Bluetooth Scatternet Routing for Wireless Home Network
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(2) All communications between different piconets are filtered through the individual piconet master devices in a scatternet. If there are 10 piconets, they contain 80 different Bluetooth devices. We could make those devices to be included in a single scatternet. Because Bluetooth only uses 79 frequencies in total, the network saturates beyond this number. However, that multiple piconets share the same physical space. They can continue to operate independently each other without interfering. This is because each piconet operates on a different frequency-hop channel. The result is that the devices in piconet number one never share the same frequency as the devices in piconet number two.. 3. Bluetooth Home Network Architecture Our proposal uses a two level piconet hierarchy. Level 1 piconets are general piconets. They are used for communication among themselves. It is called a personal area network (PAN). Level 2 piconet is used for routing of PANs. It is called a routing area network (RAN). If there is a slave belonging to a PAN and a RAN at the same time, it is called a gateway slave. A gateway slave is not only working in a PAN but also working in a RAN simultaneously. In a PAN, master and gateway slaves need to be fixed. The other devices can be mobile nodes. In a RAN, the master is a router and it has to be fixed as well. An example of the two-level Bluetooth home network architecture is shown in Figure 3-1. PAN #1. Slave #2/2 Slave #1/2. Slave #1/1. Slave #1/3 or 0/1. PAN #2 Slave #2/1 or 0/2. Master #2/x. Master #1/x RAN #0. Master #4/x. 4.1 Protocol Header Format The routing packet header format is shown in Figure 4-1. Layer 3 Header. PAN #3 Slave #4/2 or 0/4. In Figure 3-1, we can find that the PA of RAN is set to 0. Because a gateway has two masters, it has two types of address. When it gets the address from the router, its PA must be zero. Both slave #1/3 and slave #0/1 point to gateway of PAN #1. If source node and destination node are in the same piconet, the gateway will use slave #1/3 as its address. Otherwise slave #0/1 will be used. We will explain these addresses in later sections.. Router #0/x. PAN #4 Slave #4/1. get PA from the gateway. In addition, each slave of a PAN is given a number as its Device Address (DA). Then slaves could get its PA from its master, which gets the value of PA from gateway. PA and DA combined is an address in the Bluetooth Home Network. The address is used for designing a packet header, which is used in the routing protocol. It also indicates a source or destination node in the Bluetooth Home Network. The notation of Slave #m/n is used in Figure 3-1. m is the address of piconet, which is assigned by the RAN master node. The slave could get the value of m from its master and each node has the same value of m in the same piconet. n is the address of device, which is assigned by the PAN master node. Each master will collect the information of its slaves. Then the information will be sent to the router. After the router receives the information from each master, it can make a routing map and send to each master. Thus each slave can get the routing map from its master. Using the routing map, each node can get all the information about PA and DA in the scatternet.. Slave #3/2 or 0/3 Slave #3/3. Master #3/x Slave #3/1. Figure 3-1: Bluetooth Home Network Architecture. 4. Routing for Wireless Home Networks At first, a RAN will be created. Then the master of RAN will assign addresses for its slaves, which will be gateways. The address is the Piconet Address (PA) of the slave when it becomes a gateway. The slave of RAN is in parked state and it will wait for a page from the master of PAN that it belongs to. When the master of PAN pages to the gateway, the master will. M. SW. DPA. DDA. SPA. SDA. 1 Bit. 1 Bit. 3 Bit. 3 Bit. 3 Bit. 3 Bit. Layer 3 Payload. 14 Bit. Figure 4-1: Packet Header Format The first field is M (Master Flag). It indicates whether the destination node is a master in a PAN. If a packet will be transmitted to a master in a PAN, Master Flag of this packet will be set. The second field is SW (Switch Flag). It defines whether the packets need to be switched between PAN and RAN. In addition, the field also indicates whether the packets have reached destination node. If this field is 1, any slave receiving the packet will automatically detect where the packet comes from. If the packet comes from PAN, the slave will send the packet to its master of RAN. Otherwise, the slave will.
