Low-complexity optimization for large-scale MIMO

Transcription

1 Low-complexity optimization for large-scale MIMO Maria Gkizeli Telecommunications Systems Institute Technical University of Crete Chania, Greece This work was supported by the Seventh Framework Program of the European Commission under Grant PIOF-GA TRADENET

2 Signal Model and Problem Statement w data x. w 2 2 select K out of N switch 2 3. H 2 XN y1 y 2 y=hwx+n f=hw f H y w K K N Assumptions: N transmit and M = 2 receive antennas. x C: Transmitted symbol. w C N, w 0 = K: Beamforming vector. H = [ [h 1 h 2 ] H C M N h H = 1,1 h1,2... h1,n ] h2,1 h2,2... h2,n : Channel matrix (M = 2). h H m C1 N : Channel from Tx to mth Rx antenna, m = 1,2. n CN(0,σn 2 I): AWGN. 2 36

3 Signal Model and Problem Statement w data x. w 2 2 select K out of N switch 2 3. H 2 XN y1 y 2 y=hwx+n f=hw f H y w K K N Received vector: y = Hwx +n. Maximum-SNR filter: f = Hw. Filter-output SNR: { f E H Hwx 2} { E f H n 2} = E{ x 2 } σn 2 Hw

4 Signal Model and Problem Statement Objective: max Hw w 0 =K 4 36

5 Signal Model and Problem Statement Objective: where max Hw = max max H :,Iw w 0 =K I S w S = {I {1,2,...,N} : I = K} = { } I 1,I 2,...,I ( N. K) 4 36

6 Signal Model and Problem Statement Objective: where max Hw = max max H :,Iw w 0 =K I S w S = {I {1,2,...,N} : I = K} = For a given I S, w(i) = argmax H :,I w. w { } I 1,I 2,...,I ( N. K) 4 36

7 Signal Model and Problem Statement Objective: where max Hw = max max H :,Iw w 0 =K I S w S = {I {1,2,...,N} : I = K} = For a given I S, w(i) = argmax H :,I w. w Our objective becomes P : I opt = argmax I S { } max H :,Iw w { } I 1,I 2,...,I ( N. K) 4 36

8 Signal Model and Problem Statement Objective: where max Hw = max max H :,Iw w 0 =K I S w S = {I {1,2,...,N} : I = K} = For a given I S, w(i) = argmax H :,I w. w Our objective becomes P : I opt = argmax I S w opt = w(i opt ). { } max H :,Iw w { } I 1,I 2,...,I ( N. K) 4 36

9 Polynomial-Complexity Optimal TAS { } P : I opt = argmax max H :,Iw I S w In general, P examines all ( N K) antenna subset combinations 5 36

10 Polynomial-Complexity Optimal TAS { } P : I opt = argmax max H :,Iw I S w In general, P examines all ( N K) antenna subset combinations complexity O(N K ). 5 36

11 Polynomial-Complexity Optimal TAS { } P : I opt = argmax max H :,Iw I S w In general, P examines all ( N K) antenna subset combinations complexity O(N K ). Our contribution: If w = 1 or w k = 1, k = 1,2,...,K, and M = 2, then we identify with complexity O ( N 4) a subset S S that has size S = O ( N 3) and contains the optimal solution I opt of P. 5 36

12 Polynomial-Complexity Optimal TAS Notes: { } P : I opt = argmax max H :,Iw I S w S = O ( N K) O(N 4 ) S = O ( N 3). 6 36

13 Polynomial-Complexity Optimal TAS Notes: { } P : I opt = argmax max H :,Iw I S w S = O ( N K) O(N 4 ) S = O ( N 3). S is the same for (i) w = 1 or (ii) w k =

14 Polynomial-Complexity Optimal TAS Notes: { } P : I opt = argmax max H :,Iw I S w S = O ( N K) O(N 4 ) S = O ( N 3). S is the same for (i) w = 1 or (ii) w k = 1. If w = 1, then and (i) I opt = argmax I S σ max (H :,I ) = argmax H :,I 2 I S (ii) w opt = argmax w H H H Hw. w =1, w 0 =K 6 36

15 Polynomial-Complexity Optimal TAS Case 1: M = 1. { } P : I opt = argmax max H :,Iw I S w H = h H, H :,I = h H I, and { h I w(i) = h I, w = 1 e j arg(hi), w k =

