Range-Free Mobile Sensor Localization and a Novel Obstacle Detection Technique

Abstract

Mobile sensor localization is a challenging problem in wireless sensor networks. Due to mobility, it is difficult to find exact position of the sensors at any time instance. The aim of localization is to minimize positioning errors of the mobile sensors. In this paper we propose two range-free distributed localization algorithms for mobile sensors with static anchors. Both the algorithms depend on selection of beacon points. First we assume that mobile sensors move straight during localization which helps us to provide an upper bound on localization error. Certain applications may not allow sensors to move in a straight line. Obstacles may also obstruct path of sensors. Moreover beacon point selection becomes difficult in presence of obstacles. To address these issues, we propose another localization algorithm with an obstacle detection technique which selects correct beacon points for localization in presence of obstacles. Simulation results show improvements in performance over existing algorithms.

Appendix

Proof of Lemma 1 is given below.

Proof

From Fig. 2, $P_1C\le u$ and $P_2C'\le u$, where $P_1$ and $P_2$ are the points of intersection of the actual line of movement of the mobile sensor with communication circle of the anchor S. For some $h_1,h_2\in [0,1]$, we can write

$$\begin{aligned} P_1C= & {} h_1u \nonumber \\ P_2C'= & {} h_2u \end{aligned}$$

(2)

According to Fig. 2, possible approximate lines of movement are $P'_1P'_2$ and $P''_1P''_2$. Mobile sensor can measure the length $CC'$ from it’s velocity and time stamps of the beacon points. The sensor misinterprets the length $CC'$ as the length of the chords $P_1P_2$ along which it is moving. So, the actual line of movement $P_1P_2$ is projected to the approximate line of movement $P'_1P'_2$ such that $CC'=P'_1P'_2$. Due to the symmetric nature of circle, we get another possible line of movement $P''_1P''_2$. It is possible to discard $P''_1P''_2$ as a line of movement using one more communication circle of another anchor as discussed later. The error due to computing approximate line of movement is $C'P'_2$ as shown in Fig. 2. According to Fig. 2, the actual line of movement is $P_1P_2$ and the selected approximate line of movement is $P'_1P'_2$. Let T be the time stamp of receiving beacon at $C'$. With the knowledge of $P'_1P'_2$, the mobile sensor computes position of the beacon point as $P'_2$ instead of $C'$. So, the error in positioning is equal to $C'P'_2$. $P'_1P'_2=2l$, say. Now, $C'P'_2=\sqrt{(P'_2H)^2+(C'H)^2}$. Again $P'_1P'_2=CC''=P_1P_2-(h_1u+h_2u)$.

Hence, $C'H=C'P_2-HP_2=(h_2-h_1) \frac{u}{2}$.

Also, $P'_2H=MM'=SM'-SM=\sqrt{r^2-l^2}-\sqrt{r^2-\left( l+ \frac{h_1u+h_2u}{2}\right) ^2}$.

Here, $P'_2H$ and $C'H$ are the perpendicular and horizontal components of the error $C'P'_2$. We denote perpendicular component of error as perpendicular error $E_P$ and horizontal component of error as horizontal error $E_H$. The error in positioning, $C'P'_2$ is denoted as $\zeta$ which is equal to $\sqrt{E_P^2+E_H^2}$.

$$\begin{aligned} E_H= & {} (h_2-h_1) \frac{u}{2} \nonumber \\ E_P= & {} \sqrt{r^2-l^2}-\sqrt{r^2-\left( l+ \frac{h_1u+h_2u}{2}\right) ^2}\nonumber \\ \zeta= & {} \sqrt{E_P^2+E_H^2} \end{aligned}$$

(3)

$\square$

Proof of Lemma 2 is given below.

