Throughput-Oriented Water-Filling
4.2 Throughput-Oriented Water-Filling: Only Total Noise Variance is Available
Based on the adjusted approximation formula in (3.4), (3.2) becomes Tw(P ) = min
We focus on the situation that all channels are active, which means that
Pi ≥ Pth,i ∀ 1 ≤ i ≤ K. (4.7)
A straightforward approach to eliminate the minimization in (4.8) is to use the Lagrange multiplier technique and Kuhn-Tucker condition to find σi2 such that the worst-case effective throughput is achieved. However, in general, the worst-case effective throughput is not a concave function of σi2 since
∂2Tw(P )
we further constrain our problem to be concave over σi2 such that (4.9) becomes always negative. Under this constraint, we know that the σi2 to achieve Tw should be chosen as either 0 or σt2. Since the total noise variance is a given value, only one channel will be allocated the whole noise power and the rest of the channels is allocated zero noise power.
Next we consider one possible power allocation Pi†, which is defined as Pi†= 2σt2
where ν is chosen such that
K
X
i=1
Pi†= Pt.
Our aim is to prove that this power allocation performs better than any other power alloca-tion policies and hence is optimal.
From the power constraints in (4.7) and (4.10), Pi† should satisfy Pi†≥ max 2σi2
Define the minimum power required in channel i as Pth,i† , max 2σt2
Replacing Pi† by (4.11), we further deduce (4.13) as a condition on ν, 2σt2 We let νmin denote the minimum value of the choice of ν, which satisfies (4.14) for every i.
From the definition of Pi† and (4.14), we have Pi†≥ 2σt2
(4.15) equivalently implies a constraint in system SNR by taking summation over Pi† and dividing it by σt2, which is
where γth† is the threshold system SNR. From above, we have claimed that by using Pi† as power allocation, the optimal choice of σi2 is either 0 or σt2, when system SNR is greater than γth†. Using this result, the worst-case effective throughput due to {Pi†}, which is denoted by Tw† can be computed as follows.
Tw† P†
We define the noise variance that achieves Tw† as
σi2† , σt2, if i = m
By (4.18), it is noted that the worst-case effective throughput is independent of m. Thus definition of σi2† is justified. Besides, we know that Rm− e−ν is always non-negative for all possible value of m from (4.14).
The optimality of using Pi†as power allocation is proved by the method of contradiction.
We will show that the Tw obtained from any other power allocation is less than or equal to
Tw†. The proof is as follows. Consider any power allocation ˆPi = Pi†+ 4Pi†, where 4Pi†6= 0 for at least one channel and
K
X
i=1
4Pi†= 0. (4.19)
We also consider a specific noise power allocation
ˆ
An upper bound for the worst-case effective throughput of ˆP can be found as the following:
P σmini2=σ2t
The first inequality holds for the reason that the minimization over the effective throughput is always less than or equal to the effective throughput using the noise power ˆσi2 in our case.
For the second equality, it is obvious that the minimization over 1 and Aie−di ˆ2 ˆσi2Pi is always greater than zero. Moreover, if we have
min
(4.21) becomes
By comparing (4.22) with Tw† in (4.18), the only difference is in the exponential term. From the definition of k in (4.20), we know that
dkPˆk
There always exists at least a channel that has its di2σ4P2i
t being negative since ˆPi 6= Pi† and 4Pi should satisfy the constraint in (4.19). Taking minimization over (4.25), we have
1≤i≤Kmin
From the discussion above, we have proved that the worst-case effective throughput of any power allocation ˆPi is less than that of Pi†. The optimality of Pi† is then justified. Thus, we can claim that when system SNR is greater than γth† , the optimal power allocation that maximizes Tw is by using Pi† as the power allocation. It is also worth knowing that the corresponding choice of σi2 that achieves Tw is to put total noise power to any one of the channel.
The power allocation scheme can also be interpreted as a variation of water filling princi-ple. For each channel, a vessel with base width 2σ2t and base height log 1 will be used for
water filling. From our constraints in power in (4.13), each channel should be allocated at least Pth,i† . The resulting ν must be no less than νmin. The water filling inside each vessel is then the optimal power to be allotted. An example with three channels(K = 3) is illustrated in Fig. 4.2.
Example 4.2. A three channels (K = 3) system is considered. Each channel has its base height and base width as defined in the paragraph immediately above this example. At least Pth†,1, Pth†,2 and Pth†,3 should be allocated to three channels, respectively. The lowest water level νmin should then be chosen as
νmin = max
Figure 4.2: An example of the throughput-oriented water-filling with K = 3.
From the definition of Pi†, we can see that the proposed power allocation depends on σ2t. Alternatively, we can say that the proposed power allocation scheme depends on system SNR
if we look at the power allocation ratio for each channel, which is derived as the following: obtained by applying feedback technique to the system. Besides, the effect of system SNR in the power allocation can be eliminated if the code used in each channel can be chosen such that the product of Ai and Ri is the same for different channels. In this way, the ratio of power between different channels becomes a constant, which is as the following:
p†i : p†j = 1 di
: 1 dj
,
for i 6= j. Thus, the allocated power ratio for channel i becomes
p†i =
which is only related to di. Furthermore, we look at p†i when the system SNR goes without bound. From (4.26), we have
γlimt→∞
The optimal allotted power to channel i should be inversely proportional to its di, which is closed to its free distance. It coincides with the fact that the free distance dominates frame error rate and thus dominates worst-case effective throughput when SNR tends to be infinity. Finally, our proposed power allocation scheme can be simplified to the traditional equal power allocation scheme when all channels use the same code.