A-Duplex：Medium Access Control for Efficient Coexistence Between Full-Duplex and Half-Duplex Communications

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IEEE TRANSACTIONS ON WIRELESS COMMUNICATIONS, VOL. 14, NO. 10, OCTOBER 2015 5871 A-Duplex: Medium Access Control for Efficient Coexistence Between Full-Duplex and Half-Duplex Communications Aimin Tang and Xudong Wang, Senior Member, IEEE Abstract—As full-duplex wireless communication evolves into a practical technique, it will be built into communication nodes in many application scenarios. However, it is difficult to do so for legacy communication nodes. Thus, full-duplex communication nodes will coexist with half-duplex communication nodes in the same application environment. In this paper, a wireless local area network with a full-duplex access point (AP) and half-duplex clients is studied, and a media access control (MAC) protocol called asymmetrical duplex (A-Duplex) is developed to support ef- ficient coexistence between half-duplex clients and the full-duplex AP. A-Duplex explores packet-alignment-based capture effect to establish dual links between the AP and two different clients. In this way, the capability of a full-duplex AP can be utilized by half-duplex clients, which leads to much improved network throughput. Moreover, to ensure fairness of the MAC protocol, a virtual deficit round-robin algorithm is proposed for the AP to select appropriate half-duplex clients for dual-link setup. A-Duplex does not require any change in the physical layer of half-duplex clients; only an update of MAC driver is necessary. Thus, it is well suited for coexistence between half-duplex clients and a full-duplex AP. Both analysis and simulations are conducted to evaluate performance of A-Duplex. Results show that it improves the throughput by 48% and 188% and reduces the average packet delay by 26% and 22%, as compared to the IEEE 802.11 Distributed Coordination Function with and without RTS/CTS, respectively. Moreover, the throughput remains steady as the number of clients grows. A-Duplex also maintains a high level of fairness. Index Terms—Medium access control, full-duplex communication, wireless LAN, coexistence between full-duplex and halfduplex communications. I. INTRODUCTION F ULL duplex wireless communication is evolving into a more practical technique [1]–[5]. In [1], three antennas are used to achieve about 30 dB antenna cancellation. Combined with path loss attenuation, analog cancellation and digital cancellation, this design can achieve about 100 dB self-interference cancellation within 5 MHz bandwidth for IEEE 802.15.4. In [2], Balun cancellation is leveraged for an improved design of self-interference cancellation, in which only two antennas Manuscript received October 21, 2014; revised February 25, 2015 and June 6, 2015; accepted June 7, 2015. Date of publication June 12, 2015; date of current version October 8, 2015. The associate editor coordinating the review of this paper and approving it for publication was C.-F. Chiasserini. (Corresponding author: Xudong Wang.) The authors are with the UM–SJTU Joint Institute, Shanghai Jiao Tong University, Shanghai 200240, China (e-mail: wxudong@ieee.org). Color versions of one or more of the figures in this paper are available online at http://ieeexplore.ieee.org. Digital Object Identifier 10.1109/TWC.2015.2443792 are used. With the help of path loss attenuation and digital cancellation, about 110 dB cancellation can be achieved within 20 MHz. In [3], an additional transmitting chain is added to generate an invert analog signal to cancel the self-interference. So far the best design is given in [4], where only one antenna is used to achieve almost perfect self-interference cancellation for full duplex WiFi radio. In this design, a passive circuit is first used to achieve 60 dB self-interference cancellation and then both linear and nonlinear digital cancellation are used to cancel the remaining self-interference. In [5], a balanced cancellation circuit is designed for RF front-end. This design can achieve 60 dB cancellation at the RF front-end. On the other hand, it can be integrated with the design in [4] to further improve the self-interference cancellation. Full duplex can nearly double the capacity of a point-to-point communication link and thus significantly improves the spectrum efficiency. However, to fully leverage the capability of full duplex communications in a network, an efficient media access control (MAC) protocol plays a critical role. So far a few MAC protocols have been proposed to support full duplex communications [6]–[9], but they assume that all nodes in the network are equipped with full duplex radios. However, in many application scenarios full duplex and half duplex radios have to coexist in the same network for two main reasons. First, despite fast progress in the development of full duplex radios, challenges still remain when applying full duplex radios to nodes such as smart phones or laptops. For example, in [1]–[3], [5], all radios need more than one antennas to achieve full duplex communications, and the distance between antennas are more than 20 cm. In [4], only one antenna is needed, but implementation of the circuit is rather complicated. These factors make it difficult for a smart phone or personal digital assistant (PDA) to adopt a full duplex radio. Second, many legacy devices only support half duplex communications. It is fine to replace a legacy AP by a full duplex AP, but replacing all legacy devices by full duplex ones is neither economical nor acceptable. Thus, coexistence between full duplex and half duplex communications becomes indispensable. For example, supporting a wireless LAN with a full duplex AP and half duplex clients is extremely meaningful for the next generation WiFi networks. In this paper, we study a full duplex wireless LAN where the AP supports full duplex communications but all clients are half duplex nodes. The full duplex AP is assumed to have perfect self-interference cancellation [4]. Since all clients have 1536-1276 © 2015 IEEE. Personal use is permitted, but republication/redistribution requires IEEE permission. See http://www.ieee.org/publications_standards/publications/rights/index.html for more information

5s72 IEEE TRANSACTIONS ON WIRELESS COMMUNICATIONS.VOL 14.NO.10.OCTOBER 2015 Preamble PACKET1 PACKET 2 AP to Client C will be longe Fig 1.A wireless LAN with a full duplex AP. AP ne packet om Chent A to the capture effect can help establish dual links but it cann no capability of full duplex communications.only asymr netric ensure throughput improvement.Asa result,to better utilize the capture effect n in Fig.I nds less LAN with a full duplex AP and hal AP the AP C).In [6] etric dual links (i.e.,AAP- c)do first:2)the received signal strength of the packet from the AP o a client is stronger than the interference from the packet sen cap from ano ther chent to th o fully leverage the capability of the full duplex AP.we develop that throug hput can be imp oved.To this end,the AP needs re effect at the client side to collect relative signal strength between nodes to build an be properly f full h clients can selecte n 1 is de ed to Thanks to the asymmetric dual links,our MAC pr ocol sur ports efficient co-existen ce between a full duplex AP and hal Ao be hidden to cachtke fom the AP to Client B.dug ex (A-Dup utilized to form full duplex communications at the AP.even though Clients A and B are not hidden from each other. improved by 48%and 188%as compared to DCF [14]wit RTS/CT 2线 proeL-align nt for hetter cantur A-Duplex is distinct with several features.First.it leverage Sant first.In either type.eket as shown in packet o fully uti lize the capa for the fir to ba cond packet (This e links example e first packet blist the based on this map asymm nks are e ba that of the sc environment.Third.a virtual deficit round robin algorithm is effect ma asymmetric dual links are established (eA ver update is peeded.To the best of our knowledge.this is the first MAC that holds these features the transmission rate from the AP to Client C is lower than nt sign of A-Dup links in A-Duplex are elaborated in Section IV.The fairness issue is studied in Section V.Simulation results are presented in

