\documentstyle[12pt]{article}

\title{\large AN ALGORITHM FOR DETERMINING RESISTORS\\
AND CAPACITORS IN A 2-LAYER NETWORK}
\author{Neil L. York}
\date{August 13, 1993}


\setlength{\parskip}{.1in}

\begin{document}

\maketitle

\begin{abstract}
In this paper it is shown that the resistors and
capacitors of a 2-layer network 
can be determined by measuring alternating currents and voltages at 
boundary nodes.
\end{abstract}  

{\bf Key terms:}

{\em Resistor}. A resistor is the electrical device that limits current 
flow in accordance with {\it Ohm's Law}, ${\Delta}u=IR$. 
Here, ${\Delta}u$ represents 
an electric potential difference, $I$ the current, and $R$ the 
resistance.  Henceforth, resistors shall be referred to as 
{\em conductors}; numerically, a conductor's {\em conductance}, 
$\gamma$, is 
equivalent to $R^{-1}$.  Ohm's Law can thus be rewritten as 
$I={\gamma}{\Delta}u$.  All conductances to be considered are assumed
to be greater than zero.   

{\em Capacitor}.  A capacitor is the electrical device that condenses
charge.  Its {\em capacitance}, $C$, relates how much charge it can
collect for some ${\Delta}u$ across its
terminals. All capacitances will be assumed to be greater
than zero. These voltages will
be considered to be of the form
$Ae^{i{\omega}t}$, where $A$ is some arbitrary amplitude,
$i=\sqrt{-1}$, $\omega$ stands for frequency, and $t$ represents
time.  The expression $i{\omega}C$ is
unitwise equivalent to conductance; 
both $i{\omega}C$ and $\gamma$ will be termed
{\em admittances}, $a$.

{\em Network}.  Any network to be considered in this work is a graph of 
an electrical construction with conductors and
capacitors as edges.  Each of these conductors
and capacitors must be connected 
to at least one other element, but not necessarily to all others.  
Networks will be denoted by either $\Omega$ or $T$.

{\em Node}.  In a network, the nodes are those points that 
represent a terminal 
of at least one element.  A node may be shared by the terminals of 
several elements.

{\em Boundary node}.  A boundary node is a node that is arbitrarily chosen 
to have a given net current, which may be nonzero.  A node that is not 
a boundary node is called an {\em interior} node.

{\em Kirkhoff's Current Law}.  Kirkhoff's Current Law states
that an interior
node of a network must always have a
net current flow of zero.  Boundary nodes, by definition, are exempt from
Kirkhoff's Law.   

{\em 2-Layer network}.  A 2-layer network is one that consists of 
2 layers of conductors, with capacitors connecting the nodes between
them.  The conductors in the top layer lead to boundary nodes, 
while the 
lower layer conductors all meet at one interior node.  Figure 1 shows 
a 2-layer network 
consisting of three boundary nodes (denoted by 1, 2, and 3), four 
interior nodes, (denoted by 4 through 7), two layers of three conductors 
each, and two capacitors:

\input{fig1}

This specific example of a 2-layer network will now be called  
the {\em 3-prong, 2-layer network}.  An {\em n-prong, 2-layer network},
with $n\geq 2$, 
consists of $n$ boundary nodes, $n+1$ interior nodes, $2n$ 
conductors, and $n-1$ 
capacitors.  Figure 2 shows the 5-prong, 2-layer network: 

\input{fig2}

{\em Forward problem}.  Assume that a
network is known.  That is, all conductances and capacitances
are known, as well
as the connections between them.  The forward problem uses given potentials
at the network's boundary nodes to find potentials at its interior nodes
which satisfy Kirkhoff's Law.  By solving the forward problem, it is then
possible to determine boundary currents.  

{\em Inverse problem}.  For an inverse problem, we are given a graph
$\Omega$ with unknown values of admittances.
The inverse problem uses currents and potentials
at $\Omega$'s boundary nodes to determine its admittances.  
To ``solve''
an inverse problem means that all
of a network's internal components can be recovered by measuring
its boundary currents and potentials.


