Optimality analysis of range sensor placement under constrained deployment region

Abstract

Source localization is a critical issue in various wireless sensor network applications. However, communication and concealment constraints often restrict sensor placement, resulting in non-arbitrary sensor deployment regions. To further enhance localization accuracy, this paper presents an optimality analysis of range sensor placement under constrained deployment regions, focusing on optimal geometries rather than specific localization algorithms. The optimality analysis is formulated as a constrained optimisation problem that maximizes the determinant of the Fisher information matrix, also known as D-optimality, while taking into account the constraints imposed by the deployment region. To simplify the analysis, we introduce the concepts of maximum feasible angle and separation angle, which are used to express the objective function and constraints in equivalent forms. By comparing the maximum feasible angle with the optimal separation angles in unconstrained cases, our method will be applicable to both circular constrained regions and general irregular regions. The conclusions we have reached are comprehensive and intuitive, and they differ significantly from the conventional uniform angular geometry. The proposed range sensor-source geometries are verified through theoretical analysis and simulations.

Appendices

Appendix A

In the case of $\varphi <\pi /3$, an equivalent problem to the optimality criterion (21) is

$$\begin{aligned} \begin{aligned}&\max \,f=\sigma ^{-4}\left[ \sin ^2 \theta _{13}+\sin ^2 \theta _{12}+\sin ^2 \left( \theta _{13}-\theta _{12} \right) \right] \\&\text {s.t.}\left\{ \begin{array}{l} \theta _{12},\theta _{13},\theta _{13}-\theta _{12}\in \left[ 0,2\varphi \right] \\ \varphi <\pi /3\\ \end{array} \right. \end{aligned} \end{aligned}$$

(24)

Taking derivative of objective function in (24) with respect to $\theta _{13}$ and $\theta _{12}$, respectively, yields

$$\begin{aligned}{} & {} \frac{\partial f}{\partial \theta _{13}}=2\sigma ^{-4}\sin \left( 2\theta _{13}-\theta _{12} \right) \cos \theta _{12} \\{} & {} \frac{\partial f}{\partial \theta _{12}}=2\sigma ^{-4}\cos \theta _{13} \sin \left( 2\theta _{12}-\theta _{13} \right) \end{aligned}$$

$a)\,\varphi <\pi /4$

In the case of $\varphi <\pi /4$, we can easily obtain

$$\begin{aligned} \begin{aligned}&\theta _{12}\le 2\varphi<\pi /2\\&2\theta _{13}-\theta _{12}\le 2\theta _{13}\le 4\varphi <\pi \end{aligned} \end{aligned}$$

then, $\cos \theta _{12}>0$, $\sin \left( 2\theta _{13}-\theta _{12}\right) >0$. Obviously, $\frac{\partial f}{\partial \theta _{13}}>0$.

According to $\theta _{13}\le 2\varphi$, we know the objective function gets its maximum at $\theta _{13}=2\varphi$. Then the optimal geometry is that any two sensors are located at the two tangent points, respectively. Therefore, we can respectively write

$$\begin{aligned} \begin{aligned} f&=\sigma ^{-4}\left[ \sin ^2 2\varphi +\sin ^2\theta _{12}+\sin ^2\left( 2\varphi -\theta _{12}\right) \right] \\ \frac{\partial f}{\partial \theta _{12}}&= 2\sigma ^{-4}\cos 2\varphi \sin \left( 2\theta _{12}-2\varphi \right) \end{aligned} \end{aligned}$$

Applying $2\varphi <\pi /2$ yields $\cos 2\varphi >0$. Then f achieves its minimum at $\theta _{12}=\varphi$, and its maximum is obtained at $\theta _{12}=0$ or $2\varphi$, it is easy to verify that the maximum is the same, this indicates that the third sensor should be placed at any tangent point, as shown in Fig. 5(a).

The maximum for $\varphi <\pi /4$ is $f= 2\sigma ^{-4}\sin ^2 2\varphi$.

$b)\,\pi /4<\varphi <\pi /3$

In the case of $\pi /4<\varphi <\pi /3$, we have $\theta _{13}\le 2\varphi <2\pi /3$.

