 112°
π
2
1
4
, −
3
4
−
1
4
,
3
4
75.5°
, 138.6°
41.4°
, 104.5°
3π
4
3
8
, −
5
8
−
1
8
,
7
8
68.0°
, 128.7°
29.0°
, 97.2°
π
1
2
, −
1
2
0, 1 60°
, 120°
90°
, 0°

10. 9
𝑥 𝑥
𝑦 𝑦
𝑎 𝑎
−𝑎
(a) (b)
𝑆𝑟 ∝ cos2
π cos 𝜙 , 𝜒 = 0 𝑆𝑟 ∝ cos2
π cos 𝜙 − π
8
, 𝜒 = π
4
−𝑎
2𝑎 = 𝜆
𝜙

11. 10
𝑥
𝑦
𝑎
−𝑎
(c)
𝑥
𝑦
𝑎
−𝑎
(d)
𝑆𝑟 ∝ cos2
π cos 𝜙 − π
4
, 𝜒 = 0 𝑆𝑟 ∝ cos2
π cos 𝜙 − 3π
8
, 𝜒 = 3π
4
𝑥
𝑦
𝑎
−𝑎
(e)
𝑆𝑟 ∝ cos2
π cos 𝜙 − π
2
, 𝜒 = π
Figure 3: With a half wavelength dipole spacing 2𝑎 = 𝜆
2
, there are four main power lobes.

12. 11
An 𝑁 Dipole Array 𝜃 =
π
2
𝑥
𝑦
𝑧
𝜙
𝜃
𝑛 = −𝑁
𝑛 = −3
𝑛 = −2
𝑛 = −1
𝑛 = 𝑁
𝑛 = 3
𝑛 = 2
𝑛 = 1
𝑛 = 0
𝐼𝑁𝑑𝑙𝑁
𝐼3𝑑𝑙3
𝐼−3𝑑𝑙−3
𝐼−𝑁𝑑𝑙−𝑁
𝑎
𝑎 𝑟
𝑝
𝑟𝑛 = 𝑟2 + 𝑛𝑎 2 − 2𝑛𝑎 cos 𝜉
≈ 𝑟 − 𝑛𝑎 cos 𝜉
≈ 𝑟 − 𝑛𝑎 sin 𝜃 cos 𝜙
lim
𝑟≫𝑛𝑎
𝑟𝑛 ≈ 𝑟 − 𝑛𝑎 cos 𝜙 for −𝑁 ≤ 𝑛 ≤ 𝑁
𝜉
Figure 4: A linear point dipole array with 2𝑁 + 1 equally spaced dipoles.
𝑟𝑛

13. 12
𝐸𝜃 𝑟, 𝜃 =
π
2
, 𝜙 = 𝜂𝐻𝜙 𝑟, 𝜃 =
π
2
, 𝜙
=
𝑘𝜂𝑑𝑙
j4π𝑟
−𝑁
+𝑁
𝐼𝑛ej𝑘𝑛𝑎 cos 𝜙
e−j𝑘𝑟
Example:
𝐼𝑛 = 𝐼0e−j𝑛𝜒0 −𝑁 ≤ 𝑛 ≤ 𝑁
𝐴𝐹 = 𝐼0
−𝑁
+𝑁
ej𝑛 𝑘𝑎 cos 𝜙 −𝜒0
Let 𝛽 ≡ ej 𝑘𝑎 cos 𝜙 −𝜒0
𝑆 =
𝐴𝐹
𝐼0
=
−𝑁
+𝑁
𝛽𝑛
= 𝛽−𝑁
− 𝛽−𝑁+1
+ … + 𝛽−2
+ 𝛽−1
+ 1 + 𝛽 + 𝛽2
+ … + 𝛽𝑁−1
+ 𝛽𝑁
Array factor = 𝐴𝐹