{\bf 1.  Preliminaries}.  While the inverse
problem can be solved for many types of networks,
there exists a relatively simple class of networks whose
inverse problems do not yield unique solutions.  The
networks of this class 
have $n$ connected spokes, $n$ boundary nodes,
$2mn$ conductors ($m$ on each spoke, $m>1$),
and $n(m-1)+1$ interior nodes. Figure 3 shows
an example similar to those in this class, ($n=3$ and $m=1$), which $\em can$ 
be solved, and will be shortly:

\input{ffig3.tex}

\vspace{.5in}  

In this network, to be referred to as
$T_1$ (for $T$ripod 1), nodes 1, 2, and 3 are 
boundary nodes.  For networks
with a greater number of spokes, the boundary nodes will
always be those at the
end of each spoke (those that are common to only one conductor).

Figure 4 shows a network of this type with $n=3$ and $m=2$:

\input{ffig4.tex}

\vspace{.5in}

The inverse problem of the network of Figure 4, $T_2$, can
{\em not} be solved, because of the multiple conductors on each spoke.
Any of these networks with $m$ greater than 1 can not be solved.
(It is important to note that all conductivities in these networks
come from resistors.  A tripod with $m>1$ may have a solvable
inverse problem if capacitors take the places of some resistors.) 

The following network is of a different type for which the
inverse problem can not be solved:

\input{ffig5.tex}

\vspace{.5in}

If $\gamma_{4,5}$, $\gamma_{5,6}$, and $\gamma_{4,6}$ in this
network were replaced with capacitors, it {\em would} have
a solvable inverse problem.  (Such a network would be
very similar to the network in Figure 9 on page 10.  The
algorithm to be described is based on the fact that this substitution
of capacitors for conductors allows such a network to have
a solvable inverse problem.)

Two representative matrices exist for any network which play
important roles in forward and inverse problems:
the {\em Kirkhoff matrix}, $K$, and the {\em Lambda matrix},
$\Lambda$.  The Kirkhoff matrix is square, with the
number of rows (and columns) being equal to the total
number of nodes in the network.

Construction of $K$ is dependent on the indexing of the nodes.
For a network with  $p$ boundary nodes and $q$ interior nodes,
the boundary nodes must be indexed from $1$ to $p$, and the interior
nodes indexed from $p+1$ to $p+q$.  Each node will then correspond
to one row and one column in $K$.  $K$ will consist of $(p+q)^2$ entries
labeled $k_{i,j}$, with $i,j=1$ to $(p+q)$.

Now, for example, 
consider some node $n$ in $T_1$, corresponding to row
$n$ in $K$. The ${n,j}$ entry of $K$ will always have a value of zero,
unless one
of two conditions is true:  either node $n$ is directly connected to node $j$
by a conductor, or $j=n$.  If the first condition is true,
$k_{n,j}= -\gamma_{n,j}$ where $\gamma_{n,j}$ is the conductance
between nodes $n$ and $j$.  If $j=n$,
$k_{n,n}$ is equal to the sum of all conductors touching node $n$.

A result of this construction is that the sum of all entries in any row
or column of $K$ will necessarily be zero.  That is,
{$\Sigma_{i}{\gamma_{i,j}}=0$} and {$\Sigma_{j}{\gamma_{i,j}}=0$}. 
$K$ is also symmetric,
as $k_{m,n}=-\gamma_{m,n}=-\gamma_{n,m}=k_{n,m}$ for any nodes
$m$ and $n$ in the network.   

Figure 6 shows the Kirkhoff matrix for $T_1$:

\begin{center}
\begin{large}
$K=$\input{matrix1}
\end{large}

{\sc Fig. 6}
\end{center}

(With $\sigma_4=\gamma_{1,4}+\gamma_{2,4}+\gamma_{3,4}$.)

The Kirkhoff matrix is now broken into four blocks, as shown by Figure 7:

\begin{center}
\begin{large}
$K=$\input{matrix2}
\end{large}

{\sc Fig. 7}
\end{center}

The break represented by the vertical line occurs between columns $p$ and
$p+1$, while the horizontal line occurs between rows $p$ and
$p+1$.  In other words, the breaks separate boundary nodes and interior nodes.
The part of $K$ now called $K'$ represents connections between boundary
nodes, while that called $A$ shows those between interior nodes.  $B$ and
$B^T$ show connections between one boundary node and one interior node.
Returning to the example network of $T_1$,  the components are:
 
\begin{center}
\begin{large}
$K'=$\input{k1}
$B^T=$\input{bt1}
$B=$\input{b1}
$A=[\sigma_4$]
\end{large}
\end{center}

