earth transmission line equations
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In order to provide a steady-state analysis, we use the phasor representation for the electrical quantities. We denote $V_k(f,x)$ for $k\in \{a,b,c\}$, where $\{a,b,c\}$ are the labels for the three conductors, the voltage phasor associated with the generator circuit and the receiver circuit at a frequency $f$ and coordinate $x$. To simplify the notation, we do not explicitly show the frequency dependency in the following. Solving the Kirchhoff equations for the PUL circuit represented in and letting $dx \rightarrow 0$, provides the telegraph equations [@Paul:2007:AMT:1554645] $$\begin{aligned}
\frac{\partial \bm{V}(x)}{\partial x}=&-(\bm{R}+j2\pi f\bm{L})\bm{I}(x),\\
\frac{\partial \bm{I}(x)}{\partial x}=&-(\bm{G}+j2\pi f\bm{C})\bm{V}(x),
\label{eq:telegraph}\end{aligned}$$ where $\bm{V}=(V_a,V_b,V_c)^T$ is the voltage phase vector, $\bm{I}=(I_a,I_b,I_c)^T$ is the current phasor vector, and $\{\dot{}\}^T$ denotes the transposition operator.

Furthermore, $$\begin{aligned}
\bm{R}
=&
\begin{bmatrix}
r_a+r_o & r_o       & r_o \\
r_o     & r_b+r_o   & r_o \\
r_o     & r_o       & r_c +r_o \\
\end{bmatrix} \notag\\ 
\bm{L}
=&
\begin{bmatrix}
l_a     & l_{ab}        & l_{ac} \\
l_{ab}  & l_b       & l_{bc} \\
l_{ac}  & l_{bc}        & l_c \\
\end{bmatrix} \notag\\
\bm{C}
=&
\begin{bmatrix}
c_{ab}+c_{ac}+c_{ao}        & -c_{ab}                   & -c_{ac}                        \\
-c_{ab}                     & c_{ab}+c_{bc}+c_{bo}      & -c_{bc}                           \\
-c_{ac}                     & -c_{bc}                   & c_{bc}+c_{ac}+c_{co}          \\
\end{bmatrix} \notag\\
\bm{G}
=&
\begin{bmatrix}
g_{ab}+g_{ac}+g_{ao}    & -g_{ab}                   & -g_{ac}                        \\