14. 13
𝑆 1 − 𝛽 = 𝛽−𝑁
− 𝛽𝑁+1
𝑆 =
𝛽−𝑁
− 𝛽𝑁+1
1 − 𝛽
=
𝛽
−𝑁−
1
2−𝛽
𝑁+
1
2
𝛽
−
1
2−𝛽
1
2
=
sin 𝑁 +
1
2
𝑘𝑎 cos 𝜙 − 𝜒0
sin
1
2
𝑘𝑎 sin 𝜙 − 𝜒0
Maxima: 𝑘𝑎 cos 𝜙 − 𝜒0 = 2𝑛π, 𝑛 = 0, 1, 2, …
Principle maximum at , 𝑛 = 0 ⇒ cos 𝜙 =
𝜒0
𝑘𝑎
Minima: 𝑁 +
1
2
𝑘𝑎 cos 𝜙 − 𝜒0 = 𝑛π, 𝑛 = 1, 2, 3, …
Multiply by
𝛽−
1
2
𝛽−
1
2

15. 14
Demonstration: 𝑁 = 2 2 dipole array
2𝑎 =
3
2
𝜆, 𝜒0 = 0
𝐼 ∝ cos2
𝑘𝑎 cos 𝜙 = cos2
2π
𝜆
3
4
𝜆 cos 𝜙 = cos2
3π
2
cos 𝜙
Minima:
3π
2
cos 𝜙 =
π
2
⇒ cos 𝜙 =
1
3
⇒ 𝜙 = 70.5°
3π
2
cos 𝜙 =
3π
2
⇒ cos 𝜙 = 1 ⇒ 𝜙 = 0°
Maxima:
3π
2
cos 𝜙 = 0 ⇒ 𝜙 = 90°
3π
2
cos 𝜙 = π ⇒ cos 𝜙 =
2
3
⇒ 𝜙 = 48.2°

16. 15
Figure 5: Intensity pattern.
𝜙

17. 16
Figure 6: The radiation pattern for an 𝑁 dipole linear array for various values
of 𝑁, dipole spacing 2𝑎, and relative current phase 𝜒0 = 0 in the 𝜃 = π
2
plane.
𝑁 = 1 𝑁 = 2 𝑁 = 3
𝑎 =
𝜆
4
, 𝜒0 = 0

18. 17
𝑁 = 1 𝑁 = 2
𝑁 = 3
𝑎 =
𝜆
4
, 𝜒0 =
π
2
Figure 7: The radiation pattern for an 𝑁 dipole linear array for various values
of 𝑁, dipole spacing 2𝑎, and relative current phase 𝜒0 =
π
2
in the 𝜃 = π
2
plane.

19. 18
Figure 8: The radiation pattern for an 𝑁 dipole linear array for various values
of 𝑁, dipole spacing 2𝑎, and relative current phase 𝜒0 = 0 in the 𝜃 = π
2
plane.
𝑎 =
𝜆
2
, 𝜒0 = 0
𝑁 = 1 𝑁 = 2

20. 19
Figure 9: The radiation pattern for an 𝑁 dipole linear array for various values
of 𝑁, dipole spacing 2𝑎, and relative current phase 𝜒0 =
π
2
in the 𝜃 = π
2
plane.
𝑎 =
𝜆
2
, 𝜒0 =
π
2
𝑁 = 1 𝑁 = 2

21. 20
Figure 10: The radiation pattern for an 𝑁 dipole linear array for various
values of 𝑁, dipole spacing 2𝑎, and relative current phase 𝜒0 = 0 in the 𝜃 = π
2
plane.
𝑎 = 𝜆, 𝜒0 = 0
𝑁 = 1 𝑁 = 2

22. 21
Figure 11: The radiation pattern for an 𝑁 dipole linear array for various values
of 𝑁, dipole spacing 2𝑎, and relative current phase 𝜒0 =
π
2
in the 𝜃 = π
2
plane.
𝑎 = 𝜆, 𝜒0 =
π
2
𝑁 = 1 𝑁 = 2

23. 22
(1) Markus Zahn, Electromagnetics and Applications, Massachusetts
Institute of Technology: MIT Open Course Ware, 2005.
References