The other key matrix, $\Lambda$, is now calculated from $K$ by:

\begin{equation}
\Lambda=K'-B^TA^{-1}B.
\end{equation}

For $T_1$,

\begin{center}
\begin{large}
$\Lambda=$ \input{matrix3}
\end{large}

{\sc Fig. 8}
\end{center}

$\Lambda$ shares two important characteristics of $K$.  $\Lambda$
is symmetric, and all of its rows and columns sum to zero.  These
two facts mean that only the off-diagonal entries of $\Lambda$ are
of interest. Furthermore,
$\Lambda$ has the same number of rows and columns as the network  has
boundary nodes, and the ${i,j}$ entry of $\Lambda$ represents
the current flowing into boundary node $i$ due to a potential of 1
applied at boundary node $j$.

The preceding follows the steps of the forward problem for $T_1$.
Since we now know the form
of $\Lambda$, the inverse problem--the problem of getting from
boundary currents to conductances--is readily solved.  We have, for
$T_1$:

\begin{eqnarray}
\lambda_{1,2} & = & -\gamma_{1,4}\gamma_{2,4}/\sigma_4  \\
\lambda_{1,3} & = & -\gamma_{1,4}\gamma_{3,4}/\sigma_4  \\
\lambda_{2,3} & = & -\gamma_{2,4}\gamma_{3,4}/\sigma_4
\end{eqnarray}

At this point, the ratio $\gamma_{1,4}/\gamma_{2,4}$ can be found by
dividing $\lambda_{1,3}$ by $\lambda_{2,3}$; 
$\gamma_{1,4}/\gamma_{3,4}$ and $\gamma_{2,4}/\gamma_{3,4}$ are
obtained similarly.  To proceed, choose a conductor to solve for, such as
$\gamma_{1,4}$.  The first and second of the above ratios are used to
express $\gamma_{2,4}$ and $\gamma_{3,4}$ in terms of $\gamma_{1,4}$.
Specifically,
$\gamma_{2,4}=\gamma_{1,4}\lambda_{2,3}/\lambda_{1,3}$, and
$\gamma_{3,4}=\gamma_{1,4}\lambda_{2,3}/\lambda_{1,2}$.
These expressions can then be substituted into $\sigma_4$,
which then factors down to
$\gamma_{1,4}(1+\lambda_{2,3}/\lambda_{1,3}+\lambda_{2,3}/\lambda_{1,2})$.

This reformulated $\sigma_4$, and the expression for
$\gamma_{2,4}$ in terms of $\gamma_{1,4}$ are then substituted into Equation
2 above.  That equation is now strictly in terms of
$\gamma_{1,4}$ and the three lambdas.  A little more algebra yields

\begin{equation}
\gamma_{1,4}=(\lambda_{1,2}\lambda_{1,3}+\lambda_{1,2}\lambda_{2,3}+\lambda_{1,3}\lambda_{2,3})/\lambda_{2,3}.
\end{equation}

By the same method, it is found that
\begin{equation}
\gamma_{2,4}=(\lambda_{1,2}\lambda_{1,3}+\lambda_{1,2}\lambda_{2,3}+\lambda_{1,3}\lambda_{2,3})/\lambda_{1,3}
\end{equation}

and

\begin{equation}
\gamma_{3,4}=(\lambda_{1,2}\lambda_{1,3}+\lambda_{1,2}\lambda_{2,3}+\lambda_{1,3}\lambda_{2,3})/\lambda_{1,2}.
\end{equation}

Thus, this inverse problem has yielded the conductances of
the three conductors of $T_1$
from information strictly about boundary currents.  (The same
can be done for any similar network with at least 3 spokes,
provided it has only 
one conductor per spoke.) The same technique
could have been applied to $T_2$, but there is only
enough information in $\Lambda$ to find the sums
$(\gamma_{1,4}+\gamma_{4,7})$, $(\gamma_{2,5}+\gamma_{5,7})$, and
$(\gamma_{3,6}+\gamma_{6,7})$.  Specific values for the six conductances
can not be determined because there are an infinite number of combinations
that could yield each sum.

