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1. A 3D CIRCUIT MODEL FOR EASIER AN UNDERSTANDING AND
SOLUTION OF ELECTROMAGNETIC INTERFERENCE PROBLEMS
Dénes CSALA, Dan Doru MICU
Technical University Cluj-Napoca, Faculty of Electrical Engineering, Tel. +40-744772756, E-mail: csaladenes@yahoo.com
Abstract: This paper presents a new method for representing
and solving the electric circuit associated with an
electromagnetic interference problem. A 3 dimensional
approach was used to geometrically represent the equivalent
electrical circuit. Then, the emerging properties of this new
structure were used to perform a mathematical solution
analysis. The presented methodology might provide an easier
understanding of the problem as it is closely coupled with the
geometrical interpretation. The resultant mathematical structure
is fully comprehensive and valid for the most general cases. It is
also robust, not needing redesign in case of changes in geometry
of the interfering electrical system. At this point only the
graphical and mathematical representation and description of
the problem is presented, exact circuit solution is not addressed
and represents the topic of further research.
Keywords: electromagnetic interference problem, 3D circuit,
block matrix
1. Introduction
The study of electromagnetic interference problems
between high voltage overhead lines and nearby metallic
pipelines presents a great importance, given by the
induced potentials. To reduce construction costs, in most
cases, water, gas or oil pipelines share the same
distribution corridor. The electromagnetic interference
between a power system network and neighboring gas
pipeline has been traditionally divided into three
categories: capacitive (electrostatic), conductive
(resistive) and inductive (magnetic) coupling.
The equivalent circuit method is the most widely used
method for theoretically treating electromagnetic
interference problems. Thoroughly studied and analyzed
by Dawalibi et. al [1] [2], its main idea stands in replacing
the elements of the interfering geometry with circuit
components with appropriate parameters for a normalized
segment of interfering voltage line – pipeline pair. This
way, circuit theory can be used to solve the problem, as
opposed to field analysis.
However, as the interfering geometry gets more
complex and number of interfering lines increases, visual
interpretation and understanding is burdened.
This paper tries to represent the equivalent the circuit
of an electromagnetic interference problem using a 3
dimensional approach, both graphically and
mathematically and it might provide an easier
understanding of the problem. The pursuing mathematical
analysis that follows is based on 3D arrays and matrices.
It is important to be noted that in this paper only the
graphical and mathematical representation and description
of the problem is presented, circuit solution is not
addressed and represents the topic of further research.
2. 3D circuit setup
Following Dawalibi’s idea of segmentation of infinitely
long high voltage lines and assigning an equivalent circuit
to each of segments, these circuits are transferred into 3D
space as presented in Fig. 1.
The three characteristic descriptors of the circuit
created in this way are segments, sections and nodes:
 Segments represent longitudinal portions of the
interfering power line geometry
 Sections represent transversal cuts of the
interfering power line geometry.
 Nodes are intersection points of segments with
sections.
 A segment is delimited by two sections
(throughout this paper, all circuits segments
studied are delimited by two consecutive
sections, but this is not a compulsory criterion)
 All quantities above are multiplied for the
number of interfering entities, from now on
referred to as lines (line, in this case, could also
represent a metallic pipeline!).
From the theory of electromagnetic induction, one
can assume that inductive currents will be flowing along
segments and capacitive currents will appear within
sections. To solve the created 3D circuit, voltage
potentials will be assigned to nodes and complex
impedances to the portions that connect nodes between
segments and nodes within sections. For an effective
analysis of the problem, we introduce the following
notations:
 Sections are indiced and are referred to as i
 Segments are delimited by two indiced sections i
and j and are referred to as Sij
 Lines are indiced and are referred to as Li
 Let i
j
V be the electric potential measured at the
intersection of section i and line Lj
 Let j
k
i
Z be the electric impedance measured
between sections i and j (segment Sij), on line
Lk. This represents longitudinal impedance.
 Let j
k
i Z be the electric impedance measured in
section k between lines Li and Lj. This
represents transversal impedance.
3. Solution method
Having developed and represented the circuit, the
method of node potentials will be used for solution. The
reasoning behind choosing this method is that it leads to
the smallest number of equations, compared to other
methods, such as the method of loop currents or the
classic Kirchhoff method.

