In this section, we consider the power-minimizing problem and rate-maximizing problem with integer bit allocation. With the constraint of integer bit allocation, the power-minimizing problem becomes
where Z+ denotes the set of nonnegative integers. The rate-maximizing problem with integer bit allocation is formulated as
(Arate,int) The following lemma shows that for the power-minimizing problem with integer bit constraint, the inequalities in the bit rate constraint and error rate constraint become equalities when the solution is optimal. This is similar to the power minimization problem without integer constraint. Such a property does not hold for the rate maximization problem with integer constraint as we will see later.
Lemma 5.4 For the power-minimizing problem Apow,int in (5.7), the bit rate of the optimal solution is equal to B0 and the symbol error rates ǫk = ǫ for all k.
Proof: See Appendix D.
Lemma 5.4 leads to the following result that if a solution is optimal for Apow,int, it is also optimal for A .
Theorem 5.3 Consider the power-minimizing problem Apow,intwith a target trans-mission rate B0 and symbol error rate constraint ǫ. Suppose (F∗, {σ∗2sk}, {b∗k}) is optimal for Apow,int, and in this case the minimized power is P∗. Now for the problem Arate,int with transmit power constraint P0 = P∗ and error rate con-straint ǫ, the same (F∗, {σ∗2sk}, {b∗k}) also maximizes the transmission rate and the maximized rate is equal to B0.
Proof: As (F∗, {σs∗2k}, {b∗k}) is optimal for the problem Apow,int. By Lemma 5.4, the bit rate is B∗ =PM −1
k=0 b∗k = B0, and all the symbol error rates satisfy ǫ∗k = ǫ.
Now, let us consider the problem Arate,int with power constraint P0 = P∗ and error rate constraint ǫ. Suppose ( ˜F, {˜σ2sk}, {˜bk}) is optimal for the problem Arate,int and the maximal bit rate is
B =˜
M −1
X
k=0
˜bk. (5.9)
All the corresponding error rates ˜ǫk satisfy ˜ǫk ≤ ǫ and the transmit power ˜P satisfies the power constraint, i.e., ˜P ≤ P∗. Since we already know the solution of Apow,int can achieve bit rate B0 with power P∗, the maximal bit rate ˜B in Arate,int must be larger than or equal to B0, i.e., ˜B ≥ B0. We will prove the theorem by showing (i) the transmit power ˜P is equal exactly to P∗, and (ii) the maximized rate ˜B is in fact equal to B0.
(i) ˜P = P∗: Suppose ˜P < P∗. This means ˜F, {˜bk}, and ˜σs2k can achieve a smaller transmit power and still satisfy all the constraints in Apow,int. This contradicts the assumption that F∗, {b∗k}, and σ∗2sk are optimal for Apow,int. So we have ˜P = P∗.
(ii) ˜B = B0: If ˜B = B0, we get the desired result that (F∗, σ∗2sk, {b∗k}) is optimal for Arate,int. Suppose ˜B > B0. Similarly to the procedure in Lemma 5.4, we can find another system that achieves bit rate B′ = ˜B −1 ≥ B0, with transmit power P′ < P∗, and error rate ǫ′k ≤ ǫ. This contradicts the assumption that (F∗,
{σs∗2k}, {b∗k}) is optimal for Apow,int. Therefore, we conclude that the maximized bit rate for the problem Arate,int is B0 and the power used is P∗. Therefore, the solution (F∗, {σs∗2k}, {b∗k}) of Apow,int is also an optimal solution for the problem
Arate,int. △△△
In Section 5.1, we saw that the transmit power of the optimal solution for the rate-maximizing problem is equal to the power constraint P0when the bit loading is not constrained to be integer. Such a property may not hold when there is integer bit constraint as we will see later. When the symbol error rate constraint ǫ is fixed, the maximal rate for Arate,int is a function of the power constraint P0. Similarly, for a fixed ǫ, the minimal power of Apow,int is a function of target rate B0. For convenience, we use Pint∗ (x) to denote the minimal transmit power for Apow,int when the target bit rate x is given and B∗int(x) to denote as the maximal bit rate for Arate,int when the power constraint is x.
The function Bint∗ (x) and Pint∗ (x). Using theorem 3, we will see that Bint∗ (x) is not continuous. It is a staircase-like function as shown in Fig. 5.1(a). This means a nonzero increase in the power constraint does not necessarily implies a nonzero increase in the maximized bit rate. This is different from the case without integer constraint in section 5.1. To explain this, consider the problem Apow,int
with two target bit rates B1 and B1+ 1. Let P1 = Pint∗ (B1) and P2 = Pint∗ (B1+ 1).
