Optimal Power Allocation

 1 σ²_θ +

PK

k=1hkαk

2

PK

k=1|hk|²|αk|²σ_n² + σ_ν²





−1

(3.11)

The MSE Jo is a lower bound of J in (3.10) and can serve as a benchmark against which the performance of the estimator (3.7) can be compared.

3.3 Optimal Power Allocation

During the training phase, each sensor uses the same training symbol and thus consumes the same amount of training power P_t/K, where P_t is the total allocated training power.

From (3.5), it is clear that as Pt increases, the MSE in channel estimation decreases. In a sensor network, there is likely a total power constraint, that is, there is an upper bound imposed on the sum of training power and the power used to transmit data. Hence, when more power is allocated for training, less power is available for data transmission and vice versa. Under the total power constraint, the minimum MSE of ˆθ, that is, J in (3.10), depends on the training power P_t and how the remaining network power is allocated to each sensor for data transmission. In the following, we consider the optimal power allocation problem, that is, to choose Ptand data power for each sensor to minimize J under a total power constraint. For comparison, we will also consider the case when channel information is available, no training, no channel error, and all power is used for data transmission. The comparison of the two cases will show the penalty incurred due to the fact that the channel is unknown.

3.3.1 When channels are known

If the channel hk is available at the sensor k, the phase of αk can be chosen to match that of the channel, that is, ∠αk=−∠hk, so that hkαk =|hk||αk|. The MSE Jo in (3.11) can

then be rewritten as for the kth sensor. Such choices of phases make Jo smallest among αk’s of the same magnitude. Note that gk has a Rayleigh distribution with density function fg(x) = 2x exp (−x²), x≥ 0, and E[gk] =p

π/4; g_k² has an exponential distribution with density function f_g²(x) = exp (−x), x ≥ 0, and E[gk²] = 1 [37, p.51]. The signal transmitted from the kth sensor is αk(θ + nk) with power Pk = E[|αk(θ + nk)|²] = |αk|²(σ_θ²+ σ²_n). From (3.12), the optimal power allocation problem with the total network power constrained to P > 0 can be formulated as the following optimization problem



By the same reasons shown in the observations (i) to (iii) in Section 2.2 in Chapter 2, instead of solving (3.13), we consider an alternative problem in which the power is mini-mized subject to an MSE constraint:



The solution, from Proposition 2.1, can then be written as

|α^k|² =

which is also the minimum MSE of (3.13). Since the minimum MSE depends on the total network power P and the number of sensors K, we hereafter write the MSE Jo in (3.16) as Jo(P, K).

As the power P increases, we expect Jo to decrease, which is easy to see from (3.16).

For a fixed K, as P → ∞, we have

P →∞lim Jo(P, K) = σ²_θ

1 + Kβ (3.17)

where β = σ_θ²/σ²_n is the observation SNR. The limit dose not go to zero but is roughly proportional to 1/K as we would expect. On the other hand, for a fixed P > 0, as K increases, we have

where in the first equality we used the law of large numbers [39]. From (3.18), we conclude that in the coherent MAC model, the MSE decreases in the order of 1/K as K goes to infinity even though the total network power P is finite. Similar conclusion for the unit variance case, σ_θ² = σ²_n= σ²_ν = 1, appeared in [26].

3.3.2 When channels are estimated

Suppose training for channel estimation consumes power Pt, then the remaining power for data transmission is P− Pt. The power allocation problem now is to optimally choose training power Pt and data power for each sensor. The phase of αk is chosen to match that of ˆhk, i.e., ∠αk = −∠ˆh^k. Write hk = ˆhk + (hk − ˆh^k), since ˆhk and hk − ˆh^k are uncorrelated we have σ_h² = σ_ˆh²+ δ_k², where δ_k² = E[|h^k− ˆh^k|²]. Use (3.5) and σ²_ˆh = σ_h²− δk², we can express the MSE in (3.10) as

J =

where ˆg_k = |ˆhk|/σ^ˆh is the normalized estimated channel gain for the kth sensor. Since ˆhk is circular Gaussian, ˆgk and gk have identical distribution. From (3.19), the MMSE

optimization problem under a total network power constraint can be formulated as

Again instead of solving problem (3.20) directly, we consider a problem in which the roles of objective function and constraint are interchanged. The solution to problem (3.20) is given in the following proposition, the proof of which is given in Appendix D.

Proposition 3.1. For K > 1, the solution to (3.20) gives the optimal training power P_t^opt = K(ζP + 1)−p

Note that the optimal training power in (3.21) depends on the number of sensors K, the channel SNR ζ, and the total network power P . Moreover, we have

P_t^opt− P 2 =

pK(ζP + 1)−√

ζP + K2

2ζ(K− 1) > 0,

that is, for a total power constraint P , the power used for training is greater than that used for data transmission.

From (3.23), we see that the MSE decreases as the power P increases. For a fixed K, as P → ∞, we obtain

P →∞lim J(P, K) = σ²_θ

1 + Kβ (3.24)

which is the same as (3.17). This makes sense since P → ∞ implies Pt^opt → ∞ and thus the MSE of channel estimation in (3.5) approaches to zero, that is, ˆhk → hk as P → ∞ in the mean square sense. It is shown in Appendix E that, for a fixed P ,

K→∞lim J(P, K) = σ²_θ



1 + β 1 + β

pζP + 1− 12₋₁

. (3.25)

The MSE does not approach to zero. The reason is that the order of 1/K decrease in MSE in (3.18) is offset by the order of K increase in the power of the error term E[|ε|²|ˆh]

in (C.3) in Appendix C. Therefore, in the presence of the channel estimation error, the MSE reaches a finite nonzero value as K goes to infinity.

3.3.3 Comparison of two cases

If the total network power and number of sensors are fixed, with estimated channel, the estimation performance is worse than when channel information is available due to the presence of channel estimation error. To quantitative compare the two cases, we set the same MSE objective, use optimal power allocation for both cases, and determine the respective total network power that would be required. Suppose to achieve the selected MSE, total network power P^a is required when channel information is available and the required total network power is P^ewhen channels are estimated. The ratio P^a/P^egives an indication of the penalty incurred by the consumption of training power and the presence of channel estimation error. A small ratio would imply a heavy penalty. But the MSE expressions in (3.16) and (3.23) are random variables, we instead derive the condition on P^a and P^e under which the distributions of MSEs are identical. This is possible due to the fact that the random variables gk and ˆgk have identical Rayleigh distribution. From (3.16) and (3.23), the distributions of MSE expressions are identical if the deterministic terms in the denominator are equal, that is,

K− 1 pK(ζP^e+ 1)−√

ζP^e+ K

!2

= 1

ζP^a (3.26)

Rearranging (3.26), we get P^a P^e =

pK(1 + 1/(ζP^e))−p

1 + K/(ζP^e) K− 1

!2

(3.27)

Note that the ratio in (3.27) is less than one, and for P^e large P^a

P^e ≈ 1

(√

K + 1)². (3.28)

The ratio decreases as the number of sensors K increases. This means that the penalty caused by channel estimation becomes worse as the number of sensor increases.