Recharging of Flying Base Stations using Airborne RF Energy Sources
RRecharging of Flying Base Stations using AirborneRF Energy Sources
Jahan Hassan
School of Engineering and TechnologyCentral Queensland UniversityAustraliaEmail: [email protected]
Ayub Bokani
School of Engineering and TechnologyCentral Queensland UniversityAustraliaEmail: [email protected]
Salil S. Kanhere
School of Computer Scienceand Engineering, University ofNew South Wales, AustraliaEmail: [email protected]
Abstract —This paper presents a new method for recharging flyingbase stations, carried by Unmanned Aerial Vehicles (UAVs),using wireless power transfer from dedicated, airborne, RadioFrequency (RF) energy sources. In particular, we study a systemin which UAVs receive wireless power without being disruptedfrom their regular trajectory. The optimal placement of theenergy sources are studied so as to maximize received powerfrom the energy sources by the receiver UAVs flying with alinear trajectory over a square area. We find that for ourstudied scenario of two UAVs, if an even number of energysources are used, placing them in the optimal locations maximizesthe total received power, while achieving fairness among theUAVs. However, in the case of using an odd number of energysources, we can either maximize the total received power, orachieve fairness, but not both at the same time. Numerical resultsshow that placing the energy sources at the suggested optimallocations results in significant power gain compared to non-optimal placements.
Keywords—Unmanned Aerial Vehicle (UAV), wireless powertransfer, charger placement for UAV recharging.
I. I
NTRODUCTION
Recent advances in miniaturization, robotics, sensor tech-nology and communications have revolutionized UnmannedAerial Vehicles (UAVs) and brought about their adoption in awide range of applications. One such application is the use ofUAVs carrying base station equipment acting as aerial base sta-tions that can dynamically re-position themselves to improvecoverage and capacity demands of existing networks [1]–[7].Such aerial base stations could supplement terrestrial infras-tructure when it is overloaded or unavailable, as presented inthe context of G networks in [8], [9]. While the majority ofthese proposals considered non-mobile UAVs hovering overa service area, some recent works [1], [4], [10] have arguedfor the use of flying (or cruising) aerial base stations whereinthe UAVs continue to service ground nodes while in flight.The trajectory, i.e., the movement patterns of the aerial basestations is tailored so as to maximize network performancein the presence of geospatial variation in user demand, orto improve spectral efficiency. A prototype demonstrating theuse of flying aerial base stations was developed recently byEurecom [11].UAVs rely on an on-board battery for power, which limitstheir operational duration before recharging is required. Re- searchers have investigated power-efficient operations of UAVsto extend battery lifetime by reducing the energy consumed forcommunications (electronics) and mobility (mechanical), assummarized in [3]. Since extending the lifetime of the batterydoes not eliminate the need for recharging, a promising andparallel direction of research involves investigating ways forrecharging UAVs to ensure service continuity. In particular,mechanisms for replenishing energy without disrupting theUAV’s usual trajectory (in the case of flying UAVs) or de-ployed locations (in the case of non-mobile UAVs) where theUAV is not required to move to a different location to receivepower, is essential for uninterrupted service provisioning. Inthis paper, we propose an architecture for recharging cruisingUAVs using energy harvesting from received Radio Frequency(RF) transmitted by dedicated, non-mobile airborne UAVsequipped with RF transmitters referred to as transmitter UAVs( tU AV s). In particular, we study the optimum placement ofthe tU AV s to maximize the received energy by the receiverUAVs ( rU AV s).Researchers in [12] have also explored RF energy harvest-ing for recharging UAVs. However, they rely on terrestrialenergy sources for charging UAVs while we consider airbornechargers. As such, our approach can be used in a wide rangeof scenarios where deployment of terrestrial chargers may notalways be possible, for example where UAVs are deployedto monitor ground sensors in a forest. The energy