On the Optimal Phase Control in MIMO Systems with Phase Quantization

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1 On the Optimal hase Control in MIMO Systems with hase Quantization Majid Khabbazian, Kin-Kwong Leung and Mohammad A. Safari Department of Electrical and Computer Engineering University of British Columbia, Vancouver, BC V6T 1Z4, Canada {majidk, Department of Computer Science University of British Columbia, Vancouver, BC V6T 1Z4, Canada Abstract We consider the optimal phase control problem in wireless MIMO systems with quantized phases. We show that the problem is N-hard when number of receive antennas grows linearly with the number of transmit antennas. For the case where the number of receive antennas is constant, we show that the problem is no longer N-hard by providing a polynomial time algorithm. I. INTRODUCTION Transmit diversity is a key technology to improve the performance of a wireless communication system in a fading environment [1]. It is classified as open loop or closed loop, depending on whether there is any feedback information sent from the receiver to the transmitter [2]. Closed-loop diversity was first proposed by Gerlach and aulraj [4]. Based on the estimated channel gain matrix, the receiver calculates the optimal antenna weight vector and then sends it back to the transmitter. Since the feedback channel is usually of limited data rate, the optimal antenna weight vector has to be quantized before being sent back. If the quantization level is sufficiently dense, one may simply choose a quantized point that is closest in Euclidean distance to the optimal vector. In practice, however, the allowed feedback information is rather limited. To maximize the performance gain, a combinatorial optimization is needed. To find the optimal quantized vector, the brute force approach has computational complexity increasing exponentially with n T, which is the number of transmit antennas. To reduce the complexity, the Lloyd algorithm is used in [6]. However, it is suitable only when the number of transmit antennas is small, since its computational cost still grows rapidly with n T. In this paper, we consider the phase control problem, where all the feedback bits are used to adjust the transmission phases of the transmit antennas. Suboptimal solutions for this problem have been proposed in the literature. One example is the Co-phase Algorithm, which is of complexity O(n T [5]. Since its performance is far from optimal, a near-optimal solution approach called Local Search is proposed in [7]. Its complexity is O(n 2 T, which is higher than that of the Cophase algorithm. When the number of transmit antennas grows, whether efficient algorithm exists for optimal solution is still unknown. We will answer this question in this paper. Our results are summarized as follows: When the numbers of receive antennas grows linearly with transmit antennas, the problem is N-hard. When either the number of receive or transmit antennas is constant, polynomial time algorithm exists. The paper is organized as follows. The next section is the problem definition. The N-hardness of the first case is proved in Section III, by reducing the maximum cut, which is a wellknown N-complete problem in graph theory. In Section IV, we show the existence of polynomial time algorithm for the second case by construction. Simulation result is presented in Section V to show the performance differences between the optimal solution and the suboptimal solutions. The paper is concluded in Section VI. For readability, some proofs are omitted. II. ROBLEM DEFINITION We consider the phase control problem in a single user multiple-input multiple-output (MIMO system, with N transmit and M receive antennas. Let G be an M N complex matrix with entry g m,n representing the link gain between transmit antenna n and receive antenna m, and u =(u 1,...,u N be an N-dim complex column vector representing the transmit signal. Therefore the receive signal is v = Gu. (1 The phase control problem assumes that each component u n has a constant amplitude and the system is allowed to control the transmit signal phases through a feedback channel. Since the feedback channel has limited bandwidth, the phase of u n is quantized into L n discrete values, i.e., u n {exp(jθ 1 n,...,exp(jθ Ln n }, where 0 = θn 1 < < θn Ln < 2π. Letu l n = exp(jθn, l U n = {u 1,...,u 2 } and U = U 1 U N. Also let V = {Gu : U}, i.e., v V. Our goal is to maximize the /06/$20.00 (c 2006 IEEE

