Solutions 6 – Antenna arraysAntennas and Propagation, Frühjahrssemester 2011 Antenna array (linear array, constant feed magnitude, isotropic radiators) Problem 6.1 : A three-element array of isotropic sources has the phase and magnitude relationships shown in the figure below. The spacing between the elements is d = l 2 . a) b) Find the array factor. Find the nulls. z #2 -1 d #1 d #3 +1 -j y Method I : a) AF = -e + jkd cos q - j + e - jkd cos q = -2 j sin ( kd cos q ) - j To normalize the array factor so that its maximum equals unity, the normalization factor has to be -3j. AFn = 1 ( 1 + 2 sin ( kd cos q ) ) 3 = 1 ( 1 + 2 sin ( p cos q ) ) 3 b) 2 sin ( p cos qn ) = -1 p cos qn = sin-1 1 ( 2 ) = - p , - 56p , - 136p ,..., 76p , 11p , 196p = x 6 6 n æx ö qn = cos-1 ç n ÷ , ç ÷ çp÷ è ÷ ø p q1 = 99.59 6 5p q2 = 146.44 x2 = 6 x1 = - sin2 2 y ) sin 2 y 1 sin 2 y 1 1 = 3 cos2 2 y .ú = 99.2 ( 1 .sin2 2 y 1 1 = 3 2 ( 1 + cos y ) . î é 1ù cos-1 ê . the phase centre is in the physical center of the array (element #1) Using the trigonometric identities: sin(x + y ) = sin x cos y + cos x sin y 1 cos2 x = 2 ( 1 + cos 2x ) 1 sin2 x = 2 ( 1 ..59 ë 6û é 5ù cos-1 ê . û ì n = 1..ùú 2 = 2ë 2û AF = pù 1é y sin ê p cos q .. ï ï í ï ï n ¹ N .. 3.Solutions 6 – Antenna arrays Antennas and Propagation.ú = 146.44 ë 6û .cos y ) = 1 + 2 cos y = 1 + 2 cos ( kd cos q . 3N .2N . Frühjahrssemester 2011 Method II : a) uniform array with x = -p / 2 : Ny p 3 sin éê p cos q ..ú sin 2ë 2û 2 sin Here. where phase center was located at element #3. b) 2n é l -x qn = cos-1 ê p N ë 2pd n =1: n =2: ( ) ù ú.cos 2x ) the array factor can be written as AF = = 3 1 1 sin 2 y sin y cos 2 y + cos y sin 2 y = 1 1 sin 2 y sin 2 y 1 1 1 1 1 1 ( 2 sin 2 y cos 2 y ) cos 2 y + ( cos2 2 y .2.p 2 ) = 1 + 2 sin(kd cos q) AFn = 1 ( 1 + 2 sin(kd cos q) ) 3 Same result as of Method I. 98228 ) = 10. The spacing between the elements is d = l 4 . constant feed magnitude.28 d D0 = 4N l D0 = 20 dB = 100 a) ( ) 100 = 4N ( d ) = 4N ( 4ll ) = N l l = 24. since y = kd cos q kd @ kd (1 .1 )d = 99 ⋅ c) see slide 5. isotropic radiators) Problem 6. approximate half-power beamwidth (in degrees). amplitude level (compared to the maximum of the major lobe) of the first minor lobe (in dB).cos q ) 2 N 3p ( kd cos qs kd ) 2 2 The maximum of the first minor lobe approximately occurs at sin ( N y ) = 1 2 s or when .18 qh = cos-1 1 - 1.15 x = kd = 90 For small values of y the normalized array factor can be written as (Slide 5. overall length of the array (in wavelengths). and its length is much greater than the spacing.12) e) sin AFn = N y 2 ( N y ) . see slide 5.391l = cos-1 ( 0.799 Nd p HPBW = 2 ⋅ qh = 21. Determine the a) b) c) d) e) number of elements.Solutions 6 – Antenna arrays Antennas and Propagation. progressive phase between the elements (in degrees).6 ( ) d) see slide 5.2 : Design an ordinary end-fire uniform linear array with only one maximum so that its directivity is 20 dBi (above isotropic).75l 4 N = 100 b) L = ( N . Frühjahrssemester 2011 Antenna array (linear array. Solutions 6 – Antenna arrays Antennas and Propagation.212 3p which is in dB equal to æ 2 ö AFn = 20 log ç ÷ = -13.5 dB ç ÷ ç 3p ÷ è ø . Frühjahrssemester 2011 qs = cos-1 { 2ld éêë -x 3Np ùúû } p (N y ) = 2 s At that point the magnitude of AFn reduces to sin AFn = N y 2 s 2 = 0. 1 ý ï since: .3 : Determine the azimuthal and elevation angles of the grating lobes for a 10 by 10 element uniform planar array when the spacing between the elements is λ.sin q0 cos f0 ) = 2p ( sin q cos f ) = 2m p ü ï ï m = -1. 