Lecture 33 Antenna Arrays and Phase Arrays In this lecture you will learn: • Antenna arrays • Gain and radiation pattern for antenna arrays • Antenna beam steering with phase array antennas ECE 303 – Fall 2007 – Farhan Rana – Cornell University Two Hertzian Dipoles – A Two Element Array z Consider first an array of just two Hertzian dipoles: r r J1 r r 3 r r J0 r r J0()r = zˆ I0 d δ ()r − h0 h0 h1 r r 3 r r J1()r = zˆ I1 d δ ()r − h1 y x Each antenna in the array is an “element” of the array One can write the E-field in the far-field as a superposition of the E-fields produced by all the elements in the array: r j η kd r r E ()rr = θˆ o sin()θ e− j k r I e j k rˆ.h0 + I e j k rˆ.h1 ff 4π r [ 0 1 ] j η k I d ⎡ I r I r ⎤ ˆ o 0 − j k r 0 j k rˆ.h0 1 j k rˆ.h1 = θ sin()θ e ⎢ e + e ⎥ 4π r ⎣ I0 I0 ⎦ Element Factor Array Factor r = Ε()()r ,θ ,φ F θ ,φ • The “element factor” is just the E-field produced by the first element if it were sitting at the origin • The “array factor” captures all the interference effects ECE 303 – Fall 2007 – Farhan Rana – Cornell University 1 An N-Element Antenna Array - I Consider an N-element antenna array where the elements are: - all identical (all loops, or all Hertzian dipoles, or all Half-wave dipoles, etc) - all oriented in the same way - but with possibly different current phasors Let the current phasor of the m-th antenna be Im r Let the position vector of the m-th antenna be hm One can write the E-field in the far-field as a superposition of the E-fields produced by all the elements in the array: z IN−1 r h N−1 r m r r r Im j k rˆ.hm Eff ()r = Ε()r,θ ,φ ∑ e m=0 I0 Im r = Ε()()r,θ ,φ F θ ,φ y I1 z I0 x ECE 303 – Fall 2007 – Farhan Rana – Cornell University An N-Element Antenna Array - II N−1 r r r r r Im j k rˆ.hm Eff ()r = Ε ()r ∑ e m=0 I0 r = Ε()()r ,θ ,φ F θ ,φ r r S()r ,t .rˆ r 2 2 Gain: G()θ ,φ = ∝ ⎡angular part of Ε()r,θ ,φ ⎤ F()θ ,φ P 4π r 2 ⎣⎢ ⎦⎥ G()θ ,φ Pattern: p()θ ,φ = z IN−1 G()θ ,φ max r hm Im y I1 z I0 x ECE 303 – Fall 2007 – Farhan Rana – Cornell University 2 An N-Element Hertzian Dipole Phase Array z N-element Hertzian dipole array a • Current magnitude is the same for all the dipoles jα 1 • Current phase difference between j2α e e y successive dipoles is α j (N −1)α e I x m+1 = e j α I Element Factor: m r j η k I d Ε()r ,θ ,φ = θˆ o 0 sin()θ e− j k r 4π r Array Factor: N−1 r r N−1 I j k rˆ.h j k rˆ.h jα m j k a m sin()θ cos ()φ F()θ ,φ = ∑ m e m = e 0 ∑ e e m=0 I0 m=0 ⎡N ⎤ sin2 α + ka sin θ cos φ N−1 2 ⎢ ()() ()⎥ 2 j k a sin()θ cos ()φ +α m ⎣ 2 ⎦ ⇒ F()θ ,φ = ∑ ()e = ⎡1 ⎤ m=0 sin2 ()α + ka sin()θ cos ()φ ⎣⎢2 ⎦⎥ ECE 303 – Fall 2007 – Farhan Rana – Cornell University An N-Element Hertzian Dipole Phase Array z N-element Hertzian dipole phase array antenna a ηo 2 P = k I0 d N jα 1 12π j2α e e y r S rr,t .rˆ j (N −1)α ( ) 3 2 2 x e G()θ ,φ = = sin ()θ F (θ ,φ ) P 4π r 2 2 N Coming from the array factor element factor Pattern: 1 p()θ ,φ = sin2 ()θ F (θ ,φ )2 N2 ⎡N ⎤ sin2 ()α + ka sin()θ cos ()φ 1 ⎢ 2 ⎥ p()θ ,φ = sin2()θ ⎣ ⎦ 2 ⎡1 ⎤ N sin2 ()α + ka sin()θ cos ()φ ⎣⎢2 ⎦⎥ Coming from the array factor element factor ECE 303 – Fall 2007 – Farhan Rana – Cornell University 3 An N-Element Hertzian Dipole Phase Array: Maxima y Lets look at radiation in the x-y plane: a cos()φ Radiation going in the φ direction from adjacent elements will add in-phase, and one will have a big maximum in the radiation pattern, provided: φ k a cos φ + α = ±2π n x ( ) a { n = 0,1,2,3,KK 2 p()θ = π 2,φ F(θ = π 2,φ ) big maxima N = 20 (main lobes) N = 20 main side π small lobes α = − π lobes φ 4 maxima α = − (side lobes) 4 π nulls ka = π 2 ka = 2 φ (degrees) ECE 303 – Fall 2007 – Farhan