<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\section{Modal decomposition} \label{Modal decomposition}

The multi-conductor frequency domain transmission line equations are given by

\begin{equation} \label{eq:MTLequa_fin2}
\begin{array}{rcl}
\frac{\dif}{\dif z} \left(\tilde{V}\left(z\right)\right) &amp; = &amp; -\left[\tilde{Z}\right]\left(\tilde{I}\left(z\right)\right) \\[10pt]
\frac{\dif}{\dif z} \left(\tilde{I}\left(z\right)\right) &amp; = &amp; -\left[\tilde{Y}\right]\left(\tilde{V}\left(z\right)\right),
\end{array}
\end{equation}
where
\begin{equation} \label{eq:ZandY2}
\begin{array}{rcl}
\left[\tilde{Z}\right] &amp; = &amp; \left[R\right]+j\omega\left[L\right] \\[10pt]
\left[\tilde{Y}\right] &amp; = &amp; \left[G\right]+j\omega\left[C\right].
\end{array}
\end{equation}

The frequency domain solution of these equations relied on a modal decomposition method as described in section \ref{MTLtheory}
in which the $[Z][Y]$ product was diagonalised i.e. we must diagonalise the matrix

\begin{equation} \label{eq:ZYproduct}
\left[\tilde{Z}\right]\left[\tilde{Y}\right]  =  \left[ \left[R\right]+j\omega\left[L\right] \right]
                                                   \left[ \left[G\right]+j\omega\left[C\right].\right]
\end{equation}

Even if we assume that the parameters of the cable are independent of frequency then the problems with the diagonalisation 
process can be seen by looking at the limits as $j\omega \rightarrow 0$ and $j\omega \rightarrow \inf$. 
In the first case we have (if we assume that $[G]=0$

\begin{equation} \label{eq:ZYproduct_low_f}
\left[\tilde{Z}\right]\left[\tilde{Y}\right]  = \left[R\right]\left[j\omega\left[C\right]\right]
\end{equation}

whereas at high frequency we can neglect $[R]$ and $[G]$ in comparison to  to $j\omega[L]$ and $j\omega[C]$ and we have

\begin{equation} \label{eq:ZYproduct_high_f}
\left[\tilde{Z}\right]\left[\tilde{Y}\right]  =  -\omega^2 \left[ L \right] \left[ C \right]
\end{equation}

In the first case the $[Z][Y]$ matrix product is dominated by $[R]$ and $[C]$ whereas at high frequency it is
dominated by $[L]$ and $[L]$. In the general case, we must therefore diagonalise the $[Z][Y]$ matrix product at every frequency.

A model which eliminates this problem may be found by lumping the d.c. resistances of the conductors at the conductor ends i.e. place a lumped resistance with value $R_{dc}/2$ at either end of the transmission line. Once this is done, the $[Z][Y]$ matrix product is again given by equation \ref{eq:ZYproduct_high_f}. 

A justification for this is that the transmission line resistance may be attributed to the finite conductivity of the conductors. The resistance resulting from this is freqency dependent and increases as $\frac{1}{\sqrt(f)}$. Subtracting the d.c. resistance will result in a model which is still correct at d.c. and reasonably accurate provided the transmission line is short compared with the wavelength. The modal decomposition will be a weak function of frequency and the losses associated with the propagation of a mode on a lossy transmission line may be corrected by a frequency dependent loss term as described below.  

It is interesting to observe that (if we consider a single mode transmission line), if there is a d.c. resistance present then the characteristic impedance of the lossy line is given by
\begin{equation} \label{eq:Z0_with_rdc}
Z_0=\sqrt{\frac{R+j\omega L}{j\omega C}} 
\end{equation}
at low frequency
\begin{equation} \label{eq:Z0_with_rdc}
Z_0 \rightarrow \sqrt{\frac{R}{j\omega C}} 
\end{equation}
i.e. the characteristic impedance varies as $\frac{1}{\sqrt(j\omega)}$ i.e. $Z_0 \rightarrow \inf$ as $\omega \rightarrow 0$.