(3) send the packet to its PAN. Then if the filed is set to zero, it means that the packet belongs to the receiving slave. The third field is the Destination Piconet Address (DPA). This field indicates which piconet is to be routed. The fourth field is the Destination Device Address (DDA). The field indicates which node is the destination node within destination PAN. Each piconet is defined by DPA in a home network. Then each node is defined by DDA in a PAN. If the value is zero, it means that the packet needs to be broadcasted within destination PAN. The fifth field is the Source Piconet Address (SPA). The field indicates the received device which piconet transmitted the packet. The sixth field is the Source Device Address (SDA). The field indicates which node is the originating node within source PAN. SPA-SDA and DPA-DDA are used to specify a unique source and destination node. Finally the payload is the information from higher layer such as application functions, control information, voice field, data field, etc.. Start Set SW=1 Master will forward the packet to its slave node defined by the value of DPA.. ≠ LPA. DPA = LPA. =1. M. Master will treat the packet as its own.. =0. Set SW=0 Master will forward the packet to its slave node defined by the value of DDA. If the DDA node doesn't exist, the packet will be given up.. ≠0. DDA =0. Set SW=0 This is a broadcast, so master will send the packet to each slave and change the DPA of header for each packet.. Figure 4-3: Routing protocol for master of RAN. Start. =1. SW. Slave will relay the packet to the master of the other piconet.. =0. 4.2 Protocol Operation We support both intra-piconet and inter-piconet communications. In addition, unicast and broadcast communications are also supported. If a destination device address (DDA) is zero, it means that the packet is a broadcast. Otherwise, the packet is used for unicast communication. In addition, we add a variable called Local Piconet Address (LPA). If a node is located within PAN #1, its LPA will be set to 1. Figure 4-2 shows the routing protocol for the master of PAN. Figure 4-3 shows the routing protocol for the master of RAN. The routing protocol for slaves is shown in Figure 4-4.. End. Slave will treat packets as its own. Figure 4-4: Routing protocol for slave 4.2.1 Intra-piconet Communication In this section, we present an example for Intra-piconet communication. Figure 4-5 shows the example. Table 4-1 shows the packet header of Intra-piconet communication in each hop. The symbol X means “not applicable”. Slave #2/2. PAN #1. Slave #1/2. PAN #2. Slave #1/1. Start. Slave #1/3 or 0/1. DPA. End. ≠ LPA. Master #2/x. Slave #2/1 or 0/2. Master #1/x. Set SW=1 Master will send the packet to the gateway in its piconet.. Router #0/x. RAN #0. = LPA. PAN #4. M. =1. Master will treat the packet as its own.. PAN #3 Slave #4/2 or 0/4. Slave #4/1. Master #3/x. Slave #3/2 or 0/3. =0. DDA. ≠. 0. =0 Set SW=0 This is a broadcast, so master will send the packet to each slave in the same piconet.. Set SW=0 Master will forward the packet to its slave node defined by the value of DDA. If the DDA node doesn't exist, the packet will be given up.. End. Figure 4-2: Routing protocol for master of PAN. Master #4/x. Slave #3/3. Slave #3/1. Figure 4-5: An example of Intra-piconet communication Table 4-1: The packet header of Intra-piconet communication Path. M. SW. DPA. DDA. SPA. SDA. Slave #1/1 to X care Master #1 /don’t. 0. X. 1. 3. 1. 1. X Master #1 /don’t care to Slave#1/3. 0. 0. 1. 3. 1. 1.
