1.1.3 The narrowband Uniform Linear Array (ULA) with d = λ/2:

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1 Seminar 1: Signal Processing Antennas 4ED024, Sven Nordebo The narrowband Uniform Linear Array (ULA) with d = λ/2: d Array response vector: a() = e e 1 jπ sin. j(π sin )(M 1) = 1 e jω. e jω(m 1) Array output signal: = w H x(t) = s(t)w H a() = s(t)h() Array response function for spatial frequency Ω = π sin H() = w H a() = M 1 m=0 w me j(π sin )m = M 1 m=0 w me jωm (FIR response) 1

2 1.2 The adaptive beamforming problem 1. Optimum beamforming Multiple signal scenario x(t) = s(t)a( 0 ) L i=1 s i(t)a( i ) n(t) Array output signal = w H x(t) Output power E = E{ 2 } = w H E{x(t)x H (t)}w = w H Rw Covariance matrix for the input signal x(t) (all signals are uncorrelated) L R = E{x(t)x H (t)} = σsa( 2 0 )a H ( 0 ) σi 2 a( i )a H ( i ) σni 2 i=1 2

3 1.2.1 Optimization using normal equations Minimize the real cost function: f(w) = E { s(t) 2} = E { s(t) w H x(t) 2} Solution 1: normal equations E {x(t)(s (t) y (t))} = E { x(t)(s (t) x H (t)w) } = 0 yielding Rw o = p, or w o = R 1 p where R = E{x(t)x H (t)} and p = E {x(t)s (t)} = σ 2 sa( 0 ) 3

4 1.2.2 Optimization by differentiation Minimize the real cost function: f(w) = E { s(t) 2} = E { s(t) w H x(t) 2} f(w) = E { (s(t) w H x(t))(s(t) w H x(t)) } = E { (s(t) w H x(t))(s (t) x H (t)w) } = E { s(t) 2 w H x(t)x H (t)w s(t)x H (t)w w H x(t)s (t) } = w H Rw p H w w H p σ 2 s where R = E{x(t)x H (t)}, p = E {x(t)s (t)} and σ 2 s = E { s(t) 2}. 4

5 1.2.2 Optimization by differentiation Complex derivatives where w = w x jw y f w = 1 ( f j f ) f, 2 w x w y w = 1 ( f j f ) 2 w x w y p = p w wh Differentiation rules w = 0 w ph Rw = Rw w wh 5

6 1.2.2 Optimization by differentiation Minimize: f(w) = E { s(t) 2} = w H Rw p H w w H p σ 2 s Solution 2: f w = Rw p = 0 w o = R 1 p 6

7 1.2.2 Completing the square Minimize: f(w) = E { s(t) 2} = w H Rw p H w w H p σ 2 s Solution 3: Completing the squares f(w) = w H Rw p H w w H p σ 2 s = ( w R 1 p ) H R ( w R 1 p ) p H R 1 p σ 2 s Since R is positively definite, it follows that w o = R 1 p. 7

8 1.3 Linearly Constrained Minimum Variance (LCMV) Beamformer Multiple signal scenario x(t) = s(t)a( 0 ) L i=1 s i(t)a( i ) n(t) Covariance matrix for the input signal x(t) (all signals are uncorrelated) R = E{x(t)x H (t)} = σ 2 sa( 0 )a H ( 0 ) L σi 2 a( i )a H ( i ) σni 2 The power of the output signal = w H x(t) can be decomposed as i=1 E { 2} = w H Rw = σ 2 s w H a( 0 ) 2 w H R NI w 8

9 1.3 Linearly Constrained Minimum Variance (LCMV) Beamformer LCMV optimization criterion { min w wh R NI w, subject to H( 0 ) = w H a( 0 ) = g Lagrangian function L(w, λ), Lagrange multiplier λ = λ x jλ y L(w, λ) = w H R NI w Re { λ (w H a( 0 ) g) } or L(w, λ) = w H R NI w λ x Re { w H a( 0 ) g } λ y Im { w H a( 0 ) g } 9

10 1.3 Linearly Constrained Minimum Variance (LCMV) Beamformer Lagrangian: L(w, λ) = w H R NI w Re { λ (w H a( 0 ) g) } Since we get w Re { w H a } = w H a a H w w 2 w L(w, λ) = R NIw λ 1 2 a( 0) = 0 and hence w = λ 1 2 R 1 NI a( 0) = 1 2 a Constraint: w H a( 0 ) = g λ 1 2 ah ( 0 )R 1 NI a( 0) = g 2g and hence λ = a H ( 0 )R 1 NI a( 0) LCMV Optimal solution w = Minimum value w H R NI w = g a H ( 0 )R 1 NI a( 0) R 1 NI a( 0) g 2 a H ( 0 )R 1 NI a( 0) 10