2. Fig. 1. Structure of the 3D circuit
The method of node potentials has the following
general formula:
























ep
p
pp
p
p
e
p
p
e
p
p
I
V
Y
V
Y
V
Y
I
V
Y
V
Y
V
Y
I
V
Y
V
Y
V
Y




2
2
1
1
2
2
2
22
1
21
1
1
2
12
1
11
(1)
where: Yii is the total complex admittance of the circuit
arms connected to node i;
Vi is the complex potential of node i;
Yij is the total complex admittance of the circuit
arm situated between nodes i and j;
Iei is the equivalent current of node i;
p
j
i ,
1
,  ; p is number of nodes in the circuit.
Considering the notations presented in the previous
paragraph, one may write:
;
1
;
1
c
b
a
c
b
a
a
c
b
a
c
b
Z
Y
Z
Y 

Throughout this paper, node 1, 0 is considered
grounded: 0
1
0 
V , however that is just an assumption and
can be chosen arbitrarily.
In the demonstrator example a general circuit will be
considered, with n sections (1 - n) – thus n-1 segments
(S12 – Sn-1n) and m+1 lines (L0 – Lm), where line L0 is
considered to be the ground and all nodes along it nodes
have potential 0.
4. Circuit solution
4.1. General considerations
The equation that describes the first section (1,, left
terminal) of the circuit can be formulated as follows:
0
0
0
2
1
2
1
2
1
2
1
1
1
1
1
0
1
0
2
1
1
1
1
0
1
1
2
1
2
1
1
1
1
1
2
1
1
2
1
0
1
1
0
1
2
1
1
1
1
2
1
1
0
1
1
1
0
2
1
2
0
1
0
1
1
2
0
1
2
1
1
0
1
1
0
2
1
1
0
1
2
0
1
1
0
1
0



































































































































m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
Y
V
Y
V
Y
V
Y
V
Y
Y
Y
V
Y
V
Y
V
Y
V
Y
V
Y
Y
Y
Y
V
Y
V
Y
V
Y
V
Y
V
Y
Y
Y
Y
V







(2)
Section
Section

3. The equation that describes an arbitrary inner section (j,) of the circuit is:
0
0
0
1
1
1
1
1
1
0
0
1
1
1
0
1
1
1
1
1
1
1
1
1
0
1
0
1
1
1
1
1
0
1
1
1
0
1
0
0
1
1
0
0
1
0
1
1
0
0
1
0
1
0
0






 





























 

















 





































 









 





























 





























j
m
j
j
m
j
m
j
j
m
j
m
j
j
m
j
j
m
j
j
m
j
j
m
j
m
j
m
j
j
j
j
j
j
j
m
j
m
j
j
j
j
j
j
j
m
j
j
j
j
j
j
j
j
j
m
j
m
j
j
j
j
j
j
j
m
j
j
Y
V
Y
V
Y
V
Y
V
Y
Y
Y
Y
V
Y
V
Y
V
Y
V
Y
V
Y
Y
Y
Y
V
Y
V
Y
V
Y
V
Y
V
Y
Y
Y
Y
V







(3)
The equation that describes the last section n will have
the same form as (2). The equations differ for the
terminal sections and the inner sections. This is
because the first section does not have a “previous”
section and the last one does not have a “next” section,
respectively. To conserve symmetry, two extra
sections, 0 and n+1 are introduced. The m+1 nodes
m
t
Vt ,
0
;
0
 of section Σ0 will be connected only to
the nodes m
t
Vt ,
0
;
1
 of section Σ1, without any
connections within the section. The m+1 nodes
m
t
Vn
t ,
0
;
1


of section Σn+1 will be connected only
to the nodes m
t
Vn
t ,
0
;  of section Σn, also without
any connections within the section (only longitudinal
impedances). These will be the control potentials of the
system, depending on the type of a line. If the arbitrary
line Lt is an active circuit, then its control potential will
be equal to its nominal operating voltage,  t
N
t V
V 
0
.
In case of a ground wire or an earthed pipeline, the
control potential will be set to 0
0

t
V . The potentials
of the nodes of section n+1 m
t
Vn
t ,
0
;  will depend
on the load of the system and will be adjusted
correspondingly. The modified circuit is presented in
Figure 2.
Fig. 2. Structure of the extended 3D circuit
After this correction, the system can be rewritten in the following compact notation (the equations will be similar for all
sections, regardless of their type):