(4) For this example, assume that slave #1/1 wants to send a packet to the gateway of PAN #1. The source node will use slave #1/3 as destination address because this is an Intra-piconet communication. At first, slave #1/1, the source node, creates a packet header shown in the first row of Table 4-1. Because the source node is slave #1 in PAN #1, both SPA and SDA are set to one. DPA is the same as SPA because the source and destination node are in the same piconet. DDA, which is 3, tells the master about the identity of the destination node. It is slave-slave communication, so M is set to zero. The source node sends the packet to its master at first. When the master receives the packet, it will check DPA with its LPA. If the values of DPA and LPA are the same, the source node and destination node are in the same piconet. Thus SW flag of the packet is set to zero by master. Then the master checks M flag of the packet. In this case, the flag is zero, so the master continues to check DDA field of the packet. Because the field is not zero, master will forward the packet to a slave node defined by the DDA. After the slave receives the packet, it will check SW flag. If it is zero, the slave will receive the packets. Otherwise the slave will switch the packet to another network. At last, the packet will be received by slave #1/3. 4.2.2 Inter-piconet Communication In this section, we present two examples for Inter-piconet communication. The first example demonstrates how to send a packet to the router. Figure 4-6 shows the route. Table 4-2 shows the packet header in each hop. Slave #2/2. PAN #1. PAN #2. Slave #1/2. Slave #1/1 Slave #1/3 or 0/1. Master #2/x. Slave #2/1 or 0/2. In this example, the slave #1 of PAN #1 wants to transmit the packet to the master of PAN #0, the router. The operation of the protocol along each hop is listed in Table 4-2. The destination node is a master node so M is set to 1. Because the destination node is in PAN #0, DPA is set to zero. The value of DPA is irrelevant because the destination node is a master. SDA and SPA are set to 1 as they indicate the source node and the originating piconet. The source node sends the packet to its master first. When the master receives the packet, it will check DPA with its LPA. Because DPA is not equal to LPA in this case, the destination node is in other piconet. Thus the master sets SW flag of the packet to 1, and forwards the packet to its gateway. The gateway slave checks the SW flag. Because the value of SW flag is 1, a gateway node will relay the packet to the master of other piconet. When the master receives the packet, it will check DPA with its LPA. DPA is equal to LPA in this case so the destination node is in the same piconet. Then the master checks M flag of the packet. Because the flag had been set to 1 by the source node, the master of RAN will receive the packet. 4.2.3 Broadcasts Broadcast is a basic and important function. If we want to send packets to every node without using broadcast, the load in the networks will be very heavy. Also the network will have a bottleneck near a master. Our proposal not only provides a broadcast function in a piconet, which is also provided in [2], but it also has scatternet broadcast capability. Figure 4-7 shows the route when slave # 1/1 wants to broadcast in a piconet. Table 4-3 shows that packet header for each hop. Slave #1/2. Master #1/x. Slave #2/2. PAN #1. PAN #2. Slave #1/1 Router #0/x. RAN #0. Slave #1/3 or 0/1. PAN #4. PAN #3 Slave #4/2 or 0/4. Slave #4/1. Master #2/x. Master #3/x. Slave #3/2 or 0/3. RAN #0 Master #4/x. Slave #2/1 or 0/2. Master #1/x. Slave #3/3. Slave #3/1. Router #0/x. PAN #4. Figure 4-6: An Inter-piconet communication for router Table 4-2: The packet header of Intra-piconet communication for router Path. M. SW. DPA. DDA. SPA. SDA. Slave #1/1 to X Master #1/don’t care. 1. X. 0. X. 1. 1. X Master #1/don’t care to Slave#1/3. 1. 1. 0. X. 1. 1. Slave#1/3 to X care Router #0/don’t. 1. 1. 0. X. 1. 1. PAN #3 Slave #4/2 or 0/4. Slave #4/1. Master #4/x. Slave #3/2 or 0/3. Slave #3/3. Master #3/x. Slave #3/1. Figure 4-7: A broadcast in a piconet.
