\section{Per-Unit-Length Parameters} \label{PULpara}

The multi-conductor transmission line equations express the transmission line voltages and currents in terms of the inductance, capacitance, resistance and conductance matrices. 

This chapter describes the per-unit-length parameters of the transmission lines and gives some formulae which may be used for their evaluation. A detailed discussion of the per-unit-length parameters may be found in \cite{PaulMTL}

\subsection{The per-unit-length inductance matrix}

The inductance matrix relates the total magnetic flux, $\Phi$ linking the ith circuit to the currents which produce it 

\begin{equation} \label{eq:Lmat}
\begin{pmatrix} 
\Phi_1 \\ \Phi_2 \\ \vdots \\ \Phi_n
\end{pmatrix}
=
\begin{bmatrix}
L_{1,1} &  L_{1,2} & \hdots & L_{1,n} \\ 
L_{2,1} &  L_{2,2} & \hdots & L_{2,n} \\ 
\vdots  &  \vdots  & \ddots & \vdots  \\ 
L_{n,1} &  L_{n,2} & \hdots & L_{n,n} 
\end{bmatrix}
\begin{pmatrix} 
I_1 \\ I_2 \\ \vdots \\ I_n
\end{pmatrix}
\end{equation}

An element of the inductance matrix $L{ij}$ may be calculated by putting a current on conductor j and zero current on all other conductors then

\begin{equation}
L_{i,j}= \left. \frac{\Phi_i}{I_j} \right\vert_{I_{k\neq j}=0}
\end{equation}

\subsection{The per-unit-length capacitance matrix}

The capacitance matrix relates the charge , q, on conductors to the conductor voltages, V.
\begin{equation} \label{eq:Cmat}
\begin{pmatrix} 
q_1 \\ q_2 \\ \vdots \\ q_n
\end{pmatrix}
=
\begin{bmatrix}
C_{1,1} &  C_{1,2} & \hdots & C_{1,n} \\ 
C_{2,1} &  C_{2,2} & \hdots & C_{2,n} \\ 
\vdots  &  \vdots  & \ddots & \vdots  \\ 
C_{n,1} &  C_{n,2} & \hdots & C_{n,n} 
\end{bmatrix}
\begin{pmatrix} 
V_1 \\ V_2 \\ \vdots \\ V_n
\end{pmatrix}
\end{equation}

The capacitance matrix is commonly calculated from its inverse, the elements of the inverse capacitance matrix $[P]=[C]^{-1}$ are determined by placing a charge 
on conductor j and zero change on all other conductors then 

\begin{equation}
P_{i,j}= \left. \frac{V_i}{q_j} \right\vert_{I_{k\neq j}=0}
\end{equation}

then the capacitance matrix is found as
\begin{equation}
\left[C\right]=\left[P\right]^{-1}
\end{equation}

In a homogeneous medium the capacitance matrix may be found from the inductance matrix using

\begin{equation}
\left[C\right]=\mu_0 \epsilon_0 \epsilon_r \left[L\right]^{-1}
\end{equation}

\subsection{The per-unit-length resistance matrix}

The per-unit-length resistance matrix, $[R]$ incorporates the effects of conductor losses in the model i.e. the effects of the finite conductivity of the conductors.
It relates the voltage drop along the transmission line to the currents on the conductors. The conductor resistance is a function of frequency due to the 'skin effect'. The calculation of the resitance of conductors, and the associated 'internal inductance' is discussed in 
section \ref{FD_conductivity_model}. The resistance matrix relates the conductor voltages and currents at zero frequency by

\begin{equation} \label{eq:Rmat}
\frac{\partial}{\partial z}
\begin{pmatrix} 
V_1 \\ V_2 \\ \vdots \\ V_n
\end{pmatrix}
=
\begin{bmatrix}
R_{1,1} &  R_{1,2} & \hdots & R_{1,n} \\ 
R_{2,1} &  R_{2,2} & \hdots & R_{2,n} \\ 
\vdots  &  \vdots  & \ddots & \vdots  \\ 
R_{n,1} &  R_{n,2} & \hdots & R_{n,n} 
\end{bmatrix}
\begin{pmatrix} 
I_1 \\ I_2 \\ \vdots \\ I_n
\end{pmatrix}
\end{equation}

If a conductor i, $i=1 \dots n$ has a resistance $R_i$ and the reference conductor has resistance $R_0$ then the resistance matrix is given by
\begin{equation} \label{eq:Rmat2}
\left[ R \right] =
\begin{bmatrix}
R_1+R_0 &  R_0 & \hdots & R_0 \\ 
R_0 &  R_2+R_0 & \hdots & R_0 \\ 
\vdots  &  \vdots  & \ddots & \vdots  \\ 
R_0 &  R_0 & \hdots & R_n+R_0 
\end{bmatrix}
\end{equation}

\subsection{The per-unit-length conductance matrix}

The per-unit-length conductance matrix , $[G]$ incorporates the effects of losses due to conductance of the medium in which the conductors are situated and also due to polarisation losses in dielectrics. The conductance loss will be assumed to be zero therefore the only contributions to the conductance matrix are due to lossy dielectrics. In this work the conductance matrix is calculated by generalising the capacitance matrix calculation to complex permittivities.

The admittance matrix is written as
\begin{equation}
\left[Y\right]=\left[G\right]+j \omega \left[C\right]=j \omega \left[C-\frac{j}{\omega}G\right]=j \omega \left[C'\right]
\end{equation}

Thus if we determine the complex capacitance matrix $[\left[C'\right]$ then we can calculate the capacitance matrix and admittance matrices as

\begin{equation}
\left[C\right]= \Re \left\{ \left[C'\right] \right\}
\end{equation}
and 
\begin{equation}
\left[G\right]=-\omega \Im \left\{ \left[C'\right] \right\}
\end{equation}

The calculation of the capacitance and conductance matrices for configurations with lossy inhomogeneous dielectrics is discussed in section \ref{Laplace}.