5872 IEEE TRANSACTIONS ON WIRELESS COMMUNICATIONS, VOL. 14, NO. 10, OCTOBER 2015 Fig. 1. A wireless LAN with a full duplex AP. no capability of full duplex communications, only asymmetric dual links can be utilized to boost network throughput. For example, as shown in Fig. 1, while Client A sends a packet to the AP, the AP sends a packet to another client (e.g., Client C). In [6], asymmetric dual links (i.e., A → AP → C) do not allow collisions at both receivers, i.e., the signal from Client A does not interfere Client C. However, such a condition cannot be easily satisfied in a wireless LAN, so the capability of the full duplex AP cannot fully utilized. In order to fully leverage the capability of the full duplex AP, we develop a novel MAC protocol to explore capture effect at the client side to improve opportunity of full duplex communications in a wireless LAN. Capture effect [10]–[12] means, when a receiver receives two colliding packets, the stronger one can still be decoded correctly. For example, Client A and Client B may not be hidden to each other, but a packet from Client A to the AP may not corrupt the packet from the AP to Client B, due to capture effect at Client B. In this situation, capture effect is utilized to form full duplex communications at the AP, even though Clients A and B are not hidden from each other. The performance of capture effect can be significantly improved through alignment of two colliding packets. To leverage packet-alignment for better capture effect,1 there are two options: 1) the desired packet is sent first; 2) the interfering packet is sent first. In either type, if the preamble of the first packet does not collide with the second packet as shown in Fig. 2, it is relatively easy for the first packet to be decoded correctly. However, it is rather difficult to decode the second packet (This case is also called message in message (MIM) in [13]). For example, under the basic rate in IEEE 802.11a, the first packet can be decoded when its signal strength is 0 dB stronger than that of the second packet [12]. If we want to decode the second packet correctly, the signal strength of the second packet need to be 11 dB stronger than that of the first packet [12]. However, the asymmetric dual links supported via capture effect may not always improve the network performance. When asymmetric dual links are established (e.g., A → AP → C), the transmission rate from the AP to Client C is lower than that in the half duplex case since Client C is interfered by Client A. Thus, the transmission time of the packet from the 1If successive interference cancellation (SIC) is taken into account, then packet alignment is not necessary. However, the client side consists of legacy nodes, so any change to the physical layer is infeasible. Thus, SIC is not considered in this paper. Fig. 2. Capture effect without collision in preamble time. AP to Client C will be longer than that in the half duplex case. The concurrent transmission in dual links can save some time; however, if the transmission time from the AP to Client C is much longer than that of the packet from Client A to the AP, the dual links may reduce the throughput. In other words, the capture effect can help establish dual links but it cannot ensure throughput improvement. As a result, to better utilize the capture effect in a wireless LAN with a full duplex AP and half duplex clients, the MAC protocol must take into account three requirements: 1) the packet from the AP needs to be transmitted first; 2) the received signal strength of the packet from the AP to a client is stronger than the interference from the packet sent from another client to the AP; 3) the AP needs to choose a client with a proper transmission rate to establish dual links so that throughput can be improved. To this end, the AP needs to collect relative signal strength between nodes to build an information map from which clients can be properly selected to establish dual links effectively. In this paper, a dynamic information update scheme is designed to establish such a map. Thanks to the asymmetric dual links, our MAC protocol supports efficient co-existence between a full duplex AP and half duplex clients, so it is called asymmetrical-Duplex (A-Duplex) in this paper. A-Duplex not only improves the throughput but also reduces packet delay. Simulation results show that the throughput is improved by 48% and 188% as compared to IEEE 802.11 DCF [14] with RTS/CTS and without RTS/CTS, respectively. The packet delay is reduced by 26% and 22%, respectively. A-Duplex is distinct with several features. First, it leverages packet-alignment based capture effect to establish asymmetric dual links so as to fully utilize the capability of the full duplex AP. As a result, throughput is improved effectively as compared to the case of only using the hidden nodes to establish dual links. Second, a map of signal-to-interference ratio (SIR) is built at the AP; based on this map asymmetric dual links are established to take advantage of capture effect. The SIR map can be updated dynamically to capture the variable network environment. Third, a virtual deficit round robin algorithm is developed for the AP to select a downlink in dual link setup, which improves fairness of the MAC protocol. Finally, no physical layer change is needed at legacy nodes; only MAC driver update is needed. To the best of our knowledge, this is the first MAC that holds these features. The rest of the paper is organized as follows. Related work is introduced in Section II. The protocol design of A-Duplex is described in Section III. Detailed procedures of establishing dual links in A-Duplex are elaborated in Section IV. The fairness issue is studied in Section V. Simulation results are presented in Section VII. Compatibility and practicality issues are discussed in Section VIII. The paper is concluded in Section IX

TANG AND WANG:MAC FOR EFFICIENT COEXISTENCE BETWEEN FULL AND HALF-DUPLEX COMMUNICATIONS 5873 II.RELATED WORK A.Capture effect A当A mitted SFS ACK 1 destroy all the frames.If the receiving power of one of the arge eough than that other,the others NAVIRTS of receiving a frame from colliding frames is called (a) effect.The physical layer model for capture effect has been Delay ti ng de comes ear r has AP CTS DATA to node B ACK synchronized with the first frame.However.if a weaker frame B SIFS ACK 1 since the ver n be wit other be much larger than the weak one so that the receiver can re. 16].Since the receiver ta scenario:the transmis mak even e [121 the experimental results show that even two frame n time of the Howeve a pream ne th the SIRe and channel estimation which is consistent with [12].If packet alignment-based capture ofar the work in[]provides the best model for capture eftect is leveraged,then dB SIR is sufficient [12].For other des (ie. te),the pnbtknfhMaCp phor net ork It ador B.Full Duplex MAC noise(PN)sequence-based signatures toidentify the concurren links on-the-fly and explore the cha So far there exist a few mac protocols that supp full ance of exposed transmi duplex communications [6]-[8).In [6],both symmetric and ed.The ghput c are for by two no ng sig in Fig.1.A set of asymmetric dual links are esta IIL MAC PROTOCOL DESIGN ablished by three nodes in a two-hop setup.such as the links (BAP →C) In this section e the Ap full duplex C)in Fig. 11. full du ilit ork links need to be established.As explained before,to explore capture effect in an efficient way,the packet from the AP to downlin C)at the Client C:b)the A Howe since a lent is in th In 7.a reservation based MAC protocol was developed for the channel busy and cannot start a second link with the ap a network where all nodes have full du lex radios.All nodes are To resolve this issue,an RTS/CTS mechanism is added into the allocated wit tim ots by the Al In this M. pro to star dual links v re the AP-to-client lin dual lin -AP-C n00 an RTS if they are not hidden nodes,since the interference by Client Three cases are considered below.When a client gets the A is small at Client C.In this sche channel via the RTS frame,two cases need to be AP a pac ever packet-aigment based capture effect are not considered third case is that the Ap acquires the channel first.Note that

TANG AND WANG: MAC FOR EFFICIENT COEXISTENCE BETWEEN FULL- AND HALF-DUPLEX COMMUNICATIONS 5873 II. RELATED WORK A. Capture effect In random access, if more than one frames are transmitted concurrently, collision occurs. However, it does not always destroy all the frames. If the receiving power of one of the colliding frames is larger enough than that of the other, then the receiver can decode this frame correctly. Such a process of receiving a frame from colliding frames is called capture effect. The physical layer model for capture effect has been explored by various experiments in [10]–[12], [15]. Usually if the stronger frame comes earlier than other frames, it is easier to decode this frame since the receiver has already been synchronized with the first frame. However, if a weaker frame comes first, MIM is needed to decode the stronger frame. In this case, since the receiver has been synchronized with the weaker frame, the signal strength of the stronger frame must be much larger than the weak one so that the receiver can resynchronize with the stronger one to utilize capture effect [12], [13], [16]. Since the receiver takes a certain amount of time to achieve synchronization, even if the stronger frame arrives earlier, it is necessary to make sure the lead time is sufficient. In [12], the experimental results show that even two frames come at the same time, capture effect may still succeed in 802.11a networks. However, if the first frame is earlier by a preamble time, the success probability of capture effect can be improved, because a clean preamble is helpful to conduct synchronization and channel estimation. So far the work in [12] provides the best model for capture effect in 802.11a networks. It provides the measurement-based SIR capture thresholds for all the 802.11a bit rates. B. Full Duplex MAC So far there exist a few MAC protocols that support full duplex communications [6]–[8]. In [6], both symmetric and asymmetric dual links are considered. A set of symmetric dual links are formed by two nodes transmitting signal to each other simultaneously. An example is the link (A → AP → A) shown in Fig. 1. A set of asymmetric dual links are established by three nodes in a two-hop setup, such as the links (B → AP → C) shown in Fig. 1. In [6], the MAC protocol mainly explores symmetric dual links to improve network throughput, and in the asymmetric case (e.g., links (B → AP → C) in Fig. 1), dual links can be established only under the following conditions: a) the uplink transmission (B → AP) does not collide with the downlink transmission (AP → C) at the Client C; b) the AP transmits to Client C later. In [7], a reservation based MAC protocol was developed for a network where all nodes have full duplex radios. All nodes are allocated with some time slots by the AP. In this MAC protocol, asymmetric dual links are allowed even for non-hidden nodes. For example, dual links A-AP-C in Fig. 1 are allowed even if they are not hidden nodes, since the interference by Client A is small at Client C. In this scheme, AP first collects the interference information of clients and then schedules the transmission order according to the interference relationship. However, packet-alignment based capture effect are not considered Fig. 3. Dual link establishment. (a) The “AP-shorter” scenario: the transmission time of the packet from the AP to B is shorter than the packet transmission time of the other link plus a preamble time. (b) The “AP-longer” scenario: the transmission time of the packet from the AP to B is longer than the packet transmission time of the other link plus a preamble time. in this scheme. Thus, the performance of capture effect is limited. For example, in the basic operation mode (i.e., lowest rate with robust modulation/coding), the SIR needs to be 10 dB, which is consistent with [12]. If packet alignment-based capture effect is leveraged, then 1 dB SIR is sufficient [12]. For other modes (i.e., higher rate), the situation is similar. In [8], the MAC protocol addresses the exposed terminal problem in a full duplex network. It adopts the pseudo-random noise (PN) sequence-based signatures to identify the concurrent links on-the-fly and explore the chance of exposed transmissions. As a result, the throughput of a full duplex network can be improved. The protocol also assumes all nodes have full duplex capability. III. MAC PROTOCOL DESIGN In this section, a MAC protocol called A-Duplex is designed for a wireless LAN where the AP has full duplex capability, but all clients can only work in the half duplex mode. To utilize full duplex capability in this wireless LAN, asymmetrical dual links need to be established. As explained before, to explore capture effect in an efficient way, the packet from the AP to a client needs to start first. However, since a client is in the half duplex mode, once it detects such a transmission, it considers the channel busy and cannot start a second link with the AP. To resolve this issue, an RTS/CTS mechanism is added into the MAC protocol to start dual links where the AP-to-client link can be started first. As shown in Fig. 3, the procedure to set up a full duplex link always starts with an RTS frame by a client. Three cases are considered below. When a client gets the channel via the RTS frame, two cases need to be considered: 1) the AP can send a packet to another client to establish dual links; 2) the AP does not have a packet for another client. The third case is that the AP acquires the channel first. Note that

