Optimal transceivers for diagonal channels - Signal Processing and Optimization for Transceiver Systems - page 375

3. Channels with poor gains create problems. For a fixed set of noise powers σ q k ,
the error can become arbitrarily large if some of the channel gains H k get
arbitrarily small. So, there is no channel-independent upper bound on the
reconstruction error (11.17). This problem disappears if the zero-forcing
constraint is removed as shown in Sec. 11.3.

4. No inequality constraint? In the optimization we did not explicitly incorpo-
rate the inequality constraint x k ≥ 0 . The solution (11.12) however turned

out to be positive, so all is well. Formally, if we incorporate x k ≥ 0 and use

the Karush-Kuhn-Tucker conditions for optimality (Chap. 22), we obtain
the same result as above.

5. Minimumor maximum? In general, setting the derivative of the Lagrangian
to zero yields the necessary conditions for a local extremum, which can be
a minimum or a maximum. In our problem, we know from Eq. (11.8)
that E mse has really no upper bound (no finite maximum). It can be made

arbitrarily large by making α n arbitrarily small for some n (for which σ q n
is nonzero) and adjusting the remaining α k such that the power constraint
holds. Since

E mse has no upper bound, the stationary point obtained from
the Lagrange approach can only be a minimum and not a maximum. As
the solution is unique, this minimum has to be global as well.

6. Noise-to-signal ratio . From Eq. (11.17) we see that the noise-to-signal ratio
at the receiver (i.e., at the detector input) is

M− 1 2

E ZF mmse
σ s

1
p 0

σ q k
|

= .

H k |

k =0

This depends on the channel input power p 0 and the quality of the channels
σ q k / , but not on σ s , as expected.

7. Half-whitening . Consider Fig. 11.4, which shows a redrawing of the diag-
onal system of Fig. 11.3. The signals

| H k |

s k ( n ) are identical in both systems
(for the same inputs s k ( n )). The equivalent noise sources are q k /H k . We
see that

s k ( n )= s k ( n )+ e k ( n ) ,
where e k ( n ) has the variance

σ q k

1
α k

E k =

H k | 2 × |

If α k were chosen such that α k = σ q k / | H k | 2 then

E k =1

for all k , that is, the error variances would have been equalized. Such
a choice of multipliers would be called a “whitening” system. 1 But the

1 The term whitening comes about by imagining that k is a frequency index (as in OFDM),
in which case the E k are like the samples (in frequency) of an error spectrum.