16 Polynomial-Complexity Optimal TAS Case 1: M = 1. { } P : I opt = argmax max H :,Iw I S w H = h H, H :,I = h H I, and { h I w(i) = h I, w = 1 e j arg(hi), w k = 1. Then, I opt = arg max I S arg max I S h I = argmax I S h I 1 = argmax I S h i 2, w = 1 h i, w k = 1. i I i I 7 36

17 Polynomial-Complexity Optimal TAS P : I opt = arg max I S arg max I S We define function select(u;k) = argmax I S whose cost is O(N). h I = argmax I S h I 1 = argmax I S h i 2, w = 1 h i, w k = 1 i I i I u I = argmax u I 1, u C N, I S 8 36

18 Polynomial-Complexity Optimal TAS P : I opt = arg max I S arg max I S We define function select(u;k) = argmax I S h I = argmax I S h I 1 = argmax I S h i 2, w = 1 h i, w k = 1 i I i I u I = argmax u I 1, u C N, I S whose cost is O(N). For M = 1, I opt = select(h;k) complexity O(N). 8 36

19 Polynomial-Complexity Optimal TAS Case 2: M = 2. H = [h 1 h 2 ] H. { } P : max max H :,Iw I S w 9 36

20 Polynomial-Complexity Optimal TAS Case 2: M = 2. { } P : max max H :,Iw I S w H = [h 1 h 2 ] H. We introduce angles φ [ 0, π 2] and θ ( π,π] and define the unit-norm 2 1 vector c(φ,θ) = [ sin(φ) e jθ cos(φ) ], (φ, θ) Φ = [ 0, π ] ( π,π]

21 Polynomial-Complexity Optimal TAS Case 2: M = 2. { } P : max max H :,Iw I S w H = [h 1 h 2 ] H. We introduce angles φ [ 0, π 2] and θ ( π,π] and define the unit-norm 2 1 vector c(φ,θ) = [ sin(φ) e jθ cos(φ) Cauchy-Schwartz Inequality: ], (φ, θ) Φ = a H c(φ,θ) a c(φ,θ) = a }{{} =1 [ 0, π ] ( π,π]

22 Polynomial-Complexity Optimal TAS Case 2: M = 2. { } P : max max H :,Iw I S w H = [h 1 h 2 ] H. We introduce angles φ [ 0, π 2] and θ ( π,π] and define the unit-norm 2 1 vector c(φ,θ) = [ sin(φ) e jθ cos(φ) Cauchy-Schwartz Inequality: ], (φ, θ) Φ = [ 0, π ] ( π,π]. 2 a H c(φ,θ) a c(φ,θ) = a max }{{} (φ,θ) Φ ah c(φ,θ) = a. =1 9 36

23 Polynomial-Complexity Optimal TAS Case 2: M = 2. { } P : max max H :,Iw I S w H = [h 1 h 2 ] H. We introduce angles φ [ 0, π 2] and θ ( π,π] and define the unit-norm 2 1 vector c(φ,θ) = [ sin(φ) e jθ cos(φ) Cauchy-Schwartz Inequality: ], (φ, θ) Φ = [ 0, π ] ( π,π]. 2 a H c(φ,θ) a c(φ,θ) = a max }{{} (φ,θ) Φ ah c(φ,θ) = a. =1 Then, P becomes max max H :,Iw = max I S w max max I S w (φ,θ) Φ 9 36 w H H H. :,I c(φ,θ)

24 Polynomial-Complexity Optimal TAS P : max max max I S w (φ,θ) Φ w H H H :,Ic(φ,θ) We set u(φ,θ) = H H c(φ,θ)

25 Polynomial-Complexity Optimal TAS P : max max max I S w (φ,θ) Φ w H H H :,Ic(φ,θ) We set u(φ,θ) = H H c(φ,θ). Then, H H :,I c(φ,θ) = u I(φ,θ) 10 36

26 Polynomial-Complexity Optimal TAS P : max max max I S w (φ,θ) Φ w H H H :,Ic(φ,θ) We set u(φ,θ) = H H c(φ,θ). Then, H H :,I c(φ,θ) = u I(φ,θ) and P becomes max max max I S w (φ,θ) Φ w H u I (φ,θ) 10 36

27 Polynomial-Complexity Optimal TAS P : max max max I S w (φ,θ) Φ w H H H :,Ic(φ,θ) We set u(φ,θ) = H H c(φ,θ). Then, H H :,I c(φ,θ) = u I(φ,θ) and P becomes max max max I S w (φ,θ) Φ w H u I (φ,θ) = max max max (φ,θ) Φ I S w w H u I (φ,θ)

