# A Novel Method to Determine the Optimal Location for a Cellular Tower by Using LiDAR Data
## 4 Results and Discussions
### Determination of the Strength of the Signal
#### 4.1.3 Relation between Signal Strength and Terrain Parameters  
For equation fitting, an area of the RGIPT campus has been taken. Signal strength data of 1500 points are observed by \"Network cell info lite\". According to the study, distance attenuation is a factor of the logarithm of distance, and the barrier attenuation is dependent on path difference; that is, the difference between the direct path and indirect path between the source and receiver. A signal is calculated in terms of dBm and at the cell phone tower location, the signal strength is considered to be 0 (dB m). \"\(SR\)\" is denoted for signal received. \(SR_{Calculated}\) (user location) = Signal (cell phone tower) \(-\) Attenuation losses  
As per the analogy with propagation models, the attenuation losses are a function of distance and barrier attenuations. \[Attenuation\ Losses=(a\times log_{10}d)+B\cdot A \tag{7}\]  
Signal (cell phone tower) = 0 dB m  
\[SR_{Calculated}=-((a\times log_{10}d)+B\cdot A) \tag{8}\]  
\[SR_{Observed}=-((a\times log_{10}d)+B\cdot A+Error) \tag{9}\]  
Training data is 1000 points out of 1500 points. A total of 500 points are for testing purposes. For each point, _SRObserved_ (dB m) is captured from \"Network cell info lite\". For point 1, the equation is  
\[SR_{Observed1}=-((a\times log_{10}(d_{1}))+B\cdot A_{1}+Error) \tag{10}\]  
For point 2, the equation is  
\[SR_{Observed2}=-((a\times log_{10}(d_{2}))+B\cdot A_{2}+Error) \tag{11}\]  
For point 3, the equation is  
\[SR_{Observed3}=-((a\times log_{10}(d_{3}))+B\cdot A_{3}+Error) \tag{12}\]  
and so on, for point 1000, the equation is  
\[SR_{Observed1000}=-((a\times log_{10}(d_{1000}))+B\cdot A_{1000}+Error) \tag{13}\]  
where \(d_{1}\), \(d_{2}\), \(d_{3}\)... \(d_{1000}\) are the direct distance in meters from the source tower to the receiver location. \(B\cdot A_{1}\), \(B\cdot A_{2}\), \(B\cdot A_{3}\)..., \(B\cdot A_{1000}\) are the barrier attenuation (dB) that occurred due to buildings in between the source tower and receiver. We can write the above equations as  
\[\mathrm{Y}=M\mathrm{X}+C \tag{14}\]  
where \(\mathrm{Y}=SR\) Observed (user location) + \(B\cdot A\)  
\[\mathrm{X}=-log_{10}(d) \tag{15}\]  
\[C=-Error \tag{16}\]  
\(\mathrm{Y}\), \(\mathrm{X}\) are known values, and \(M\), \(C\) are unknowns. All the equations from 10 to 13 are now written as:  
For point 1,  
\[SR_{Calculated}=-((a\times log_{10}d)+B\cdot A) \tag{8}\]0  
For point 2,  
\[SR_{Calculated}=-((a\times log_{10}d)+B\cdot A) \tag{8}\]1For point 3,  
\[SR_{Calculated}=-((a\times log_{10}d)+B\cdot A) \tag{8}\]2  
...  
For point 1000,  
\[SR_{Calculated}=-((a\times log_{10}d)+B\cdot A) \tag{8}\]3  
To find the best-fit line, we try to solve the above equations (from 17 to 20) with the unknowns \(M\) and \(C\). As the 1000 points do not lie on a line, there is no actual solution, so instead, we compute a least-squares solution putting our linear equation into matrix form. Computing:  
\[SR_{Calculated}=-((a\times log_{10}d)+B\cdot A) \tag{8}\]4  
\[A=\left( \begin{array}{cc}\mathrm{X}_{1}&1\\ \mathrm{X}_{2}&1\\ \mathrm{X}_{3}&1\\ \vdots&\vdots\\ \mathrm{X}_{1000}&1\end{array} \right),\;B=\left( \begin{array}{c}\mathrm{Y }_{1}\\ \mathrm{Y}_{2}\\ \mathrm{Y}_{3}\\ \vdots\\ \mathrm{Y}_{1000}\end{array} \right),\;\mathrm{and}\;x=\left( \begin{array}{ ccc}M\\ C\end{array} \right)A^{T}=\left( \begin{array}{cccccc}\mathrm{X}_{1}&\mathrm{X}_{2}& \mathrm{X}_{3}&\cdots&\mathrm{X}_{1000}\\ 1&1&1&\cdots&1\end{array} \right) \tag{22}\]  
where \(A^{T}\) is the transpose of \(A\). Multiplying \(A^{T}\) to both sides of Equation (21), the equation becomes:  
\[A^{T}\;A\cdot x=A^{T}\cdot b \tag{23}\]  
\[\left( \begin{array}{ccccc}\mathrm{X}_{1}&\mathrm{X}_{2}&\mathrm{X}_{3}& \cdots&\mathrm{X}_{1000}\\ 1&1&\cdots&1\end{array} \right)\times\left( \begin{array}{c}\mathrm{X}_{1}&1 \\ \mathrm{X}_{2}&1\\ \mathrm{X}_{3}&1\\ \vdots&\vdots\\ \mathrm{X}_{1000}&1\end{array} \right)\times\left( \begin{array}{c}M\\ C\end{array} \right)=\left(