5s74 LEEE TRANSACTIONS ON WIRELESS COMMUNICATIONS,VOL 14.NO.10.OCTOBER 2015 RTS is only initiated by a client.so the AP does not g nerate A DATA to AP any RTS frame but respondstoaclient's RTS with a CTS frame an RTS cas Fig. CAP find AP others_ A and then transmits the packet to client B immediately.Note that in each frame the calle whic Fig.4.An example showing no dual link frame,it finds out that the duration in the CTS frame is longer The delay for Client A serves two purposes.First,for a legacy the pa time.The delay time can be computed by Eg.2).There are the packet to Client B is sent first two scenarios in this case.If the transmission time of the packet protected without interference rom th n the above procedur Client A in Fig.3 vaits a tim b busytone signal to ensure two transmissions finish at the same Frame Spacing (DIFS)time.Thus,the ACK may collide with me Otherwis as denoted by the "AP-longer h3(b).Client the packet from the other client.In our design.the CTS frame problem.I dura中 n in CI covers the enti and the AP finish their missions,Client B first returns an CTS fran update the NAV value ordins to the duration in ACK frame to the AP and then the AP returns an ACK frame the CTS frame.As a result.the ACK from the AP can also be to Client A upon re protected the aph L rly.Sin in the beeinnin the Client A does not know whether or not the AP will establish dual links.it computes the duration as if the A hepCteoraohercietomypiesaCTS es not set up From the cts frame Client A finds out that the duration in CT up a li he half me w A knows th duplex case.If the AP does set up dual links.it just uses th time and then starts a packet transmissio .When the packet is duration in the RTS frame to compute the duration of the CTS expla ed be transmission cove In the th and the AP As be set up i ot support full duplex communications.so symmetrical dual inks (1) selve the queue status as well as the length of the first packet in 3()the nC as follows.In the the AP knows exactly which client is about to send a pa et and the DurationcTs DurationgTS-TCTs-2TsIFS+Tp TACK. es how to establish dual links with another client.In [6] an业 time.In the case of Fig.3(b),the the packet is transmitted by OFDM symbols such as in 802.11a network,the AP cannot decode the packet header to get the Durationcrs =T2+TSIFs +2TACK ransm itting ent ad dress until it receives the wl e packet where t is the transmission time of the packet from the ap to AP Client B.When A receive the CTS frame,it decides the delay the packet is not interfered by the packet from Client A to the time according to can decod e the packet more easily throug Delay=DurationcTs-T-TsIFs-2TACK 21

5874 IEEE TRANSACTIONS ON WIRELESS COMMUNICATIONS, VOL. 14, NO. 10, OCTOBER 2015 RTS is only initiated by a client, so the AP does not generate any RTS frame but responds to a client’s RTS with a CTS frame. The first case is shown in Fig. 3. Client A first transmits an RTS frame to the AP. When the AP finds that it has a suitable packet for Client B, it replies a CTS frame to Client A and then transmits the packet to client B immediately. Note that in each frame there is a field in the MAC header called duration, which is used for other nodes to update the network allocation vector (NAV) value. When Client A receives the CTS frame, it finds out that the duration in the CTS frame is longer than its duration. Thus, Client A knows that the AP intends to establish dual links. It then transmits the packet to the AP after a certain delay, which is at least longer than a preamble time. The delay time can be computed by Eq. (2). There are two scenarios in this case. If the transmission time of the packet from the AP to Client B is shorter than the packet transmission time of the other link plus a preamble time, as denoted by the “AP-shorter” scenario in Fig. 3(a), then the AP transmits a busytone signal to ensure two transmissions finish at the same time. Otherwise, as denoted by the “AP-longer” scenario in Fig. 3(b), Client A delays its transmission for enough time such that two transmissions finish simultaneously. When Client A and the AP finish their transmissions, Client B first returns an ACK frame to the AP and then the AP returns an ACK frame to Client A upon receiving the ACK frame from Client B. To ensure correct operation of A-Duplex, the duration in RTS and CTS needs to be computed properly. Since in the beginning Client A does not know whether or not the AP will establish dual links, it computes the duration as if the AP does not set up dual links. In case that the AP does not set up dual links, the duration field in the RTS frame works exactly as that in the half duplex case. If the AP does set up dual links, it just uses the duration in the RTS frame to compute the duration of the CTS frame as explained below. Since the AP’s transmission covers all clients, the duration field in the CTS frame can gracefully protect the transmissions between Client A and the AP. As a result, the duration in RTS frame is computed as DurationRTS = 3TSIFS + TCTS + T1 + TACK, (1) where TSIFS is the Short Inter-Frame Spacing (SIFS) time, TCTS is the time of CTS frame, TACK is the time of ACK frame, and T1 is the time of data packet to the AP. With the duration from RTS, the AP computes the duration in CTS as follows. In the case of Fig. 3(a), the duration in CTS is computed as DurationCTS = DurationRTS − TCTS − 2TSIFS + Tp + TACK, where Tp is the preamble time. In the case of Fig. 3(b), the duration in CTS is computed as DurationCTS = T2 + TSIFS + 2TACK, where T2 is the transmission time of the packet from the AP to Client B. When A receive the CTS frame, it decides the delay time according to Delay = DurationCTS − T1 − TSIFS − 2TACK. (2) Fig. 4. An example showing no dual links. The delay for Client A serves two purposes. First, for a legacy node, when it finishes a data packet transmission, it waits a fix time for ACK. If no ACK is received after the preset timeout, it starts the retransmission procedure. Second, the delay ensures the packet to Client B is sent first and its preamble can be protected without interference. In the above procedure, Client A in Fig. 3 waits a time period of TSIFS + TACK before it receives its ACK. This time period is longer than Distributed Coordination Function Inter Frame Spacing (DIFS) time. Thus, the ACK may collide with the packet from the other client. In our design, the CTS frame can solve this problem. The duration in CTS covers the entire period of the above procedure, so all other clients receiving the CTS frame update the NAV value according to the duration in the CTS frame. As a result, the ACK from the AP can also be protected. In the second case when the AP has no packet for a client, the operation procedure is shown in Fig. 4. When AP finds no suitable packet for all other clients, it only replies a CTS frame. From the CTS frame, Client A finds out that the duration in CTS frame is the same with its duration. Thus, Client A knows that the AP will not establish dual links, so it only delays an SIFS time and then starts a packet transmission. When the packet is received correctly, the AP returns an ACK to Client A. In the third case, the AP may get the channel first by sending a data packet without RTS/CTS exchange. No dual link can be set up in this case for two reasons. First, a client does not support full duplex communications, so symmetrical dual links cannot be formed. Second, clients cannot decide among themselves about which one can start asymmetrical links to explore capture effect. If the decision is done by the AP, then the queue status as well as the length of the first packet in queue of each client must be reported to the AP. To avoid such complexity, dual link setup is ignored in the third case. In A-Duplex, the RTS/CTS frame exchange procedures ful- fill the following functions. Firstly, from the RTS frame, the AP knows exactly which client is about to send a packet and then decides how to establish dual links with another client. In [6], the AP transmits packet to another client to establish dual links when it decodes the header of the received packet. However, if the packet is transmitted by OFDM symbols such as in 802.11a network, the AP cannot decode the packet header to get the transmitting client address until it receives the whole packet. Secondly, the RTS/CTS mechanism guarantees that the packet from the AP to Client B is transmitted first and the preamble of the packet is not interfered by the packet from Client A to the AP. Thus, Client B can decode the packet more easily through capture effect. Thirdly, the RTS frame contains signal strength information that can be used by the AP to select a proper client