With the addition of two
capacitors, possibly unknown, across the middle nodes of
$T_2$, the resulting
network is the 3-prong, 2-layer network of Figure 1 (now
called $\Omega_1$).
Lambda matrices
{\em can} be used to uniquely determine the six conductances
of $\Omega_1$.

{\bf 2.  2-Layer Networks vs. Spoked Networks}.  
All 2-layer networks, as defined,
are structured essentially the same as the spoked networks 
$T_1$ and $T_2$, 
with the addition of $n-1$ 
capacitors. ($n$ is the number of spokes in $T$.)  
All examples of these networks could have 
included a last, $n$th, capacitor, but the algorithm to be outlined
works just as well if it is not present.  
Furthermore, by using the 2-layer networks
as will be done, a geophysical application is suggested.  The network
$\Omega_1$, or perhaps better the network of Figure 2, 
could be used to model the
structure of a section of the Earth, with conductivities being analogous
to depth, and capacitances being the analog of magnetic 
induction\footnote{These interpretations 
were brought to the author's attention
by a paper titled {\em The Inverse Problem of Electromagnetic Induction:
Existence and Construction of Solutions Based On Incomplete Data} by 
Robert L. Parker. Published in the {\em Journal of Geophysical Research,
vol. 85, no. B8}, pp. 4421-4428, 1980.}. 

The property of capacitors that allows inverse problems containing them
to be solved is this: when admittance is expressed in terms of
capacitance, it is dependent on frequency. That is, 
$a=f(C,\omega)$ or, in this case, $a=i{\omega}C$.  
This implies that the Kirkhoff
matrix of a network containing capacitors, as well as its Lambda matrix,
also become functions of $\omega$.  (In the previous example of
$T_1$, $\Lambda$ had a guaranteed existence because $A$, being
composed of one real entry, was invertible.  An $A$ matrix
containing capacitances is not so obviously invertible, 
but will soon be shown to be so.)

A network consisting strictly of conductors is analyzed with the
use of direct currents; therefore, there is no $\omega$ dependence and
the network has only one Kirkhoff matrix 
and one Lambda matrix.  A network
containing capacitors and their resulting $\omega$ dependence,
on the other hand, has as many $K$'s and $\Lambda$'s as an analyst
has frequencies to choose from.

The algorithm to follow is a method of comparing these multitudes of
Lambda matrices to determine the composition of 2-layer networks.

{\bf 3.  Forward Problem}.  Before dealing with the inverse problem
of a 2-layer network, it needs to be shown that
the forward problem can be solved.  This, and all future calculations
and proofs, will be demonstrated using the 3-prong, 2-layer
network of Figure 9, to be called $\Omega_2$:

\input{fig8}

$\Omega_2$ has the same structure as $\Omega_1$, except that rather than
having six different conductors, it has only two types, with each
layer consisting of one type of conductor.  Anything
to be done with this network can be similarly done with $\Omega_1$. 
$\Omega_2$ is preferable strictly because it will have fewer unknowns
to solve for.  This model
is perhaps a little simple, but will suffice for our purposes.

The Kirkhoff matrix for $\Omega_2$ is shown in Figure 10:

\vspace{1in}

\begin{center}
\begin{large}
$K(\omega)=$ \input{matrix4}
\end{large}

{\sc Fig. 10}
\end{center}

{\noindent}where:

\begin{eqnarray}
\sigma_4 & = & \gamma_1 + \gamma_2 + i{\omega}C_{4,5} \\
\sigma_5 & = & \gamma_1 + \gamma_2 + i{\omega}(C_{4,5}+C_{5,6}) \\
\sigma_6 & = & \gamma_1 + \gamma_2 + i{\omega}C_{5,6}
\end{eqnarray}

The lines in $K(\omega)$ represent the borders between $K'$, $B^T$,
$B$, and $A$, as in Figure 7.

The question of whether or not $\Lambda(\omega)$ exists for
$\Omega_2$, or any other 2-layer network,
comes down to the existence or non-existence of $A^{-1}$.  
In the example of 
$T_1$, this was easily verified because $A$ was simply one 
real entry.  Here, this question of $A(\omega)$'s invertibility 
is not so straightforward.

