Generalized Sidelobe Canceller and MVDR Power Spectrum Estimation Bhaskar D Rao University of California, San Diego Email: brao@ucsd.edu
Reference Books 1. Optimum Array Processing, H. L. Van Trees 2. Stoica, P., & Moses, R. L. (2005). Spectral analysis of signals (Vol. 1). Upper Saddle River, NJ: Pearson Prentice Hall.
Narrow-band Signals Assumptions D 1 x(ω c, n) = V(ω c, k s )F s [n] + V(ω c, k l )F l [n] + Z[n] l=1 F s [n], F l [n], l = 1,.., D 1, and Z[n] are zero mean E( F s [n] 2 ) = p s, and E( F l [n] 2 = p l, l = 1,..., D 1, and E(Z[n]Z H [n]) = σ 2 z I All the signals/sources are uncorrelated with each other and over time: E(F l [n]f m[p]) = p l δ[l m]δ[n p] and E(F l [n]f s [p]) = 0 The sources are uncorrelated with the noise: E(Z[n]F l [m]) = 0
Relationship between MVDR, MPDR, and LMMSE beamformers W mvdr = 1 D 1 x(n) = V(k s )F s [n] + V(k l )F l [n] + Z[n] S 1 Vs H S 1 n V s V s W lmmse = p s S 1 x l=1 n V s, W mpdr = 1 S 1 Vs H S 1 x V s x V s and For Uncorrelated sources W mvdr = W mpdr, W lmmse = cw mpdr with c = p s (V H s S 1 x V s ) = c > 0. Output of MPDR beamformer is of the form r[n] = W H mpdr x[n] = F s [n] + q[n], where q[n] = W H mpdr I[n] Now we find a LMMSE estimate of F s [n] given r[n]. with c = p s (V H s S 1 x V s ) ˆF s [n] = c r[n]
Special Cases Case 1: σ 2 z dominates the interference W mvdr V s This is often referred to as coherent combining. Case 2: Z[n] dominates and E(Z[n]Z H [n]) = S z = diag(σ 2 0, σ2 1,..., σ2 N 1 ) W mvdr S 1 z V s Sensors weighted by the quality of the measurements they provide Case 3: (D 1) Interferers dominate Interferers are cancelled. r[n] = W H mvdr x[n] = F s [n]
Generalized Sidelobe Cancelers (GSC) Efficient Structure to implement a beamformer when there are constraints. Generalize the MPDR to include M constraints, where M N and g C M 1. C H W = g If C = V s and g = 1, we get the MPDR/MVDR distortionless constraint. Beamformer Design: Minimize beamformer output power subject to multiple constraints. min W W H S x W subject to C H W = g. Using Lagrange multipliers, we can show W o = S 1 x C(C H S 1 C) 1 g x
Singular Value Decomposition (SVD) Assumptions: C C N M, N M (constraints M fewer than sensors N) and rank(c) = M (columns of C are linearly independent) [ ] C = UΣV H S = [U 1, U 2 ] V H = U 0 1 SV H 1. U C N N and V C M M are orthonormal matrices, i.e. UU H = U H U = I N N and VV H = V H V = I M M 2. R(C) = R(U 1 ), R(C H ) = R(V ), and N (C H ) = R(U 2 ). 3. S = diag(σ) C M M and σ 1 σ 2.. σ M > 0. Since the beamformer W C N 1, W = U 1 α + U 2 β = U 1 α BW a = W q + W q where B is any orthonormal set with R(B) = R(U 2 ) and W a C (N M) 1 vector. Will show α C M 1 or W q can be pre-determined by the constraints and only β C (N M) 1 or Wq be determined from data. has to
GSC structure Computing W q C H W = VSU H 1 W = VSU H 1 (U 1 α + U 2 β) = VSα = g So α = S 1 V H g leading to W q = U 1 α = U 1 S 1 V H g = (C H ) + g Looking at the beamformer output r[n] = W H x[n] = W H q x[n] + (W q ) H x[n] = r q [n] W H a B H x[n] = r q [n] W H a r B [n] r q [n] and r B [n] can be computed from x[n] knowing the constraints, and hence can be assumed to be readily available. Problem is one of predicting r q [n] using r B [n]. Choose W a to minimize the mean squared error W a = Σ 1 r B r B Σ rb r q
GSC block diagram Computing W a LMMSE estimate W a = Σ 1 r B r B Σ rb r q. Can use an adaptive algorithm like the LMS or RLS