TANG AND WANG:MAC FOR EFFICIENT COEXISTENCE BETWEEN FULL AND HALF-DUPLEX COMMUNICATIONS 5875 to set up a dual link Details of this functionre explained in the then they compute the SIR value with the two recorded values an A-Duplex may experience two types of collisions.In the first Client B)contends the channel successfully,it transr mits its type of collisions.more than on clients send an Rohe recorded SIR value via the RTS frame to the AP.Since the AP the last roun colision ha and do n the RTSfm ws th SIR In the second type of collisions,the AP starts a transmission given the interference from Client A.Thus,it updates such the SIR MAP.If the same client llican sim eive,it the AP stop cha value is igno and let the client's RTS frame pr channel successfully In this scenario all client just update ceed.so the client can still contend the channel successfullv thethe channe clents pdate the IV.ESTABLISHMENT OF ASYMMETRICAL DUAL LINKS the the The critical step of setting up dual links is to ensure capture STR value in the AP's SIR MAP By building up SIR MAP the effect to be properly utilized.Forexampe if du whe interfere other words,the AP needs to find out the signal-to-interference RTS frame from the interfering client.To achieve this goa ent ve use the transmit the RTS that othe an of SIR for all client what follow h is effec an information map is called the sir map wireless LAN the transmit power ofan RTS frame is usuall set the same as that of a regular data packet.Sec ond,an RT A.Establishment of SIR MAP rame uses the rate for tra n,so it The SIR MAP contains the SIR information of all clients a busy channel but can nnot receive an rTS frame it relies or Considering Clients A and B in a wireless LAN.SIR at Client wce from repre nted by sh le that ere was an RT nal str Cto A.A to B.Cto B.A to C.and B toC.because there are 6 interference signal strength en two clients all clients can find out whether 10 the a clie on happens via kI ge.M cally,if ing an RTS frame from a client,it knows collision has happened and thus skips updating its interference signal strength. ansmits sa packet toa client (B).th ue to channel f .an A)tre AP,the current client (i.e.Client B)can also get the inter- the value instead a moving average strategy is necessary one erence strength by o erhearing the pack et.Sinc we make n casy approach is to use the average of the most recent value change to the physic signal str the Al store to the Strength Indicator (RSSD when a client succeeds in receiving a needs to stores 5 SIR values for each client.In order to reduce packet in the MAC layer. the storage of AP.a simple moving average scheme is given as AC layer m follows d help buld SIR=SIROm x (1-0)+ (3 the channel.it first transmits an KTS frame to the AP All where 6 is a weight strength nter re ing the R 4,the the current the signal strength from AP according to the CTS frame,and scheme.the AP only needs to store one SIR value for each

TANG AND WANG: MAC FOR EFFICIENT COEXISTENCE BETWEEN FULL- AND HALF-DUPLEX COMMUNICATIONS 5875 to set up a dual link. Details of this function are explained in the next section. Fourthly, ACK frames in dual links are protected via the duration in RTS/CTS frames (i.e., the NAV mechanism). A-Duplex may experience two types of collisions. In the first type of collisions, more than one clients send an RTS to the AP. Without correct reception of an RTS, AP does not reply a CTS frame. As a result, the competing clients know that collision happens and do not start data packet transmission. In the second type of collisions, the AP starts a transmission simultaneously with an RTS frame from a client. Since the AP can simultaneously transmit and receive, it can find out collisions while it is transmitting. In this case, the AP stops transmitting immediately and let the client’s RTS frame proceed, so the client can still contend the channel successfully. IV. ESTABLISHMENT OF ASYMMETRICAL DUAL LINKS The critical step of setting up dual links is to ensure capture effect to be properly utilized. For example, if dual links (from A to AP and from AP to C) are to be set up, the AP needs to make sure that the signal strength at Client C is stronger enough than interference from Client A according to the transmission rate. In other words, the AP needs to find out the signal-to-interference ratio (SIR) for Client C and decides the transmission rate based on the SIR value. To this end, it is necessary for the AP to build an information map of SIR for all clients. In what follows, such an information map is called the SIR MAP. A. Establishment of SIR MAP The SIR MAP contains the SIR information of all clients. Considering Clients A and B in a wireless LAN, SIR at Client A given the interference from B can be represented by SIR of B to A. Thus, if a wireless LAN has 3 clients: clients A, B, and C, then SIR map table must have 6 items such as SIR of B to A, C to A, A to B, C to B, A to C, and B to C, because there are 6 different scenarios of interference between two clients. To measure the SIR value experienced by a client, two values are needed: the signal strength from the AP and that from the interfering client. However, an AP cannot get such information by itself, so it has to rely on clients to collect such information. When the AP transmits a packet to a client (e.g., Client B), the client can get the signal strength of this on-going transmission. When another client (e.g., Client A) transmits a packet to the AP, the current client (i.e., Client B) can also get the interference strength by overhearing the packet. Since we make no change to the physical layer of legacy nodes, the signal strength that can be collected in the MAC layer is the Received Signal Strength Indicator (RSSI) when a client succeeds in receiving a packet in the MAC layer. MAC layer messages are used to support SIR measurements at clients and help build the SIR MAP at the AP. As explained in the previous section, when a client (e.g., Client A) accesses the channel, it first transmits an RTS frame to the AP. All other clients who receive this RTS frame record the interference strength. After receiving the RTS frame, the AP replies a CTS frame and records this client address. All other clients update the signal strength from AP according to the CTS frame, and then they compute the SIR value with the two recorded values: the signal strength from the AP and interference strength. In the next round of RTS/CTS/Data/ACK, if one client (e.g., Client B) contends the channel successfully, it transmits its recorded SIR value via the RTS frame to the AP. Since the AP records which client (i.e., Client A) started the last round of RTS/CTS/Data/ACK, it knows that the SIR value carried in the RTS frame is the SIR of a new client (i.e., Client B) given the interference from Client A. Thus, it updates such an SIR value (of A to B) in its SIR MAP. If the same client contends the channel successfully, the SIR value is ignored. As explained in the protocol design, the AP may contend the channel successfully. In this scenario, all clients just update the signal strength from the AP and recompute the SIR value. If the AP continuously gets the channel, all clients update the SIR value until a client contends the channel successfully. Since then, the above procedure is followed to record or update the SIR value in the AP’s SIR MAP. By building up SIR MAP, the AP can have a dynamic view of SIR for each client. In the MAC layer, RSSI information is only available when a MAC packet is received. Given a client, in order to detect signal strength from an interfering client, it needs to get the RTS frame from the interfering client. To achieve this goal, we use the basic rate to transmit the RTS frame so that other clients can rely on this frame to detect strength of an interfering signal. This approach is effective for two reasons. First, in a wireless LAN the transmit power of an RTS frame is usually set the same as that of a regular data packet. Second, an RTS frame uses the basic rate for transmission, so it can mostly cover the large area of a wireless LAN. In case a client can sense a busy channel but cannot receive an RTS frame, it relies on a CTS frame to get RSSI. More specifically, when the client receives a CTS frame, it can conclude that there was an RTS frame before. Thus, the client can use the minimum required signal strength for decoding an RTS frame to approximate the interference signal strength. As for the collision case, all clients can find out whether a collision happens via RTS/CTS exchange. More specifically, if a client does not receive a CTS frame from the AP after receiving an RTS frame from a client, it knows collision has happened and thus skips updating its interference signal strength. Due to channel fading, an instantaneous SIR value cannot always accurately reflect the channel quality very well. Thus, when the AP updates the MAP table, it cannot simply substitute the value. Instead, a moving average strategy is necessary. One easy approach is to use the average of the most recent values. However, this demands the AP to store too many SIR values. For example, if we want to get the average of 5 SIR values, AP needs to stores 5 SIR values for each client. In order to reduce the storage of AP, a simple moving average scheme is given as follows: SIRNew = SIROld × (1 − θ ) + SIRupdate × θ , (3) where θ is a weight factor from 0 to 1, the value SIROld is from the current SIR MAP, SIRUpdate is newly learned from frame exchange, and SIRNew is the new SIR in the MAP. With this scheme, the AP only needs to store one SIR value for each

