In this section, the fault location algorithm is proposed. The algorithm aims at pinpointing fault locations for multiple simultaneous faults. The algorithm is certainly able to locate a single fault as well.

The basic idea of the proposed fault location algorithm is as follows. Without loss of generality, consider simultaneous faults that occur in a transmission system. Fictitious fault buses will be added. Then the bus impedance matrix with fault buses of the transmission system can be established. Furthermore, the measured voltages at each bus during the fault are functions of the relevant driving point impedances and transfer impedances. The driving point impedances are the principal diagonal elements of the bus impedance matrix, and the transfer impedances are the off-diagonal elements [29]. Moreover, the driving point impedances and the transfer impedances are derived in terms of the fault locations. Therefore, the measured voltages during the fault can be derived in terms of the fault locations as well. Consequently, the fault locations can be solved based on the wide area voltage measurements.

The method for deriving the driving point impedances and the transfer impedances during the fault is described in detail in Section 3.1. The fault location algorithm is proposed in Section 3.2.

### Derivation of driving point impedances and transfer impedances

In this section, the driving point impedances and the transfer impedances during the fault will be derived. A the fault point, we add non-existent nodes, or called fictitious nodes. These nodes are called fictitious fault nodes. In contrast, the nodes at other buses are called non-fault nodes. The impedances between non-fault nodes and fault nodes will be derived first in Section 3.1.1. The impedances between fault nodes will be derived in Section 3.1.2.

Figure 1 depicts the one-line diagram of a three-phase transmission line segment. Symbols *p* and *q* represent the buses of the line. Bus *p* comprises nodes *p*_{1}, *p*_{2}, and *p*_{3}. Bus *q* comprises nodes *q*_{1}, *q*_{2}, and *q*_{3}. Define the fictitious fault bus to be *r*, consisting of nodes *r*_{1}, *r*_{2}, and *r*_{3}. The remaining notations in Fig. 1 are explained as follows:

*E*_{p}: node voltage vector during the fault for bus *p*. *E*_{p} = [*E*_{p1}, *E*_{p2}, *E*_{p3}]^{T}, with T denoting vector transpose;

*E*_{q}: node voltage vector during the fault for bus *q*. *E*_{q} = [*E*_{q1}, *E*_{q2}, *E*_{q3}]^{T};

*E*_{r}: node voltage vector during the fault for bus *r*. *E*_{r} = [*E*_{r1}, *E*_{r2}, *E*_{r3}]^{T};

*z*_{pr}, *z*_{qr}: the equivalent series impedance matrix of the line segment *pr* and *qr*, respectively;

*y*_{pr}, *y*_{qr}: the equivalent shunt admittance matrix of the line segment *pr* and *qr*, respectively.

The method is based on distributed parameter line model to accurately represent long line effects. The parameters of the equivalent PI model [30] of line segment *pr* are as follows:

$$ {\boldsymbol{z}}_{pr}=\boldsymbol{zB}\mathit{\operatorname{diag}}\left[\sinh \left(\boldsymbol{\gamma} {l}_{pr}\right)./\boldsymbol{\gamma} \right]{\boldsymbol{B}}^{-1} $$

(3)

$$ {\boldsymbol{y}}_{pr}=2\boldsymbol{B}\mathit{\operatorname{diag}}\left[\tanh \left(\boldsymbol{\gamma} {l}_{pr}/2\right)./\boldsymbol{\gamma} \right]{\boldsymbol{B}}^{-1}\boldsymbol{y} $$

(4)

where,

*diag*(.): a diagonal matrix with input vector as its diagonal element;

./: element-wise division;

*z*: series impedance matrix of the line *pq* in per unit length;

*y*: shunt admittance matrix of the line *pq* in per unit length;

*B*: eigenvector of (*yz*);

*γ*: vector consisting of *γ*_{j} that is the square root of the *jth* eigenvalue of (*yz*).

*l*_{pr}: length of the line segment *pr*.

Similarly, the parameters of the equivalent PI model of line segment *qr* can be derived. It should be noted that (3) and (4) hold for both transposed and untransposed lines. Moreover, they are applicable to single-circuit lines with *j* = 1, 2, 3 as well as to double-circuit lines with *j* = 1, 2, … , 6.