The individual conductor currents and voltages within a domain are transformed into the modal
currents and voltages. The conductor voltages and currents are solutions of the coupled 
first order equations
%
\begin{equation} \label{eq:md_mtl_equations_1}
\begin{array}{rcl}
\frac{\dif}{\dif z} \left(\tilde{V}\left(z\right)\right) &amp; = &amp; -\left[L\right]\frac{\dif}{\dif t} \left(\tilde{I}\left(z\right)\right) \\[10pt]
\frac{\dif}{\dif z} \left(\tilde{I}\left(z\right)\right) &amp; = &amp; -\left[C\right]\frac{\dif}{\dif t} \left(\tilde{V}\left(z\right)\right) \\[10pt]
\end{array}
\end{equation}
%
Two uncoupled second order equations may be written:
%
\begin{equation} \label{eq:md_mtl_equations_2}
\begin{array}{rcl}
\frac{\dif^2}{\dif z^2} \left(\tilde{V}\left(z\right)\right) &amp; = &amp; \left[\tilde{L}\right]\left[\tilde{C}\right]\frac{\dif^2}{\dif t^2}\left(\tilde{V}\left(z\right)\right) \\[10pt]
\frac{\dif^2}{\dif z^2} \left(\tilde{I}\left(z\right)\right) &amp; = &amp; \left[\tilde{C}\right]\left[\tilde{L}\right]\frac{\dif^2}{\dif t^2}\left(\tilde{I}\left(z\right)\right).
\end{array}
\end{equation}
%
A change of variables via a similarity transformation to modal quantities allows the decoupling
(diagonalisation) of the matrix system. The conductor voltages and currents are related to their
corresponding modal quantities by voltage and current transformation matrices
%
\begin{equation} 
\begin{array}{rcl}
\left(\tilde{V}\left(z\right)\right) &amp; = &amp; \left[\tilde{T_{V}}\right]\left(\tilde{V_{m}}\left(z\right)\right) \\[10pt]
\left(\tilde{I}\left(z\right)\right) &amp; = &amp; \left[\tilde{T_{I}}\right]\left(\tilde{I_{m}}\left(z\right)\right) \\[10pt]
\end{array}
\end{equation}
%
Thus the coupled first order equations \ref{eq:md_mtl_equations_1} become
%
\begin{equation} \label{eq:md_mtl_equations_1_decoupled}
\begin{array}{rcl}
\frac{\dif}{\dif z} \left(\tilde{V_{m}}\left(z\right)\right) &amp; = &amp; -\left[T_{V}\right]^{-1}\left[L\right]\left[T_{I}\right]\frac{\dif}{\dif t} \left(\tilde{I_{m}}\left(z\right)\right) \\[10pt]
\frac{\dif}{\dif z} \left(\tilde{I_{m}}\left(z\right)\right) &amp; = &amp; -\left[T_{I}\right]^{-1}\left[C\right]\left[T_{V}\right]\frac{\dif}{\dif t} \left(\tilde{V_{m}}\left(z\right)\right) \\[10pt]
\end{array}
\end{equation}
%
and the uncoupled second order equations \ref{eq:md_mtl_equations_2} become

%
\begin{equation} \label{eq:md_mtl_equations_2_decoupled}
\begin{array}{rcl}
\frac{\dif^2}{\dif z^2} \left(\tilde{V_{m}}\left(z\right)\right) &amp; = &amp; 
\left[T_{V}\right]^{-1}\left[\tilde{L}\right]\left[\tilde{C}\right]\left[T_{V}\right]\frac{\dif^2}{\dif t^2}\left(\tilde{V_{m}}\left(z\right)\right) \\[10pt]
\frac{\dif^2}{\dif z^2} \left(\tilde{I_{m}}\left(z\right)\right) &amp; = &amp; 
\left[T_{I}\right]^{-1}\left[\tilde{C}\right]\left[\tilde{L}\right]\left[T_{I}\right]\frac{\dif^2}{\dif t^2}\left(\tilde{I_{m}}\left(z\right)\right).
\end{array}
\end{equation}
%
In the case of a lossless inhomogeneous transmission line it is possible to find matrices $\left[T_{V}\right]$ and $ \left[T_{I}\right]$ which 
simultaneoously diagonalise both $\left[L\right]$ and  $\left[C\right]$ i.e.
%
\begin{equation} \label{eq:Lm_Cm}
\begin{array}{rcl}
\left[T_{V}\right]^{-1}\left[L\right]\left[T_{I}\right] $ = $  \left[L_{m}\right] \\[10pt]
\left[T_{I}\right]^{-1}\left[C\right]\left[T_{V}\right] $ = $  \left[C_{m}\right] \\[10pt]
\end{array}
\end{equation}
%
where $\left[L_{m}\right]$ and $\left[C_{m}\right]$ are diagonal matrices.
Thus we have an uncoupled system for modal propagation where the mode characteristic impedance of the ith mode is
%
\begin{equation} \label{eq:Zcm}
Z_{cmi}=\sqrt{ \frac{l_{mi}}{c_{mi}} }
\end{equation}
%
and the mode velocity is
%
\begin{equation} \label{eq:Vcm}
V_{mi}=\frac{1}{\sqrt{ l_{mi}c_{mi} } }
\end{equation}
%
The matrices $\left[L\right]$ and  $\left[C\right]$ are real, symmetric and positive definite and we may obtain the diagonalisation of these
matrices as follows:

First diagonalise the real,symmetric matrix, $\left[C\right]$ with the orthogonal transformation matrix $\left[U\right]$ 
%
\begin{equation} \label{eq:diagonalise_C}
\left[U\right]^{t}\left[C\right]\left[U\right]=\left[\theta^{2}\right]
\end{equation}
%
where
%
\begin{equation} \label{eq:Ut_UI}
\left[U\right]^{t}=\left[U\right]^{-1}
\end{equation}
%
The capacitance matrix, $\left[C\right]$ is positive definite so the square root of the diagonal eigenvalue matrix, $\left[\theta\right]$
is real.

