201245746 /、 發明說明： 【發明所屬之技術領域】 本發明係關於一種無線射頻辨 Identification，RFID)定位，特別是指_ ^®^Kad10 Frequency 個天線量測距離為半徑的兩個球體Y二f$位過，中，運用兩 時之一種RFID標籤定位演算法。’热父集或父集範圍過大 【先前技術】 RPID天線定位適用於室内且建 僅採用單一 RFID天線進行定位，所 -，但二維空間若 球面’若增加一個_天線，可將d = 置為一 天線，將會與此-圓弧有兩個交= Π增加第三個 個可能位置，為求得合理解，—般需要4個^疋位目標的兩 -維讀概念如圖丨所示， — 號發射塔且發射塔位置已知，假訊盔如Α/尤心主乂而要二個汛 MU臟—η，若將此概念運用於三維空間定位，則3= 以上訊號發射塔。 』疋诅則而要四個 97134617 ««^SPAl.O) , 以二念’以定位空間中心做為起始點， 5稱為sPAi適合處理大空間細定位問題，這個 出最可能之立方體網格。另口 201245746 99104CM7號），結合sp 1〇與2 〇的方法加速座標定位的計算 速率。第四申請案SPA4.〇(申請專利案第99121147號)運用地 理資訊系統定位RPID標籤位置。 、上述RFID標籤定位技術，皆為求取誤差最小之座標位 置，並未考慮各個天線所量測之距離為半徑的球體是否有交 集，然而由於標籤回饋接收訊號強度(Recdved Signal汾出 Indication，RSSI)受定位空間的障礙物、人員走動、設備運轉 因素衫響，依據RSSI圖表所得的距離，就可能會有某些 f對應的距離可靠’某些方位較不可靠(有較大的誤差）的問201245746 /, invention description: [Technical field of the invention] The present invention relates to a radio frequency identification (RFID) positioning, in particular to two spheres Y 2 of the radius of the _ ^^^Kad10 Frequency antenna measurement distance $ bit over, medium, using a two-time RFID tag location algorithm. 'Hot parent set or parent set range is too large [previous technique] RPID antenna positioning is suitable for indoor use and only uses a single RFID antenna for positioning, - but if the two-dimensional space is spherical, if you add one _ antenna, you can set d = For an antenna, there will be two intersections with this-arc = Π increase the third possible position. For the sake of understanding, the two-dimensional reading concept that requires 4 ^ target is as shown in the figure. Show, the number of towers and the location of the tower, the false helmets such as Α / especially the main 乂 要 汛 MU dirty - η, if this concept is applied to three-dimensional positioning, then 3 = above signal tower .疋诅 而 要 971 971 971 971 971 971 971 971 971 971 971 971 971 971 971 971 971 971 971 971 971 971 971 971 971 971 971 971 971 971 971 971 971 971 971 971 971 971 971 971 971 971 971 971 971 971 971 971 971 grid. Another mouth 201245746 99104CM7), combined with the method of sp 1〇 and 2 加速 to accelerate the calculation rate of coordinate positioning. The fourth application, SPA4.〇 (Patent Application No. 99121147), uses the geographic information system to locate the RPID tag location. The above RFID tag positioning technology is to obtain the coordinate position with the smallest error, and does not consider whether the spheres whose radius is measured by each antenna has an intersection, but because of the tag feedback, the signal strength (Recdved Signal, Indication, RSSI) The obstacles in the positioning space, the people walking, the equipment running factors, the distance according to the RSSI chart, there may be some f corresponding distance is reliable 'some directions are less reliable (have a larger error) ask
H tit自，n線^心’可能使天線所對應的距 雔亚,、，、父木或父集範圍過大，使用前述的方法，可能導致收斂 速度慢或甚至無法收歛，因而增加求解時間。本發明 以解決上述的問題。 eJ 【發明内容】 本發明揭露-種無線射頻辨識卿D)標藥定位之方法 天線量測距離為半徑的兩個球體，並無交集或交 驟=四個天線A、B、C、D於一直角定位空間的 =訊號衰減曲線_距離關係圖;依據該訊號衰減曲線·距_ ，求取該些天線對—目標mD標籤量測值，並^此二 半徑’以該些天線所在位置為圓心，分別在x_y平面 y-Z平面㈣複_投侧；求取該目標紐j)在 ， 之投影座標；平均鮮投碰標。 彳戌在該些千面 其田中上述之投影Η之間容許不她、相交於—點 相重，其中一種或二種或三種同時存在。當投影圓 不相父時，預估投影座標為該些圓其巾之―的圓心至^ ί接=該平面兩歡鄉。#投之間彼此相交於 =預估财座縣·___平面 外 影因之間彼此健疊時，難投影鋪為該些一 = 5 201245746 心至重純之半龍平面兩轴之投影 【實施方式】 的流ΐ示鄉座獅猶，進行定位 目標標籤之轉。驟2q的計算距離’即丽"天線與 RFID^^g It位空間内設置複數個已知位置的 兮此個%FT:, 土再之定點所設置複數個腿0天線逐一讀取 関係圖。參標鐵的咖值。