transferefficiency is influenced by both distance and the presenceof obstacles (line-of-sight vs no-line-of-sight). Our approachoffers flexibility to address both these issues. We can positionthe airborne energy sources in a way that would minimizethis distance and improve line-of-sight RF links thus increaseenergy transfer efficiency. Moreover, our work is the first toconsider the optimal positioning of tU AV s which maximizesthe total received energy in the rU AV s.Our contributions in this paper are as follows: (i) we proposean UAV re-charging architecture using wireless power transferfrom carefully positioned, airborne, stationary energy sourcesthat provide power to the UAVs without disrupting their trajec-tories, (ii) we provide a mathematical model to derive optimalplacement of the energy sources to maximize the total receivedenergy in the system, and (iii) we consider a specific scenario a r X i v : . [ c s . N I] M a r ig. 1. The system model for flying base station recharging. The rU AV srepresent the power-receiver U AV s, and the tU AV s represent the RF energysources. Only the ( x , y ) coordinates of the tU AV s are shown, since these areplaced at the same height as those of the rU AV s. of two rU AV s moving along a linear trajectory servicingground nodes stationed within a square region and use ourmodel to determine the optimal locations for two tU AV s. Fromour solutions to the optimal placement problem, we observedthat for this specific scenario, the optimal placement of an evennumber of energy sources will also result in fairness in termsof equal amount of received energy by all rU AV s. However,we found that if we used an odd number of energy sources,either the fairness could be achieved or the total amount ofreceived energy could be maximized, but not both at the sametime. Our numerical results revealed that placing the chargingnodes at the suggested optimal locations resulted in significantpower gain compared to non-optimal placements.The rest of the paper is organized as follows. Section IIpresents our proposed UAV recharging architecture. We firstpresent a general case of any number of tU AV s and rU AV s,followed by a specific case of two tU AV s and two rU AV s.We solve the specific case of energy source placement inSection III. We provide implications of our solutions inSection IV, ending our paper with some numerical results andconclusion in Sections V and VI.II. UAV R ECHARGING A RCHITECTURE : S
YSTEM M ODEL
Our UAV charging architecture is shown in Figure 1, whichis used for charging a number of cruising UAVs that fly backand forth with a linear trajectory over a square area of sidelength l . The trajectories of the cruising UAVs can be ofany form, e.g., of geometric form (circular, linear), or canbe of other forms, as shown in [1], [4]. In our work, we haveassumed a linear trajectory for the cruising UAVs. These arethe RF energy receiver UAVs- the rU AV s, and fly back andforth on a path parallel to the horizontal axis of the square area,with a constant speed V . The rU AV s harvest energy from thereceived RF signals while in service, from airborne, dedicatedenergy sources. We assume that these energy sources arespecialized UAVs, equipped with wireless power transmitters, and are referred to as transmitter UAVs- tU AV s. The tU AV sare placed at fixed locations (i.e., non-mobile) over this areawith their ( x , y ) coordinates given by ( x , y ) & ( x , y ) , etc.,and their z coordinate (the height) is the same as those of the rU AV s. The heights of the tU AV s and the rU AV s being sameimproves the amount of received RF power, which we canadjust since we are using airborne energy sources, as opposedto using terrestrial energy sources with non-adjustable heights.The tU AV s are assumed to have a wire-line connection to theground for a constant power supply [13]. In order to avoidthe chance of collision, the tU AV s must be placed outside thecollision zone R , anywhere in zone R as shown in Figure 1.By careful positioning of the tU AV s in terms of their ( x , y ) coordinates, this architecture aims to maximize the receivedenergy by the rU AV s during their flight time to travel oneside of the square, achieving service continuity by the rU AV swithout disrupting their trajectory. We provide a general modelfor this architecture next.Time taken to travel one side length of the square area bya rU AV