\subsection{Exact analytic formulae for per-unit-length parameters}

Exact analytical formulae are available for the inductance and capacitance of some conductor configurations. There are also approximations which may be made which allow the derivation of approximate analytic formulae. Where possible these formulae are used in the SACAMOS software. The derivation of the formulae presented in the following sections may be found in \cite{PaulMTL}.

Figure \ref{fig:coax} shows a coaxial cable

\begin{figure}[h]
\centering
\includegraphics[scale=1]{./Imgs/coax.eps}
\caption{Coaxial cable}
\label{fig:coax}
\end{figure}

In SACAMOS, the inductance and capacitance of the coaxial cable is always calculated using the analytic formulae

\begin{equation}
L=\frac{\mu_0}{2 \pi} \ln\left(\frac{r_s}{r_w}\right)
\end{equation}
\begin{equation}
C=\frac{2 \pi \epsilon_0 \epsilon_r \left(j\omega)\right)}{ \ln\left(\frac{r_s}{r_w}\right)}
\end{equation}

Figure \ref{fig:two_wire} shows a two wire transmission line. 

\begin{figure}[h]
\centering
\includegraphics[scale=1]{./Imgs/two_wire.eps}
\caption{Two wire cable}
\label{fig:two_wire}
\end{figure}


The analytic formulae used to calculate the inductance and capacitance of the two wire transmission line are

\begin{equation}
C=\frac{ \pi \epsilon_0 }{ \ln\left( \frac{s}{2r_w}+\sqrt{ \frac{s}{2r_w}^2-1 }  \right)}
\end{equation}
\begin{equation}
L=\frac{\mu_0 \epsilon_0}{C}
\end{equation}

These formulae are used for the calculation of the differential mode inductance and capacitance for twisted pairs.

\subsection{Approximate analytic formulae for per-unit-length parameters}

Approximate analytic formulae are available for multi-conductor configurations where the assumption made is that the conductor radii are small compared to the conductor separation. It is also assumed that the conductors are perfectly conducting and are situated in a homogeneous dielectric (free space here).

Figure \ref{fig:multi-wire} shows part of a multi-conductor configuration.

\begin{figure}[h]
\centering
\includegraphics[scale=1]{./Imgs/multi-wire.eps}
\caption{Multi-wire cable}
\label{fig:multi-wire}
\end{figure}

The analytic formulae used to calculate the inductance and capacitance of the multi-wire transmission line are
as follows:

For the diagonal elements of the inductance matrix
\begin{equation}
L_{ii}=\frac{\mu_0}{2 \pi} \ln\left(\frac{d_{i0}^2}{r_{w0}r_{wi}}\right) 
\end{equation}

The off diagonal elements are given by
\begin{equation}
L_{ij}=\frac{\mu_0}{2 \pi} \ln\left(\frac{d_{i0}d_{j0}}{d_{ij}r_{w0}}\right) 
\end{equation}

Since the conductors are in a homogeneous dielectric medium, the capacitance matrix is found from the inverse of the inductance matrix.
\begin{equation}
\left[ C \right]=\mu_0 \epsilon_0 \left[L\right]^{-1}
\end{equation}


Figure \ref{fig:multi-wire_gnd} shows part of a multi-conductor configuration above a ground plane. Here the ground plane is the reference conductor


\begin{figure}[h]
\centering
\includegraphics[scale=1]{./Imgs/multi-wire_gnd.eps}
\caption{Multi-wire cable over a ground plane}
\label{fig:multi-wire_gnd}
\end{figure}

For the diagonal elements of the inductance matrix
\begin{equation}
L_{ii}=\frac{\mu_0}{2 \pi} \ln\left(\frac{2h_{i}}{r_{wi}}\right) 
\end{equation}

The off diagonal elements are given by
\begin{equation}
L_{ij}=\frac{\mu_0}{4 \pi} \ln\left(1+\frac{4h_{i}h_{j}}{d_{ij}^2}\right) 
\end{equation}

Since the conductors are in a homogeneous dielectric medium, the capacitance matrix may be found from the inverse of the inductance matrix.
\begin{equation}
\left[ C \right]=\mu_0 \epsilon_0 \left[L\right]^{-1}
\end{equation}

Figure \ref{fig:multi-wire_shield} shows part of a multi-conductor configuration within a cylindrical shield. Here the shield is the reference conductor

\begin{figure}[h]
\centering
\includegraphics[scale=1]{./Imgs/multi-wire-shield.eps}
\caption{Multi-wire cable inside a cylindrical shield}
\label{fig:multi-wire_shield}
\end{figure}


For the diagonal elements of the inductance matrix
\begin{equation}
L_{ii}=\frac{\mu_0}{2 \pi} \ln\left(\frac{r_{s}^2-d_i^2}{r_{s}r_{wi}}\right) 
\end{equation}

The off diagonal elements are given by
\begin{equation}
L_{ij}=\frac{\mu_0}{2 \pi} \ln\left( \frac{d_j}{r_s} \sqrt{ \left( \frac{d_{i}^2 d_{j}^2 +r_s^4-2d_id_jr_s^2\cos(\theta_{ij})}{d_{i}^2 d_{j}^2 +d_j^4-2d_i d_j^3\cos(\theta_{ij})}   \right) } \right) 
\end{equation}

Since the conductors are in a homogeneous dielectric medium, the capacitance matrix is found from the inverse of the inductance matrix.
\begin{equation}
\left[ C \right]=\mu_0 \epsilon_0 \left[L\right]^{-1}
\end{equation}


\cleardoublepage