Proof

Perpendicular component of positioning error is equal to $MM'$ as shown in Fig. 2. Hence, maximum possible value of $MM'$ for all possible chords is the maximum possible perpendicular component of error in localization. As the values of $P_1C$ and $P_2C'$ increases, $MM'$ also increases. Without considering the irregularity in signal propagation, [0, u] is the range of possible values of $P_1C$ and $C'P_2$, where u is the beacon distance. When $P_1C=u=C'P_2$, then the length of the actual chord along which the sensor is moving becomes $2l+2u$, where 2l is the calculated chord length.

So, length of $MM'=MaxE_P(2l)=\sqrt{r^2-l^2}-\sqrt{r^2-(l+u)^2}$.

This expression gives maximum value when $u+l=r$. That is, if the sensor moves along the diameter and computes the chord length as $2(r-u)$ instead of 2r, then the perpendicular error becomes maximum. Hence the maximum possible perpendicular error is $\sqrt{r^2-(r-u)^2}$. $\square$

Proof of Lemma 3 is given below.

Proof

We have already shown in Sect. 3.2 that for any calculated chord length 2l, the horizontal component of error is $E_H=|h_2-h_1|\frac{u}{2}$. Let $|h_2-h_1|=f$. So $f\in [0,1]$. If horizontal error is nonzero then $f>0$. Then value of horizontal error is $\frac{fu}{2}$. Without loss of generality, let $h_2=h_1+f$. Perpendicular error increases as $h_1+h_2$ increases, i.e., calculated chord length decreases. So, maximum possible value of $h_1=1-f$ and $h_2=1$. Hence using Eq. 3, for any l, perpendicular error becomes $E_P=\sqrt{r^2-l^2}-\sqrt{r^2-\left( l+\frac{(2-f)u}{2}\right) ^2}$.

From Lemma 2, for any f, $E_P$ is maximum when the original line passes along the diameter, i.e., $l+\frac{(2-f)u}{2}=r$. So, for any f, $\zeta$ also attains maximum if $l+\frac{(2-f)u}{2}=r$ and the expression of maximum value of $\zeta$ becomes $\sqrt{r^2-\left( r-\frac{(2-f)u}{2}\right) ^2+(\frac{uf}{2})^2}$, which is a decreasing function of $f\in (0,1]$. Using limit of function and existence of $\sqrt{r^2-\left( r-\frac{(2-f)u}{2}\right) ^2+(\frac{uf}{2})^2}$ at $f=0$, we can say, it attains maximum at $f=0$. So, least upper bound of error $\zeta$ is $\sqrt{r^2-(r-u)^2}=\epsilon$. $\square$

Proof of Theorem 1 is given below.

Proof

For simplicity, we prove the theorem assuming that the mobile sensor is moving along a line parallel to x-axis. Let coordinates of two anchors, S and $S'$, be (a, b) and (c, d) respectively where $a\ge 0$, $b\ge 0$, $c\ge 0$, $d>0$ and $(d-b)\ge \epsilon$. There are two possible cases:

Fig. 22

Anchors are at same side of the line of movement

Case 1: If actual line of movement passes through same side of two anchors.According to Fig. 22 the actual line of movement is $P_1P_4$, passes through upper side of the anchors S and $S'$. Let equation of the actual line of movement which passes through both the circles be

$$\begin{aligned} y=dk \,\mathrm{where}\, k\ge 1 \end{aligned}$$

(4)

Let the chord length corresponding to the communication circles of the anchors S and $S'$ be 2l and $2l'$ respectively. According to Lemma 2, sensor computes equation of the line $P'_1P'_2$ and $P''_1P''_2$ as,

$$\begin{aligned} y= & {} dk+\frac{MaxE_P(2l)}{q_1} \,\hbox {where}\, q_1\ge 1\,\hbox {and} \end{aligned}$$

(5)

$$\begin{aligned} y= & {} 2b-dk-\frac{MaxE_P(2l)}{q_1} \,\hbox {where}\, q_1\ge 1 \end{aligned}$$

