MMSE transceivers with zero-forcing equalizers - Signal Processing and Optimization for Transceiver Systems - page 425

if we have a 0 ≥ a 1 ≥ ... ≥ a M− 1 ≥ 0 . Thus, E mse is Schur-concave in the vector

[ B 00 B 11 ...B M− 1 ,M− 1 ] . (12 . 90)

12.B.4 Optimal choice of

U f

Consider now the choice of the matrix U b . Given any B whatsoever, suppose
we choose U b such that B new in Eq. (12.86) is diagonal. Then diag( B new )
majorizes diag( B ) (Sec. 21.5.1). So, if we want to minimize a Schur-concave
function of diag( B ) the best thing to do would be to choose U b (i.e., choose U f )
such that B is diagonal. Having established that
B = ( U f Σ h U f ) M − 1

has to be diagonal, we now use Lemma 12.2 to see that U f can be chosen such
that U f Σ h U f itself is diagonal. Thus, we finally obtain

M− 1 M− 1

B kk
σ f,k

1
σ h,k σ f,k

E mse = σ q = σ q (12 . 91)

k =0 k =0

For any choice of σ f,k whatsoever it is clear that the M values of σ h,k should
be the largest M singular values of the channel, otherwise E mse can be reduced

further by making such a choice. So it follows that the best U f should have its
columns ordered such that the largest singular values of the channel are the first
M diagonal elements of Σ h . The optimal choice of σ f,k in Eq. (12.91) is done
precisely as in Sec. 12.4.3.

12.C Rectangular channel

So far in this chapter we have assumed that the channel matrix H is a square
matrix. For the case where H is a K P (rectangular) matrix as shown in Fig.

12.11, the results extend readily as we shall show next. Thus consider Problem
1 described by Eqs. (12.12)-(12.14) again. The ZF constraint GHF = I can
again be eliminated by defining G to be the minimum-norm left inverse of HF
as in (12.21). Then the problem reduces to the form shown in Eqs. (12.22) and
(12.23). Thus the problem is again stated entirely in terms of the P ×M matrix
F and the P

×

P matrix H † H . The integer K has disappeared, and does not
affect the optimum precoder F , which therefore still has the form (12.55), that
is,

×

F = U f Σ f
0

,

where U f is as in Eq. (12.59), that is,

U f = V h .