4. known.
are
,
,
Y
,
Y
;
,
0
;
,
1
,
,
,
0
0
1
0
1
1
0
1
1
0
1
1
1
1
1
1
1
1
0
1
1
1
1
1
1
0
1
1
1
1
0




















































 






















 




















 










n
k
k
n
k
n
k
m
k
t
t
j
k
k
t
t
j
k
j
k
k
j
k
j
j
k
j
j
k
j
j
k
j
j
k
k
k
j
k
j
j
j
k
k
j
j
j
k
m
k
t
j
t
k
j
t
k
t
j
t
k
j
t
k
j
j
k
k
j
j
j
k
k
j
j
j
k
m
k
t
j
t
k
j
t
k
t
j
t
k
j
t
k
j
j
k
j
j
m
k
t
t
j
k
k
t
t
j
k
j
k
V
V
m
k
n
j
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
V
Y
V
Y
V
Y
V
Y
V
or
Y
V
Y
V
Y
V
Y
V
Y
Y
Y
Y
V
(4)
This equation system has n·(m+1) equations, and the corresponding m
k
n
j
V j
k ,
0
;
,
1 
 unknowns, which means is
fully compatible and has a unique solution. If represented spatially, this system of equation would look like:
0
0
0
0
0
2
1
2
1
0
0
1
1
1
1
0
1
1
2
1
1
0
1
1
1
0
1
1
2
1
2
1
0
0
1
1
1
1
0
1
1
0
1
1
0
1
1
1
0
1
1
0
2
1
2
0
0
1
0
0
0
1
0
1
0
1
1
0
0
1
0
1
0
2
1
0
1
0
1
0
1
0
1
0
0
1
0
1
0
0
1
1
0
0
1
1
0
1
0
0
1
0
0
0
0
1
0
1
1
0
0
1
0
0
0
0
0
1
1
0
0
1
1
0
1
0
0
1
0
0
0
0
1
0
1
1
0
1
0
0
0
0




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







m
m
m
m
m
m
t
t
m
t
m
t
t
m
t
m
m
m
m
t
t
m
m
t
t
m
m
k
k
k
k
m
k
t
t
k
t
k
t
t
k
t
k
k
m
k
t
t
k
k
t
t
k
k
m
t
t
t
t
t
t
m
t
t
t
t
j
j
j
j
j
m
t
j
t
j
t
t
j
t
j
t
j
j
j
j
m
t
t
j
t
t
j
j
n
n
n
n
n
n
m
t
n
t
n
t
t
n
t
n
t
n
n
n
n
m
t
t
n
k
t
t
n
n
Y
V
Y
V
Y
V
Y
V
Y
Y
Y
Y
V
Y
V
Y
V
Y
V
Y
V
Y
Y
Y
Y
V
Y
V
Y
V
Y
V
Y
V
Y
Y
Y
Y
V
Y
V
Y
V
Y
V
Y
V
Y
Y
Y
Y
V
Y
V
Y
V
Y
V
Y
V
Y
Y
Y
Y
V








(5)
At this point the equation system can be solved and
a solution can be obtained by expanding the total jk
relations. However, a more compact form can be
obtained with just a few manipulations. In order to
achieve this, let us examine the variables encountered
and their types.
V is the unknown matrix and has the size of
n·(m+1):





















n
m
j
m
m
m
n
k
j
k
k
k
n
j
n
j
V
V
V
V
V
V
V
V
V
V
V
V
V
V
V
V




















2
1
2
1
1
1
2
1
1
1
0
0
2
0
1
0
V (6)
After the introduction of the symmetry-conserving
sections, the extended Vext will be a matrix of size
(n+2)·(m+1):

























1
1
1
1
1
0
2
1
2
1
1
1
2
1
1
1
0
0
2
0
1
0
0
0
0
1
0
0
n
m
n
k
n
n
n
m
j
m
m
m
n
k
j
k
k
k
n
j
n
j
m
k
V
V
V
V
V
V
V
V
V
V
V
V
V
V
V
V
V
V
V
V
V
V
V
V
























ext
V (7)
The transversal impedances can be arranged 3D array,
with the size of (m+1)·(m+1)·n:
































n
j
j
k
m
j
m
j
m
j
m
k
j
k
j
k
j
m
j
k
j
j
m
j
k
j
j
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
T
T
T
T
T
Z
Z
Z
Z
Z






