0.0.866 » £ sin q sin f £ » 0.47): é 3 ù nú ê ú f = tan-1 êê 2 m úú ê êë úû for 0 £ f £ 360 (1) é 3 ù nú ê é m ù ú q = sin-1 ê ú = sin-1 êê 2 sin f úú êë cos f úû ê êë úû for 0 £ q £ 180 (2) With dx = dy = λ follows kdx = kdy = 2π and the limits of m and n are given as: kdx ( sin q cos f .46) Ψ x = kd x sin θ cos φ + ξ x = ±2mπ Ψ y = kd y sin θ cos φ + ξ y = ±2nπ in the directions given by (Slide 5. Frühjahrssemester 2011 Antenna array (square array..3 ï -1. The main radiation occurs in the following direction: q0 = 60 . constant feed magnitude. .1339 ï ï ï 2 2 2 þ This means that the main radiation lobe and the grating lobes can occur in the directions given by (1) and (2) for m = -1.0 (i.e. isotropic radiators) Problem 6.1 £ sin q cos f £ 1 ï þ æ ö ü 3÷ ï ç ï kdy ( sin q sin f .sin q0 sin f0 ) = 2p ç sin q sin f ÷ = 2n p ï ç 2 ÷ è ø ï ý n = -1. six combinations).Solutions 6 – Antenna arrays Antennas and Propagation. f0 = 90 ì 3 ï ï sin q0 sin f0 = ï 2 í ï sin q cos f = 0 ï 0 0 ï î Grating lobes occur if the following conditions are fulfilled: (Slide 5. f0 = 90 and the array is located on the xy-plane. The maximum of the main beam is directed toward q0 = 60 . 0 ï 2+ 3 3 2.1 and n = -1. Case II: n = 0 . m = -1 : From (1): From (2): ï é 3 / 2 .11 f = tan-1 ê ú =ï í êë 2 úû ï139.3228 ) ï ï í -1 ï sin ( -1. Case IV: n = -1 .89 ï î æ 1 ö ÷ ç q = sin-1 ç ÷= ÷ è cos f ø ì sin-1 ( 1.3228 ) ÷ ï ç q = sin-1 ç ÷ = í -1 ç sin f ÷ ï sin ( +1.1066 ï î ì sin-1 ( -1. m = 1 : From (1): ï é 3 ù ì 40. respectively.3228 ) ï ï î . respectively.866 ) = 60 or 120 ï ï q = sin í -1 ï sin ( -0.009 ) » 270 ï -1 æ -1 ö ÷ ï q = sin ç ÷ ï ç cos f ø = í -1 ÷ è ï sin ( 1. m = -1 : From (1): From (2): ï é 3 ù ì -40.63 ï f = tan-1 ê ú=í -1 êë úû ï187. m = 0 ( main radiation) : ì 90 ï -1 ( ¥) = ï f = tan í ï 270 ï î From (1): 0 q = sin-1 0 .866 ) = -60 or . the main radiation occurs Because of the limits of at f = 90 and q = 60 and (because of the symmetry) at f = 90 and q = 120 .3228 ) ï î and -1 ï æ 3 / 2 ö ì sin ( -1.009 ) » 90 ï î and .8934 f = tan-1 ê ú =ï í êë 2 úû ï 220. Frühjahrssemester 2011 Case I: n = 0 .63 ï î -1 ì sin ( -1. -1 æ ç 3 /2 ö ÷ ç sin f ÷ = ÷ ç è ø Case III: n = 0 .Solutions 6 – Antenna arrays Antennas and Propagation.1 ù ì 7.3228 ) ÷=ï ÷ í ÷ sin f ø ï sin-1 ( -1. From (2): æ q = sin-1 ç ç ç è This means that there is no solution in case III.3228 ) ï ï î and -1 ì 3 / 2 ö ï sin ( 1.3228 ) è ø ï ï î . This means that there is no solution for case I. meaning that the term " 0 / 0 " can be anything. and From (2): () ì sin-1 ( 0.8934 319.3228 ) ï -1 æ -1 ö ï q = sin ç ÷ ï ç cos f ÷ = í -1 ÷ è ø ï sin ( +1.120 ï ï î q ( 0 £ q £ 180 ). and From (2): ì -7.Solutions 6 – Antenna arrays Antennas and Propagation.1ö ÷ ç sin f ÷ = ÷ ç è ø ì -1 ( ) ï ï sin -1.009 ) » 90 -1 æ -1 ö ÷=ï q = sin ç ÷ í ç cos f ø ï -1 ÷ è ï sin ( -1. q = 7.37 ëê ï î -1 ì ï sin ( 1.37 and q = 90 .172. -1 ( Case V: n = -1 . but rather an “almost constructive interference”. a large and significant “almost grating” lobe does exist.31 ï -1 æ 3 / 2 . a large and significant “almost grating” lobe does exist. . m = 1 : -1 From (1): From (2): ï é 3 / 2 .009 ) » 270 and ï î ì -1 ï æ ö ï sin ( 1.009 ) » 90 ï î The approximate “solution” of the arcsin-fuction means that there is no perfectly constructive interference of the waves originating from the respective array elements. ì 90 ï ¥) = ï f = tan í ï 270 ï î From (1): 0 q = sin-1 0 . a grating lobe occurs at f = 187.1 ÷ q = sin ç ÷ ï ç sin f ÷ = í sin-1 ( -1.37 f = tan ê ú =ï í 1 úû ï172.63 and q = 90 . a grating lobe occurs at f = 357. but rather an “almost constructive interference”.009 » 270 í -1 ï ï sin ( 1.1 ù ì -7.63 352. two grating lobes occur Because of the limits of q ( at f = 270 .009 ) » 270 è ø ï ï î The approximate “solution” of the arcsin-fuction means that there is no perfectly constructive interference of the waves originating from the respective array elements. As shown by the plots below.009 ) » 90 -1 ç 3 / 2 .69 or 172. m = 0 : () Case VI: n = -1 . As shown by the plots below.31 . q = 172.31 ÷ ï ç è ø ï î 0 £ q £ 180 ).1 ö ï q = sin ç ÷ ç sin f ÷ = í 7. Because of the limits of q ( 0 £ q £ 180 ).69 and at f = 270 . Frühjahrssemester 2011 -1 æ ç q = sin 3 /2 . Because of the limits of q ( 0 £ q £ 180 ). where the term " 0 / 0 " can be anything.69 or . Frühjahrssemester 2011 The 3D array factor is shown in the figure below in linear and logarithmic scale. linear z grating lobe (case V) grating lobe (case IV) main beam (case II) grating lobe (case VI) y x symmetric main beam (case II) grating lobe (case V) dB z y x .Solutions 6 – Antenna arrays Antennas and Propagation. Determine the nulls of the radiation pattern in the case d = l / 2 . d Feed a) Thanks to image theory the corner reflector problem can be transformed into an array consisting of four elements: r2 (a) (b) y Image #2 Plate #1 Image #2 r3 r r1 d d Feed #1 Image #3 d d d d Feed #1 r2 x Image #4 Plate #2 . constant feed magnitude. Frühjahrssemester 2011 Antenna array (non-linear array. a) b) Determine the far-field if the input current of the dipole is I 0 . The minimum distance to each plane from the dipole is equal. A l / 2 dipole is placed parallel to the intersection line of the two planes and at the distance d from the same line.4 : A corner reflector consists of two semi-infinite perfectly conducting planes at angle of 90 to each other.Solutions 6 – Antenna arrays Antennas and Propagation. non-isotropic radiators) Problem 6. the total electric field is: E tot p j hI 0e .cos ( kd sin q sin f ) ] 2pr sin q ( ) b) Here.jkr cos 2 cos q = ⋅ ⋅ 2 [ cos ( kd sin q cos f ) .cos ( kd sin q sin f ) = 0 cos A .jkr cos 2 cos q . the nulls are occurring for: cos ( kd sin q cos f ) .cos B = 2 sin A+B A-B ⋅ sin =0 2 2 where: A = kd sin q cos f = p sin q cos f B = kd sin q sin f = p sin q sin f .jkd sin q sin f ùû = AE ⋅ AF with the single element pattern being the one of a l / 2 dipole: p j hI 0e .e jkd sin q sin f + e .cos ( kd sin q sin f ) ] . For the array factor AF . Thus. AE = E 0 = 2pr sin q ( ) and the array factor: AF = 2 [ cos ( kd sin q cos f ) .d sin f sin q r3 = r + d cos f sin q r4 = r + d sin f sin q and the total electric field of the configuration is given by: E tot = E1 + E2 + E 3 + E 4 = E 0 éë e jkd sin q cos f .e .d cos f sin q r2 = r .Solutions 6 – Antenna arrays Antennas and Propagation.jkd sin q cos f . Frühjahrssemester 2011 The distances from the four dipoles to the observation point r are r1 = r . the zeros of the single element pattern and of the array factor have to be determined. For the element AE . This result was to-be-expected given the geometry of the problem: the dipole polarized parallel to the corner mirror metal plates shall indeed give a radiation null in the direction of the plates (that is. q =0 This result was to-be-expected given the geometry of the problem: the dipole has a null in the direction θ=0°. and thus the rule of l'Hospital has to be 0 applied to check if a null of the radiation pattern is occuring at q = 0 For q = 0 the expression has the form p p sin cos q sin q f (q) f ¢(0) 2 = lim = 2 lim q 0 g (q) q 0 g ¢(0) cos q ( ) = 0. 