A Combinatorial Optimization Problem in Wireless Communications and Its Analysis

Transcription

1 A Combinatorial Optimization Problem in Wireless Communications and Its Analysis Ralf R. Müller EE Dept, NTNU Benjamin Zaidel Tel Aviv Rodrigo de Miguel SINTEF, Trondheim Vesna Gardašević EE Dept, NTNU Dongning Guo EECS Dept, Northwestern Aris Moustakas Physics Dept, NCUA Finn Knudsen Math Dept, NTNU

2 Introduction 2 Let with x C K and J C K K. Example (sphere): The Problem E := K min x X x Jx X = {x : x x = K} = E = minλ(j) Example 2 (cube): X = {+, } K =??? Example 3 (vector precoding): X = (4Z+) K =???

3 Introduction 2 Let with x C K and J C K K. Example (sphere): The Problem E := K min x X x Jx X = {x : x x = K} = E = minλ(j) Example 2 (cube): X = {+, } K =??? Example 3 (vector precoding): X = (4Z+) K =???

4 Introduction 2 Let with x C K and J C K K. Example (sphere): The Problem E := K min x X x Jx X = {x : x x = K} = E = minλ(j) Example 2 (cube): X = {+, } K =??? Example 3 (vector precoding): X = (4Z+) K =???

5 Introduction 2 Let with x C K and J C K K. Example (sphere): The Problem E := K min x X x Jx X = {x : x x = K} = E = minλ(j) Example 2 (cube): X = {+, } K =??? Example 3 (vector precoding): X = (4Z+) K =???

6 Application 3 The Gaussian Vector Channel Let the received vector be given by where t is the transmitted vector y = Ht+n n is uncorrelated (white) Gaussian noise H is a coupling matrix accounting for crosstalk In many applications, e.g. antenna arrays, code-division multiple-access, the coupling matrix is modelled as a random matrix with independent identically distributed entries (i.i.d. model). Crosstalk can be processed either at receiver or transmitter

7 Application 4

8 Application 5 Processing at Transmitter If the transmitter is a base-station and the receiver is a hand-held device one would prefer to have the complexity at the transmitter. E.g. let the transmitted vector be where x = s is the data to be sent. t = H (HH ) x Then, y = s+n. No crosstalk anymore due to channel inversion.

9 Application 6 Problems of Simple Channel Inversion Channel inversion implies a significant power amplification, i.e. x ( HH ) x > x x. In particular, let α = K N ; the entries of H are i.i.d. with variance /N. Then, for fixed aspect ratio α with probability. x ( HH ) x lim K x x = α

10 Vector Precoding 7 Lattice-Relaxation Precoding Tomlinson 7, Harashima & Miyakawa 72 Choose that representation that gives the smallest transmit power.

11 Vector Precoding 7 Lattice-Relaxation Precoding Tomlinson 7, Harashima & Miyakawa 72 Choose that representation that gives the smallest transmit power.

12 Vector Precoding 7 Lattice-Relaxation Precoding Tomlinson 7; Harashima & Miyakawa 72 Choose that representation that gives the smallest transmit power.

13 Vector Precoding 7 Lattice-Relaxation Precoding Tomlinson 7, Harashima & Miyakawa 72 Instead of representing the logical 0 by +, we present it by any element of the set {..., 7, 3,+,+5,...} = 4Z+. Correspondingly, the logical is represented by any element of the set 4Z. Choose that representation that gives the smallest transmit power.

14 Vector Precoding 8 General Relaxation Precoding Let B 0 and B denote the sets presenting 0 and, resp. Let (s,s 2,s 3,...,s K ) {0,} K denote the data to be transmitted. Then, the transmitted energy per data symbol is given by with and E = K min x X x Jx X = B s B s2 B sk J = (HH ). What is a smart choice for B 0 and B?

15 Replica Method 9 Zero Temperature Formulation Quadratic programming is the problem of finding the zero temperature limit of a quadratic energy potential. The transmitted power is written as a zero temperature limit E = lim β lim β lim K E J e βtr(x Jx) βk log x X βk log x X e βtr(jxx ) with β denoting temperature.

16 Replica Method 9 Zero Temperature Formulation Quadratic programming is the problem of finding the zero temperature limit of a quadratic energy potential. The transmitted power is written as a zero temperature limit E = lim β lim β lim K E J e βtr(x Jx) βk log x X βk log x X e βtr(jxx ) with β denoting temperature.