is given by T = l / V . Locations of the rU AV and rU AV that are on the parallel edges of the square at time t is given by ( Vt , ) and ( Vt , l ) respectively, for t ∈ [ , T ] . Thereceived power of far-field RF transmission attenuates as perthe reciprocal of the squared distance between the transmitterand the receiver. Therefore, the harvested RF power ( P R ) at thereceiver can be calculated using Frii’s free space propagationmodel [14] as: P R = P T G T G R λ ( π R ) (1)where P T is the transmit power, G T and G R are the antennagains of the transmitter and the receiver, λ is the power transferwavelength, and R is the distance between the transmitter andthe receiver. Without the loss of generality, we can say thatthe received power varies inversely with the square of thedistance between the transmitter and the receiver, which ourmodel is based on. In Section V, we use specific values of theother parameters of Frii’s equation to estimate received power.Distance of rU AV from tU AV at time t is (cid:113) ( Vt − x ) + y .So, energy received by rU AV from the tU AV over [ , T ] is ∝ ∫ T dt ( Vt − x ) + y . (2)For a general rU AV path ( x ( t ) , y ( t )) , ≤ t ≤ T , i.e.,the rU AV located anywhere in the considered area, energyreceived by an rU AV from a tU AV located at ( x , y ) is ∝ ∫ T dt ( x ( t ) − x ) + ( y ( t ) − y ) . (3)Let E rU AV k , tU AV j be the energy received by rU AV k from tU AV j over time ≤ t ≤ T . Then the total energy received by rU AV k is: E k = (cid:213) j E rU AV k , tU AV j ∝ (cid:213) j ∫ T dt ( x k ( t ) − x j ) + ( y k ( t ) − y j ) . (4)he total energy received by all rU AV s from all tU AV s isgiven by: E total = (cid:213) k E k = (cid:213) k (cid:213) j E rU AV k , tU AV j (5)where ( x k ( t ) , y k ( t )) is the flight path of rU AV k for ≤ t ≤ T and ( x j , y j ) is the j th transmitter UAV’s ( tU AV j ) location. The E total can also be calculated by summing up the given energyby all tU AV s to all rU AV s, as: E total = (cid:213) j (cid:213) k E rU AV k , tU AV j (6)where (cid:205) k E rU AV k , tU AV j is the energy provided by tU AV j toall rU AV s. In order to gain an insight into solving the energysource placement problem, we focus on a specific case of twotransmitters and two receivers next. A. The Case of Two tU AVs and Two rU AVs
In this section, we consider a scenario where two tU AV s( tU AV and tU AV ) are placed at locations ( a , b ) & ( a , b ) ,at the same height level of two rU AV s ( rU AV , and rU AV ).The rU AV s fly back and forth over straight-line paths of twoparallel edges of the square, separated from each other witha distance l which is the length of the square. Note that thepaths of the two rU AV s are given by ( x ( t ) , y ( t )) = ( Vt , ) ,and ( x ( t ) , y ( t )) = ( Vt , l ) for ≤ t ≤ T . Total energy receivedby rU AV is given by E ∝ ∫ T dt ( x ( t ) − a ) + ( y ( t ) − b ) + ∫ T dt ( x ( t ) − a ) + ( y ( t ) − b ) . (7)Replacing the path position values of rU AV , we get E ∝ ∫ T dt ( Vt − a ) + b + ∫ T dt ( Vt − a ) + b . (8)Similarly, E ∝ ∫ T dt ( Vt − a ) + ( l − b ) + ∫ T dt ( Vt − a ) + ( l − b ) . (9)So, our objective is to maximize E + E , or equivalently P : max a , b , a , b E + E s.t. ≤ a j ≤ l j = , ε ≤ b j ≤ l − ε j = , (10)Here, ε ∈ ( , l / ) is the width of the region R as in Figure 1,representing the collision area width of each rU AV withinwhich no tU AV s are to be placed. Fig. 2. Optimal placement of two tU AV s to recharge two rU AV s. III. O
PTIMAL S OLUTION TO P ROBLEM
PIn this section, we solve the tU AV placement problem forthe specific case of two tU AV s and two rU AV s as per thedescription in the previous section, with the restriction of notplacing any tU AV in the region R of each rU AV . We showthat the optimal placement of the tU AV s are one transmitteron each boundary of the restricted zones of each rU AV , and inthe middle of the horizontal axis of the zone boundary which isshown in Figure 2. Our main result is the following Theorem. Theorem 1.
The physically unique solution to Problem ( P ) is ( a , b ) = ( l / , ε ) , ( a , b ) = ( l / , l − ε ) . In order to help prove the theorem, we state a series ofuseful Lemmas that we prove in the Appendix.
Lemma 1.
Let c : [ , l ] → R be a (strictly) concave function.Let g : [ , l ] → R , g ( x ) = c ( x ) + c ( l − x ) ∀ x ∈ [ , l ] . Then g is (strictly) maximized at x = l . Proof.
