# Introduction

In recent years great research effort has been directed toward fault diagnosis of Electrical Power Generation Systems. Many papers have been published on this subject [1, 2], and even if we restrict ourselves to the theoretical treatment of faults, there is a variety of problem formulations depending on fault models, measurement conditions, the final object of the fault diagnosis, and so forth. Key factors in the classification of diagnostic problems are the following.

1. Subsystems under diagnosis:

$$\quad$$ a. Elements contained in the subsystem: 1) “RLC or RC”, 2) “controlled sources, i. e. by definition, a two-port element which describes the energy transference or energy conversion between two subsystems.
$$\quad$$ b. Excitation of the system:

$$\quad \quad$$ 1) DC Power Supply.
$$\quad \quad$$ 2) AC Power Supply – Single-frequency or multi-frequency.

$$\quad$$ c. Number of exciting sources (independent sources or inputs):

$$\quad \quad$$ 1) Single exciting source.
$$\quad \quad$$ 2) Multiple exciting sources.
2. Fault models:

$$\quad$$ a) Short circuits or open circuits (hard faults).
$$\quad$$ b) Malfunction of an element or subsystem that occurs at intervals, usually irregular, in an element or subsystem that functions normally at other times (intermittent faults).
$$\quad$$ c) Element or parameter-value deviations outside the tolerance bounds (soft faults)
3. Measurements:
$$\quad$$ a) Voltage,
$$\quad$$ b) Current,
$$\quad$$ c) Frequency,
$$\quad$$ d) Phase.
4. Final object of fault diagnosis:
$$\quad$$ a) To determine voltages and currents,
$$\quad$$ b) To identify the faulty elements,
$$\quad$$ c) to locate and isolate the faults within a subsystem.
In addition to the problem formulation directly related to electrical and mechanical elements, attempts have been made to apply techniques of system diagnosis to Electrical Power Diagnosis (EPD).

In diagnosis of large- scale system, use of the computers is inevitable. One special feature of EPD strategy which is quite different from those of Energy System analysis is that the information for diagnosis or knowledge about the subpart of the EP system is very restricted. Therefore diagnosability, or whether or not the diagnostic problem formulated is solvable under the specific requirements and the expected performance indexes are achievable, must be preliminary researched in order to discriminate instability intrinsic to the problem from that due to the computational method and errors. Since there are several problem formulations, diagnosability must be described as structural as functional according to them (Gracios-Marin 2014)

In the following sections, the diagnosability of soft faults in a EP system which is in a “Fuzzy MLD Fault Model” is consider. An advantage of assuming soft faults only is that the EP model structural and functional description is known, and the equation derived for the Mixed Logical Dynamic (MLD) definition included simple, directed and co-directed subsystem, relationship can be inferred.

It is possible to assume that the system contains as active elements, controlled-transferring conversion subsystems and several element-values relationship in terms of passive and active energy conversion in the different steps of the model. Most elements and function conversion can be represented by their equivalent equation conversion model and at last, a complete dynamic and recursive equation. Therefore, this approach not suffers loss of generality.

For simplicity of discussion we further assume that the system contains direct transfer energy conversions in just one direction. Either if a cross-coupling effect is required to be expressed then an inverse transfer function will be used to describe it.

Computability of the parameters in the system from measured and known values is developed first, since it is the basis of fault diagnosis. Then the result is applied to fault location by the fault verification or assume-and-check method. It is shown that the diagnosability of an EP system, like the system diagnosability, depends on the connectivity of the system under test. Finally, a brief discussion is given on diagnosis by multiple taps of evaluation mode.

# Preliminaries to problem formulation.

## System Configuration

The system under consideration is denoted by n, and its structural and functional description by $$G$$. In order to overcome the difficulty arising from mutual coupling in controlled sources we use two subsystems $$N_v$$ and $$N_i$$ and two graphs $$G_v$$ and $$G_i$$.

Then $$G_v$$ and $$G_i$$ are called the voltages and current graphs, respectively (Shifang 2007), and Kirchhoff’s voltage and current laws are applied to $$G_v$$ and $$G_i$$, respectively. These subsystems and graphs are derived from $$N$$ and $$G$$ in the following way. Nv is derived from N by contracting the current sensors and nullators and deleting dependent sources and noratos. $$G_v$$ is the graph of $$N_v$$. Next, $$N_i$$ is the circuit which is derived from $$N$$ by deleting voltage sensors and nullators and contracting dependent voltage sources and norators. $$G_i$$ is the graph of $$N_i$$. Note that either the sensor or the dependent source of a controlled source remains in $$N_v$$ or $$N_i$$, and a two-port element in $$N$$ is represented by a two-terminal element in $$N_v$$ and a two-terminal element in $$N_i$$, in the same way as a resistor. Thus there is no need to distinguish a controlled source from a resistor in $$N_v$$ and $$N_i$$. The element in $$N_v$$ and the element in $$N_i$$ representing the same element in $$N$$ are given the same label. Then $$N_v$$ and $$N_i$$ have a common element set, which is denoted by $$E$$. $$G_v$$ and $$G_i$$ also have a common edge set. Strictly speaking, circuits $$N$$, $$N_v$$, and $$N_i$$ are constituted by “elements” which correspond to “edges” of graphs $$G$$, $$G_v$$, and $$G_i$$. For simplicity we will not distinguish edges from elements hereafter. Thus the common edge set of $$G_v$$ and $$G_i$$ is also denoted by $$E$$.

## Formulation of Measurement Conditions.

Next let us formulate the measurement on $$N$$ [4]. The element set $$E$$ of $$N_v$$ and $$N_i$$ is partitioned into $$E_b$$, $$E_e$$, $$E_j$$, $$E_k$$ and $$E_u$$, which are defined as follows.

$$E_b$$ => the set of elements whose voltages and currents can be both measured and know.
$$E_e$$ => the set of elements whose voltages only can be measured and know.
$$E_j$$ => the set of elements whose currents only can be measured or know.
$$E_k$$ => the set of elements whose voltages and currents are unknown but whose element values are known.
$$E_u$$ =>the set of elements whose voltages and currents are unknown and whose element values are also unknown.
More specifically, Table 9.1 applies to a resistor, inductor, capacitor, or controlled source. Table 9.2 applies to an independent voltage or current source.