If $\theta _{13}\le \pi /2$, then $\cos \theta _{13}>0$, the objective function f achieves its minimum at $\theta _{12}=\theta _{13}/2$, and gets the same maximum $2\sigma ^{-4}\sin ^2\theta _{13}$ at $\theta _{12}=0$ or $\theta _{13}$, the maximum is $f= 2\sigma ^{-4}$.

That is, when $\theta _{13}\le \pi /2$, the optimality criterion gets the maximum at $\theta _{13}=\pi /2$, $\theta _{12}=0$ or $\pi /2$, the maximum is $2\sigma ^{-4}$.

If $\pi /2<\theta _{13}<2\pi /3$, then $\cos \theta _{13}<0$, the objective function f achieves its maximum at $\theta _{12}=\theta _{13}/2$. Then

$$\begin{aligned} \frac{\partial f}{\partial \theta _{13}}=2\sigma ^{-4}\left[ \sin (3\theta _{13}/2) \cos (\theta _{13}/2)+\sin \theta _{13}\right] \end{aligned}$$

It follows from $\pi /2<\theta _{13}<2\pi /3$ that

$$\begin{aligned} \sin (3\theta _{13}/2)>0,\cos (\theta _{13}/2)>0,\sin \theta _{13}>0 \end{aligned}$$

That is, when $\pi /2<\theta _{13}<2\pi /3$, the optimality criterion gets the maximum at $\theta _{13}=2\varphi$, $\theta _{12}=\theta _{13}/2=\varphi$, the maximum is $\sigma ^{-4}\left[ \sin ^2 2\varphi +2\sin ^2\varphi \right]$.

We have to compare these two maximums to get the maximum when $\pi /4<\varphi <\pi /3$.

Denote $g=\sigma ^{-4}\left[ \sin ^2 2\varphi +2\sin ^2\varphi \right] -2\sigma ^{-4}$, then

$$\begin{aligned} \frac{\partial g}{\partial \varphi }=\sigma ^{-4}\left[ \sin 4\varphi +2\sin 2\varphi \right] >0 \end{aligned}$$

and $g>g(\pi /4)=0$.

The maximum for $\pi /4<\varphi <\pi /3$ is

$$\begin{aligned} \sigma ^{-4}\left[ \sin ^2 2\varphi +2\sin ^2\varphi \right] \end{aligned}$$

The optimal geometry is that any two sensors are placed at two tangent points, respectively, the third sensor is placed at line SO, as shown in Fig. 5(b).

$c)\,\varphi =\pi /4$

In the case of $\varphi =\pi /4$, we have

$$\begin{aligned} \begin{aligned}&2\sigma ^{-4}\sin ^2 2\varphi =2\sigma ^{-4}\\&\sigma ^{-4}\left[ \sin ^2 2\varphi +2\sin ^2\varphi \right] =2\sigma ^{-4} \end{aligned} \end{aligned}$$

Then the optimal geometries in Fig. 5(a) and (b) are equivalent.

Appendix B

As described in Sect. 5, the n sensors, $s_1,s_2,\ldots ,s_n$, are divided into n/2 or $(n+1)/2$ groups according to whether n is even or odd.

$$\begin{aligned} {\mathcal {G}}_j = \left\{ {\begin{array}{ll} \left\{ {{s_{2j - 1}},{s_{2j}}} \right\} 1 \le j \le n/2 &\quad\text{n is even}\\ {\left. {\begin{array}{l} {\left\{ {{s_{2j - 1}},{s_{2j}}} \right\} }{1 \le j \le (n - 1)/2}\\ {\left\{ {{s_n}} \right\} }\qquad{j = (n + 1)/2} \end{array}} \right\} }&\quad\text{n is odd} \end{array}} \right. \end{aligned}$$

a) n is even

If every group of ${\mathcal {G}}_1,{\mathcal {G}}_2,\ldots ,{\mathcal {G}}_{n/2}$ is placed as $C_2$ in Table 1, the separation angle $\theta _{2j-1,2j}=\pi /2$, and ${\varvec{v}}_{2j-1}+{\varvec{v}}_{2j}={\varvec{0}}$ holds for every group.

