ECE 5318/6352 Antenna Engineering. Spring 2006 Dr. Stuart Long. Chapter 6. Part 1 Array Basics

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Transcript:

ECE 5318/635 Antenna Engineering Spring 006 Dr. Stuart Long Chapter 6 Part 1 Array Basics 1

Array Rationale Single elements Usually broad beamwidth Relatively low directivity Often system requirements demand higher directivities Can increase electrical size or form an assembly of elements in an electrical and geometrical configuration called an ARRAY

Assumptions for our coverage 1. All elements are identical. No coupling between elements 3. Total field is vector sum of individual radiation patterns Use array to control overall pattern shape by: 1. Geometrical configuration of overall array (linear, circular, rectangular, spherical, ). Relative displacement between radiators 3. Amplitude of excitement of each element 4. Excitation phase of each element 5. Individual element pattern 3

Two element array simplest case ψ 4

Two element array Two identical horizontal sources along the z axis, a distance d apart, constant amplitudes; upper element leads lower one in phase by amount β look at [ y - z ] plane (φ 90 ) only there sin ψ cos θ Similar case of horizontal dipole above a ground plane, except nd. source not necessarily 180 out of phase but similar analysis 5

Two element array Total Field r 1 d r cos θ [6-] β j( k r1 ) η k Io e E1 j cosθ1 4π r 1 E t d r r + cosθ j( k r β ) η k Io e E j cosθ 4π r E 1 + E 6

Two element array Total Field E t β β j kr1 j kr η kio e e ˆ aθ j cosθ1 cosθ 4π + r1 r [6-3] E t aˆ θ η k j Io e 4π r jkr cosθ e j kd cosθ + β + e kd cosθ + β j 7

Two element array E t aˆ θ η k j Io e 4π r j k r cosθ kd cosθ + β cos [6-3] Element Factor [Ee] Array Factor [AF] 8

Two element array [Ee] η k j Io e 4π r j k r cosθ [6-4] [AF] cos kd cosθ + β (AF) n cos kd cosθ + β as function of kd and β 9

Two element array Example (a) β 0 d λ 4 k d π λ π λ 4 π AF cos cosθ n 4 ( ) ( Ee ) cosθ n 10

Two element array Example (b) k d β d λ 4 π π λ π λ 4 π 4 ( AF) cos ( cosθ + 1) n ( E e ) cos θ n 11

Two element array Example (c) β π k d d λ 4 π λ π λ 4 π 4 ( AF) cos ( cosθ 1) n ( Ee ) cosθ n 1

General example for Two-Element Array Defining Nulls kd cosθ + β ( AF) cos n nulls when kd cosθ + β cos or 0 for not phase shift for first null due to [AF] to exist d ( β 0) λ 1 n + 1 n ( kd cosθ + β ) ± π λ θ cos 1 n ) π d n 0,1,, 3, [ β ± (n + 1 π] 13

General example for Two-Element Array Defining Nulls ( Ee ) cosθ n n null at θ n 90 14

ECE 5318/635 Antenna Engineering Spring 006 Dr. Stuart Long Chapter 6 Part N-Element Uniform Linear Array 15

N-element uniform linear array Assumptions constant amplitude constant phase shift elements along a line I n I n1 e jβ [6-6] (e.g. # 3 leads # by β radians) for convenience assume isotropic sources 16

N-element uniform linear array AF 1 + e + j ( kd cosθ + β ) + j( kd cosθ + β ) + j( N 1) ( kd cosθ + β ) + e + + e AF N n1 e + j ( n1)[ kd cosθ + β ] [6-6] let jn ( 1) [ kd cos ] AF e ψ θ + β N n 1 ψ 17

N-element uniform linear array Multiply RHS by e j ψ and subtract from [AF] [6-7] [AF] jψ jψ jψ e e + e + + e jnψ [AF] ( jψ ) jn ψ 1 e 1 e [AF] e e jnψ jψ 1 1 18

N-element uniform linear array Example Look at N 5 AF 1 + e + + e jψ + j 4ψ [AF] e e + e + + e + e jψ jψ jψ j4ψ j5ψ [AF] [AF] jψ j5ψ ( 1 e ) 1 e e e j5ψ jψ 1 1 19

0 N-element uniform linear array element uniform linear array this term due to asymmetric location of elements; shifting to center or taking magnitude eliminates it 1 1 1 [AF] ψ ψ ψ ψ ψ ψ ψ j j N j N j N j j jn e e e e e e e sin sin [AF] 1 ψ ψ ψ N e N j [6-10]