To insert the Tables 9.1 and 9.2...

 Voltage Current Value Class Known Known Known $$E_b$$ Known Known Unknow $$E_b$$ Known Unkown Known $$E_b$$ Unknown Known Known $$E_b$$ Known Unkown Unkown $$E_e$$ Unkown Known Unknown $$E_j$$ Unkown Unkown Known $$E_k$$ Unkown Unknown Known $$E_k$$ Unkown Unknown UnKnown $$E_k$$

The above classification of elements can be interpreted in terms of degrees of freedom as follows. Each element in the circuit originally has two degrees of freedom, because two variables, a voltage and a current, are associated with it. The degrees of freedom are decreased by the information obtained for the element or the restriction imposed on it. Thus an element in $$E_b$$ has zero; one in $$E_e$$, $$E_j$$ or $$E_k$$ has one; and one in $$E_u$$ has two degrees of freedom. Note that in an ordinary circuit analysis each element in the circuit has one degree of freedom. The total number of KVL and KCL equations is equal to the number of elements, and thus the voltages and currents can be uniquely determined. The measurement condition regarding a node can be expressed by an element which is connected to the node and the reference node, if such an element exists, or by an imaginary element which is added between the node and the reference node.

 Voltage Current Class Known Known $$E_b$$ Known Unknown $$E_e$$ Unkown Known $$E_j$$ Unknown Unknown $$E_u$$

## Graph Notation and Basic Proposition

For any graph $$G$$ with edge set $$E$$ and subset $$E_s$$ of $$G \cdot E_s$$ (respectively $$G x E_s$$ ) is the graph obtained from $$G$$ by deleting (respectively contracting) the edges of $$\overline E_s:= E - E_s$$. The rank and nullity of $$G$$ are denoted by $$r(G)$$ and $$n(G)$$, respectively. $$\parallel E_s \parallel$$ is the cardinality of $$E_s$$ and + denotes the union of disjoint sets.

Regarding trees and cotrees of $$G$$ we have the following propositions. For simplicity we use the same notation for a tree or cotree and its edge set.

Proposition 2.1 Let $$T_a$$ and $$T_b$$ be trees of $$G \cdot E_s$$ and $$G x E_s$$, respectively. Then the union of $$T_a$$ and $$T_b$$ constitutes a tree of G, and

$r(G) = r(G \cdot E_s) + r(G x \overline E_s), n(G) =n(G \cdot E_s) + n(G \cdot E_s + n(G x \overline E_s) \tag{2.1}$

Let $$T$$ be a tree of $$G$$ and $$K$$ be the cotree of $$T$$

Proposition 2.2 Given a subset $$E_s$$ of $$E$$. Then

$\stackrel{\hbox{max}}{\hbox{all T}} \parallel T \cap E_s \parallel =r(G \cdot E_s), \stackrel{\hbox{max}}{\hbox{all K}} \parallel K \cap E_s \parallel =n(G x E_s) \tag{2.2}$

where the maximums are taken over all the trees of $$G$$. The maximum in the first (second) equation is attained if and only if $$T \cap E_s$$ is a tree of $$G \cdot E_s (K \cap E_s$$ is a cotree of $$G x E_s$$).

# Diagnosability... Al last!!!

## Computability

We first consider which voltages and currents in $$N$$ can be computed from measurement data and know data [4]. This is because element or parameter values can be computed, in general, only from voltages and currents, and also it may be possible to diagnose $$N$$ from voltages and currents. Note that the voltage of an element in $$E_u$$ and $$E_j$$ can be obtained from a KVL equation only, since its element value is unknown.
Suppose we use the fundamental tiesets determined by a cotree of $$G_v$$ to derive KVL equations. Then the voltage of such an element can be computed if and only if the element is included n the cotree and all the voltages of the tree elements which belong to the fundamental tieset determined by this element are known. Therefore we want as many elements of $$E_u$$ and $$E_j$$ to be included i the cotree as possible, and as many elements of $$E_b$$ and $$E_e$$ to be included in the tree as possible.
Dually, the current of an element in $$E_e$$ and $$E_u$$ can be obtained from a KCL equation only. The current of such an element can be compute if and only if it is included in a tree of $$G_l$$ and all the currents of the cotree elements which belong to the fundamental cutset determined by it are known. Thus we want as many elements of $$E_u$$ and $$E_e$$ to be included in the tree as possible, and as many elements of $$E_b$$ and $$E_j$$ to be included in the cotree as possible. The above situation is the same as that of determining the voltages of current sources or the currents of voltages sources in conventional circuit analysis. These currents and voltages can be determined only after the currents and voltages of elements other than the sources are determined. For the elements in $$E_k$$ we can use their voltages versus current relations, that is, $i_k=Y_kv_k \tag{3.1}$

where $$i_k$$ and $$v_k$$ are the current and voltage vectors of the elements in $$E_k$$, respectively, and $$Y_k$$ is a diagonal matrix with admittances of the elements on the diagonal. We can get KVL and KCL equations for the $$E_k$$ usingn the cotree of $$G_v$$ and the tree of $$G_i$$ as stated above. They can be written as:

$B_kv_k=e_k \tag{3.2}$

$Q_ki_k=j_k \tag{3.3}$

where $$e_k$$ and $$j_k$$ are vectors due to the known voltages and currents, respectively, and $$B_k$$ and $$Q_k$$ are the fundamental tieset and cutset matrices, respectively, which can be derived from graphs $$G_vk$$ and $$G_ik$$ defined as follows:

$G_{vk}:=G_v x (E_u + E_j + E_k) \cdot E_k \tag{3.4}$

$G_{ik}:=G_i \cdot (E_u + E_j + E_k) x E_k \tag{3.5}$

In order to get $$G_{vk}(G_{ik})$$ we contract the elements in $$E_b$$ and $$E_e$$ ($$E_u$$ and $$E_e$$) which we want to be in the tree of $$G_v (G_i)$$, and we delete those in $$E_u$$ and $$E_j$$ ($$E_b$$ and $$E_j$$) which we want to be in the cotree.