{\sc Theorem 1}.  {\em $A(\omega)$ is invertible for any 2-layer network.}

{\em Proof}.  Recall that $A(\omega)$ is the lower-right 
block of $K(\omega)$, as is shown by Figure 11:   

\begin{center}
\begin{large}
$A(\omega)=$\input{matrix5}

\end{large}
{\sc Fig. 11}
\end{center}

To prove the existence of $A^{-1}(\omega)$, it is sufficient to show
that the only solution to the equation:

\begin{equation}
A(\omega)u(\omega)=0
\end{equation}

{\noindent}is $u=0$; that is, $A(\omega)$ is nonsingular.  
Assume that $A(\omega)u(\omega)=0$.

Consider a column vector $x(\omega)$:

\begin{center}
\begin{large}
$x(\omega)=$\input{matrix6}
\end{large}

{\sc Fig. 12}
\end{center}

{\noindent}whose entries stand for potentials at a network's nodes. 
The zero in $x(\omega)$ is a 
column vector representing a network's boundary
potentials, all zero, and $u(\omega)$ is another column vector of interior
potentials.  Now, 

\begin{center}
\begin{large}
$K(\omega)x(\omega)=$\input{matrix7}$=$\input{matrix8}
\end{large}
\end{center}
 
{\noindent}by our assumption that $A(\omega)u(\omega)=0$.

It can be verified that: 

\begin{equation}
\bar{x}^{T}(\omega){K(\omega)}{x}(\omega)=0, 
\end{equation}

{\noindent}with $\bar{x}(\omega)$ being the complex 
conjugate of $x(\omega)$.

Now consider a matrix $C(\omega)$, for some 
network, to be constructed as follows.
Number its rows according to the network's nodes, and number
its columns according to its edges (conducting elements).  In each
column $l$, corresponding to edge $l$, there will be two entries. 
One will appear in each row corresponding to one
of the nodes that edge $l$ connects, differing only by sign.  
The entries will be of the form
$\pm\sqrt{a_{m,n}}$, with $a_{m,n}$ being the value of the admittance
of edge $l$ between connected nodes $m$ and $n$ in the network.
The sign of each element does not matter, as long as one entry in each
column is positive and the other is negative.  Any $C(\omega)$  
will look something like Figure 13, with appropriate dimensions: 

\begin{center}
\begin{large}
$C(\omega)=$\input{matrix9}
\end{large}

{\sc Fig. 13}
\end{center}
 
It can be verified that:

\begin{equation}
K(\omega)={C(\omega)}{C^T{(\omega)}}.
\end{equation}

This is substituted into Equation 12 above, resulting in:

\begin{equation}
({\bar{x}^{T}}C)({C^T}x)=0.
\end{equation}

$C^T{(\omega)}{x}$ looks like Figure 14:

\begin{center}
\begin{large}
$C^T{(\omega)}{x}=$\input{matrix10}
\end{large}

{\sc Fig. 14}
\end{center}
 
\vspace{1in}

Therefore,

\begin{equation}
0={\bar{x}^{T}}{K}{x}={\bar{x}^{T}}{C}{C^T}{x}={\Sigma}{a_{i,j}}{|{x_i}-{x_j}|}^2.
\end{equation}

This can be broken down to:

\begin{equation}
0=\Sigma Re (a_{i,j}){{|{x_i}-{x_j}|}^2}+i\Sigma Im (a_{i,j}){{|{x_i}-{x_j}|}^2}.
\end{equation}
 
All real parts of $A$ are non-negative, 
since all conductivities were originally
assumed to be greater than zero.  The same applies 
to the imaginary parts.  From Equation 16, it follows
that $x_i=x_j$ for all $i,j$ where $\gamma_{i,j}\neq 0$.
Hence, all the $x_i$'s must be equal to
all the $x_j$'s (for all $i,j$), which will 
then be equal to some constant.  But since
the $x$ matrix (Figure 12) already has some entries equal to zero,
{\em all} entries of $x$ must be equal to zero.  Therefore, $u=0$
and $A$ is invertible, concluding the proof. $\bullet$

It is significant that 
the only conditions necessary for the above to hold is
that all conductivities be greater than zero and that likewise all
capacitors be greater than zero.  The shape of the network and size
of $A$ make no difference.

The forward problem can therefore
always be solved for 2-layer networks as considered here.  
$\Lambda (\omega)$ can in fact be computed for $\Omega_2$.
The entries of $\Lambda (\omega)$ will consist of ratios of
polynomials of degree $n-1$ (where $n$ is the number of spokes in
the network) in $\omega$.