5s76 IEEE TRANSACTIONS ON WIRELESS COMMUNICATIONS.VOL 14.NO.10.OCTOBER 2015 Octets 2 6 6 1 AP to Client B,then the throughput can be improved.When the Frame Duration RA TA SIR FCS AP transmits a packet to a client,the transmission time TAp is given by Fig 5.RTS frames ructu TAP=Tdata TsIFS TACK TDIFS. client.Hov thever.if o where is the DIFS time and Ta for a specific environment.as discussed in performance evaluation. is given b establishment pro cedure of SIR MAP we kn the Ap us is not th rea (5 this SIR i links,the capture link may fail.From the simulation results where T is th in Section VII-B2,we know that the failure rate is less than n the 0% s that th Wot perfe Fig.3(a).the transmission time T is given by dual links.the receiving client fwe try toa ieve 100 fal。he r ne e re where T is the preamble time.If the AP establishes dual links hish overhead.In contrast.the overhead from setting up the SIR in the"AP-longer"scenario in Fig.3(b),the transmission time A2 is given by MAP in A-Duplex is small. TA2=TRTS +2TsIFs +Tcrs +T2 +2TAck +TDIFS.(T) B.RTS Frame Structure and SIR Value Since the SIR value is sent in an RTS frame,we make a wheres time of the packet from the APto slight changetothe RTS frame in the TEE acket from APto Clent B (i .we rame determined by the co onding capture rate,which is furthe and I represents negative.The last 7 bits give an abolute determined by the SINR Thus .when the AP stablishes the SIR value.Since 64 dB SIR is enough for a lual-link S tup.it is ne els can be (ie repre 127 Eas.(5)and (6)we get C.Setup of Asvmmetric Dual Links Tadd =TAI-TA=Tp-TsIEs+TACk for do re set un The hasie condition to set un such a downlink is Thus.we know that T<TAp can always be satisfied in this that the capture effect at the selected client exists. case.In the case of"AP-longer"in Fig.3(b).from Eqs.(5)and (7)we ge at the must exce capture thre Todd=TA2-TA=T3-TI-TsIEs+TACk. to send packet I to the AP.the AP may establish a dowr nlink and e the s e.then the w sel the TAP AP decrease the throughput.Thus.when AP sets up dual links can set a condition T <to improve the throughpu it needs to make sure dual links can improve the throughput. where B1.Note that B needs to be determined properly: that the AP s the clier such that the ngle link in hal be eas. [1 This the has up cann of T can be viewed in another way.The packet sent from the AP to protocol takes the following procedure to select a client for s to the ost of than the time s.select the clien T.8

5876 IEEE TRANSACTIONS ON WIRELESS COMMUNICATIONS, VOL. 14, NO. 10, OCTOBER 2015 Fig. 5. RTS frame structure. client. However, if θ is too small, the SIR cannot reflect the change of the channel in time. Thus, we need to find a suitable θ for a specific environment, as discussed in performance evaluation. From the establishment procedure of SIR MAP, we know that the SIR stored in the MAP is not the real time SIR. Thus, when the AP uses this SIR information to establish dual links, the capture link may fail. From the simulation results in Section VII-B2, we know that the failure rate is less than 20%, which reveals that the SIR is not perfect but is effective to capture dual-link opportunities. When capture effect fails in dual links, the receiving client cannot successfully receive a packet from the AP. Thus, the AP will retransmit the packet later, which is acceptable. However, if we try to achieve 100% success rate for the capture link, the AP needs to collect the real time SIR of all clients for all dual links, which will lead to very high overhead. In contrast, the overhead from setting up the SIR MAP in A-Duplex is small. B. RTS Frame Structure and SIR Values Since the SIR value is sent in an RTS frame, we make a slight change to the RTS frame in the IEEE 802.11 standard. As shown in Fig. 5, we add one byte in the RTS frame. The first bit is used to indicate the sign of the SIR, i.e., 0 represents positive and 1 represents negative. The last 7 bits give an absolute SIR value. Since 64 dB SIR is enough for a practical system and considering a granularity of 0.5 dB, all SIR levels can be represented by 7 bits (i.e., from level 0 till level 127). C. Setup of Asymmetric Dual Links When the AP receives an RTS frame from a client, it needs to select a client for downlink transmission so that dual links are set up. The basic condition to set up such a downlink is that the capture effect at the selected client exists, i.e., the SIR at the selected client must exceeds a capture threshold. However, this basic condition is insufficient to ensure throughput improvement. For example, when Client A uses the highest rate to send packet 1 to the AP, the AP may establish a downlink and send packet 2 to Client B with the lowest rate. If the payloads of the two packets are the same, then the time needed by packet 2 is much longer than packet 1. As a result, the dual links will decrease the throughput. Thus, when AP sets up dual links, it needs to make sure dual links can improve the throughput. One solution is that the AP chooses the client such that the overall transmission rate of dual links is higher than the case of a single link in half duplex communications [17]. This problem can be viewed in another way. The packet sent from the AP to Client B under dual-link setup leads to the cost of additional transmission time. If the additional time is less than the time needed for sending the packet in the half duplex link from the AP to Client B, then the throughput can be improved. When the AP transmits a packet to a client, the transmission time TAP is given by TAP = Tdata + TSIFS + TACK + TDIFS, (4) where TDIFS is the DIFS time and Tdata is the data frame time. When a client gets the channel and AP does not establish dual links as shown in Fig. 4, the transmission time TA is given by TA = TRTS + 3TSIFS + TCTS + T1 + TACK + TDIFS, (5) where T1 is the time of data packet to the AP. However, when the AP establishes dual links in the “AP-shorter” scenario in Fig. 3(a), the transmission time TA1 is given by TA1 = TRTS +2TSIFS + TCTS +Tp + T1 +2TACK +TDIFS, (6) where Tp is the preamble time. If the AP establishes dual links in the “AP-longer” scenario in Fig. 3(b), the transmission time TA2 is given by TA2 =TRTS + 2TSIFS + TCTS + T2 + 2TACK + TDIFS, (7) where T2 is the transmission time of the packet from the AP to Client B. Note that, in both “AP-shoter” and “AP-longer”, the transmission time of the packet from AP to Client B (i.e., T2) is determined by the corresponding capture rate, which is further determined by the SINR. Thus, when the AP establishes the downlink for dual-link setup, it is necessary that the additional time Tadd (i.e., Tadd = TA1 − TA or TA2 − TA) is less than TAP. Considering the case of “AP-shoter” in Fig. 3(a), from Eqs. (5) and (6) we get Tadd = TA1 − TA = Tp − TSIFS + TACK. Thus, we know that Tadd < TAP can always be satisfied in this case. In the case of “AP-longer” in Fig. 3(b), from Eqs. (5) and (7) we get Tadd = TA2 − TA = T2 − T1 − TSIFS + TACK. Thus, Tadd < TAP is not always guaranteed. To ensure throughput improvement, the AP needs to make sure Tadd < TAP is satisfied when selecting a downlink for dual-link setup. The AP can set a condition Tadd ≤ TAP/β to improve the throughput, where β ≥ 1. Note that β needs to be determined properly; a larger β potentially increases throughput, but the condition of dual-link setup cannot be easily satisfied. To satisfy both the basic condition and the condition of Tadd ≤ TAP/β, our protocol takes the following procedure to select a client for downlink transmissions: 1) determine a set of clients that satisfy the capture threshold; 2) from this set of clients, select the client that meets the requirement of Tadd ≤ TAP/β

