Energy-efficient link adaptation in frequency-selective channels
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Citations
A survey of energy-efficient wireless communications
Energy-efficient wireless communications: tutorial, survey, and open issues
Wireless Information and Power Transfer: Energy Efficiency Optimization in OFDMA Systems
Energy-Efficient Resource Allocation in OFDMA Systems with Large Numbers of Base Station Antennas
Energy-Efficient Resource Allocation for Downlink Non-Orthogonal Multiple Access Network
References
Convex Optimization
Fundamentals of Wireless Communication
Communication in the presence of noise
Information Theory and Reliable Communication
Variable-rate variable-power MQAM for fading channels
Related Papers (5)
Frequently Asked Questions (15)
Q2. How does adaptive modulation increase energy efficiency?
By increasing the transmit power from 15 dBm to 25 dBm,the throughput of adaptive modulation increases, however, the energy efficiency first increased and then decreases.
Q3. What is the energy efficiency of a subchannel?
With infinite number of subchannels, the highest energy efficiency, 1P ′ T (0) , is obtained by transmitting with infinite small data rate.
Q4. What is the BER for a multipath frequency-selector?
The International Telecommunication Union (ITU) pedestrian channel model B [26] is used to implement the multipath frequency-selective fading.
Q5. How many iterations can be found within a rate r?
A rate r, which satisfies |r − r∗| ≤ ², can be found within at most M iterations, where M is the minimum integer such that M ≥ log2( (α−1)r ∗² − 1).
Q6. What are the three ways to improve energy efficiency?
Propositions 1, 2, and 3 discover three ways to improve energy efficiency: increasing channel power gain, reducing circuit power, and allocating more subchannels.
Q7. What is the meaning of the term dynamic energy-efficient water-filling?
Since the water level is determined by the optimal energy efficiency, the authors refer to their scheme as dynamic energy-efficient water-filling.
Q8. What is the optimal data rate vector without constraint?
Then R̂ ∗ can be easily obtained via the Lagrangian technique [23] and isr̂∗k = max { f−1k (λ), 0 } (20)for k = 1, · · · ,K, where λ is determined by K∑k=1r̂∗k = Γ. (21)When the channel capacity is achieved on each subchannel, the corresponding optimal power allocation is a water-filling allocation, which achieves the sum channel capacity Γ.Similarly, with a maximum transmit power constraint, the problem is to findR̃∗ = arg max RRPC + PT (R) , (22a)subject to PT (R) ≤ Pm. (22b)If the optimal data rate vector without constraint in (10) satisfies PT (R∗) ≤
Q9. What is the optimal step size for a line search?
a line search of the optimal step size needs to cover a large range to assure global convergence on all subchannels, which is computationally expensive.
Q10. What is the optimal step size for the multiple subchannel case?
With sufficiently small step size, U(R[i+1]) will be always bigger than U(R[i]) except when ∇U(R[i]) = 0 that indicates the optimality of R[i] [23].
Q11. What is the energy-efficient link adaptation for OFDM?
In OFDM systems with subchannelization, subcarriers are grouped into subchannels and the subcarriers forming one subchannel may, but not necessarily be adjacent, such as the contiguous and distributed subchannelization schemes in 802.16e [9].
Q12. What is the optimal data rate vector for the multiple subchannel case?
do r̂← r2+r12 ; ĥ ← dU(r)dr ∣∣∣ r̂To find the optimal data rate vector for the multiple subchannel case, the authors design a gradient ascent method to produce a maximizing sequence R[i], n = 0, 1, · · · , andR[i+1] = [ R[i] + µ∇U(R[i]) ]+ , (26)where [R]+ sets the negative part of the vector R to be zero, µ > 0 is the search step size, and ∇U(R[i]) is the gradient at iteration i.
Q13. What is the partial derivative of U(R) with ri?
The partial derivative of U(R) with ri is∂U(R) ∂ri= PC + PT (R)−RP ′T (R)(PC + PT (R))2 , β(ri) (PC + PT (R))2 ,(I.35)where P ′ T (R) is the first partial derivative of PT (R) with respect to ri.
Q14. what is the PT (R) if it is strictly convex?
a unique globally optimal transmission rate vector always exists and its characteristics are summarized in Theorem 1 according to the proofs in Appendix I.Theorem 1. If PT (R) is strictly convex, there exists a unique globally optimal transmission data rate vector R∗ = [r∗1 , r ∗ 2 , · · · , r∗K ]T for (10), where r∗i is given by(i) when PC+PT (R (0) i )R (0) i≥ ∂PT (R)∂ri ∣∣∣ R=R(0) i, ∂U(R)∂ri ∣∣∣ R=R∗ =0, i.e. 1∂PT (R∗) ∂r∗i= R ∗ PC+PT (R∗) = U(R∗);(ii) when PC+PT (R (0) i )R (0) i< ∂PT (R)∂ri ∣∣∣ R=R(0) i, r∗i = 0,where R(0)i = [r ∗ 1 , r ∗ 2 , · · · , r∗i−1, 0, r∗i+1, · · · , r∗K ] and R(0)i =∑j 6=i r ∗ j , i.e. the overall data rate on all other subchannelsexcept i.
Q15. What is the optimal solution to Problem (22)?
it is also the solution to Problem (22), i.e. R̃∗ = R∗. Otherwise, via the the Lagrangian technique again, the authors have the unique optimal solution as followsr̃∗k = max { f−1k (λ), 0 } , k = 1, · · · ,K, (23)where λ is determined byPT (R̃∗) = Pm. (24)When channel capacity is achieved on each subchannel, the power allocation is the classical water-filling where the water level is determined by Pm [1].