\section{Twisted Pair Model} \label{twisted_pair_model}

The twisted pair model assumes that any crosstalk or coupling in the cable
bundle (as opposed to the terminations) will be to the common mode of the twisted pair. The
differential mode propagates without any coupling to or from other conductors in the bundle.  
This twisted pair model behaves in a similar manner to a shielded cable with a perfect 
shield where the differential mode corresponds to the internal (shielded) mode which does not interact with its surrounding conductors and the common mode corresponds to the external propagation on the shield which does interact with the surrounding
conductors. 
This correspondence allows a circuit model of a twisted pair to be constructed based on
the domain decomposition method used for shielded cables.
The model decomposes the twisted pair system into domains. The domains consist of a common
mode domain which is exposed to the external EM environment and a differential mode domain which
is perfectly shielded from the external environment. In an internal domain the natural reference
conductor is the shield enclosing the domain. The model may be simply extended to the case of a
shielded twisted pair. Note that for a twisted pair in isolation i.e. with no other return conductor then
only the differential mode should be included in the model.

Figure\ref{fig:twisted_pair_domain_decom} shows a twisted pair cable with an additional external conductor (ground plane). The
external voltages and currents at the termination (LHS) are reference to ground plane (global
reference) whilst the voltages and currents used in the model are the common and differential mode
shown to the right.

\begin{figure}[h]
\centering
\includegraphics[scale=0.6]{./Imgs/Twisted_pair_domain_decomp.eps}
\caption{Domain decomposition for twisted pairs}
\label{fig:twisted_pair_domain_decom}
\end{figure}
%
For the twisted pair we can relate the differential mode and common mode voltages to the conductor voltages by
%
\begin{equation} \label{eq:dd_v1a}
\begin{array}{c}
V_{d}=V_{1}-V_{2} \\[10pt]
V_{c}=\frac{\left( V_{1}+V_{2}\right) }{2}
\end{array}
\end{equation}
%
therefore the domain decomposition matrix ${M_{V}}$ is given by
%
%
\begin{equation}
\left(\begin{array}{c} V_{d} \\[10pt] V_{c} \end{array}\right)
=\left[M_{V}\right] 
\left(\begin{array}{c} V_{1} \\[10pt] V_{2} \end{array}\right)
=\left[\begin{array}{cc} 1 & -1 \\[10pt] \frac{1}{2} & \frac{1}{2} \end{array} \right] 
\left(\begin{array}{c} V_{1} \\[10pt] V_{2} \end{array}\right)
\end{equation}
%
%
Similarly we can relate the differential mode and common mode currents to the conductor current by
%
\begin{equation} \label{eq:dd_v1a}
\begin{array}{c}
I_{d}=\frac{\left( I_{1}-I_{2}\right) }{2} \\[10pt]
I_{c}=I_{1}+I_{2}
\end{array}
\end{equation}
%
therefore the domain decomposition matrix ${M_{I}}$ is given by
%
%
\begin{equation}
\left(\begin{array}{c} I_{d} \\[10pt] I_{c} \end{array}\right)
=\left[M_{I}\right] 
\left(\begin{array}{c} I_{1} \\[10pt] I_{2} \end{array}\right)
=\left[\begin{array}{cc} \frac{1}{2} & -\frac{1}{2} \\[10pt] 1 & 1  \end{array} \right] 
\left(\begin{array}{c} I_{1} \\[10pt] I_{2} \end{array}\right)
\end{equation}
%

