This article has been accepted for publication in a future issue of this journal,but has not been fully edited.Content may change prior to final publication.Citation information:DOI 10.1109/TMC.2018.2857812.IEEE Transactions on Mobile Computing 5 distances,the profiles of the correspond depth histogram is the semiperimeter of the triangle,i.e.,s (d1+d+d) are very similar to each other in most cases.In particular, Moreover,since the area of this triangle can also be com- when the object is deployed at a distance very close to the puted as A =h x d,we can thus compute the vertical depth camera,e.g.,50cm,the profile may be distorted to a distanceh Then,according to the certain degree.When the object is deployed at a distance of 450cm,the depths over 475cm are no longer illustrated since Apollonius'theorem [31],for a triangle composed of point they are out of the effective scanning distance.Therefore, A1,A2,and T,the length of median TO bisecting the the experiment results show that the depth camera is able to side A1A2 is equal to m =v2d+2d2-d2.Hence,the horizontal distance between the tag and the midpoint of the extract the depth information of the objects at a distance as two antennas,i.e.,T'O,should be vm2-h2.Therefore,if far as 475cm. we build a local coordinate system with the origin set to 4.2 Extract the Phase Value from RF-Signals the midpoint of the two antennas,the coordinate (z',y)is Phase is a basic attribute of a signal along with amplitude computed as follows: and frequency.The phase value of an RF signal describes the degree that the received signal offsets from the sent signal, √d+-d2-h2 d1≥d2 (2) ranging from 0 to 360 degrees.Let d be the distance between -(V2+-d2-h2) d1<d2 the RFID antenna and the tag,the signal traverses a round- trip with a distance of 2d in each backscatter communica- y=h. (3) tion.Therefore,the phase value output by the RFID reader Therefore,the next problem we need to address is to es- can be expressed as [20,29]: timate the absolute distance between the tag and antenna 0=0 2m according to the extracted phase value from RF-signals. ,×2d+4)）mod2m, (1) Suppose the RFID system respectively obtains two phase where A is the wave length.u is a diversity term which values 01 and 02 from two separated RFID antennas,then, is related with additional phase rotation introduced by according to the definition in Eq.(1),the possible distances the reader's transmitter/receiver and the tag's reflection from the tag to the two antennas are:di=支·(+&i):入， characteristic.According to the previous study [4],as u is and d2=支·（经+2）·X.Here,ki andk2 are integers rather stable,we can record u for different tags in advance. ranging from 0 to +o0.Due to the multiple solutions of k Then,according to each tag's response,we can calibrate the and k2,there could be multiple candidate positions for the phase by offsetting the diversity term.Thus,the phase value tag.However,since the difference of the lengths of two sides can be used as an accurate and stable metric to measure is smaller than the length of the third side in a triangle,i.e., distance. ld-d2<d,we can leverage this constraint to effectively According to the definition in Eq.(1),the phase is a eliminate many infeasible solutions of k and k2.Besides, periodical function of the distance.Hence,given a speci- due to the limited scanning range of the RFID system(the fied phase value from the RF-signal,there can be multiple maximum scanning range l is usually smaller than 10 m), solutions for estimating the distance between the tag and the value of ki and k2 should be upper bounded by a certain antenna.Therefore,we can deploy an RFID antenna array threshold,ie,头 to scan the tags from slightly different positions,so as to Fig.5 shows an example of feasible positions of the target figure out the unique solution of the distance.Without loss tag according to the obtained phase values 01 and 02.The of generality,in this paper,we separate two RFID antennas feasible solutions include multiple positions like A ~D, with a distance of d,use them to scan the RFID tags,and which respectively belong to two hyperbolas Hi and H2. respectively obtain their phase values from the RF-signals, Due to the existence of multiple solutions,we can use these as shown in Fig.4. hyperbolas to denote a superset of these feasible positions in a straightforward approach. -axis HyPerbola H: HyPerbola H: Vertical distance A2 -axis 2 2 Fig.4.Compute the(,y)coordinate of the tag Fig.5.Estimate the distance from phase values If we respectively use A1 and A2 to denote the midpoint of Antenna 1 and Antenna 2,and use T to denote the 5 MATCH THE STATIONARY TAGGED OBJECTS VIA position of the tag,as a matter of fact,the three sides ROTATE SCANNING of (T,A1),(T,A2),and (A1,A2)form a triangle.Since Antenna A1 and Antenna A2 are separated with a fixed 5.1 Motivation distance d,according to Heron's formula [30],the area of To identify and distinguish the multiple tagged objects, this triangle is A =Vs(s-d1)(s-d2)(s-d),where s a straightforward solution is to scan the tags in a static 1536-1233(c)2018 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.1536-1233 (c) 2018 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. This article has been accepted for publication in a future issue of this journal, but has not been fully edited. Content may change