CN105512502B - One kind is based on the normalized weight function the least square estimation method of residual error - Google Patents

One kind is based on the normalized weight function the least square estimation method of residual error Download PDF

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CN105512502B
CN105512502B CN201610019294.1A CN201610019294A CN105512502B CN 105512502 B CN105512502 B CN 105512502B CN 201610019294 A CN201610019294 A CN 201610019294A CN 105512502 B CN105512502 B CN 105512502B
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颜伟
王茜
赵霞
陈文超
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Chongqing University
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Abstract

本发明公开了一种基于残差归一化的权函数最小二乘状态估计方法,属于电力系统调度自动化领域。本发明方法利用计算机,通过程序,首先输入数据终端采集到的任一时间断面的SCADA数据、网络结构及参数信息并初始化,然后计算网络的节点导纳矩阵,形成零注入等式约束方程,接着综合考虑电压幅值量测方程、注入功率量测方程、支路功率量测方程,以节点电压的幅值与相角为状态变量,计算相应残差、雅可比矩阵以及权函数,最后更新状态变量,进行收敛性判断,来实现电网的状态估计。本发明能有效抑制不良杠杆量测,具有强抗差性和良好的收敛性,且计算效率很高,具有良好的工程应用前景。

The invention discloses a weight function least square state estimation method based on residual normalization, which belongs to the field of electric power system scheduling automation. The method of the present invention uses a computer to first input and initialize the SCADA data, network structure and parameter information of any time section collected by the data terminal through a program, and then calculate the node admittance matrix of the network to form a zero-injection equality constraint equation, and then Comprehensively consider the voltage amplitude measurement equation, injection power measurement equation, and branch power measurement equation, and use the node voltage amplitude and phase angle as state variables to calculate the corresponding residual, Jacobian matrix, and weight function, and finally update the state Variables, to judge the convergence, to realize the state estimation of the power grid. The invention can effectively suppress bad lever measurement, has strong error resistance and good convergence, and has high calculation efficiency, and has good engineering application prospect.

Description

一种基于残差归一化的权函数最小二乘状态估计方法A Weighted Function Least Squares State Estimation Method Based on Residual Normalization

技术领域technical field

本发明属于电力系统调度自动化领域,具体涉及一种基于残差归一化的权函数最小二乘状态估计方法。The invention belongs to the field of power system scheduling automation, and in particular relates to a weight function least square state estimation method based on residual normalization.

背景技术Background technique

电力系统状态估计是能量管理系统的重要组成部分,是运行电力系统其它应用软件的基础,其结果直接影响电网调度的智能化分析与决策。自1970年Schweppe首次将状态估计引入电力系统以来,国内外学者和工程人员对状态估计进行了大量、深入的研究和实践,这期间出现了各种各样的状态估计方法。Power system state estimation is an important part of the energy management system and the basis for running other application software in the power system. The results directly affect the intelligent analysis and decision-making of power grid dispatching. Since Schweppe introduced the state estimation into the power system for the first time in 1970, scholars and engineers at home and abroad have conducted a large number of in-depth research and practice on state estimation, and various state estimation methods have emerged during this period.

目前,加权最小二乘估计法(Weighted least squares,WLS)作为在状态估计中应用最为广泛的方法之一,其模型简单、易于求解,但不具有抗干扰性,因此出现了抗差状态估计。现有的抗差状态估计方法包括指数型目标函数抗差状态估计法(Maximumexponential square,MES)、基于标准化残差倒数的权函数最小二乘状态估计、以及指数加权最小二乘抗差状态估计法(Exponential function weighted least squares,EFWLS)等。但是这些方法在计算中由于标准化残差的计算会使每次迭代消耗大量时间,从而影响了它们在实际系统中的应用。At present, Weighted least squares (WLS) is one of the most widely used methods in state estimation. Its model is simple and easy to solve, but it is not anti-interference, so robust state estimation appears. The existing robust state estimation methods include the exponential objective function robust state estimation method (Maximum exponential square, MES), the weighted function least squares state estimation method based on the reciprocal of the standardized residual, and the exponentially weighted least squares robust state estimation method (Exponential function weighted least squares, EFWLS) and so on. However, the calculation of these methods consumes a lot of time for each iteration because of the calculation of the standardized residual, which affects their application in the actual system.

基于此,曾有提出的方法可在一定程度上提高计算效率,但求解过程复杂,对一些大型的电力系统计算量太大,因此不易得到广泛运用。也有人提出了指数权函数的状态估计算法(Exponential least absolute value,E-LAV),提高了计算效率,但抗不良杠杆量测的能力较低。因此,需要为电力系统状态估计提供一种既能满足计算效率要求,又有良好抗差性能的估计方法,以满足实际工程运用的需要。Based on this, there have been proposed methods that can improve computational efficiency to a certain extent, but the solution process is complex, and the amount of calculation for some large power systems is too large, so it is not easy to be widely used. Some people have also proposed a state estimation algorithm (Exponential least absolute value, E-LAV) of the exponential weight function, which improves the calculation efficiency, but the ability to resist bad leverage measurement is low. Therefore, it is necessary to provide an estimation method for power system state estimation that can not only meet the requirements of computational efficiency, but also have good tolerance performance, so as to meet the needs of practical engineering applications.

发明内容Contents of the invention

本发明的目的是针对现有状态估计研究方法的不足,提出一种计算效率高、收敛性好、且能有效抑制不良杠杠量测的基于残差归一化的权函数最小二乘状态估计方法(Residuals normalized weighted least squares,RNWLS)。The purpose of the present invention is to address the shortcomings of existing state estimation research methods, and propose a weight function least squares state estimation method based on residual normalization that has high computational efficiency, good convergence, and can effectively suppress bad leverage measurements. (Residuals normalized weighted least squares, RNWLS).

实现本发明的技术方案是:一种基于残差归一化的权函数最小二乘状态估计方法,利用计算机,通过程序,首先输入数据终端采集到的任一时间断面的SCADA数据、网络结构及参数信息并初始化,然后计算网络的节点导纳矩阵,形成零注入等式约束方程,接着综合考虑电压幅值量测方程、注入功率量测方程、支路功率量测方程,以节点电压的幅值与相角为状态变量,计算相应残差、雅克比矩阵以及权函数,最后更新状态变量,进行收敛性判断,来实现电网的状态估计。所述方法的具体步骤如下:The technical scheme for realizing the present invention is: a kind of weight function least square state estimation method based on residual normalization, utilize computer, through program, at first input the SCADA data of any time section collected by data terminal, network structure and The parameter information is initialized, and then the node admittance matrix of the network is calculated to form a zero-injection equality constraint equation, and then the voltage amplitude measurement equation, injection power measurement equation, and branch power measurement equation are comprehensively considered, and the node voltage amplitude The value and phase angle are state variables, and the corresponding residuals, Jacobian matrix, and weight function are calculated, and finally the state variables are updated to judge the convergence, so as to realize the state estimation of the power grid. The concrete steps of described method are as follows:

(1)输入基础数据及初始化(1) Input basic data and initialization

1)输入基础数据1) Enter basic data

首先输入任一时间断面下电网的数据采集与监视控制系统即SCADA数据、网络结构及参数信息;所述电网的SCADA数据包括节点电压幅值、节点注入功率以及支路功率;所述网络结构及参数信息包括网络元件的电阻、电抗、电纳、额定电压以及功率基准;所述网络元件包括线路以及变压器。First input the data acquisition and monitoring control system of the power grid under any time section, namely SCADA data, network structure and parameter information; the SCADA data of the power grid includes node voltage amplitude, node injection power and branch power; the network structure and The parameter information includes resistance, reactance, susceptance, rated voltage and power reference of network elements; said network elements include lines and transformers.

2)参数初始化2) Parameter initialization

设置m阶单位矩阵R-1,所述单位矩阵R-1对角线元素全为1,非对角线元素全为0,m为状态估计中实际的量测变量个数;设置状态估计时节点电压幅值的初值均为标幺值1,相角均为0;所述网络元件的阻抗和电纳均归算为标幺值;节点注入功率以及支路功率也归算至标幺值;初始化最大迭代次数Tmax为40~60;收敛精度ε为10-3~10-5,以及权重计算时小残差的检测门槛值rmin=0.002,并设置迭代次数time=1。Set the m-order identity matrix R -1 , the diagonal elements of the identity matrix R -1 are all 1, and the off-diagonal elements are all 0, m is the actual number of measured variables in the state estimation; when setting the state estimation The initial value of the node voltage amplitude is 1 p.u., and the phase angle is 0. The impedance and susceptance of the network elements are all reduced to the p.u. value; the node injection power and branch power are also reduced to the p.u. value; initialize the maximum number of iterations Tmax to 40~60; the convergence precision ε is 10 -3 to 10 -5 , and the detection threshold r min =0.002 for small residual error in weight calculation, and set the number of iterations time=1.

(2)计算节点导纳矩阵(2) Calculate node admittance matrix

第(1)步完成后,利用公式(1)计算节点导纳矩阵Y:After step (1) is completed, use the formula (1) to calculate the node admittance matrix Y:

式中,节点导纳矩阵的对角线元素表示为Yii,节点导纳矩阵的非对角线元素表示为Yij,均由上式计算得到;yij为节点i和节点j之间的支路阻抗zij的倒数;符号j∈i表示节点j和节点i直接相连,且当节点i有接地支路时还应包括j=0时的情况。In the formula, the diagonal element of the node admittance matrix is expressed as Y ii , and the off-diagonal element of the node admittance matrix is expressed as Y ij , both of which are calculated by the above formula; y ij is the distance between node i and node j Reciprocal of branch impedance z ij ; symbol j∈i means that node j is directly connected to node i, and when node i has a grounding branch, it should also include the case of j=0.

(3)形成零注入等式约束方程(3) Form the zero-injection equality constraint equation

第(2)步完成后,选取量测量中既不是发电机节点也不是负荷节点的节点,将所述节点的注入功率作为零注入等式约束方程,表示为:After step (2) is completed, select a node that is neither a generator node nor a load node in the quantity measurement, and use the injected power of the node as the zero injection equality constraint equation, expressed as:

c(x)=0 (2)c(x)=0 (2)

式中,c(x)是零注入节点的量测方程,其表示的节点注入有功功率和无功功率值可由公式(3)计算得到;x是n维的状态量,n是状态估计中实际的状态变量个数。c(x)的计算公式为:In the formula, c(x) is the measurement equation of the zero-injected node, and the injected active power and reactive power value of the node represented by it can be calculated by formula (3); x is the n-dimensional state quantity, and n is the actual The number of state variables of . The calculation formula of c(x) is:

式中,Pi′为节点i注入的有功功率;Qi′为节点i注入的无功功率;ui为节点i的电压幅值,uj为节点j的电压幅值;θi为节点i的电压相角,θj为节点j的电压相角,且θij=θij,表示节点i和节点j的电压相角差;Gij、Bij分别为节点导纳矩阵中对应节点i和j之间元素的实部和虚部;N为电网的节点个数。In the formula, P i ′ is the active power injected by node i; Q i ′ is the reactive power injected by node i; u i is the voltage amplitude of node i, u j is the voltage amplitude of node j ; The voltage phase angle of i, θ j is the voltage phase angle of node j, and θ ij = θ ij , which means the voltage phase angle difference between node i and node j; G ij and B ij are respectively Corresponding to the real part and imaginary part of the elements between nodes i and j; N is the number of nodes in the grid.

(4)计算残差、雅克比矩阵及权函数(4) Calculation of residuals, Jacobian matrix and weight function

第(3)步完成后,以电网中节点电压幅值、除去零注入节点后的节点注入功率以及支路功率为量测量,计算电网中量测量的残差、雅克比矩阵及权函数。具体步骤内容如下:After the step (3) is completed, the node voltage amplitude in the grid, the node injection power after removing the zero injection node and the branch power are measured, and the residual error, Jacobian matrix and weight function of the grid measurement are calculated. The specific steps are as follows:

I)计算残差I) Calculation of residuals

基于公式(4)计算状态估计中各量测量的参数:Calculate the parameters measured by each quantity in state estimation based on formula (4):

r=z-h(x) (4)r=z-h(x) (4)

式中,z是m维的量测量,m是状态估计中实际的量测变量个数;x是n维的状态量,n是状态估计中实际的状态变量个数。h(x)是量测方程,包括注入功率对应的量测方程、线路支路功率对应的量测方程、变压器支路功率对应的量测方程和节点电压对应的量测方程;h(x)可由公式(5)-公式(8)计算得到;r是量测残差。In the formula, z is the m-dimensional quantity measurement, m is the actual number of measured variables in the state estimation; x is the n-dimensional state quantity, and n is the actual number of state variables in the state estimation. h(x) is the measurement equation, including the measurement equation corresponding to the injected power, the measurement equation corresponding to the line branch power, the measurement equation corresponding to the transformer branch power and the measurement equation corresponding to the node voltage; h(x) It can be calculated by Formula (5)-Formula (8); r is the measurement residual.

基于公式(5)得到除去零注入节点后的节点注入功率对应的量测方程:Based on formula (5), the measurement equation corresponding to the node injection power after removing the zero injection node is obtained:

式中,Pi为节点i注入的有功功率;Qi为节点i注入的无功功率;ui为节点i的电压幅值,uj为节点j的电压幅值;θi为节点i的电压相角,θj为节点j的电压相角,且θij=θij,表示节点i和节点j的电压相角差;Gij、Bij分别为节点导纳矩阵中对应节点i和j之间元素的实部和虚部。In the formula, P i is the active power injected by node i; Q i is the reactive power injected by node i; u i is the voltage amplitude of node i, u j is the voltage amplitude of node j; θ i is the voltage amplitude of node i Voltage phase angle, θ j is the voltage phase angle of node j, and θ ij = θ ij , which means the voltage phase angle difference between node i and node j; G ij and B ij are the corresponding nodes in the node admittance matrix The real and imaginary parts of the elements between i and j.

基于公式(6)得到线路支路功率对应的量测方程:Based on the formula (6), the measurement equation corresponding to the line branch power is obtained:

式中,Pij、Qij分别为线路支路节点i侧的有功功率、无功功率,Pji、Qji为支路节点j侧的有功功率、无功功率。ui为节点i的电压幅值,uj为节点j的电压幅值;θi为节点i的电压相角,θj为节点j的电压相角,且θij=θij,表示节点i和节点j的电压相角差。g为线路电导,b为线路电纳,yc为线路对地电纳。In the formula, P ij and Q ij are the active power and reactive power of the line branch node i side respectively, and P ji and Q ji are the active power and reactive power of the branch node j side. u i is the voltage amplitude of node i, u j is the voltage amplitude of node j; θ i is the voltage phase angle of node i, θ j is the voltage phase angle of node j, and θ ij = θ ij , Indicates the voltage phase angle difference between node i and node j. g is the line conductance, b is the line susceptance, and y c is the line-to-ground susceptance.

基于公式(7)得到变压器支路功率对应的量测方程:Based on formula (7), the measurement equation corresponding to the transformer branch power is obtained:

式中,Pijk、Qijk分别为变压器支路节点i侧的有功功率、无功功率,Pjik、Qjik分别为变压器支路节点j侧的有功功率、无功功率。ui为节点i的电压幅值,uj为节点j的电压幅值。θi为节点i的电压相角,θj为节点j的电压相角,且θij=θij,表示节点i和节点j的电压相角差。K为变压器非标准变比:j为标准侧,变比为1,i为非标准侧,变比为K;bT为变压器标准侧的电纳。In the formula, P ijk , Q ijk are the active power and reactive power at the node i side of the transformer branch respectively, and P jik and Q jik are the active power and reactive power at the node j side of the transformer branch respectively. u i is the voltage amplitude of node i, and u j is the voltage amplitude of node j. θ i is the voltage phase angle of node i, θ j is the voltage phase angle of node j, and θ ij = θ i - θ j represents the voltage phase angle difference between node i and node j. K is the non-standard transformation ratio of the transformer: j is the standard side, the transformation ratio is 1, i is the non-standard side, and the transformation ratio is K; b T is the susceptance of the standard side of the transformer.

