# Prediction of RF-EMF Exposure by Outdoor Drive Test Measurements
## 4 Results
### ANN-Based Prediction of RF-EMF Exposure  
The proposed models (a) and (b) have, respectively, \(N+2\) and \(4N+2\) inputs. The number of outputs for both models (a) and (b) is one, e.g., total Eq900. Both ANN models were \begin{table}
\begin{tabular}{l c c c c c c c} \hline \hline  & **1 st BS** & **2 nd BS** & **3 rd BS** & **4 th BS** & **5 th BS** & **6 th BS** & **7 th BS** \\ \hline Mean Distance (m) & 80.693 & 127.296 & 163.649 & 192.726 & 219.452 & 242.019 & 262.701 \\ \hline \hline \end{tabular}
\end{table} 
Table 1: Mean distance between each measurement point and neighboring BS. Figure 6: (**Left**): The variation of the total Eq900 along the drive path. (**Right**): CDF of total Eq900 and Eq900 from each operator. tuned using the grid-search method with a 5-fold cross-validation. The hyper-parameters used for the two models in the current paper are shown in Table 2. The exponential decay learning rate is adopted with \(k=0.001\) when training the ANN model. The neural network is constructed with the Python interface Keras in the TensorFlow environment. The PC is equipped with GPU from NVIDIA Quadro RTX 8000. After the ANN model is trained, the model is then applied to the test data with 10-fold averaging to obtain performance metrics. The obtained metrics, i.e., mean square error (MSE) and \(R^{2}\), are shown in Table 3. Here, \(R^{2}\) measures how close the two distributions are. A higher \(R^{2}\) indicates a better prediction performance. The results of both models are shown in Table 3. In the spatial reconstruction of total Eq900, we compared the prediction performance by considering different values of the \(N\) nearest BSs with \(N=\{3,5,7\}\). For model (a), there is a significant improvement comparing \(N=5\) to \(N=3\). While when \(N=7\), there is a slight improvement in prediction quality in terms of \(R^{2}\). When \(N\) increases, the complexity of the model and the data pre-processing time increase, as well as the time of the model training. In model (b), total Eq900 E-field is predicted by considering \(N\) nearest BSs from each operator, as shown in Table 3. Increasing \(N\) from three to seven yields a limited improvement in the prediction performance but increases the computation time and the model complexity. Consequently, \(N=3\) is the adequate choice for this model. Nonetheless, if we compare horizontally in Table 3, we observe a clear improved prediction accuracy under the same value of \(N\). This could be explained by the higher correlation between E-fields from the same operator, as shown in Figure 4.  
From Figures 7 and 8, we observe that predictions (figure on the right) from the ANN model can provide a good reconstruction performance with ground truth (figure on the left), which agrees with the metrics in Table 3. The proposed ANN model is able to predict the total Eq900 E-field level with MSE = 0.000805 and \(R^{2}=0.805\) for \(N=3\). From the \begin{table}
\begin{tabular}{c c c} \hline \hline
**Hyper-Parameters** & **Model (a)** & **Model (b)** \\ \hline Number of layers & 5 & 3 (LC) + 3 (FC) \\ \hline Number of neurons (Hidden layers) & 40 & 3\(N\) (LC) + 50 (FC) \\ \hline Optimizer & \multicolumn{2}{c}{Adamax} \\ \hline Activation function & \multicolumn{2}{c}{Relu} \\ \hline Learning rate \(l_{ini}\) & 0.01 & 0.005 \\ \hline Learning Rate decay & \(l=l_{ini}\exp^{-kEpoch}\) \\ \hline Number of epoch & \multicolumn{2}{c}{100} \\ \hline Train:Validation:Test & 0.49:0.21:0.3 \\ \hline Loss function & \multicolumn{2}{c}{mean squared error} \\ \hline \hline \end{tabular}
\end{table} 
Table 2: Hyper-parameters of the proposed ANN model. \begin{table}
\begin{tabular}{c c c c c} \hline \hline \multirow{2}{*}{**N**} & \multicolumn{2}{c}{**Model (a)**} & \multicolumn{2}{c}{**Model (b)**} \\ \cline{2-5}  & \(\mathbf{R^{2}}\) & **MSE** & \(\mathbf{R^{2}}\) & **MSE** \\ \hline
3 & 0.66522 & 0.00148 & 0.805 & 0.000805 \\ \hline
5 & 0.74810 & 0.00111 & 0.806985 & 0.00079 \\ \hline
7 & 0.75016 & 0.0011 & 0.813376 & 0.000764 \\ \hline \hline \end{tabular}
\end{table}