(5) Table 4-3: Packet header for a broadcast in a piconet Path. M. SW. DPA. DDA. SPA. SDA. Slave #1/1 to X care Master #1/don’t. 0. X. 1. 0. 1. 1. X Master #1/don’t care to Slave#1/1. 0. 0. 1. 0. 1. 1. X Master #1/don’t care to Slave#1/2. 0. 0. 1. 0. 1. 1. X Master #1/don’t care to Slave#1/3. 0. 0. 1. 0. 1. 1. Comparing to [2], our proposal adds scatternet broadcast function to improve the performance. In our proposal, a source node just sends a scatternet broadcast packet to router. Then the router will copy the packet and send to each master within the scatternet. Figure 4-8 shows the route for a scatternet broadcast. Table 4-4 lists the header of the packet for each hop. Slave #1/2. Slave #2/2. PAN #1. 5. Analysis and Results At first, we evaluate the Ideal Throughput of our proposal. Let the scatternet contain P+1 piconets (P ≥ 1) and each piconet contains S slaves (including Both non-gateway slaves and gateways). So the total number of devices in the Bluetooth scatternet is S ⋅ P + P + 1 . We shall derive the average number of hops distance that a data packet needs to travel before reaching its destination. This depends on the role of the source, which can be a slave, a master, or a router. Figure 5-1 shows each kind of nodes in our Bluetooth scatternet. Slave. Slave Master. Master #2/x. Master. PAN #3 Slave #4/2 or 0/4. Master #3/x. Slave #3/2 or 0/3. Master #4/x. Slave #3/3. Slave #3/1. Figure 4-8: Our proposal to implement a scatternet broadcast Table 4-6: the header of the packet for each hop. P A N #1. P A N #2. P A N #3. P A N #4. Slave. Figure 5-1: Each kind of nodes in our Bluetooth scatternet (S=3, P=4). Router #0/x. PAN #4 Slave #4/1. Master. Slave Slave. Master #1/x. RAN #0. Master. Slave (gateway) Slave (gateway). Slave. Slave #2/1 or 0/2. Slave (gateway). Slave (gateway). Router. PAN #2. Slave #1/1 Slave #1/3 or 0/1. Slave. Slave. P ath. M. SW. DPA. DDA. SPA. SDA. S lave #1 /1 to M aster #1 /x. 0. X. 0. 0. 1. 1. M aster #1 /x to Slave# 1/3. 0. 1. 0. 0. 1. 1. S lave# 1/3 to R ou ter # 0 /x. 0. 1. 0. 0. 1. 1. R ou ter # 0 /x to Slave# 0/1. 0. 1. 1. 0. 1. 1. S lave #0 /1 to M aster #1 /x. 0. 1. 1. 0. 1. 1. M aster #1 /x to Slave# 1/1. 0. 0. 1. 0. 1. 1. M aster #1 /x to Slave# 1/2. 0. 0. 1. 0. 1. 1. M aster #1 /x to Slave# 1/3. 0. 0. 1. 0. 1. 1. R ou ter # 0 /x to Slave# 0/2. 0. 1. 2. 0. 1. 1. S lave #0 /2 to M aster #2 /x. 0. 1. 2. 0. 1. 1. M aster #2 /x to Slave# 2/1. 0. 0. 2. 0. 1. 1. M aster #2 /x to Slave# 2/2. 0. 0. 2. 0. 1. 1. R ou ter # 0 /x to Slave# 0/3. 0. 1. 3. 0. 1. 1. S lave #0 /3 to M aster #3 /x. 0. 1. 3. 0. 1. 1. M aster #3 /x to Slave# 3/1. 0. 0. 3. 0. 1. 1. M aster #3 /x to Slave# 3/2. 0. 0. 3. 0. 1. 1. M aster #3 /x to Slave# 3/3. 0. 0. 3. 0. 1. 1. R ou ter # 0 /x to Slave # 0/4. 0. 1. 4. 0. 1. 1. S la ve #0 /4 to M a ster #4 /x. 0. 1. 4. 0. 1. 1. M aster #4 /x to Slave# 4/1. 0. 0. 4. 0. 1. 1. M aster #4 /x to Slave# 4/2. 0. 0. 4. 0. 1. 1. Table 5-1 summarizes the case when the source is a slave. In this table, destinations are classified into intra-piconet slave (including its gateway), intra-piconet master, inter-piconet slave (non-Gateway), inter-piconet master, router, or gateway. Table 5-1: Source node is a slave. Table 5-2 summarizes the case when the source is a master. In this table, destinations are classified into intra-piconet slave (including its gateway), inter-piconet slave (non-Gateway), inter-piconet master, router, or gateway. Table 5-2: Source node is a master. Table 5-3 summarizes the case when the source is the router. In this table, destinations are classified into inter-piconet slave (non-Gateway), inter-piconet master, or gateway..