基于公式(8)得到节点电压对应的量测方程:Based on the formula (8), the measurement equation corresponding to the node voltage is obtained:

Ui=ui (8)U i =u i (8)

式中,Ui、ui均表示为节点i的电压幅值。In the formula, U i and u i both represent the voltage amplitude of node i.

II)计算雅克比矩阵II) Calculate the Jacobian matrix

基于公式(9)-公式(19)形成雅克比矩阵H和C。其中H为量测量的雅克比矩阵,包括注入功率、线路支路功率、变压器支路功率以及节点电压幅值所形成的雅克比矩阵元素。C为零注入等式约束的雅克比矩阵,由零注入节点的注入功率所形成的雅克比矩阵元素组成。The Jacobian matrices H and C are formed based on formula (9) - formula (19). Among them, H is the Jacobian matrix of quantity measurement, including the Jacobian matrix elements formed by injection power, line branch power, transformer branch power and node voltage amplitude. C is the Jacobian matrix constrained by the zero-injection equality, which is composed of elements of the Jacobian matrix formed by the injection power of the zero-injection node.

对于注入功率,基于公式(9)和(10)形成其雅克比矩阵元素:For the injected power, its Jacobian matrix elements are formed based on equations (9) and (10):

式中,Pi为节点i注入的有功功率;Qi为节点i注入的无功功率。ui为节点i的电压幅值,uj为节点j的电压幅值。θi为节点i的电压相角,θj为节点j的电压相角,且θij=θij,表示节点i和节点j的电压相角差。Gij、Bij分别为节点导纳矩阵中对应节点i和j之间元素的实部和虚部。Gii、Bii分别为节点导纳矩阵中对应节点i处主对角线上的元素的实部和虚部。In the formula, P i is the active power injected by node i; Q i is the reactive power injected by node i. u i is the voltage amplitude of node i, and u j is the voltage amplitude of node j. θ i is the voltage phase angle of node i, θ j is the voltage phase angle of node j, and θ ijij represents the voltage phase angle difference between node i and node j. G ij , B ij are the real part and imaginary part of the elements between corresponding nodes i and j in the node admittance matrix, respectively. G ii and B ii are the real part and imaginary part of the elements on the main diagonal corresponding to node i in the nodal admittance matrix, respectively.

对于线路i侧支路功率,基于公式(11)和(12)形成其雅克比矩阵元素:For the power of the branch on the line i side, its Jacobian matrix elements are formed based on formulas (11) and (12):

式中,Pij、Qij为线路支路节点i侧的有功功率、无功功率。ui为节点i的电压幅值,uj为节点j的电压幅值。θi为节点i的电压相角,θj为节点j的电压相角,且θij=θij,表示节点i和节点j的电压相角差。g为线路电导,b为线路电纳,yc为线路对地电纳。In the formula, P ij and Q ij are the active power and reactive power of the line branch node i side. u i is the voltage amplitude of node i, and u j is the voltage amplitude of node j. θ i is the voltage phase angle of node i, θ j is the voltage phase angle of node j, and θ ijij represents the voltage phase angle difference between node i and node j. g is the line conductance, b is the line susceptance, and y c is the line-to-ground susceptance.

对于线路j侧支路功率,基于公式(13)和(14)形成其雅克比矩阵元素:For the branch power on the side of line j, its Jacobian matrix elements are formed based on formulas (13) and (14):

式中,Pji、Qji为线路支路节点j侧的有功功率、无功功率。ui为节点i的电压幅值,uj为节点j的电压幅值。θi为节点i的电压相角,θj为节点j的电压相角,且θij=θij,表示节点i和节点j的电压相角差。g为线路电导,b为线路电纳,yc为线路对地电纳。In the formula, P ji and Q ji are the active power and reactive power at node j side of the line branch. u i is the voltage amplitude of node i, and u j is the voltage amplitude of node j. θ i is the voltage phase angle of node i, θ j is the voltage phase angle of node j, and θ ijij represents the voltage phase angle difference between node i and node j. g is the line conductance, b is the line susceptance, and y c is the line-to-ground susceptance.

对于变压器i侧支路功率,基于公式(15)和(16)形成其雅克比矩阵元素:For the branch power on the i side of the transformer, its Jacobian matrix elements are formed based on formulas (15) and (16):

式中,Pijk、Qijk为变压器支路节点i侧的有功功率、无功功率。ui为节点i的电压幅值,uj为节点j的电压幅值。θi为节点i的电压相角,θj为节点j的电压相角,θij=θij,表示节点i和节点j的电压相角差。K为变压器非标准变比:j为标准侧,变比为1,i为非标准侧,变比为K;bT为变压器标准侧的电纳。In the formula, P ijk and Q ijk are the active power and reactive power of the transformer branch node i side. u i is the voltage amplitude of node i, and u j is the voltage amplitude of node j. θ i is the voltage phase angle of node i, θ j is the voltage phase angle of node j, and θ ij = θ i - θ j represents the voltage phase angle difference between node i and node j. K is the non-standard transformation ratio of the transformer: j is the standard side, the transformation ratio is 1, i is the non-standard side, the transformation ratio is K; b T is the susceptance of the standard side of the transformer.

对于变压器j侧支路功率,基于公式(17)和(18)形成其雅克比矩阵元素:For the power of the side branch of the transformer j, its Jacobian matrix elements are formed based on formulas (17) and (18):

式中,Pjik、Qjik为变压器支路节点j侧的有功功率、无功功率。ui为节点i的电压幅值,uj为节点j的电压幅值。θi为节点i的电压相角,θj为节点j的电压相角,θij=θij,表示节点i和节点j的电压相角差。K为变压器非标准变比:j为标准侧,变比为1,i为非标准侧,变比为K;bT为变压器标准侧的电纳。In the formula, P jik and Q jik are the active power and reactive power at the node j side of the transformer branch. u i is the voltage amplitude of node i, and u j is the voltage amplitude of node j. θ i is the voltage phase angle of node i, θ j is the voltage phase angle of node j, and θ ij = θ i - θ j represents the voltage phase angle difference between node i and node j. K is the non-standard transformation ratio of the transformer: j is the standard side, the transformation ratio is 1, i is the non-standard side, and the transformation ratio is K; b T is the susceptance of the standard side of the transformer.

对于节点电压,基于公式(19)形成其雅克比矩阵元素:For the node voltages, the elements of its Jacobian matrix are formed based on formula (19):

式中,Ui、ui均为节点i的电压幅值,uj为节点j的电压幅值,θi为节点i的电压相角,θj为节点j的电压相角。In the formula, U i and u i are the voltage amplitude of node i, u j is the voltage amplitude of node j, θ i is the voltage phase angle of node i, and θ j is the voltage phase angle of node j.

零注入等式约束的雅可比矩阵为其中c(x)是零注入节点的量测方程,其表示节点的注入有功功率和无功功率值;x是n维的状态量,n是状态估计中实际的状态变量个数。The Jacobian matrix of the zero-injection equality constraint is Among them, c(x) is the measurement equation of the zero injection node, which represents the injected active power and reactive power value of the node; x is the n-dimensional state quantity, and n is the actual number of state variables in the state estimation.

III)计算权函数III) Calculate the weight function

基于公式(20)计算权函数W的对角阵元素,计算公式为:Based on the formula (20) to calculate the diagonal matrix elements of the weight function W, the calculation formula is:

式中,rmin为小残差的检测门槛值,取0.002,wi *为量测i的权函数,为量测i的固定权重;ri为量测i的残差。In the formula, r min is the detection threshold of small residual error, take 0.002, w i * is the weight function of measurement i, is the fixed weight of measurement i; r i is the residual error of measurement i.

(5)状态变量更新和收敛性判断(5) State variable update and convergence judgment

i)状态变量更新i) State variable update

第(4)步完成后,根据公式(21)计算状态变量的修正量Δx(time),然后更新状态变量,得到状态变量新值x(time+1)=x(time)+Δx(time),time=time+1。After step (4) is completed, calculate the correction amount Δx (time) of the state variable according to the formula (21), and then update the state variable to obtain the new value of the state variable x (time+1) = x (time) + Δx (time) , time=time+1.

式中,time为计算迭代次数;x(time)为第time次迭代时的状态量;W为权函数对角阵,其对角元素等于权函数,即Wii=wi *为量测量的雅可比矩阵,HT为其转置。c(x(time))为迭代值是x(time)时的零注入等式约束,为零注入等式约束的雅可比矩阵,CT为其转置。z-h(x(time))表示迭代值为x(time)时的残差;λ(time)为第time次迭代时的拉格朗日乘子向量。In the formula, time is the number of calculation iterations; x (time) is the state quantity at the time-th iteration; W is the diagonal matrix of the weight function, and its diagonal elements are equal to the weight function, that is, W ii =w i * . is the Jacobian matrix of quantity measurement, and H T is its transpose. c(x (time) ) is the zero injection equality constraint when the iteration value is x (time) , Inject the equality-constrained Jacobian matrix for zero, and C T its transpose. zh(x (time) ) indicates the residual error when the iteration value is x (time) ; λ (time) is the Lagrangian multiplier vector at the time-th iteration.

ii)收敛性判断ii) Judgment of convergence

当状态变量的修正量Δx(time)满足max(|Δx(time)|)<ε,则结束迭代计算,输出结果;当max(|Δx(time)|)≥ε且迭代次数time≥Tmax,则停止迭代,输出“不收敛!”。When the correction amount Δx (time) of the state variable satisfies max(|Δx (time) |)<ε, the iterative calculation is ended and the result is output; when max(|Δx (time) |)≥ε and the number of iterations time≥Tmax, Then stop the iteration and output "does not converge!".

当max(|Δx(time)|)≥ε且迭代次数time<Tmax,使迭代次数time增加1,返回第(3)步,进行重新迭代计算。When max(|Δx (time) |)≥ε and the number of iterations time<Tmax, increase the number of iterations time by 1, and return to step (3) for re-iteration calculation.

本发明采用上述技术方案后,主要有以下效果:After the present invention adopts above-mentioned technical scheme, mainly have following effect:

与现有技术的权系数赋值方式比较,本发明采用基于残差归一化的权函数,在迭代过程中只需对权重进行改变,便可以自动消除坏数据的影响。抗差性能良好,且能有效抑制不良杠杆量测对估计结果的不良影响。Compared with the weight coefficient assignment method in the prior art, the present invention adopts a weight function based on residual normalization, and only needs to change the weight during the iterative process to automatically eliminate the influence of bad data. The robustness performance is good, and it can effectively suppress the bad influence of the bad leverage measurement on the estimation result.

本发明在权函数计算时引入门槛值判据,即对于残差小于门槛值的量测其权重均取为1,而残差大于门槛值的量测其权重均小于1。从而充分利用了量测量中的有效信息,避免了因量测量权重的波动太大而导致状态估计不收敛,因此极大改善了坏数据与量测误差共存时的收敛性和数值稳定性。The present invention introduces a threshold criterion in the calculation of the weight function, that is, the weights of the measurements whose residuals are smaller than the threshold are taken as 1, and the weights of the measurements whose residuals are greater than the threshold are all less than 1. Therefore, the effective information in the quantity measurement is fully utilized, and the state estimation does not converge due to the large fluctuation of the quantity measurement weight, so the convergence and numerical stability when bad data and measurement errors coexist are greatly improved.

本发明在权函数计算时直接以量测量的残差作为变量,避免了标准化残差的计算,因此大大提高了计算速度,节省了计算时间,非常适宜于工程实际的应用。The present invention directly uses the residual error of the quantity measurement as a variable when calculating the weight function, avoids the calculation of the standardized residual error, thus greatly improves the calculation speed, saves calculation time, and is very suitable for engineering practical application.

本发明的附加方面和优点将在下面的描述中部分给出,部分将从下面的描述中变得明显,或通过本发明的实践了解到。Additional aspects and advantages of the invention will be set forth in the description which follows, and in part will be obvious from the description, or may be learned by practice of the invention.

附图说明Description of drawings

图1为本发明的基于残差归一化的权函数最小二乘状态估计方法的流程示意图;Fig. 1 is a schematic flow chart of the weight function least squares state estimation method based on residual normalization of the present invention;

图2为3节点系统的网络参数和量测分布;Figure 2 shows the network parameters and measurement distribution of the 3-node system;

图3为IEEE-9节点系统的接线图。Figure 3 is a wiring diagram of the IEEE-9 node system.

具体实施方式Detailed ways

下面结合附图和实施例对本发明作进一步说明,但不应该理解为本发明上述主题范围仅限于下述实施例。在不脱离本发明上述技术思想的情况下,根据本领域普通技术知识和惯用手段,做出各种替换和变更,均应包括在本发明的保护范围内。The present invention will be further described below in conjunction with the accompanying drawings and embodiments, but it should not be understood that the scope of the subject matter of the present invention is limited to the following embodiments. Without departing from the above-mentioned technical ideas of the present invention, various replacements and changes made according to common technical knowledge and conventional means in this field shall be included in the protection scope of the present invention.

如1所示,一种基于残差归一化的权函数最小二乘状态估计方法的具体步骤如下:As shown in 1, the specific steps of a weight function least squares state estimation method based on residual normalization are as follows:

(1)输入基础数据及初始化(1) Input basic data and initialization

1)输入基础数据1) Enter basic data

首先输入任一时间断面下电网的SCADA数据、网络结构及参数信息。即输入电网在任一时间断面下的节点电压幅值、节点注入功率以及支路功率,网络结构及参数信息为网络元件(包括线路、变压器)的电阻、电抗和电纳参数、以及额定电压、功率基准。First, input the SCADA data, network structure and parameter information of the power grid under any time section. That is, the node voltage amplitude, node injection power and branch power of the input grid at any time section, the network structure and parameter information are the resistance, reactance and susceptance parameters of network components (including lines and transformers), as well as rated voltage and power benchmark.

2)参数初始化2) Parameter initialization

根据Anderson P M与Fouad A A所著“Power system control and stability”一文关于IEEE-9节点系统的标准数据,输入网络结构、负荷数据及相关参数信息。以潮流计算所得节点电压幅值、节点注入功率、支路功率为量测量,假设量测中无不良数据,且正常量测均带有一定的量测误差,该误差符合正态分布。设置m阶单位矩阵R-1(即m阶单位矩阵R-1的对角线元素全为1,非对角线元素全为0),m为状态估计中实际的量测变量个数,为57;置状态估计时节点电压幅值的初值均为标幺值1,相角均为0;网络元件的阻抗、电纳均归算为标幺值;节点注入功率以及支路功率也归算至标幺值,基准功率设为100MVA;以及权重计算时小残差的检测门槛值rmin=0.002,并设置迭代次数time=1。对于初始化最大迭代次数和收敛精度的选取采用,本实施例优选Tmax=40、ε=10-4According to the standard data of IEEE-9 node system in the article "Power system control and stability" written by Anderson PM and Fouad AA, input the network structure, load data and related parameter information. The node voltage amplitude, node injection power, and branch power obtained from the power flow calculation are measured. Assuming that there is no bad data in the measurement, and the normal measurement has a certain measurement error, the error conforms to the normal distribution. Set the m-order identity matrix R -1 (that is, the diagonal elements of the m-order identity matrix R -1 are all 1, and the off-diagonal elements are all 0), m is the actual number of measured variables in the state estimation, which is 57; The initial value of the node voltage amplitude when setting the state estimation is 1 per unit, and the phase angle is 0; the impedance and susceptance of the network elements are all reduced to the per unit value; the node injection power and branch power are also normalized to Calculated to the per unit value, the reference power is set to 100MVA; and the detection threshold value r min =0.002 of the small residual error during weight calculation, and the iteration number time=1 is set. For the selection and adoption of the maximum number of initialization iterations and convergence accuracy, Tmax=40 and ε=10 −4 are preferred in this embodiment.