依據這些數據，建立RSSI_ 鮮咖天_目標賴讀取腿值，依H titse, n-line ^heart' may make the range of the distance corresponding to the antenna, the parent tree, or the parent set too large. Using the foregoing method may cause the convergence speed to be slow or even unable to converge, thus increasing the solution time. The present invention solves the above problems. eJ SUMMARY OF THE INVENTION The present invention discloses a radio frequency identification (D) standard method of positioning a sample. The antenna measures two spheres with a radius of radius, and there is no intersection or intersection = four antennas A, B, C, and D. The signal attenuation curve _ distance relationship graph of the straight-angle positioning space; according to the signal attenuation curve·distance _, the antenna pair-target mD label measurement value is obtained, and the two radii 'the position of the antennas are The center of the circle is respectively on the x_y plane yZ plane (four) complex _ projection side; the target of the target j), the projection coordinates; the average fresh touch target.容许 Between the above projections in the thousands of fields, the above-mentioned projections are allowed to be separated from each other, and one or two or three of them are simultaneously present. When the projection circle is not the same as the parent, the projection coordinates are estimated to be the center of the circle of the circle to ^ ί 接 = the plane two happy. #投Between each other in the intersection = = estimated fiscal seat county ___ plane shadow between the two sides of each other, difficult to project the shop for the one = 5 201245746 heart to the pure half-dragon plane projection of two axes [ The implementation method] shows the rogue of the township lion, and carries the positioning target label. The calculation distance of step 2q is 'Yi Li' and the antenna and RFID^^g It bit space are set in a plurality of known positions. This %FT:, the set of multiple legs and 0 antennas are set one by one. . The value of the standard iron. Based on these data, establish RSSI_ fresh coffee days _ target 赖 read leg value, according to
λ r值 距離関係圖獲取目標標籤P與各個RFID 天線A、C之距離分別為pA、pc。 …矛Γ巨f的關係，第一個腦天線C為球體球 :罢為Γΐ，釗—球體，球體表面皆為目標標籤可能之 位置，苐一個RFID天線A為球體圓心，pA為半徑^，劃一 球體’所劃之球體表面亦為目標標鐵可能之位置。 ;、、i而C天線所|W之球體1〇〇與A天線所繪之球體2〇〇 亚無父集’如圖3所示，為解決兩球體無交集的問題，依據本 發明的-實關，進行步驟21，.計算投影座標。為此，仍請來 考圖3，將包含天線A、c與其它的天線β、D為定位空間的 四個角落’劃出一定位空間300。請注意，Ac、BC、Dc分別 為定位空間300各面之對角線。同樣，处、AD、bd、也在對 角線。圖3中，雖只繪出制_體A、c，但，圖3同時示a、 B、C、D的天線的位置。以一較佳實例而言，位空 空間六面體，但不以此為限，例如，圓形劇場、或者__般建物 或特殊建物包含局部·空間，本發_方法也都顧。以下 201245746 我們將以定位空間為一長方體為說明。 請參考圖4，其圖示兩個_ A、c在y_z平面的投影。 兩圓心距離為L’ Μ為兩圓半徑的連接線以勺長。連接線中點 m是目標標籤ρ在y-z平面位置的最可能投影位置。 ” 因此’由圓心C至連接線你中點爪的線段&可表示為 1丄从 "、 a = r I + —_ 2 ⑴ 其中IV[可運用式(2)計算： (2) ⑶ 如圖5所示。 M = L - (rl + r2 ) 將式(2)代入式(1)可獲得式(3): 2 中點位置m可藉由式(4)計算y與ζ座標λ r value The distance map obtains the distance between the target tag P and each of the RFID antennas A and C as pA and pc, respectively. ...the relationship between the spear and the giant f, the first brain antenna C is the sphere ball: Γΐ Γΐ, 钊-sphere, the surface of the sphere is the possible position of the target label, 苐 an RFID antenna A is the center of the sphere, pA is the radius ^, The surface of the sphere drawn by the "ball" is also the possible position of the target. ;,, i, and C antennas | W sphere 1 〇〇 and A antenna painted sphere 2 〇〇 sub-no father set ' As shown in Figure 3, in order to solve the problem