Example 3.1 For the circuit shown in Figure 1.1 a suppose that the voltages $$e_1$$ and $$v_6$$ (the subscripts correspond to the numbers on the edges in the figure) are measurable and that conductances $$g_3$$, $$g_4$$, $$g_5$$ and $$g_6$$ are known but $$g_2$$ is unknown. Then $$E_b=\{6\}$$, $$E_e=\{1\}$$, $$E_j=\emptyset$$, $$E_k=\{3, 4, 5\}$$ and $$E_u=\{2\}$$.
Voltage graph $$G_v$$ is shown in Figure 1.1b together with a tree $$T_v$$ which is indicated by the dashed lines. $$T_v$$ is chosen so that $$T_v \ni$$1, 6 and its cotree $$K_v(:=E-T_v) \ni$$2. Current graph $$G_i$$ is shown in Figure 1.1c together with a tree ]$$T_i$$. $$T_i$$ is chosen so that $$T_i \ni$$1, 2 and its cotree $$K_i(:=E-T_i)\ni$$6. $$G_{vk}$$ and $$G_{ik}$$ are given in Figures 1.1d and 1.1e, respectively. The KVL and KCL equations for $$E_k$$ are $\left[ {\begin{array}{cc} 1 & 0 & 1\\ 0 & 1 & 0 \ \end{array} } \right] \left[ {\begin{array}{cc} v_3\\v_4\\v_5 \end{array} } \right]= \left[ {\begin{array}{cc} v_6\\e_1 - v_6\ \end{array} } \right] \tag{3.6}$

Replace this text with your caption

$\left[ {\begin{array}{cc} 0 & -1 & 1 \ \end{array} } \right] \left[ {\begin{array}{cc} i_3\\i_4\\i_5 \end{array} } \right]= \left[ {\begin{array}{cc} -i_6\ \end{array} } \right] \tag{3.7}$

respectively. We also have

$i_3=g_3v_3, i_4=g_4v_k, i_5=g_kv_k \tag{3.8}$

Solving Esq. (3.6) - (3.8) we can determine all the voltages and currents of the elements in $$E_k$$ and then all the remaining voltages and currents in the circuit.

Equations (3.1) - (3.3) may or may not be solvable. We will consider topological conditions on the solvability of these equations, assuming that there exists no special relation among the element values, or all the elements values are independent. This assumption is called the generality assumption on element values. If we eliminate $$i_k$$ using Eq. (3.1) we get a set of equations for $$v_k$$ whose coefficient matrix is

$\left[ {\begin{array}{cc} B_k\\Q_k Y_k \end{array} } \right] \tag{3.9}$

It is known that this matrix can be, in general, transformed to a block diagonal matrix [xx] as

$\left[ {\begin{array}{cc} M_{++} & 0 & 0 \\ M_{0+} & M_{00} & 0 \\ M_{-+} & M_{-0} & M_{--} \end{array} } \right] \tag{3.10}$

and $$E_k$$ is tripartitioned accordingly into $$E_{k+}$$, $$E_{k0}$$, and $$E_{k-}$$. In (3.10) $$M_{++}$$ is a rectangular matrix with more rows than columns, $$M_{00}$$ is a square matrix, and $$M_{--}$$ is a rectangular matrix with fewer rows than columns; $$M_{++}$$, $$M_{00}$$, and $$M_{--}$$ correspond to $$E_{k+}$$, $$E_{k0}$$ and $$E_{k-}$$, respectively; and 0 is a zero matrix. Let $$v_{k+}$$, $$v_{k0}$$, and $$v_{k-}$$ be the voltage vectors of the elements in $$E_{k+}$$, $$E_{k0}$$, and $$E_{k-}$$, respectively. Then $$M_{++}$$ gives an overdetermined system of equations for $$v_{k+}$$; that is, there are more equations than unknown variables of $$v_{k+}$$. If $$e_k$$ in Eq. (3.2) and $$j_k$$ in Eq. (3.3) are obtained from measured voltages and currents, these equations for $$v_{k+}$$ must be consistent, and we can determine $$v_{k+}$$. Then we can determine $$v_{k0}$$, since $$M_{00}$$ is a square matrix and is nonsigular under the generality assumption on element values. We cannot, however, determine $$v_{k-}$$. In this example we had a special case of $$E_{k+}=E_{k-}=0$$. The partition of the matrix in (3.10) or the partition of $$E_k$$ can be determined by Algorithm PART+ given below. This algorithm determines the “principal partition” of graphs $$G_{vk}$$ and $$G_{ik}$$ together with a special pair of trees, $$T_{vk}$$ of $$G_{vk}$$ and $$T_{ik}$$ of $$G_{ik}$$, which have as many common edges as possible. Such a pair of trees is called a pair of maximally common trees (MCT). The cotrees of $$T_{vk}$$ and $$T_{ik}$$ are denoted by $$K_{vk}$$ and $$K_{ik}$$ respectively, that is, $$K_{vk}:=E_{k}-T_{vk}$$ and $$K_{ik}:=E_{k}-T_{ik}$$. We first give some definitions. Let $$e$$ be a cotree edge in $$K_{vk}$$. We denote by $$L_{v}(e/T_{vk})$$ the fundamental tieset uniquely determined by $$e$$ with respect to $$T_{vk}$$ and define

$L_{v}(E_{s}/T_{vk}):=\bigcup_{e \epsilon E_{s}} L_{v}(e/T_{vk}) \tag{3.11}$ Similarly, if $$e$$ is a tree edge in $$T_{ik}$$, $$C_{i}(e/K_{ik})$$ is the fundamental cutset uniquely determined by $$e$$ with respect to $$K_{ik}$$, and

$C_{i}(E_{s}/K_{ik}):=\bigcup_{e \epsilon C_{i}} L_{v}(e/K_{ik}) \tag{3.12}$

Likewise, for $$e$$ in $$K_{ik}$$, $$L_{i}(e/T_{ik})$$ [respectively $$e$$ in $$T_{vk}$$ $$C_{v}(e/K_{vk})$$] denotes the fundamental tieset(cutset) uniquely determined by $$e$$ with respect to $$T_{ik}(K_{vk})$$. $$L_{i}(E_{s}/T_{ik})$$ and $$C_{v}(E_{s}/K_{vk})$$ are defined similarly to Eqs. (3.11) and (3.12).