28 Polynomial-Complexity Optimal TAS P : max max max I S w (φ,θ) Φ w H H H :,Ic(φ,θ) We set u(φ,θ) = H H c(φ,θ). Then, H H :,I c(φ,θ) = u I(φ,θ) and P becomes max max max I S w (φ,θ) Φ w H u I (φ,θ) = max w H u I (φ,θ). } {{ } O(N) max max (φ,θ) Φ I S w 10 36

29 Polynomial-Complexity Optimal TAS P : max max max I S w (φ,θ) Φ w H H H :,Ic(φ,θ) We set u(φ,θ) = H H c(φ,θ). Then, H H :,I c(φ,θ) = u I(φ,θ) and P becomes max max max I S w (φ,θ) Φ w H u I (φ,θ) = max w H u I (φ,θ). } {{ } O(N) max max (φ,θ) Φ I S w I opt (φ,θ) I(φ, θ) }{{} O(N) = (φ,θ) select(u(φ, θ); K). }{{} O(N) 10 36

30 Polynomial-Complexity Optimal TAS P : For fixed (φ,θ), max max (φ,θ) Φ max I S w w H u I (φ,θ) I(φ,θ) = select(u(φ,θ);k) = select u 1 (φ,θ) u 2 (φ,θ). u N (φ,θ) ;K

31 Polynomial-Complexity Optimal TAS P : For fixed (φ,θ), max max (φ,θ) Φ max I S w w H u I (φ,θ) I(φ,θ) = select(u(φ,θ);k) = select u 1 (φ,θ) u 2 (φ,θ). u N (φ,θ) Recall that, for n = 1,...,N, u n (φ,θ) = H H :,n c(φ,θ) H = 1,n sinφ+h2,n ejθ cosφ. ;K

32 Polynomial-Complexity Optimal TAS 12 36

33 Polynomial-Complexity Optimal TAS L 1,4 L 1,3 π/2 L 2,3 φ π/4 L 2,4 L 1,2 L 3,4 0 π π/2 0 π/2 π θ Intersection of two surfaces: L n,m = {(φ,θ) Φ : un (φ,θ) = u m (φ,θ) }

34 Polynomial-Complexity Optimal TAS π π/2 0 π/2 π 0 π/4 π/2 θ φ L 2,3 L 2,4 L 3, C B L 1,4 L 1,3 L 1,2 A Three-curve intersection Three-surface intersection: u n (φ,θ) = u m (φ,θ) = u l (φ,θ), n m, n l, & m l

35 Input: H C 2 N (channel matrix), K {1,...,N} (desired selection) S { } (set of candidates) Ln,m 0 n,m {1,...,N}, n m for{j1,j2,j3} {1,2,...,N}: j1 j2,j1 j3, j2 j3, d [H:,j1 H:,j3] 1 H:,j2 D 1 d 2 2 d1d2 if D 1 then Lj1,j2 1, Lj1,j3 1, Lj2,j3 1 ψ angle(d1d 2 )±cos 1 D λ angle ([ 1 e jψ] d ) µ ψ +λ c null ejλ H H :,j1 HH :,j2 e jµ H H :,j1 HH :,j3 u Hc I = select(u;k) if multiple entries equal the Kth order element, include all in I if I = K S S {I} elseif I = K +1 else I I {j1,j2,j3} { } S S I {j1,j2}, I {j1,j3}, I {j2,j3} I I {j1,j2,j3} { } S S I {j1}, I {j2}, I {j3} for{n,m}: Ln,m = 0, when a curve does not intersect with any other curve c null ([ H H :,n H H ]) :,m u Hc I = select(u;k) if multiple entries equal the Kth order element, include all in I if I = K else S S {I} I I {n,m} { } S S I {m}, I {n} Output: S

36 Simulation Results BER Random (M=2) Optimal (M=1) Proposed (M=2) Total Number of Antennas N Figure : Bit error rate versus total number of transmit antennas N for M = 2 receive antennas and selection of K = 6 transmit antennas. Unimodular beamforming

37 Simulation Results Number of Antenna Selection Sets Exhaustive search Proposed (bound) Proposed (actual) Total Number of Antennas N Figure : Complexity versus total number of transmit antennas N for M = 2 receive antennas and selection of K = 6 transmit antennas. Unimodular beamforming

