Optimization with equality and inequality constraints - Signal Processing and Optimization for Transceiver Systems - page 740

q ( n )
0

β 0

α 0

y ( n )
0

s ( n )
0

s ( n )
0

α 1

y ( n )
1

β 1

s ( n )
1

s ( n )
1

q ( n )
N − 1

α N −1 β N −1

s ( n )
N −1

s ( n )
N − 1

Figure 22.4 . A parallel connection of N independent channels.

Observe that if the noise-to-signal ratio σ q k /σ s is small then this reduces to

β k ≈ 1 /α k as expected. The preceding expression holds for any fixed set of

transmitting multipliers α k . With this choice the mean square error depends
only on α k , which can be optimized to minimize the MSE further. For this we
first show that the expression for the MSE with the above choice of β k is

N− 1

σ q k

E mse = (22 . 37)

α k | 2 + σ q k
σ s

|

k =0

Proof. Themeansquareerrorinthe k th channel with optimal β k is

E [ e k e k ]= E [ e k ( s k − s k )]

E [ e k ( y k β k −

= s k )]

E [ e k s k ]

= − (from Eq. (22.35))

E ( β k α k − 1) s k + β k q k ∗ s k (from Eq. (22.33))

= −

σ s (1 β k α k ) ∗ = − (from Eq. (22.29))

Substituting the optimal value of β k from Eq. (22.36) and adding for all k ,
Eq. (22.37) follows immediately.

Now consider the problem of optimizing the distribution of multipliers such

that Eq. (22.37) is minimized. For simplicity we write Eq. (22.37) in the form

{ α k }

N− 1

A k
x k + B k

E mse = (22 . 38)

k =0

α k | 2 . The problem therefore is to minimize this quantity under the

where x k = |