VN-APIT: virtual nodes-based range-free APIT localization scheme for WSN

Abstract

As an important supporting technology, the localization technology has become the basis of the practical applications of wireless sensor networks for information acquisition and processing . APIT (approximate Point-In-Triangulation) localization scheme is widely used for localization estimation in wireless sensor networks due to the advantages of the high localization accuracy and easy to deploy. However, there are some inherent defects caused by the uneven distribution of sensor nodes in APIT scheme. To overcome the problem, in this paper a novel virtual nodes-based range-free localization scheme, i.e., VN-APIT, is proposed to improve the APIT scheme. By rationally deploying virtual nodes in sensor network according to the proposed VN-APIT test theory, VN-APIT localization scheme can determine independently that whether a target node is inside or outside the triangle formed by three certain anchor nodes. Therefore, it will not subject to the effect of the density and distribution of sensor nodes and there is no the problem of the In-to-Out error and the Out-to-In error in VN-APIT scheme. Simulation evaluation shows that the proposed VN-APIT scheme is more robust and has a lower average localization error (ALE) than the conventional APIT scheme.

Appendix

1.1 Proofs of Proposition 1 and 2

Proposition 1

If M is inside triangle $\triangle ABC$, when M is shifted in any directions which are perpendicular to the trilateral of triangle $\triangle ABC$ respectively, the new position of M mustn’t be nearer to or further from all three anchors A, B and C simultaneously.

Proof

As shown in Fig. 14(a), M is inside triangle $\triangle ABC$ and assuming ${\mathrm{MM}}_{{1}}^{\mathrm{'}}$, ${\mathrm{MM}}_{{2}}^{\mathrm{'}}$ and ${\mathrm{MM}}_{{3}}^{\mathrm{'}}$ are perpendicular to the trilateral BC, AC and AB of $\triangle ABC$ respectively. We first prove that M can’t be closer to the three vertexes simultaneously when M moves to ${\mathrm{M}}_{{1}}^{\mathrm{'}}$. Consider the two cases:

Case 1: A, M and ${\mathrm{M}}_{{1}}^{\mathrm{'}}$ are on the same line. Clearly, AM $<{\mathrm{AM}}_{{1}}^{\mathrm{'}}$. And in view of the hypotenuse is greater than the right-angle side, we easily get that ${\mathrm{BM}}_{{1}}^{\mathrm{'}}<{\mathrm{BM}}$ and ${\mathrm{CM}}_{{1}}^{\mathrm{'}} <{\mathrm{CM}}$.

Case 2: A, M and ${\mathrm{M}}_{\mathrm{1}}^{\mathrm{'}}$ are not on the same line. From A draw a perpendicular line to ${\mathrm{MM}}^{\mathrm{'}}$ meeting in at ${\mathrm{D}}_{\mathrm{1}}$. When when M moves to ${\mathrm{M}}_{{1}}^{\mathrm{'}}$, because of ${\mathrm{D}}_{{1}}<{\mathrm{MD}}_{{1}}{\mathrm{M}}_{{1}}^{\mathrm{'}}$, by Pythagorean Theorem, ${\mathrm{AM}}<{\mathrm{AM}}_{{1}}^{\mathrm{'}}$. As same as case 1, ${\mathrm{BM}}_{{1}}^{\mathrm{'}}<{\mathrm{BM}}$ and ${\mathrm{CM}}_{{1}}^{\mathrm{'}}<\hbox {CM}$.

Hence, M can’t be closer to the three vertexes simultaneously when M moves to ${\mathrm{M}}_{{1}}^{\mathrm{'}}$. Similarly, when M moves to ${\mathrm{M}}_{{2}}^{\mathrm{'}}$ and ${\mathrm{M}}_{{3}}^{\mathrm{'}}$ respectively, we can prove that M can’t be closer to the three vertexes simultaneously.

Fig. 14

Proofs for Propositions. a Proposition 3. b Proposition 4

Proposition 2

If M is outside triangle $\triangle ABC$, when M is shifted in any directions which are perpendicular to the trilateral of triangle $\triangle ABC$ respectively, there must exist a direction in which the new position of M is closer to or further from all three anchors A, B and C simultaneously.

Proof

As shown in Fig. 14(a), we prove this proposition by a special case. For any point M exterior to $\triangle ABC$, there is always a point ${\mathrm{M}}^{\mathrm{'}}$ on the side (or the extension line of the side) closest to M and M${\mathrm{M}}^{\mathrm{'}}$ is perpendicular to the side. Without loss of generality, We might assume that ${\mathrm{M}}^{\mathrm{'}}$ is on the side of BC and M${\mathrm{M}}^{\mathrm{'}}$ is perpendicular to BC. As same as Proposition 1, it also has two case. In the case 1, the proof is clear as proved in Proposition 1. Here we only consider the case 2. From a draw perpendicular line to M${\mathrm{M}}^{\mathrm{'}}$ meeting in at ${\mathrm{D}}_{\mathrm{1}}$. When when M moves to ${\mathrm{M}}^{\mathrm{'}}$, because of ${\mathrm{D}}_{\mathrm{1}}{\mathrm{M}}^{\mathrm{'}} < {\mathrm{D}}_{\mathrm{1}}$M, by Pythagorean Theorem, ${\mathrm{AM}}_{\mathrm{1}}^{\mathrm{'}} <{\mathrm{AM}}$. And similarly in view of the hypotenuse is greater than the right-angle side, we easily get that ${\mathrm{BM}}^{\mathrm{'}} <{\mathrm{BM}}$ and ${\mathrm{CM}}^{\mathrm{'}} <\hbox {CM}$.

Hence, when M moves to ${\mathrm{M}}^{\mathrm{'}}$, we proved that M is closer to all three anchors A, B and C simultaneously. In other words, when M is outside triangle $\triangle ABC$, there must exist a perpendicular direction in which the shifted new position of M is closer to or further from all three anchors A, B and C simultaneously.