1
1
0
1
0
1
1
0
1
0
0
1
0
0
0
0
0
(8)
where
n
1,
=
j
,
m
0,
=
q
p,
;
n
1,
=
j
,
m
0,
=
p
;
0
p
j
q
q
j
p
p
j
p
Z
Z
Z


1←j→n
1←k→m

5. Correspondingly, the admittance matrix is also a
(m+1)·(m+1)·n 3D array:

























0
0
0
0
1
0
1
0
1
1
0
1
0
0
1
0




















j
k
m
j
m
j
m
j
m
k
j
k
j
k
j
m
j
k
j
j
m
j
k
j
j
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
T
Y (9)
where
n
1,
=
j
,
m
0,
=
p
;
0
n
1,
=
j
,
p
,
m
0,
=
q
p,
;
1












p
j
p
n
q
p
n
q
p
Y
q
Z
Y
One must observe that j
j
T
T Z
Y and are always
symmetrical matrices and also that
usually n
1,
=
j
,
1







 j
j
T
T Z
Y ,
but n
1,
=
j
,
1









j
j
T
T Z
Y , using o to note
element-by-element operation.
The longitudinal impedances can be arranged in a
matrix, with the size of (n-1)·(m+1).

































m
n
n
m
j
j
m
m
k
n
n
k
j
j
k
k
n
n
j
j
n
n
j
j
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
1
1
3
2
2
1
1
1
3
2
2
1
1
1
1
1
1
3
2
1
2
1
0
1
0
1
0
3
2
0
2
1




















L
Z (10)
where m
0,
=
k
p;
q
if 
k
q
p
Z
After the introduction of symmetry-conserving
sections, the extended impedance matrix Zext
L will
become a matrix of size (n+1)·(m+1):
And the corresponding extended longitudinal complex
admittance matrix Yext
L is:





































1
1
1
1
1
0
1
1
3
2
2
1
1
1
3
2
2
1
1
1
1
1
1
3
2
1
2
1
0
1
0
1
0
3
2
0
2
1
1
0
1
0
1
1
0
0
1
0
n
m
n
n
k
n
n
n
n
n
m
n
n
m
j
j
m
m
k
n
n
k
j
j
k
k
n
n
j
j
n
n
j
j
m
k
ext
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
























L
Y (11)
0
,
0
if
m
0,
=
k
,
n
0,
=
j
;
where
1
1
1
1
1














k
j
j
k
j
j
k
j
j
k
j
j
Y
Z
Z
Y
There are two possible expansions of the equation
system presented in relation (5), after j and after k,
respectively.
Fig. 3. Structure of the 3D equation system
4.2. Analysis after j (per section analysis)
Hereby, we present the expansion after j, using
vector notation. Deducing an expansion after k (per
line analysis) is very similar in steps and difficulty, so
this paper will not be addressing it.
These expressions have the following general form:
known.
are
m
0,
=
k
,
,
,
Y
,
Y
,
,
,
0
k
;
,
1
,
1
0
1
1
0
1
1
0
1
1
1
1
1
1
1
0
1
0













 






 






















 

































n
k
k
n
k
n
k
m
k
t
t
j
k
k
t
t
j
k
j
k
k
j
k
j
j
k
j
j
k
j
j
k
j
k
j
k
j
k
j
j
k
j
j
k
j
j
m
j
k
j
j
j
m
k
j
k
k
j
k
j
k
V
V
Y
Y
Y
Y
Y
Y
m
n
j
V
V
V
Y
Y
Y
V
V
V
V
Y
Y
Y
Y




(12)
An arbitrary section can be described by the following doubly-recursive formulation:

6. 



























































































































1
1
1
1
1
1
1
1
1
0
1
0
1
1
1
1
1
1
1
1
1
0
0
1
1
0
1
0
1
0
1
1
1
0
1
0
0
1
0
0
j
m
j
m
j
j
k
j
k
j
j
j
j
j
j
j
j
m
j
m
j
j
k
j
k
j
j
j
j
j
j
j
j
m
j
k
j
j
j
m
j
k
m
j
m
j
m
j
m
k
j
k
j
k
j
k
j
m
j
k
j
j
j
m
j
k
j
j
V
Y
V
Y
V
Y
V
Y
V
Y
V
Y
V
Y
V
Y
V
V
V
V
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y


