2 receive signal v = M m=1 v m 2, given the link gain matrix G and quantization set U. Together with (1, the phase control problem can be expressed as maximize subject to Gu u U. We denote f(u = Gu. To simplify our model and improve readability, we assume L n = L for all n {1,...,N}. However, all our results apply to the general model with different L n s. III. N-HARDNESS OF THE HASE CONTROL ROBLEM In this section, we will prove that when the number of receive antennas grows linearly with number of transmit antennas, the phase control problem is N-Hard. We will show the N-hardness by reducing the maximum cut (MAX CUT problem [3] to the phase control problem with two quantized phases, i.e., U n = {+1, 1} for all n {1,...,N}. Let g n be the n-th column of G and define the function { 1, if x =0; δ(x = (3 0, otherwise. Then we have Gu = = (2 N u n g n (4 n=1 N n=1 u n g n + m<n u m u n [g H mg n + g H n g m ](5 = C 2 m<n δ(u m + u n [g H mg n + g H n g m ] (6 = C +4 m<n δ(u m + u n Re { g H mg n }, (7 where x H is the Hermitian transpose of x, C = N n=1 u ng n + m<n [gh mg n + g H n g m ] and Re {z} is the real part of the complex number z. Given the link gain matrix G, the phase control problem (2 can be expressed as maximize m<n δ(u m + u n Re { } g H mg n (8 subject to u U. Therefore, any algorithm A that solves the phase control problem (2 solves (8. On the other hand, consider an undirected graph with the vertex set N = {1,...,N}.LetW be a real symmetric matrix where the entry W mn represnts the weight between vertices m and n. The MAX CUT problem is to find a subset S N such that W mn (9 m S,n S is maximized, where S = N S. We define { +1, if n S; u n = 1, if n S. The MAX CUT problem can be written as maximize m<n δ(u m + u n w mn subject to u U. (10 (11 Since W is a symmetric matrix, we can diagonalize it such that W = D H, where is a real orthonormal matrix composed of the eigenvectors of W and D is a real diagonal matrix. Any MAX CUT problem with N vertices can be transformed to the optimal phase control problem in an N N MIMO system with link gain matrix G = D H, U n = { 1, +1}. Note that some entries of D may be negative and hence G is a complex matrix in general. With the above reduction, any algorithm A that solves such a phase control problem can solve the MAX CUT problem and we have the following theorem: Theorem 1: The optimal phase control problem in an N N MIMO system is N-hard. Corollary 2: The optimal phase control problem in an N M MIMO system is N-hard when M grows linearly with N. roof: The case of M N is trivial. For M < N, consider that only M of the receive antennas are active. The link gains are represented by an αn αn matrix and the problem is at least as difficult as MAX CUT problem with αn vertices. Hence the problem is still N-hard. IV. OLYNOMIAL TIME ALGORITHM FOR SECIAL CASES We proved in the previous section that when M grows linearly with N, we do not have any efficient algorithms to solve the maximization problem (2. In this section, we investigate the cases of M or N being constant. The latter case, N being constant, is simple since we can exhaustively calculate all L N possible combinations and choose the optimal one. The complexity is O(M. However, for the other case, where M is fixed while N grows, whether polynomial time algorithms exist is still unknown. In the literature, some algorithms for suboptimal solutions were proposed, like [5], [7]. We will show by construction that polynomial time algorithm for optimal solution does exist. Before introducing the algorithm, we look at some properties of the discrete phase control algorithm. Since U is finite, we immediately have the following existence theorem: Theorem 3: An optimal transmit vector u = (u 1,...,u N U exists. That is, f(u f(u for all u U. Let v = (v 1,...,v M = Gu be the optimal receive vector and recall that g n is the n-th column of G. Define d(x, y =Re { x H y }. We remark that for unit vectors x, y, d(x, y measures how close they are. d(x, y =1if and only if x = y whereas d(x, y = 1 if and only if x = y, which means x and y are farthest apart. Lemma 4: d(u ng n, v >d(u n g n, v for all u n n n, u n U n, n {1,...,N}. 4273