Algorithm PART $$+$$
$$Step 0$$: {Initialization}
$$\quad$$Obtain an arbitrary pair of trees $$T_{vk}$$ of $$G_{vk}$$ and $$T_{ik}$$ of $$G_{ik}$$ (their cotrees are $$K_{vk}$$ and $$K_{ik}$$, respectively).
$$Step 1:$$
$$\quad$$(1.1) Set $$X_{1}:=K_{vk} \cap T_{ik}$$, and $$m:=1;$$
$$\quad$$(1.2) If $$X_{1}:=\varnothing$$, go to step 7;
$$Step 2:$$ {Beginning of the search for tree transformation}
$$\quad$$(2.1) Set $$m:=2$$, and $$X_{2}:=C_{i}(X_{1}/K_{ik});$$
$$\quad$$(2.2) If $$X_{2}\cap (T_{vk}\cap K_{ik}) \neq \varnothing$$, go to step 5;
$$Step 3:$$ {Search in G_vk}
$$\quad$$(3.1) Set $$m:=m+1$$, and obtain $X_{m}:=L_{v}(X_{m-1}/T_{vk})-\bigcup_{n=1}^{m-1} X_{n} \notag{}$

$$\quad$$(3.2) If $$X_{m}:=\varnothing$$, go to step 7;
$$\quad$$(3.3) If $$X_{m}\cap (T_{vk}\cap K_{ik}):=\varnothing$$, go to step 6;
$$Step 4:$$ {Search in G_ik}
$$\quad$$(4.1) Set $$m:=m+1$$, and obtain $X_{m}:=C_{i}(X_{m-1}/K_{ik})-\bigcup_{n=1}^{m-1} X_{n} \notag{}$
$$\quad$$ (4.2) If $$X_{m}=\varnothing$$, go to step 7;
$$\quad$$ (4.3) If $$X_{m} \cap (T_{vk} \cap K_{ik})=\varnothing$$, go to step 3;
$$Step 5$$:{Tree transformation when search ends in $$G_{ik}$$}
$$\quad$$ There must be a sequence of edges {$$e_{j},..., e_{n-1}, e_{n},..., e_{m}$$} such that $$j=1$$ of 2, $$e_{j}\in K_{vk} \cap T_{ik}$$, $$e_{m}\in T_{vk} \cap K_{ik}$$, $$e_{m} \in X_{n}$$ and $$e_{n} \in C_{i}(e_{n-1}/K_{ik})$$ or $$L_{v}(e_{n-1}/K_{ik})$$. Perform either of the following tree transformations:

(5.1) If $$j=1$$, $$\quad T_{vk}:=T_{vk} \cup \{e_2, e_4,.... e_{m-2}\}-\{e_3, e_5,..., e_{m-1}\}$$
$$\quad T_{ik}:=T_{ik} \cup \{e_2, e_4,.... e_m\}-\{e_1, e_3,..., e_{m-1}\}$$

(5.2) If $$j=2$$, $$\quad T_{vk}:=T_{vk} \cup \{e_2, e_4,.... e_{m-2}\}-\{e_3, e_5,..., e_{m-1}\}$$
$$\quad T_{ik}:=T_{ik} \cup \{e_4, e_6,.... e_m\}-\{e_3, e_5,..., e_{m-1}\}$$
Goto step 1;
$$Step 6$$:{Tree transformation when search ends in $$G_{vk}$$}
There must be a sequence of edges $$\{e_j,..., e_{n-1},e_n,..., e_m\}$$ such that j=1 or 2, $$e_j \in K_{vk} \cap T_{ik}$$, $$e_m \in T_{vk} \cap K_{ik}$$, $$e_n \in X_n$$, and $$e_n \in C_i(e_{n-1}/T_{ik})$$ or $$L_v(e_{n-1}/T_{vk})$$. Perform either of the following tree transformations:

(6.1) If $$j=1$$, $$\quad T_{vk}:=T_{vk} \cup \{e_2, e_4,.... e_{m-1}\}-\{e_3, e_5,..., e_{m}\}$$
$$\quad T_{ik}:=T_{ik} \cup \{e_2, e_4,.... e_{m-1}\}-\{e_1, e_3,..., e_m\}$$

(6.2) If $$j=2$$, $$\quad T_{vk}:=T_{vk} \cup \{e_2, e_4,.... e_{m-1}\}-\{e_3, e_5,..., e_m\}$$
$$\quad T_{ik}:=T_{ik} \cup \{e_4, e_6,.... e_{m-1}\}-\{e_3, e_5,..., e_{m-2}\}$$
Go to step 1.
$$Step 7:$$ {Search ends without breakthrough. $$T_{vk}$$ and $$T_{ik}$$ obtained are a pair of MCT.}
Set $E_{k+}:=\bigcup_{n=1}^{m-1} X_{n}\notag{}$
Return $$T_{ik}$$, $$T_{vk}$$ and $$E_+$$.
Algorithm PART+ begins, at step 0, with an arbitrary pair of trees, $$T_{vk}$$ and $$G_{ik}$$ of $$G_{ik}$$, and tries to increase $$\mid T_{vk} \cap T_{ik}\mid$$ by decreasing $$\mid K_{vk} \cap T_{ik}\mid$$ and $$\mid T_{vk} \cap K_{ik}\mid$$. To this end PART+ performs a breadth-first search for tree transformation of $$T_{vk}$$ and $$T_{ik}$$. This search starts from the edges of $$K_{vk}\cap T_{ik}$$ (steps 1 and 2), aiming at the breakthrough to $$T_{vk}\cap K_{ik}$$, and is performed in $$G_{ik}$$ and $$G_{vk}$$ alternating (steps 3 and 4, respectively). If a breakthrough is obtained, tree transformation of $$T_{vk}$$ and $$T_{ik}$$ is performed in steps 5 and 6. In case of no breakthrough a pair of MCT and $$E_+$$ are obtained at step 7.
The search for tree transformation may be initiated in either $$G_{ik}$$ or $$G_{vk}$$. We choose $$G_{ik}$$ in Algorithm PART+, as given in step 2. Note that X_2 includes $$K_{vk} \cap T_{ik}$$, and if $$X_2 \cap (T_{vk} \cap K_{ik})=\varnothing$$, the fundamental tiesets determined by the edges of $$K_{vk} \cap T_{ik}$$ are sought in $$G_{vk}$$. The sequence of edges obtained by the search initiated in this way begins with $$e_2$$, which corresponds to $$j=2$$ in step 5 and step 6. $$X_2, X_3,..., X_{m-1}$$ are mutually disjoint, and thus we only have to find new fundamental tiesets or cutsets.
Once a pair of MCT is obatined by Algorithm PART+, $$E_{k-}$$ can be determined by the following algorithm PART-.