(2)计算节点导纳矩阵(2) Calculate node admittance matrix

第(1)步完成后,计算该网络的节点导纳矩阵,计算公式为技术方案中的公式(1)。After step (1) is completed, the node admittance matrix of the network is calculated, and the calculation formula is the formula (1) in the technical scheme.

根据IEEE-9节点系统的网络结构和参数信息,按照技术方案中的公式(1),计算得到该网络的节点导纳矩阵Y为:According to the network structure and parameter information of the IEEE-9 node system, according to the formula (1) in the technical solution, the node admittance matrix Y of the network is calculated as:

(3)形成零注入等式约束方程(3) Form the zero-injection equality constraint equation

第(2)步完成后,选取量测量中既不是发电机节点也不是负荷节点的节点,其注入功率作为零注入等式约束方程,按照技术方案中的公式(2),以第1次迭代计算的结果举例,计算得到零注入功率c(x)为:After step (2) is completed, the nodes that are neither generator nodes nor load nodes in the quantity measurement are selected, and their injected power is used as the zero injection equality constraint equation. According to the formula (2) in the technical plan, the first iteration As an example of the calculation results, the calculated zero injection power c(x) is:

(4)计算残差、雅克比矩阵及权函数(4) Calculation of residuals, Jacobian matrix and weight function

第(3)步完成后,以电网中节点电压幅值、除去零注入节点后的节点注入功率、以及支路功率为量测量,计算电网中量测量的残差、雅可比矩阵及权函数,计算公式为技术方案中的公式(4)-公式(20)。After step (3) is completed, take the node voltage amplitude in the power grid, the node injection power after removing the zero injection node, and the branch power as the measurement, and calculate the residual error, Jacobian matrix and weight function of the power grid measurement, The calculation formula is formula (4)-formula (20) in the technical scheme.

I)计算残差I) Calculation of residuals

根据残差及量测方程的定义,按照技术方案中的公式(4)-公式(8),以第1次迭代计算的结果举例,计算得到量测量的残差r为:According to the definition of residual error and measurement equation, according to the formula (4)-formula (8) in the technical solution, taking the result of the first iterative calculation as an example, the calculated residual r of the quantity measurement is:

II)计算雅克比矩阵II) Calculate the Jacobian matrix

根据雅克比矩阵元素的定义,按照技术方案中的公式(9)-公式(19)计算雅可比矩阵H和C,以第1次迭代计算的结果举例,计算得到雅可比矩阵H和C为:According to the definition of the elements of the Jacobian matrix, calculate the Jacobian matrix H and C according to the formula (9)-formula (19) in the technical solution, and take the result of the first iterative calculation as an example, the calculated Jacobian matrix H and C are:

III)计算权函数III) Calculate the weight function

根据权函数的定义,按照技术方案中的公式(20),以第1次迭代计算的结果举例,计算得到权函数W为:According to the definition of the weight function, according to the formula (20) in the technical solution, taking the result of the first iteration calculation as an example, the calculated weight function W is:

(5)状态变量更新和收敛性判断(5) State variable update and convergence judgment

i)状态变量更新i) State variable update

第(4)步完成后,按照技术方案中的公式(21)计算状态变量的修正量Δx(time),然后更新状态变量,得到状态变量新值,即:x(time+1)=x(time)+Δx(time),time=time+1。After the 4th step is finished, calculate the correction amount Δx (time) of state variable according to formula (21) in the technical scheme, then update state variable, obtain the new value of state variable, namely: x (time+1) =x ( time) +Δx (time) , time=time+1.

以第1次迭代计算的结果举例,按技术方案中的公式(21),计算得到状态变量的修正量Δx(1)为:Taking the result of the first iterative calculation as an example, according to the formula (21) in the technical solution, the correction amount Δx (1) of the state variable is calculated as:

ii)收敛性判断ii) Judgment of convergence

当状态变量的修正量Δx(time)满足max(|Δx(time)|)<ε,则结束迭代计算,输出结果;当max(|Δx(time)|)≥ε且迭代次数time≥Tmax,则停止迭代,输出“不收敛!”。当max(|Δx(time)|)≥ε且迭代次数time<Tmax,使迭代次数time增加1,返回第(3)步,进行重新迭代计算。When the correction amount Δx (time) of the state variable satisfies max(|Δx (time) |)<ε, the iterative calculation is ended and the result is output; when max(|Δx (time) |)≥ε and the number of iterations time≥Tmax, Then stop the iteration and output "does not converge!". When max(|Δx (time) |)≥ε and the number of iterations time<Tmax, increase the number of iterations time by 1, and return to step (3) for re-iteration calculation.

以第1次迭代计算的结果举例,此时,ε=10-4、Tmax=40、max(|Δx(1)|)=0.16419>ε,time=1<Tmax。根据收敛性判断,进行如下步骤:time=time+1=2,返回第(3)步,重新进行迭代计算。Taking the result of the first iterative calculation as an example, at this time, ε=10 −4 , Tmax=40, max(|Δx (1) |)=0.16419>ε, time=1<Tmax. According to the convergence judgment, the following steps are performed: time=time+1=2, return to step (3), and perform iterative calculation again.

根据前面的步骤,迭代3次后满足收敛条件,此时max(|Δx(3)|)=7.4934E-05<ε,状态变量的估计结果如下表所示:According to the previous steps, the convergence condition is satisfied after 3 iterations, at this time max(|Δx (3) |)=7.4934E -05 <ε, the estimated results of the state variables are shown in the following table:

表1 IEEE-9节点系统状态变量估计结果Table 1 Estimation results of IEEE-9 node system state variables

实验效果比较分析Comparative Analysis of Experimental Effects

为使本领域技术人员更好地理解本发明以及了解本发明相对现有技术的优点,申请人结合具体实施例进行进一步的阐释。In order to enable those skilled in the art to better understand the present invention and understand the advantages of the present invention over the prior art, the applicant further explains in conjunction with specific embodiments.

(1)抗差性能的比较(1) Comparison of tolerance performance

发明人将本发明的RNWLS与其他状态估计器进行比较,来测试RNWLS的抗差性。The inventors compared the RNWLS of the present invention with other state estimators to test the robustness of the RNWLS.

1)3节点系统1) 3-node system

以2所示3节点系统为例,1号量测为杠杆量测,并设置为不良数据,其他量测值取真值(潮流计算值),其中P、Q分别表示有功和无功功率。用四种状态估计器(RNWLS、EFWLS、E-LAV、WLS)对上述算例进行仿真分析,结果如下所示:Taking the 3-node system shown in 2 as an example, the measurement No. 1 is lever measurement and is set as bad data, and the other measurement values take the true value (calculated value of power flow), where P and Q represent active and reactive power, respectively. Using four state estimators (RNWLS, EFWLS, E-LAV, WLS) to simulate and analyze the above example, the results are as follows:

表2 3节点系统中各方法估计结果的比较Table 2 Comparison of estimation results of various methods in 3-node system

其中,粗体字标识的数据为量测坏数据。由上述计算结果可见,在处理不良杠杆量测的能力上,本发明方法RNWLS和EFWLS估计的结果最接近于潮流真值,说明其能准确地估计出系统的真实状态,且1号不良杠杆量测能够被辨识并自动排除;而E-LAV估计的有功功率较接近潮流真值,无功功率则相差较大;WLS估计结果与潮流真值相差最大,表明其抗不良杠杆量测的能力有限,无法估计出系统的真实状态。Among them, the data marked in bold is bad measurement data. It can be seen from the above calculation results that in terms of the ability to deal with bad leverage measurement, the results estimated by the method RNWLS and EFWLS of the present invention are closest to the true value of power flow, indicating that it can accurately estimate the real state of the system, and the amount of bad leverage of No. 1 However, the active power estimated by E-LAV is closer to the true value of the power flow, while the reactive power has a large difference; the difference between the estimated result of WLS and the true value of the power flow is the largest, indicating that its ability to resist bad leverage measurement is limited , the true state of the system cannot be estimated.

2)IEEE-118节点系统2) IEEE-118 node system

为分析本发明方法RNWLS在规模更大系统上的抗差性能,以IEEE-118节点系统为例,分别对坏数据比例为0%~10%共11种情况进行试验。其中正常量测均带有一定的量测误差,且该误差符合正态分布;坏数据为潮流真值中相应比例电压幅值、有功功率、无功功率值的相反数(其中,有功功率、无功功率均在最大值处设置坏数据)。用四种状态估计器(RNWLS、EFWLS、E-LAV、WLS)对上述算例进行仿真分析,结果如下所示:In order to analyze the tolerance performance of the RNWLS method of the present invention on a larger scale system, taking the IEEE-118 node system as an example, 11 cases in which the proportion of bad data is 0% to 10% are tested respectively. Among them, the normal measurement has a certain measurement error, and the error conforms to the normal distribution; the bad data is the opposite number of the corresponding proportional voltage amplitude, active power, and reactive power value in the true value of the power flow (among them, active power, The reactive powers are all set at the maximum (bad data). Using four state estimators (RNWLS, EFWLS, E-LAV, WLS) to simulate and analyze the above example, the results are as follows:

表3 IEEE-118节点系统不同估计方法电压幅值误差比较Table 3 Voltage amplitude error comparison of different estimation methods for IEEE-118 node systems

其中,ΔUmax与ΔUmean分别表示节点电压幅值的最大估计误差与平均估计误差,考虑到由SCADA所获取的电网潮流信息中节点的电压相角均为0,因此只对节点电压的幅值进行仿真分析。由上述计算结果可见,在不良数据从0%~10%变化的过程中,E-LAV与WLS所得结果误差较大;而RNWLS与EFWLS的估计结果,其误差都接近于0,表明在规模较大系统中,这两种估计方法都能排除不良数据,获得准确的估计结果,抗差性能良好。Among them, ΔU max and ΔU mean represent the maximum estimation error and average estimation error of the node voltage amplitude respectively. Considering that the voltage phase angle of the node in the power flow information obtained by SCADA is 0, only the node voltage amplitude Perform simulation analysis. From the above calculation results, it can be seen that in the process of bad data changing from 0% to 10%, the errors of the results obtained by E-LAV and WLS are relatively large; while the errors of the estimated results of RNWLS and EFWLS are close to 0, indicating that in the process of large-scale In a large system, these two estimation methods can eliminate bad data, obtain accurate estimation results, and have good tolerance performance.

(2)计算效率的比较(2) Comparison of computational efficiency

发明人为了进行效率比较,以IEEE-118节点系统为例,假设量测中无不良数据,且正常量测均带有一定的量测误差,该误差符合正态分布。用四种状态估计器(RNWLS、EFWLS、E-LAV、WLS)对上述算例进行仿真分析,得到平均每次迭代时间和总计算时间如下表所示:In order to compare the efficiency, the inventor took the IEEE-118 node system as an example, assuming that there is no bad data in the measurement, and the normal measurement has a certain measurement error, and the error conforms to the normal distribution. Using four state estimators (RNWLS, EFWLS, E-LAV, WLS) to simulate and analyze the above example, the average time of each iteration and the total calculation time are shown in the following table:

表4 IEEE-118节点系统中各方法的计算效率比较Table 4 Computational efficiency comparison of each method in IEEE-118 node system

由上述计算结果可见,本发明方法RNWLS和EFWLS估计收敛性更好,但针对平均每次迭代时间而言,WLS估计最短,但其不具有抗差性;EFWLS估计最长;RNWLS估计与E-LAV估计居中。It can be seen from the above calculation results that the RNWLS and EFWLS estimation methods of the present invention have better convergence, but for the average time of each iteration, the WLS estimation is the shortest, but it does not have tolerance; the EFWLS estimation is the longest; the RNWLS estimation and E- The LAV estimate is intermediate.

综上所述,本发明所提出的RNWLS估计在抗差性能上近似于EFWLS估计,抗差性能良好,且能有效抑制不良杠杆量测对结果的不良影响;而在计算效率上近似于E-LAV估计,计算效率高,收敛性良好。从而有效集中了现有状态估计方法在抗差性能和计算效率上的优点,非常适宜于实际工程的应用。To sum up, the RNWLS estimation proposed by the present invention is similar to the EFWLS estimation in terms of robustness, has good robustness, and can effectively suppress the adverse effects of bad leverage measurements on the results; and is similar in computational efficiency to E- LAV estimation has high computational efficiency and good convergence. Therefore, the advantages of existing state estimation methods in robustness and calculation efficiency are effectively concentrated, and it is very suitable for practical engineering applications.

Claims (1)

