# Antenna Arrays. EE-4382/ Antenna Engineering

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1 Antenna Arrays EE-4382/ Antenna Engineering

2 Outline Introduction Two Element Array Rectangular-to-Polar Graphical Solution N-Element Linear Array: Uniform Spacing and Amplitude Theory of N-Element Linear Array Rectangular to Polar Graphical Solution Broadside Array Ordinary End-Fire Array Phased Array Hansen-Woodyard End-Fire Array N-Element Linear Array: Directivity Design Procedure Radio Observatory Antenna Arrays Linear Antenna Arrays 2

3 Introduction 3

4 Antenna Arrays - Introduction Antenna arrays are a configuration of multiple radiating elements in a geometrical order. Antenna arrays are an efficient way to freely change the pattern of an antenna, making it more directive and therefore increasing the gain. Electronically adjusting the excitation of individual elements leads to a phased (scanning) array, which enables greater degrees of freedom. Linear Antenna Arrays Slide 4

5 Antenna Arrays - Introduction In an array of identical radiating elements, there are at least five factors that can be controlled to shape the overall pattern: 1. The geometrical configuration of the array (linear, circular, rectangular, elliptical, etc.) 2. The relative displacement between the elements 3. The excitation amplitude of the individual elements 4. The excitation phase of the individual elements 5. The relative pattern of the individual elements Linear Antenna Arrays Slide 5

6 Two-Element Array 6

7 Two-Element Array Two infinitesimal dipoles are placed along the z-axis. The total field radiated assuming no mutual coupling, is equal to the sum of the two elements. In the y-z plane: E t = E 1 + E 2 = a θ jη ki 0l 4π Where the β is the difference in the phase excitation between elements. Assuming far-field observations: θ 1 θ 2 θ r 1 r d 2 cos(θ) r 2 r + d cos θ 2 r 1 r 2 r e j kr 1 β 2 cos θ r 1 + e j kr 2 β 2 cos θ 1 r 2 2 Linear Antenna Arrays Slide 7

8 Two-Element Array Assuming far-field observations, the total field becomes E t = a θ jη ki 0le jkr cos θ e +j kd cos θ +β /2 + e +j kd cos θ +β /2 4πr E t = a θ jη ki 0le jkr 4πr cos θ 2 cos 1 2 kd cos θ + β Field of single element Array Factor AF = 2cos 1 (kd cos θ + β 2 (AF) n = cos 1 2 (kd cos θ + β E total = E single element at ref. point [array factor] Linear Antenna Arrays Slide 8

9 Two-Element Array - Examples Given the array shown for two identical isotropic sources, find the total field when d = λ/2 and β = 0. Linear Antenna Arrays Slide 9

10 Two-Element Array - Examples Linear Antenna Arrays Slide 10

11 Two-Element Array - Examples Given the array shown for two identical isotropic sources, find the normalized total field when d = λ/4 and β = 90. Linear Antenna Arrays Slide 11

12 Two-Element Array - Examples Linear Antenna Arrays Slide 12

13 Two-Element Array - Examples Given the array shown for two identical isotropic sources, find the normalized total field when d = λ and β = 0. Linear Antenna Arrays Slide 13

14 Two-Element Array Examples Linear Antenna Arrays Slide 14

15 Isotropic Point Sources Array Factor for two elements Linear Antenna Arrays Slide 15

16 Isotropic Point Sources Array Factor for two elements Linear Antenna Arrays Slide 16

17 Two-Element Array - Examples Given the array shown for two identical infinitesimal dipoles, find by the nulls of the total field when d = λ/4 and a. β = 0 b. β = +π/2 c. β = π/2 Linear Antenna Arrays Slide 17

18 Two-Element Array - Examples Linear Antenna Arrays Slide 18

19 Two-Element Array - Examples Linear Antenna Arrays Slide 19

20 Two-Element Array - Examples Linear Antenna Arrays Slide 20

21 Antenna Array Scanning Array Linear Antenna Arrays Slide 21

22 N-Element Linear Array: Uniform Amplitude and Spacing 22

23 Linear Array: Uniform Amplitude and Spacing An uniform array is an array of elements, all with identical magnitude, and each with a progressive phase. Linear Antenna Arrays Slide 23

24 Linear Array: Uniform Amplitude and Spacing Linear Antenna Arrays Slide 24

25 Linear Array: Uniform Amplitude and Spacing AF = 1 + e +j(kd cos θ +β) + e +j2(kd cos θ +β) + + e +j N 1 N AF = n=1 +j n 1 e N AF = n=1 kd cos θ +β +j n 1 Ψ e Ψ = kd cos(θ) + β kd cos θ +β Another useful expression is the closed form expression of the array factor. Multiply by e jψ Subtract AF summation AF e jψ = e jψ + e j2ψ + e j3ψ + + e jnψ AF e jψ 1 = ( 1 + e jnψ ) Simplify AF = ejnψ e jψ 1 = ej N 1 2 Ψ e j N 2 Ψ e j N 2 Ψ e j 1 2 Ψ e j 1 2 Ψ = e j N 1 2 Ψ sin N 2 Ψ sin 1 2 Ψ Linear Antenna Arrays Slide 25