17 Replica Method 0 The Harish-Chandra Integral (also called the Itzykson-Zuber integral) Let P be any positive semi-definite matrix of bounded rank n and let J be bi-unitarily invariant. Then, lim K K log E J e KtrJP = n a= λ a (P) 0 R J ( w)dw with λ a denoting the positive eigenvalues of P and R J (w) denoting the R-transform of the spectral measure of J (Marinari et al. 94; Guionnet & Maïda 05).

18 Replica Method We want lim K with K E J log x X The Replica Method e βtr(jxx ) = lim K lim = lim K lim = lim K lim = lim K lim nk log E J nk log E J Q ab := K x ax b. ( x X n a= x a X nk log E J x X nk log E Q exp e βtr(jxx ) ) n ( e βtr Jx a x ) a K x n X n a= ( e tr Jβ n ) x a x a a= βλ a (Q) 0 R J ( w)dw

19 Replica Method We want lim K with K E J log x X The Replica Method e βtr(jxx ) = lim K lim = lim K lim = lim K lim = lim K lim nk log E J nk log E J Q ab := K x ax b. ( n x X a= x a X nk log E J x X nk log E Q exp e βtr(jxx ) ) n ( e βtr Jx a x a ) K x n X n a= ( e tr Jβ n ) x a x a a= βλ a (Q) 0 R J ( w)dw

20 Replica Method We want lim K with K E J log x X The Replica Method e βtr(jxx ) = lim K lim = lim K lim = lim K lim = lim K lim nk log E J nk log E J Q ab := K x ax b. ( n x X a= x a X nk log E J x X nk log E Q exp e βtr(jxx ) ) n ( e βtr Jx a x a ) K x n X n a= ( e tr Jβ n ) x a x a a= βλ a (Q) 0 R J ( w)dw

21 Replica Method We want lim K with K E J log x X The Replica Method e βtr(jxx ) = lim K lim = lim K lim = lim K lim = lim K lim nk log E J nk log E J Q ab := K x ax b. ( n x X a= x a X nk log E J x X nk log E Q exp e βtr(jxx ) ) n ( e βtr Jx a x ) a K x n X n a= ( e tr Jβ n ) x a x a a= βλ a (Q) 0 R J ( w)dw

22 Replica Method 2 We find lim K K E J log x X e βtr(jxx ) Laplace Integration = lim K lim = min Q:Pr(Q)>0 lim min Q:Pr(Q)>0 lim nk log E Q exp n log K n a= n a= βλ a (Q) n tr[qr J( βq)]. 0 βλ a (Q) 0 R J ( w)dw R J ( w)dw How to optimize over Q?

23 Replica Method 2 We find lim K K E J log x X e βtr(jxx ) Laplace Integration = lim K lim = min Q:Pr(Q)>0 lim min Q:Pr(Q)>0 lim nk log E Q exp n log K n a= n a= βλ a (Q) n tr[qr J( βq)]. 0 βλ a (Q) 0 R J ( w)dw R J ( w)dw How to optimize over Q?

24 Replica Method 2 We find lim K K E J log x X e βtr(jxx ) Laplace Integration = lim K lim = min Q:Pr(Q)>0 lim min Q:Pr(Q)>0 lim nk log E Q exp n log K n a= n a= βλ a (Q) n tr[qr J( βq)]. 0 βλ a (Q) 0 R J ( w)dw R J ( w)dw How to optimize over Q?

25 Replica Symmetry 3 Replica Symmetric (RS) Ansatz We assume a certain structure for a matrix Q. The easiest try is q + χ β q q q q q + χ... β q q Q := q q... q + χ β q q q q q + χ β with some parameters q and χ. This is a critical step. Sometimes, the structure of Q is more complicated.

