## 4 Network Interference Probability Analysis
### Heterogeneous Node Deployment Probability Analysis
#### 4.2.1 The Node Distance is \(r/2\leq d\leq r\)  
**Theorem 4. **_In the heterogeneous WSNs, if the distance between node \(u\) and node \(v\) is smaller than \(r\) and larger than \(r/2\), the relationship between the overlapping coverage area \(s\) and the node distance \(d\) is approximately \(s=-1.57\cdot d^{2}+3.23\cdot rd-0.44\cdot r^{2}\)._  
Figure 10: Different node positions in heterogeneous networks: (**a**) \(r/2\leq d\leq r\); (**b**) \(0\leq d\leq r/2\). **Proof. ** As shown in Figure 10a, when the transmission ranges of two nodes are different, the overlapping coverage area can be calculated as:  
\[s=r^{2}\arccos\frac{r}{2d}+2d^{2}\arcsin\frac{r}{2d}-r\sqrt{d^{2}-\left(\frac{r}{ 2}\right)^{2}}, \tag{22}\]  
where \(r\) is the transmission range of node \(u\), and \(d\) is the distance between two nodes. Based on the Taylor Formula, the Taylor Expansion of Equation (22) is:  
\[s_{t}=\frac{\pi}{2}d^{2}-\frac{d}{2r}+o[(\frac{d}{2r})^{3}], \tag{23}\]  
where \(\frac{d}{2r}\in(\frac{1}{4},\frac{1}{2})\); since the \(o[(\frac{d}{2r})^{3}]\) is small, so the relationship between the overlapping coverage area \(s\) and the node distance \(d\) is approximately quadratic. For simplifying the calculation of Equation (22), we assume:  
\[f=ad^{2}+brd+cr^{2}, \tag{24}\]  
Additionally, we define the difference between \(s\) and \(f\) as:  
\[y =s-f \tag{25}\] \[=r^{2}\arccos\frac{r}{2d}+2d^{2}\arcsin\frac{r}{2d}-r\sqrt{d^{2}- \left(\frac{r}{2}\right)^{2}}-\left(ad^{2}+brd+cr^{2}\right)^{\prime}.\]  
The variance of the difference \(y\) is:  
\[D(y)=E(y^{2})-\left[E(y)\right]^{2}, \tag{26}\]  
\[D(y)\left|{}_{a}^{\prime}=E(y^{2})\left|{}_{a}^{\prime}-\left[E(y)\right]^{2 }\left|{}_{a}^{\prime},\right.\right. \tag{27}\]  
In Equation (27), according to the values \((r/2,s_{1})\) and \((r,s_{2})\) (which can be got from Equation (22)), we can confirm the relationship between \(a\), \(b\) and \(c\). The values of \(s_{1}\) and \(s_{2}\) are:  
\[\left\{ \begin{array}{l}s_{1}=\pi d^{2}\\ s_{2}=(\frac{\pi}{3}+\frac{\sqrt{3}}{2})r^{2}+\frac{\pi}{3}d^{2}\end{array}  \right.^{\prime}. \tag{28}\]  
In Equation (27), when \(-\infty<a<-1.57\), \(D(y)\left|{}_{a}^{\prime}<0\right.\); when \(-1.57<a<\infty\), \(D(y)\left|{}_{a}^{\prime}>0\right.\). Therefore, when \(a=-1.57\), the variance of \(y\) is minimum. This is also can be found in Figure 11. Thus, Equation (24) can be rewritten as:  
\[f=-1.57\cdot d^{2}+3.23\cdot rd-0.44\cdot r^{2}, \tag{29}\]  
For investigating the evaluation error between \(s\) and \(f\), we define the evaluation error is:  
\[e=\frac{y}{s}=1-\frac{f}{s}, \tag{30}\]  
The evaluation error between \(s\) and \(f\) is shown in Figure 12. In Figure 12, we can conclude that the maximum evaluation error between \(s\) and \(f\) is only 0.025 and the mean absolute value is only 0.0139. This means that the error between \(s\) and \(f\) can be ignored during the calculation of the overlapping coverage area in heterogeneous networks, i.e., the Equation (29) can be used to calculate the overlapping coverage area approximately. The results of \(s\) and \(f\) are shown in Figure 13. In Figure 13, the curve of \(s\) and\(f\) can be fitted in small error. Thus, in the calculation of the overlapping coverage area in heterogeneous WSNs, the Equation (22) can approximately be replaced by the quadratic function. Figure 11: The variance of difference between \(s\) and \(f\) under different value of \(a\). Figure 12: The evaluation error between \(s\) and \(f\). Figure 13: The fitting results of Equations (22) and (29).