3 The idea of Lemma 4 is that if there were another u n u n closer to v than u n,thisu n would produce another a greater v, which contradicts to the assumption v being optimal. Let E l n be the set {v V: d(u l ng n, v >d(u n g n, v, u n U n }. So v E l n implies that u n u l n. From the definition of E l n, we immediately have the following lemma. Lemma 5: E l n E l n = φ for any l l. For convenience in notation, we extend the sets of {θn} l and {u l n} for l beyond L with θn l+l = θn l and un l+l = u l n for all n, l. For each n {1,...,N},l {1,...,L}, letr l n is an M- dim row vector defined as r l n =(u l n u l 1 n g H n. (12 Let R n be an L M matrix, with r l n as the l-th row, and R be a block matrix: R 1 R =.. (13 R N Therefore, R is an M matrix. Define the sign function for matrices as follows: if Y = sign(x, (14 then Y is a matrix with the same dimension as X and the (i, j-th entry of Y is defined as 1, Re {X ij } > 0; Y ij = 0, Re {X ij } =0; (15 1, Re {X ij } < 0. Let For any v V, consider the vector σ v = sign(rv. (16 Σ={σ v : v V}. (17 Lemma 6: If v E l n, v E l n and l l, then sign(rv sign(rv. From Lemma 5 and 6, we see that if u u, then σ v σ v,where v = Gu and v = Gu. Therefore the function that maps u to the corresponding σ v is injective and we can define a function ξ :Σ U as ξ(σ(v = (u l1 1,...,ulN N (18 where v En ln for all n {1,...,N}. Although σ(v is an -dim vector, we will see later that Σ has only O(( 2M elements, among which are the ones corresponding to the optimal transmit phase u, namely ξ 1 (u 1. Therefore, we can find out u by comparing f(u 1 ξ 1 is a point to set function. for some potential candidates ξ 1 (u Σ, and the complexity is polynomial in N. To figure out the potential candidates in Σ, we need the following two theorems. The notation ρ 0 means that all components of the matrix ρ are non-zero. Lemma 7: There exists v V such that σ v 0 and ξ(σ v =u. Definition 1: Let rspan(φ be the set of vectors that are linear combinations of the rows of Φ. Definition 2: Let Φ be a p q matrix and A be a subset of {1,...,p}, A = p p. We construct a row submatrix of Φ, written as Φ[A], by taking the w-th row from Φ for all w A. Therefore, Φ[A] is a p q matrix. Lemma 8: For any p q real matrix Φ with rank-k and q-dim column real vector x, there are a subset A {1,...,p} and x rspan(φ such that sign(φ[ā]x =sign(φ[ā]x (19 and Φ[A]x =0, (20 where Ā = {1,...,p} A and Φ[A] has rank k 1. Moreover, the choice of x is unique for each A, up to scalar multiples. Lemma 7 ensures that the optimal solution is in Σ = {v Σ:v 0} and we will see that Lemma 8 is useful in figuring out all elements of Σ. While Lemma 8 applies to real matrices, R is a M complex matrix. However since the operation sign(rv only considers the real part of the product, we can view R in (16 as a 2M real matrix. In this sense, the effective rank of R can be 2M. We will interprete the rank of a complex matrix in this sense in the rest of the paper. i.e., a 2n n complex matrix can have rank at most 2n. Before going into the general optimal phase control algorithm, we look at a special case, which gives us more insight on how the general algorithm works. A. A Special Case If we assume that any 2M rows of R are linearly independent, we can solve our maximization problem (2 as follows. 1 Evaluate All v(i s: ( From Lemma 8, find out all by solving the equation 2M 1 possible v A s R[A]v A =0 (21 for each of the subset A with 2M 1 elements. Label them as v(i, and the corresponding ( subset as A(i, for i {1,...,N v }, where N v = 2M 1. 2 Define Σ (i: For i {1,...,N v }, define subset Σ (i of Σ as follows: Σ (i ={σ v Σ : σ v [Ā(i] = σv(i[ā(i] 0 and σ v [A(i] = 0.} (

4 Since {A(i} i {1,...,Nv} are all the possible subset with cardinality 2M 1, N v i=1 Σ (i =Σ. (23 3 Evaluate the Elements in Σ (i s: For each of i {1,...,N v }, the j-th entries of all elements in Σ (i are the same, which equal to the j- th entry of σ v(i, for all j Ā. These components can hence be determined since v(i is known from Step 1. Each of the remaining 2M 1 components can only be +1 or 1, implying that Σ (i 2 2M 1. We find out all 2 2M 1 possible elements of Σ (i. 4 Find Out the Optimal Transmit hase: Totally, we have ( 2M 1 2 2M 1 σ v s, which is polynomial in N. The solution to the maximization problem (2 can be obtained by comparing f(ξ(σ v for all these candidates σ v and this algorithm has a complexity polynomial to N. B. General Cases The above conclusion is obtained under the assumption that any 2M