In addition, define the following variables:

*m*: per unit fault distance from bus *p* to the fault bus *r*;

*n*: the total number of nodes of the transmission system without counting fictitious fault nodes;

*Z*_{0}: the bus impedance matrix in phase domain of the network preceding the fault, excluding fictitious fault nodes;

*Z*_{0, kl}: the element in the *kth* row and *lth* column of *Z*_{0};

*Z*: the bus impedance matrix in phase domain of the network during the fault, including fictitious fault nodes;

*Z*_{kl}: the element in the *kth* row and *lth* column of *Z*;

Based on the definition, the size of *Z*_{0} is *n* by *n*. The size of *Z* is (*n* + 3*n*_{f}) by (*n* + 3*n*_{f}), where *n*_{f} represents the total number of faults in the transmission system.

Matrix *Z*_{0} can be readily developed following the established method in [29]. It can be shown that the first *n* rows and *n* columns of *Z* are identical to *Z*_{0}, and the rest of rows and columns of *Z* consist of driving point and transfer impedances related to the fault nodes.

#### Derivation of the transfer impedance between a non-fault node and fault nodes

Remove all the sources in the network shown in Fig. 1. Then inject 1 Ampere current into node *k* [31]. Applying Kirchhoff’s Current Law (KCL) at bus *r*, it is obtained that

$$ \frac{\left({\boldsymbol{y}}_{pr}+{\boldsymbol{y}}_{qr}\right)}{2}{\boldsymbol{E}}_r+{\boldsymbol{z}}_{pr}^{-1}\left({\boldsymbol{E}}_r-{\boldsymbol{E}}_p\right)+{\boldsymbol{z}}_{qr}^{-1}\left({\boldsymbol{E}}_r-{\boldsymbol{E}}_q\right)=0 $$

(5)

From (5), *E*_{r} can be written as

$$ {\boldsymbol{E}}_r={\left[\frac{\left({\boldsymbol{y}}_{pr}+{\boldsymbol{y}}_{qr}\right)}{2}+{\boldsymbol{z}}_{pr}^{-1}+{\boldsymbol{z}}_{qr}^{-1}\right]}^{-1}\left({\boldsymbol{z}}_{pr}^{-1}{\boldsymbol{E}}_p+{\boldsymbol{z}}_{qr}^{-1}{\boldsymbol{E}}_q\right) $$

(6)

Based on the definition of transfer impedance, the value of the transfer impedance between a non-fault node *k* and a fault node is equal to the voltage at the fault node, when 1 Ampere current is injected into node *k*, with all the sources in the network being removed [31]. Therefore, the values of *E*_{p} and *E*_{q} are equal to *Z*_{kp} and *Z*_{kq}, respectively. Hence, the transfer impedance between node *k* and fault nodes are:

$$ {\boldsymbol{Z}}_{kr}={\left[\frac{\left({\boldsymbol{y}}_{pr}+{\boldsymbol{y}}_{qr}\right)}{2}+{\boldsymbol{z}}_{pr}^{-1}+{\boldsymbol{z}}_{qr}^{-1}\right]}^{-1}\left({\boldsymbol{z}}_{pr}^{-1}{\boldsymbol{Z}}_{kp}+{\boldsymbol{z}}_{qr}^{-1}{\boldsymbol{Z}}_{kq}\right) $$

(7)

where,

\( {\boldsymbol{Z}}_{kr}={\left[{Z}_{k{r}_1},{Z}_{k{r}_2},{Z}_{k{r}_3}\right]}^T \), with T denotes vector transpose;

\( {\boldsymbol{Z}}_{kp}={\left[{Z}_{k{p}_1},{Z}_{k{p}_2},{Z}_{k{p}_3}\right]}^T \);

\( {\boldsymbol{Z}}_{kq}={\left[{Z}_{k{q}_1},{Z}_{k{q}_2},{Z}_{k{q}_3}\right]}^T \);

\( {Z}_{k{r}_1} \), \( {Z}_{k{r}_2} \), \( {Z}_{k{r}_3} \): the transfer impedance between node *k* and fault node *r*_{1}, *r*_{2}, and *r*_{3}, respectively;

\( {Z}_{k{p}_1} \), \( {Z}_{k{p}_2} \), \( {Z}_{k{p}_3} \): the transfer impedance between node *k* and node *p*_{1}, *p*_{2}, and *p*_{3}, respectively;

\( {Z}_{k{q}_1} \), \( {Z}_{k{q}_2} \), \( {Z}_{k{q}_3} \): the transfer impedance between node *k* and node *q*_{1}, *q*_{2}, and *q*_{3}, respectively.

