## 4 Channel Models
### Relationship between the Transmitter Power and SNR  
The goal of the first part is to determine the relationship between the transmitter power and SNR. We first define the transmitter power as \(P\) and the SNR as \(\gamma_{dB}\). Note that the unit of SNR is dB in this paper. The SNR can be divided into four parts [44], and be expressed as follows:  
\[\gamma_{dB}=S_{level,dB}-T_{loss,dB}-N_{level,dB}+D_{index,dB}, \tag{1}\]  
where \(S_{level,dB}\), \(T_{loss,dB}\), \(N_{level,dB}\), and \(D_{index,dB}\) are the source level, transmission loss, noise level, and directivity index, respectively. For the rest of the paper, we simplify the notation. Specifically, we replace \(\gamma_{dB}\), \(S_{level,dB}\), \(T_{loss,dB}\), \(N_{level,dB}\), and \(D_{index,dB}\) with \(\gamma\), \(S_{level}\), \(T_{loss}\), \(N_{level}\), and \(D_{index}\), respectively. Accordingly, the SNR can be expressed as follows:  
\[\gamma=S_{level}-T_{loss}-N_{level}+D_{index} \tag{2}\]  
Note that the unit of each factor (i.e., \(\gamma\), \(S_{level}\), \(T_{loss}\), \(N_{level}\), and \(D_{index}\)) is dB. \begin{table}
\begin{tabular}{c c c} \hline \hline
**Features** & **TWSNs** & **UWSNs** \\ \hline Transmission Media & Radio Wave & Sound Wave \\ Propagation Speed & 300,000,000 m/s & 1500 m/s \\ Transmission Range & 10 m–100 m & 100 m–10,000 m \\ Transmission Rate & \(\sim\)250 kbps & \(\sim\)10 kbps \\ Difficulty to Readarge & Depend on Applications & Difficult \\ Mobility (of Nodes) & Depend on Applications & High \\ Reliability (of Links) & Depend on Applications & Low \\ \hline \hline \end{tabular}
\end{table} 
Table 1: The differences between TWSNs and UWSNs. First, the source level \(S_{level}\) is defined as the effective level of sound. To calculate \(S_{level}\), we need to start from the relationship between \(S_{level}\) and transmitted signal intensity \(I\)[45]. The relationship can be expressed by  
\[S_{level}=10\Big{[}\log(I)-\log\Big{(}0.67\times 10^{-18}\Big{)}\Big{]}. \tag{3}\]  
To calculate \(I\), we also need the relationship between the transmitter power \(P\) and the transmitted signal intensity \(I\). The relationship can be expressed as follows  
\[P=4\pi r^{2}\times I, \tag{4}\]  
where \(r\) is the transmission range (i.e., the radius of an imaginary sphere), and the surface area of a sphere is: \(A=4\pi r^{2}\). Specifically, we consider that the sound source is omnidirectional, and its power is uniformly distributed over the surface area of an imaginary sphere of radius \(r\). Note that the units of \(P\) and \(r\) are W and m, respectively. Using Equation (4), we can first compute the intensity \(I\) by  
\[I=\frac{P}{4\pi r^{2}}. \tag{5}\]  
Using Equations (3) and (5), we can first compute the source level \(S_{level}\) according to the transmitter power \(P\) and its range \(r\) as follows  
\[ \begin{array}{ll}S_{level}&=10\Big{[}\log\Big{(}\frac{P}{4\pi r^{2}}\Big{)} -\log(0.67\times 10^{-18})\Big{]}\\ &=10\big{[}\log(P)-\log(4\pi r^{2})-\log(0.67\times 10^{-18})\big{]}.\end{array}  \tag{6}\]  
Second, the transmission loss \(T_{loss}\) is defined as a measure of the rate at which sound energy is lost. The underwater transmission loss \(T_{loss}\)[46] over a distance \(d\) (in m) for a signal of frequency \(f\) (in kHz) can be derived by  
\[T_{loss}=20\log d+\alpha(f)\times d\times 10^{-3}, \tag{7}\]  
where \(\alpha(f)\) is the absorption coefficient (of sound in seawater) in dB/km. To calculate \(\alpha(f)\), we can use [PERSON]'s formula [47], as follows:  
\[a(f)=\frac{0.11f^{2}}{1+f^{2}}+\frac{44f^{2}}{4100+f^{2}}+2.75\times 10^{-4}f^{2 }+0.003. \tag{8}\]  
Note that \(20\log d\) in Equation (7) is an approximation used to describe how sound level decreases when a sound wave propagates away from a sound source (uniformly in all directions), and the approximation is called spherical spreading. For spherical spreading, the rate at which energy decreases can be obtained by using the signal intensity.