Second form the product
%
\begin{equation} \label{eq:define_M}
\left[M\right] = \left[\theta\right] \left[U\right]^{t}  \left[L\right]  \left[U\right] \left[\theta\right]
\end{equation}
%
$\left[M\right] $ is a real symmetric matrix which may again be diagonalised with an orthogonal matrix.
%
\begin{equation} \label{eq:diagonalise_M}
\left[S\right]^{t}\left[M\right]\left[S\right]=\left[\Gamma^{2}\right]
\end{equation}
%
where
%
\begin{equation} \label{eq:St_SI}
\left[S\right]^{t}=\left[S\right]^{-1}
\end{equation}
%
Define the normalised matrix $\left[T_{norm}\right]$ as
%
\begin{equation} \label{eq:T_definition}
\left[T_{norm}\right]= \left[U\right] \left[\theta\right] \left[S\right] \left[\alpha\right]
\end{equation}
%
where the diagonal matrix $\left[\alpha\right]$ normalises the columns of $\left[T_{norm}\right]$ to length 1.

The matrices which diagonalise $\left[L\right]$ and  $\left[C\right]$ are then
%
\begin{equation} \label{eq:TV}
\left[T_{V}\right]^{-1}= \left[T_{norm}\right]^{t}
\end{equation}
%
and 
%
\begin{equation} \label{eq:TI}
\left[T_{I}\right]^{-1}= \left[T_{norm}\right]^{-1}
\end{equation}
%
Thus the voltage and current modal transformation matrices are related by
%
\begin{equation} 
\left[\tilde{T_{I}}\right]^{T} =\left[\tilde{T_{V}}\right]^{-1} 
\end{equation}
%
The mode impedance and velocity for a mode i are
%
\begin{equation} \label{eq:Zcm2}
Z_{cmi}=\sqrt{ \frac{l_{mi}}{c_{mi}} }=\alpha_{ii}^{2}\Gamma_i
\end{equation}
%
%
\begin{equation} \label{eq:Vcm2}
V_{mi}=\frac{1}{\sqrt{ l_{mi}c_{mi} } }=\frac{1}{\Gamma_i}
\end{equation}
%
Mode propagation will underly the method of characteristics based propogation algorithm in the Spice model.

This leads to an efficient algorithm for both a.c. and transient models of multi-conductor propagation
provided that the modal decomposition matrices are independent of frequency. Discussion of the
incorporation of loss and frequency dependent properties will be discussed in sections 7.4 and 7.5.

For the example of a lossless 3 conductor cable the conductor currents are related to the modal
currents by matrices of dimension 2:
%
\begin{equation}
\left(\begin{array}{c} I_{1} \\[10pt] I_{2} \end{array}\right)
=\left[\tilde{T_{I}}\right] 
 \left(\begin{array}{c} I_{m1} \\[10pt] I_{m2} \end{array}\right)
\end{equation}
%
Hence the modal currents are related to the conductor currents by
%
\begin{equation}\label{MD_I}
\left(\begin{array}{c} I_{m1} \\[10pt] I_{m2} \end{array}\right)
=\left[\tilde{T_{I}}\right] ^{-1}
 \left(\begin{array}{c} I_{1} \\[10pt] I_{2} \end{array}\right)
\end{equation}
%
And the conductor voltages are related to the modal voltages by
%
\begin{equation}\label{MD_V}
\left(\begin{array}{c} V_{1} \\[10pt] V_{2} \end{array}\right)
=\left[\tilde{T_{V}}\right] 
 \left(\begin{array}{c} V_{m1} \\[10pt] V_{m2} \end{array}\right)
\end{equation}
%

A modal decomposition for a three conductor (two mode) transmission line is implemented by the network model in Figure \ref{fig:spice_modal_decomposition}. The model implements equations \ref{MD_I} and \ref{MD_V}. As for the domain decomposition matrix implementation, the voltage sources on the left hand side of the figure have a value of zero volts, they are only required as Spice uses such voltage sources as current sensing elements for the current controlled voltage sources.

\begin{figure}[h]
\centering
\includegraphics[scale=0.7]{./Imgs/Modal_decomposition_example.eps}
\caption{Spice circuit model which implements modal decomposition for a three conductor system}
\label{fig:spice_modal_decomposition}
\end{figure}


This model may be implemented in any of the Spice versions considered in this study and is suitable
for a.c. and transient analysis. It may be extended to an arbitrary number of conductors under the
assumption that the multi-conductor is lossless and non-dispersive.


\clearpage
</pre></body></html>