Note that *Z*_{kp1}, *Z*_{kq1} and so on are elements of the bus impedance matrix of the prefault network. Furthermore, **y**_{pr}, **y**_{qr}, **z**_{pr}, and **z**_{qr} are functions of the fault location *m*. Therefore, the transfer impedance *Z*_{kr} is a function of the fault location *m* as well.

The prefault voltage at fictitious fault nodes can be calculated using the prefault terminal voltages based on (6):

$$ {\boldsymbol{E}}_r^0={\left[\frac{\left({\boldsymbol{y}}_{pr}+{\boldsymbol{y}}_{qr}\right)}{2}+{\boldsymbol{z}}_{pr}^{-1}+{\boldsymbol{z}}_{qr}^{-1}\right]}^{-1}\left({\boldsymbol{z}}_{pr}^{-1}{\boldsymbol{E}}_p^0+{\boldsymbol{z}}_{qr}^{-1}{\boldsymbol{E}}_q^0\right) $$

(8)

where,

\( {\boldsymbol{E}}_r^0 \): node voltage vector preceding the fault at fictitious bus *r*. \( {\boldsymbol{E}}_r^0={\left[{E}_{r1}^0,{E}_{r2}^0,{E}_{r3}^0\right]}^T \), with T denotes vector transpose;

*E*_{p0}: node voltage vector preceding the fault at bus *p*. \( {\boldsymbol{E}}_p^0={\left[{E}_{p1}^0,{E}_{p2}^0,{E}_{p3}^0\right]}^T \),

*E*_{q0}: node voltage vector preceding the fault at bus *q*. \( {\boldsymbol{E}}_q={\left[{E}_{q1}^0,{E}_{q2}^0,{E}_{q3}^0\right]}^T \).

#### Derivation of the driving point impedance at fault nodes and transfer impedance between fault nodes

Remove all the sources in the network shown in Fig. 1. Then inject 1 Ampere current into node *γ*_{i}, where *i* = 1, 2, 3 for a single circuit line, and *i* = 1, 2, … , 6 for a double-circuit line. Applying KCL at bus *r* yields

$$ \frac{\left({\boldsymbol{y}}_{pr}+{\boldsymbol{y}}_{qr}\right)}{2}{\boldsymbol{E}}_r+{\boldsymbol{z}}_{pr}^{-1}\left({\boldsymbol{E}}_r-{\boldsymbol{E}}_p\right)+{\boldsymbol{z}}_{qr}^{-1}\left({\boldsymbol{E}}_r-{\boldsymbol{E}}_q\right)={\boldsymbol{u}}_i $$

(9)

where *u*_{i} is the *ith* column of a three by three or six by six identity matrix. From (9), *E*_{r} can be written as

$$ {\boldsymbol{E}}_r={\left[\frac{\left({\boldsymbol{y}}_{pr}+{\boldsymbol{y}}_{qr}\right)}{2}+{\boldsymbol{z}}_{pr}^{-1}+{\boldsymbol{z}}_{qr}^{-1}\right]}^{-1}\left({\boldsymbol{z}}_{pr}^{-1}{\boldsymbol{E}}_p+{\boldsymbol{z}}_{qr}^{-1}{\boldsymbol{E}}_q+{\boldsymbol{u}}_i\right) $$

(10)

Based on the definition, the value of the transfer impedance between a fault node and other fault nodes is equal to the voltage at other fault nodes, and the value of the driving point impedance at a fault node is equal to the voltage at the fault node [31]. Therefore, the values of *E*_{r}, *E*_{p}, and *E*_{q} are equal to \( {\boldsymbol{Z}}_{r{r}_i} \), \( {\boldsymbol{Z}}_{p{r}_i} \), and \( {\boldsymbol{Z}}_{q{r}_i} \), respectively. Hence, the transfer impedance and driving point impedance at fault nodes are:

$$ {\boldsymbol{Z}}_{r{r}_i}={\left[\frac{\left({\boldsymbol{y}}_{pr}+{\boldsymbol{y}}_{qr}\right)}{2}+{\boldsymbol{z}}_{pr}^{-1}+{\boldsymbol{z}}_{qr}^{-1}\right]}^{-1}\bullet \left({\boldsymbol{z}}_{pr}^{-1}{\boldsymbol{Z}}_{p{r}_i}+{\boldsymbol{z}}_{qr}^{-1}{\boldsymbol{Z}}_{q{r}_i}+{\boldsymbol{u}}_i\right) $$

(11)

where,

\( {\boldsymbol{Z}}_{r{r}_i}={\left[{Z}_{r_1{r}_i},{Z}_{r_2{r}_i},{Z}_{r_3{r}_i}\right]}^T \)*,* with T denotes vector transpose*;*

\( {\boldsymbol{Z}}_{p{r}_i}={\left[{Z}_{p_1{r}_i},{Z}_{p_2{r}_i},{Z}_{p_3{r}_i}\right]}^T \);

\( {\boldsymbol{Z}}_{q{r}_i}={\left[{Z}_{q_1{r}_i},{Z}_{q_2{r}_i},{Z}_{q_3{r}_i}\right]}^T \).