(6)

respectively. Similarly, sensor computes equation of the line $P'_3P'_4$ and $P''_3P''_4$ as

$$\begin{aligned} y= & {} dk+\frac{MaxE_P(2l')}{q_2} \,\hbox {where}\, q_2\ge 1\,\hbox {and} \end{aligned}$$

(7)

$$\begin{aligned} y= & {} 2d-dk-\frac{MaxE_P(2l')}{q_2} \,\hbox {where}\, q_2\ge 1 \end{aligned}$$

(8)

respectively. The distance between the lines (Eqs. 5 and 7) is equal to

$|\frac{MaxE_P(2l')}{q_2}-\frac{MaxE_P(2l)}{q_1}|\le \max (\frac{MaxE_P(2l)}{q_1}, \frac{MaxE_P(2l')}{q_2})$.

We denote distance between lines (equation i) and (equation j) by D(i, j) hereafter.

So, $D(5,7)\le \max (\frac{MaxE_P(2l)}{q_1}, \frac{MaxE_P(2l')}{q_2})$.

Let it be the smallest distance among those six possible distances D(i, j). If we choose Eqs. 5 or 7, we are choosing lines within $\max (\frac{MaxE_P(2l)}{q_1}, \frac{MaxE_P(2l')}{q_2})$ error, since $D(4,5)=\frac{MaxE_P(2l)}{q_1}$ as well as $D(4,5)=\frac{MaxE_P(2l')}{q_2}$. Actually we are finding distances between correct line (Eq. 4) and the line we are choosing according to our technique and checking it is less or equal to $\epsilon$ or not. Equations 5 and 7 make sure that there is at least one pair of lines whose perpendicular distance is less than or equal to $\max (\frac{MaxE_P(2l)}{q_1}, \frac{MaxE_P(2l')}{q_2})$.

Let D(5, 6) be the least. We have to check D(4, 6) only.

$D(4,6)=2dk-2b+\frac{MaxE_P(2l)}{q_1} \le 2dk-2b+2\frac{MaxE_P(2l)}{q_1} = D(5,6) < D(5,7)=|\frac{MaxE_P(2l')}{q_2}-\frac{MaxE_P(2l)}{q_1}| \le \max (\frac{MaxE_P(2l)}{q_1}, \frac{MaxE_P(2l')}{q_2})$.

Let D(5, 8) be the least. We have to check D(4, 8) only.

$D(4,8)=2dk-2d+\frac{MaxE_P(2l')}{q_2} \le (2dk-2d+\frac{MaxE_P(2l)}{q_1}+\frac{MaxE_P(2l')}{q_2}) = D(5,8) < D(5,7)=|\frac{MaxE_P(2l')}{q_2}-\frac{MaxE_P(2l)}{q_1}| \le \max (\frac{MaxE_P(2l)}{q_1}, \frac{MaxE_P(2l')}{q_2})$.

$D(6,7)=2dk-2b+\frac{MaxE_P(2l)}{q_1}+\frac{MaxE_P(2l')}{q_2}>2dk-2d+\frac{MaxE_P(2l)}{q_1}+\frac{MaxE_P(2l')}{q_2}= D(5,8)$, since $d>b$.

So, it cannot be the least.

$D(6,8)=2d-2b+\frac{MaxE_P(2l)}{q_1}-\frac{MaxE_P(2l')}{q_2}\ge \epsilon$, since $d\ge (b+\epsilon )$.

So it cannot be the least.

Let D(7, 8) be least. We have to check D(4, 8) only. $D(4,8)=2dk-2d+\frac{MaxE_P(2l')}{q_2} \le 2dk-2d+2\frac{MaxE_P(2l')}{q_2} = D(7,8) \le |\frac{MaxE_P(2l')}{q_2}-\frac{MaxE_P(2l)}{q_1}| \le \max (\frac{MaxE_P(2l)}{q_1}, \frac{MaxE_P(2l')}{q_2})$.