(6) Table 5-3: Source node is a Router. Based on the above-mentioned tables, we derive the average distances d s , d m , or d r for packets originated at a slave, a master, or a router respectively. It is shown in the following. c s1 ⋅ d s1 + c s 2 ⋅ d s 2 + c s 3 ⋅ d s 3 + c s 4 ⋅ d s 4 + c s 5 ⋅ d s 5 + c s 6 ⋅ d s 6 c s1 + c s 2 + c s 3 + c s 4 + c s 5 + c s 6. ds =. =. 6 SP − 4 S + 3 P − 1 SP + P. dm =. dr =. (1). cm1 ⋅ d m1 + cm 2 ⋅ d m 2 + cm3 ⋅ d m3 + cm 4 ⋅ d m 4 + cm 5 ⋅ d m5. =. cm1 + cm 2 + cm3 + cm 4 + cm5. 5SP − 4S + 2P SP + P. c r1 ⋅ d r1 + c r 2 ⋅ d r 2 + cr 3 ⋅ d r 3 3SP = cr 1 + c r 2 + c r 3 SP + P. (2) (3). a single piconet under an interference-free environment is T. Extending to a scatternet, different piconets that choose the same FH (Frequency Hopping) channel in the same time slot result in a collision. If we consider the worst case, each piconet is in the radio range of each other. Give that 79 frequencies are available in Bluetooth, the probability that a time slot of a piconet suffers no interference can be approximated by PermP79+1 / 79 P +1 , where P is the number of piconets connected to router. The available network bandwidth is T ⋅ ( P + 1) ⋅ Pr emP79+1 / 79 P +1 . Dividing this by the average traveling distance Davg , we obtain the maximum achievable throughput Tmax as [1]. T ⋅ ( P + 1) ⋅. Because there are S ⋅ P slaves, P masters, and one router, the average traveling distance can be derived as follow. S ⋅ P ⋅ d s + P ⋅ d m + 1 ⋅ d r 6S 2 P − 4S 2 + 8SP − 2S + 2 P = SP + P + 1 S 2 P + 2SP + S + P + 1. Davg =. For 1 ≤ S ≤ 7 and 1 ≤ P ≤ 7 , the value of the average traveling distance Davg is shown in Table 5-4. Figure 5-2 is Table 5-4 in graphical form. Table 5-4: Average traveling distance for different P’s and S’s S. P 1 2. 1. 2. 3. 4. 5. 6. 7. 1.666667. 1.833333. 1.9. 1.933333. 1.952381. 1.964286. 1.972222. 2.6. 3. 3. 4. 3.222222. 5. 3.363636. 3.047619. 3.277778. 3.533333. 3.807692. 3.794872. 4.088235. 3.958333. 4.261905. 3.418182. 3.512821. 3.975. 4.087719. 4.266667. 4.386667. 4.446154. 4.569892. 3.580952 4.168831 4.472906 4.65873. 3.632353 4.23. Tmax =. PermP79+1 79 P +1. Davg. We use the Bluetooth nominal bandwidth T=1 Mbps and assume 1 ≤ S ≤ 7 and 1 ≤ P ≤ 7 . The Ideal Throughput Tmax is shown in Table 5-5. Figure 5-3 is Table 5-5 in graphical form. Table 5-5: Ideal throughput Tmax by varying P and S (Kbps) S. 3. 4. 5. 6. 7. 1. 1184.81. 1077.1. 1039.307. 1021.388. 1011.423. 1005.293. 1001.248. 2. 1110.399. 947.3091. 880.7911. 844.612. 821.8573. 806.2205. 794.8119. 3. 1234.401. 1048.077. 972.5586. 931.6236. 905.934. 888.3075. 875.4619. 4. 1363.849. 1158.043. 1074.944. 1029.99. 1001.814. 982.4988. 968.4316. 5. 1468.584. 1247.945. 1159.055. 1111.023. 1080.94. 1060.328. 1045.321. 6. 1538.442. 1308.39. 1215.839. 1165.866. 1134.581. 1113.151. 1097.554. 7. 1569.867. 1336.106. 1242.155. 1191.451. 1159.717. 1137.985. 1122.17. P. 1. 2. 4.537879 4.72561. 6. 3.461538. 4.070175. 4.38. 4.567742. 4.693694. 4.784053. 4.852041. 7. 3.533333. 4.151515. 4.465517. 4.655556. 4.782946. 4.874286. 4.942982. 1800 6. average number of hops. 5 4. S=1 S=2. 3. S=3 S=4. 2. S=5 1. S=6 S=7. Ideal throughput (Kbps). 1600 1400 S=1. 1200. S=2. 1000. S=3. 800. S=4. 600. S=5. 400. S=6. 200. S=7. 0 1. 0 1. 2. 3. 4. 5. the number of piconets. 6. 7. Figure 5-2: Average traveling distance for different P’s and S’s Next we consider interferences between piconets. Assume that the maximum throughput of. 2. 3. 4. 5. The number of piconets. 6. 7. Figure 5-3: Ideal throughput Tmax for different P’s and S’s We compare the performance of our proposal with that of BlueRing [1]. The ideal throughput.