rows of R are linearly independent. If this assumption does ( not hold, the above algorithm may end up with more than 2M 1 2 2M 1 possible σ v s. For example, if there are set A with N/2 elements and R[A] has rank 2M 1, then in Step 3, we can determine only N/2 components of σ v(i and the rest would produce 2 N/2 different cases. The complexity of the algorithm then becomes exponential in N. In the discussion above, we note that in Step 3 of the algorithm, the number of undetermined components for the elements in Σ (i is actually 2 α, where α = A(i. Also, Lemma 8 is used to reduce a rank 2M matrix, R, to a set of rank 2M 1 matrices, R[A(i]. We see that we can do the operation again to yield a set of rank (2M 2 matrices, R[A(i 1,i 2 ]. When we do the operation recursively for 2M times, we end up with a set of rank-0 matrices, R[A(i 1,...,i 2M ]. In other words, A(i 1,...,i 2M s are empty sets and each of the corresponding Σ (i 1,...,i 2M has only one element. Therefore, the exponential term in the discussion disappears. Before proceeding to the general algorithm, we have to introduce the following lemma, which makes sure that we only have polynomial number of Σ (i 1,...,i 2M s. Lemma 9: Let R[A] be a rank-k matrix and R[A]v =0. Then there exists a subset B A such that B has exactly k elements, R[B] has rank k and R[B]v =0. When the linear independence assumption is not satisfied, the number of elements of the subset A in Lemma 8 may range from K 1 to. However, Lemma 9 ensures there ( are no more than K 1 distinct v A s since there are only ( K 1 choices of subset B. This will ensure that number of Σ (i 1,...,i 2M is polynomial. The general algorithm for solving the maximization problem (2 follows. Optimal hase Control Algorithm GivenalinkgainmatrixG and the phase quantizations set U, we construct matrix R from (12 and (13. Let K be the rank of R, where K 2M. The algorithm is composed of K levels. In each level, the ranks of R[A] is reduced by 1. 1 At level 1, we want to find out all possible v A. Each v A corresponds to a subset ( A {1,...,}. By Lemma 9 there are at most K 1 distinct v A s. We label each distinct v A as v(i 1, the corresponding A as A(i 1, and define Σ (i as in (22. For each σ v Σ (i 1,thewth component, w Ā(i 1, can be determined since it equals to the w-th component of σ v(ii. The rest of the components will be determined in the succeeding levels. Note that Σ (i =Σ. (24 all i i 2 At level k, k 2. For any σ v Σ (i 1,...,i k 1, Lemma 8 states that there are A(i 1,...,i k A(i 1,...,i k 1 and v(i 1,...,i k rspan(a(i 1,...,i k 1 such that and σ v(i1,...,i k [Ã] =σ v[ã(i 1,...,i k ] 0 (25 R[A(i 1,...,i k ]v(i 1,...,i k =0, (26 where Ã(i 1,...,i k =A(i 1,...,i k 1 A(i 1,...,i k and R[A(i 1,...,i k ] has rank K k. Therefore, for each A(i 1,...,i k 1, we have the following: Let = A(i 1,...,i k 1. We consider all the subsets of A(i 1,...,i k 1 with cardinality K k and then a set of distinct v(i 1,...,i k rspan(r[a(i 1,...,i k 1 ] are obtained. We also have A(i 1,...,i k A(i 1,...,i k 1 and Σ (i 1,...,i k Σ (i 1,...,i k 1, defined in a similar way as (22. As Step 3 in the algorithm for the special case, for any σ v Σ (i 1,...,i k,thej-th component of σ v equals to the j-th component of σ v(i1,...,i k, j Ã(i 1,...,i k. Moreover, since σ v Σ (i 1,...,i k Σ (i 1,...,i k 1, components belonging to A(i 1,...,i k 1 have been obtained in the preceding levels. The remaining unknown components are those in A(i 1,...,i k, and they will be ( determined in the succeeding levels. Since there are K k possible subsets of A(i 1,...,i k 1,wehave N v i k =1 ( where N v = Σ (i 1,...,i k =Σ (i 1,...,i k 1, (27 K k. 4275

5 3 At level K, R[A(i 1,...,i K ] has rank 0, which implies that A(i 1,...,i K is an empty set. All components of σ v has been determined. (There is only one element in every Σ (i 1,...,i K. Moreover, by induction, we have all i 1,...,i K Σ (i 1,...,i K =Σ. (28 Then Lemma 7 implies that the maximization problem 2 can be solved by comparing all f(ξ(σ v s, where σ v s are given by the algorithm at level K. Complexity Analysis Obviously, the construction of R and the evaluation of its rank can be done in O(M 2 computations. The complexity of level 1 follows. We decompose level 1 into different steps. The first step is to( find out all distinct v(i 1. There are at most N v = 2M 1 choices of subset B (refer to Lemma 9. For each of