prior to final publication. Citation information: DOI 10.1109/TMC.2018.2857812, IEEE Transactions on Mobile Computing 5 distances, the profiles of the correspond depth histogram are very similar to each other in most cases. In particular, when the object is deployed at a distance very close to the depth camera, e.g., 50cm, the profile may be distorted to a certain degree. When the object is deployed at a distance of 450cm, the depths over 475cm are no longer illustrated since they are out of the effective scanning distance. Therefore, the experiment results show that the depth camera is able to extract the depth information of the objects at a distance as far as 475cm. 4.2 Extract the Phase Value from RF-Signals Phase is a basic attribute of a signal along with amplitude and frequency. The phase value of an RF signal describes the degree that the received signal offsets from the sent signal, ranging from 0 to 360 degrees. Let d be the distance between the RFID antenna and the tag, the signal traverses a round￾trip with a distance of 2d in each backscatter communica￾tion. Therefore, the phase value θ output by the RFID reader can be expressed as [20, 29]: θ = (2π λ × 2d + µ) mod 2π, (1) where λ is the wave length. µ is a diversity term which is related with additional phase rotation introduced by the reader’s transmitter/receiver and the tag’s reflection characteristic. According to the previous study [4], as µ is rather stable, we can record µ for different tags in advance. Then, according to each tag’s response, we can calibrate the phase by offsetting the diversity term. Thus, the phase value can be used as an accurate and stable metric to measure distance. According to the definition in Eq. (1), the phase is a periodical function of the distance. Hence, given a speci- fied phase value from the RF-signal, there can be multiple solutions for estimating the distance between the tag and antenna. Therefore, we can deploy an RFID antenna array to scan the tags from slightly different positions, so as to figure out the unique solution of the distance. Without loss of generality, in this paper, we separate two RFID antennas with a distance of d, use them to scan the RFID tags, and respectively obtain their phase values from the RF-signals, as shown in Fig. 4. Antenna 1 Antenna 2 d Tag G  G  O m h X-axis Y-axis Vertical distance T A1 A2 Tÿ Fig. 4. Compute the (x, y) coordinate of the tag If we respectively use A1 and A2 to denote the midpoint of Antenna 1 and Antenna 2, and use T to denote the position of the tag, as a matter of fact, the three sides of hT, A1i, hT, A2i, and hA1, A2i form a triangle. Since Antenna A1 and Antenna A2 are separated with a fixed distance d, according to Heron’s formula [30], the area of this triangle is A = p s(s − d1)(s − d2)(s − d), where s is the semiperimeter of the triangle, i.e., s = (d1+d2+d) 2 . Moreover, since the area of this triangle can also be com￾puted as A = 1 2 h × d, we can thus compute the vertical distance h = 2 √ s(s−d1)(s−d2)(s−d) d . Then, according to the Apollonius’ theorem [31], for a triangle composed of point A1, A2, and T, the length of median T O bisecting the side A1A2 is equal to m = 1 2 p 2d 2 1 + 2d 2 2 − d 2. Hence, the horizontal distance between the tag and the midpoint of the two antennas, i.e., T 0O, should be √ m2 − h 2. Therefore, if we build a local coordinate system with the origin set to the midpoint of the two antennas, the coordinate (x 0 , y0 ) is computed as follows: x 0 =    q 1 2 d 2 1 + 1 2 d 2 2 − 1 4 d 2 − h 2 d1 ≥ d2 −( q 1 2 d 2 1 + 1 2 d 2 2 − 1 4 d 2 − h 2) d1 < d2 (2) y 0 = h. (3) Therefore, the next problem we need to address is to es￾timate the absolute distance between the tag and antenna according to the extracted phase value from RF-signals. Suppose the RFID system respectively obtains two phase values θ1 and θ2 from two separated RFID antennas, then, according to the definition in Eq. (1), the possible distances from the tag to the two antennas are: d1 = 1 2 · ( θ1 2π + k1) · λ, and d2 = 1 2 · ( θ2 2π + k2) · λ. Here, k1 and k2 are integers ranging from 0 to +∞. Due to the multiple solutions of k1 and k2, there could be multiple candidate positions for the tag. However, since the difference of the lengths of two sides is smaller than the length of the third side in a triangle, i.e., |d1 − d2| < d, we can leverage this constraint to effectively eliminate many infeasible solutions of k1 and k2. Besides, due to the limited scanning range of the RFID system (the maximum scanning range l is usually smaller than 10 m), the value of k1 and k2 should be upper bounded by a certain threshold, i.e., 2l λ . Fig. 5 shows an example of feasible positions of the target tag according to the obtained phase values θ1 and θ2. The feasible solutions include multiple positions like A ∼ D, which respectively belong to two hyperbolas H1 and H2. Due to the existence of multiple solutions, we can use these hyperbolas to denote a superset of these feasible positions in a straightforward approach. Antenna 1 Antenna 2 d<Ȝ/2 Target A B D C Ȝ/2 G T T T T  T HyPerbola H1 HyPerbola H2 G Fig. 5. Estimate the distance from phase values 5 MATCH THE STATIONARY TAGGED OBJECTS VIA ROTATE SCANNING 5.1 Motivation To identify and distinguish the multiple tagged objects, a straightforward solution is to scan the tags in a static