1 N-element uniform linear array element uniform linear array sin sin [AF] ψ N ψ or Normalized sin sin 1 [AF] ψ Nψ N n

N-element uniform linear array [Note: for small arguments lim sin x x ; thus, Nψ x 0 for small x (but not necessarily small)] [AF] Nψ sin ψ or [AF] n 1 N Nψ sin ψ as ψ 0 [AF] N [AF] 1 or n

Nulls N-element uniform linear array [AF] n sin x x sin x x nulls occur when 1.0 N sin ψ 0 (where N ψ ± nπ ).707.63 N ( kd cos θ n + β ) ± nπ λ θn cos 1 π d β ± n π N π 1.39 π 3π π.17 x n 1,,3, but not n 0 [6-11] 3

N-element uniform linear array Absolute maxima Absolute max. when ψ ± mπ m 0,1,,3, θ λ cos 1 π d [ β ± mπ ] m [6-1] No matter how small ψ, always have max at ψ 0, (m 0) θ m cos 1 λβ π d [6-13] 4

N-element uniform linear array Half Power Point for small ψ : [AF] n sin x x (see appendix) 3 db point sin x when N 0.707 ψ ± 1. 391 x Half Power Point θ h λ cos 1 π d β ±.78 N [6-14] 5

N-element uniform linear array Secondary maxima Secondary maxima approx. when sin N ψ is max ± 1 (minor lobes) Nψ ± s π ( + 1) 1 λ s 1 θs cos + β ± π π d N [6-15] S 1,,3, 6

N-element uniform linear array First minor lobe maxima Maxima value of first minor lobe occurs N 3π π at ψ ± NOT at λ θs cos 1 π d β ± 3π N [6-16] 7

N-element uniform linear array First minor lobe maxima At this point the magnitude of the AF [AF] n N sin ψ ψ N 3π θ θs 1 3π 0.1 [6-17] in [ db] 0 log ( 0.1) 13.5 [ db] (note this side lobe is independent of N ) 8

ECE 5318/635 Antenna Engineering Spring 006 Dr. Stuart Long Chapter 6 Part 3 Broadside Arrays 9

BROADSIDE arrays β 0 maximum radiation in direction normal to line of array (θ 90 ) Max. when ψ 0 kd cosθ + β 0 [6-18] for θ 90 kd cosθ + β 0 β 0 Broadside radiation pattern when all elements are in phase ( independent of separation dist d ) 30

BROADSIDE arrays β 0 Nulls θ n cos 1 ± nλ N d n n 1,,3, N,N,3N, Maxima θ m cos 1 ± mλ d m 0,1,,3, Half Power Points π d assuming << 1 λ θ h cos 1 ± 1.391λ π N d Side Lobe Maxima π d assuming << 1 λ θ s cos 1 ± λ s + 1 d N s 1,,3, [Table 6.1] 31

BROADSIDE arrays β 0 First Null Beamwidth Θ n π 1 cos λ N d Half Power Beamwidth Θ h π 391λ 1 1. cos π N d [Table 6.] NOTE FIRST SIDE LOBE BEAMWIDTH (not the width of any beam) ΘS 3

BROADSIDE arrays β 0 Example λ N 10 ; d ; β 4 0 Nulls n λ N d n λ λ 10 4 n 5 θn θn θ n cos cos cos 1 1 1 ± ± ± n 5 5 4 5 66 37 n n 1,,3, N,N,3N, Absolute Maxima m λ 4m d θ m cos 1 1 [ ± 4 m] cos ( 0) 90 m 0,1,,3, 33

BROADSIDE arrays β 0 Example Half Power Points 1.391λ π Nd 1.391 π 5 θ h 1.391 cos 1 ± 5 π 79.8 Half Power Beamwidth Θ h ( ) 90 79.8 0.4 34

BROADSIDE arrays β 0 Example Side Lobes Maxima λ d s + 10 1 s + 10 1 θs θs θ s cos cos cos 1 1 1 ± ± ± s + 1 10 6 10 10 10 53 0 s 1,,3, 35

BROADSIDE arrays Example β 0 Array Factor Magnitude at θ 53 N 10 π λ 5π sin ψ sin cosθ sin cosθ λ 4 1 [AF] n ψ π λ π 0.45 sin sin cosθ sin cosθ λ 4.19 1 10 [ AF].19 0.19 0 log ( 0.19) 13. [ db n ] 36

BROADSIDE arrays Example β 0 Array Factor Magnitude at θ 0 5π sin 1 [AF] 1.414 π 0.707 sin 4 1 10 [ AF] 1.414 0.1414 0 log ( 0.1414) 17.0 [ db n ] 37

