Consider a
-receiver
parallel Gaussian broadcast channel as shown in Figure 1
where
is the transmitted signal,
represents a complex noise with variance
, and
corresponds to the channel seen by receiver
on tone
. The maximum common information rate that can be supported by the channel is given by
with
the variance of the input signal on channel
,
the power allocation among all subchannels,
the total power constraint, and
the SNR gap which measures the loss with respect to theoretically optimum performance [22]. To achieve the maximum common information rate, the common message codebook cannot be broken into different codebooks for each channel, that is, joint encoding and joint decoding must be performed across all subchannels [23]. This transmission scheme is referred to as "single codebook, variable power" transmission [24].
The expression in (2) is the maximization of the minimum of a set of sums of concave functions of
. Since the sum and the minimum operations preserve concavity, the objective is concave, and maximizing a concave function yields a convex optimization problem. This max-min optimization problem can be efficiently solved by the approach based on minimax hypothesis testing given in [14, 17, 18]. For two receivers, the optimal power allocation algorithm is given by three steps
Step 1.
Find
given by
with
If
, then the optimal power allocation is
and finish.
Step 2.
Find
given by
If
then the optimal power allocation is
and finish.
Step 3.
For a given set of weights
corresponding to the index
with
, find
given by
Search over all
to find
that satisfies
, then the optimal power allocation is
and finish.
First consider the optimization problem of Steps 1 and 2. As the objective function is concave, the power allocation can be derived by the standard Karush-Kuhn-Tucker (KKT) conditions [25]. The modified Lagrangian function for Steps 1 and 2 is given by
with
the Lagrange multiplier associated with the total power constraint. By taking the derivative of the modified Lagrangian function with respect to
, we can solve the KKT system of the optimization problem. The derivative with respect to
is given by
Nulling the derivative gives
The optimal power allocation corresponds to Gallager's water-filling strategy for single-user parallel Gaussian channels [11].
Step 1.
Step 2.
We now consider the optimization problem of Step 3. As the objective is a weighted sum of concave functions, the power allocation can also be derived by the standard KKT conditions. The modified Lagrangian function for Step 3 is given by
with
the Lagrange multiplier associated with the total power constraint. By taking the derivative of the modified Lagrangian function with respect to
, we can solve the KKT system of the optimization problem. The derivative with respect to
is given by
Nulling the derivative gives
The quadratic equation to be solved is
The discriminant is given by
The power allocation is given by the positive root
In this formula, the optimal power allocation takes into account the difference between the water-fill functions and the weights of the different receivers.
For more than two receivers, the optimal power allocation algorithm is driven by the solutions of higher-degree polynomials and involves more steps under multiple hypothesis testing. For instance, with three receivers
, the optimal power allocation algorithm is given by seven steps involving the solutions of three linear equations, three quadratic equations and a cubic equation [26]. Therefore, for three receivers
and four receivers
, the optimal power allocation is a type of water-filling strategy given by the solutions up to a cubic and a quartic equation, respectively. The optimal power allocation can also be found analytically (the solution is not given in this paper due to space limitations). With
, the optimal power allocation is given by the solutions of polynomial equations up to degree
from the formula
In general, the roots for polynomials with a higher degree than four cannot be expressed analytically but can be solved numerically. Note that to reduce the complexity in a practical algorithm, the weights
are taken from a given data set in interval
with
samples, leading to a possible exhaustive search over
possibilities. Therefore, to satisfy the conditions requiring the rates of the different receivers to be equal, the optimal value
should minimize the dispersion of the rates
Figure 2 shows the power control for a single tactical radio network. An inner loop determines the power allocation maximizing the common rate subject to a total power constraint. Then, an outer loop minimizes the power such that a common rate constraint
is achieved. Algorithm 1 provides the proposed power allocation for power minimization subject to a common rate constraint. The inner loop and the outer loop correspond to lines 13–21 and 6–30, respectively. Note that if all the steps in the multiple hypothesis testing are needed, the complexity of the algorithm increases exponentially with the number of receivers
.
Algorithm 1: Minimization of the power subject to a common rate constraint.
(1) init 
(2) init 
(3) init 
(4) init 
(5) while 
(6) for all steps
(7) init
according to step
(8) init 
(9) init 
(10) init 
(11) init 
(12) while 
(13) Calculate
according to the roots of(18)
(14) if 
(15) 
(16) 
(17) 
(18) end if
(19) 
(20) end while
(21) If condition satisfied on
exit step loop
(22) end for
(23) if 
(24) 
(25) 
(26) 
(27) end if
(28) 
(29) end while