See Appendix A. (cid:3)
Lemma 2.
Let f : [ ε, l − ε ] → R be (strictly) convex, where ε < l / .Let h : [ ε, l − ε ] → R , h ( x ) = f ( x ) + f ( l − x ) ∀ x ∈ [ ε, l − ε ] . Then h is (strictly) maximized at the endpoints (i.e., at x = ε and x = l − ε ).Proof. See Appendix B. (cid:3)
Below we provide the proof of Theorem 1.
Proof.
We have to maximize total energy received by the rU AV s from tU AV and tU AV , which is given byTotal Energy = (Energy provided by tU AV ) + ( Energy provided by tU AV ) . Energy provided by tU AV j is = ∫ T dt ( Vt − a j ) + b j + ∫ T dt ( Vt − a j ) + ( l − b j ) = φ ( a j , b j ) + φ ( a j , l − b j ) , here φ ( a , b ) : = ∫ T dt ( Vt − a ) + b , ≤ a ≤ l , ε ≤ b ≤ l − ε and so, E total ∝ φ ( a , b ) + φ ( a , l − b ) + φ ( a , b ) + φ ( a , l − b ) . Since total energy received are additions of energy con-tributed by each tU AV , which come from the same function φ ( a , b ) + φ ( a , l − b ) with independent values of ( a , b ) , itsuffices to find how to maximize φ ( a , b ) + φ ( a , l − b ) for ≤ a ≤ l , and ε ≤ b ≤ l − ε . Let F ( a , b ) = φ ( a , b ) + φ ( a , l − b ) .Note that F ( a , b ) is proportional to total energy received byboth rU AV s in travelling one side length of the square, fromone tU AV located at ( a , b ) . We want to maximize F over ( a , b ) .We prove the following two properties of F ( a , b ) :Property 1. arg max a ∈[ , l ] F ( a , b ) = l ∀ b ∈ [ ε, l − ε ] We have φ ( a , b ) = ∫ T dt ( Vt − a ) + b = V b (cid:16) tan − (cid:16) ab (cid:17) + tan − (cid:16) l − ab (cid:17)(cid:17) . Recall, F ( a , b ) = φ ( a , b ) + φ ( a , l − b ) . Hence to show arg max a ∈[ , l ] F ( a , b ) = l ∀ b ∈ [ ε, l − ε ] , it suffices to showthat arg max a ∈[ , l ] φ ( a , b ) = l ∀ b ∈ [ ε, l − ε ] . (11)Because ∀ b ∈ [ ε, l − ε ] we have V b > and tan − (cid:16) ab (cid:17) is strictly concave in a for a ∈ [ , l ] , Lemma 1 impliesthe desired result of Equation 11.Property 2. arg max b ∈[ ε, l − ε ] F ( a , b ) = { ε, l − ε } ∀ a ∈ [ , l ] We first show for any constant k > , V b tan − (cid:16) kb (cid:17) isstrictly convex in b for b ∈ [ ε, l − ε ] . Observe V b tan − (cid:16) kb (cid:17) = f ( g ( b )) where f ( x ) = xV tan − (cid:16) k x (cid:17) and g ( b ) = V b (for b ∈ [ ε, l − ε ] ). Since if f is a convex and strictly increasing function,and g is a strictly convex function, then f ( g ( b )) is strictlyconvex, it suffices to show: • g is strictly convex. It is clear that g ( b ) is strictlyconvex. • f is convex, and strictly increasing. We see that f isstrictly increasing because f ( x ) is a product of twostrictly increasing positive functions (for x > ). Toshow f is convex, we observe that its second derivative f (cid:48)(cid:48) ( x ) is kV (cid:16) + k x (cid:17) > for k > .Since φ ( a , b ) = V b tan − (cid:16) ab (cid:17) + V b tan − (cid:16) l − ab (cid:17) the above implies that φ ( a , b ) is the sum of two strictlyconvex functions in b for fixed a ∈ ( , l ) . For a = or l , φ ( a , b ) = V b tan − (cid:16) lb (cid:17) , which is also a strictlyconvex function of b . Therefore, for any a ∈ [ , l ] , φ ( a , b ) is strictly convex in b . Since F ( a , b ) = φ ( a , b ) + φ ( a , l − b ) , Lemma 2 now implies Property 2.These properties prove F ( a , b ) is maximized precisely at ( l / , ε ) and ( l / , l − ε ) . Recall F ( a , b ) is proportional to totalenergy received by both rU AV s in travelling one side lengthof the square, from one tU AV located at ( a , b ) . Total receivedenergy is maximized if we place the (one) tU AV at either ofthese two points. As such, if we have two tU AV s, we need toplace one tU AV at ( l / , ε ) and the other tU AV at ( l / , l − ε ) for the total received energy to be maximized, as total energyis the sum of energy received by the rU AV s from both of the tU AV s.Thus, Theorem 1 is proved. (cid:3) IV. R
AMIFICATIONS OF THE M ODEL