(13)
One must observe that the right hand side of the equation can be expressed in matrix form:




































































































































































1
1
1
1
1
0
1
1
1
1
1
0
1
1
1
1
1
0
1
1
1
1
0
1
1
0
1
0
1
0
1
1
1
0
1
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
j
m
j
k
j
j
j
m
j
j
k
j
j
j
j
j
j
m
j
k
j
j
j
m
j
j
k
j
j
j
j
j
j
m
j
k
j
j
j
m
j
k
m
j
m
j
m
j
m
k
j
k
j
k
j
k
j
m
j
k
j
j
j
m
j
k
j
j
V
V
V
V
Y
Y
Y
Y
V
V
V
V
Y
Y
Y
Y
V
V
V
V
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y


































































(14)
Stacking the three V matrices and collating (augmenting) the Y matrices, the previous sum of matrix equations (14) can
be represented as a single matrix equation:
0

























































































































1
1
1
1
1
0
1
0
1
1
1
1
1
0
1
1
1
1
1
0
1
0
1
0
1
1
1
0
1
0
0
1
0
0
1
1
1
1
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
j
m
j
k
j
j
j
m
j
k
j
j
j
m
j
k
j
j
j
m
j
j
k
j
j
j
j
j
j
m
j
k
m
j
m
j
m
j
m
k
j
k
j
k
j
k
j
m
j
k
j
j
j
m
j
k
j
j
j
m
j
j
k
j
j
j
j
j
V
V
V
V
V
V
V
V
V
V
V
V
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y


































































(15)
If the elements of the previous set (triade) section are
attached to relation (15), the general rule which governs
the system of equations emerges. Due to space constraints
of this sheet, the following compact notations are
introduced:
;
,
0
;
,
1
,
,
,
where
1
1
0
1
1
m
k
n
j
Y
Y
Y
Y
Y
Y
Y
Y
Y
m
k
t
t
j
k
k
t
t
j
k
j
k
k
j
k
j
j
k
j
j
k
j
j
k
j
j
k
k
k
j













































j
m
j
k
j
j
j
Y
Y
Y
Y




















0
0
0
0
0
0
0
0
0
0
0
0
1
0
D
Y (16)

7. ;
,
1
,
1
0
1
0
1
1
1
0
1
0
0
1
0
0
n
j
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
j
m
j
k
m
j
m
j
m
j
m
k
j
k
j
k
j
k
j
m
j
k
j
j
j
m
j
k
j
j
j
j




























































T
D Y
Y (17)
Let us rename the resultant matrix with YA:
arrays.
3D
are
that
knowing
,
;
,
1
,
A
D
T
A
T
D
A
T
D
Y
,
Y
,
Y
Y
Y
Y
Y
Y
Y





or
n
j
j
j
j
For the case of the longitudinal impedances, we introduce
the following matrices (arranged after j):
1
+
n
1,
=
j
,
0
0
0
0
0
0
0
0
0
0
0
0
1
1
1
1
0
1











































j
j
j
m
j
j
k
j
j
j
j
j
j
diag
Y
Y
Y
Y
ext
L
B
B
Y
Y
Y




















(18)
Then, we introduce block matrices Φ and Φext for the
potential matrices (column vectors):








































































 1
1
0
1
1
0
1
,
n
n
j
n
n
j
n
j
ext
ext
ext
ext
ext
ext
ext
ext
V
V
V
V
V
V
V
V
V
V
Φ
V
V
V
Φ






(19)
1
,
1
0
0
0
0
0
0
0
0
0
1



































































n
m
m O
O
O
O
O
O




 


 

















(20)
With these notations, relation (15) can be generalized to
the following form (block matrix equation)




















































































O
O
O
O
O
V
V
V
V
V
Y
Y
Y
O
O
O
O
O
O
O
Y
Y
O
O
O
O
O
O
O
O
O
Y
Y
O
O
O
O
O
O
O
Y
Y
Y
O
O
O
O
O
O
O
Y
Y
O
O
O
O
O
O
O
O
O
Y
Y
O
O
O
O
O
O
O
Y
Y
Y
ext
ext
ext
ext
ext
B
A
B
B
A
A
B
B
A
B
B
A
A
B
B
A
B








































