Optimal transceivers for diagonal channels - Signal Processing and Optimization for Transceiver Systems - page 377

q ( n )
k

H k

α k

β k

s ( n )
k

s ( n )
k

(a)

q ( n )
k

1 /H k

α k H k

β k

(b)

s ( n )
k

s ( n )
k

q ( n )
k

1 /H k

α k H k

β k

(c)

s ( n )
k

s ( n )
k

q ( n )
k

α k

β k

s ( n )
k

s ( n )
k

(d)

Figure 11.5 . Redrawing the k th channel path for further simplification.

In Fig. 11.5(a) we show the k th branch of the system with channel multiplier
H k . Assuming H k =0wecanalwayswrite β k in the form 2

β k = β k

H k ,

(11 . 18)

as shown in Fig. 11.5(b). We now move the noise source to the left of H k
to obtain an equivalent noise source q k ( n )= q k ( n ) /H k . This is shown in Fig.

11.5(c). Note that q k ( n ) has variance

σ q k

=

σ q k (11 . 19)

| H k | 2

The equivalent system of Fig. 11.5(c) can now be redrawn as in Fig. 11.5(d). In
this system we have a channel with unit gain and a noise source with a certain
variance

σ q k . With each branch of Fig. 11.2 modified in this way, the results of
Chap. 22 can be applied here. For example, for fixed α k the optimal β k that
minimizes the mean square error

| 2 ] E k = E [ | s k ( n ) − s k ( n ) (11 . 20)

2 If H k =0forsome k, then no matter how we choose α k and β k , the quantity

s k ( n )would
only be the noise amplified by β k . So we take the solution to be α k = β k =0when H k =0
for a particular k.