\subsection{Incorporating terminal conditions} \label{TermCond}

The solution in terms of voltages and currents of the MTL equations
are given by (\ref{eq:CurrentSol}) and (\ref{eq:VoltageSol}).
These expressions allow computing the state, i.e. current and voltage, 
of each conductor at any point of the transmission line, after 
determining the constants $\tilde{I}^+_m$ and
$\tilde{I}^-_m$. Therefore the terminal conditions have to be incorporated,
which can be done via the generalised Th\'{e}venin equivalent:
%
\begin{equation} \label{eq:Thevenin}
\begin{array}{rcl}
\left[\tilde{V}\left(0\right)\right] & = & \left[\tilde{V}_S\right] - \left[\tilde{Z}_S\right] \left(\tilde{I}\left(0\right)\right) \\[10pt]
\left[\tilde{V}\left(L\right)\right] & = & \left[\tilde{V}_L\right] + \left[\tilde{Z}_L\right] \left(\tilde{I}\left(L\right)\right),
\end{array}
\end{equation}
where $\left[\tilde{Z}_S\right]$ and $\left[\tilde{Z}_L\right]$
are the matrices with termination impedances on respectively the
source and load side of the multi conductor transmission line, i.e.
respectively at $z=0$ and $z=L$.
$\tilde{V}_S$ and $\tilde{V}_L$ are the vectors containing the
voltage sources on respectively source and load side of the
transmission line.

By evaluating equations (\ref{eq:CurrentSol}) and (\ref{eq:VoltageSol})
at $z=0$ and $z=L$ and substituting this into the Thevenin equivalent
given by equation \ref{eq:Thevenin}, we obtain:
\begin{equation}
\left[
\begin{array}{cc}
\left(\left[\tilde{Z}_C\right] + \left[\tilde{Z}_S\right]\right) \left[\tilde{T}_I\right] & \left(\left[\tilde{Z}_C\right] - \left[\tilde{Z}_S\right]\right) \left[\tilde{T}_I\right] \\[10pt]
\left(\left[\tilde{Z}_C\right] - \left[\tilde{Z}_L\right]\right) \left[\tilde{T}_I\right] \rm e^{-\gamma L} & \left(\left[\tilde{Z}_C\right] + \left[\tilde{Z}_L\right]\right) \left[\tilde{T}_I\right]\rm e^{\gamma L}
\end{array}
\right]
\left(
\begin{array}{c}
\tilde{I}_m^+ \\[10pt]
\tilde{I}_m^-
\end{array}
\right)
=
\left(
\begin{array}{c}
\tilde{V}_S \\[10pt]
\tilde{V}_L
\end{array}
\right)
\end{equation}
%
This matrix equation is readily solverd for the current waves,  $\tilde{I}^+_m$ and
$\tilde{I}^-_m$ and then equations (\ref{eq:CurrentSol}) and (\ref{eq:VoltageSol}) 
lead to a final solution for the conductor voltages and currents.

The detail of the solution as it is implemented in the SACAMOS software is described here.
In order to keep the solution clear, these equations are rewritten as
%
\begin{equation}
\left[
\begin{array}{cc}
\left[\tilde{M_{11}}\right]  &  \left[\tilde{M_{12}}\right] \\[10pt]
\left[\tilde{M_{21}}\right]  &  \left[\tilde{M_{22}}\right] 
\end{array}
\right]
\left(
\begin{array}{c}
\tilde{I}_m^+ \\[10pt]
\tilde{I}_m^-
\end{array}
\right)
=
\left(
\begin{array}{c}
\tilde{V}_S \\[10pt]
\tilde{V}_L
\end{array}
\right)
\end{equation}
These equations are solved for $\tilde{I}^+_m$ and
$\tilde{I}^-_m$  first by eliminating $\tilde{I}^-_m$ to give

\begin{equation}
\left[ 
\left[\tilde{M_{11}}\right] - 
\left[\tilde{M_{12}}\right] \left[\tilde{M_{22}}\right]^{-1} \left[\tilde{M_{21}}\right] 
\right] \left( \tilde{I}_m^+ \right) =
 \left( \tilde{V}_S \right) - 
 \left[\tilde{M_{12}}\right] \left[\tilde{M_{22}}\right]^{-1} \left( \tilde{V}_L \right)
\end{equation}