1.一种基于残差归一化的权函数最小二乘状态估计方法,利用计算机,通过程序,实现电网的状态估计,其特征在于:所述方法的具体步骤包括以下内容;1. A weight function least squares state estimation method based on residual normalization, utilize computer, by program, realize the state estimation of power grid, it is characterized in that: the concrete steps of described method comprise the following contents; (1)输入基础数据及初始化(1) Input basic data and initialization 1)输入基础数据1) Enter basic data 首先输入任一时间断面下电网的数据采集与监视控制系统即SCADA数据、网络结构及参数信息;所述电网的SCADA数据包括节点电压幅值、节点注入功率以及支路功率;所述网络结构及参数信息包括网络元件的电阻、电抗、电纳、额定电压以及功率基准;所述网络元件包括线路以及变压器;First input the data acquisition and monitoring control system of the power grid under any time section, namely SCADA data, network structure and parameter information; the SCADA data of the power grid includes node voltage amplitude, node injection power and branch power; the network structure and The parameter information includes resistance, reactance, susceptance, rated voltage and power reference of network elements; the network elements include lines and transformers; 2)参数初始化2) Parameter initialization 设置m阶单位矩阵R-1,所述单位矩阵R-1对角线元素全为1,非对角线元素全为0,m为状态估计中实际的量测变量个数;设置状态估计时节点电压幅值的初值均为标幺值1,相角均为0;所述网络元件的阻抗和电纳均归算为标幺值;节点注入功率以及支路功率也归算至标幺值;初始化最大迭代次数Tmax为40~60;收敛精度ε为10-3~10-5,以及权重计算时小残差的检测门槛值rmin=0.002,并设置迭代次数time=1;Set the m-order identity matrix R -1 , the diagonal elements of the identity matrix R -1 are all 1, and the off-diagonal elements are all 0, m is the actual number of measured variables in the state estimation; when setting the state estimation The initial value of the node voltage amplitude is 1 p.u., and the phase angle is 0. The impedance and susceptance of the network elements are all reduced to the p.u. value; the node injection power and branch power are also reduced to the p.u. value; initialize the maximum number of iterations Tmax to 40~60; the convergence precision ε is 10 -3 to 10 -5 , and the detection threshold of small residual error r min =0.002 during weight calculation, and set the number of iterations time=1; (2)计算节点导纳矩阵(2) Calculate node admittance matrix 第(1)步完成后,利用公式(1)计算节点导纳矩阵Y;After step (1) is completed, use the formula (1) to calculate the node admittance matrix Y; <mrow> <mfenced open = "{" close = ""> <mtable> <mtr> <mtd> <mrow> <msub> <mi>Y</mi> <mrow> <mi>i</mi> <mi>i</mi> </mrow> </msub> <mo>=</mo> <munder> <mo>&amp;Sigma;</mo> <mrow> <mi>j</mi> <mo>&amp;Element;</mo> <mi>i</mi> </mrow> </munder> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <msub> <mi>Y</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <mo>=</mo> <mo>-</mo> <msub> <mi>y</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> </mrow> </mtd> </mtr> </mtable> </mfenced> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>1</mn> <mo>)</mo> </mrow> </mrow> <mrow><mfenced open = "{" close = ""><mtable><mtr><mtd><mrow><msub><mi>Y</mi><mrow><mi>i</mi><mi>i</mi></mrow></msub><mo>=</mo><munder><mo>&amp;Sigma;</mo><mrow><mi>j</mi><mo>&amp;Element;</mo><mi>i</mi></mrow></munder><msub><mi>y</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub></mrow></mtd></mtr><mtr><mtd><mrow><msub><mi>Y</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub><mo>=</mo><mo>-</mo><msub><mi>y</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub></mrow></mtd></mtr></mtable></mfenced><mo>-</mo><mo>-</mo><mo>-</mo><mrow><mo>(</mo><mn>1</mn><mo>)</mo></mrow></mrow> 式中,节点导纳矩阵的对角线元素表示为Yii,节点导纳矩阵的非对角线元素表示为Yij,均由上式计算得到;yij为节点i和节点j之间的支路阻抗zij的倒数;符号j∈i表示节点j和节点i直接相连,且当节点i有接地支路时还应包括j=0时的情况;In the formula, the diagonal element of the node admittance matrix is expressed as Y ii , and the off-diagonal element of the node admittance matrix is expressed as Y ij , both of which are calculated by the above formula; y ij is the distance between node i and node j The reciprocal of the branch impedance z ij ; the symbol j∈i means that the node j is directly connected to the node i, and when the node i has a grounding branch, it should also include the case of j=0; (3)形成零注入等式约束方程(3) Form the zero-injection equality constraint equation 第(2)步完成后,选取量测量中既不是发电机节点也不是负荷节点的节点,将所述节点的注入功率作为零注入等式约束方程,表示为;After the (2) step is completed, select a node that is neither a generator node nor a load node in the quantity measurement, and use the injected power of the node as a zero injection equation constraint equation, expressed as; c(x)=0 (2)c(x)=0 (2) 式中,c(x)是零注入节点的量测方程,其表示的节点注入有功功率和无功功率值可由公式(3)计算得到;x是n维的状态量,n是状态估计中实际的状态变量个数;c(x)的计算公式为;In the formula, c(x) is the measurement equation of the zero-injected node, and the injected active power and reactive power value of the node represented by it can be calculated by formula (3); x is the n-dimensional state quantity, and n is the actual The number of state variables; the calculation formula of c(x) is; <mrow> <mfenced open = "{" close = ""> <mtable> <mtr> <mtd> <mrow> <msubsup> <mi>P</mi> <mi>i</mi> <mo>&amp;prime;</mo> </msubsup> <mo>=</mo> <msub> <mi>u</mi> <mi>i</mi> </msub> <munderover> <mo>&amp;Sigma;</mo> <mrow> <mi>j</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>N</mi> </munderover> <mo>&amp;lsqb;</mo> <msub> <mi>u</mi> <mi>j</mi> </msub> <mrow> <mo>(</mo> <msub> <mi>G</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <msub> <mi>cos&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <mo>+</mo> <msub> <mi>B</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <msub> <mi>sin&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>&amp;rsqb;</mo> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <msubsup> <mi>Q</mi> <mi>i</mi> <mo>&amp;prime;</mo> </msubsup> <mo>=</mo> <msub> <mi>u</mi> <mi>i</mi> </msub> <munderover> <mo>&amp;Sigma;</mo> <mrow> <mi>j</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>N</mi> </munderover> <mo>&amp;lsqb;</mo> <msub> <mi>u</mi> <mi>j</mi> </msub> <mrow> <mo>(</mo> <msub> <mi>G</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <msub> <mi>sin&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <mo>-</mo> <msub> <mi>B</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <msub> <mi>cos&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>&amp;rsqb;</mo> </mrow> </mtd> </mtr> </mtable> </mfenced> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>3</mn> <mo>)</mo> </mrow> </mrow> <mrow><mfenced open = "{" close = ""><mtable><mtr><mtd><mrow><msubsup><mi>P</mi><mi>i</mi><mo>&amp;prime;</mo></msubsup><mo>=</mo><msub><mi>u</mi><mi>i</mi></msub><munderover><mo>&amp;Sigma;</mo><mrow><mi>j</mi><mo>=</mo><mn>1</mn></mrow><mi>N</mi></munderover><mo>&amp;lsqb;</mo><msub><mi>u</mi><mi>j</mi></msub><mrow><mo>(</mo><msub><mi>G</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub><msub><mi>cos&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub><mo>+</mo><msub><mi>B</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub><msub><mi>sin&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub><mo>)</mo></mrow><mo>&amp;rsqb;</mo></mrow></mtd></mtr><mtr><mtd><mrow><msubsup><mi>Q</mi><mi>i</mi><mo>&amp;prime;</mo></msubsup><mo>=</mo><msub><mi>u</mi><mi>i</mi></msub><munderover><mo>&amp;Sigma;</mo><mrow><mi>j</mi><mo>=</mo><mn>1</mn></mrow><mi>N</mi></munderover><mo>&amp;lsqb;</mo><msub><mi>u</mi><mi>j</mi></msub><mrow><mo>(</mo><msub><mi>G</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub><msub><mi>sin&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub><mo>-</mo><msub><mi>B</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub><msub><mi>cos&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub><mo>)</mo></mrow><mo>&amp;rsqb;</mo></mrow></mtd></mtr></mtable></mfenced><mo>-</mo><mo>-</mo><mo>-</mo><mrow><mo>(</mo><mn>3</mn><mo>)</mo></mrow></mrow> 式中,P′i为节点i注入的有功功率;Q′i为节点i注入的无功功率;ui为节点i的电压幅值,uj为节点j的电压幅值;θi为节点i的电压相角,θj为节点j的电压相角,且θij=θij,表示节点i和节点j的电压相角差;Gij、Bij分别为节点导纳矩阵中对应节点i和j之间元素的实部和虚部;N为电网的节点个数;In the formula, P′ i is the active power injected into node i; Q′ i is the reactive power injected into node i; u i is the voltage amplitude of node i, u j is the voltage amplitude of node j ; The voltage phase angle of i, θ j is the voltage phase angle of node j, and θ ij = θ ij , which means the voltage phase angle difference between node i and node j; G ij and B ij are respectively Corresponding to the real part and imaginary part of the elements between nodes i and j; N is the number of nodes in the grid; (4)计算残差、雅克比矩阵及权函数(4) Calculation of residuals, Jacobian matrix and weight function 第(3)步完成后,以电网中节点电压幅值、除去零注入节点后的节点注入功率以及支路功率为量测量,计算电网中量测量的残差、雅克比矩阵及权函数;具体步骤内容如下;After step (3) is completed, take the node voltage amplitude in the power grid, the node injection power after removing the zero injection node, and the branch power as the measurement, and calculate the residual error, Jacobian matrix and weight function of the power grid measurement; specifically The steps are as follows; I)计算残差I) Calculation of residuals 基于公式(4)计算状态估计中各量测量的参数;Calculate the parameters measured by each quantity in the state estimation based on formula (4); r=z-h(x) (4)r=z-h(x) (4) 式中,z是m维的量测量,m是状态估计中实际的量测变量个数;x是n维的状态量,n是状态估计中实际的状态变量个数;h(x)是量测方程,包括注入功率对应的量测方程、线路支路功率对应的量测方程、变压器支路功率对应的量测方程和节点电压对应的量测方程;h(x)可由公式(5)-公式(8)计算得到;r是量测残差;In the formula, z is the m-dimensional quantity measurement, m is the actual number of measurement variables in the state estimation; x is the n-dimensional state quantity, n is the actual number of state variables in the state estimation; h(x) is the quantity The measurement equations include the measurement equations corresponding to the injected power, the measurement equations corresponding to the line branch power, the measurement equations corresponding to the transformer branch power and the measurement equations corresponding to the node voltage; h(x) can be expressed by the formula (5)- Calculated by formula (8); r is the measurement residual; 基于公式(5)得到除去零注入节点后的节点注入功率对应的量测方程;Based on formula (5), the measurement equation corresponding to the node injection power after removing the zero injection node is obtained; <mrow> <mfenced open = "{" close = ""> <mtable> <mtr> <mtd> <mrow> <msub> <mi>P</mi> <mi>i</mi> </msub> <mo>=</mo> <msub> <mi>u</mi> <mi>i</mi> </msub> <munderover> <mo>&amp;Sigma;</mo> <mrow> <mi>j</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>N</mi> </munderover> <mo>&amp;lsqb;</mo> <msub> <mi>u</mi> <mi>j</mi> </msub> <mrow> <mo>(</mo> <msub> <mi>G</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <msub> <mi>cos&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <mo>+</mo> <msub> <mi>B</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <msub> <mi>sin&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>&amp;rsqb;</mo> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <msub> <mi>Q</mi> <mi>i</mi> </msub> <mo>=</mo> <msub> <mi>u</mi> <mi>i</mi> </msub> <munderover> <mo>&amp;Sigma;</mo> <mrow> <mi>j</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>N</mi> </munderover> <mo>&amp;lsqb;</mo> <msub> <mi>u</mi> <mi>j</mi> </msub> <mrow> <mo>(</mo> <msub> <mi>G</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <msub> <mi>sin&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <mo>-</mo> <msub> <mi>B</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <msub> <mi>cos&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>&amp;rsqb;</mo> </mrow> </mtd> </mtr> </mtable> </mfenced> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>5</mn> <mo>)</mo> </mrow> </mrow> <mrow><mfenced open = "{" close = ""><mtable><mtr><mtd><mrow><msub><mi>P</mi><mi>i</mi></msub><mo>=</mo><msub><mi>u</mi><mi>i</mi></msub><munderover><mo>&amp;Sigma;</mo><mrow><mi>j</mi><mo>=</mo><mn>1</mn></mrow><mi>N</mi></munderover><mo>&amp;lsqb;</mo><msub><mi>u</mi><mi>j</mi></msub><mrow><mo>(</mo><msub><mi>G</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub><msub><mi>cos&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub><mo>+</mo><msub><mi>B</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub><msub><mi>sin&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub><mo>)</mo></mrow><mo>&amp;rsqb;</mo></mrow></mtd></mtr><mtr><mtd><mrow><msub><mi>Q</mi><mi>i</mi></msub><mo>=</mo><msub><mi>u</mi><mi>i</mi></msub><munderover><mo>&amp;Sigma;</mo><mrow><mi>j</mi><mo>=</mo><mn>1</mn></mrow><mi>N</mi></munderover><mo>&amp;lsqb;</mo><msub><mi>u</mi><mi>j</mi></msub><mrow><mo>(</mo><msub><mi>G</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub><msub><mi>sin&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub><mo>-</mo><msub><mi>B</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub><msub><mi>cos&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub><mo>)</mo></mrow><mo>&amp;rsqb;</mo></mrow></mtd></mtr></mtable></mfenced><mo>-</mo><mo>-</mo><mo>-</mo><mrow><mo>(</mo><mn>5</mn><mo>)</mo></mrow></mrow> 