BROADSIDE arrays β 0 Nulls Maxima Summary at θ n 66 & 37 θm 90 53 [AF]n 1.0 0.19 θ 37 53 Half Power 0 0.1414 θ h 79.8 Θh (90-79.8) 0.4 66 0.4 79.8 0.4 38

ECE 5318/635 Antenna Engineering Spring 006 Dr. Stuart Long Chapter 6 Part 4 Ordinary Endfire Array 39

ORDINARY ENDFIRE ARRAYS Maximum radiation in direction along line of array β ± kd for max. in θ 180 ψ 0 for θ 180 kd cos θ + β 0 β + kd [6-0] for max. in θ 0 0 ψ 0 for θ 0 kd cos θ + β 0 β - kd 40

ORDINARY ENDFIRE arrays β ± kd Nulls θ n cos 1 1 nλ N d n n 1,,3, N,N,3N, Maxima θ m cos 1 1 mλ d m 0,1,,3, Half Power Points π d assuming << 1 λ θ h cos 1 1 1.391λ π N d Side Lobe Maxima π d assuming << 1 λ θ s cos 1 1 λ s + 1 d N s 1,,3, [Table 6.3] 41

ORDINARY ENDFIRE arrays β ± kd First Null Beamwidth Θ n cos 1 1 λ N d [Table 6.4] Half Power Beamwidth Θ h cos 1 1 1.391λ π N d HPBW Θh θ h Example λ N 10 ; d ; β 4 kd 4

ECE 5318/635 Antenna Engineering Spring 006 Dr. Stuart Long Chapter 6 Part 5 Scanning Phased Array 43

SCANNING PHASED Arrays - kd < β < kd θo Can scan beam between endfire and broadside by allowing phase shift β to be 0 β kd or -kd β 0 44

PHASED Arrays - kd < β < kd for maximium radiation at θ θo ψ 0 for θ θo kd cos θo + β 0 β - kd cos θo [6-1] θo 45

PHASED Arrays - kd < β < kd (use general expressions for nulls, max, sidelobes) For HPBW the main beam is no longer symmetric about θ 90 or θ 0 ; subtracting the two θ h (+ and sign) Θ h cos 1 cosθo.78 N k d cos 1 cosθo +.78 N k d 46

PHASED Arrays - kd < β < kd L L (N-1)d L + d Nd can also write Θh as Θ h cos 1 cosθo 0.443λ L + d cos 1 cosθo + 0.443λ L + d [6-] 47

48 Example Example 60 ; 4 ; 10 o d N θ λ 0.785 4 cos60 cos60 4 cos π π λ λ π θ β o kd 3.7 47.4 71..5 0.443 1 cos.5 0.443 1 cos 1 1 + Θ h PHASED Arrays PHASED Arrays

PHASED Arrays Example 47.4-1.6 +11. 71. ALSO NOTE ROTATIONAL SYMMETRY GIVES 3-D3 PATTERN OF CONICAL BEAM NOTE: MAIN BEAM IS NOT SYMMETRIC ABOUT ITS MAX AT θ 60 49

ECE 5318/635 Antenna Engineering Spring 006 Dr. Stuart Long Chapter 6 Part 5 Directivity 50

DIRECTIVITY the ratio of the radiation intensity in a given direction from the t antenna to the radiation intensity averaged over all directions D o U U o U 4π P max rad Fig..13 3-D directivity pattern of a λ/ dipole 51

DIRECTIVITY For BROADSIDE and small spacing ( d << λ ) where N Z kd cosθ [AF] n [AF] N sin ψ N ψ n sin Z Z For ORDINARY ENDFIRE and small spacing ( d << λ ) where N Z kd cosθ 1 [AF] n N sin kd cosθ N kd cosθ [AF] n N sin kd cosθ 1 N kd cosθ 1 [6-38] [6-45] 5

DIRECTIVITY For BROADSIDE and small spacing ( d << λ ) sin Z U ( θ ) [( AF) ] n Z For ORDINARY ENDFIRE and small spacing ( d << λ ) U max 1 at θ 90 D o U 4π P max rad U max 1 at θ 0 or 180 with approx. that N k d large; can calculate Prad 53

DIRECTIVITY P rad Ω U dω π 0 dφ π 0 U sinθ dθ π 0 dφ π 0 sin Z Z sinθ dθ π π 0 sin Z Z sinθ dθ [-13], [6-39] 54