2 A + B = 0 and therewith p sin q cos f + p sin q sin f = 0 Thus. 2 sin q 0 .B = 0 and therewith p sin q cos f . sin q = 0 or cos f . Frühjahrssemester 2011 This means that sin This gives A+B =0 2 or sin A-B = 0.p sin q sin f = 0 Thus. so does a planar group of four parallel dipoles.sin f = 0 and thus f = 45 .Solutions 6 – Antenna arrays Antennas and Propagation.225 . φ=45° and φ=315°). A . . 315 . sin q = 0 or cos f + sin f = 0 and thus f = 135 . the nulls are occurring for: cos ( p cos q ) = 0 . Solutions 6 – Antenna arrays Antennas and Propagation. The currents on the dipoles are constant along the wire and have the following time dependence: D1: D2: D3: I 1 ( t ) = I 0 sin ( wt ) I 2 ( t ) = 2I 0 cos ( wt ) I 3 ( t ) = -I 0 sin ( wt ) a) b) c) Calculate the array factor AF (q. As well the phase difference between D3 and D2 is 90 . dipole radiators) Problem 6. non-constant feed magnitude. Which angle q maximizes the AF .jkr cos q 4pr The excitations can be written as . having distance d of half a wavelength? Calculate distance d for which the array radiates in end-fire direction. d. d ) and the complete far field EFF (q.5 : Assume an antenna array composed of 3 infinitesimal horizontal dipoles positioned along the z -axis. So we have an array with uniform spacing d and progressive phase x ( = 90 ). q and r . The far field radiation of one single infinitesimal dipole is located in the origin and oriented like shown in figure is Eq = j h I 0le . a) The phase difference between D2 and D1 is 90 . Frühjahrssemester 2011 Antenna array (linear array. r ) of this antenna array in function of d . as shown in figure below. Solutions 6 – Antenna arrays Antennas and Propagation.jkr 4pr AF = 2 + e -j p 2 é e jkd cos q ... æ æ 1ö 1ö qmax = arccos ç 2n + ÷ = arccos ç + ÷ = 60 ÷ ÷ ç ç ÷ ç ç 2÷ 2ø è è ø This array factor is shown in the following figure (note: direction for positive z / θ=0° pointing horizontally in the picture) . .90 ) The sum of the far field contributions of the 3 dipoles can be written as Eq = j h I 0l cos q 4pr p é .e-jkd cos q ù êë úû AF = 2 + 2 sin ( kd cos q ) The total field becomes EFF = j h I 0l .90 ) I 2 = 2I 0 cos ( wt ) I 3 = -I 0 cos ( wt . 2 where n = 0.p ) .j (kr -kd cos q + p ) . 1..j (kr +kd cos q + ) ù 2 + 2e . solving for q we get æ 2n p + p ÷ ö æ 2n p + p ÷ ö ç ÷ ç ç 2 ÷ = arccos ç 2 ÷ = arccos 2n + 1 ÷ ç = arccos ç ÷ ÷ ç kd ç ÷ ÷ p 2 ç ç ÷ ÷ ÷ ÷ ç ç è ø è ø qmax ( ) A real solution occurs only for arguments of the arcos whose modulus is smaller or equal to one.e 2 ú êe êë úû p é j (kd cos q . Frühjahrssemester 2011 I 1 = I 0 cos ( wt .jkr . 2...j (kd cos q + ) ù 2 + 2 -e 2 ú êe êë úû Il = j h 0 cos q e . Therefore the only possible n is n = 0 .jkr e cos q ( 2 + 2 sin ( kd cos q ) ) 4pr b) The maximum of the AF (AFmax = 4) appears if the argument of the sine function is kd cos q = p + 2n p. ... AF(d .. 2. Frühjahrssemester 2011 90 120 3 4 60 150 2 1 30 180 0 210 330 240 270 300 c) The endfire direction of the given array is either q = 0 or q = 180 . q = 180 : p p + 2n p kd = -2n p 2 2 kd cos q = -kd = d = nl l . q) = 2 + 2 sin ( kd cos q ) = 4 for p + 2n p 2 where n = 0.. In order to achieve radiation in the endfire direction the argument of the cosine function has to be maximized. q) = 2 + 2 sin ( kd cos q ) = 4 for kd cos q = kd = d = nl + l .Solutions 6 – Antenna arrays Antennas and Propagation.1.. The following picture shows the AF for d = λ / 4... (note: direction for positive z / θ=0° pointing horizontally in the picture) 90 120 3 150 2 1 30 4 60 180 0 210 330 240 270 300 . 4 q = 0 : AF(d . 2. 4 where n = 1.