\( {Z}_{r_1{r}_i} \), \( {Z}_{r_2{r}_i} \), \( {Z}_{r_3{r}_i} \): the transfer impedance between fault nodes and driving point impedance at fault nodes;

\( {Z}_{p_1{r}_i} \), \( {Z}_{p_2{r}_i} \), \( {Z}_{p_3{r}_i} \): the transfer impedance between nodes of bus *p* and fault nodes;

\( {Z}_{q_1{r}_i} \), \( {Z}_{q_2{r}_i} \), \( {Z}_{q_3{r}_i} \): the transfer impedance between nodes of bus *q* and fault nodes;

Setting *i* = 1, 2, 3 for single circuit lines and *i* = 1, 2, … , 6 for double-circuit lines will give all relevant driving point and transfer impedances related to fault nodes.

Similarly, as discussed in Section 3.1.1, it is revealed that these driving point and transfer impedances are functions of fault locations.

### Fault location algorithm

This subsection presents the proposed fault location algorithm to locate multiple simultaneous faults in transmission systems. Figure 2 illustrates a scenario involving two simultaneous faults. A phase A to ground fault occurs at point *F*_{1} on line segment *P*_{1}*Q*_{1} with fault location *m*_{1}, and a phase C to ground fault occurs simultaneously at point *F*_{2} on line segment *P*_{2}*Q*_{2} with fault location *m*_{2}.

Voltage measurements from specified locations are utilized to locate the unknown fault location *m*_{1} and *m*_{2}. The voltage at any bus *l* during the fault can be expressed as

$$ {\boldsymbol{E}}_l={\boldsymbol{E}}_l^0-\left[\begin{array}{cc}{\boldsymbol{Z}}_{l{F}_1}& {\boldsymbol{Z}}_{l{F}_2}\end{array}\right]{\left[\begin{array}{cc}{\boldsymbol{I}}_{F_1}& {\boldsymbol{I}}_{F_2}\end{array}\right]}^T $$

(12)

where,

*E*_{l}: the voltage at bus *l* during the fault;

\( {\boldsymbol{E}}_l^0 \): the voltage at bus *l* preceding the fault;

\( {\boldsymbol{Z}}_{l{F}_1} \), \( {\boldsymbol{Z}}_{l{F}_2} \): the transfer impedance between bus *l* and fault bus *F*_{1}, bus *l* and fault bus *F*_{2}, respectively;

*I*_{F1}, *I*_{F2}: the fault currents at the point *F*_{1} and *F*_{2}.

Based on the measurements from two buses *L*_{1} and *L*_{2}, The following two equations are obtained:

$$ {\boldsymbol{E}}_{L_1}={\boldsymbol{E}}_{L_1}^0-\left[\begin{array}{cc}{\boldsymbol{Z}}_{L_1{F}_1}& {\boldsymbol{Z}}_{L_1{F}_2}\end{array}\right]{\left[\begin{array}{cc}{\boldsymbol{I}}_{F_1}& {\boldsymbol{I}}_{F_2}\end{array}\right]}^T $$

(13)

$$ {\boldsymbol{E}}_{L_2}={\boldsymbol{E}}_{L_2}^0-\left[\begin{array}{cc}{\boldsymbol{Z}}_{L_2{F}_1}& {\boldsymbol{Z}}_{L_2{F}_2}\end{array}\right]{\left[\begin{array}{cc}{\boldsymbol{I}}_{F_1}& {\boldsymbol{I}}_{F_2}\end{array}\right]}^T $$