1
1
0
1
1
1
1
1
1
2
2
2
1
1
n
n
j
n
n
n
n
n
j
j
j
j
j
j
j
(21)
This equation can be further simplified. If one notes:
Ψ
Y
Y
O
O
O
O
O
Y
Y
O
O
O
O
O
O
O
Y
Y
O
O
O
O
O
Y
Y
Y
O
O
O
O
O
Y
Y
O
O
O
O
O
O
O
Y
Y
O
O
O
O
O
Y
Y
A
B
B
A
A
B
B
A
B
B
A
A
B
B
A






































n
n
n
n
j
j
j
j
j
j
j
































1
1
1
1
1
2
2
2
1
(22)

8. 







































 Φ
V
Ψ
O
O
O
O
O
O
Y
ext
B
0
1
,










































 Φ
V
Ψ
Y
O
O
O
O
O
O
ext
B
1
1
,
n
n


(23)
the whole system of equations defined by relation (24)
and can be expressed in the following compact form:








 Φ
Ψ
Φ
Ψ
Φ
Ψ (24)
4.2 Choosing the method for solving the system
Both expansions (after j and after k) result in a
tridiagonal matrix. In case of the expansion after j this is a
block tridiagonal matrix. One might either choose the
expansion after j, because it contains simpler terms and
square matrices, but also it has only a “matrix-in-matrix”
form, which may present difficulties, or the expansion
after k, which has an explicit tridiagonal matrix form for
all k elements, however, the individual terms of the Ψ
block matrix have more complicated forms and it has two
more columns than rows.
The sytem can be solved with a variety of methods,
for example the tridiagonal (Thomas) algorithm, which is
briefly presented in the following.
The Thomas algorithm is a simplified Gauss
algorithm, having significantly less steps for a certain
system. (and complexity of O(n) instead of O(n3
) for a
system with n equations).
























































n
n
n
n
n
n
d
d
d
d
x
x
x
x
b
a
c
b
b
a
c
b
a
c
b













3
2
1
3
2
1
1
1
3
3
2
2
2
1
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
(25)






























n
i
a
c
b
a
d
d
i
b
d
d
n
i
a
c
b
c
i
b
c
c
i
i
i
i
i
i
i
i
i
i
i
i
,
2
,
'
'
1
,
'
1
,
2
,
'
1
,
'
1
1
1
1
1
1
1
(26)
1
,
1
,
'
'
'
1 





 n
i
x
c
d
x
d
x
i
i
i
i
n
n
(27)
To apply he Thomas algorithm for the expansion after j,
the following analogies can be made:
1
,
2
i
,
1
,
1
i
,
,
2
i
,
,
1
i
,
1
1
0
1
1
1



















n
d
d
d
n
c
n
a
n
b
i
n
n
n
i
i
i
i
i
i
O
V
Y
V
Y
Y
Y
Y
ext
B
ext
B
B
B
A
(28)





















































n
i
c
d
n
i
c
d
i
d
n
i
c
i
c
n
n
n
n
n
i
i
i
i
i
i
i
i
i
i
i
,
'
'
1
,
2
,
'
'
1
,
'
1
,
2
,
'
1
,
'
1
1
1
0
1
1
1
1
1
0
1
1
1
1
2
B
A
B
ext
B
B
A
B
A
ext
B
B
A
B
A
B
Y
Y
Y
V
Y
Y
Y
Y
O
Y
V
Y
Y
Y
Y
Y
Y
(29)
By studying and analysing the condition number of
the equation system presented in this paper, one can
deduce the system’s resistance to perturbations and since
the transversal and longitudinal mutual impedances highly
depend on the system geometry, an optimization
possibility of the geometry becomes possible.
It also must be noted that the presented 3D circuit
does not need redesign in case of a change in geometry of
the studied interfering system as only the components of
the equation system would be changing.
5. Conclusion
In this a paper presented a new approach to apply the
ubiquitous equivalent circuit method, extensively
analyzed by Dawalibi et al. A 3 dimensional approach was
used to geometrically represent the equivalent electrical
circuit. Then, the emerging properties of this new
structure were used to perform a mathematical solution
analysis. The presented methodology might provide an
easier understanding of the problem as it is closely
coupled with the geometrical interpretation
It is important to be noted that in this paper only the
graphical and mathematical representation and description
of the problem is presented, circuit solution is not
addressed and represents the topic of further research.
References