11 1.3 Comparison with the adaptive beamforming problem LCMV Optimal solution w = g a H ( 0 )R 1 NI a( 0) R 1 NI a( 0) Adaptive beamforming problem w o = R 1 p Matrix inversion lemma: If B 1 = A 1 C H D 1 C then B = A AC H ( CAC H D ) 1 CA. Use B 1 = R, A 1 = R NI, C H = a( 0 ) and D 1 = σ 2 s w o = R 1 p = ( σ 2 sa( 0 )a H ( 0 ) R NI ) 1 σ 2 s a( 0 ) = R 1 NI a( 0)σ 2 s (1 σ2 sa H ( 0 )R 1 NI a( ) 0) 1 σsa 2 H ( 0 )R 1 NI a( 0) 11

12 1.4 Optimum beamforming Multiple signal scenario x(t) = s(t)a( 0 ) L i=1 s i(t)a( i ) n(t) Covariance matrix for the input signal x(t) (all signals are uncorrelated) R = E{x(t)x H (t)} = σ 2 sa( 0 )a H ( 0 ) L σi 2 a( i )a H ( i ) σni 2 The power of the output signal = w H x(t) can be decomposed as i=1 E { 2} = w H Rw = σ 2 s w H a( 0 ) 2 w H R NI w 12

13 1.4 Optimum beamforming Maximize the signal to noise and interference ratio SNIR SNIR = σ2 s w H a( 0 ) 2 w H R NI w Introduce: y = R 1 2 NI w and w = R 1 2 NI y Rayleigh quotient SNIR = σ 2 s = σ2 s w H a( 0 )a( 0 ) H w w H R 1/2 NI R1/2 NI w. y H R 1 2 NI a( 0)a H ( 0 )R 1 2 NI y y H y The optimum y is the maximum eigenvector of R 1 2 NI a( 0)a H ( 0 )R 1 2 NI Since R 1 2 NI a( 0)a H ( 0 )R 1 2 NI R 1 2 NI a( 0) = a H ( 0 )R 1 NI a( 0) R 1 2 NI a( 0) it is concluded that y = R 1 2 NI a( 0) is the optimum solution Optimal weight vector w = R 1 2 NI y = R 1 NI a( 0) Consistent with the previous adaptive and LCMV beamforming problems 13

14 1.5.3 Matlab programming tasks: Optimum beamformer ULA with d = λ/2 and array response vector given by 1 jπ sin() a() = e. jπ sin()(m 1) e Multiple signal scenario x(t) = s(t)a( 0 ) L i=1 s i(t)a( i ) n(t) Covariance matrix for the input signal x(t) (all signals are uncorrelated) R = E{x(t)x H (t)} = σ 2 sa( 0 )a H ( 0 ) Correlation vector p = E {x(t)s (t)} = σ 2 sa( 0 ) Optimum beamforming w o = R 1 p Resulting response function H() = w H o a() L σi 2 a( i )a H ( i ) σni 2 i=1 14

15 1.1.4 Matrix representation: Seminar 1: Signal Processing Antennas 4ED024, Sven Nordebo Consider I discrete spatial spatial angles i, i = 1,..., I The array response function at these spatial angles is then given by or, in matrix notation H( i ) = w H a( i ) H = [ w H a( 1 ),..., w H a( I ) ] = w H [a( 1 ),..., a( I )] = w H A where H is an 1 I row vector of discrete responses H = [H( 1 ),..., H( I )] and A is an M I response matrix given by A = [a( 1 ),..., a( I )] 15

16 1.5.3 Matlab programming tasks: Optimum beamformer Matlab Tips: The case with 0 = 10 degrees and 1 = 60 degrees above can be solved using the following Matlab code: M=5; phi0=10*pi/180; phi1=60*pi/180; sigman2=10ˆ(-30/10); a0=exp(-j*pi*(0:m-1) *sin(phi0)); a1=exp(-j*pi*(0:m-1) *sin(phi1)); R=a0*a0 a1*a1 sigman2*eye(m); p=a0; w0=inv(r)*p; phi=(-90:0.1:90)*pi/180; A=exp(-j*pi*(0:M-1) *sin(phi)); figure(1) plot(phi*180/pi,20*log10(abs(w0 *A))) axis([ ]) figure(2) polar(phi,max(0,6020*log10(abs(w0 *A)))) 16

17 Seminar 1: Signal Processing Antennas 4ED024, Sven Nordebo Matlab programming tasks: Optimum beamformer Plot of 20 log w H o a(θ) in Decibel (db) M = 5 0 = 10 1 = 60 M = 5 0 = 20 1 = 60 M = 5 0 = 30 1 = 60 M = 15 0 = 10 1 = 60 M = 15 0 = 20 1 = 60 M = 15 0 = 30 1 =