(7) Table 5-6: Broadcast distance for packets originated at a slave, a master, or a router.. of our proposal is better than BlueRing. Figure 5-4 shows the calculation results. S=5, 6, or 7 Ideal Throughput (Kbps). 1200 1000 800 600. Our proposal. 400. BlueRing. Based on Table 5-6, we derive the average distances Bavg , as follows:. 200 B avg =. 0 Node=25, S=5. Node=36, S=6. cb1 ⋅ d b1 + c b 2 ⋅ d b 2 + cb 3 ⋅ d b 3 S 2 ⋅ P 2 + 2 P 2 S + 4 SP + 4 P + 2 P 2 + SP 2 = c b1 + c b 2 + c b 3 P ⋅ S + P +1. Node=49, S=7. S=2. Figure 5-4: Our proposal compares with BlueRing for the ideal throughput. Number of hops. We compare the performance of our proposal with that of [2]. The average traveling distance of our proposal is better than [2]. Figure 5-5 shows the calculation results.. 30 20 10 0 1. 2. 3. 5. S=7. 4 3. Our proposal 2 1 0 7. (a) Each piconet has 2 slaves. Number of hops. In [2]. 2 3 4 5 6 Number of piconets. 4. 5. 6. Number of piconets. 7. (a) Each piconet has 2 slaves. S=2. 1. In [2]. 40. 6. Number of hops. Our proposal. 50. Our proposal. In [2]. 90 80 70 60 50 40 30 20 10 0 1. 2. 3. 4. 5. 6. 7. Number of piconets S=7. 6. (b) Each piconet has 7 slaves. Number of hops. 5 4 3. In [2]. 2. Our proposal. Figure 5-6: Our proposal compares with [2] about the number of hops in a broadcast. 6. Conclusions. 1 0 1. 2. 3 4 5 6 Number of piconets. 7. (b) Each piconet has 7 slaves Figure 5-5: Our proposal compares with [2] about the number of hops in a unicast Although the routing protocol of broadcast has been presented in [2], it is only for piconet. In our proposal, we present a new routing protocol for scatternet broadcast. Besides, we still support the piconet broadcast. We want to derive the average number of hops that a data packet needs to travel before reaching its destination in a broadcast. This will depend on the role of the source, which can be a slave, a master, or a router. Table 5-6 summarizes these three cases.. In this paper, we propose a method to extend and improve the performances of Bluetooth devices for the wireless home network in [2]. We reduce the packet header length and the number of hops needed for some routing paths. In addition, we add a scatternet broadcast function. In [2], the broadcast function is only designed for piconet. If some device needs to send a scatternet broadcast, it must send a lot of packets to each master within the same scatternet. Our proposal retains piconet broadcast function and supports a scatternet broadcast function. If a device needs to send a scatternet broadcast in our proposal, it only needs to send one packet to the router. Then the router will copy the packet to other masters in the same scatternet. We can reduce the number of packets in a scatternet broadcast. Although our protocol can reduce the packet traveling distance and increase the throughput, a bottleneck may form in some.
(8) gateways or the router. It will affect the inter-piconet communication. How to avoid the bottleneck is an important problem. If all the Bluetooth enabled devices are in each other’s radio range, we can find a lot of scatternet formation protocols in the literature. However, those protocols can’t be used in our home networks. In general situation, the communication distance of a Bluetooth device is 10 meters. Its covered area superficially is π ⋅ r where r is 10 meters. Thus the maximum area of a piconet is 314 square meters. However if all the Bluetooth nodes are in each other’s radio range, the maximum area of a scatternet will be reduced to 78.5 (5*5*3.14) square meters. It is too small for our wireless home network. 2. Therefore, an interesting research topic would be to devise scatternet formation protocol that has no limits on the distances between Bluetooth nodes.. References [1]. Lin, T.Yu., Tseng, Y. C., and Chang, K. M., and Tu, C. L., ”Formation, routing, and maintenance protocols for the bluering scatternet of bluetooths,” Proceedings of the 36th Annual Hawaii International Conference on System Sciences, pp.313 –322, 2003.. [2]. Lilakiatsakun, W., and Seneviratne, A., “Wireless Home networks based on a Hierarchical Bluetooth Scatternet Architecture,” IEEE International Conference on Networks, Ninth, pp.481 –485, 2001..
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