TANG AND WANG:MAC FOR EFFICIENT COEXISTENCE BETWEEN FULL AND HALF-DUPLEX COMMUNICATIONS 587 A called deficit.The value of deficit for Client k at round n. denoted by T(n).is updated as (AP) T(n)quantum+Tx (n -1)I(n-1). (8) (B the packet in the p empty,and quantum is a constant number.If the deficit() Fig.6.An example of the faimess issue. chent k.ther the AP turns (n)is en a packet is sent.the transmission time is deducted from( This process is repeated until the deficit T)is insufficient for When selecting a client for the downlink of dual-link setup. sending anoth the AP pr nts tha thave more hidden nodesre clos to the AP which example,when the AP finishes its transmission to Client A.it gives thedo D.which Ap dual-lin ink ch link setup,as selecting Client B does not satisfy the condtio of capture effect.Thus,the deficit round robin D gets the channel,only half duplex link is allowed,i.e..the nt to s algorith ing pro leads a)When the AP contends the channel succ ssfully it follows the deficit round robin scheme to send a packet.After On the other hand,for each chance the throughput improve each transmission,it needs to find and record the nex ding to the For am when B ge Suppose the ent on if the inks Howe r.since Client D is very close to the AP and has ugh to packet.stl recorded s the higher SIR than Client C.more thr be a eved if the AP choc nt D The sa a pa As resu Client D is the best choice to get the chance each time,which makes the fairness worse.Thus,there exists a tradeoff between repeated until a client with enough deficit is found and is the AP in differe needs to select hy conside In this paper,it is evaluated on the channel access time 7..e. the channel access time of a downlink is expected to be equal effect.In this case,it does not follow the order of deficit e fairness,tw s the internal queueing system gives a higher priority to clients that the capture condition.After the downlink transmission ave a lower ch the deficit is deducted by the time that is needed when and B in ha conducted when to clients that have a lower chance of being selected for dual h link setup.To fulfill the above two mechanisms,the AP needs will give more opportunities to other client o look i o the dual-link s client and the ncluding etup c nces for eac e chances. tho arucipate in asymn an appropriate client for downlink transmission. a).In this situation.the AP needs to find a new next client in the similar way as lone in case a). 18 eve fa It should he noted that the Ap in notaclient can get a chance is determined by a time parameter if the deficit is enough to transmit the packet,so the deficit

TANG AND WANG: MAC FOR EFFICIENT COEXISTENCE BETWEEN FULL- AND HALF-DUPLEX COMMUNICATIONS 5877 Fig. 6. An example of the fairness issue. V. FAIRNESS: THE VIRTUAL DEFICIT ROUND ROBIN ALGORITHM When selecting a client for the downlink of dual-link setup, the AP prefers clients that have more hidden nodes or are closer to the AP, which favors throughput improvement. However, a fairness issue arises. An example is shown in Fig. 6, where there are 4 clients in the network. When Clients A or B gets the channel (for uplink transmission), the AP can establish dual links with Client C or D, which is called a dual-link chance. When Client C gets the channel, the AP can establish the downlink with any of three other nodes. However, when Client D gets the channel, only half duplex link is allowed, i.e., the AP cannot select any other client to set up the downlink for dual-link setup. As a result, Client D has more chances of receiving packets from the AP, which leads to unfairness in downlink. On the other hand, for each chance the throughput improvement is also different. For example, when Clients A or B gets the channel, the AP can select either Client C or D for dual links. However, since Client D is very close to the AP and has higher SIR than Client C, more throughput improvement can be achieved if the AP choose Client D for dual links. The same situation can occur when Client C gets the channel. As a result, if the AP wants to achieve higher throughput improvement, Client D is the best choice to get the chance each time, which makes the fairness worse. Thus, there exists a tradeoff between fairness and throughput. Fairness in channel access can be defined in different ways. In this paper, it is evaluated on the channel access time [7], i.e., the channel access time of a downlink is expected to be equal for all clients. In order to improve fairness, two mechanisms are considered. First, when the AP accesses a channel for data transmission, its internal queueing system gives a higher priority to clients that have a lower chance of being selected for dual-link setup, e.g., Clients A and B in Fig. 6. Second, when the AP establishes a downlink during dual-link setup, it also gives a higher priority to clients that have a lower chance of being selected for duallink setup. To fulfill the above two mechanisms, the AP needs to look into the dual-link setup chances for each client and then balance the utilization of these chances. To this end, a virtual deficit round robin algorithm is developed for the AP to select an appropriate client for downlink transmission. For a network with variable packet transmission time, the deficit round robin algorithm [18] can achieve fairness with low complexity. More specifically, during round robin, whether or not a client can get a chance is determined by a time parameter called deficit. The value of deficit for Client k at round n, denoted by Tk(n), is updated as Tk(n) = quantum + Tk(n − 1) I(n − 1), (8) where I(n − 1) is 1 if the deficit is not enough to transmit the packet in the previous round or 0 if the queue becomes empty, and quantum is a constant number. If the deficit Tk(n) is insufficient to send a packet from the AP to Client k, then the AP turns to Client k + 1. If the deficit Tk(n) is enough to transmit a packet, then Client k can be selected in round n. After a packet is sent, the transmission time is deducted from Tk(n). This process is repeated until the deficit Tk(n) is insufficient for sending another packet. However, due to capture effect, the above algorithm is not applicable to client selection for downlink transmission. For example, when the AP finishes its transmission to Client A, it gives the downlink chance to Client B according to round robin. However, if Client A gets the channel successfully, then the AP can only give the downlink chance to Client C or D for duallink setup, as selecting Client B does not satisfy the condition of capture effect. Thus, the deficit round robin scheme is not working in this case. To this end, a virtual deficit round robin algorithm is developed. It works with the following procedures: a) When the AP contends the channel successfully, it follows the deficit round robin scheme to send a packet. After each transmission, it needs to find and record the next client according to the deficit. Suppose the Client k is the recorded next client in deficit round robin. After this packet transmission, if the remaining deficit is enough to send another packet, Client k is still recorded as the next client. Otherwise, the deficit round robin goes to Client k + 1. If the deficit of Client k + 1 is enough for a packet, Client k + 1 is recorded as the next client. If the deficit is insufficient, Client k + 2 is considered. This process is repeated until a client with enough deficit is found and is recorded as the next client. b) If a client wins the channel access, the AP needs to select a client for downlink transmission by considering capture effect. In this case, it does not follow the order of deficit round robin, but it still relies on the deficit to select a client. More specifically, the AP selects the client that has the maximum deficit among all clients that satisfy the capture condition. After the downlink transmission, the deficit is deducted by the time that is needed when half-duplex communication is conducted. The reason for deducting such a time is to make sure the deficit is updated consistently with case a). As a result, winning a downlink transmission in full duplex communications by a client will give more opportunities to other clients including those that cannot participate in asymmetrical dual links. It is possible that the client selected by the AP is the one recorded as the next client in case a). In this situation, the AP needs to find a new next client in the similar way as done in case a). It should be noted that the AP in case b) does not check if the deficit is enough to transmit the packet, so the deficit

