4.4 BA system with Unconstrained Bit Allocation
4.4.1 Optimal Bit Allocation
We first consider the case when the precoder F is a fixed Mt× M unitary matrix.
The number of bits transmitted per channel use is Rb and b0+b1+· · ·+bM −1= Rb. Assume the transmission rate is high and bkis large enough so that 1−2bk2 ≈ 1 and 1 − 2bk ≈ 1, then the symbol error rate expression in (2.10) can be approximated by
SERk ≈ 4Q
r 3 2bkβk
!
. (4.16)
With the high bit rate assumption, bk > 0, for all k and thus Rs = P0/MIM. For the convenience of derivation, we define the function
f (y) = Q(√1y), y > 0. (4.17)
The function f (y) is monotone increasing and it can be verified that f (y) is convex for y ≤ 1/3. Using f(·), we can express SERk as
SERk≈ 4f(2bk 3βk
). (4.18)
Therefore the average BER in (4.3) can be written as
BER(b) ≈ 4
where we have dropped the dependence of BER function on the channel H for convenience. Assume the arguments of f (·) are smaller than 1/3 so that the convexity of f (·) holds (we will see later why this assumption is reasonable).
Using the convexity of f (·), we have 1
and the AM-GM (arithmetic mean-geometric mean) inequality 1
and also using the monotone increasing property of f (·).
Notice that the lower bound in (4.23) is independent of bit allocation. The optimal bit allocation is such that the two inequalities in (4.22) and (4.23) become
equalities. Due to convexity of f (·), the first inequality (4.22) holds if and only if 2bk/(3βk) are of the same value for all k. The same set of conditions is also necessary and sufficient for equality to hold in the second inequality as f (·) is monotone increasing. When both inequalities hold, the lower bound in (4.23) is achieved. Therefore the optimal bit allocation for minimizing the BER is such that 2bk/βk= 2Rb/M(QM −1
We can see that the symbols with larger SNR βk are allocated with more bits.
We have denoted the BER lower bound in (4.23) as BERBA, where the subscript is a reminder which notifies that it is the BER of the BA system. Note that BERBA is obtained when the bits are allocated as in (4.26) and there is no in-teger constraint on bit allocation in the above derivation. The bit allocation bk
computed in (4.26) are not integers in general. Nonetheless BERBA gives useful insight on the performance of the BA system and connections with other system as we will see in Section 4.4.2
Remarks
1. In the above derivation, we have assumed that the argument of f (·) in (4.19) is larger than 1/3 so that the convexity of f (·) can be used in (4.20).
We now examine the validity of such an assumption. When the argument 2bk/(3βk)= 1/3, the corresponding SERkis SERk≈ 4Q(√
3) ≈ 0.17, a large symbol error rate that may not be useful. In practical applications, it is more reasonable to have smaller error rate, which requires 2bk/(3βk) < 1/3.
2. When bits allocated optimally as in (4.26), 2bk/βk are the same for all k.
This means the symbol error rates are equalized for all transmitted symbols.
3. The actual number of symbols transmitted may be smaller than M if some symbols are allocates with 0 bits. However the number of bits transmitted
Now let us consider the case F is not fixed, but an Mt× Mt augmented pre-coder F′ (implicitly M < Mt in this case). The imput s′ is an augmented Mt× 1 vector and bit allocation vector b′ is Mt× 1 as in Section 4.2.2. For a given M, we can choose M columns out of F′ to form the actual Mt× M precoder. As we choose M columns of F′, there are C(Mt, M) possible choices. For each of these choices, we can compute the optimal bit allocation and the corresponding BER using (4.23), and choose the best precoder. In this case the BA system with augmented precoder F′ is always better than the BA system with a fixed precoder F if F us a submatrix of F′.
Bit allocation for optimal number of substreams
In the above discussion of optimal bit allocation, we assumed all symbols carry nonzero bits and transmission power is loaded on all M symbols. In the end some of the symbols may be assigned zero bits while take up 1/M of the total power. To make efficient use of power, we can allocate power to only the symbols that carry nonzero bits. To do this, we can compute the optimal bit allocation for all possible number of symbols with nonzero bits and choose the best one. To be more specific, let us illustrate this in another viewpoint. We start out with an Mt× Mt initial precoder F′ as before. The precoder F can be any Mt× M0 submatrix of F′, where M0 = 1, 2, · · · , M. There are PM
M0=1C(Mt, M0) possible precoders. We collect all these possible precoder in a set SF. For each F ∈ SF, we can use (4.29) to compute the BER under optimal bit allocation. The error rate BERBA given in (4.23) depends on the precoder used. For convenience let us use the notation BERBA(F) to indicate the dependence on F. The best F is
Fopt = arg min
F∈SFBERBA(F). (4.27)
The resulting minimum BER is given by BERBA,opt = min
F∈SF
BERBA(F). (4.28)
When the optimal precoder is obtained this way, all the symbols will carry nonzero bits. The reason is as follows: Let the optimal precoder Fopt be Mt× l, and the
optimal l × 1 bit allocation be bopt. Suppose one of the symbols is assigned with zero bits. The actual number of symbols transmitted is l − 1. Let us remove from Fopt the column corresponding to the symbol with zero bit and call the remaining Mt× (l − 1) submatrix F0. Also remove from bopt the element equal to zero and call the reduced vector b0. Then using precoder F0 with bit allocation b0 gives a smaller BER for the same transmission power as the power is now distributed among (l − 1) symbols instead of l symbols. So Fopt can not be optimal if one symbol is assigned 0 bits. We can therefore conclude that all symbols carry nonzero bits in the optimal system that uses Fopt as precoder.