{\em Note}: This $\Lambda (\omega)$ is 
actually a ``pseudo-Lambda'' matrix.  The true
$\Lambda$ has an additional coefficient ${e^{i\omega t}}$ in front of
each entry because of the alternating voltage being applied.  (In 
networks with strictly real conductivities, the ${i,j}$th
entry of $\Lambda$ signified the 
current into boundary node $i$ due to a potential
of 1 at boundary node $j$.  With the addition of capacitors, 
${\lambda_{i,j}}$ now represents the current into boundary node $i$ 
caused by a potential of ${e^{i\omega t}}$ at boundary node $j$.)  
The term $e_{i{\omega}t}$ will be divided out of each
entry of $\Lambda (\omega)$ to produce the Lambda
matrix to be used.  
(This has the additional benefit of making$\Lambda (\omega)$
independent of $t$).   

\vspace{1in}

{\bf 4.  Inverse Problem and Algorithm.}  The 
before-mentioned polynomials in 
$\omega$, even those of small degrees,  
will turn out to have many more terms than are comfortable
to work with, such as this entry, ${\lambda_{1,2}} (\omega)$, from
the Lambda matrix of $\Omega_2$:

\begin{tiny}
\begin{equation}
\frac{-(\gamma_1^2(\gamma_1\gamma_2^2+\gamma_2^3+3iC_{4,5}\gamma_1\gamma_2\omega+2iC_{4,5}\gamma_2^2\omega+2iC_{5,6}\gamma_2^2\omega-3C_{4,5}C_{5,6}\gamma_2\omega^2))}{(3(\gamma_1^3\gamma_2+2\gamma_1^2\gamma_2^2+\gamma_1\gamma_2^3+2iC_{4,5}\gamma_1^2\gamm
a_2\omega+2iC_{5,6}\gamma_1^2\gamma_2\omega+2iC_{4,5}\gamma_1\gamma_2^2\omega+2iC_{5,6}\gamma_1\gamma_2^2\omega-3C_{4,5}C_{5,6}\gamma_1\gamma_2\omega^2))}
\end{equation}
\end{tiny}

While this ratio is only quadratic in $\omega$, it is 
not a very simple expression.  It is theoretically possible that if 
given one Lambda matrix of a 2-layer network,
a form like the one above could be used for the  
$\lambda_{i,j} (\omega)$'s to calculate the network's 
constituent conductors and capacitors.
This is not necessary, given that an alternate method exists.

This method, soon to be described, takes advantage
of the fact that $\Lambda$ is dependent on $\omega$.  Since the 
entries of $\Lambda$ depend on $\omega$, it is reasonable
to think that there may be some values of $\omega$ that provide
useful results.  

The key aspect of the algorithm is that it manipulates these
useful frequencies to solve for a network's components.
Specifically, it involves taking the limits
of $\Lambda (\omega)$ as
$\omega$ goes to zero and infinity to find the admittances of a
2-layer network of known shape and unknown composition.  The steps
of the algorithm, as applied to $\Omega_2$, are summarized as follows:

\begin{enumerate}
\item Compute $\Lambda (\infty)$.  From this, $\gamma_1$ can be \
determined.
\item Look at $\Lambda(0)$.  From this, $\gamma_2$ can be determined.
\item Calculate $C_{4,5}$ and $C_{5,6}$ from known potentials at \
$\Omega$'s interior nodes (solved from the known conductors).
\end{enumerate}

Each step will now be considered in detail:

{\bf 4.1}.  Inspection of Equation 17 reveals that as $\omega$ goes to infinity,
$\lambda_{1,2} (\omega)$ of $\Omega_2$ looks like:
  
\begin{equation}
{\lambda_{1,2} (\omega)}=\frac{-3C_{4,5}C_{5,6}\gamma_1^2\gamma_2}{9C_{4,5}C_{5,6}\gamma_1\gamma_2}
\end{equation}

{\noindent}which, after canceling terms, leaves:

\begin{equation}
\gamma_1=-3\lambda_{1,2} (\omega)
\end{equation}

{\noindent}(as $\omega$\ approaches $\infty$).