Algorithm PART $$-$$
$$Step 1$$:
$$\quad$$(1.1) Using a pair of MCT $$T_{vk}$$ and $$T_{ik}$$, set $$X_{1}:=T_{vk} \cap K_{ik}$$, $$m:=1;$$
$$\quad$$(1.2) If $$X_{1}:=\varnothing$$, go to step 5;
$$Step 2.$$ Set $$m:=2$$, and obtain $$X_{2}:=C_{v}(X_{1}/K_{vk});$$
$$Step 3:$$
$$\quad$$(3.1) Set $$m:=m+1$$, and obtain $X_{m}:=L_{i}(X_{m-1}/T_{ik})-\bigcup_{n=1}^{m-1} X_{n} \notag{}$

$$\quad$$(3.2) If $$X_{m}:=\varnothing$$, go to step 5;
$$Step 4:$$
$$\quad$$(4.1) Set $$m:=m+1$$, and obtain $X_{m}:=C_{v}(X_{m-1}/K_{vk})-\bigcup_{n=1}^{m-1} X_{n} \notag{}$
$$\quad$$ (4.2) If $$X_{m}=\varnothing$$, go to step 3;
$$Step 5$$:
$$\quad$$ Set $E_{k-}:=\bigcup_{n=1}^{m-1} X_{n} \notag{}$

Return $$E_{k-}$$.
We have:
Theorem 3.1 $$T_{ik}$$ and $$T_{vk}$$ obtained at the termination of Algorithm PART+ is a pair of MCT.
$$Proof$$. For any tree pair $$T_{vk}$$ and $$T_{ik}$$ and an arbitrary edge subset $$E_{ks}$$ of $$E_k$$, we have
\begin{aligned} \mid K_{vk}\cap T_{ik}\mid & \geq \mid K_{vk}\cap T_{ik} \cap E_{ks} \mid \\ & =\mid E_{ks} \mid -\mid T_{vk}\cap E_{ks} \mid -\mid K_{ik}\cap E_{ks}\mid +\mid T_{vk}\cap K_{ik}\cap E_{ks}\mid \\ & \geq \mid E_{ks} \mid -\mid T_{vk}\cap E_{ks} \mid -\mid K_{ik}\cap E_{ks}\mid \\ & \geq \mid E_{ks} \mid -r(G_{vk}\cdot E_{ks})-n(G_{ik} \times E_{ks}) \\ & = r(G_{ik} \times E_s)-r(G_{vk} \cdot E_s)\end{aligned}
The last “$$\geq$$” in the above equation is due to Proposition 2.2. In last equation, the equalities hold if and only if
$K_{vk}\cap T_{ik}\subseteq E_{ks}, \quad T_{vk}\cap K_{ik}\cap E_{ks}=\varnothing \notag{}$

$\mid K_{vk}\cap E_{ks}\mid =r(G_v\cdot E_{ks}), \quad \mid T_{ik}\cap E_{ks}\mid =n(G_{ik}\times E_{ks}) \notag{}$

It can be seen that $$T_{vk}$$, $$T_{ik}$$, and $$E_{k+}$$ obtained at the termination of PART+ satisfy the last conditions for the equalities. Thus $$\mid K_{vk}\cap T_{ik} \mid$$ and $$\mid T_{vk}\cap K_{ik} \mid$$ are the minimum, and $$\mid T_{vk}\cap T_{ik} \mid$$ is the maximum.
From the discussion in the above proof and its dual we can easily get the following minimax theorem.

Theorem 3.2
$\min_{K_{vk},T_{ik}} \mid K_{vk}\cap T_{ik}\mid = \max_{E_{ks} \subseteq E_{k}} {r(G_{ik}\times E_{ks})-r(G_{vk}\cdot E_{ks})}:=\delta_{k+} \tag{3.22}$

$\min_{T_{vk},K_{ik}} \mid T_{vk}\cap K_{ik}\mid = \max_{E_{ks} \subseteq E_{k}} {r(G_{vk}\times E_{ks})-r(G_{ik}\cdot E_{ks})}:=\delta_{k-} \tag{3.23}$

$$\delta_{k+}$$ and $$\delta_{k-}$$ defined in eqs. (3.22) and (3.23) respectively are called the deficiencies. In fact, E_k+ and E_k- are minimal subsets of E_k giving the deficiencies, that is, minimal subsets such that

${r(G_{ik}\times E_{k+})-r(G_{vk}\cdot E_{k+})}:=\delta_{k+} \tag{3.24}$
${r(G_{vk}\times E_{k-})-r(G_{ik}\cdot E_{k-})}:=\delta_{k-} \tag{3.25}$

E_k+ and E_k- are uniquely determined irrespective of a pair of MCT obtained in PART+ and used in PART-. From Eqs. (3.24) and (3.25) we get

$\delta_{k+}={r(G_{ik}\times E_{k+})+n(G_{vk}\cdot E_{k+})}- \mid E_{k+}\mid \tag{3.26}$
$\delta_{k-}=\mid E_{k-}\mid-{r(G_{ik}\cdot E_{k-})-n(G_{vk}\times E_{k-})} \tag{3.27}$
From Proposition 2.2 we see that $$r(G_{ik} \times E_{k+})$$ and $$n(G_{vk} \cdot E_{k+})$$ are the minimum numbers of KCL and KVL equations, respectively, which are obtained for $$E_{k+}$$, and $$r(G_{ik} \cdot E_{k-})$$ and $$n(G_{vk} \times E_{k-})$$ are the maximum numbers of KCL and KVL equations, respectively, which can be obtained for $$E_{k-}$$. Therefore Eq. 3.26 [respectively Eq. (3.27)] indicates that no matter how we choose trees of $$G_{vk}$$ and $$G_{ik}$$, there are excess equations for $$v_{k+}$$ (excess unknown variables in $$v_{k-}$$).
Suppose a pair of MCT is obtained, $$T_{vk}$$ of $$G_{vk}$$ and $$T_{ik}$$ of $$G_{ik}$$. Using Proposition 2.1, then a expression is constructed $$T_v$$ of $$G_v$$ and $$T_i$$ of $$G_i$$ as follows. Let

\begin{aligned} T_{vb}:=\text{a tree of } G_v \cdot(E_b + E_e), & T_{vb}:=\text{a tree of }G_i \times(E_b + E_j) \\ T_{vu}:=\text{a tree of }G_v \times(E_u + E_j), & T_{vb}:=\text{a tree of }G_i \cdot(E_u + E_e)\end{aligned}