5s78 LEEE TRANSACTIONS ON WIRELESS COMMUNICATIONS,VOL 14.NO.10.OCTOBER 2015 can become negative after it is deducted by the downlink as the AP transmits successfully when all clients keep silent transmiss Following [9.the throughput S is defined as the fraction of sending payload bits.It can be be negative.When the AP in case a)finds a client has ne ative deficit.it just skips this client to the next one.Actually we can (P+P.+PPEUL) negative value of the deficit.In othe (13 ca where the parameters are defined as follows.Pr is the prob- case b).This limit improves fairess,but results in loss of dual ability of transmission from at least one node.PA and Peare Ias the proba ion at the Al e for orobability and Pis the collision and T several reasons:D Although the deficit is updated in the same are the average time of successful transmission for the AP and way as that in the deficit round robin algorithm,the deficit in case bin way:2)The deficit in case not strictly followed (due to dual links via The above quation reflects that the throughput is contributed effect).perfect faimness like deficit round robin cannot always by three components,i.e.,data transmission from the AP to be ensured. a VI PERFORMANCE ANALYSIS re effect.but it d from RTS. CTS.ACK.etc.Such overhead is captured by Ta and T as In this follows nt ITs=E(TAP)=E(Tdaal TsIEs TAck TDIES T2=E(TA]=TRTS+3TsIFs+Tcrs+E(+TAcK+TDIrs el.The nalysis can be adopted her (14) Since there are N clients and one AP.P is given by A-Duplex Por=1-(1-P)(1-P )W (15) s of the Ap and a p n be e for are different.We set the minimum contention window Wo and maximum 2Wo for the AP and the minimum contention for clients.F llowing simila PA Po(1-P) (16) can get th fully when 1+W+pW∑m(2p (9) active.Thus,Pis given by: In Pe=NP:(I-Pr)(N-D) 17) one of the clients gets the channel no matter whether the AP The collision probability is given by: gets the channel or not.Thus,p is given by P=P-P-P. (18) p=1-1-P)N- (10) where n is the nur agation mode for the aP in a slot as with deterministic power and Rayleigh fading.the (11)instantaneous powerPis exponentially distributed as: 1+W+pWo∑(2po e的y he AP and D)=元元，P>0, (19 m=1-(1-P,)

5878 IEEE TRANSACTIONS ON WIRELESS COMMUNICATIONS, VOL. 14, NO. 10, OCTOBER 2015 can become negative after it is deducted by the downlink transmission time. Even after it is updated according to Eq. (8), it may still be negative. As a result, even if the procedure in case a) is similar to the deficit round robin algorithm, the deficit can be negative. When the AP in case a) finds a client has negative deficit, it just skips this client to the next one. Actually we can set a limit to the negative value of the deficit. In other words, if the deficit of a client is not enough for a packet transmission (by considering the negative limit), the client cannot be selected in case b). This limit improves fairness, but results in loss of dual link chances. Thus, fine-tuning such a limit provides a tradeoff between fairness and throughput. We call our scheme a virtual deficit round robin scheme for several reasons: 1) Although the deficit is updated in the same way as that in the deficit round robin algorithm, the deficit in case b) is not used in a round robin way; 2) The deficit in cases a) and b) can be negative; 3) Since the round robin order is not strictly followed (due to exploring dual links via capture effect), perfect fairness like deficit round robin cannot always be ensured. VI. PERFORMANCE ANALYSIS In this section we analyze saturation throughput of A-Duplex to study its throughput improvement over 802.11 DCF. “Saturation” means the AP and clients always have packets for transmission. In [19], the performance of 802.11 DCF is analyzed via a Markov model. The same analysis can be adopted here, but two major differences in A-Duplex need to be considered. First, when a client and the AP collide, the client can still get the channel successfully due to full duplex capability at the AP. Second, capture effect exists in A-Duplex. In a practical system, the Markov chains of the AP and clients can be different since the contention windows for them are different. We set the minimum contention window W0 and maximum 2m0W0 for the AP and the minimum contention window W and maximum 2mW for clients. Following similar Markov chain analysis in [19], we can get the transmission probability Pt for a client in a slot as Pt = 2 1 + W + pW m−1 i=0 (2p)i , (9) where p is the conditional collision probability for a client. In A-Duplex, a successful transmission from a client occurs when one of the clients gets the channel no matter whether the AP gets the channel or not. Thus, p is given by p = 1 − (1 − Pt) N−1, (10) where N is the number of clients. In the same way, we can get the transmission probability Pt0 for the AP in a slot as Pt0 = 2 1 + W0 + p0W0 m0−1 i=0 (2p0)i , (11) where p0 is the conditional collision probability for the AP and is given by p0 = 1 − (1 − Pt) N, (12) as the AP transmits successfully when all clients keep silent. Following [19], the throughput S is defined as the fraction of time for successfully sending payload bits. It can be calculated as follows S = (PA + Pc + PcPca)E{Lp} (1 − Ptr)Tslot + PATs1 + PcTs2 + PcPcaTadd + PcolTc , (13) where the parameters are defined as follows. Ptr is the probability of transmission from at least one node. PA and Pc are defined as the probability of successful transmission at the AP and a client, respectively. Pca is the average conditional capture probability, and Pcol is the collision probability. Ts1 and Ts2 are the average time of successful transmission for the AP and for a client, respectively. Tadd is the additional time for capture effect, as defined in Section IV-C. Moreover, Tc is the collision time and E{Lp} is the average payload size in a data frame. The above equation reflects that the throughput is contributed by three components, i.e., data transmission from the AP to clients without utilizing capture effect, data transmission from clients to the AP, and data transmission from the AP to clients with capture effect, but it must exclude overhead from RTS, CTS, ACK, etc. Such overhead is captured by Ts1 and Ts2 as follows ⎧ ⎪⎨ ⎪⎩ Ts1 = E{TAP} = E{Tdata} + TSIFS + TACK + TDIFS Ts2=E{TA}=TRTS+3TSIFS+TCTS+E{Tdata}+TACK+TDIFS Tc = TRTS + TDIFS. (14) Since there are N clients and one AP, Ptr is given by: Ptr = 1 − (1 − Pt0)(1 − Pt) N. (15) When all clients keep silent and only the AP sends packet, the AP sends packet successfully. Thus, PA is given by: PA = Pt0(1 − Pt) N. (16) The client transmits packet successfully when only one of the clients transmits no matter whether the AP keeps silent or active. Thus, Pc is given by: Pc = NPt(1 − Pt) (N−1) . (17) The collision probability is given by: Pcol = Ptr − PA − Pc. (18) The average conditional capture probability Pca is associated with the capture threshold, i.e., Pca decreases as the capture threshold increases. Considering a propagation model with deterministic power attenuation and Rayleigh fading, the instantaneous power P is exponentially distributed as: fp(P) = 1 P0 e − P P0 ,P > 0, (19) where P0 is the received local-mean power and is determined by P0 = Ar−nPtx, where Ar−n is the deterministic path-loss