Receive versus Transmit Beamforming So far we have considered receive beamforming: Source transmits F s [n] and e jωc τ l F s [n] is received by sensor l Transmit Beamforming: How to transmit a signal s[n] from the array to a single sensor receiver? Imagine we transmit s[n] from sensor l. Signal received by receiver is e jωc τ l s[n] If all sensors transmit the same signal s[n], the transmitted signal is 1 1 x[n] =. s[n]. 1 The transmit beamformer is W = [1, 1,..., 1] T The signal received is by superposition principle N 1 l=0 τ l s[n] + z[n] = [1, 1,.., 1] V e jωc s s[n] + z[n] = }{{} xt [n]v s + z l [n] W T
Transmit BF If sensor l transmits w l s[n], the signal transmitted by the array is w 0 w 1 x[n] =. s[n]. w N 1 The transmit beamformer is W = [w 0, w 1,..., w N 1 ] T The signal received is, by superposition principle, x T [n]v s + z l [n] = W T V s s[n] + z[n] = N 1 l=0 w le jωc τ l s[n] + z[n] To maximize the signal received, we try to insure the components add up coherently, i.e. w l e jωc τ l Due to power constraints, we may have to divide the power among the sensors Summary: Transmmit signal is V s s[n] and transmit BF is V s
SNR gain due to Transmit BF Single sensor (at origin) transmitting: Received signal is s[n] + z[n]. SNR = σ2 s σ 2 z Assuming transmit BF is 1 N V s, the received signal is given by N 1 l=0 1 N e +jωc τ l e jωc τ l s[n] + z[n] = Ns[n] + z[n] SNR gain due to transmit BF is N. SNR BF = Nσ2 s σ 2 z How about transmitting to multiple users?
Transmitting to D users Challenge: Minimize interference, i.e. signal s l [n] meant for receiver l is not received by receiver k, k l. We need their location k l. Let us assume we know that. Transmitted signal D 1 x[n] = W l s l [n] = W s[n] l=0 where W = [W 0, W 1,.., W D 1 ] and s[n] = [s 0 [n], s 1 [n],..., s D 1 ] T. W l chosen to insure receiver l receives signal s l [n] and not by the others. Zero-Forcing Precoder W V (V T V ) 1 where V = [V 0, V 1,..., V D 1 ] Signal received by user l is r l [n] = x T [n]v l + z l [n] = s T [n]w T V l + z l [n] = s T [n](v H V) 1 V H V l + z l [n] = s T [n](v H V) 1 V H Ve l + z l [n] where e l = [0,....0. }{{} 1, 0,..., 0] T position l = s l [n] + z l [n]
Power Spectrum Estimation of a Time Series Quality of Power Spectrum estimate given a time series of length N depends on Quality of bandpass filter (filter length L, assume FIR) Quality of power estimate, i.e. number of samples at the output of the bandpass filter.
MVDR Power Spectrum FIR filter of order L: H(z) = h 0 + h 1 z 1... h L 1 z (L 1) Impulse response vector h = [h 0, h 1,..., h L 1 ] T. Bandpass Filter defined by a distortionless constraint, i.e. H(e jω0 ) = 1. Here ω 0 is the frequency where the PSD is of interest H(e jωo ) = 1 or h H V(e jω0 ) = 1 where V(e jω0 ) = [1, e jω0,..., e j(l 1)ω0 ] T. Bandpass filter design: Minimize the filter output power E( y[n] 2 ), where L 1 y[n] = hl x[n l] = h H x[n], where x[n] = [x[n], x[n 1],..., x[n L+1]] T. l=1 Note that E( y[n] 2 ) = h H R xx h, where R xx = E(x[n]x H [n]) is a Toeplitz correlation matrix MVDR filter: h = 1 V H (e jω 0 )R 1 xx V(e jω 0 ) R 1 xx V(e jω0 ) MVDR Power spectrum: P(e jω0 ) = 1 V H (e jω 0 )R 1 xx V(e jω 0 )
Direction of Arrival (DOA) Estimation D 1 x[n] = V l F l [n] + Z[n], n = 0, 1,..., L 1 l=0 Here we have chosen to denote V s by V 0 for simplicity of notation. Note V l = V(k l ). We will refer to the L vector measurements as L snapshots. Goal: Estimate the direction of arrival k l or equivalently (θ l, φ l ), l = 0,..., D 1. Can also consider estimating the source powers p l, l = 0, 1,..., D 1, and noise variance σ 2 z. ULA : N 1 j V(ψ) = [e 2 ψ,..., e j N 1 2 ψ ] T N 1 j = e 2 ψ [1, e jψ, e j2ψ,..., e j(n 1)ψ ] T Note for a ULA, V(ψ) = JV (ψ), where J is all zeros except for ones along the anti-diagonal. Also note J = J 2.
