Alternative proof for claim in [1]

Transcription

1 Alternative proof for claim in [1] Ritesh Kolte and Ayfer Özgür Aydin The problem addressed in [1] is described in Section 1 and the solution is given in Section. In the proof in [1], it seems that obtaining the conclusion of λ i P i being equal for all i = 1,..., k from equations 15 and 16 uses the fallacious argument: If fx gx and x = x satisfies this with equality, then x maximizes fx. While the final result is correct, the proof presented in Section arrives at it in a careful manner. 1. Problem Consider a multiple antenna system with t transmit and r receive antennas. The channel may be modeled as y = Hx + z, where z CN I r, and H C r t. Assuming that H is known at the transmitter and the receiver and that HH has eigenvalues λ 1,..., λ n, n = min{r, t}, suppose that by moving the antennas we are able to choose λ 1,..., λ n. The only constraint, other than λ i 0 for all i, is n λ i = rt. We assume, as usual, that trk x P, i.e. n P i P, where P i is the power allocated to the subchannel corresponding to eigenvalue λ i. Characterize the optimal eigenvalues and power allocations.. Solution If the eigenvalues are fixed, then the capacity is given by the water-filling solution: C = log 1 + λ i P i + where P i = µ 1 λ i and n P i = P. Assume that the optimal eigenvalues are denoted by λ 1, λ,..., λ n and similarly, the optimal power allocations are denoted by P1, P,..., Pn. We will first note that λ i = 0 Pi = 0. Proof. Assume λ i > 0 when P i = 0 for some i. Then by decreasing λ i to zero and increasing some λ j for which P j > 0, we can increase capacity. Hence, λ 1,..., λ n cannot be optimal if λ i > 0 when P i = 0 for some i. Similarly, we can not have Pi > 0 when λ i = 0 for the optimal power allocation P1..., Pn. We conclude that λ i = 0 iff Pi = 0. The problem at hand is: Maximize log 1 + λ i P i 1

2 subject to P i = P, λ i = rt, P i 0, λ i 0. Introducing Lagrangian multipliers a, b, {c i }, {d i } for the constraints respectively, the Lagrangian is: k k k L λ, P, a, b, c, d = log1 + λ i P i + a λ i rt + b P i P c T λ d T P. The KKT conditions which are necessary for an optimal solution but not sufficient since the problem is not convex in this case are given by P i 1 + λ i P i + a c i = 0, c i 0, c i λ i = 0, λ i 1 + λ i P i + b d i = 0, d i 0, d i P i = 0 and the feasilibity and sum-constraint satisfaction of λ i and P i. Eliminating c i s and d i s, we obtain P i λ i a = and b =, if λ i > 0 and P i > λ i P i 1 + λ i P i These two equalities imply that P i = aλ b i when λ i > 0 and P i > 0, which also holds trivially for the zero λ i and P i s. Summing over all i gives P = a rt. Hence, b P i = P rt λ i i = 1,..., n This is necessarily going to be true for the optimal λ i and P i, so we can substitute for P i in the original objective function to obtain the following problem: Maximize subject to log 1 + P rt λ i λ i = rt, λ i 0. We will first solve this problem for the case n =. In this case, we can denote the two eigenvalues by λ and rt λ where 0 λ rt. The objective function is log 1 + Prt λ + log 1 + Prt rt λ. 3 Claim: The optimal λ, rt λ is either 0, rt, rt, 0 or rt, rt. Proof. Note the following two facts about the objective function: It is symmetric around λ = rt. It goes to + as λ ±.

3 Setting the derivative of the function to zero gives us the following cubic equation: P rt λ3 3P λ + P L + λ L = 0. 4 λ = rt/ is always a root of this equation. There are going to be other roots placed symmetrically around λ = rt/, call them λ x and λ y. Case i λ = rt/ is a minimum. Then λ x and λ y can t be maxima; otherwise the function will either have to fall off to as λ ± or need to have more extremal points. See Fig. 1. Note that rt = 4 in all the figures, so the feasible domain for λ is [0, 4], marked by vertical dotted lines. This means that for 0 λ rt, 3 will attain the maximum value at the endpoints, i.e. 0 and rt. Case ii λ = rt/ is a maximum. Then λ x and λ y have to be minima, so that 3 as λ ±. Then the maximum value for 0 λ rt is going to be either at the endpoints 0 and rt Fig. or at λ = rt/ Fig rt = 4, P = Figure 1: Case i Now let s return to. Let λ 1,..., λ n be the optimal solution. Claim: All non-zero λ i have to be equal. Proof. by contradiction Assume that there are non-zero λ i not equal to each other; let them be λ 1 and λ without loss of generality. Now keeping all other λ i s fixed, consider the part of the objective function that depends only on λ 1 and λ : fλ 1, λ = log 1 + P rt λ 1 + log 1 + P rt λ From the previous claim, we know that either fλ 1 + λ λ, 0 or f 1 +λ, λ 1 +λ has to be larger than this. Choose λ1, λ = λ 1 + λ λ, 0 or 1 +λ, λ 1 +λ accordingly. Then, λ1, λ, λ 3,..., λ n satisifes the sum constraint and achieves a higher value for the objective function in, which is a contradiction. 3

4 3 rt = 4, P = Figure : Case ii a 4.4 rt = 4, P = Figure 3: Case ii b Hence, we have proved that λ 1,..., λ n should contain k non-zero eigenvalues all equal to rt/k for some 1 k n. The value of the optimal objective function is given by Ck = k log 1 + P rt. 5 k as a function of the non-zero eigenvalues k. To find the value of k, We can maximize 5 by setting the derivative to zero, which gives the following equation: ln 1 + P rt = k This needs to be solved numerically to get: P rt k + P rt. k = P rt. 6 4

5 We can check from the value of the second derivative of 5 at k to confirm that it is indeed a maximum. Using the expressions for the first and second derivative, it can also be shown that 5 with k 0, is a quasiconcave function. This allows us to conclude that if k is not an integer, either k or k will result in optimality. In particular, when P 0, 6 tells us that k = 1, and for P, k = n = min{r, t}. References [1] Nicolae Chiurtu and Bixio Rimoldi, Varying the antenna locations to optimize the capacity of multi-antenna Gaussian channels, Proceedings of the International Conference on Acoustics, Speech and Signal Processing ICASSP, 000, pp