Then $$T_v$$ and $$T_i$$ and their cotrees are

\begin{aligned} T_{v}:=T_{vk}+T_{vb}+T_{vu}, K_v:=E-T_v & \\ T_i:=T_{ik}+T_{ib}+T_{iu}, K_i:=E-T_i\end{aligned}

Now the voltage of an element in $$K_{vu}$$ can be computed if and only if the fundamental tieset determined by it consists of this element and elements from $$E_b+E_e+E_{k+}+E_{k0}$$ only. The set of all such voltage-computable elements in $$K_{vu}$$ is denoted by $$E_{vd}$$. Likewise, the current of an element in $$T_{iu}$$ can be computed if and only if the fundamental cutset determined by it consists of this element and elements from $$E_b+E_j+E_{k+}+E_{k0}$$ only. The set of all such current-computable elements in $$T_{iu}$$ is denoted by $$E_{id}$$.
If the object of diagnosis is the voltage values for each element in set $$F_v$$ and the currents of elements in set $$F_i$$, we have the following theorem for diagnosability [4].

Theorem 3.3 The circuit is diagnosable if and only if $$F_v \subseteq E_b+E_e+E_{k+}+E_{k0}+E_{vd}$$ and $$F_i \subseteq E_b+E_j+E_{k+}+E_{k0}+E_{id}$$.

Example 3.2 Let us consider the circuit shown in Figure 9.2a. Element 13 is an imaginary one representing the voltage of node x with respect to the reference mode. Suppose $$E_b$$={11, 12, 13}, $$E_e$$={1}, $$E_j=\{\varnothing\}$$, $$E_u$$={2, 3, 4} and $$E_k$$={5, 6, 7, 8, 9,10}. From $$G_v$$=$$G_i$$, which is shown in Figure 9.2b, we get $$G_{vk}$$ and $$G_{ik}$$, shown in Figures 9.2c and 9.2d, respectively. We apply PART+ to these graphs.

$$Step 0$$: {Initialization}
$$\quad$$ Choosing $$T_{vk}$$={5,7} and $$T_{ik}$$={6, 8, 10} then $$K_{vk} \cap T_{ik}$$={6, 8, 10} and $$T_{vk} \cap K_{ik}=\{5, 7\}$$.

$$Step 1$$:
$$X_1=\{6, 8, 10\}.$$
$$Step 2$$:
$$X_2=\{6, 8, 7, 9\}$$ 7.

Replace this text with your caption

$$Step 5$$:
$$\quad$$ $$e_1$$=6, e_2=7. Getting new $$T_{vk}=\{5,7\}$$ and $$T_{ik}=\{7, 8, 10\}$$ and sets $$K_{vk} \cap T_{ik}= \{8, 10\}$$ and $$T_{vk} \cap K_{ik}=\{5\}$$.
$$Step 1$$:
$$\quad$$ $$X_1=\{8, 10\}$$.
$$Step 2$$:
$$\quad$$ $$X_2=\{8, 6\}$$. $$X_2 \cap \{5\}= \varnothing$$.
$$Step 3$$:
$$\quad$$ $$X_3=\{6, 5, 7\}-X_1-X_2=\{5, 7\} \ni 5$$.
$$Step 6$$:
$$\quad$$ $$e_1=8$$, $$e_2=6$$, $$e_3=5$$. Getting $$T_{vk}=\{6, 7\}$$ and $$T_{ik}=\{6, 7, 10\}$$. $$K_{vk} \cap T_{ik}=\{10\}, T_{vk} \cap K_{ik}=\{\varnothing \}$$
$$Step 1$$:
$$\quad$$ $$X_1=\{10\}$$.
$$Step 2$$:
$$\quad$$ $$X_2=\{10\}$$.
$$Step 3$$:
$$\quad$$ $$X_3=\{10\}-X_1-X_2=\varnothing$$.
$$Step 7$$:
$$\quad$$ $$E_{k+}=\{10\}$$.

Since $$T_{vk} \cap K_{ik}=\varnothing$$, we have $$E_{k-}=\varnothing$$. Then $$E_{k0}=\{5, 6, 7, 8, 9\}$$. Next we choose $$T_{vb}=\{11, 12, 13\}$$ and $$T_{iu}=\{2, 3\}$$. We have $$T_{vu}=T_{ib}=\varnothing$$. Finally, we get $$T_v$$ of $$G_v$$ and $$T_i$$ of $$G_i$$ as shown by the dashed lines and thin lines respectively in Figure 9.2b, and thus we get $$E_{vd}=\{2, 3, 4\}$$ and $$E_{id}=\{2\}$$.

## Fault Localization

For an element in $$E_{vd} \cap E_{id}$$ we can calculate its element value, since its voltage and current are both known. Then we can determine whether or not it is faulty. Next, suppose we have only to locate the fault within a portion of the circuit. A Classical but powerful way to do this is to assume the fault is localized in a certain subcircuit and then to check this assumption by some means or other. If the assumption is not verified, another part is assumed to be faulty and the process is repeated(Recursive procedure). Then it can be said the faulty within a portion is diagnosable if the assumption on it can be verified.
If the part assumed to be faulty is replaceable, the simplest way to check the assumption would be to replace it with a new good part and see if the circuit works correctly. In many cases this is not possible. The assumption must be checked by the measurement and equations we can get for the circuit. This assume-and-check method has been rigorously formulated and studied by several authors [6-11]. In the following we will apply the discussion of Section 3.1 to the assume-and-check method [10, 11].
Element sets $$E_b$$, $$E_e$$ and $$E_j$$ are determined from the measurement condition. The remaining elements are partitioned into $$E_k$$ and $$E_u$$, which are now redefined as follows:
$$E_k$$:= the set of elements which are assumed to be faulty-free
$$E_u$$:= the element set which is assumed to contain faulty elements
We assume the effect of element value deviations is negligible and the nominal values can be used for the elements in $$E_k$$. The element values of those in $$E_u$$ are unknown. Then we can derive exactly the same sets of equations as (3.1), (3.2) and (3.3) and get the principal partition of $$E_k$$ into $$E_{k+}$$, $$E_{k0}$$ and $$E_{k-}$$.
We have more equations than unknown variables for $$E_{k+}$$ and under the generality assumption on element values, these equations are consistent if and only if $$E_{k+}$$ contains no faulty element. Thus we have the following theorem.