26 Linear Array: Uniform Amplitude and Spacing AF = sin N 2 Ψ sin 1 2 Ψ sin N 2 Ψ Ψ 2 AF n = 1 N sin N 2 Ψ Ψ 2 sin N 2 Ψ N 2 Ψ for small values of Ψ Ψ = kd cos(θ) + β The nulls are given by setting the array factor to 0. sin N 2 Ψ = 0 > > N 2 Ψ ቚ θ=θ n = ±nπ > > θ n = cos 1 λ 2πd β ± 2n N π n = 1,2,3, (null) n N, 2N, 3N, (maximum) The number of nulls that can exist will be a function of the element separation d and phase excitation difference β. Linear Antenna Arrays Slide 26

27 Linear Array: Rectangular Plot First main maximum occurs when ψ 2 = 0 > > Ψ = 0 The principal maxima occurs when θ m = cos 1 λβ 2πd Other main maxima occurs when Ψ = ±2mπ, m = 1,2,3, Linear Antenna Arrays Slide 27

28 Linear Array: Rectangular Plot Linear Antenna Arrays Slide 28

29 Linear Array: Rectangular Plot Linear Antenna Arrays Slide 29

30 Linear Array: Rectangular Plot Linear Antenna Arrays Slide 30

31 Linear Array: Rectangular Plot Observations for rectangular plots of linear arrays with elements that are equally spaced, uniformly excited: 1. As N increases, the main lobe narrows 2. As N increases, there are more side lobes in one period of f(ψ). In fact, the number of full lobes (one main lobe and the side lobes) in one period of f(ψ) equals N 1. There are N 2 side lobes in each period. 3. The minor lobes are of width 2π/N in the variable Ψ and the major lobes are twice this width. 4. The side lobe peaks decrease with increasing N. 5. f Ψ is symmetric about π. Linear Antenna Arrays Slide 31

32 Rectangular to Polar Graphical Solution Linear Antenna Arrays Slide 32

33 Rectangular to Polar Graphical Solution In antenna theory, many solutions are of the form f ζ = f(c cos γ + δ) Where C and δ are constants and γ is a variable. The approximate array factor of an N-element, uniform amplitude linear array is a sin ζ ζ where ζ = C cos(γ) + δ = N 2 Ψ = N 2 kd cos θ + β C = N 2 kd δ = N 2 β The f ζ function can be plotted in rectilinear coordinates, and transferred to a polar graph. Linear Antenna Arrays Slide 33

34 Rectangular to Polar Graphical Solution The procedure that must be followed in the construction of the polar graph is as follows: 1. Plot, using rectilinear coordinates, the function f ζ. 2. a) Draw a circle with radius C and its center on the abscissa at ζ = δ b) Draw vertical lines to the abscissa so that they will intersect the circle. c) From the center of the circle, draw radial lines through the points of the circle intersected by the vertical lines. d) Along radial lines, mark off corresponding magnitudes from the linear plot. e) Connect all points to form a continuous graph. Linear Antenna Arrays Slide 34

35 Rectangular to Polar Graphical Solution Linear Antenna Arrays Slide 35

36 Four element linear array - Example Find and plot the array factor of a four-element, uniformly excited, equally spaced array. The spacing is λ/2 and 90 interelement phasing (i.e. β = π/2). Linear Antenna Arrays Slide 36

37 Four element linear array - Example Linear Antenna Arrays Slide 37

38 Two element linear array - Examples Find and plot the array factor of a two-element, isotropic, equally spaced array with distance d = λ/2 and uniform phase excitation α = 0 Find and plot the same array factor of a two-element, isotropic, equally spaced array with distance d = λ/2 but with phase excitation α = 180 Find and plot the same array factor of a two-element, isotropic, equally spaced array with distance d = λ/4 but with phase excitation α = 90 Linear Antenna Arrays Slide 38

39 Two element linear array - Examples Linear Antenna Arrays Slide 39

40 Five element Endfire Linear Array - Examples Find and plot the array factor of a five-element, isotropic, equally spaced array with distance d = 0.45λ and uniform phase excitation α = 0.9π Find and plot the array factor of a five-element, isotropic, equally spaced array with distance d = 0.5λ and uniform phase excitation α = π Linear Antenna Arrays Slide 40

41 Five element linear array - Examples Linear Antenna Arrays Slide 41

42 Broadside Array In many applications it is desirable to have the maximum radiation of an array directed normal to the axis of the array (θ = 90 ). To optimize this design, both the maxima of the single element and the array factor should be both directed toward θ = 90. Recall the maximum of the array factor occurs when Ψ = kd cos(θ) + β = 0 Since it is desired to have the first maximum directed toward θ = 90 Ψ = kd cos(θ) + β ቚ θ=90 = β = 0 To have the maximum of the array factor in an uniform linear array directed to the broadside to the axis, all elements need to have the same phase excitation. Linear Antenna Arrays Slide 42