We can plot the minimal transmit power as a function of target bit rate as in Fig. 5.1(b). By Theorem 5.3, we know Bint∗ (P1) = B1and Bint∗ (P2) = B1+1. Now suppose the power constraint P0 for Arate,int is such that P1 < P0 < P2. Then the maximal bit rate Bint∗ (P0) for Arate,int is equal to B1 as we will see next. Since we already know that the maximal bit rate is B1 when the power constraint is P1, we have Bint∗ (P0) ≥ B1. Suppose Bint∗ (P0) > B1. This contradicts the fact that P2 is the minimal power for Apow,int when the target bit rate is B1+ 1. Hence we have Bint∗ (P0) = B1. This implies that for any power constraint P that satisfies P ≤ P < P , the maximal bit rate is B∗ (P ) = B . When the power constraint
P0 = P2, the maximal bit rate is increased to B1 + 1. Therefore, Bint∗ (x) is the staircase like function in Fig. 5.1(a).
From the plot of Bint∗ (P0) in Fig. 5.1(a) we can see that for Arate,int there can be many solutions that achieve the same maximal bit rate, but with transmit power smaller than P0. Hence for the problem Arate,int, the results in Lemma 5.3 is not true any more and the results of the real bit allocation case do not carry over to the the integer bit allocation case. To establish the duality with Apow,int, we will consider the solution that achieve the maximal rate B with the smallest transmit power among all possible solutions.
B1+ 1 B1+ 2
B1
P2 P1 P3 Maximal bit rate
Transmit power constraint Bãint(P0)
P0
(a)
B1+ 1 B1+ 2 B1
P2 P1 P3
Minimal transmit power
Target bit rate B0 Pãint(B0)
(b)
Figure 5.1: (a) Maximal bit rate as a function of power constraint for Arate,int. (b) Minimal transmit power as a function of target bit rate for Apow,int.
Theorem 5.4 Consider the problem Arate,int with power constraint P0 and sym-bol error rate constraint ǫ. Suppose (F∗, {σ∗2sk}, {b∗k}) forms the solution that has the smallest transmit power P∗ among all possible solutions. Let the maximized rate beB∗. Given target rateB0 = B∗ and error rate constraintǫ for the problem Apow,int, the same solution also minimizes the transmit power and the minimal power is P∗.
Proof: As (F∗, {b∗k}, {σ∗2sk}) is optimal for Arate,int, the maximized rate is B∗ = PM −1
k=0 b∗k. The transmit power is P∗ ≤ P0, and all the error rates satisfy ǫ∗k ≤ ǫ.
Consider the power minimizing problem Apow,intwith target bit rate B0 = B∗and the same error rate constraint ǫ. Suppose ( ˜F, {˜bk}, {˜σs2k}) is optimal for Apow,int and the minimized power is ˜P . By Lemma 5.4, the bit rate PM −1
k=0 ˜bk is equal to the target bit rate B∗. Since we already know (F∗, {b∗k}, {σs∗2k}) can achieve bit rate B∗ with transmit power P∗, the minimal power ˜P must be smaller than or equal to P∗, i.e., ˜P ≤ P∗. If ˜P is equal to P∗, we get the desired result that (F∗, {b∗k}, {σs∗2k}) is an optimal solution for Apow,int. Assume ˜P is smaller, i.e., P < P˜ ∗. This means ( ˜F, {˜bk}, ˜σ2sk) can achieve bit rate B∗ with a smaller power P . It contradicts the assumption that (F˜ ∗, {b∗k}, σs∗2k) is the optimal solution for the problem Arate,int that has the smallest transmit power. Hence we have P = P˜ ∗ and the solution (F∗, {b∗k}, σs∗2k) is optimal for Apow,int. △△△
Theorem 5.3 shows that the optimal solution obtained in the power-minimizing problem is also an optimal solution in the rate-maximizing problem. Theorem 5.4 shows that the solution with the smallest transmit power in the rate-maximizing problem is also optimal in the power-minimizing problem.
Remark on ZF receiver: The derivations in Section 5.1 and Section 5.2 are considered for the MMSE receiver. Duality between the power minimization and rate maximization problems also hold for the ZF case. For the MMSE case, we have used the results in Lemmas 5.1, 5.3, and 5.4 to prove the main results
in Theorems 5.1-5.4. Lemma 5.2 is used in the proof of Lemmas 5.3 and 5.4. For the ZF case, Lemma 5.2 is not needed as the MSE matrix of the ZF receiver in (2.14) is independent of the power allocation. Using the methods of MMSE case, we can prove the results in Lemmas 5.1, 5.3, 5.4, and also Theorems 5.1-5.4 for the ZF case.