Theorem 3.3 If $$E_{k0}+E_{k-}=\varnothing$$, then the assumption that the fault is located in $$E_u$$ can be checked (the assumption is correct if and only if the equations for $$E_{k+}$$ are consistent) and the fault within $$E_u$$ is diagnosable.
If $$E_{k0}+E_{k-}\neq$$, we can redefine $$E_u$$, that is, replace it with $$E_u+E_{k0}+E_{k-}$$.

Corollary The fault in $$E_u+E_{k-}+E{k0}$$ is diagnosable.
Suppose that another fault assumption is made, and $$E-E_b-E_e-E_j$$ is now partitioned int $$E_u ^*$$ and $$E_k ^*$$. Suppose, for this partition, that $$E_k ^*$$ is further partitioned into $$E_{k+} ^*$$, $$E_{k0} ^*$$, and $$E_{k-} ^*$$. It may be possible that $$E_{k+} ^*=E_{k+}$$. Then faults in $$E_u$$ and those in $$E_u ^*$$ cannot be distinguished by fault verification. In such a case $$E_u$$ and $$E_u ^*$$ are said to belong to an ambiguity set.
Now a node is defined to be faulty if any element connected to it is faulty [9]. To find out whether or not a particular node is faulty, that is, whether or not the fault is located within the set of elements which are incident on this node, we have only to set $$E_u$$ equal to this element set and apply the above discussion. Node fault diagnosability can easily be obtained using Theorem 3.3. A disadvantage of considering node faults is that a single faulty branch not connected to the reference node will cause a two-node fault.
Next let us study now an ambiguity set arises. We consider the graphs obtained from $$G_v$$ and $$G_i$$ by contracting the elements of $$E_e$$ and deleting those of $$E_j$$. For simplicity, the new pair of graphs is also denoted by $$G_v$$ and $$G_i$$. It is known that the principal partition of $$G_{vk}$$ and $$G_{ik}$$ can be related to the partition with respect to electrical connections as follows [4]. The electrical interactions are defined for $$E_b$$ and $$E_u$$ as [12]
\begin{aligned} k_b& := \min_{E_b \subseteq E_t \subseteq E-E_u} {r(G_v\cdot E_t)-r(G_i \times E_t)} \\ k_u& := \min_{E_u \subseteq E_t \subseteq E-E_b} {r(G_i\cdot E_t)-r(G_v \times E_t)}\end{aligned}
where the minimums are taken as all possible subsets $$E_i$$ such that $$E_b \subseteq E_t \subseteq (E-E_u)$$ and $$E_u \subseteq E_t \subseteq (E-E_b)$$, respectively. The minimal subsets which give $$k_b$$ and $$k_u$$ are denoted by $$E_c$$ and $$E_w$$, respectively; that is, $$E_c$$ and $$E_w$$ are the terminal subsets such that \begin{aligned} r(G_v\cdot E_c)-r(G_i \times E_c)=k_b \\ r(G_i\cdot E_w)-r(G_v \times E_w)=k_u\end{aligned} respectevely.
If $$G_v$$ and $$G_i$$ have a common tree, that is, a tree which is a tree of $$G_v$$ and a tree of $$G_i$$ at the same time, it can be shown that $$k_b=k_u$$ and that $$E-E_w(E-E_c)$$ is the maximal subset which gives $$k_b(k_u)$$.
If $$G_v=G_i-G$$, it is rather easy to see the relation between the electrical connectivities and the conventional vertex connectivity. For easier understanding, suppose that $$E_b$$ and $$E_u$$ are the mutually disjoint sets of edges incident on certain vertices b and u, respectively. We note that $$r(G \cdot E_t)$$ and $$r(G \times E_t)$$ are equal to the numbers of edges belonging to trees of $$G \cdot E_t$$ and $$G \times E_t$$, respectively. The rank difference is equal to the is equal to the difference between the numbers of these tree edges. By the contraction of the edges in $$E - E_t$$ to obtain $$G \times E_t$$, the connection vertices which connect $$G \cdot E_t$$ with the remaining part of $$G$$ become a single vertex [for simplicity we assume that $$G \cdot(E - E_t)$$ is connected]. Thus $$\rho$$ is equal to the number of the connection vertices minus one. The vertex connectivity between vertices $$b$$ and $$u$$ is defined as the minimum number of such connection vertices. For an electrical circuit a vertex connectivity of one is meaningless, hence the name electrical connectivity for $$K_b$$ or $$K_u$$.

$$\rho := r(G \cdot E_t)-r(G \times E_t); E_b \subseteq E_t \subseteq E-E_u \tag{3.34}$$

Example 3.3 In graph $$G$$ shown in Figure 9.3a let $$E_t$$ be the set constituted by the edges marked $$t$$. The connection vertices are marked $$c$$. $$G\cdot E_t$$ and $$G \times E$$ are shown in Figures 9.3b and 9.3c, respectively, and trees of these graphs are indicated by the dashed lines. We see that $$r(G \cdot E_t) - r(G \times E_t)=7-5=2=3-l$$.