them, evaluation of v requires solving a system of 2M 1 equations, which has complexity of O(M 3. To label distinct v s as v(i 1, we do a sorting on all v s. This requires O(N v log N v computations. Hence, the overall complexity in this step is O(N v log N v. Since different A s may correspond to the same v(i 1, we have to find out the corresponding A(i 1 for each v(i 1, we need to figure out which row r i of R satisfies the equation r i v(i 1 0. This requires O( computations. Since {v(i 1 } mayhaveat most N v members, this step requires O(N v computations. To determine the components of σ v Σ (i 1,we need O( computations for each i 1.Thisstep requires O(N v computations in total. ( Since N v = 2M 1 ( 2M 1 and M is assumed to be constant, the overall complexity of level 1 is O(( 2M. When we apply the above arguments to level 2, all the complexity upper bounds are satisfied except that the cardinality of {v(i 1,i 2 } is higher than {v(i 1 }. To find out a suitable upper bound, we focus on a particular v(i 1. Let A(i 1 = K 1. Note that any subset B A(i 1 with rank of B equal K 1 would have yielded the same v(i 1, up to a scalar, by solving the equation R[B]v =0 (29 ( for v. At level 1, we have at most evaluated ( K 1 subsets of {1,...,} and K 1 of them have produced v(i 1. To ( evaluate vector v(i 1,i 2 at level 2, there are at most K 2 subsets of A(i 1. Since this relation holds for any other v(i 1, the ratio between the number of distinct A(i 1,i 2 to the number of A(i 1 must average output power optimal co phase number of antennas Fig. 1. solutions. The differences between the optimal solution and suboptimal be bounded by ( max 2M 1 N v /( K 2 K 1 (30 K 1 (K (31 < K (32 2M. (33 Therefore the number of v(i 1,i 2 is less than 2MN v, and the overall complexity of level 2 can be proved to be O(M( 2M. Similarly the complexity of level k is O(M k 1 ( 2M for k {1,...,K}. Since the number of levels is less than 2M, the overall complexity of the algorithm is O((1+ +M 2M 1 ( 2M O((M 2M. Keeping M to be constant, the complexity is O(( 2M, which is polynomial in N and L. V. SIMULATION In this section, we give the readers some idea about the performance difference between the optimal solution and suboptimal solution given by Co-phase[5] algorithm. We run a simple simulation with different numbers of transmit antennas. The number of receive antenna is 4. Different transmit-receive antenna pairs are subject to independent Rayleigh fadings. We generate 1000 fading scenarios. For each of them, we compute the receive signal strengths given by Co-phase algorithm and the optimal algorithm. Figure 1 shows that there is significant difference between the two when the number of transmit antennas is high. VI. CONCLUSION We consider the problem of phase control for closed-loop transmit diversity with limited feedback. The consideration of quantized antenna phase is of practical importance, since 4276

6 the bandwidth of the feedback channel is limited and the control commands need to be sent faster than the rate of change of the channel. We show that the problem is N-hard in general. For the cases of constant number of transmit or receive antennas, polynomial time algorithms for the optimal solution are proposed. REFERENCES [1] S. M. Alamouti, A simple transmit diversity technique for wireless communications, J. Select. Commun. vol. 16, no 8, pp , Oct [2] R. T. Derryberry, S. D. Gray, D. M. Ionescu, G. Mandyam and B. Raghothaman, Transmit diversity in 3G CDMA systems, IEEE Commun. Mag., pp , vol. 40, no. 4, Apr [3] M. R. Garey and D. S. Johnson, Computers and intractability. A guide to the theory of N-completeness, W.H. Freeman and Company, New York-San Francisco, [4] D. Gerlach and A. aulraj, Adaptive transmitting antenna arrays with feedback, IEEE Sig. roc. Lett., pp , vol. 1, no. 10, Oct [5] J. Hämäläinen, R. Wichman, erformance analysis of closed-loop transmit diversity in the presence of feedback errors, IEEE roc. IMRC, pp , [6] A. Narula, M. J. Lopez, M. D. Trott and G. W. Wornell, Efficient use of side information in multiple-antenna data transmission over fading channels, IEEE J. Select Commun., vol. 16, no. 8, Oct [7] C. W. Sung, S. C. Ip and K. K. Leung, A simple algorithm for closed-loop transmit diversity with a low-data-rate feedback channel, IEEE roc. acific Rim Conf. Communications, Computers, and Signal rocessing, Victoria, B.C., Canada, Aug