MMSE transceivers without zero forcing - Signal Processing and Optimization for Transceiver Systems - page 444

3. Rank of the channel. Note finally that in the pure-MMSE problem there
is no need to assume that the channel has rank

M . This is unlike in
the ZF-MMSE case, where the zero-forcing property necessitated that the
rank of the channel be at least equal to M.
4. Diagonal representation. Since the cascade of V h , H , and U h in Fig. 13.3 is
the diagonal matrix Σ h of channel singular values, the optimal transceiver
can be represented in diagonal form as shown in Fig. 13.4. The quantities
σ f,k and σ g,k , which depend only on σ h,k (and σ s ,σ q , and p 0 )canbe
calculated as described previously in this section.

≥

q ( n )
0

σ g, 0

σ f, 0

σ h, 0

s ( n )
0
s ( n )
1

s ( n )
0
s ( n )
1

σ h, 1

σ f, 1 σ g, 1

q ( n )
M − 1

σ f,M −1 σ h,M −1

σ g,M −1

s ( n )
M

s ( n )
M − 1

− 1

precoder equalizer

channel

Figure 13.4 . Diagonalized representation of the MMSE transceiver.

Example 13.1: Pure MMSE versus ZF-MMSE

Consider the case where M = K, which happens if the power is large enough
to make q kk > 0 for all k in Eq. (13.64). Assuming σ s =1 , the mean square
error (13.66) for the pure-MMSE system becomes

2

.

M− 1

E pure = σ q
p 0

1
σ h,k

1

(13 . 67)

M− 1

1+ σ q
p 0

k =0

1 /σ h,k

k =0

For the ZF-MMSE system of Chap. 12, the minimum error was given by

2 .

M− 1

E zf = σ q
p 0

1
σ h,k

(13 . 68)

k =0

The gain obtained from giving up the ZF constraint is therefore

M− 1

=1+ σ q
p 0

1 /σ h,k . G (13 . 69)

k =0