of no intersection of two spheres, according to the present invention - To perform the actual operation, proceed to step 21. Calculate the projection coordinates. For this reason, still refer to Fig. 3, and a positioning space 300 is drawn by including four antennas A and c and other antennas β and D as the four corners of the positioning space. Please note that Ac, BC, and Dc are the diagonal lines of each side of the positioning space 300, respectively. Similarly, the location, AD, bd, are also diagonal. In Fig. 3, only the body A and c are drawn, but Fig. 3 also shows the positions of the antennas of a, B, C, and D. In a preferred embodiment, the space is a space hexahedron, but is not limited thereto. For example, an amphitheater, or a __like building or a special building includes a partial space, and the present method is also considered. The following 201245746 we will use the positioning space as a rectangular box. Please refer to FIG. 4, which illustrates the projection of two _A, c in the y_z plane. The connecting line with the distance between the two centers of L' Μ is the radius of two circles is scooped. The midpoint of the connecting line m is the most likely projection position of the target label ρ at the y-z plane position. Therefore, the line segment & from the center C to the connecting line of your midpoint claw can be expressed as 1丄 from ", a = r I + —_ 2 (1) where IV [apparatus (2) can be calculated: (2) (3) As shown in Fig. 5. M = L - (rl + r2 ) Substituting equation (2) into equation (1) can obtain equation (3): 2 The midpoint position m can be calculated by equation (4) for y and ζ coordinates
=J^c + Ω X COS Θy =-^c + a x cos Θz (4) 相似於(1)至式(4) ’分別求得各個天線連線所在平面的中 點座標位置。 緊接著，前進到步驟22，計算其它天線間之平面的投影 座標。例如，天線C與D所形成的兩球體1〇〇,4〇〇，可能也是 不相交’也可能重疊，再一種可能是相交於一點。 例如，當天線C與D所形成的兩球體1〇〇,4〇〇相重疊， 201245746 請參考圖6。上述對於天線C與A所形成的兩球體1〇〇,200不 相交時的計算式(1)至(4)同樣適用，只是稍作些變化而已。 此時，目標標籤在x_z平面投影的可能位置(Χ2，Ζ2) (χ2，ζ其中，=J^c + Ω X COS Θy =-^c + a x cos Θz (4) Similar to (1) to (4) ′, find the coordinates of the midpoint coordinates of the plane where each antenna is connected. Next, proceeding to step 22, the projection coordinates of the plane between the other antennas are calculated. For example, the two spheres formed by antennas C and D, 1〇〇, 4〇〇, may also be disjointed, or may overlap, and the other may intersect at a point. For example, when the antennas C and D form two spheres 1〇〇, 4〇〇 overlap, 201245746 please refer to FIG. 6. The above calculations (1) to (4) for the two spheres 1 and 200 formed by the antennas C and A do not intersect, but only slightly changed. At this time, the possible position of the target label projected on the x_z plane (Χ2, Ζ2) (χ2, ζ where,
=xc + C X COS Φ x =Zc + C X cos φ z (5) ⑹ 當天線C與D所形成的兩球體i〇〇,4〇〇不相交時(未圖 示）’則一如上述(5)，但 c = rl + ^-~ (rl_±ii) 2 ⑺ 當天線C與D所形成的兩球體可能相交於一點(即 相切）’則目標標籤在X-Z平面投影的可能位置(Χ2，Ζ2)即在交集 點上。 此時c = rl代入式(5)即可求得二維座標點。=xc + CX COS Φ x =Zc + CX cos φ z (5) (6) When the two spheres i 〇〇 and 4 形成 formed by the antennas C and D do not intersect (not shown), the same as above (5) ), but c = rl + ^-~ (rl_±ii) 2 (7) When the two spheres formed by antennas C and D may intersect at one point (ie, tangent), then the target label is projected at the XZ plane (Χ2, Ζ 2) is at the intersection. At this time, c = rl is substituted into equation (5) to obtain a two-dimensional coordinate point.
