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1 EE 542 Antennas and Propagation for Wireless Communications Array Antennas

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O. Kilic EE 542 2 Array Antennas An antenna made up of an array of individual antennas Motivations to use array antennas: –High gain more directive pattern –Steerability of the main beam Linear array: elements arranged on a line 2-D planar arrays: rectangular, square, circular,… Conformal arrays: non-planar, conform to surface such as aircraft

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O. Kilic EE 542 3 Radiation Pattern for Arrays Depends on: The type of the individual elements Their orientation Their position in space The amplitude and phase of the current feeding them The total number of elements

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O. Kilic EE 542 4 Array Factor The pattern of an array by neglecting the patterns of the individual elements; i.e. assume individual elements are isotropic

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O. Kilic EE 542 5 Linear Receive Array + Receiver A

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O. Kilic EE 542 6 Case 1: Array Factor for Two Isotropic Sources with Identical Amplitude and Phase (d = /2) d P(x,y,z) x z (1) (2) r r1r1 r2r2 (I 0, 0 ) Isotropic sources are assumed for AF calculations. The radiated fields are uniform over a sphere surrounding the source.

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O. Kilic EE 542 7 Radiation from an Isotropic Source r

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O. Kilic EE 542 8 Case 1: Total E Field where

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O. Kilic EE 542 9 Case 1: Far Field Approximation In the far field, r>>d or (d/r) <<1

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O. Kilic EE 542 10 Case 1: Far Field Approximation Similarly, Thus, in the far field

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O. Kilic EE 542 11 Case 1: Far Field Geometry d P(x,y,z) x z (1) (2) r r1r1 r2r2 dcos If the observation point r is much larger than the separation d, the vectors r1, r and r2 can be assumed to be approximately parallel. The path lengths from the sources to the observation point are slightly different.

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O. Kilic EE 542 12 Case 1: Total E in the Far Field The slight difference in path length can NOT be neglected for the exponential term!!

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O. Kilic EE 542 13 Case 1: Total E for d= /2 Note that d= 2

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O. Kilic EE 542 14 Case 1: Array Factor The array factor is described as the magnitude of E at a constant distance r from the antenna (i.e. unit V) AF n 00 /2 1 0 /2 1 Normalized values

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O. Kilic EE 542 15 Case 1: Radiation Pattern x z (1) (2) (I 0, 0 ) Notice how the two element array is more directive than the single element; which is an isotropic source.

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O. Kilic EE 542 16 Case 2: Array Factor for Two Isotropic Sources with Identical Amplitude and Opposite Phase d P(x,y,z) x z (1) (2) r r1r1 r2r2 (I 1, 1 ) (I 2, 2 )

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O. Kilic EE 542 17 Case 2 – Far Field Geometry d P(x,y,z) x z (1) (2) r r1r1 r2r2 dcos (I 1, 1 ) (I 2, 2 )

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O. Kilic EE 542 18 Case 2: Total E in the Far Field

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O. Kilic EE 542 19 Case 2: Radiation Pattern Note that d= 2 x z (1) (2) (I 0, 0 ) (I 0, + 0 ) Observe how the pattern is rotated compared to Case1 by simply changing the phase of element 2 AF n 01 /2 0 1 /2 0

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O. Kilic EE 542 20 Case 3: Array Factor for Two Isotropic Sources with Identical Amplitudes and 90 o Phase Shift Homework: Show that:

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O. Kilic EE 542 21 Case 3 AF n 00 /2cos( /4) 1 /2cos( /4) x z (1) (2) (I 0, 0 ) (I 0, + 0 )

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O. Kilic EE 542 22 Generalization to N Equally Spaced Elements 0123N-1 ddd dcos r

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O. Kilic EE 542 23 General Case for Linear Array Total E field: Array Factor:

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O. Kilic EE 542 24 Special Case (A) Fourier series Equally Spaced Linear Array with Linear Phase Progression

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O. Kilic EE 542 25 Some Observations

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O. Kilic EE 542 26 Special Case (B) Uniformly Excited, Equally Spaced Linear Array with Linear Phase Progression

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O. Kilic EE 542 27 Observations AF similar to the sinc function (i.e. sinx/x) with a major difference: Sidelobes do not die off for increasing values because the denominator is a sine function, and does not increase beyond a value of 1. AF is periodic with 2 Maximum value (=I o ) occurs at k

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O. Kilic EE 542 28 N=4 Case Period: nulls /2 /2

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O. Kilic EE 542 29 More Observations Zeroes (Nulls) @ N /2 = k k =2k k=0,1,2, … This implies that as N increases there are more sidelobes (i.e. more secondary null points) in one period. Sidelobe widths are 2 First null at 1 =2 Within one period, N null points N-2 sidelobes (Because we discard k = N case, which corresponds to the second peak. Also 2 nulls create one sidelobe.) This implies that as N increases, the main beam narrows. Main lobe width is 2*2 twice the width of sidelobes Max value ( = NI o ) @ =2k , k=0,1,2, … For all k values except when /2 becomes an integer multiple of

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O. Kilic EE 542 30 Effect of Increasing N HW: Regenerate this plot.

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O. Kilic EE 542 31 Construction of Polar Plot from AF( ) The angle is not a physical quantity. We are more interested in observing the AF as a function of angles in real space; i.e. . Since linear arrays are rotationally symmetric wrt , we are concerned with only.