(14)

or in a compact format,

$$ \left[\begin{array}{c}{\boldsymbol{E}}_{L_1}\\ {}{\boldsymbol{E}}_{L_2}\end{array}\right]=\left[\begin{array}{c}{\boldsymbol{E}}_{{\mathrm{L}}_1}^0\\ {}{\boldsymbol{E}}_{L_2}^0\end{array}\right]-\left[\begin{array}{cc}{\boldsymbol{Z}}_{L_1{F}_1}& {\boldsymbol{Z}}_{L_1{F}_2}\\ {}{\boldsymbol{Z}}_{L_2{F}_1}& {\boldsymbol{Z}}_{L_2{F}_2}\end{array}\right]\left[\begin{array}{c}{\boldsymbol{I}}_{F_1}\\ {}{\boldsymbol{I}}_{F_2}\end{array}\right] $$

(15)

where,

\( {\boldsymbol{E}}_{L_1} \), \( {\boldsymbol{E}}_{L_2} \): the voltage during the fault at bus *L*_{1}, *L*_{2}, respectively;

\( {\boldsymbol{E}}_{L_1}^0 \), \( {\boldsymbol{E}}_{L_2}^0 \): the voltage preceding the fault at bus *L*_{1}, *L*_{2}, respectively;

\( {\boldsymbol{Z}}_{L_1{F}_1} \), \( {\boldsymbol{Z}}_{L_1{F}_2} \): transfer impedance matrix between bus *L*_{1} and *F*_{1}, bus *L*_{1} and *F*_{2}, respectively;

\( {\boldsymbol{Z}}_{L_2{F}_1} \), \( {\boldsymbol{Z}}_{L_2{F}_2} \): transfer impedance matrix between bus *L*_{2} and *F*_{1}, bus *L*_{2} and *F*_{2}, respectively.

Equation (15) can be written in a more compact form as

$$ {\boldsymbol{E}}_{L_1{L}_2}={\boldsymbol{E}}_{L_1{L}_2}^0-{\boldsymbol{Z}}_{L_1{L}_2{F}_1{F}_2}{\boldsymbol{I}}_{F_1{F}_2} $$

(16)

The superimposed quantity, or the voltage change due to a fault, is

$$ \Delta {\boldsymbol{E}}_{L_1{L}_2}=-{\boldsymbol{Z}}_{L_1{L}_2{F}_1{F}_2}{\boldsymbol{I}}_{F_1{F}_2} $$

(17)

From (17), the fault current vector is obtained as

$$ {\boldsymbol{I}}_{F_1{F}_2}=-{\left({\boldsymbol{Z}}_{L_1{L}_2{F}_1{F}_2}^T{\boldsymbol{Z}}_{L_1{L}_2{F}_1{F}_2}\right)}^{-\mathbf{1}}\left({\boldsymbol{Z}}_{L_1{L}_2{F}_1{F}_2}^T\Delta {\boldsymbol{E}}_{L_1{L}_2}\right) $$

(18)

Furthermore, the voltage during the fault at fault buses are given by

$$ \left[\begin{array}{c}{\boldsymbol{E}}_{F_1}\\ {}{\boldsymbol{E}}_{F_2}\end{array}\right]=\left[\begin{array}{c}{\boldsymbol{E}}_{F_1}^0\\ {}{\boldsymbol{E}}_{F_2}^0\end{array}\right]-\left[\begin{array}{cc}{\boldsymbol{Z}}_{F_1{F}_1}& {\boldsymbol{Z}}_{F_1{F}_2}\\ {}{\boldsymbol{Z}}_{F_2{F}_1}& {\boldsymbol{Z}}_{F_2{F}_2}\end{array}\right]\left[\begin{array}{c}{\boldsymbol{I}}_{F_1}\\ {}{\boldsymbol{I}}_{F_2}\end{array}\right] $$

(19)

where,

\( {\boldsymbol{E}}_{F_1} \), \( {\boldsymbol{E}}_{F_2} \): the voltage during the fault at fault bus *F*_{1}, *F*_{2}, respectively;

\( {\boldsymbol{E}}_{F_1}^0 \), \( {\boldsymbol{E}}_{F_2}^0 \): the voltage preceding the fault at fault bus *F*_{1}, *F*_{2}, respectively;

\( {\boldsymbol{Z}}_{F_1{F}_1} \), \( {\boldsymbol{Z}}_{F_2{F}_2} \): driving point impedance matrix at *F*_{1}, *F*_{2}, respectively;

\( {\boldsymbol{Z}}_{F_1{F}_2} \), \( {\boldsymbol{Z}}_{F_2{F}_1} \): the transfer impedance matrix between *F*_{1} and *F*_{2}.