式中,Pi为节点i注入的有功功率;Qi为节点i注入的无功功率;ui为节点i的电压幅值,uj为节点j的电压幅值;θi为节点i的电压相角,θj为节点j的电压相角,且θij=θij,表示节点i和节点j的电压相角差;Gij、Bij分别为节点导纳矩阵中对应节点i和j之间元素的实部和虚部;In the formula, P i is the active power injected by node i; Q i is the reactive power injected by node i; u i is the voltage amplitude of node i, u j is the voltage amplitude of node j; θ i is the voltage amplitude of node i Voltage phase angle, θ j is the voltage phase angle of node j, and θ ij = θ i - θ j , which means the voltage phase angle difference between node i and node j; G ij and B ij are the corresponding nodes in the node admittance matrix the real and imaginary parts of the elements between i and j; 基于公式(6)得到线路支路功率对应的量测方程;Obtain the measurement equation corresponding to the line branch power based on formula (6); <mrow> <mfenced open = "{" close = ""> <mtable> <mtr> <mtd> <mrow> <msub> <mi>P</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <mo>=</mo> <msup> <msub> <mi>u</mi> <mi>i</mi> </msub> <mn>2</mn> </msup> <mi>g</mi> <mo>-</mo> <msub> <mi>u</mi> <mi>i</mi> </msub> <msub> <mi>u</mi> <mi>j</mi> </msub> <mi>g</mi> <mi> </mi> <msub> <mi>cos&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <mo>-</mo> <msub> <mi>u</mi> <mi>i</mi> </msub> <msub> <mi>u</mi> <mi>j</mi> </msub> <mi>b</mi> <mi> </mi> <msub> <mi>sin&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <msub> <mi>Q</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <mo>=</mo> <mo>-</mo> <msup> <msub> <mi>u</mi> <mi>i</mi> </msub> <mn>2</mn> </msup> <mrow> <mo>(</mo> <mi>b</mi> <mo>+</mo> <msub> <mi>y</mi> <mi>c</mi> </msub> <mo>)</mo> </mrow> <mo>-</mo> <msub> <mi>u</mi> <mi>i</mi> </msub> <msub> <mi>u</mi> <mi>j</mi> </msub> <mi>g</mi> <mi> </mi> <msub> <mi>sin&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <mo>+</mo> <msub> <mi>u</mi> <mi>i</mi> </msub> <msub> <mi>u</mi> <mi>j</mi> </msub> <mi>b</mi> <mi> </mi> <msub> <mi>cos&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <msub> <mi>P</mi> <mrow> <mi>j</mi> <mi>i</mi> </mrow> </msub> <mo>=</mo> <msup> <msub> <mi>u</mi> <mi>j</mi> </msub> <mn>2</mn> </msup> <mi>g</mi> <mo>+</mo> <msub> <mi>u</mi> <mi>i</mi> </msub> <msub> <mi>u</mi> <mi>j</mi> </msub> <mrow> <mo>(</mo> <mo>-</mo> <mi>g</mi> <mi> </mi> <msub> <mi>cos&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <mo>+</mo> <mi>b</mi> <mi> </mi> <msub> <mi>sin&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <msub> <mi>Q</mi> <mrow> <mi>j</mi> <mi>i</mi> </mrow> </msub> <mo>=</mo> <mo>-</mo> <msup> <msub> <mi>u</mi> <mi>j</mi> </msub> <mn>2</mn> </msup> <mrow> <mo>(</mo> <mi>b</mi> <mo>-</mo> <msub> <mi>y</mi> <mi>c</mi> </msub> <mo>)</mo> </mrow> <mo>+</mo> <msub> <mi>u</mi> <mi>i</mi> </msub> <msub> <mi>u</mi> <mi>j</mi> </msub> <mrow> <mo>(</mo> <mi>g</mi> <mi> </mi> <msub> <mi>sin&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <mo>+</mo> <mi>b</mi> <mi> </mi> <msub> <mi>cos&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> </mtable> </mfenced> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>6</mn> <mo>)</mo> </mrow> </mrow> <mrow><mfenced open = "{" close = ""><mtable><mtr><mtd><mrow><msub><mi>P</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub><mo>=</mo><msup><msub><mi>u</mi><mi>i</mi></msub><mn>2</mn></msup><mi>g</mi><mo>-</mo><msub><mi>u</mi><mi>i</mi></msub><msub><mi>u</mi><mi>j</mi></msub><mi>g</mi><mi></mi><msub><mi>cos&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub><mo>-</mo><msub><mi>u</mi><mi>i</mi></msub><msub><mi>u</mi><mi>j</mi></msub><mi>b</mi><mi></mi><msub><mi>sin&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub></mrow></mtd></mtr><mtr><mtd><mrow><msub><mi>Q</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub><mo>=</mo><mo>-</mo><msup><msub><mi>u</mi><mi>i</mi></msub><mn>2</mn></msup><mrow><mo>(</mo><mi>b</mi><mo>+</mo><msub><mi>y</mi><mi>c</mi></msub><mo>)</mo></mrow><mo>-</mo><msub><mi>u</mi><mi>i</mi></msub><msub><mi>u</mi><mi>j</mi></msub><mi>g</mi><mi></mi><msub><mi>sin&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub><mo>+</mo><msub><mi>u</mi><mi>i</mi></msub><msub><mi>u</mi><mi>j</mi></msub><mi>b</mi><mi></mi><msub><mi>cos&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub></mrow></mtd></mtr><mtr><mtd><mrow><msub><mi>P</mi><mrow><mi>j</mi><mi>i</mi></mrow></msub><mo>=</mo><msup><msub><mi>u</mi><mi>j</mi></msub><mn>2</mn></msup><mi>g</mi><mo>+</mo><msub><mi>u</mi><mi>i</mi></msub><msub><mi>u</mi><mi>j</mi></msub><mrow><mo>(</mo><mo>-</mo><mi>g</mi><mi></mi><msub><mi>cos&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub><mo>+</mo><mi>b</mi><mi></mi><msub><mi>sin&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub><mo>)</mo></mrow></mrow></mtd></mtr><mtr><mtd><mrow><msub><mi>Q</mi><mrow><mi>j</mi><mi>i</mi></mrow></msub><mo>=</mo><mo>-</mo><msup><msub><mi>u</mi><mi>j</mi></msub><mn>2</mn></msup><mrow><mo>(</mo><mi>b</mi><mo>-</mo><msub><mi>y</mi><mi>c</mi></msub><mo>)</mo></mrow><mo>+</mo><msub><mi>u</mi><mi>i</mi></msub><msub><mi>u</mi><mi>j</mi></msub><mrow><mo>(</mo><mi>g</mi><mi></mi><msub><mi>sin&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub><mo>+</mo><mi>b</mi><mi></mi><msub><mi>cos&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub><mo>)</mo></mrow></mrow></mtd></mtr></mtable></mfenced><mo>-</mo><mo>-</mo><mo>-</mo><mrow><mo>(</mo><mn>6</mn><mo>)</mo></mrow></mrow> 式中,Pij、Qij分别为线路支路节点i侧的有功功率、无功功率,Pji、Qji分别为支路节点j侧的有功功率、无功功率;ui为节点i的电压幅值,uj为节点j的电压幅值;θi为节点i的电压相角,θj为节点j的电压相角,且θij=θij,表示节点i和节点j的电压相角差;g为线路电导,b为线路电纳,yc为线路对地电纳;In the formula, P ij , Q ij are the active power and reactive power of the line branch node i side respectively, P ji and Q ji are the active power and reactive power of the branch node j side respectively; u i is the power of node i Voltage amplitude, u j is the voltage amplitude of node j; θ i is the voltage phase angle of node i, θ j is the voltage phase angle of node j, and θ ij = θ i - θ j means node i and node j The voltage phase angle difference; g is the conductance of the line, b is the susceptance of the line, and y c is the susceptance of the line to the ground; 基于公式(7)得到变压器支路功率对应的量测方程;Based on the formula (7), the corresponding measurement equation of the transformer branch power is obtained; <mrow> <mfenced open = "{" close = ""> <mtable> <mtr> <mtd> <mrow> <msub> <mi>P</mi> <mrow> <mi>j</mi> <mi>k</mi> </mrow> </msub> <mo>=</mo> <mfrac> <mn>1</mn> <mi>K</mi> </mfrac> <msub> <mi>u</mi> <mi>i</mi> </msub> <msub> <mi>u</mi> <mi>j</mi> </msub> <msub> <mi>b</mi> <mi>T</mi> </msub> <msub> <mi>sin&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <msub> <mi>Q</mi> <mrow> <mi>i</mi> <mi>j</mi> <mi>k</mi> </mrow> </msub> <mo>=</mo> <mfrac> <mn>1</mn> <msup> <mi>K</mi> <mn>2</mn> </msup> </mfrac> <msup> <msub> <mi>u</mi> <mi>i</mi> </msub> <mn>2</mn> </msup> <msub> <mi>b</mi> <mi>T</mi> </msub> <mo>+</mo> <mfrac> <mn>1</mn> <mi>K</mi> </mfrac> <msub> <mi>u</mi> <mi>i</mi> </msub> <msub> <mi>u</mi> <mi>j</mi> </msub> <msub> <mi>b</mi> <mi>T</mi> </msub> <msub> <mi>cos&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <msub> <mi>P</mi> <mrow> <mi>j</mi> <mi>i</mi> <mi>k</mi> </mrow> </msub> <mo>=</mo> <mfrac> <mn>1</mn> <mi>K</mi> </mfrac> <msub> <mi>u</mi> <mi>i</mi> </msub> <msub> <mi>u</mi> <mi>j</mi> </msub> <msub> <mi>b</mi> <mi>T</mi> </msub> <msub> <mi>sin&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <msub> <mi>Q</mi> <mrow> <mi>j</mi> <mi>i</mi> <mi>k</mi> </mrow> </msub> <mo>=</mo> <mo>-</mo> <msub> <mi>b</mi> <mi>T</mi> </msub> <msup> <msub> <mi>u</mi> <mi>j</mi> </msub> <mn>2</mn> </msup> <mo>+</mo> <mfrac> <mn>1</mn> <mi>K</mi> </mfrac> <msub> <mi>u</mi> <mi>i</mi> </msub> <msub> <mi>u</mi> <mi>j</mi> </msub> <msub> <mi>b</mi> <mi>T</mi> </msub> <msub> <mi>cos&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> </mrow> </mtd> </mtr> </mtable> </mfenced> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>7</mn> <mo>)</mo> </mrow> </mrow> <mrow><mfenced open = "{" close = ""><mtable><mtr><mtd><mrow><msub><mi>P</mi><mrow><mi>j</mi><mi>k</mi></mrow></msub><mo>=</mo><mfrac><mn>1</mn><mi>K</mi></mfrac><msub><mi>u</mi><mi>i</mi></msub><msub><mi>u</mi><mi>j</mi></msub><msub><mi>b</mi><mi>T</mi></msub><msub><mi>sin&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mi>mrow></msub></mrow></mtd></mtr><mtr><mtd><mrow><msub><mi>Q</mi><mrow><mi>i</mi><mi>j</mi><mi>k</mi></mrow></msub><mo>=</mo><mfrac><mn>1</mn><msup><mi>K</mi><mn>2</mn></msup></mfrac><msup><msub><mi>u</mi><mi>i</mi></msub><mn>2</mn></msup><msub><mi>b</mi><mi>T</mi></msub><mo>+</mo><mfrac><mn>1</mn><mi>K</mi></mfrac><msub><mi>u</mi><mi>i</mi></msub><msub><mi>u</mi><mi>j</mi></msub><msub><mi>b</mi><mi>T</mi></msub><msub><mi>cos&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub></mrow></mtd></mtr><mtr><mtd><mrow><msub><mi>P</mi><mrow><mi>j</mi><mi>i</mi><mi>k</mi></mrow></msub><mo>=</mo><mfrac><mn>1</mn><mi>K</mi></mfrac><msub><mi>u</mi><mi>i</mi></msub><msub><mi>u</mi><mi>j</mi></msub><msub><mi>b</mi><mi>T</mi></msub><msub><mi>sin&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub></mrow></mtd></mtr><mtr><mtd><mrow><msub><mi>Q</mi><mrow><mi>j</mi><mi>i</mi><mi>k</mi></mrow></msub><mo>=</mo><mo>-</mo><msub><mi>b</mi><mi>T</mi></msub><msup><msub><mi>u</mi><mi>j</mi></msub><mn>2</mn></msup><mo>+</mo><mfrac><mn>1</mn><mi>K</mi></mfrac><msub><mi>u</mi><mi>i</mi></msub><msub><mi>u</mi><mi>j</mi></msub><msub><mi>b</mi><mi>T</mi></msub><msub><mi>cos&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub></mrow></mtd></mtr></mtable></mfenced><mo>-</mo><mo>-</mo><mo>-</mo><mrow><mo>(</mo><mn>7</mn><mo>)</mo>mo></mrow></mrow> 式中,Pijk、Qijk分别为变压器支路节点i侧的有功功率、无功功率,Pjik、Qjik分别为变压器支路节点j侧的有功功率、无功功率;ui为节点i的电压幅值,uj为节点j的电压幅值;θi为节点i的电压相角,θj为节点j的电压相角,且θij=θij,表示节点i和节点j的电压相角差;K为变压器非标准变比:j为标准侧,变比为1,i为非标准侧,变比为K;bT为变压器标准侧的电纳;In the formula, P ijk , Q ijk are the active power and reactive power of the transformer branch node i side respectively, P jik , Q jik are the active power and reactive power of the transformer branch node j side respectively; u i is the node i , u j is the voltage amplitude of node j; θ i is the voltage phase angle of node i, θ j is the voltage phase angle of node j, and θ ij = θ ij , which means node i and node The voltage phase angle difference of j; K is the non-standard transformation ratio of the transformer: j is the standard side, the transformation ratio is 1, i is the non-standard side, the transformation ratio is K; b T is the susceptance of the standard side of the transformer; 基于公式(8)得到节点电压对应的量测方程;Obtain the measurement equation corresponding to the node voltage based on formula (8); Ui=ui (8)U i =u i (8) 式中,Ui、ui均表示为节点i的电压幅值;In the formula, U i and u i both represent the voltage amplitude of node i; II)计算雅克比矩阵II) Calculate the Jacobian matrix 基于公式(9)-公式(19)形成雅克比矩阵H和C;其中H为量测量的雅克比矩阵,包括注入功率、线路支路功率、变压器支路功率以及节点电压幅值所形成的雅克比矩阵元素;C为零注入等式约束的雅克比矩阵,由零注入节点的注入功率所形成的雅克比矩阵元素组成;Based on formula (9)-formula (19), the Jacobian matrices H and C are formed; where H is the Jacobian matrix of quantity measurement, including the Jacobian matrix formed by injected power, line branch power, transformer branch power and node voltage amplitude Ratio matrix elements; C is the Jacobian matrix constrained by the zero injection equality, which is composed of the Jacobian matrix elements formed by the injection power of the zero injection node; 对于注入功率,基于公式(9)和(10)形成其雅克比矩阵元素;For the injected power, its Jacobian matrix elements are formed based on equations (9) and (10); <mrow> <mo>{</mo> <mtable> <mtr> <mtd> <mrow> <mfrac> <mrow> <mo>&amp;part;</mo> <msub> <mi>P</mi> <mi>i</mi> </msub> </mrow> <mrow> <mo>&amp;part;</mo> <msub> <mi>u</mi> <mi>i</mi> </msub> </mrow> </mfrac> <mo>=</mo> <mfrac> <mn>1</mn> <msub> <mi>u</mi> <mi>i</mi> </msub> </mfrac> <mrow> <mo>(</mo> <msub> <mi>G</mi> <mrow> <mi>i</mi> <mi>i</mi> </mrow> </msub> <msup> <msub> <mi>u</mi> <mi>i</mi> </msub> <mn>2</mn> </msup> <mo>+</mo> <msub> <mi>P</mi> <mi>i</mi> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mfrac> <mrow> <mo>&amp;part;</mo> <msub> <mi>P</mi> <mi>i</mi> </msub> </mrow> <mrow> <mo>&amp;part;</mo> <msub> <mi>&amp;theta;</mi> <mi>i</mi> </msub> </mrow> </mfrac> <mo>=</mo> <mo>-</mo> <msub> <mi>B</mi> <mrow> <mi>i</mi> <mi>i</mi> </mrow> </msub> <msup> <msub> <mi>u</mi> <mi>i</mi> </msub> <mn>2</mn> </msup> <mo>-</mo> <msub> <mi>Q</mi> <mi>i</mi> </msub> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mfrac> <mrow> <mo>&amp;part;</mo> <msub> <mi>P</mi> <mi>i</mi> </msub> </mrow> <mrow> <mo>&amp;part;</mo> <msub> <mi>u</mi> <mi>j</mi> </msub> </mrow> </mfrac> <mo>=</mo> <msub> <mi>u</mi> <mi>i</mi> </msub> <mrow> <mo>(</mo> <msub> <mi>G</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <msub> <mi>cos&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <mo>+</mo> <msub> <mi>B</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <msub> <mi>sin&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mfrac> <mrow> <mo>&amp;part;</mo> <msub> <mi>P</mi> <mi>i</mi> </msub> </mrow> <mrow> <mo>&amp;part;</mo> <msub> <mi>&amp;theta;</mi> <mi>j</mi> </msub> </mrow> </mfrac> <mo>=</mo> <msub> <mi>u</mi> <mi>i</mi> </msub> <msub> <mi>u</mi> <mi>j</mi> </msub> <mrow> <mo>(</mo> <msub> <mi>G</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <msub> <mi>sin&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <mo>-</mo> <msub> <mi>B</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <msub> <mi>cos&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> </mtable> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>9</mn> <mo>)</mo> </mrow> </mrow> <mrow><mo>{</mo><mtable><mtr><mtd><mrow><mfrac><mrow><mo>&amp;part;</mo><msub><mi>P</mi><mi>i</mi></msub></mrow><mrow><mo>&amp;part;</mo><msub><mi>u</mi><mi>i</mi></msub></mrow></mfrac><mo>=</mo><mfrac><mn>1</mn><msub><mi>u</mi><mi>i</mi></msub></mfrac><mrow><mo>(</mo><msub><mi>G</mi><mrow><mi>i</mi><mi>i</mi></mrow></msub><msup><msub><mi>u</mi><mi>i</mi></msub><mn>2</mn></msup><mo>+</mo><msub><mi>P</mi><mi>i</mi></msub><mo>)</mo></mrow></mrow></mtd></mtr><mtr><mtd><mrow><mfrac><mrow><mo>&amp;part;</mo><msub><mi>P</mi><mi>i</mi></msub></mrow><mrow><mo>&amp;part;</mo><msub><mi>&amp;theta;</mi><mi>i</mi></msub></mrow></mfrac><mo>=</mo><mo>-</mo><msub><mi>B</mi><mrow><mi>i</mi><mi>i</mi></mrow></msub><msup><msub><mi>u</mi><mi>i</mi></msub><mn>2</mn></msup><mo>-</mo><msub><mi>Q</mi><mi>i</mi></msub></mrow></mtd></mtr><mtr><mtd><mrow><mfrac><mrow><mo>&amp;part;</mo><msub><mi>P</mi><mi>i</mi></msub></mrow><mrow><mo>&amp;part;</mo><msub><mi>u</mi><mi>j</mi></msub></mrow></mfrac><mo>=</mo><msub><mi>u</mi><mi>i</mi></msub><mrow><mo>(</mo><msub><mi>G</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub><msub><mi>cos&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub><mo>+</mo><msub><mi>B</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub><msub><mi>sin&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub><mo>)</mo></mrow></mrow></mtd></mtr><mtr><mtd><mrow><mfrac><mrow><mo>&amp;part;</mo><msub><mi>P</mi><mi>i</mi></msub></mrow><mrow><mo>&amp;part;</mo><msub><mi>&amp;theta;</mi><mi>j</mi></msub></mrow></mfrac><mo>=</mo><msub><mi>u</mi><mi>i</mi></msub><msub><mi>u</mi><mi>j</mi></msub><mrow><mo>(</mo><msub><mi>G</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub><msub><mi>sin&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub><mo>-</mo><msub><mi>B</mi><mrow><mi>i</mi><mi>j</mi></mrow></mi>msub><msub><mi>cos&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub><mo>)</mo></mrow></mrow></mtd></mtr></mtable><mo>-</mo><mo>-</mo><mo>-</mo><mrow><mo>(</mo><mn>9</mn><mo>)</mo></mrow></mrow> <mrow> <mfenced open = "{" close = ""> <mtable> <mtr> <mtd> <mrow> <mfrac> <mrow> <mo>&amp;part;</mo> <msub> <mi>Q</mi> <mi>i</mi> </msub> </mrow> <mrow> <mo>&amp;part;</mo> <msub> <mi>u</mi> <mi>i</mi> </msub> </mrow> </mfrac> <mo>=</mo> <mfrac> <mn>1</mn> <msub> <mi>u</mi> <mi>i</mi> </msub> </mfrac> <mrow> <mo>(</mo> <mo>-</mo> <msub> <mi>B</mi> <mrow> <mi>i</mi> <mi>i</mi> </mrow> </msub> <msup> <msub> <mi>u</mi> <mi>i</mi> </msub> <mn>2</mn> </msup> <mo>+</mo> <msub> <mi>Q</mi> <mi>i</mi> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mfrac> <mrow> <mo>&amp;part;</mo> <msub> <mi>Q</mi> <mi>i</mi> </msub> </mrow> <mrow> <mo>&amp;part;</mo> <msub> <mi>&amp;theta;</mi> <mi>i</mi> </msub> </mrow> </mfrac> <mo>=</mo> <mo>-</mo> <msub> <mi>G</mi> <mrow> <mi>i</mi> <mi>i</mi> </mrow> </msub> <msup> <msub> <mi>u</mi> <mi>i</mi> </msub> <mn>2</mn> </msup> <mo>+</mo> <msub> <mi>P</mi> <mi>i</mi> </msub> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mfrac> <mrow> <mo>&amp;part;</mo> <msub> <mi>Q</mi> <mi>i</mi> </msub> </mrow> <mrow> <mo>&amp;part;</mo> <msub> <mi>u</mi> <mi>j</mi> </msub> </mrow> </mfrac> <mo>=</mo> <msub> <mi>u</mi> <mi>i</mi> </msub> <mrow> <mo>(</mo> <msub> <mi>G</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <msub> <mi>sin&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <mo>-</mo> <msub> <mi>B</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <msub> <mi>cos&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mfrac> <mrow> <mo>&amp;part;</mo> <msub> <mi>Q</mi> <mi>i</mi> </msub> </mrow> <mrow> <mo>&amp;part;</mo> <msub> <mi>&amp;theta;</mi> <mi>j</mi> </msub> </mrow> </mfrac> <mo>=</mo> <mo>-</mo> <msub> <mi>u</mi> <mi>i</mi> </msub> <msub> <mi>u</mi> <mi>j</mi> </msub> <mrow> <mo>(</mo> <msub> <mi>G</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <msub> <mi>cos&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <mo>+</mo> <msub> <mi>B</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <msub> <mi>sin&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> </mtable> </mfenced> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>10</mn> <mo>)</mo> </mrow> </mrow> <mrow><mfenced open = "{" close = ""><mtable><mtr><mtd><mrow><mfrac><mrow><mo>&amp;part;</mo><msub><mi>Q</mi><mi>i</mi></msub></mrow><mrow><mo>&amp;part;</mo><msub><mi>u</mi><mi>i</mi></msub></mrow></mfrac><mo>=</mo><mfrac><mn>1</mn><msub><mi>u</mi><mi>i</mi></msub></mfrac><mrow><mo>(</mo><mo>-</mo><msub><mi>B</mi><mrow><mi>i</mi><mi>i</mi></mrow></msub><msup><msub><mi>u</mi><mi>i</mi></msub><mn>2</mn></msup><mo>+</mo><msub><mi>Q</mi><mi>i</mi></msub><mo>)</mo></mrow></mrow></mtd></mtr><mtr><mtd><mrow><mfrac><mrow><mo>&amp;part;</mo><msub><mi>Q</mi><mi>i</mi></msub></mrow><mrow><mo>&amp;part;</mo><msub><mi>&amp;theta;</mi><mi>i</mi></msub></mrow></mfrac><mo>=</mo><mo>-</mo><msub><mi>G</mi><mrow><mi>i</mi><mi>i</mi></mrow></msub><msup><msub><mi>u</mi><mi>i</mi></msub><mn>2</mn></msup><mo>+</mo><msub><mi>P</mi><mi>i</mi></msub></mrow></mtd></mtr><mtr><mtd><mrow><mfrac><mrow><mo>&amp;part;</mo><msub><mi>Q</mi><mi>i</mi></msub></mrow><mrow><mo>&amp;part;</mo><msub><mi>u</mi><mi>j</mi></msub></mrow></mfrac><mo>=</mo><msub><mi>u</mi><mi>i</mi></msub><mrow><mo>(</mo><msub><mi>G</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub><msub><mi>sin&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub><mo>-</mo><msub><mi>B</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub><msub><mi>cos&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub><mo>)</mo></mrow></mrow></mtd></mtr><mtr><mtd><mrow><mfrac><mrow><mo>&amp;part;</mo><msub><mi>Q</mi><mi>i</mi></msub></mrow><mrow><mo>&amp;part;</mo><msub><mi>&amp;theta;</mi><mi>j</mi></msub></mrow></mfrac><mo>=</mo><mo>-</mo><msub><mi>u</mi><mi>i</mi></msub><msub><mi>u</mi><mi>j</mi></msub><mrow><mo>(</mo><msub><mi>G</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub><msub><mi>cos&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub><mo>+</mo><msub><mi>B</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub><msub><mi>sin&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub><mo>)</mo></mrow></mrow></mtd></mtr></mtable></mi>mfenced><mo>-</mo><mo>-</mo><mo>-</mo><mrow><mo>(</mo><mn>10</mn><mo>)</mo></mrow></mrow> 式中,Pi为节点i注入的有功功率;Qi为节点i注入的无功功率;ui为节点i的电压幅值,uj为节点j的电压幅值;θi为节点i的电压相角,θj为节点j的电压相角,且θij=θij,表示节点i和节点j的电压相角差;Gij、Bij分别为节点导纳矩阵中对应节点i和j之间元素的实部和虚部;Gii、Bii分别为节点导纳矩阵中对应节点i处主对角线上的元素的实部和虚部;In the formula, P i is the active power injected by node i; Q i is the reactive power injected by node i; u i is the voltage amplitude of node i, u j is the voltage amplitude of node j; θ i is the voltage amplitude of node i Voltage phase angle, θ j is the voltage phase angle of node j, and θ ij = θ ij , which means the voltage phase angle difference between node i and node j; G ij and B ij are the corresponding nodes in the node admittance matrix The real part and imaginary part of the element between i and j; G ii and B ii are respectively the real part and imaginary part of the element on the main diagonal at the corresponding node i in the node admittance matrix; 对于线路i侧支路功率,基于公式(11)和(12)形成其雅克比矩阵元素;For the branch power of the line i side, the Jacobian matrix elements are formed based on formulas (11) and (12); <mrow> <mfenced open = "{" close = ""> <mtable> <mtr> <mtd> <mrow> <mfrac> <mrow> <mo>&amp;part;</mo> <msub> <mi>P</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> </mrow> <mrow> <mo>&amp;part;</mo> <msub> <mi>u</mi> <mi>i</mi> </msub> </mrow> </mfrac> <mo>=</mo> <mn>2</mn> <msub> <mi>u</mi> <mi>i</mi> </msub> <mi>g</mi> <mo>-</mo> <msub> <mi>u</mi> <mi>j</mi> </msub> <mi>g</mi> <mi> </mi> <msub> <mi>cos&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <mo>-</mo> <msub> <mi>u</mi> <mi>j</mi> </msub> <mi>b</mi> <mi> </mi> <msub> <mi>sin&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mfrac> <mrow> <mo>&amp;part;</mo> <msub> <mi>P</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> </mrow> <mrow> <mo>&amp;part;</mo> <msub> <mi>&amp;theta;</mi> <mi>i</mi> </msub> </mrow> </mfrac> <mo>=</mo> <msub> <mi>u</mi> <mi>i</mi> </msub> <msub> <mi>u</mi> <mi>j</mi> </msub> <mrow> <mo>(</mo> <mi>g</mi> <mi> </mi> <msub> <mi>sin&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <mo>-</mo> <mi>b</mi> <mi> </mi> <msub> <mi>cos&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mfrac> <mrow> <mo>&amp;part;</mo> <msub> <mi>P</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> </mrow> <mrow> <mo>&amp;part;</mo> <msub> <mi>u</mi> <mi>j</mi> </msub> </mrow> </mfrac> <mo>=</mo> <mo>-</mo> <msub> <mi>u</mi> <mi>i</mi> </msub> <mrow> <mo>(</mo> <mi>g</mi> <mi> </mi> <msub> <mi>cos&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <mo>+</mo> <mi>b</mi> <mi> </mi> <msub> <mi>sin&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mfrac> <mrow> <mo>&amp;part;</mo> <msub> <mi>P</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> </mrow> <mrow> <mo>&amp;part;</mo> <msub> <mi>&amp;theta;</mi> <mi>j</mi> </msub> </mrow> </mfrac> <mo>=</mo> <mo>-</mo> <msub> <mi>u</mi> <mi>i</mi> </msub> <msub> <mi>u</mi> <mi>j</mi> </msub> <mrow> <mo>(</mo> <mi>g</mi> <mi> </mi> <msub> <mi>sin&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <mo>-</mo> <mi>b</mi> <mi> </mi> <msub> <mi>cos&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> </mtable> </mfenced> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>11</mn> <mo>)</mo> </mrow> </mrow> <mrow><mfenced open = "{" close = ""><mtable><mtr><mtd><mrow><mfrac><mrow><mo>&amp;part;</mo><msub><mi>P</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub></mrow><mrow><mo>&amp;part;</mo><msub><mi>u</mi><mi>i</mi></msub></mrow></mfrac><mo>=</mo><mn>2</mn><msub><mi>u</mi><mi>i</mi></msub><mi>g</mi><mo>-</mo><msub><mi>u</mi><mi>j</mi></msub><mi>g</mi><mi></mi><msub><mi>cos&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub><mo>-</mo><msub><mi>u</mi><mi>j</mi></msub><mi>b</mi><mi></mi><msub><mi>sin&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></mi>msub></mrow></mtd></mtr><mtr><mtd><mrow><mfrac><mrow><mo>&amp;part;</mo><msub><mi>P</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub></mrow><mrow><mo>&amp;part;</mo><msub><mi>&amp;theta;</mi><mi>i</mi></msub></mrow></mfrac><mo>=</mo><msub><mi>u</mi><mi>i</mi></msub><msub><mi>u</mi><mi>j</mi></msub><mrow><mo>(</mo><mi>g</mi><mi></mi><msub><mi>sin&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub><mo>-</mo><mi>b</mi><mi></mi><msub><mi>cos&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub><mo>)</mo></mrow></mrow></mtd></mtr><mtr><mtd><mrow><mfrac><mrow><mo>&amp;part;</mo><msub><mi>P</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub></mrow><mrow><mo>&amp;part;</mo><msub><mi>u</mi><mi>j</mi></msub></mrow></mfrac><mo>=</mo><mo>-</mo><msub><mi>u</mi><mi>i</mi></msub><mrow><mo>(</mo><mi>g</mi><mi></mi><msub><mi>cos&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub><mo>+</mo><mi>b</mi><mi></mi><msub><mi>sin&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub><mo>)</mo></mrow></mrow></mtd></mtr><mtr><mtd><mrow><mfrac><mrow><mo>&amp;part;</mo><msub><mi>P</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub></mrow><mrow><mo>&amp;part;</mo><msub><mi>&amp;theta;</mi><mi>j</mi></msub></mrow></mfrac><mo>=</mo><mo>-</mo><msub><mi>u</mi><mi>i</mi></msub><msub><mi>u</mi><mi>j</mi></msub><mrow><mo>(</mo><mi>g</mi><mi></mi><msub><mi>sin&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub><mo>-</mo><mi>b</mi><mi></mi><msub><mi>cos&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub><mo>)</mo></mrow></mrow></mtd></mtr></mtable></mfenced><mo>-</mo><mo>-</mo><mo>-</mo><mrow><mo>(</mo><mn>11</mn><mo>)</mo></mrow></mrow> <mrow> <mfenced