Bit allocation and power minimization - Signal Processing and Optimization for Transceiver Systems - page 460

2. Choose the precoder F as the minimum-norm right inverse of GH ,thatis,
F = H † G † GHH † G † − 1 .

(14 . 29)

With G and F so chosen, the minimized transmitted power is given by
P min = c 2 b

1 /M

1
M− 1
k =0

, (14 . 30)

σ h,k

where b = k b k /M is the average number of bits. The constant c = M k ( c k ) 1 /M ,
where c k is as in Eq. (14.9), that is,

Q − 1 P e ( k )
4

2

c k = σ q
3

, (14 . 31)

where σ q

is the variance of the channel noise components and P e ( k )isthe

acceptable error probability specified for the k th user. ♦

With G and F chosen optimally as above, it can be shown (Sec. 14.6.2.B) that
the optimal bit allocation formula (14.11) can be rewritten as

σ q
3

2 +log
2

Q − 1 P e ( k )
4

σ h,k .

b k = D − log

2

Thus, bit allocation is determined essentially by the channel singular values
σ h,k and the desired error probabilties

P e ( k ) . For completeness, note that the

| 2 ] should be taken to be as in Eq. (14.5), so

k th user power P k = E [ | s k ( n )

P e ( k ) (QoS constraints) are satisfied. Since σ e k

that the specified values of =

σ q [ GG † ] kk = σ q (using GG † = I ), this simplifies to

2

Q − 1 P e ( k )
4

σ q 2 b k
3

P k ≈ (14 . 32)

for 0 ≤ k ≤ M − 1 .

14.5 Optimal transceiver in terms of channel SVD

It is convenient to express the solution for the optimal transceiver in terms of
the SVD of the J × P channel H . Thus, let

V h
P×P

H = U h Σ h , (14 . 33)

J×J J×P