DIRECTIVITY For BROADSIDE and small spacing ( d << λ ) For ORDINARY ENDFIRE and small spacing ( d << λ ) Z N kd cosθ [6-40] [6-47] Z N kd (cosθ 1) d Z N kd sinθ dθ d Z N kd sinθ dθ π Nkd / sin Z Prad N k d Nkd / Z d Z π Nkd sin Z Prad N k d 0 Z d Z 55

DIRECTIVITY For BROADSIDE and small spacing ( d << λ ) with approx. that Ν k d large For ORDINARY ENDFIRE and small spacing ( d << λ ) P rad 4π N k d + 4π P rad π N k d U max Do 4 π P rad sin Z Z N d λ d Z [6-4] P rad P rad D o 4π N k d + sin Z Z 0 4π π N k d U max 4 π P rad d Z 4 N d λ [6-49] 56

DIRECTIVITY For BROADSIDE and small spacing ( d << λ ) For ORDINARY ENDFIRE and small spacing ( d << λ ) D o for large array D o U max 4 π P L λ rad N d λ [6-4] L >> d L Nd D o for large array D o 4 U max 4 π P L λ rad 4 N d λ [6-49] L >> d L Nd [6-44] [6-49] 57

DIRECTIVITY For BROADSIDE and small spacing ( d << λ ) Example λ N 10; d 4 L D o 4.5 λ Nd Do 5.0 λ Example λ N 0; d 19 or 0 Do HPBW 5 Pozar D o 0.3 For ORDINARY ENDFIRE and small spacing ( d << λ ) Note: The directivity for the ENDFIRE case is exactly twice that for the broadside case since it is unidirectional instead of bidirectional Example λ N 0; d 38 or 40 D o HPBW 3 Pozar D o 40 58

DIRECTIVITY Fig. 6.1 Half-Power Beamwidth for Broadside, Ordinary End- Fire, and Scanning Uniform Linear Arrays 59

GRATING LOBES When visible range (VR) includes ψ -π or ψ π,, a second major lobe of same magnitude N is produced - This is called a grating lobe - 60

AF( ψ ) Nψ sin ψ sin 61

GRATING LOBES Broadside Endfire Other phases For complete grating lobe kd π kd π kd ( π β ) But to keep second lobe from being grater than side lobe need kd π π N kd π π N π kd π N β For no part of grating lobe kd π π N kd π π N kd π π N β 6

GRAPHICAL SOLUTION FOR ARRAYS Will see several functions like f ( ζ ) f ( c cosγ + δ ) Example ( AF) n sin (ζ ) ζ where N ζ ( kd cosθ + β ) c N kd δ N β 63

GRAPHICAL SOLUTION FOR ARRAYS Often easier to plot f (ζ ) in rectangular form and then convert to polar plot for actual function of real angle θ 64

GRAPHICAL SOLUTION FOR ARRAYS Procedure 1.First, draw rectangular plot of AF(ψ ) vs. ψ.then, draw a semicircle of radius underneath, which is offset by an amount β kd 3.Draw vertical lines to intersect semicircle 4.Draw radial lines to points of intersection 5.Mark corresponding magnitudes on radial lines 6.Connect points 65

GRAPHICAL SOLUTION FOR ARRAYS Example 5 element Uniform Linear array for N 5 AF( ψ ) kd 1.π β - 0.π Nψ sin ψ sin ψ kd cosθ + β Prin. Max. at ψ 0, mag. 5 First null at ψ ± 0.4π Nulls at ψ ± n π 5 Secondary Max. at 1 st. Side Lobe mag. ψ ± (±0.8π, ±1.π, ±1.6π) (n + 1) π 5 1 1.3 sin (0.3π ) (±0.6π, ±π, ±1.4π) 66

GRAPHICAL SOLUTION FOR ARRAYS (First, draw rectangular plot of AF(ψ ) ψ ) vs. 67

Rectangular to Polar 1.3 Semicircular radius kd 1.π Offset by β -0.π 68

Rectangular to Polar Fig. 6.16 Rectangular to polar plot graphical solution 69

GRAPHICAL REPRESENTATION FOR BROADSIDE Broadside β 0 (no offset) Max. always along θ 90º VISIBLE RANGE: values of ψ which correspond to real angles θ kd ψ kd kd θ NOTES kd θ Smaller kd gives broader beamwidth Larger kd gives narrower beamwidth If kd approaches π additional main lobes can appear called kd GRATING LOBES θ 70

GRAPHICAL REPRESENTATION FOR ENDFIRE Example for no part of grating lobe kd (π -π/n) kd β - kd kd β kd 71