open = "{" close = ""> <mtable> <mtr> <mtd> <mrow> <mfrac> <mrow> <mo>&amp;part;</mo> <msub> <mi>Q</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> </mrow> <mrow> <mo>&amp;part;</mo> <msub> <mi>u</mi> <mi>i</mi> </msub> </mrow> </mfrac> <mo>=</mo> <mo>-</mo> <mn>2</mn> <msub> <mi>u</mi> <mi>i</mi> </msub> <mrow> <mo>(</mo> <mi>b</mi> <mo>+</mo> <msub> <mi>y</mi> <mi>c</mi> </msub> <mo>)</mo> </mrow> <mo>-</mo> <msub> <mi>u</mi> <mi>j</mi> </msub> <mrow> <mo>(</mo> <mi>g</mi> <mi> </mi> <msub> <mi>sin&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <mo>-</mo> <mi>b</mi> <mi> </mi> <msub> <mi>cos&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mfrac> <mrow> <mo>&amp;part;</mo> <msub> <mi>Q</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> </mrow> <mrow> <mo>&amp;part;</mo> <msub> <mi>&amp;theta;</mi> <mi>i</mi> </msub> </mrow> </mfrac> <mo>=</mo> <mo>-</mo> <msub> <mi>u</mi> <mi>i</mi> </msub> <msub> <mi>u</mi> <mi>j</mi> </msub> <mrow> <mo>(</mo> <mi>g</mi> <mi> </mi> <msub> <mi>cos&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <mo>+</mo> <mi>b</mi> <mi> </mi> <msub> <mi>sin&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mfrac> <mrow> <mo>&amp;part;</mo> <msub> <mi>Q</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> </mrow> <mrow> <mo>&amp;part;</mo> <msub> <mi>u</mi> <mi>j</mi> </msub> </mrow> </mfrac> <mo>=</mo> <mo>-</mo> <msub> <mi>u</mi> <mi>i</mi> </msub> <mrow> <mo>(</mo> <mi>g</mi> <mi> </mi> <msub> <mi>sin&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <mo>-</mo> <mi>b</mi> <mi> </mi> <msub> <mi>cos&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mfrac> <mrow> <mo>&amp;part;</mo> <msub> <mi>Q</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> </mrow> <mrow> <mo>&amp;part;</mo> <msub> <mi>&amp;theta;</mi> <mi>j</mi> </msub> </mrow> </mfrac> <mo>=</mo> <msub> <mi>u</mi> <mi>i</mi> </msub> <msub> <mi>u</mi> <mi>j</mi> </msub> <mrow> <mo>(</mo> <mi>g</mi> <mi> </mi> <msub> <mi>cos&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <mo>+</mo> <mi>b</mi> <mi> </mi> <msub> <mi>sin&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> </mtable> </mfenced> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>12</mn> <mo>)</mo> </mrow> </mrow> <mrow><mfenced open = "{" close = ""><mtable><mtr><mtd><mrow><mfrac><mrow><mo>&amp;part;</mo><msub><mi>Q</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub></mrow><mrow><mo>&amp;part;</mo><msub><mi>u</mi><mi>i</mi></msub></mrow></mfrac><mo>=</mo><mo>-</mo><mn>2</mn><msub><mi>u</mi><mi>i</mi></msub><mrow><mo>(</mo><mi>b</mi><mo>+</mo><msub><mi>y</mi><mi>c</mi></msub><mo>)</mo></mrow><mo>-</mo><msub><mi>u</mi><mi>j</mi></msub><mrow><mo>(</mo><mi>g</mi><mi></mi><msub><mi>sin&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub><mo>-</mo><mi>b</mi><mi></mi><msub><mi>cos&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></mi>msub><mo>)</mo></mrow></mrow></mtd></mtr><mtr><mtd><mrow><mfrac><mrow><mo>&amp;part;</mo><msub><mi>Q</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub></mrow><mrow><mo>&amp;part;</mo><msub><mi>&amp;theta;</mi><mi>i</mi></msub></mrow></mfrac><mo>=</mo><mo>-</mo><msub><mi>u</mi><mi>i</mi></msub><msub><mi>u</mi><mi>j</mi></msub><mrow><mo>(</mo><mi>g</mi><mi></mi><msub><mi>cos&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub><mo>+</mi>mo><mi>b</mi><mi></mi><msub><mi>sin&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub><mo>)</mo></mrow></mrow></mtd></mtr><mtr><mtd><mrow><mfrac><mrow><mo>&amp;part;</mo><msub><mi>Q</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub></mrow><mrow><mo>&amp;part;</mo><msub><mi>u</mi><mi>j</mi></msub></mrow></mfrac><mo>=</mo><mo>-</mo><msub><mi>u</mi><mi>i</mi></msub><mrow><mo>(</mo><mi>g</mi><mi></mi><msub><mi>sin&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub><mo>-</mo><mi>b</mi><mi></mi><msub><mi>cos&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub><mo>)</mo></mrow></mrow></mtd></mtr><mtr><mtd><mrow><mfrac><mrow><mo>&amp;part;</mo><msub><mi>Q</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub></mrow><mrow><mo>&amp;part;</mo><msub><mi>&amp;theta;</mi><mi>j</mi></msub></mrow></mfrac><mo>=</mo><msub><mi>u</mi><mi>i</mi></msub><msub><mi>u</mi><mi>j</mi></msub><mrow><mo>(</mo><mi>g</mi><mi></mi><msub><mi>cos&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub><mo>+</mo><mi>b</mi><mi></mi><msub><mi>sin&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub><mo>)</mo></mrow></mrow></mtd></mtr></mtable></mfenced><mo>-</mo><mo>-</mo><mo>-</mo><mrow><mo>(</mo><mn>12</mn><mo>)</mo></mrow></mrow> 式中,Pij、Qij分别为线路支路节点i侧的有功功率、无功功率;ui为节点i的电压幅值,uj为节点j的电压幅值;θi为节点i的电压相角,θj为节点j的电压相角,且θij=θij,表示节点i和节点j的电压相角差;g为线路电导,b为线路电纳,yc为线路对地电纳;In the formula, P ij and Q ij are the active power and reactive power of node i side of the line branch respectively; u i is the voltage amplitude of node i, u j is the voltage amplitude of node j; θ i is the voltage amplitude of node i Voltage phase angle, θ j is the voltage phase angle of node j, and θ ij = θ ij , which means the voltage phase angle difference between node i and node j; g is the line conductance, b is the line susceptance, y c is line-to-ground susceptance; 对于线路j侧支路功率,基于公式(13)和(14)形成其雅克比矩阵元素;For the branch power on the side of line j, its Jacobian matrix elements are formed based on formulas (13) and (14); <mrow> <mfenced open = "{" close = ""> <mtable> <mtr> <mtd> <mrow> <mfrac> <mrow> <mo>&amp;part;</mo> <msub> <mi>P</mi> <mrow> <mi>j</mi> <mi>i</mi> </mrow> </msub> </mrow> <mrow> <mo>&amp;part;</mo> <msub> <mi>u</mi> <mi>i</mi> </msub> </mrow> </mfrac> <mo>=</mo> <msub> <mi>u</mi> <mi>j</mi> </msub> <mrow> <mo>(</mo> <mo>-</mo> <mi>g</mi> <mi> </mi> <msub> <mi>cos&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <mo>+</mo> <mi>b</mi> <mi> </mi> <msub> <mi>sin&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mfrac> <mrow> <mo>&amp;part;</mo> <msub> <mi>P</mi> <mrow> <mi>j</mi> <mi>i</mi> </mrow> </msub> </mrow> <mrow> <mo>&amp;part;</mo> <msub> <mi>&amp;theta;</mi> <mi>i</mi> </msub> </mrow> </mfrac> <mo>=</mo> <msub> <mi>u</mi> <mi>i</mi> </msub> <msub> <mi>u</mi> <mi>j</mi> </msub> <mrow> <mo>(</mo> <mi>g</mi> <mi> </mi> <msub> <mi>sin&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <mo>+</mo> <mi>b</mi> <mi> </mi> <msub> <mi>cos&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mfrac> <mrow> <mo>&amp;part;</mo> <msub> <mi>P</mi> <mrow> <mi>j</mi> <mi>i</mi> </mrow> </msub> </mrow> <mrow> <mo>&amp;part;</mo> <msub> <mi>u</mi> <mi>j</mi> </msub> </mrow> </mfrac> <mo>=</mo> <mn>2</mn> <msub> <mi>u</mi> <mi>j</mi> </msub> <mi>g</mi> <mo>+</mo> <msub> <mi>u</mi> <mi>i</mi> </msub> <mrow> <mo>(</mo> <mo>-</mo> <mi>g</mi> <mi> </mi> <msub> <mi>cos&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <mo>+</mo> <mi>b</mi> <mi> </mi> <msub> <mi>sin&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mfrac> <mrow> <mo>&amp;part;</mo> <msub> <mi>P</mi> <mrow> <mi>j</mi> <mi>i</mi> </mrow> </msub> </mrow> <mrow> <mo>&amp;part;</mo> <msub> <mi>&amp;theta;</mi> <mi>j</mi> </msub> </mrow> </mfrac> <mo>=</mo> <mo>-</mo> <msub> <mi>u</mi> <mi>i</mi> </msub> <msub> <mi>u</mi> <mi>j</mi> </msub> <mrow> <mo>(</mo> <mi>g</mi> <mi> </mi> <msub> <mi>sin&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <mo>+</mo> <mi>b</mi> <mi> </mi> <msub> <mi>cos&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> </mtable> </mfenced> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>13</mn> <mo>)</mo> </mrow> </mrow> <mrow><mfenced open = "{" close = ""><mtable><mtr><mtd><mrow><mfrac><mrow><mo>&amp;part;</mo><msub><mi>P</mi><mrow><mi>j</mi><mi>i</mi></mrow></msub></mrow><mrow><mo>&amp;part;</mo><msub><mi>u</mi><mi>i</mi></msub></mrow></mfrac><mo>=</mo><msub><mi>u</mi><mi>j</mi></msub><mrow><mo>(</mo><mo>-</mo><mi>g</mi><mi></mi><msub><mi>cos&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub><mo>+</mo><mi>b</mi><mi></mi><msub><mi>sin&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub><mo>)</mo></mrow></mrow></mtd></mtr><mtr><mtd><mrow><mfrac><mrow><mo>&amp;part;</mo><msub><mi>P</mi><mrow><mi>j</mi><mi>i</mi></mrow></msub></mrow><mrow><mo>&amp;part;</mo><msub><mi>&amp;theta;</mi><mi>i</mi></msub></mrow></mfrac><mo>=</mo><msub><mi>u</mi><mi>i</mi></msub><msub><mi>u</mi><mi>j</mi></msub><mrow><mo>(</mo><mi>g</mi><mi></mi><msub><mi>sin&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub><mo>+</mo><mi>b</mi><mi></mi><msub><mi>cos&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub><mo>)</mo></mrow></mrow></mtd></mtr><mtr><mtd><mrow><mfrac><mrow><mo>&amp;part;</mo><msub><mi>P</mi><mrow><mi>j</mi><mi>i</mi></mrow></msub></mrow><mrow><mo>&amp;part;</mo><msub><mi>u</mi><mi>j</mi></msub></mrow></mfrac><mo>=</mo><mn>2</mn><msub><mi>u</mi><mi>j</mi></msub><mi>g</mi><mo>+</mo><msub><mi>u</mi><mi>i</mi></msub><mrow><mo>(</mo><mo>-</mo><mi>g</mi><mi></mi><msub><mi>cos&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub><mo>+</mo><mi>b</mi><mi></mi><msub><mi>sin&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub><mo>)</mo></mrow></mrow></mtd></mtr><mtr><mtd><mrow><mfrac><mrow><mo>&amp;part;</mo><msub><mi>P</mi><mrow><mi>j</mi><mi>i</mi></mrow></msub></mrow><mrow><mo>&amp;part;</mo><msub><mi>&amp;theta;</mi><mi>j</mi></msub></mrow></mfrac><mo>=</mo><mo>-</mo><msub><mi>u</mi><mi>i</mi></msub><msub><mi>u</mi><mi>j</mi></msub><mrow><mo>(</mo><mi>g</mi><mi></mi><msub><mi>sin&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub><mo>+</mo><mi>b</mi><mi></mi><msub><mi>cos&amp;theta;</mi><mrow><mi>i</mi>mi><mi>j</mi></mrow></msub><mo>)</mo></mrow></mrow></mtd></mtr></mtable></mfenced><mo>-</mo><mo>-</mo><mo>-</mo><mrow><mo>(</mo><mn>13</mn><mo>)</mo></mrow></mrow> <mrow> <mfenced open = "{" close = ""> <mtable> <mtr> <mtd> <mrow> <mfrac> <mrow> <mo>&amp;part;</mo> <msub> <mi>Q</mi> <mrow> <mi>j</mi> <mi>i</mi> </mrow> </msub> </mrow> <mrow> <mo>&amp;part;</mo> <msub> <mi>u</mi> <mi>i</mi> </msub> </mrow> </mfrac> <mo>=</mo> <msub> <mi>u</mi> <mi>j</mi> </msub> <mrow> <mo>(</mo> <mi>g</mi> <mi> </mi> <msub> <mi>sin&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <mo>+</mo> <mi>b</mi> <mi> </mi> <msub> <mi>cos&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mfrac> <mrow> <mo>&amp;part;</mo> <msub> <mi>Q</mi> <mrow> <mi>j</mi> <mi>i</mi> </mrow> </msub> </mrow> <mrow> <mo>&amp;part;</mo> <msub> <mi>&amp;theta;</mi> <mi>i</mi> </msub> </mrow> </mfrac> <mo>=</mo> <msub> <mi>u</mi> <mi>i</mi> </msub> <msub> <mi>u</mi> <mi>j</mi> </msub> <mrow> <mo>(</mo> <mi>g</mi> <mi> </mi> <msub> <mi>cos&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <mo>-</mo> <mi>b</mi> <mi> </mi> <msub> <mi>sin&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mfrac> <mrow> <mo>&amp;part;</mo> <msub> <mi>Q</mi> <mrow> <mi>j</mi> <mi>i</mi> </mrow> </msub> </mrow> <mrow> <mo>&amp;part;</mo> <msub> <mi>u</mi> <mi>j</mi> </msub> </mrow> </mfrac> <mo>=</mo> <mo>-</mo> <mn>2</mn> <msub> <mi>u</mi> <mi>j</mi> </msub> <mrow> <mo>(</mo> <mi>b</mi> <mo>+</mo> <msub> <mi>y</mi> <mi>c</mi> </msub> <mo>)</mo> </mrow> <mo>+</mo> <msub> <mi>u</mi> <mi>i</mi> </msub> <mrow> <mo>(</mo> <mi>g</mi> <mi> </mi> <msub> <mi>sin&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <mo>+</mo> <mi>b</mi> <mi> </mi> <msub> <mi>cos&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mfrac> <mrow> <mo>&amp;part;</mo> <msub> <mi>Q</mi> <mrow> <mi>j</mi> <mi>i</mi> </mrow> </msub> </mrow> <mrow> <mo>&amp;part;</mo> <msub> <mi>&amp;theta;</mi> <mi>j</mi> </msub> </mrow> </mfrac> <mo>=</mo> <mo>-</mo> <msub> <mi>u</mi> <mi>i</mi> </msub> <msub> <mi>u</mi> <mi>j</mi> </msub> <mrow> <mo>(</mo> <mi>g</mi> <mi> </mi> <msub> <mi>cos&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <mo>-</mo> <mi>b</mi> <mi> </mi> <msub> <mi>sin&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> </mtable> </mfenced> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>14</mn> <mo>)</mo> </mrow> </mrow> <mrow><mfenced