Replace this text with your caption
Replace this text with your caption

Example 3.4 Let us consider the circuit shown in Figure 9.5a. Elements 4 and 10 are voltage-controlled current sources. We assume the following measurement conditions. (1) The voltages of nodes $$a$$, $$d$$, and $$h$$ with respect to the reference node $$g$$ are measurable. (2) The current of the exciting voltage source 1 is measurable. (3) Resistor 14 is fault-free, and thus its resistance is known.
In order to represent the measurement condition for node $$h$$ we add an imaginary element 15 between nodes $$h$$ and $$g$$. Voltage graph $$G_v$$ and current graph G_i for this circuit are shown in Figure 9.5b. From the measurement conditions we have $$E_b = \{1, 14, 15\}$$. Let us first assume that the voltage-controlled current sources 4 and 10 are faulty but all the other elements are good. Then $$E_u=\{4, 10\}$$ and $$E_k=\{2, 3, 5, 6, 7, 8, 9, 11, 12, 13\}$$. We get $$G_{vk}$$ and $$G_{ik}$$ as shown in Figure 9.5c. We apply Algorithm PART + to these graphs and get a pair of MCT, which are indicated by the dashed lines in Figure 9.5c. $$K_{vk} \cap T_{ik} = \{6\}$$ and from this we get $$C_i(6/K_{ik}) = \{6, 12\}, L_v/T_{vk})=\{6, 3\}, L_v(12/t_{vk}) = \{12\}, C_i(3/K_{ik}) = \{3,2\}, L_v(2/T_{ik}) = \{2\}$$. Thus $$E_{k+} = \{2, 3, 6, 12\}. T_{vk}\cap K_{ik} = \varnothing$$ and thus $$E_{k-} = 0$$. Then $$E_{k0} = \{5, 7, 8, 9, 11, 13\}$$. We get $$\delta_{k+} = 1, \rho_b = 3, \delta_{k-}=0, \rho_u = 2, K_b = K_u = 2$$.
Next let us assume that $$E^*_u= {4, 5, 7, 8, 9} \subseteq E_u + E_{ko}$$ is faulty. Then $$E^*_k = \{2, 3, 6, 10, 11, 12, 13\}$$ and we get $$G^*_{vk}$$ and $$G^*_{ik}$$ as shown in Figure 9.5d. A pair of MCT, $$T^*_{vk}$$ and $$T^*_{ik}$$, are indicated by the dashed lines in the graphs.
$$K^*_{vk} \cap T^*_{ik} = \{6\}$$ and $$T^*_{vk} \cap V^*_{ik} = \{13\}$$. From these sets we get $$E^*_{k+} = \{2, 3, 6, 12\} = E_{k+}$$, $$E^*_{k-} = \{10, 11, 13\}$$, and $$E^*_{k0} = \varnothing$$. Note that $$E_u + E_{k0} + E_{k-} = E^*_{ku} + E^*_{k0} + E^*_{k-}$$, and $$K_b = K^*_b$$. Therefore $$E_u$$ and $$E^*_u$$ belong to an ambiguity set.
Finally, let us assume that $$E^o_u = \{ 10, 11\}$$ is faulty. $$G^o_{vk}$$ and $$G^o_{ik}$$ for this assumption are shown in Figure 9.5e. Again a pair of MCT is indicated by the dashed lines in the graphs: $$K^o_{vk} \cap T^o_{ik} = \{4, 6\}$$, and from this set we get $$E^o_{k+}=E^o_{k}= \{2, 3, 4, 5, 6, 7, 8, 9, 12, 13\}$$. It is possible to check the fault assumption by the consistency of equations for $$E_{k+}$$·

# Diagnosis by multiple excitation

The discussions in the previous sections concern a circuit with a particular excitation, that is, particular source voltages and currents with a fixed frequency. If the circuit has multiple exciting sources or the frequency of the excitation can be varied, we can get more measurement data. Even so, computable voltages and currents are the same if the measurable voltages and currents, or $$E_b$$, $$E_e$$, and $$E_i$$, are the same. Thus the following discussion is also based on the partition of the elements obtained in the previous sections.
Suppose all the elements in $$E_b + E_{k+}$$ are found to be fault-free. We assume the elements in $$E_{k0}$$ are fault-free and the fault is localized in $$E_w = E_u + E_{k-}$$· We want to check this assumption by calculating element or parameter values which are independent of excitation or, equivalently, by checking the consistency of equations for such element or parameter values. Let $$E_x:= E - E_w$$.
Ex:= E - Ew. The voltages and currents of the elements in Ek+ + Eko are computed using their nominal element values. Thus $$E_b$$ in the previous sections is now replaced with $$E_x$$. For each excitation we get a set of equations for E_w which can be written in the same form as Eqs. (3.1), (3.2), and (3.3) except that the subscript $$k$$ is now replaced with $$w$$. The coefficient matrix $$B_w$$ and $$Q_w$$ can be obtained from graphs.

$$G_{vw} := G_v \times E_w \tag{4.1}$$
$$G_{iw} := G_v \cdot E_w \tag{4.2}$$

respectively. The equations for E_w contains the voltages, currents, and element values of the elements in E_w as unknowns. Thus they are nonlinear equations for unknown variables.
Suppose the frequency of the excitation can be varied. One way to check the fault assumption is use the method of Rapisarda and DeCarlo [13] (see also Chapter 11 in this book) and compute directly, if possible, the values of elements in E_w in more than one way and check their consistency. Another way is to check the consistency of parameter values in the equations which relate the terminal voltages and currents of the subcircuit constituted by E_w [11]. To get such equations, the internal voltages and currents of the subcircuit must be eliminated. This can be done if the principal partition of $$G_{vw}$$ and $$G_{iw}$$ (the partition can be obtained by Algorithms PART + and PART-) gives $$E_{w+} = E_w$$ and $$E_{wo} = E_{w-} = \varnothing$$. There is no need to express the parameters in terms of element values. They are rational functions of the frequency, and the coefficients of the functions are the constants wnich are independent ofthe excitation and whose consistency is to be checked.

# Concluding remarks

Diagnosability of analog circuits is investigated based on the computability of voltages and currents in the circuits. The results presented are graph theoretical and therefore give a necessary condition for diagnosability of practica! circuits, for which measurement errors and element value deviations within tolerance bounds must be taken into consideration.
The results are applicable to circuit tuning or design where the graph of the circuit and some voltages and currents in the circuit are specified and element values are to be determined (thus unknown). For example, Germán_et al. [14] used the results in water distribution network expansion planning. The branches of the existing network and those of the network to be added constitute $$E_k$$ and $$E_u$$, respectively. Waterheads (voltages) and waterflows (currents) are specified at certain points in the network which determines $$E_b$$, $$E_e$$ and $$E_j$$. Thus, whether or not this planning is possÍble can be checked by using the diagnosability theorem given in Section 3.
Diagnosis by calculating element values is not discussed in detail. Multiple circuit excitations are inevitable in this case. Diagnosability or computability of element values has been investigated by many authors [15-22].

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