因此’依據上述(1)至⑺將可以求出CB、ab、AD、BD ^間之平面投影座標，以分別求得(^抓，〇，(〜2抓，士 最後將所有之目標標狀投影位置依χ，y, 2各㈣分量各自平 均，即為所求之目標標籤座標。 其中 S ~ (X1 + X2 + ... + /6 , ' =(少1 +少2 +…+少6 ) / 6 = (Z1 + + …+ Z6)/6 是球财相交時的解決赃，且容許投影 0之間不相父、相父於一點，或投影圓相重疊的其中一種或二 201245746 種或三種同時存在。 本發明雖以較佳實例闡明如上，然其並非用以限定 本發明精神與發明實體僅止於上述實施例。凡熟悉此項 技術者，當可輕易了解並利用其它元件或方式來產生相 同的功效。是以，在不脫離本發明之精神與範疇内所作 之修改，均應包含在下述之申請專利範圍内。 201245746 【圖式簡單說明】 示與精神可以藉由以下詳細描述及•圖 塔 位置圖定位概念至少需要三個訊號發射塔且發射 圖2為本㈣空間定位的流程圖。 ^3示意當兩天線球體並沒有交 粒空間，將另兩個天線安排在放in方 形成各個天線都是在對角線位置。咕疋位工間的角落， 圖4示將兩天線球體投影於 上之可能投影位置。 y卞囬及目钴私紙在该平面 =5，兩天線球體投影於 :投影位置以極座標表示:‘圖 圖。㈤τ兩天線球體投影於χ_ζ平面時，兩圓重疊時之示意 【主要元件符號說明】 20 ' 2卜22流程圖步驟 100、200、4〇〇定位天線 300定位空間 乂偵測到的距離為半徑所形成的球體 Μ球體間連接線 ®連接線之中點Therefore, based on the above (1) to (7), the plane projection coordinates between CB, ab, AD, and BD ^ can be obtained to obtain (^ grab, 〇, (~2 grab, and finally all target marks) The projection position depends on y, y, 2 each (four) component is averaged, that is, the target label coordinates sought. Among them, S ~ (X1 + X2 + ... + /6 , ' = (less 1 + less 2 +... + less) 6 ) / 6 = (Z1 + + ... + Z6)/6 is the solution for the intersection of the financial assets, and allows one or two of the projections to be unfamiliar, the father is at a point, or the projection circle overlaps one or two 201245746 The present invention is described above by way of a preferred example. However, it is not intended to limit the spirit of the invention and the inventive entity is only limited to the above embodiments. Those skilled in the art can easily understand and utilize other components. And the manner in which the same effect is obtained, and modifications made without departing from the spirit and scope of the invention are included in the scope of the following patent application. 201245746 [Simple description of the schema] The spirit and spirit can be as follows Detailed description and • Tuta location map positioning concept requires at least three signal transmission towers Figure 2 is a flow chart of (4) spatial positioning. ^3 indicates that when two antenna spheres do not have a space for intersection, the other two antennas are arranged in the in-side to form each antenna at a diagonal position. Figure 4 shows the possible projection positions for projecting the two antenna spheres. y卞回和目Co's private paper is on this plane=5, and the two antenna spheres are projected on: the projection position is represented by polar coordinates: '图图. (5)τ When the two antenna spheres are projected on the χ_ζ plane, the two circles overlap. [Main component symbol description] 20 ' 2 Bu 22 Flowchart steps 100, 200, 4〇〇 Positioning antenna 300 positioning space 乂 The detected distance is the radius The midpoint of the formed sphere Μ ball connecting line® connecting line