Delay-and-Sum Beamforming for Plane Waves

Transcription

1 Delay-and-Sum Beamforming for Plane Waves ECE 6279: Spatial Array Processing Spring 2011 Lecture 6 Prof. Aaron D. Lanterman School of Electrical & Computer Engineering Georgia Institute of Technology AL: <lanterma@ece.gatech.edu>

2 Where We Are in J&D Lecture material drawn from: Secs. 4.1, 4.1.2, (up to but not including Point Focusing part on p. 123), Next lecture: Secs , 4.1.3, ( Point Focusing part on p. 123)

3 Integrating Across Apertures Here s one way aperture smoothing functions show up Typically integrate across the aperture z(t) = w( x ) f ( x,t) d x Input a monochromatic plane wave to the system f ( x,t) = exp{ j(ω 0 t k 0 x )} z(t) = exp( jω 0 t) w( x )exp( jk 0 x ) W ( k 0 ) d x

4 Delay-and-Sum Beamforming x 0 x Array of M sensors at positions For convenience, put the phase center at the origin m = 0 x m = 0 Delay-and-sum beamforming m = 0 z(t) w m y m (t Δ m )

5 Beamforming for Plane Waves m = 0 m = 0 f ( x,t) = s(t α 0 x ) α 0 = ζ 0 /c y (t) = s(t α 0 x ) m m z(t) = w m y m (t Δ m ) = w m s(t Δ m α 0 x m )

6 z(t) = If we pick When Things Line Up w m m =0 s(t Δ m α 0 x m ) Δ = α 0 x = ζ 0 m m then we get the signal back! z(t) = w m m =0 c m =0 x m s(t) = s(t) w m

7 z(t) = When They Don t w m m =0 s(t Δ m α 0 x m ) More generally, if we pick Δ = α x = ζ m m then we get a degraded version of the signal z(t) = m =0 w m c x m s(t + ( α α 0 ) x m )

8 Strategy for Parameter Estimation m =0 z(t) w m y m (t Δ m ) Δ m = α x m = ζ x m c Find parameter that maximizes energy in z(t) Radar and sonar: If you know c, sweep ζ to find direction of arrival Seismology: If you know ζ, sweep c to find wave speed (determines material properties)

9 Monochromatic Plane Waves (1) f ( x,t) = exp{ jω 0 (t α 0 x )} = s(t α 0 x ) s(t) = exp( jω 0 t) Plane wave delay-and-sum beamformer response z(t) = m =0 w m = w m m =0 where s(t + ( α α 0 ) x m ) exp( jω 0 [t + ( α α 0 ) x m ])

10 = w m m =0 = w m m =0 Monochromatic Plane Waves (2) z(t) = w m m =0 exp( jω 0 [t + ( α α 0 ) x m ]) exp( jω 0 ( α α 0 ) x ) m exp( jω 0 t) Recall k 0 = ω 0 α 0 exp( j(ω 0 α k 0 ) x ) m exp( jω 0 t)

11 Monochromatic Plane Waves (3) z(t) = m =0 w m exp( j(ω 0 α k 0 ) x ) m exp( jω 0 t) = W (ω 0 α k 0 )exp( jω 0 t) where the aperture smoothing function is W ( k ) = w m m =0 exp( j k x m ) Also called the array pattern

12 General Wavefields f ( x,t) = 1 (2π) 4 F( k,ω)exp{ j(ωt k x )}dk dω Delay-and-sum beamformer focused on α z(t) = 1 (2π) 4 F( k,ω)w (ωα k )exp( jωt)dk dω

13 z(t) = 1 (2π) 4 = 1 2π General Plane Waves (1) f ( x,t) = s(t α 0 x ) F( k,ω) = S(ω)(2π) 3 δ( k ω α 0 ) F( k,ω)w (ωα k )exp( jωt)dk dω S(ω)W (ω[ α α 0 ])exp( jωt)dω Z(ω) = S(ω)W (ω[ α α 0 ])

14 If we pick General Plane Waves (2) Z(ω) = S(ω)W (ω[ α α 0 ]) α = α 0 Z(ω) = S(ω)W (0) z(t) = s(t)w (0) we get the original signal back! α α 0 If we pick we get a filtered version

15 Uniform Linear Array (1) From earlier slide, the response of delayand-sum beamformer (tuned to α ) to a monochromatic plane wave is z(t) = W (ω 0 α k 0 )exp( jω 0 t) For a linear uniform array from the last lecture W ( k ) = sin(mk x d /2) sin(k d /2) x W (ω 0 α k 0 ) = sin(m[ω 0 α x k x 0 ]d /2) sin([ω 0 α x k x 0 ]d /2)

16 Using Uniform Linear Array (2) k = ω 0 α x x W (k k 0 ) = sin(m[k x k 0 ]d /2) x x x sin([k k 0 ]d /2) x x In terms of angles, let k x = (2π /λ)sin(φ) W (k x k x 0 ) = sin M π λ [sinφ 0 sinφ]d sin π λ [sinφ 0 sinφ]d

17 Beam Pattern (Boresight) sin(m[k x k x 0 ]d /2) sin([k x k x 0 ]d /2) k x = 0 M =12

18 Beam Pattern (60 ) sin(m[k x k x 0 ]d /2) sin([k x k x 0 ]d /2) k x = (2π /λ)sin(φ) φ = π /3 M =12

19 Beam Pattern (Boresight) sin M π λ [sinφ 0 sinφ]d sin π λ [sinφ 0 sinφ]d φ = 0 M =12

20 Beam Pattern (60 ) sin M π λ [sinφ 0 sinφ]d sin π λ [sinφ 0 sinφ]d φ = π /3 M =12

21 Beampattern: fix If Steered response: fix Terminology α = k /ω = k func(ω 0, k 0 ) = W (ω 0 α k 0 ) If ω 0, k 0 func( α ) = W (ω 0 α k 0 ) ζ /ω, k k = W ω 0 k ω k 0 k = k 0 ζ, ω = ω 0 : func( ζ 0 ) = W (k 0 [ ζ ζ 0 ]) k = k 0 ζ, ω = ω 0 : = W ω 0 k ω k 0 func( ζ ) = W (k 0 [ ζ ζ 0 ])