38 Summary - We developed an algorithm that has complexity O(N 4 ) and identifies a candidate set of size O(N 3 ) that contains the optimal solution of the following problems. Maximum-SNR TAS for N 2 systems with unimodular beamforming. Maximum-SNR TAS for N 2 systems with unit-norm beamforming. Maximum-SNR RAS for 2 N systems with unit-norm beamforming. I opt = argmax I S w opt = argmax w =1, w 0 =K σ max (H :,I ) = argmax H :,I 2, for H 2 N. I S w H H H Hw, for H 2 N. - For all the above problems, the new candidate set is identical! - The principles of our algorithm may be extended to the M > 2 case

39 Signal Model -Stage I (Multiple-Access Stage) Consider FSK modulation. Transmitted signals at the nth transmission, n = 1,2,...,N: Node A [ ] d A (n) [ ] {0,1}, [ ] 0 1 da (n) x A (n) = or = d A (n) Node B [ ] d B (n) [ ] {0,1}, [ ] 0 1 db (n) x B (n) = or = d B (n) Received vector at the nth transmission, n = 1,2,...,N: y R (n) = h AR x A (n)+h BR x B (n)+w(n)

40 Signal Model - Stage I (Multiple-Access Stage) Concatenate signals from N transmissions. Transmitted signal matrix: x A (1) x B (1) X = [ ] x A (2) x B (2) x A x B =.. x A (N) x B (N) 2N 2 Received vector at the relay: y R (1) y R =. = [ ] [ ] h x A x AR B +w = Xh+w. h BR y R (N)

41 Signal Model - Stage I (Multiple-Access Stage) GLRT-optimal detection: min X,h y R Xh 21 36

42 Signal Model - Stage I (Multiple-Access Stage) GLRT-optimal detection: min X,h y R Xh = P : X = argmin{ y R Xh GLRT } X where h GLRT = argmin y R Xh. h 21 36

43 Signal Model - Stage II (Broadcast Stage) ˆX = (x A,x B ) (d A,d B ) XOR d R FSK x R Received signals at nodes A and B: y A = h RA x R +w A, y B = h RB x R +w B. GLRT-optimal detection at node A: ˆx R = argmin xr {min hra y A h RA x R 2} = Q : arg max x R x T R y A ˆx R ˆd R XOR(,d A ) ˆd B = ˆd R d A 22 36

44 Problem Statement P : X = argmin y R Xh GLRT X Q : ˆx R = argmax x R x T R y A Optimal solution of Q complexity O(NlogN). [Alevizos-Fountzoulas-Karystinos-Bletsas, IEEE T-COM, 2016] In general, optimal solution of P O(2 2N ). Correlation between bit sequences: ρ = x T A x B = d T A d B. x A = x B, iff ρ = N. x A x B, iff 0 ρ N 1. Solve P for each ρ: X 0, X 1,..., X N

45 Case ρ = N Case ρ = N: X = [x A x A ] = x A [1 1]. h GLRT = 1 2N [1 1 ]xt A y R = P N : X N = argmin y R 1 x A N x Ax T A y R = argmax x T A y R x A Therefore, P N is equivalent to Q (noncoherent FSK detection) O(NlogN)

46 Case ρ = 0 Case ρ = 0: x B = x c A = 1 N x A P 0 : X0 = argmax x A [x A x c A ]T y R We have shown that P 0 : b max T z b {±1} N where b = 2d A 1 and z(n) = y R (2n 1) y R (2n), n = 1,...,N. Therefore, P 0 is equivalent to noncoherent PSK detection O(NlogN). [Mackenthun, IEEE T-COM, 1994] 25 36

47 Case ρ = 1,2,...,N 1 Case ρ = 1,2,...,N 1: h GLRT = ( X T X ) 1 X T y R = P ρ : ( ) 1 X ρ = argmax X T 2 X X T y R. x T A x B=ρ 26 36

48 Case ρ = 1,2,...,N 1 Case ρ = 1,2,...,N 1: h GLRT = ( X T X ) 1 X T y R = P ρ : ( ) 1 X ρ = argmax X T 2 X X T y R. x T A x B=ρ Consider c C 2, c = 1. Cauchy-Schwarz Inequality: R{c H a} a H c a c = a. { ( ) 1 } P ρ : max max R c H X T 2 X X T y R. c =1 x T A x B=ρ 26 36