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O. Kilic EE 542 32 Case 1: Construction of Polar Plot N = 2, d = /2, = 0 (uniform phase) Using the general representation from Page 24 /2 x z I o, =0 r Compare to page 62

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O. Kilic EE 542 33 Normalized AF for Case1 Period = 2

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O. Kilic EE 542 34 Normalized AF for Case1 – Polar Plot Visible range: : [0- ] : [-kd,kd] = kdcos = cos kd f( ) f x Circle of radius kd f kd Visually relate to

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O. Kilic EE 542 35 Constructing the Polar Plot kd f( ) f x Circle of radius kd f f f

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O. Kilic EE 542 36 Case 2 N = 2, d = /2, = Note: AF( ) same for all N=2 Value of different, depends on , d

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O. Kilic EE 542 37 Case 2: Polar Format = kdcos + kd f( ) f x 0 22 Shifted by Circle of radius kd

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O. Kilic EE 542 38 Normalized AF for Case 2 – Polar Plot f( ) f x Circle of radius kd f f f

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O. Kilic EE 542 39 Shift by kdcos kd kd kd

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O. Kilic EE 542 40 Generalize to Arbitrary N Shift by Visible Range:

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O. Kilic EE 542 41 General Rule AF plot with respect to is identical for all cases with identical N. The polar plot is determined by shifting the unit circle by , the linear phase progression amount. Visible range is always the 2kd range centered around that point.

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O. Kilic EE 542 42 Shift and construct f( ) f Observe the dependence of main beam direction on , the phase progression. Main beam + kd - kd peak cos( peak ) = /kd

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O. Kilic EE 542 43 Shift and construct f( ) Observe the dependence of main beam direction on , the phase progression. Main beam + kd - kd

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O. Kilic EE 542 44 Array Pattern vs kd If kd > 2 ; i.e. d> /2 multiple peaks can occur in the visible range. These are known as grating lobes, and are often undesirable. Why?? –Will cause reduced directivity as power will be shared among all peaks –Likely to cause interference

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O. Kilic EE 542 45 Grating Lobes Three main beams. kd , x -kd

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O. Kilic EE 542 46 Pattern Multiplication So far only isotropic elements were considered. Actual arrays are made up of nearly identical antennas AF still plays a major role in the pattern Normalized Array Pattern Normalized element pattern Normalized Array factor

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O. Kilic EE 542 47 Validation with Dipoles Consider the case of an ideal dipole array as below. I0I0 (N-1)d ddd dcos r I1I1 I2I2 I3I3 0

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O. Kilic EE 542 48 Sum of the E fields For the center dipole, assuming z << Normalized pattern

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O. Kilic EE 542 49 Vector Potentials for Each Dipole

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O. Kilic EE 542 50 Total Vector Potential

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O. Kilic EE 542 51 Total E field Array pattern Normalized element pattern Array factor

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O. Kilic EE 542 52 Directivity of Linear Arrays where

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O. Kilic EE 542 53 Radiation Intensity

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O. Kilic EE 542 54 Total Radiated Power

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O. Kilic EE 542 55 Directivity

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O. Kilic EE 542 56 Directivity for Arrays with Isotropic Elements Easier to calculate Represents an approximate solution for elements with broad patterns Uniform amplitude and equal spacing will be assumed. Using

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O. Kilic EE 542 57 Directivity: Isotropic Elements, Linear Phase Progression, Uniform Spacing, Uniform Amplitude

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O. Kilic EE 542 58 Non-Uniformly Excited Linear Arrays We have seen the effects of phase shifting on the beam direction. We can also shape the beam and control the level of sidelobes by adjusting the amplitude of the currents in an array.

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O. Kilic EE 542 59 Array Factor for Non-Uniform Excitation

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O. Kilic EE 542 60 Can we eliminate the sidelobes??? Yes! First consider the 1x2 element array as in case 1 we studied. Recall that the AF did not have any sidelobes AF = |1+e j | =

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O. Kilic EE 542 61 Binomial Series Coefficients If the amplitudes are equal to the coefficients of the binomial series, no sidelobes. Consider the array factor, which is the square of Case 1: AF = (1+Z)(1+Z)=1 + 2Z + Z 2 This corresponds to a three element array with current amplitudes in the ratio of 1:2:1 Since this array factor is simply the square of an array factor with no sidelobes there are no sidelobes.

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O. Kilic EE 542 62 2-Dimensional Arrays The elements lie on a plane instead of a line. Many geometric shapes are possible; circle, square, rectangle, hexagon, etc. Will consider rectangular arrays

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O. Kilic EE 542 63 Rectangular Array Geometry dx x(m) y(n) dy mn r r mn z

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O. Kilic EE 542 64 Individual Fields

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O. Kilic EE 542 65 Total Field Element pattern Array factor

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O. Kilic EE 542 66 Array Factor

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O. Kilic EE 542 67 Array Factor : Linear Phase, Uniform Amplitude

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O. Kilic EE 542 68 Factors of Planar AF

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O. Kilic EE 542 69 Homework, Problem 1 Show that the Array Factor for two isotropic sources with identical amplitudes and 90 o phase shift is given by

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O. Kilic EE 542 70 HW Problem 2 Construct by hand after plotting the AF for N=4, = /2, d = /2 Hint: The AF vs should look like this:

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O. Kilic EE 542 71 References Stutzman, et. al. “Antenna Theory” provides an excellent discussion on array antennas!!!

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