26 Replica Symmetry 4 RS Solution Let P(s) denote the limit of the empirical distribution of the information symbols s,s 2,...,s K as K. Let q and χ be the simultaneous solutions to 2 q = argmin z 2qR ( χ) 2xR( χ) DzdP(s) x B s z χ = argmin 2qR ( χ) 2xR( χ) z DzdP(s) 2qR ( χ) x B s where Dz = exp( z 2 /2)dz/ 2π, R( ) is the R-transform of the limiting eigenvalue spectrum of J, and 0 < χ <. Then, replica symmetry (RS) implies as K. K min x X x Jx q χ χr( χ)

27 Replica Symmetry 5 Some R-Transforms I : R(w) = HH : R(w) = αw Marchenko-Pastur (MP) law (HH ) : R(w) = α ( α) 2 4αw 2αw inv. MP

28 Replica Symmetry 6 Odd Integer Quadrature Lattice

29 Replica Symmetry 7 Complex Lattice Precoding 8 RS Solution Lower Bound 7 E [db] π ( α) α α

30 Replica Symmetry 8 8 RS Solution Lower Bound K=27 K= E [db] α

31 Replica Symmetry Breaking 9 -Step Replica Symmetry Breaking Q := µ β { columns }}{ ai aaaaaaaaaaaaaaaa q +p+ χ β q +p q q q q q +p q +p+ χ β q q q q q q q +p+ χ β q +p... q q q q q +p q +p+ χ β q q q q q q q +p+ χ β q +p q q q q q +p q +p+ χ β with the macroscopic parameters q,p and χ and the blocksize µ β.

32 Replica Symmetry Breaking 20 -Step Replica Symmetry Breaking E = ( q +p+ χ ) R( χ µp) χ µ µ R( χ) q(µp+χ)r ( χ µp) The macroscopic parameters q, p, χ and µ are given by 4 coupled non-linear equations. Solving those equations numerically is a tedious and tricky task.

33 Replica Symmetry Breaking 2 8 RS Solution Lower Bound RSB Solution 7 6 K=27 K=64 E [db] α

34 Convex Precoding 22 Complex Convex Relaxation... allows for convex programming (and is replica symmetric).

35 Convex Precoding Energy Penalty Comparison (QPSK) RS Solution Lower Bound RSB Solution L=2 RSB Solution L=3 Discrete Lattice (L=2) Simulation Results: K=27 Discrete Lattice (L=2) Simulation Results: K=64 CR QPSK RS Solution CR QPSK Simulation Results: K=64 E [db] α

36 Inverting Singular Channels 24 Inverting Singular Channels What happens if the MP-law has a mass point at zero (K > N)? Can we precode without interference? The precoder produces The received signal becomes lim argmin ǫ 0 x X x (HH +ǫi) x K y = lim ǫ 0 HH (HH +ǫi) x+n. If the energy is finite, there is no interference.

37 Inverting Singular Channels 24 Inverting Singular Channels What happens if the MP-law has a mass point at zero (K > N)? Can we precode without interference? The precoder produces The received signal becomes lim argmin ǫ 0 x X x (HH +ǫi) x K y = lim ǫ 0 HH (HH +ǫi) x+n. If the energy is finite, there is no interference.

38 Inverting Singular Channels 24 Inverting Singular Channels What happens if the MP-law has a mass point at zero (K > N)? Can we precode without interference? The precoder produces The received signal becomes lim argmin ǫ 0 x X x (HH +ǫi) x K y = lim ǫ 0 HH (HH +ǫi) x+n. If the energy is finite, there is no interference.

39 Inverting Singular Channels 25 Overloaded Convex Precoding. quadrant The probability that a random N dimensional subspace in K real dimensions intersects the. K-tant is N P(K,N) = 2 K l=0 ( K l ) subspace As K,N to infinity, we get K < 2N P(K,N) = /2 K = 2N 0 K > 2N

40 Inverting Singular Channels 25 Overloaded Convex Precoding. quadrant The probability that a random N dimensional subspace in K real dimensions intersects the. K-tant is N P(K,N) = 2 K l=0 ( K l ) subspace As K,N to infinity, we get K < 2N P(K,N) = /2 K = 2N 0 K > 2N

41 Inverting Singular Channels 25 Overloaded Convex Precoding. quadrant The probability that a random N dimensional subspace in K complex dimensions intersects the. K-tant is 2N P(K,N) = 2 2K l=0 ( 2K l ) subspace As K,N to infinity, we get K < 2N P(K,N) = /2 K = 2N 0 K > 2N

42 Inverting Singular Channels 26 Overloaded Convex Precoding P(K,N) N=4 N=8 N=6 N=32 N=64 N= α

43 Open Problems 27 Wanted lim K K log E A,B e KtrAPBP = f {R A ( ),R B ( ),...}... or other more complicated exponents.

44 Appendix 28 Negative Entropy 0 4 entropy 0 3 RSB S = χr( χ) χ R( w)dw S RS α The closer the entropy is to zero, the better the RSB approximation.