TANG AND WANG: MAC FOR EFFICIENT COEXISTENCE BETWEEN FULL- AND HALF-DUPLEX COMMUNICATIONS 5879 law [20], n is the path loss index (usually with a range 2 − 4), Ptx is the transmit power at the transmitter, and r is the distance between the transmitter and the receiver. We assume A, n, and Ptx are the same for all nodes. To guarantee success of capture effect, the SIR needs to exceed the capture threshold z. In our design, the SIR (γ = Pu/Pi) at a client u is the ratio of signal power from the AP over that from an interfering Client i. The conditional capture probability is given by [21] Pca(γ > z|i) = 1 1 + z(ri/ru)−n , (20) where ru is the distance between the AP and the receiving client u, and ri is the distance between the interfering client i and the receiving client u. Considering the clients are uniformly distributed around the AP, the probability density function (PDF) of ru is given by h(ru) = 2ru, 0 < ru ≤ 1. (21) The PDF of ri is calculated by [22] h(ri) = 1 2 1 B(2, 2.5) ri 2 1 − ri 2 3 2 , 0 < ri ≤ 2, (22) where B(·,·)is the Beta function. The approximate average conditional capture probability can be computed as follows [22]: Pca = 1 0 2 0 h(ri)h(ru) 1 + z ri ru −n dridru. (23) From the above equation we know that Pca is determined by the capture threshold z, and z is determined by the capture link rate. As explained in Section IV-C, setting up a dual link needs to check an additional condition, i.e., Tadd ≤ TAP/β. If the β is set larger, then the threshold z is higher, which means it demands a higher capture link rate. When the link rate and data frame length are variable, it is difficult to derive a closed-form relationship between β and z. However, given a fixed link rate and data frame length, the value of z can be determined from β as follows. First, the allowed additional time is determined, i.e., Tadd = TAP/β, where TAP = TS1 since link rate and data frame length are fixed. Following that, the transmission time of the capture link T2 is determined as T2 = Tadd + T1 + TSIFS − TACK, where T1 is the transmission time for data frame in half duplex link. Since the link rate and the data frame length are fixed, T1 is also a fixed value. With T2 and the data frame length, we can get the transmission rate of downlink. Finally, the capture threshold z is determined by the transmission rate. Substituting Eqs. (14)–(18) and (23) into Eq. (13) and let Tadd = Ts1/β, we can derive the throughput. Note the above derivation assumes the maximum additional time that satis- fies Tadd ≤ TAP/β, so the derived throughput is the minimum throughput that can be achieved by our protocol. Fig. 7. The probability of collision. Fig. 8. The probability of successful transmission. VII. PERFORMANCE EVALUATION In this section the performance of A-Duplex is first evaluated based on a protocol model, and then a physical model is used to evaluate A-Duplex in a more practical setup. A. Performance Results Based on the Protocol Model We set the minimum contention window to 32 for both the AP and clients and the maximum contention window to 128 for the AP and 1024 for clients. The collision probabilities (Eq. (18)) of our protocol and 802.11 DCF are shown in Fig. 7, where the result of 802.11 DCF follows the analysis in [19]. The probability of successful transmission (PA + Pc given by Eqs. (16) and (17)) are shown in Fig. 8. Thus, the collision probability of A-Duplex is lower than 802.11 DCF and the successful transmission probability is higher than 802.11 DCF, which leads to higher throughput in A-Duplex even if the capture effect is not considered as shown in Fig. 9. However, without capture effect, the improvement over 802.11 DCF with RTS/CTS is small. To evaluate the saturation throughput, the system parameters are selected according to Table I. We set the link rate to 18 Mbps. We set β = 2.2. Considering the requirement of Tadd = TAP/2.2, the link rate considering capture effect needs

880 LEEE TRANSACTIONS ON WIRELESS COMMUNICATIONS,VOL 14.NO.10.OCTOBER 2015 Mode Modulation Coding rate (dB)7 (dB) BPSK 12 5-6 1-4 BPSK 4 67 45 QPSK 12 7-9 57 9-13 7-12 16-QAM 12 13-17 12-16 616-QAM 3/417-2016-21 64-QAM 320-2221-23 864-QAM 34≥222223 Number of The path loss in the channel model is given by Fig.9.The throughput results under the protocol model PL(d=PL(do)+l0×n×log(d/do)[dBl, where we set do I m and n =3.and let PL(1)=48 dB to re Parameter Value Parameter flect the indoor environment.Moreover.the minimum required SNK to nt modu PHY header (PH) tion and coding schemes fo 20 PHY P amble RTS(802.1 effect((B))in Table I.where is adopted from PHY ion [23]andy is from experimental results in [12]. Pavload idefault 14Byte +PH me adopted in our simulatio Link rate 18 Mbps 16 three times Capture rate 12 Mbps Wo 16 next lower level.When a client succeeds in 5 consecutive Slot time 9 us m 6 SIFS 16us mo transmissions or the rate is not changed over 10 times.the DIFS 345 capture threshold 5dB rate rises to AP does not establish dual links when the SIR MAP is empty in the beginning can get n Throughput an Fa app in Ea (3).Since it is e for in s valid.as simu ractical s stem.we consider discrete values of=n.(n ts are con y23w res up .2.3. .and then determine a proper value of n.When we ans the new SIR e and 246 e is the latest u and 54%over 802.11 DCF without RTS/CTS for 5 clients and dual links is about 55%.When we set =3.the succe 40clients.respectively. nt can be achieved.As a result.we set n 3 in B.Performance Results Based on the Physical Model In the fr of 1)Plysical Model Setup:The physical model simulation is considered,and 40%clients(sl.s2.s3.s4)have hidden nodes conducted in Matlab.In addition to thr oughput,other perfo The frame length is set to1500 Bytes.In this.8 case s such .packet re sh ind compare own in la The fir parameters are given in Table I except that link rate.capture rat 802.11 DCF without using RTS/CTS.The second case is the (i.e.the link rate under capture effect).and capture threshold same as the first case except that RTS/CTS is enabled.The third und the AP.W and eh fad study the ing.Thus,the channel is a time varying channel.However.we A-Duplex when the parameter B is set to 1.2,and 3.respec tively.The last case is that the APal ways chooses the client with (in dua nin Fig.I

5880 IEEE TRANSACTIONS ON WIRELESS COMMUNICATIONS, VOL. 14, NO. 10, OCTOBER 2015 Fig. 9. The throughput results under the protocol model. TABLE I PARAMETERS USED IN SIMULATIONS to be 12 Mbps. To support such a rate, the capture threshold is 5 dB [12]. From Eq. (23) we can get Pca = 0.4371. Both simulation (marked by symbols) and analytical (represented by line) results of throughput are shown in Fig. 9. From these results, we know that: 1) The analytical model is valid, as simulation results are consistent with analytical results; 2) A-Duplex improves throughput by 23% and 24% over 802.11 DCF with RTS/CTS for 5 clients and 40 clients, respectively, and 24% and 54% over 802.11 DCF without RTS/CTS for 5 clients and 40 clients, respectively. B. Performance Results Based on the Physical Model 1) Physical Model Setup: The physical model simulation is conducted in Matlab. In addition to throughput, other performance metrics such as fairness, packet loss, and packet delay are also considered in performance evaluation. The system parameters are given in Table I except that link rate, capture rate (i.e., the link rate under capture effect), and capture threshold are determined dynamically. The clients are set in a wide area uniformly distributed around the AP. We consider a propagation model with deterministic power attenuation and Rayleigh fading. Thus, the channel is a time varying channel. However, we assume that the channel only changes between packets, i.e., the channel parameters remain the same during the transmission of each packet. TABLE II THE MINIMUM REQUIRED SNR FOR A CLEAN CHANNEL (γ1 (dB)) AND THE MINIMUM SIR FOR CAPTURE EFFECT (γ2 (dB)) The path loss in the channel model is given by PL(d) = PL(d0) + 10 × n × log(d/d0) [dB], where we set d0 = 1 m and n = 3, and let PL(1) = 48 dB to re- flect the indoor environment. Moreover, the minimum required SNR to decode different modulation and coding schemes for a clean channel (γ1 (dB)) and the minimum SIR for capture effect (γ2 (dB)) are given in Table II, where γ1 is adopted from [23] and γ2 is from experimental results in [12]. The rate adaptation scheme adopted in our simulation is ARF [24]. More specifically, when a client misses an ACK three times consecutively, its transmission rate moves to the next lower level. When a client succeeds in 5 consecutive transmissions or the rate is not changed over 10 times, the transmission rate rises to the next higher level. The capture rate is determined according to the SIR MAP table. Note that the AP does not establish dual links when the SIR MAP is empty in the beginning. 2) Saturation Throughput and Fairness: To conduct experiments, we first need to determine an appropriate value for θ in Eq. (3). Since it is hard to find an optimal value for θ in a practical system, we consider discrete values of θ = 1/n, (n = 1, 2, 3 ···) and then determine a proper value of n. When we set n = 1, which means the new SIR value is the latest updated value SIRnew = SIRupdate, the successful transmission ratio of dual links is about 55%. When we set n = 3, the successful ratio increases to 80%. However, when we increase n further, little improvement can be achieved. As a result, we set n = 3 in the following simulations. In the first set of experiments, 10 clients shown in Fig. 10 are considered, and 40% clients (s1, s2, s3, s4) have hidden nodes. The frame length is set to 1500 Bytes. In this scenario, 8 cases are simulated and compared as shown in Table III. The first case is that all clients and the AP transmit packets based on the 802.11 DCF without using RTS/CTS. The second case is the same as the first case except that RTS/CTS is enabled. The third case is our MAC but setting no limit to the additional time Tadd. The fourth case is that the AP only establishes dual links with hidden nodes. The next three cases study the performance of A-Duplex when the parameter β is set to 1, 2, and 3, respectively. The last case is that the AP always chooses the client with the best capture rate to establish dual links. As shown in Fig. 11, if we set no limit to Tadd (i.e., case three), the throughput is

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