\subsection{Per-unit-length parameters for the twisted pair model}

The decomposition into the differential mode and the common mode for the twisted pair model requires the per-unit-length inductance and capacitance to be determined for these modes. The differentail mode of a twisted pair in a bundle with other conductors and/ or a ground plane is assumed not to interact with the other conductors in the bundle, the differential mode inductance and capacitance are calculated for the two conductors in isolation i.e. with all other conductors removed. The differential mode inductance and capcitance are then independent of the orientation of the 
conductors in the analysis. 
The common mode inductance and capacitance are calculated by assuming that the common mode current is carried by an equivalent cylindrical conductor placed on the axis of the twisted pair. This equivalent conductor is then used in the analysis of the bundle cross section in the calculation of the twisted pair common mode contribution to the inductance and capacitance matrices. This process is illustrated in figure \ref{fig:twisted_pair_LC}

\begin{figure}[h]
\centering
\includegraphics[scale=0.6]{./Imgs/Twisted_pair_LC.eps}
\caption{Twisted pair inductance and capacitance matrix calculation}
\label{fig:twisted_pair_LC}
\end{figure}

\subsection{Per-unit-length parameters for the shielded twisted pair model}

The per-unit-length parameters for a shielded twisted pair are determined in a slightly different manner than for a twisted pair in a bundle with other conductors. In the case of a shielded twisted pair the orientation of the conductors is not of any significance due to the symmetry of the shield conductor. The inductance and capacitance of the three conductor system formed by the twisted pair and the shield can be calculated using the Laplace solver with the shield chosen as the reference conductor. This gives the conductor based transmission line equations of the following form

\begin{equation} \label{eq:L_stp1}
\frac{\partial}{\partial z}\left( \begin{array}{c}  V_1 \\[10pt] V_2 \end{array} \right) = 
-\left[
L \right]
\frac{\partial}{\partial t}\left( \begin{array}{c}  I_1 \\[10pt] I_2 \end{array} \right) 
\end{equation}
where
\begin{equation} \label{eq:L_stp2}
\left[ L \right] = \left[
\begin{array}{cc} 
L_{A} & L_{B} \\[10pt]
L_{B} & L_{A} 
\end{array} \right]
\end{equation}
and 
\begin{equation} \label{eq:C_stp1}
\frac{\partial}{\partial z}\left( \begin{array}{c}  I_1 \\[10pt] I_2 \end{array} \right) = 
-\left[C \right]
\frac{\partial}{\partial t}\left( \begin{array}{c}  V_1 \\[10pt] V_2 \end{array} \right) 
\end{equation}
where
\begin{equation} \label{eq:C_stp2}
\left[ C \right] = \left[
\begin{array}{cc} 
C_{A} & C_{B} \\[10pt]
C_{B} & C_{A} 
\end{array} \right]
\end{equation}

The common and differential mode inductance and capacitance are then found by applying the domain decomposition matrices
to transform the conductor based voltage and current variables to common and differential mode based voltages and currents i.e.

\begin{equation} \label{eq:L_stp3}
\frac{\partial}{\partial z}\left( \begin{array}{c}  V_d \\[10pt] V_c \end{array} \right) = 
-\left[ M_V \right] \left[ L \right] \left[ M_I \right]^{-1}
\frac{\partial}{\partial t}\left( \begin{array}{c}  I_d \\[10pt] I_c \end{array} \right) ,
\left[ L \right] = \left[
\begin{array}{cc} 
2\left(L_{A}-L_{B} \right) & 0 \\[10pt]
0 & \frac{1}{2}\left(L_{A}+L_{B} \right) 
\end{array} \right]
\end{equation}
and 
\begin{equation} \label{eq:C_stp3}
\frac{\partial}{\partial z}\left( \begin{array}{c}  I_d \\[10pt] I_c \end{array} \right) = 
-\left[ M_I \right] \left[ C \right] \left[ M_V \right]^{-1}
\frac{\partial}{\partial t}\left( \begin{array}{c}  V_d \\[10pt] V_c \end{array} \right) ,
\left[ C \right] = \left[
\begin{array}{cc} 
\frac{1}{2}\left(C_{A}+C_{B} \right)  & 0 \\[10pt]
0 & 2\left(C_{A}-C_{B} \right)
\end{array} \right]
\end{equation}
 
from which we identify the common and differential mode inductance and capacitances. 

\clearpage