As a result,

$$\begin{aligned} \sum _{i=1}^n{{\varvec{v}}_i}=\sum _{j=1}^{n/2}\sum _{i\in {\mathcal {G}}_j}{{\varvec{v}}_i}={\varvec{0}} \end{aligned}$$

(25)

it is shown in (25) that the geometry is optimal.

b) n is odd

We can conclude from Sect. 4.2 that $\sum _{i=1}^3{{\varvec{v}}_i}\ne {\varvec{0}}$ for any three sensors. For the objective function (8), we can get the following inequality after some calculations [29].

$$\begin{aligned} \begin{aligned} \det \left( {\textbf{F}} \right)&\le \cos ^2\theta _n\sum _{i=1}^{n-1}{\sin ^2\theta _i}+\sin ^2\theta _n\sum _{i=1}^{n-1}{\cos ^2\theta _i}\\&\quad +\left( \sum _{i=1}^{n-1}{\cos ^2\theta _i} \right) \left( \sum _{i=1}^{n-1}{\sin ^2\theta _i} \right) \end{aligned} \end{aligned}$$

(26)

and the equality holds when $\sum _{i=1}^{n-1}{\sin 2\theta _i}=0$.

It is worth noting that if any two sensors $s_{2j-1}$ and $s_{2j}$ are placed symmetrically about the x-axis, we have $\theta _{2j-1}=-\theta _{2j}$, and $\sin 2\theta _{2j-1}+\sin 2\theta _{2j}=0$, this means that if we place the two sensors in every group of ${\mathcal {G}}_1,{\mathcal {G}}_2,\ldots ,{\mathcal {G}}_{(n-1)/2}$ in this way, $\sum _{i=1}^{n-1}{\sin 2\theta _i}=0$ can be guaranteed.

In this case, the modulus and azimuth angle of $\varvec{v'}=\sum _{i=1}^{n-1}{{\varvec{v}}_i}$ are

$$\begin{aligned} \left\| \varvec{v'} \right\| \in {\left\{ \begin{array}{ll} \left[ 0,-\left( n-1 \right) \cos 2\varphi \right] &{} \angle \varvec{v'}=0\\ \left[ 0,n-1 \right] &{} \angle \varvec{v'}=\pi \\ \end{array}\right. } \end{aligned}$$

(27)

By the definition of ${\varvec{v}}_i$, we know that

$$\begin{aligned} \begin{aligned}&\left\| {\varvec{v}}_n \right\| =1\\&\angle {\varvec{v}}_n \in \left[ \pi -2\varphi ,\pi \right] \,\text {or}\,\left( -\pi , 2\varphi -\pi \right] \end{aligned} \end{aligned}$$

(28)

It can be proved that the minimum is achieved when $\varvec{v'}$ and ${\varvec{v}}_n$ are in the opposite direction, i.e., $\angle \varvec{v'}=0$, $\angle {\varvec{v}}_n=\pi$, then $s_n$ should be placed at line SO, the positions of the sensors are determined by comparing 1 with $-\left( n-1 \right) \cos 2\varphi$.

Let $-\left( n-1 \right) \cos 2\varphi =1$, then

$$\begin{aligned} \varphi = \frac{1}{2}\text{arc}\cos \left( -\frac{1}{n-1} \right) \end{aligned}$$

Hence, if $\pi /4<\varphi \le \frac{1}{2}\text{arc}\cos \left( -\frac{1}{n-1} \right)$, we have $-\left( n-1 \right) \cos 2\varphi <1$, that is

$$\begin{aligned} \left\| {\varvec{v}}_n+\varvec{v'}\right\| =\left\| {\varvec{v}}_n\right\| -\left\| \varvec{v'}\right\| \ne {\varvec{0}} \end{aligned}$$

(29)

To achieve the minimum, $\left\| \varvec{v'}\right\|$ should be equal to $-\left( n-1 \right) \cos 2\varphi$, then the two sensors in ${\mathcal {G}}_1,{\mathcal {G}}_2,\ldots ,{\mathcal {G}}_{(n-1)/2}$ should be placed as $C_1$ in Table 1, and the minimum is

$$\begin{aligned} \left\| {\varvec{v}}_n+\varvec{v'}\right\| _{\min }=1+\left( n-1 \right) \cos 2\varphi \end{aligned}$$