Hence we show that if D(i, j) is the least and we choose any one of the $i-$th and $j-$th equation, then we are choosing a line of movement which is within $\max (\frac{MaxE_P(2l)}{q_1},\frac{MaxE_P(2l')}{q_2})$ perpendicular error with the actual line of movement.

Fig. 23

Anchors are at different sides of the line of movement

Case 2: If actual line of movement passes through different sides of two anchors. According to Fig. 23, the actual line of movement is $P_1P_4$ passes through the different sides of S and $S'$. Let the equation of the actual line of movement which passes through both the circles be

$$\begin{aligned} y=\frac{d}{k} \,\mathrm{where}\, k\ge 1 \end{aligned}$$

(9)

Let the chord length corresponding to the communication circles of the anchors S and $S'$ be 2l and $2l'$ respectively. According to Lemma 2, sensor computes equation of the line $P'_1P'_2$ and $P''_1P''_2$ as,

$$\begin{aligned} y= & {} \frac{d}{k}+\frac{MaxE_P(2l)}{q_1} \,\mathrm{where}\, q_1\ge 1\,\mathrm{and}\end{aligned}$$

(10)

$$\begin{aligned} y= & {} 2b-\frac{d}{k}-\frac{MaxE_P(2l)}{q_1} \,\mathrm{where}\, q_1\ge 1 \end{aligned}$$

(11)

respectively. Similarly, sensor computes equation of the line $P'_3P'_4$ and $P''_3P''_4$ as,

$$\begin{aligned} y= & {} \frac{d}{k}-\frac{MaxE_P(2l')}{q_2} \,\mathrm{where}\, q_2\ge 1\,\mathrm{and}\end{aligned}$$

(12)

$$\begin{aligned} y= & {} 2d-\frac{d}{k}+\frac{MaxE_P(2l')}{q_2} \,\mathrm{where}\, q_2\ge 1 \end{aligned}$$

(13)

respectively. Now perpendicular distance between Eqs. 10 and 12 is equal to

$|\frac{MaxE_P(2l')}{q_2}+\frac{MaxE_P(2l)}{q_1}|\le 2\max (\frac{MaxE_P(2l)}{q_1}, \frac{MaxE_P(2l')}{q_2})$.

So, $D(10,12)=\frac{MaxE_P(2l)}{q_1}+\frac{MaxE_P(2l')}{q_2}\le 2\max (\frac{MaxE_P(2l)}{q_1}, \frac{MaxE_P(2l')}{q_2})$.

Let it be the smallest distance among those six possible distances. If we choose Eqs. 10 or 12, we are choosing lines within $\max (\frac{MaxE_P(2l)}{q_1}, \frac{MaxE_P(2l')}{q_2})$ error, since $D(9,10)=\frac{MaxE_P(2l)}{q_1}$ as well as $D(9,12)=\frac{MaxE_P(2l')}{q_2}$. Equations 10 and 12 make sure that there is at least one pair of lines whose perpendicular distance is less than or equal to $[2\max (\frac{MaxE_P(2l)}{q_1}, \frac{MaxE_P(2l')}{q_2})]$.

Let D(10, 11) be least. We have to check D(9, 11) only.

$D(9,11)=2\frac{d}{k}-2b+\frac{MaxE_P(2l)}{q_1}\le 2\frac{d}{k}-2b+2\frac{MaxE_P(2l)}{q_1} =D(10,11)<\frac{MaxE_P(2l)}{q_1}+\frac{MaxE_P(2l')}{q_2}=D(10,12)$, implies, $D(9,11)=2\frac{d}{k}-2b+\frac{MaxE_P(2l)}{q_1} \le \frac{MaxE_P(2l')}{q_2}$.

Let D(10, 13) be least. We have to check D(9, 13) only.