Rana – Cornell University An N-Element Hertzian Dipole Phase Array: Maxima y Condition for a big maximum in the x-y a cos()φ plane: k a cos(φ ) + α = ±2π n { n = 0,1,2,3, φ KK At a big maximum the value of the array x a factor is: 2 ⎡N ⎤ 2 sin ()α + ka cos()φ F()θ = π 2,φ 2 ⎣⎢ 2 ⎦⎥ F()θ = π 2,φ max = 2 ⎡1 ⎤ big sin ⎢ ()α + ka cos()φ ⎥ maxima ⎣2 ⎦ (main lobes) N = 20 = N2 π small α = − 4 maxima And the value of the antenna gain is: (side lobes) π nulls ka = 3 2 G()θ ,φ = sin2()θ F (θ ,φ )2 2 N ⎛ π ⎞ 3 ⇒ G⎜θ = ,φ ⎟ = N φ (degrees) ⎝ 2 ⎠max 2 ECE 303 – Fall 2007 – Farhan Rana – Cornell University 4 An N-Element Hertzian Dipole Phase Array: Nulls y Radiation in the x-y plane: a cos()φ The array factor will give a null in the radiation pattern provided: 2π n φ k a cos()φ + α = ± N x { n ≠ 0,N,2N,3N, a KK F(θ = π 2,φ )2 big 2 ⎡N ⎤ maxima sin ()α + ka cos()φ (main lobes) 2 ⎣⎢ 2 ⎦⎥ N = 20 F()θ = π 2,φ = π small 1 α = − 2 ⎡ ⎤ 4 maxima sin ()α + ka cos()φ (side lobes) ⎢ ⎥ π nulls ⎣2 ⎦ ka = 2 φ (degrees) ECE 303 – Fall 2007 – Farhan Rana – Cornell University An N-Element Hertzian Dipole Phase Array: Width of Big Maxima y F(θ = π 2,φ )2 N = 20 π φ α = − 4 π x ka = a 2 φ Angular width of a lobe associated with a big maxima: ∆φ nN For a big maxima: k a cos()φ + α = 2π n = 2π { n = 0,±1,±2,±3, N KK At the nulls nearest to the above maxima we must have: ⎛ ∆φ ⎞ nN ± 1 k a cos⎜φ + ⎟ + α = 2π ⎝ 2 ⎠ N Which can be solved for ∆φ – and for N large (N >> 1) 4π 1 ∆φ ≈ one gets approximately: N ka sin()φ ECE 303 – Fall 2007 – Farhan Rana – Cornell University 5 An N-Element Hertzian Dipole Phase Array: Examples z N-element Hertzian dipole array a ka = π Phase difference between successive elements is α jα 1 j2α e y e j (N −1)α p()θ ,φ x e p()θ ,φ N = 2 N = 10 ECE 303 – Fall 2007 – Farhan Rana – Cornell University An N-Element Hertzian Dipole Phase Array: Examples z N-element Hertzian dipole array a ka = π / 2 Phase difference between successive elements is α jα 1 j2α e y e j (N −1)α p()θ ,φ x e p()θ ,φ N = 2 N = 10 ECE 303 – Fall 2007 – Farhan Rana – Cornell University 6 An (N+1)-Element Hertzian Dipole Binomial Array λ z (N+1)-element Hertzian dipole array a = 2 a • Current phase is the same for all the ⇒ ka = π dipoles • Currents magnitudes in the dipoles y follow a binomial distribution I N! x m = I0 m! ()N − m ! Element Factor: r j η k I d Ε()r ,θ ,φ = θˆ o 0 sin()θ e− j k r 4π r Array Factor: N r r N I j k rˆ.h j k rˆ.h N! j k a m sin()θ cos ()φ F()θ ,φ = ∑ m e m = e 0 ∑ e m=0 I0 m=0 m! ()N − m ! 2 2N ⎡k ⎤ ⇒ F()θ ,φ = 1+ e j k a sin()θ cos ()φ = 22N cos2N a sin()θ cos ()φ ⎣⎢2 ⎦⎥ ECE 303 – Fall 2007 – Farhan Rana – Cornell University An (N+1)-Element Hertzian Dipole Binomial Array: Maxima and Nulls y Lets look at radiation in the x-y plane: a cos()φ Radiation going in the φ direction from adjacent elements will add in-phase, φ and one will have a big maximum in the radiation pattern, provided: x a k a cos(φ ) = ±2π n 2 2N 2N ⎡π ⎤ { n = 0,1,2,3, F()θ = π 2,φ = 2 cos ⎢ cos()φ ⎥ KK ⎣ 2 ⎦ π ⎡π ⎤ ⇒ cos()φ = ±π n p()θ = π 2,φ = cos2N cos()φ 2 ⎣⎢ 2 ⎦⎥ p()θ = π 2,φ p(θ = π 2,φ ) N+1=40 Note: N+1=40 N+1=2 φ (degrees) N+1=2 • No side lobes • Two nulls at φ = 0,π φ (degrees) ECE 303 – Fall 2007 – Farhan Rana – Cornell University 7 Applications of Antenna Arrays - I Phased Array TRack to Intercept Of Target Alaska ECE 303 – Fall 2007 – Farhan Rana – Cornell University Applications of Antenna Arrays - II Communication satellite with two phase array MRI system with phase array antenna panels RF receivers Jicamarca facility (Peru) with an array of half-wave dipoles SETI ECE 303 – Fall 2007 – Farhan Rana – Cornell University 8.