## 3 Basics of Acoustic Target Tracking
### A New Proposed Method for Solving Simultaneous Equations  
The simultaneous equations of target tracking in Equation (2) are degree two and by differencing these equations, as in Equation (3), we can eliminate the degree two factors. This approach is similar to the mathematical solution for the lateration problem in sensor nodes' localization but with major differences [28]. In the lateration problem, the distance of localizing more from anchor points are known, but in target tracking the distances of target from sensing notes are unknown. The number of unknown variables in target tracking is three but in localization it is two [28]:  
\[\begin{cases}(x-x_{1})^{2}+(y-y_{1})^{2}-(x-x_{2})^{2}-(y-y_{2})^{2}=((t_{1}-t )^{2}-(t_{2}-t)^{2})v^{2}\\ (x-x_{1})^{2}+(y-y_{1})^{2}-(x-x_{3})^{2}-(y-y_{3})^{2}=((t_{1}-t)^{2}-(t_{3}- t)^{2})v^{2}\end{cases} \tag{3}\]  
Equation (3) is simplified to:  
\[\begin{cases}2(x_{2}-x_{1})x+2(y_{2}-y_{1})y=2(t_{2}-t_{1})tv^{2}+({t_{1}}^{2 }-{t_{2}}^{2})v^{2}+({x_{2}}^{2}-{x_{1}}^{2})+({y_{2}}^{2}-{y_{1}}^{2})\\ 2(x_{3}-x_{1})x+2(y_{3}-y_{1})y=2(t_{3}-t_{1})tv^{2}+({t_{1}}^{2}-{t_{3}}^{2}) v^{2}+({x_{3}}^{2}-{x_{1}}^{2})+({y_{3}}^{2}-{y_{1}}^{2})\end{cases} \tag{4}\]  
Equation (4) is converted into the matrix form as follows:  
\[\begin{bmatrix}2(x_{2}-x_{1})&2(y_{2}-y_{1})\\ 2(x_{3}-x_{1})&2(y_{3}-y_{1})\end{bmatrix}\begin{bmatrix}x\\ y\end{bmatrix} \tag{5}\] \[=\begin{bmatrix}2(t_{2}-t_{1})v^{2}\\ 2(t_{3}-t_{1})v^{2}\end{bmatrix}[t]+\begin{bmatrix}({t_{1}}^{2}-{t_{2}}^{2})v^{ 2}+({x_{2}}^{2}-{x_{1}}^{2})+({y_{2}}^{2}-{y_{1}}^{2})\\ ({t_{1}}^{2}-{t_{3}}^{2})v^{2}+({x_{3}}^{2}-{x_{1}}^{2})+({y_{3}}^{2}-{y_{1}}^{ 2})\end{bmatrix}\]  
Equation (5) is formulated as follows:  
\[m\begin{bmatrix}x\\ y\end{bmatrix}=b[t]+c \tag{6}\]  
Equation (6) is solved by using the inverse matrix of \(m\) as follows [29]:  
\[m^{-1}m\begin{bmatrix}x\\ y\end{bmatrix}=m^{-1}b[t]+m^{-1}c \tag{7}\]  
Equation (7) is further simplified into Equation (8):  
\[\begin{bmatrix}x\\ y\end{bmatrix}=\begin{bmatrix}mib(0)\\ mib(1)\end{bmatrix}[t]+\begin{bmatrix}mic(0)\\ mic(1)\end{bmatrix} \tag{8}\]There are only two simultaneous equations in Equation (8), but three unknown variables \(x\), \(y\) and \(t\), so these simultaneous equations have unlimited answers. We need extra equations, so we use the first equation of simultaneous equations of Equation (2) as follows:  
\[(x-x_{1})^{2}+(y-y_{1})^{2}=(t_{1}-t)^{2}v^{2} \tag{9}\]  
We rewrite Equation (9) in the following form:  
\[(x-x_{1})^{2}+(y-y_{1})^{2}-(t_{1}-t)^{2}v^{2}=0 \tag{10}\]  
We represent unknown variables \(x\) and \(y\) in Equation (8) based on the unknown variable \(t\) and then substitute them in Equation (10) as follows:  
\[(mib(0)t+mic(0)-x_{1})^{2}+(mib(1)t+mic(1)-y_{1})^{2}-(t_{1}-t)^{2}v^{2}=0 \tag{11}\]  
Factorizing Equation (11) with respect to \(t\), yields a degree two equation as in Equation (12) that can be solved with delta rule:  
\[\begin{split}(mib(0)^{2}+& mib(1)^{2}-v^{2})t^{2}\\ &+~{}(2 mib(0)(mic(0)-x_{1})+2 mib(1)(mic(1)-y_{1})+2v^{2}t_{1})t\\ &+(mic(0)-x_{1})^{2}+(mic(1)-y_{1})-{t_{1}}^{2}v^{2}=0\end{split} \tag{12}\]  
Equation (12) can be formulated as follows:  
\[at^{2}+bt+c=0 \tag{13}\]  
The inherent structure of the problem causes the delta of Equation (13) never to become negative. We summarize the result as follows:  
\[\begin{split} t=&\frac{-b+\sqrt{\Delta}}{2a}~{}, \quad\quad x=mib(0)*t+mic(0),y=mib(1)*t+mic(1)~{},\\ t=&\frac{-b-\sqrt{\Delta}}{2a}~{},\quad x=mib(0)*t +mic(0),y=mib(1)*t+mic(1)\end{split} \tag{14}\]  
Equation (13) gives two different values for variable \(t\) when delta is greater than zero. Values of \(x\) and \(y\) variables are computed by replacing the computed value of \(t\) variable in Equation (8).