Optimal transceivers for diagonal channels - Signal Processing and Optimization for Transceiver Systems - page 374

mization problem define the Lagrangian function

+ λ σ s

,

M− 1 M− 1

σ q k
x k |

ψ = x k − p 0 (11 . 11)

H k | 2

k =0 k =0

where λ is the Lagrange multiplier. Setting ∂ψ/∂x k = 0 we get
− σ q k

+ λσ s =0 ,

x k | H k | 2

which yields

c 2 σ q k

x k = , (11 . 12)

| H k | σ s

where the constant c 2 =1 / √ λ should be such that the power constraint (11.9) is
satisfied, that is,

M− 1 − 1

p σ s σ q m

c 2 = . (11 . 13)

| H m |

m =0

The optimum multipliers α k therefore can be taken to be the nonnegative num-
bers defined by

α k,opt = c σ q k
|

1 / 2

. (11 . 14)

H k | σ s

From Eq. (11.4) we have

1
H k α k,opt

β k,opt = (11 . 15)

which therefore yields

σ s
σ q k

1 / 2

1

= 1
c

1

| β k,opt | = × (11 . 16)

| H k α k,opt | | H k | 1 / 2

The minimized objective function E mse is obtained by substituting the optimal

α k into Eq. (11.8). The result is

M− 1 2

E ZF mmse = σ s
p 0

σ q k
|

. (11 . 17)

H k |

k =0

A few remarks are in order:

1. Effective noise. Note that the quantity σ q k / H k | 2 appears in all the ex-

pressions. This can be regarded as the effective noise variance of the k th
channel. This will become more clear when we discuss Fig. 11.4 below.

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2. Power allocation. The signal power at the output of α k is the power input
to the k th channel H k . This is given by
P k = σ s α k,opt = γσ q k
| H k |

where γ is a temporary notation for the constant part. Thus channels with
large effective noise σ q k / H k | 2 get more power in the optimal system.

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