43 Broadside Array To ensure that there are no other maxima in other directions (grating lobes), the separation between the elements should not be equal to multiples of a wavelength (d nλ, n = 1,2,3, ) when β = 0. If d = nλ, n = 1,2,3, and β = 0, then Ψ = kd cos θ + βȁ d=nλ β=0 n=1,2,3, = 2πn cos θ ȁ θ=0,π = ±2nπ avoid this! To avoid any grating lobes, the largest spacing between the elements should be less than one wavelength (d = λ) Linear Antenna Arrays Slide 43

44 Broadside Array Linear Antenna Arrays Slide 44

45 Broadside Array Linear Antenna Arrays Slide 45

46 Broadside Array Introduction to Antennas Slide 46

47 Broadside Array Linear Antenna Arrays Slide 47

48 Ordinary End-Fire Array Instead of having the maximum radiation broadside to the axis of an array, it may be desirable to direct it along the axis of the array (endfire). Sometimes it may be desirable that it radiates toward only one direction (θ = 0, 180 ) For the maximum toward θ = 0 : Ψ = kd cos(θ) + β ቚ θ=0 = kd + β = 0 > > β = kd For the maximum toward θ = 180 : Ψ = kd cos(θ) + β ቚ θ=180 = kd + β = 0 > > β = kd Linear Antenna Arrays Slide 48

49 Ordinary End-Fire Array Linear Antenna Arrays Slide 49

50 Ordinary End-Fire Array Linear Antenna Arrays Slide 50

51 Ordinary End-Fire Array Linear Antenna Arrays Slide 51

52 Ordinary End-Fire Array Linear Antenna Arrays Slide 52

53 Linear Antenna Arrays Slide 53

54 Scanning/Phased Array Linear Antenna Arrays Slide 54

55 Scanning/Phased Array Linear Antenna Arrays Slide 55

56 Scanning/Phased Array Linear Antenna Arrays Slide 56

57 Hansen-Woodyard End-Fire Array We discussed the conditions to have an ordinary end-fire array in the previous sections. In order to enhance the directivity of an end-fire array without destroying any of the other characteristics, Hansen and Woodyard proposed in 1938 proposed that the required phase shift between closely spaced elements of a very long array should be For the maximum toward θ = 0 : β = kd N kd + π N For the maximum toward θ = 180 : β = kd N + kd + π N For both directions, spacing should be d = N 1 λ N 4 λ for large N 4 Linear Antenna Arrays Slide 57

58 Hansen-Woodyard End-Fire Array Linear Antenna Arrays Slide 58

59 Hansen-Woodyard End-Fire Array Linear Antenna Arrays Slide 59

60 Linear Arrays - Summary Linear Antenna Arrays Slide 60

61 Linear Arrays - Summary Linear Antenna Arrays Slide 61

62 N-Element Linear Arrays: Directivity Linear Antenna Arrays Slide 62

63 Antenna Array Directivity For a linear antenna array, determine total length by L = N 1 d For a large broadside array (L d), directivity reduces to D 0 2N d λ = L d d λ 2 L λ For a large ordinary end-fire array (L d), directivity reduces to D 0 4N d λ = L d d λ 4 L λ For a Hansen-Woodyard end-fire array (L d), directivity reduces to D N d = L d L λ d λ λ Linear Antenna Arrays Slide 63

64 Linear Arrays: Design Procedure Linear Antenna Arrays Slide 64

65 Linear Arrays: Design Procedure N = L + d d L = N 1 d Linear Antenna Arrays Slide 65

66 Linear Arrays: Design Procedure Example Design an uniform linear scanning array whose maximum array factor is 30 from the axis of the array θ = 30. The desired halfpower beamwidth is 2 while the spacing of the elements is λ/4. Determine the phase excitation of the elements, length of the array (in wavelengths), number of the elements, and directivity (in db). Linear Antenna Arrays Slide 66

67 Radio Observatory Antenna Arrays Linear Antenna Arrays Slide 67

68 Karl G. Jansky Very Large Array (VLA) It s a cm-wavelength radio astronomy observatory located 50 miles west of Socorro, NM The radio telescope comprises 27 independent antennae, each of which has a dish diameter of 25 meters and weighs 209 metric tons. The antennae are distributed along the three arms of a track, shaped in a wye-configuration, (each of which measures 21 km). The frequency coverage is 74 MHz to 50 GHz (400 to 0.7 cm) Linear Antenna Arrays Slide 68

69 Very Long Baseline Array Linear Antenna Arrays Slide 69 /sites/

70 Very Long Baseline Array (VLBA) and High Sensitivity Array (HSA) VLBA is an interferometer consisting of 10 identical antennas on transcontinental baselines up to 8000 km (Mauna Kea, Hawaii to St. Croix, Virgin Islands). The VLBA is controlled remotely from the Science Operations Center in Socorro, New Mexico. The VLBA observes at wavelengths of 28 cm to 3 mm (1.2 GHz to 96 GHz) It is part of the High Sensitivity Array (HSA), which comprises the VLBA, phased Very Large Array (VLA), Green Bank Telescope (GBT), Effelsberg, and Arecibo telescopes, and subsets thereof. This array spans around 12,000 km in length. Linear Antenna Arrays Slide 70

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