Equation (19) can be written in a more compact form as

$$ {\boldsymbol{E}}_{F_1{F}_2}={\boldsymbol{E}}_{F_1{F}_2}^0-{\boldsymbol{Z}}_{F_1{F}_2{F}_1{F}_2}{\boldsymbol{I}}_{F_1{F}_2} $$

(20)

Based on (8), prefault voltages at the fault bus *F*_{1} and *F*_{2} can be expressed in terms of fault locations and the prefault voltages at the bus *P*_{1} and *Q*_{1}, bus *P*_{2} and *Q*_{2}. For example, \( {\boldsymbol{E}}_{F_1}^0 \) is derived as follows:

$$ {\boldsymbol{E}}_{F_1}^0={\left[\frac{\left({\boldsymbol{y}}_{P_1{F}_1}+{\boldsymbol{y}}_{Q_1{F}_1}\right)}{2}+{\boldsymbol{z}}_{P_1{F}_1}^{-1}+{\boldsymbol{z}}_{Q_1{F}_1}^{-1}\right]}^{-1}\bullet \left({\boldsymbol{z}}_{P_1{F}_1}^{-1}{\boldsymbol{E}}_{P_1}^0+{\boldsymbol{z}}_{Q_1{F}_1}^{-1}{\boldsymbol{E}}_{Q_1}^0\right) $$

(21)

where,

\( {\boldsymbol{z}}_{P_1{F}_1} \), \( {\boldsymbol{z}}_{Q_1{F}_1} \): equivalent series impedance matrix of segment *P*_{1}*F*_{1} *and Q*_{1}*F*_{1};

\( {\boldsymbol{y}}_{P_1{F}_1} \), \( {\boldsymbol{y}}_{Q_1{F}_1} \): equivalent shunt admittance matrix of segment *P*_{1}*F*_{1} *and Q*_{1}*F*_{1};

\( {\boldsymbol{E}}_{P_1}^0 \), \( {\boldsymbol{E}}_{Q_1}^0 \): prefault voltages at *P*_{1} and *Q*_{1}, which can be obtained by WAMS.

Since the fault resistances are purely resistive, the reactive power consumed by fault resistances at the two fault locations is zero, i.e.,

$$ Imag\left(\left[\begin{array}{c}{\boldsymbol{E}}_{F_1}^T{\boldsymbol{I}}_{F_1}^{\ast}\\ {}{\boldsymbol{E}}_{F_2}^T{\boldsymbol{I}}_{F_2}^{\ast}\end{array}\right]\right)=0 $$

(22)

Solving (22) will yield the two unknown variables *m*_{1} and *m*_{2}.

An alternative approach is described as follows when more measurements are available. Assuming that measurements from another two buses *L*_{3} and *L*_{4} are known. The following equation is obtained in a similar way to (18),

$$ {\boldsymbol{I}}_{F_1{F}_2}=-{\left({\boldsymbol{Z}}_{L_3{L}_4{F}_1{F}_2}^T{\boldsymbol{Z}}_{L_3{L}_4{\mathrm{F}}_1{F}_2}\right)}^{-\mathbf{1}}\left({\boldsymbol{Z}}_{L_3{L}_4{F}_1{F}_2}^T\Delta {\boldsymbol{E}}_{L_3{L}_4}\right) $$

(23)

Equating (18) and (23), it is obtained that

$$ {\displaystyle \begin{array}{c}{\left({\boldsymbol{Z}}_{L_1{L}_2{F}_1{F}_2}^T{\boldsymbol{Z}}_{L_1{L}_2{F}_1{F}_2}\right)}^{-\mathbf{1}}\left({\boldsymbol{Z}}_{L_1{L}_2{F}_1{F}_2}^T\Delta {\boldsymbol{E}}_{L_1{L}_2}\right)\\ {}={\left({\boldsymbol{Z}}_{L_3{L}_4{F}_1{F}_2}^T{\boldsymbol{Z}}_{L_3{L}_4{F}_1{F}_2}\right)}^{-\mathbf{1}}\left({\boldsymbol{Z}}_{L_3{L}_4{F}_1{F}_2}^T\Delta {\boldsymbol{E}}_{L_3{L}_4}\right)\end{array}} $$

(24)

Equation (24) contains two unknown variables *m*_{1} and *m*_{2}. Separating the equation into two real equations, from which the fault location can be obtained using Newton-Raphson method.

It is noted that the derivation does not assume the values of fault resistances, and thus the method is essentially immune to fault resistances.