ULA and Time Series x 0 [n] x 1 [n]. x N 1 [n] D 1 = l=0 1 e jψ l. e j(n 1)ψ l e j N 1 2 ψ l F l [n]+z[n], n = 0, 1,..., L 1 Considerable similarity with a time series problems composed of complex exponentials x[n] = D 1 l=0 c le jω l n + z[n], n = 0, 1,..., (M 1). In vector form x[0] 1 z[0] x[1] e jω l z[1]. x[m 1] D 1 = l=0. e j(m 1)ω l c l +. z[m 1] The time series, sum of exponentials, problem with M samples is equivalent to a ULA with M sensors and one (vector) measurement or one snapshot
Estimation of the Frequencies of the Exponential In array processing we have a limited number of sensors N but can compensate potentially with L snapshots. The time series problem is equivalent to one snapshot but a potentially large array if the number of samples M is large. One complex exponential (Noise free): x[n] = c 0 e jω0n, n = 0, 1,..., M 1. Interested in the frequency ω 0. Potential solution: Take the FFT and look at the peak. Another option is to note that x[n] = c 0 e jω0(n 1) e jω0 = e jω0 x[n 1]. In the noise free case, 2 samples is enough. Can solve a least squares problem min a M 1 n=1 x[n] ax[n 1] 2 and find ω 0 from the estimated a.
More than one exponentials Suppose we have two exponentials: x[n] = c 0 e jω0n + c 1 e jω1n, n = 0, 1,.., M 1. A FFT based approach may fail to resolve the frequencies if they are close and the number of samples M is small. Thumb rule ω 0 ω 1 > 2π M. Can show x[n] = a 1 x[n 1] + a 2 x[n 2] and so one can solve a 1 and a 2 using a least squares technique. In the absence of noise we just need 4 samples to set a system of equations to solve for the parameters. [ ] [ ] [ ] x[n] x[n 1] x[n 2] a1 = x[n 1] x[n 2] x[n 3] a 2 How to get the frequencies from the parameters a 1 and a 2? Can show that the zeros of the filter H(z) = 1 a 1 z 1 a 2 z 2 are located at e jω0 and e jω1.
D exponentials Suppose we have a series composed of D exponentials: x[n] = D 1 l=0 c le jω l n, n = 0, 1,.., M 1. A FFT based approach can be used with drawbacks as before: resolution and large number of samples may be needed. Can show x[n] = D k=1 a kx[n k] and so one can solve a k, k = 1,.., D using a least squares technique. In the absence of noise we just need 2D samples to set up a system of equations to solve for the parameters. x[n] x[n 1]... x[n D] a 1 x[n 1] x[n 2]... x[n D 1] a 2. x[n D + 1] =..... x[n D]... x[n 2D 1] How to get the frequencies from the parameters a k, k = 1,..D? Can show that the zeros of the filter H(z) = 1 D k=1 a kz k are located at e jω l, l = 0,.., D 1. a D
Relationship of Zeros of Filters to e jω l Claim: A signal with D exponentials, i.e. x[n] = D 1 l=0 c le jω l n, is exactly predictable from its D past samples, i.e. x[n] = D k=1 a kx[n k] or x[n] D k=1 a kx[n k] = 0 Consider a FIR filter H(z) = 1 P k=1 b kx[n k], P D, with input x[n] and denote its output by e[n] e[n] = x[n] P b k x[n k] = k=1 If we minimize n e[n] 2, what is the result? D H(e ω k )c k e jω k n The output will be zero. This is achieved by making H(e ω k ) = 0 at the D frequencies. Since the filter is of order P D, we can select the filter zeros to meet this criteria and minimize the output. k=1 The excess (P D) zeros can be placed arbitrarily.