Figure 3 shows the values of F ( a , b ) , which is proportionalto the received energy by two rU AV s for various placementsof one tU AV within zone R . We observe that placing the tU AV anywhere but at either of the two mid-points of theboundaries of R and R results in lower values of F ( a , b ) . Assuch, placing the tU AV s at these optimal locations will resultin maximum total received energy by the two rU AV s. If weplace an even number of tU AV s by equally distributing theseat these two locations, we will achieve an equal amount ofreceived energy in each rU AV (fairness). However, if we havean odd number of tU AV s, there will be an imbalance of energyreceived individually by the rU AV s, since after distributing the tU AV s equally in these two locations, if we place the leftover tU AV in one of these two locations, received total energy willbe maximized, however, this will also mean that the rU AV closer to the last tU AV will receive more energy than the other(unfair). If we place the last tU AV in the middle position fromboth the rU AV s, they will receive an equal amount of energybut the total received energy will not be maximized since thelast tU AV is at a non-optimal location. Thus, we make thefollowing observations: Observation 1.
To recharge two rU AV s with an even numberof tU AV s, it is possible to place the tU AV s in such a waythat maximizes E total , i.e., E + E , and also achieves fairness,i.e., E = E . Observation 2.
To recharge two rU AV s with an odd numberof tU AV s, it is possible to achieve either maximized E total ,or fairness (E = E ), but not both at the same time. Observation 3.
The optimal placement locations are validfor recharging two rU AV s by any number of tU AV s, i.e.,not only by two tU AV s, since new tU AV s contribute to thetotal energy in an additive manner.
V. N
UMERICAL R ESULTS
In order to gain an insight into the average power thatthe optimal placement of one tU AV can provide to the two ig. 3. F ( a , b ) within R , for l = m , ε = m , V = m / s . The two rU AV s are flying over the opposite horizontal axes of the square area.TABLE IP ARAMETER V ALUES
Parameter Value f MHz c m/s P t kW G r , G t dBi V m/s l m ε m rU AV s compared to the non-optimal placements, we reportthe numerical results for a similar scenario as used in ourmodel but with one tU AV . The parameter values are listed inTable I. The calculation is based on Equation 1, however, wereplaced λ with cf where c is the speed of light, and f is thefrequency.For the tU AV power transmission frequency, we have used MHz, which belongs to the non-licensed ISM band. Thisfrequency is commonly found to be generated by garage dooropeners, however, we use this in our numerical experiments forthe RF power transmission. Using this frequency in a remotelocation should not cause interference with other devices. Notethat the commercially available RF transmitters that are usedfor charging low-power devices use higher frequencies, forexample,
Powercaster transmitters [15] use mHz. Ourrequirement of charging UAVs demanding higher power, the tU AV s need to transmit power at lower frequencies, sincelower frequencies result in higher received power. Moreover,due to the significant reduction of RF power at the receivercompared to the transmitted power, we also need the tU AV totransmit at a higher power, which we have taken to be kW.This can be justified by our assumption that the tU AV s haveground power supply connections, thus it can transmit at thislevel. Results discussed below assumes full energy conversionefficiency.Figure 4 shows the average power over the time takento traverse one side of the square (for the rU AV ) in dBm,received by the two rU AV s from one tU AV placed at different ( x , y ) coordinates within the safe placement zone R of the Fig. 4. Average power (dBm) received by two rU AV s from one tU AV located at ( a , b ) , over the time interval [ , T ] . The two rU AV s are flyingover the opposite horizontal axes of the square area. a b ( w i t h i n R ) X: 40Y: 5Z: 25.42