If $\frac{1}{2}\text{arc}\cos \left( -\frac{1}{n-1} \right)<\varphi <\pi /3$, then $\varvec{v'}+{\varvec{v}}_n={\varvec{0}}$ can be achieved under certain conditions. We only give a sufficient but not necessary condition here. The two sensors in ${\mathcal {G}}_1,{\mathcal {G}}_2,\ldots ,{\mathcal {G}}_{(n-1)/2}$ can be placed with bearing angles $\varphi _0=\pm \frac{1}{2}\text{arc}\cos \left( -\frac{1}{n-1} \right)$, respectively. Straightforwardly, we can easily obtain $\left\| \varvec{v'} \right\| =1$, and

$$\begin{aligned} \left\| {\varvec{v}}_n+\varvec{v'}\right\| =\left\| {\varvec{v}}_n\right\| -\left\| \varvec{v'}\right\| = {\varvec{0}} \end{aligned}$$

(30)

It can be seen from (30) that the geometry is optimal, and the minimum is $\left\| {\varvec{v}}_n+\varvec{v'}\right\| _{\min }=0$.

Appendix C

a) n is even

Similar to (26), we have the following inequality,

$$\begin{aligned} \det \left( {\textbf{F}} \right) \leqslant \sum _{i=1}^n{\cos ^2\theta _i}\sum _{i=1}^n{\sin ^2\theta _i} \end{aligned}$$

(31)

the equality holds when $\sum _{i=1}^n{\sin 2\theta _i}=0$.

We can verify that if the two sensors in ${\mathcal {G}}_1,{\mathcal {G}}_2,\ldots ,{\mathcal {G}}_{n/2}$ are placed symmetrically about the x-axis, then $\theta _{2j-1}=-\theta _{2j}$ and $\sum _{i=1}^n{\sin 2\theta _i}=0$.

Since $\varphi \le \pi /4$, the azimuth angle of ${\varvec{v}}_i$ is $2\theta _i \le 2\varphi \le \pi /2$, and the modulus and azimuth angle of $\varvec{v''}=\sum _{i=1}^n{{\varvec{v}}_i}$ are

$$\begin{aligned} \left\| \varvec{v''} \right\| \in \left[ n\cos 2\varphi ,n\right] ,\quad \angle \varvec{v''}= \pi \end{aligned}$$

(32)

We can conclude directly that the minimum can be achieved when the two sensors in every group are placed as $C_1$ in Table 1. The minimum is $\left\| \varvec{v''} \right\| _{\min } =n\cos 2\varphi$.

b) n is odd

Similarly, the two sensors in groups ${\mathcal {G}}_1,{\mathcal {G}}_2,\ldots ,{\mathcal {G}}_{(n-1)/2}$ should be placed symmetrically about the x-axis, and

$$\begin{aligned} \left\| \varvec{v'} \right\| \in \left[ (n-1)\cos 2\varphi ,n-1\right] ,\quad \angle \varvec{v'}= \pi \end{aligned}$$

Because $\varvec{v'}$ is independent of ${\varvec{v}}_n$, then $\left\| \varvec{v'} \right\|$ should be as small as possible, and its angle with ${\varvec{v}}_n$ should be as large as possible to achieve the minimum. At this time, $\left\| \varvec{v'} \right\| =(n-1)\cos 2\varphi$, $\angle {\varvec{v}}_n= \pm 2\varphi$. This indicates that the optimal geometry is the two sensors in ${\mathcal {G}}_1,{\mathcal {G}}_2,\ldots ,{\mathcal {G}}_{(n-1)/2}$ should be placed as $C_1$ in Table 1, the sensor in ${\mathcal {G}}_{(n+1)/2}$ is placed at any tangent point. The minimum is

$$\begin{aligned} \left\| {\varvec{v}}_n+\varvec{v'} \right\| _{\min } =\sqrt{1+(n^2-1)\cos ^2 {2\varphi }} \end{aligned}$$