Thus

\begin{equation} \label{solution_Imp}
 \left( \tilde{I}_m^+ \right) =
\left[ 
\left[\tilde{M_{11}}\right] - 
\left[\tilde{M_{12}}\right] \left[\tilde{M_{22}}\right]^{-1} \left[\tilde{M_{21}}\right]\right]^{-1}
\left( \left( \tilde{V}_S \right) - 
\left[\tilde{M_{12}}\right] \left[\tilde{M_{22}}\right]^{-1} \left( \tilde{V}_L \right) \right)
\end{equation}

 $\tilde{I}^-_m$ may then be found as
 
\begin{equation} \label{solution_Imm}
\left( \tilde{I}_m^- \right) =
\left[\tilde{M_{22}}\right]^{-1} 
\left( \left( \tilde{V}_L \right) \right) -
\left[\tilde{M_{21}}\right] \left( \tilde{I}_m^+ \right)
\end{equation}

substituting these expressions in
equations (\ref{eq:CurrentSol}) and (\ref{eq:VoltageSol}) 
leads to a final solution for the conductor voltages
and currents along the entire length of the multi conductor
transmission line.

\subsection{Time domain solution of the transmission line equations}

The frequency domain solution of the transmission line equations may be used as the basis for calculating the time domain solution by the Inverse Fourier Transform (IFT) method. If we assume that we have specified termination source voltages as a function of time $V_S(t)$ and $V_L(t)$ then we wish to calculate the time response of the conductor voltages $V(0,t)$, $V(L,t)$.

The solution proceeds by finding the Fourier Transform of the source voltage functions 

\begin{equation} \label{FT_Vs}
\tilde{V}_S\left(j\omega\right)=\int_{-\inf}^{\inf} V_S\left(t\right)e^{j\omega t} dt
\end{equation}

\begin{equation} \label{FT_Vl}
\tilde{V}_L\left(j\omega\right)=\int_{-\inf}^{\inf} V_L\left(t\right)e^{j\omega t} dt
\end{equation}

The Frequency domain solution is then obtained for the transmission line current waves (equations \ref{solution_Imp}, \ref{solution_Imm}) and then the termination voltages and currents using equations \ref{eq:VoltageSol} \ref{eq:CurrentSol}.

The time domain solution for the transmission line voltages and currents may then be calculated by the inverse Fourier Transform

\begin{equation} \label{IFT_V}
\tilde{V}\left(z,t\right)=\frac{1}{2\pi}\int_{-\inf}^{\inf} V\left(z,j\omega\right)e^{-j\omega t} d\omega
\end{equation}

\begin{equation} \label{IFT_I}
\tilde{I}\left(z,t\right)=\frac{1}{2\pi}\int_{-\inf}^{\inf} I\left(z,j\omega\right)e^{-j\omega t} d\omega
\end{equation}

In practice we use the Fast Fourier Transform (FFT) algorithm to transform between the time and frequency domains \cite{Brigham}. The FFT algorithm assumes that the time response is discrete in time and periodic with period $T$ i.e. we have a finite number of samples of the time response. These assumptions also make the frequency response periodic and discrete thus the analysis outlined above may be performed at a discrete set of frequencies. We assume that the time response may be adequately sampled at a sampling rate of $\delta t$ and the period for the time response is chosen to be $T=n\delta t$ where $n$ is conveniently chosen to be an integer power of 2. It is important that the period $T$ is chosen to be long enough such that the time response of the transmission line has decayed to an insignificant level, otherwise the solution will be affected by aliasing i.e. the time response within a period $T$ is corrupted by the response from adjacent periods.

\cleardoublepage