Fig. 5. Average power (dBm) received by one rU AV s from one tU AV located at ( a , b ) , over the time interval [ , T ] . The rU AV is flying over thelower horizontal axis of the square area. square area. The optimal placement of the tU AV as per ourmodel, which is the middle of either of the boundary of R and R (the mid point of the horizontal lines of the coloured zone)resulted in the maximum average received power of . dBm by the two rU AV s. If the tU AV is placed at the middleof the area, i.e, at ( , ) , the average received power became . dBm. Placing the tU AV at the middle of the verticalaxes, i.e., at ( , ) , or ( , ) resulted in the average receivedpower of . dBm. As we can see, the optimal placementof the tU AV resulted in a power gain of . dBm and . dBm over the reported two non-optimal placements.In order to observe the level of power received by one rU AV from one power source, in Figure 5 we have reported theaverage power over the time taken to traverse one side of thesquare when the tU AV is placed at different ( x , y ) coordinates.In this case, the average power received by the rU AV is foundto be . dBm when the tU AV is placed at the optimallocation which is at ( , ) in this case, and . dBmwhen it is located at the middle of the vertical axes. A gain of . dBm was achieved in this case by placing the tU AV at the optimal location.Clearly the optimal placement of the tU AV must be usedin our considered scenario for the best outcome of the en-rgy transmission system, however, received level of energyharvested from RF signals is still quite low, and will requiremultiple dedicated RF sources to power the UAVs [16]. For thefixed-wing drones consuming much less power than the rotary-wing drones since the fixed-wing drones only need energy tomove forward but not to keep afloat in air, our system cancertainly be considered to extend the operation duration ofsuch drones using multiple tU AV s, all placed at (or closer to)the suggested optimal locations. For the placements of multiple tU AV s, Section IV provided some insights.VI. C ONCLUSION
This paper considered the problem of in-situ rechargingaerial base stations without disrupting their regular trajec-tory. We proposed a solution that leverages wireless powertransfer via carefully positioned airborne but stationary energysources. We presented a mathematical model for solving theoptimal placement of these energy sources so as to maximizethe total received power at the UAVs while simultaneouslyachieving fairness. Our numerical results showed that placingthe charging nodes at the suggested optimal locations resultedin significant power gain compared to non-optimal placements.In our future work, we will consider an optimization modelthat takes the number of rU AV s and tU AV s as well as theirmultidimensional trajectories as tuning parameters to find theoptimum solution. A
PPENDIX
A. Proof of Lemma 1Since c is a strictly concave function, therefore c ( l − x ) is also strictly concave. As such, g being the sum of twostrictly concave functions, g is also strictly concave. For g to be (strictly) maximized at x = l / , we have to prove that g ( x ) < g ( l / ) ∀ x ∈ [ , l ] , and x (cid:44) l / . Let x ∈ [ , l ] , and x (cid:44) l / . We have g ( x ) = c ( x ) + c ( l − x ) = (cid:16) c ( x ) + c ( l − x ) (cid:17) < c (cid:16) x + ( l − x ) (cid:17) by strict concavity,and since x (cid:44) l − x = c (cid:16) l (cid:17) = g (cid:16) l (cid:17) . Thus, Lemma 1 is proved.B. Proof of Lemma 2Since f is a strictly convex function, therefore f ( l − x ) isalso strictly convex. As such, h being the sum of two strictlyconvex functions, h is also strictly convex. We have to showthat if ε < x < l − ε then h ( x ) < h ( ε ) (note h ( ε ) = h ( l − ε ) ).Suppose ε < x < l − ε . Then there exists θ ∈ ( , ) such that x = θε + ( − θ )( l − ε ) . So, h ( x ) = h (cid:16) θε + ( − θ )( l − ε ) (cid:17) < θ h ( ε ) + ( l − θ ) h ( l − ε ) by strict convexity,and since ε (cid:44) l − ε = h ( ε ) . Thus, Lemma 2 is proved.R
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