$D(9,13)=2\frac{d}{k}-2d-\frac{MaxE_P(2l')}{q_2}\le 2\frac{d}{k}-2d+\frac{MaxE_P(2l)}{q_1}-\frac{MaxE_P(2l')}{q_2} = D(10,13) < \frac{MaxE_P(2l)}{q_1}+\frac{MaxE_P(2l')}{q_2}=D(10,12)$,

implies, $D(9,13)=2\frac{d}{k}-2d-\frac{MaxE_P(2l')}{q_2}\le \frac{MaxE_P(2l')}{q_2}$.

$D(11,12)=2\frac{d}{k}-2b+\frac{MaxE_P(2l)}{q_1}+\frac{MaxE_P(2l')}{q_2}>2\frac{d}{k}-2d+\frac{MaxE_P(2l)}{q_1}+\frac{MaxE_P(2l')}{q_2}= D(10,13)$, since $d>b$.

So, it cannot be the least.

$D(11,13)=2d-2b+\frac{MaxE_P(2l)}{q_1}-\frac{MaxE_P(2l')}{q_2}\ge 2\epsilon$, since $d\ge (b+\epsilon )$.

So it cannot be the least.

Let D(12, 13) be least. We have to check D(10, 13) only.

$D(10,13)=2\frac{d}{k}-2d+\frac{MaxE_P(2l')}{q_2} \le 2\frac{d}{k}-2d+2\frac{MaxE_P(2l')}{q_2} = D(12,13) < D(10,12)=\frac{MaxE_P(2l)}{q_1}+\frac{MaxE_P(2l')}{q_2}$, implies, $D(9,13)=2\frac{d}{k}-2d+\frac{MaxE_P(2l')}{q_2} \le \frac{MaxE_P(2l)}{q_1}$.

Hence in this case also, if D(i, j) is the least and we choose any one of the $i-$th or $j-$th equation, then we are choosing a line of movement which is within $\max (\frac{MaxE_P(2l)}{q_1}, \frac{MaxE_P(2l')}{q_2})$ perpendicular error with the actual line of movement.

Without loss of generality, let $i-$th equation is chosen. Now $i-$th equation is generated from first(second) communication circle. Then according to Lemma 3, if sensor localizes itself corresponding to the last beacon received from the first(second) communication circle, then the error will be less than equal to $\epsilon$. $\square$

Proof of Lemma 4 is given below.

Proof

Sensor localizes itself using the theorem 1 within error bound $\epsilon$. There is a possibility of reducing error by updating the line of movement whenever it passes through communication circle of some anchor as explained below. Suppose a sensor computed it’s approximate line of movement $y=mx+c$. Currently the sensor is passing through the communication circle of an anchor $S''$, where the actual line of movement is $P_1P_2$ and the approximate line of movement is $P'_1P'_2$ as shown in Fig. 24, whose equation is $y=mx+c$. Error can be reduced by finding a suitable approximate line of movement between $P'_1P'_2$ and $P_1P_2$. To do this, the sensor approaches as follows: The sensor computes length l (say) of $P'_1P'_2$ from the equations of the communication circle and the already computed line $y=mx+c$. The sensor also computes length L (say) of the chord along which it is moving using two beacon points C, $C'$ corresponding to the anchor $S''$. If it finds $L>l$, error usually reduces by updating the approximate line of movement $P'_1P'_2$ to $y=mx+c'$, which is equation of the chord of length L, where $c'=\left[ ma-b\pm \frac{\sqrt{(4r^2-l^2)(m^2+1)}}{2}\right]$ and (a, b) is the position of $S''$. Among these two lines $y=mx+c'$ (for different values of $c'$), which is closer to $P'_1P'_2$ is chosen as the new approximate line of movement. According to Fig. 24, $P''_1P''_2$ is the updated line. If the sensor finds $L\le l$, no improvement in error is possible, hence it does not change the approximate line of movement. $\square$

Fig. 24

Sensor minimizes error by updating line of movement from $P'_1P'_2$ to $P''_1P''_2$