Physically, what is occurring in $\Omega_2$ 
as $\omega$ approaches infinity is
that the two capacitors are short circuiting, and the
network takes the form and 
has the same properties as the tripod networks with three
conductors discussed earlier.

In fact, each non-diagonal entry of $\Lambda (\infty)$ can be used
to compute $\gamma_1$.
For any $n$-prong, 2-layer network with only two types
of conductor, one in the top layer and one in the lower, as 
$\omega$ goes to infinity:

\begin{equation}
\gamma_1=-n\lambda_{i,j} (\omega)
\end{equation}
  
{\noindent}with $i\neq j$.

For a network like $\Omega_2$, with three different
conductors in the top layer, a situation would occur
similar to that of the tripod network previously solved.  That is,
some algebraic manipulation would still need to be performed
to isolate the three top layer conductors from $\Lambda$.
 
The problem with this method is that it is a physical impossibility
to actually have $\omega$ approach infinity.  Therefore, an alternate
approach is required.  Since Equation 17 is quadratic in $\omega$,
it can be expressed in the form:

\begin{equation}
\lambda_{1,2} (\omega)=p(\omega)=\frac{1+b\omega+c\omega^2}{\alpha+\beta\omega+\gamma\omega^2}
\end{equation}

{\noindent}with $b,c,\alpha,\beta$, and $\gamma$ being five 
undetermined coefficients.  
For an $n$-prong, 2-layer network in general, 
$\lambda_{i,j} (\omega)$ is a ratio of 
polynomials of degree $n-1$ with $2n-1$ coefficients.  If, in the case of 
$\Omega_2$, $c$ and $\gamma$ could be determined from other data,
$p(\omega)$ would be known and the $\lambda_{1,2} (\omega)$ 
(for $\omega = \infty$) would be 
found as the ratio $c/\gamma$.

The values $p (\omega_i)$ for five
different frequencies will yield a system of
five linear equations in the five unknowns
$b,c,\alpha,\beta$, and $\gamma$.  
(In the case of the general 2-layer network,
$2n-1$ different values of $\omega$ would be needed.) 
That system is:

\begin{equation}
1+b\omega_i+c\omega_i^2=p(\omega_i)(\alpha+\beta\omega_i+\gamma\omega_i^2) \ i=1,\cdots,5
\end{equation}

Let:

\begin{center}
\begin{large}
$M=$\input{matrix11}
\end{large}

{\sc Fig. 15}
\end{center}

Unique solutions for b,c,$\alpha$,$\beta$,
and $\gamma$ 
can be found if the determinant of $M$ does not equal zero.

The ratio of quadratic expressions $p(\omega_i)$ can be rewritten as:

\begin{equation}
p(\omega_i)=tq(\omega_i)=\frac{t(\omega_i-\delta_1)(\omega_i-\delta_2)}{(\omega_i-\epsilon_1)(\omega_i-\epsilon_2)}
\end{equation}

{\noindent}(with some non-zero constant $t$).

%\vspace{2in}

$M$ can therefore be written as:

\begin{center}
\begin{large}
$M=$\input{matrix12}
\end{large}

{\sc Fig. 16}
\end{center}

The determinant of $M$ can be shown to be:

\begin{equation}
t^3\frac{\Pi_{i>j}(\omega_i-\omega_j)\Pi_{i=1}^2\Pi_{j=1}^2(\delta_i-\epsilon_j)}{\Pi_{i=1}^5\Pi_{j=1}^2(\omega_i-\epsilon_j)}
\end{equation}

If $\omega_i\neq\epsilon_j$,
$\delta_i\neq\epsilon_j$, and $\omega_i\neq\omega_j$, for all $i,j$,
det($M$)$\neq$0.
Therefore, as long as these conditions hold, $b,c,\alpha,\beta$, and 
$\gamma$ can be determined.
This allows the limit of $\lambda_{1,2}$ as 
$\omega$ approaches infinity to be calculated.  
Equation 19 then allows $\gamma_1$ to be computed. 

In the general case of the $n$-prong, 2-layer network (where
all conductors are different), each entry of $\Lambda (\omega)$ would
have a different form, and consist of a ratio of different polynomials.
Here, enough coefficients (such as b,c,$\alpha, \beta,$ and $\gamma$)
would have to be found to find each entry.  From this point,
the conductor in the network's top layer
can be recovered in a similar way to those of the earlier
tripod example.
  