49 Case ρ = 1,2,...,N 1 = P ρ : max c =1 { max d T A d B=ρ n=1 } N d A (n)λ n,1 +d B (n)λ n,2 with ] λ n,1 = R {z(n)c H}[ 1 N+ρ 1, λ n,2 = R {z(n)c H}[ 1 N ρ Recall that N+ρ 1 N ρ z(n) = y R (2n 1) y R (2n), n = 1,2,...,N, ρ indicates the number of equal symbols between d A and d B. ]

50 Case ρ = 1,2,...,N 1 = P ρ : max max c =1 I =ρ max d A (1),...,d A (N) { n I (λ n,1 +λ n,2 ) d A (n) + A (n)λ n,1 +(1 d A (n))λ n,2 n Id where I = { n {1,2,...,N} : d A (n) = d B (n) }. = P ρ : max max c =1 I =ρ n I λ n,1 +λ n,2 + + n I max(λ n,1, λ n,2 )

51 Simulation Results 10 0 Coherent Optimal (N=4) Optimal (N=8) Optimal (N=16) Suboptimal (K=200, N=4) Suboptimal (K=200, N=8) Suboptimal (K=200, N=16) Suboptimal (K=200, N=32) Suboptimal (K=200, N=64) BER at node R SNR (db) Figure : BER at relay node R versus SNR for GLRT-optimal and suboptimal noncoherent PNC with sequence length N = 4,8,16,32, and 64 and optimal coherent PNC

52 Conclusions Large-scale optimization problems and algorithms. MIMO: Large number of antennas antenna selection. Bidirectional relaying: Unknown channels processing of long sequences. Polynomial-complexity algorithms for optimal antenna selection and sequence detection. Collection of linear-complexity problems over continuous variables. Approximate algorithms with linear complexity. Strong relation with large-size linear algebra and combinatorics

53 Computation of the Intersections Three-curve intersection Three-surface intersection: u n (φ,θ) = u m (φ,θ) = u l (φ,θ), where n,m,l {1,...,N} with n m, n l, and m l. H H :,n c(φ,θ) = H H :,m c(φ,θ) H = H :,l c(φ,θ) or, equivalently, [ e jλ H H :,n HH :,m e jµ H H :,n H H :,l ] [ 0 c(φ,θ) = 0 ]

54 Unimodular Beamforming Unimodular beamforming approximation algorithm Consider the relaxed version of P max w H w 1,w 2 Ω K 1 H H :,IH :,I w 2. For fixed w 2, w 1 = e j arg(h I H H I w 2), while, for fixed w 1, w 2 = e j arg(h I H H I w 1). Cyclic maximization, starting from an initial w (0), w (t+1) = e j arg(hh :,I H :,Iw (t) ). Continue until w (t+1) w (t) ǫ or for a predefined number of steps

55 Linear-Complexity Suboptimal TAS π/2 L 2,3 L 1,4 L 1,3 B C A φ π/4 L 2,4 L 1,2 L 3,4 0 -π -π/2 0 π/2 π θ Sample (φ,θ) at P points O(PN)

56 Simulation Results 10 2 BER Random (M=2) Optimal (M=1) [25] (M=2) Optimal (M=2) Suboptimal (M=2, P=15) Suboptimal (M=2, P=81) Total Number of Receive Antennas N Figure : Bit error rate versus total number of receive antennas N for M = 2 transmit antennas and selection of K = 6 receive antennas. Total power constraint

57 Low-Complexity TAS for M > 2 Receive Antennas M > 2 receive antennas: - Define c(φ,θ) = sinφ 1 e jθ 1 cosφ 1 sinφ 2. e jθ M 2 cosφ 1...cosφ M 2 sinφ M 1 e jθ M 1 cosφ 1...cosφ M 2 cosφ M 1. - Generate P samples c(φ p,θ p ), p = 1,2,...,P - Form the N 1 vector u(φ p,θ p ) = H H c(φ p,θ p ) - Call select(u(φ p,θ p );K) (returns with O(N) the indices of the K largest in magnitude elements of u(φ p,θ p )). - Obtain P antenna selection subset candidates. - Overall complexity O(NP)

58 Simulation Results BER Random (M=3) Optimal (M=1) [25] (M=3) Suboptimal (M=3, P=24) Suboptimal (M=3, P=225) Suboptimal (M=3, P=784) Suboptimal (M=3, P=1600) Total Number of Received Antennas N Figure : Bit error rate versus total number of receive antennas N for M = 3 transmit antennas and selection of K = 6 receive antennas. Total power constraint