## 2 Theoretical Approach
Figure 1: (**a**) Problem setup, and (**b**) expected range profile. Figure 2: Five-layer model of the cardboard box. Curved arrows represent reflection. Materials with \(\sigma>0\) will dissipate energy when the wave propagates. Each of the five stages contains a different amount of paper or quantity of lossy material inside the box, for a unique amount of energy dissipation at each stage. For low-lossy material, the propagation constant, \(\beta\), and the attenuation constant, \(\alpha\), are found using the Binomial expansion:  
\[\alpha=\frac{\omega e^{\omega}}{2}\sqrt{\frac{\mu}{\epsilon^{\prime}}}=\frac{ \sigma}{2}\sqrt{\frac{\mu}{\epsilon^{\prime}}},\ \beta=\omega\sqrt{\mu\epsilon^{\prime}}\cdot\left(1+\frac{1}{8}\left(\frac{ \epsilon^{\prime\prime}}{\epsilon^{\prime}}\right)^{2}\right) \tag{3}\]  
As we have all the material parameters, we can calculate the magnitude of two peaks. The magnitude of the first peak, \(A_{1}\), is calculated by summing the two echoes. The first echo is from the boundary between air (1) and the cardboard box (2), and the second echo is from the boundary between the cardboard box (2) and the material inside (3). These two echoes overlap in the range profile because the thickness of the cardboard box (1 cm) is less than the range resolution (5.02 cm) of the radar. It should be noted that the phase delay and attenuation factor are considered in the equation:  
\[A_{1}=M\cdot\Gamma_{12}+M\cdot T_{12}\cdot\Gamma_{23}\cdot T_{21}\cdot e^{-j \beta_{box}2d_{1}}\cdot e^{-\alpha_{box}2d_{1}} \tag{4}\]  
where \(M\) is the magnitude of the incident wave, \(d_{1}\) is the thickness of the box, and \(d_{2}\) is the length of the box inside. The magnitude, \(A_{2}\), of the second peak in the range profile comes from the third boundary between the material inside (3) and the cardboard box (4), as well as the fourth boundary between the cardboard box (4) and air (5). Again, because the thickness of the cardboard is less than the range resolution of the radar, we observe only one peak from the two reflections. \(A_{2}\) can be calculated as:  
\[ \begin{split}& A_{2}=M\cdot T_{12}\cdot T_{23}\cdot\Gamma_{34} \cdot T_{32}\cdot T_{21}\cdot e^{-j\beta_{box}2d_{1}}\cdot e^{-\beta_{box}2d_ {1}}\cdot e^{-j\beta_{m}2d_{2}}\cdot e^{-\alpha_{m}2d_{2}}\\ &+M\cdot T_{12}\cdot T_{23}\cdot T_{34}\cdot T_{45}\cdot T_{43} \cdot T_{32}\cdot T_{21}\cdot e^{-j\beta_{box}4d_{1}}\cdot e^{-\alpha_{box}4d _{1}}\cdot e^{-\beta_{m}2d_{2}}\cdot e^{-\alpha_{m}2d_{2}}\end{split}  \tag{5}\]  
where the subscript of m means the material inside. It is expected that the difference between \(A_{1}\) and \(A_{2}\) comes mainly from the attenuation in the material, as the reflection coefficient between air and box is small. Based on the material properties at each stage and the physical size of the cardboard box, the magnitudes of the two peaks are calculated. The results for attenuation per stage, the ratio between the second peak and the first peak, were obtained, as shown in Table 2. It is observed that the higher conductance results in a higher attenuation constant. These ratios of attenuation are compared with the measurement results in the next section. \begin{table}
\begin{tabular}{c c c c} \hline
**Stage** & **Paper Quantity** & **Relative Permittivity** & **Conductivity (S/m)** \\ \hline
1 & 44 & 1.09 & 0.000063 \\
2 & 75 & 1.15 & 0.000107 \\
3 & 126 & 1.25 & 0.000180 \\
4 & 197 & 1.39 & 0.000282 \\
5 & 327 & 1.65 & 0.000468 \\ \hline \end{tabular}
\end{table} 
Table 1: Quantity, permittivity, and conductivity per stage. \begin{table}
\begin{tabular}{c c c} \hline
**Stage** & **Simulated Attenuation** & **Measured Attenuation** \\ \hline I & 0.89281 & 0.89205 \\ II & 0.57376 & 0.46242 \\ III & 0.25906 & 0.29256 \\ IV & 0.07659 & 0.06288 \\ V & 0.00661 & 0.03114 \\ \hline \end{tabular}
\end{table} 
Table 2: Attenuation for all five stages using simulation and measurement.