{\bf 4.2}.  Another look at Equation 17 shows that the limit of 
$\lambda_{1,2} (\omega)$ as $\omega$ approaches zero will be:

\begin{equation}
\lambda_{1,2}(0)=\frac{-\gamma_1^2(\gamma_1\gamma_2^2+\gamma_2^3)}{3(\gamma_1^3\gamma_2+2\gamma_1^2\gamma_2^2+\gamma_1\gamma_2^3)}
\end{equation}

{\noindent}This reduces to:

\begin{equation}
\lambda_{1,2}(0)=\frac{-\gamma_1\gamma_2}{3(\gamma_1+\gamma_2)}
\end{equation}

This equation can be solved for $\gamma_2$:

\begin{equation}
\gamma_2=\frac{-3\lambda_{1,2}(0)\gamma_1}{\gamma_1+3\lambda_{1,2}(0)}
\end{equation}

Taking the limit of $\Lambda$ as $\omega$ goes to zero is {\em not}
a physical impossibility--this is the equivalent of applying direct current
to the network.  Therefore $\lambda_{1,2}(0)$ can be 
found much easier than $\Lambda(\infty)$.  As long as
we have $\Lambda(0)$, $\gamma_2$ is found simply by solving 
Equation 27, using the previously obtained $\gamma_1$.  (In the case
where not all lower layer conductors are the same, a substitution 
method similar
to that used for solving $T_1$ (Figure 3) will have to be used to arrive
at an equation like Equation 27.)

A more general form exists for $\gamma_2$ in the $n$-prong, 
2-layer network with all top-layer conductors the same
and all lower-layer conductors the same:

\begin{equation}
\gamma_2=\frac{-n\lambda_{i,j}(0)\gamma_1}{\gamma_1+n\lambda_{i,j}(0)}
\end{equation}

{\noindent}with $i\neq j$.

At this point, the algorithm has recovered $\gamma_1$ and $\gamma_2$.  
Therefore, we know all of the 
conductors in the network.  All that remains to be done 
is to find the capacitances.

{\bf 4.3.}  All of the conductors in $\Omega_2$, shown in
Figure 9, are known.  The same is true for 
the $\omega$-dependent boundary currents, 
by the nature of inverse problems.  Ohm's Law can
now be used to find interior potentials throughout the network.  
With the potentials
at each node known, the currents across each conductor are also known, 
again via Ohm's Law.

To solve for $C_{4,5}$, consider the current from node 1 to node 4,
$I_{1,4}$.  Then
consider the current from node 4 to node 7, $I_{4,7}$.  The difference between
$I_{1,4}$ and $I_{4,7}$ must go across $C_{4,5}$
(by Kirkhoff's Law).  With this, and
knowing the potential difference across nodes 4 and 5, Ohm's Law will give
the admittance between nodes 4 and 5.  This 
then allows $C_{4,5}$ to be calculated
from $a=i{\omega}C$, with any positive, finite $\omega$.  
The same method is used to find $C_{5,6}$, and is also
used to find the capacitances of a more general network such as
$\Omega_1$ once all of its conductors are known. 

This completes the reconstruction algorithm for $n$-prong, 2-layer
networks.

{\bf  5.  Commentary}.  This reconstruction algorithm works only for 
2-layer networks.  The
inverse problem for a  3-layer (or more)
network, with an extra layer (or more) of conductors and one additional
row (or more)
of capacitors, can $\em not$ be solved this way.  The limits of
$\Lambda$ as $\omega$ approaches zero and infinity provide
only enough information to solve for the conductors in the top layer,
and a combined conductivity for each pair of connected conductors
in the bottom two layers--the same problem
facing the original tripod with 2 conductors on each spoke.

It may be that knowing $\Lambda (\omega)$ for more values of
$\omega$ allows the inverse problems of such networks to be solved.
However, it is beyond the scope of this article to examine
this possibility\footnote{For a more detailed look
at the inverse resistor problem, 
see {\em Finding the
Conductors in Circular Networks From Boundary Measurements} by Edward
Curtis and James Morrow, published in the {\em SIAM Journal 
of Applied Mathematics, vol. 50, no. 3}, pp. 918-930, June 1990.}.

\end{document}
