First, we will find the power allocation that maximizes the bit rate for a given zero-forcing transceiver. To this end, we use the Karush-Kuhn-Tucker (KKT) condition [77]. Let σs2∗k be a local maximum for the optimization problem in (4.6). Then there exists constants α and µk, for k = 0, 1, · · · , M − 1 such that:
By solving the above conditions, the optimal power allocation is given by σs2k = P0
M[F†F]kk
. (3.3)
From (3.3), we can see that the power allocation depends only on the transmitter
Next, we will design the optimal zero-forcing transceiver that maximizes the bit rate in (3.5). Suppose the P × N channel matrix H has rank K. Let the singular value decomposition of H be
H = U Λ 0 0 0
V†, (3.6)
where the K × K diagonal matrix Λ contains the nonzero singular values of H.
The P × P matrix U and the N × N matrix V are unitary. We assume that the elements of Λ are in nonincreasing order and K ≥ M so that solutions of zero-forcing transceivers exist.
Lemma 3.1 Without loss of generality, we can express F to be of the following form:
F = V A 0
, (3.7)
for appropriate K × M matrix A of rank M.
Proof: Suppose (G, F) is a transceiver pair that satisfies the zero-forcing condi-tion. As V is unitary, F can always be represented as
F = V A A1
, (3.8)
where A is a K × M matrix, A1 is an (N − K) × M matrix, and the notation T denotes the transpose. Define a new transceiver F′ as
F′ = V A 0
. (3.9)
Then we have
GHF′ = GHF. (3.10)
Therefore, when we replace the transmitter by F′, the new system still satisfies the zero-forcing condition GHF = IM. As the receiver is not changed, the new system has the same subchannel noise variances and hence the same bit rate performance. Now, let us compare the transmit power of F and F′ for the same Λs. The transmit power when we use F is
Tr(FΛsF†) = Tr(AΛsA†) + Tr(A1ΛsA†1). (3.11) Note that the transmit power with F′is Tr(F′ΛsF′†) = Tr(AΛsA†) ≤ Tr(FΛsF†).
This means a transmitter of the form in (3.7) is no loss of generality. △△△
Using Lemma 3.1 and Lemma 2.1, the receiver G is given by
G = (A†Λ2A)−1[ A†Λ 0 ]U†, (3.12) where A is the matrix given in (3.7). In this case, the noise variance at the k-th subchannel becomes
σe2k = N0[GG†]kk = N0[(A†Λ2A)−1]kk. (3.13) Note that the transmitter and receiver in (3.7) and (3.12) have the same form as those in [28] and [14]. The transceivers in [28] and [14] are designed for minimizing the transmit power for a given bit allocation, while we jointly design the optimal transceiver and bit allocation for maximizing the transmission rate. Lemma 3.1 lead us to conclude that the matrix A in (3.7) is the only part of the transceiver left to be designed. Using the expression of F in Lemma 3.1 and the expression of σ2ek in (3.13), the bit rate in (3.5) becomes
B = log2
( P0
MN0Γ)M1 Φ
, (3.14)
where Φ =QM −1
k=0 [A†A]kk[(A†Λ2A)−1]kk. To maximize b, we need to find A that
Optimal structure of A. Applying the Hadamard inequality [41], we have The equality holds if and only if the matrix A satisfies the following two con-ditions: 1) A†A is diagonal, and 2) A†Λ2A is diagonal. The first condition means that the columns of A are orthogonal, while the second means that the columns of ΛA are orthogonal. As ΛA is orthogonal, we can express it as ΛA = QD, for some K × M unitary matrix Q such that Q†Q = IM, and some M × M nonsingular diagonal matrix D. As Λ is nonsingular, we can write A = Λ−1QD. Then the product of the two determinant quantities in (3.16) becomes det[A†A]det[(A†Λ2A)−1] = det(Q†Λ−2Q). Hence the bit rate in (3.14)
Note that the bit rate in (3.17) is independent of D. Without loss of generality, we can choose D to be any M × M nonsingular diagonal matrix. For example, we can choose D = IM. To achieve the maximal bit rate, we need to find Q that minimizes det(Q†Λ−2Q).
Optimal Q: We can find Q with the help of the Poincar´e separation theorem [41], which is presented below for convenience.
Poincar´e separation theorem [41]: Let B be an n × n Hermitian matrix and C be an n × r unitary matrix with C†C = Ir. Then we have ρi(B) ≤ ρi(C†BC) ≤ ρn−r+i(B), i = 0, 1, · · · , r − 1, where the notation ρi(X) denotes the i-th smallest eigenvalue of X.
By the Poincar´e separation theorem, we have [Λ−2]ii ≤ ρi(Q†Λ−2Q), i = 0, 1, · · · , M − 1, where we have used the property that the diagonal elements of Λ are in nonincreasing order. Since the diagonal matrix Λ is nonsingular,
we know that Λ−2 is positive definite and [Λ−2]ii > 0 for i = 0, 1, · · · , K − 1.
where ΛM is an M × M diagonal matrix whose diagonal elements consist of the M largest singular values of H. The inequality in (3.19) becomes an equality when we choose
Using (3.7) and (3.12), the optimal transceiver is given by F = V Λ−1M
0
, G = [ IM 0 ]U† . (3.22) The maximal bit rate in (3.17) is given by b = log2[(M NP00Γ)Mdet(Λ2M)]. Substitut-ing (3.22) into (3.3) and (3.13), we have
σs2k = P0
M[Λ2M]kk and σe2k = N0. (3.23) Using (2.19), the number of bits allocated to the k-th subchannel becomes
bk = log2
We can see that more bits are assigned to subchannels that correspond to larger singular values of the channel.
Remarks:
1. Note that if we choose D = ΛM, we have F = V IM
0
, G = [ Λ−1M 0 ]U†. (3.25)
In this case, σs2k = P0/M and all subchannels are assigned the same power.
The bit allocation and bit rate are the same as the case when we choose D = IM. This is because the signal to noise ratio σs2k/σ2ek is not affected by D. Therefore, bit assignment and hence bit rate performance will be the same.
2. In (4.50), the bits are not integers in general. We can use truncation, i.e., ˜bk = ⌊bk⌋, where the notation ⌊z⌋ denotes the largest integer that is less than or equal to z. Zero bits may be assigned to some subchannels (˜bk = 0 if P0[Λ2M]kk < MN0Γ) and the power allocated to these subchannels is wasted. In this case, we will remove the worst subchannel and compute bit and power allocation in the remaining subchannel. We continue like this until all the power is used by subchannels with nonzero bits. The iterative bit allocation algorithm is given below.
Integer bit allocation algorithm:
Let M0 be the number of subchannels that will be assigned nonzero bits.
Initially, set M0 = M.
(a) Compute ξk = PM0[Λ0N2M0]Γkk for k = 0, 1, · · · , M0− 1.
(b) If ξk ≥ 1 for k = 0, 1, · · · , M0 − 1, then go to step (c). Else, if ξk < 1 for some subchannels, set M0 = M0− 1 and go to step (a).
(c) Compute the bit allocation by bk = ⌊log2(1 + ξk)⌋ for 0 ≤ k < M0. For M0 ≤ k < M, we set bk= 0.