open = "{" close = ""><mtable><mtr><mtd><mrow><mfrac><mrow><mo>&amp;part;</mo><msub><mi>Q</mi><mrow><mi>j</mi><mi>i</mi></mrow></msub></mrow><mrow><mo>&amp;part;</mo><msub><mi>u</mi><mi>i</mi></msub></mrow></mfrac><mo>=</mo><msub><mi>u</mi><mi>j</mi></msub><mrow><mo>(</mo><mi>g</mi><mi></mi><msub><mi>sin&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub><mo>+</mo><mi>b</mi><mi></mi><msub><mi>cos&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub><mo>)</mo></mrow></mrow></mtd></mtr><mtr><mtd><mrow><mfrac><mrow><mo>&amp;part;</mo><msub><mi>Q</mi><mrow><mi>j</mi><mi>i</mi></mrow></msub></mrow><mrow><mo>&amp;part;</mo><msub><mi>&amp;theta;</mi><mi>i</mi></msub></mrow></mfrac><mo>=</mo><msub><mi>u</mi><mi>i</mi></msub><msub><mi>u</mi><mi>j</mi></msub><mrow><mo>(</mo><mi>g</mi><mi></mi><msub><mi>cos&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub><mo>-</mo><mi>b</mi><mi></mi><msub><mi>sin&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub><mo>)</mo></mrow></mrow></mtd></mtr><mtr><mtd><mrow><mfrac><mrow><mo>&amp;part;</mo><msub><mi>Q</mi><mrow><mi>j</mi><mi>i</mi></mrow></msub></mrow><mrow><mo>&amp;part;</mo><msub><mi>u</mi><mi>j</mi></msub></mrow></mfrac><mo>=</mo><mo>-</mo><mn>2</mn><msub><mi>u</mi><mi>j</mi></msub><mrow><mo>(</mo><mi>b</mi><mo>+</mo><msub><mi>y</mi><mi>c</mi></msub><mo>)</mo></mrow><mo>+</mo><msub><mi>u</mi><mi>i</mi></msub><mrow><mo>(</mo><mi>g</mi><mi></mi><msub><mi>sin&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub><mo>+</mo><mi>b</mi><mi></mi><msub><mi>cos&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub><mo>)</mo></mrow></mrow></mtd></mtr><mtr><mtd><mrow><mfrac><mrow><mo>&amp;part;</mo><msub><mi>Q</mi><mrow><mi>j</mi><mi>i</mi></mrow></msub></mrow><mrow><mo>&amp;part;</mo><msub><mi>&amp;theta;</mi><mi>j</mi></msub></mrow></mfrac><mo>=</mo><mo>-</mo><msub><mi>u</mi><mi>i</mi></msub><msub><mi>u</mi><mi>j</mi></msub><mrow><mo>(</mo><mi>g</mi><mi></mi><msub><mi>cos&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub><mo>-</mo><mi>b</mi><mi></mi><msub><mi>sin&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub><mo>)</mo></mrow></mrow></mtd></mtr></mtable></mfenced><mo>-</mo><mo>-</mo><mo>-</mo><mrow><mo>(</mo><mn>14</mn><mo>)</mo></mrow></mrow> 式中,Pji、Qji分别为线路支路节点j侧的有功功率、无功功率;ui为节点i的电压幅值,uj为节点j的电压幅值;θi为节点i的电压相角,θj为节点j的电压相角,且θij=θij,表示节点i和节点j的电压相角差;g为线路电导,b为线路电纳,yc为线路对地电纳;In the formula, P ji and Q ji are the active power and reactive power of node j side of the line branch respectively; u i is the voltage amplitude of node i, u j is the voltage amplitude of node j; θ i is the voltage amplitude of node i Voltage phase angle, θ j is the voltage phase angle of node j, and θ ij = θ ij , which means the voltage phase angle difference between node i and node j; g is the line conductance, b is the line susceptance, y c is line-to-ground susceptance; 对于变压器i侧支路功率,基于公式(15)和(16)形成其雅克比矩阵元素;For the branch power of the transformer i side, the Jacobian matrix elements are formed based on formulas (15) and (16); <mrow> <mfenced open = "{" close = ""> <mtable> <mtr> <mtd> <mrow> <mfrac> <mrow> <mo>&amp;part;</mo> <msub> <mi>P</mi> <mrow> <mi>i</mi> <mi>j</mi> <mi>k</mi> </mrow> </msub> </mrow> <mrow> <mo>&amp;part;</mo> <msub> <mi>u</mi> <mi>i</mi> </msub> </mrow> </mfrac> <mo>=</mo> <mo>-</mo> <mfrac> <mn>1</mn> <mi>K</mi> </mfrac> <msub> <mi>u</mi> <mi>j</mi> </msub> <msub> <mi>b</mi> <mi>T</mi> </msub> <msub> <mi>sin&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mfrac> <mrow> <mo>&amp;part;</mo> <msub> <mi>P</mi> <mrow> <mi>i</mi> <mi>j</mi> <mi>k</mi> </mrow> </msub> </mrow> <mrow> <mo>&amp;part;</mo> <msub> <mi>&amp;theta;</mi> <mi>i</mi> </msub> </mrow> </mfrac> <mo>=</mo> <mo>-</mo> <mfrac> <mn>1</mn> <mi>K</mi> </mfrac> <msub> <mi>u</mi> <mi>i</mi> </msub> <msub> <mi>u</mi> <mi>j</mi> </msub> <msub> <mi>b</mi> <mi>T</mi> </msub> <msub> <mi>cos&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mfrac> <mrow> <mo>&amp;part;</mo> <msub> <mi>P</mi> <mrow> <mi>i</mi> <mi>j</mi> <mi>k</mi> </mrow> </msub> </mrow> <mrow> <mo>&amp;part;</mo> <msub> <mi>u</mi> <mi>j</mi> </msub> </mrow> </mfrac> <mo>=</mo> <mo>-</mo> <mfrac> <mn>1</mn> <mi>K</mi> </mfrac> <msub> <mi>u</mi> <mi>i</mi> </msub> <msub> <mi>b</mi> <mi>T</mi> </msub> <msub> <mi>sin&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mfrac> <mrow> <mo>&amp;part;</mo> <msub> <mi>P</mi> <mrow> <mi>i</mi> <mi>j</mi> <mi>k</mi> </mrow> </msub> </mrow> <mrow> <mo>&amp;part;</mo> <msub> <mi>&amp;theta;</mi> <mi>j</mi> </msub> </mrow> </mfrac> <mo>=</mo> <mfrac> <mn>1</mn> <mi>K</mi> </mfrac> <msub> <mi>u</mi> <mi>i</mi> </msub> <msub> <mi>u</mi> <mi>j</mi> </msub> <msub> <mi>b</mi> <mi>T</mi> </msub> <msub> <mi>cos&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> </mrow> </mtd> </mtr> </mtable> </mfenced> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>15</mn> <mo>)</mo> </mrow> </mrow> <mrow><mfenced open = "{" close = ""><mtable><mtr><mtd><mrow><mfrac><mrow><mo>&amp;part;</mo><msub><mi>P</mi><mrow><mi>i</mi><mi>j</mi><mi>k</mi></mrow></msub></mrow><mrow><mo>&amp;part;</mo><msub><mi>u</mi><mi>i</mi></msub></mrow></mfrac><mo>=</mo><mo>-</mo><mfrac><mn>1</mn><mi>K</mi></mfrac><msub><mi>u</mi><mi>j</mi></msub><msub><mi>b</mi><mi>T</mi></msub><msub><mi>sin&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub></mrow></mtd></mtr><mtr><mtd><mrow><mfrac><mrow><mo>&amp;part;</mo><msub><mi>P</mi><mrow><mi>i</mi><mi>j</mi><mi>k</mi></mrow></msub></mrow><mrow><mo>&amp;part;</mo><msub><mi>&amp;theta;</mi><mi>i</mi></msub></mrow></mfrac><mo>=</mo><mo>-</mo><mfrac><mn>1</mn><mi>K</mi></mfrac><msub><mi>u</mi><mi>i</mi></msub><msub><mi>u</mi><mi>j</mi></msub><msub><mi>b</mi><mi>T</mi></msub><msub><mi>cos&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub></mrow></mtd></mtr><mtr><mtd><mrow><mfrac><mrow><mo>&amp;part;</mo><msub><mi>P</mi><mrow><mi>i</mi><mi>j</mi><mi>k</mi></mrow></msub></mrow><mrow><mo>&amp;part;</mo><msub><mi>u</mi><mi>j</mi></msub></mrow></mfrac><mo>=</mo><mo>-</mo><mfrac><mn>1</mn><mi>K</mi></mfrac><msub><mi>u</mi><mi>i</mi></msub><msub><mi>b</mi><mi>T</mi></msub><msub><mi>sin&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub></mrow></mi>mtd></mtr><mtr><mtd><mrow><mfrac><mrow><mo>&amp;part;</mo><msub><mi>P</mi><mrow><mi>i</mi><mi>j</mi><mi>k</mi></mrow></msub></mrow><mrow><mo>&amp;part;</mo><msub><mi>&amp;theta;</mi><mi>j</mi></msub></mrow></mfrac><mo>=</mo><mfrac><mn>1</mn><mi>K</mi></mfrac><msub><mi>u</mi><mi>i</mi></msub><msub><mi>u</mi><mi>j</mi></msub><msub><mi>b</mi><mi>T</mi></msub><msub><mi>cos&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub></mrow></mtd></mtr></mtable></mfenced><mo>-</mo><mo>-</mo><mo>-</mo><mrow><mo>(</mo><mn>15</mn><mo>)</mo></mrow></mrow> <mrow> <mfenced open = "{" close = ""> <mtable> <mtr> <mtd> <mrow> <mfrac> <mrow> <mo>&amp;part;</mo> <msub> <mi>Q</mi> <mrow> <mi>i</mi> <mi>j</mi> <mi>k</mi> </mrow> </msub> </mrow> <mrow> <mo>&amp;part;</mo> <msub> <mi>u</mi> <mi>i</mi> </msub> </mrow> </mfrac> <mo>=</mo> <mo>-</mo> <mfrac> <mn>2</mn> <msup> <mi>K</mi> <mn>2</mn> </msup> </mfrac> <msub> <mi>u</mi> <mi>i</mi> </msub> <msub> <mi>b</mi> <mi>T</mi> </msub> <mo>+</mo> <mfrac> <mn>1</mn> <mi>K</mi> </mfrac> <msub> <mi>u</mi> <mi>j</mi> </msub> <msub> <mi>b</mi> <mi>T</mi> </msub> <msub> <mi>cos&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mfrac> <mrow> <mo>&amp;part;</mo> <msub> <mi>Q</mi> <mrow> <mi>i</mi> <mi>j</mi> <mi>k</mi> </mrow> </msub> </mrow> <mrow> <mo>&amp;part;</mo> <msub> <mi>&amp;theta;</mi> <mi>i</mi> </msub> </mrow> </mfrac> <mo>=</mo> <mo>-</mo> <mfrac> <mn>1</mn> <mi>K</mi> </mfrac> <msub> <mi>u</mi> <mi>i</mi> </msub> <msub> <mi>u</mi> <mi>j</mi> </msub> <msub> <mi>b</mi> <mi>T</mi> </msub> <msub> <mi>sin&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mfrac> <mrow> <mo>&amp;part;</mo> <msub> <mi>Q</mi> <mrow> <mi>i</mi> <mi>j</mi> <mi>k</mi> </mrow> </msub> </mrow> <mrow> <mo>&amp;part;</mo> <msub> <mi>u</mi> <mi>j</mi> </msub> </mrow> </mfrac> <mo>=</mo> <mfrac> <mn>1</mn> <mi>K</mi> </mfrac> <msub> <mi>u</mi> <mi>i</mi> </msub> <msub> <mi>b</mi> <mi>T</mi> </msub> <msub> <mi>cos&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mfrac> <mrow> <mo>&amp;part;</mo> <msub> <mi>Q</mi> <mrow> <mi>i</mi> <mi>j</mi> <mi>k</mi> </mrow> </msub> </mrow> <mrow> <mo>&amp;part;</mo> <mi>&amp;theta;</mi> </mrow> </mfrac> <mo>=</mo> <mfrac> <mn>1</mn> <mi>K</mi> </mfrac> <msub> <mi>u</mi> <mi>i</mi> </msub> <msub> <mi>u</mi> <mi>j</mi> </msub> <msub> <mi>b</mi> <mi>T</mi> </msub> <msub> <mi>sin&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> </mrow> </mtd> </mtr> </mtable> </mfenced> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>16</mn> <mo>)</mo> </mrow> </mrow> <mrow><mfenced open = "{" close = ""><mtable><mtr><mtd><mrow><mfrac><mrow><mo>&amp;part;</mo><msub><mi>Q</mi><mrow><mi>i</mi><mi>j</mi><mi>k</mi></mrow></msub></mrow><mrow><mo>&amp;part;</mo><msub><mi>u</mi><mi>i</mi></msub></mrow></mfrac><mo>=</mo><mo>-</mo><mfrac><mn>2</mn><msup><mi>K</mi><mn>2</mn></msup></mfrac><msub><mi>u</mi><mi>i</mi></msub><msub><mi>b</mi><mi>T</mi></msub><mo>+</mo><mfrac><mn>1</mn><mi>K</mi></mfrac><msub><mi>u</mi><mi>j</mi></msub><msub><mi>b</mi><mi>T</mi></msub><msub><mi>cos&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub></mrow></mtd></mtr><mtr><mtd><mrow><mfrac><mrow><mo>&amp;part;</mo><msub><mi>Q</mi><mrow><mi>i</mi><mi>j</mi><mi>k</mi></mrow></msub></mrow><mrow><mo>&amp;part;</mo><msub><mi>&amp;theta;</mi><mi>i</mi></msub></mrow></mfrac><mo>=</mo><mo>-</mo><mfrac><mn>1</mn><mi>K</mi></mfrac><msub><mi>u</mi><mi>i</mi></msub><msub><mi>u</mi><mi>j</mi></msub><msub><mi>b</mi><mi>T</mi></msub><msub><mi>sin&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub></mrow></mtd></mtr><mtr><mtd><mrow><mfrac><mrow><mo>&amp;part;</mo><msub><mi>Q</mi><mrow><mi>i</mi><mi>j</mi><mi>k</mi></mrow></msub></mrow><mrow><mo>&amp;part;</mo><msub><mi>u</mi><mi>j</mi></msub></mrow></mo>mfrac><mo>=</mo><mfrac><mn>1</mn><mi>K</mi></mfrac><msub><mi>u</mi><mi>i</mi></msub><msub><mi>b</mi><mi>T</mi></msub><msub><mi>cos&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub></mrow></mtd></mtr><mtr><mtd><mrow><mfrac><mrow><mo>&amp;part;</mo><msub><mi>Q</mi><mrow><mi>i</mi><mi>j</mi><mi>k</mi></mrow></msub></mrow><mrow><mo>&amp;part;</mo><mi>&amp;theta;</mi></mrow></mfrac><mo>=</mo><mfrac><mn>1</mn><mi>K</mi></mfrac><msub><mi>u</mi><mi>i</mi></msub><msub><mi>u</mi><mi>j</mi></msub><msub><mi>b</mi><mi>T</mi></msub><msub><mi>sin&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub></mrow></mtd></mtr></mtable></mfenced><mo>-</mo><mo>-</mo><mo>-</mo><mrow><mo>(</mo><mn>16</mn><mo>)</mo></mrow></mrow> 式中,Pijk、Qijk分别为变压器支路节点i侧的有功功率、无功功率;ui为节点i的电压幅值,uj为节点j的电压幅值;θi为节点i的电压相角,θj为节点j的电压相角,θij=θij,表示节点i和节点j的电压相角差;K为变压器非标准变比:j为标准侧,变比为1,i为非标准侧,变比为K;bT为变压器标准侧的电纳;In the formula, P ijk and Q ijk are the active power and reactive power of node i side of the transformer branch respectively; u i is the voltage amplitude of node i, u j is the voltage amplitude of node j; θ i is the voltage amplitude of node i Voltage phase angle, θ j is the voltage phase angle of node j, θ ij = θ i - θ j , indicating the voltage phase angle difference between node i and node j; K is the non-standard transformation ratio of the transformer: j is the standard side, and the transformation ratio is 1, i is the non-standard side, and the transformation ratio is K; b T is the susceptance of the standard side of the transformer; 对于变压器j侧支路功率,基于公式(17)和(18)形成其雅克比矩阵元素;For the power of the side branch of the transformer j, its Jacobian matrix elements are formed based on formulas (17) and (18); <mrow> <mfenced open = "{" close = ""> <mtable> <mtr> <mtd> <mrow> <mfrac> <mrow> <mo>&amp;part;</mo> <msub> <mi>P</mi> <mrow> <mi>j</mi> <mi>i</mi> <mi>k</mi> </mrow> </msub> </mrow> <mrow> <mo>&amp;part;</mo> <msub> <mi>u</mi> <mi>i</mi> </msub> </mrow> </mfrac> <mo>=</mo> <mfrac> <mn>1</mn> <mi>K</mi> </mfrac> <msub> <mi>u</mi> <mi>j</mi> </msub> <msub> <mi>b</mi> <mi>T</mi> </msub> <msub> <mi>sin&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mfrac> <mrow> <mo>&amp;part;</mo> <msub> <mi>P</mi> <mrow> <mi>j</mi> <mi>i</mi> <mi>k</mi> </mrow> </msub> </mrow> <mrow> <mo>&amp;part;</mo> <msub> <mi>&amp;theta;</mi> <mi>i</mi> </msub> </mrow> </mfrac> <mo>=</mo> <mfrac> <mn>1</mn> <mi>K</mi> </mfrac> <msub> <mi>u</mi> <mi>i</mi> </msub> <msub> <mi>u</mi> <mi>j</mi> </msub> <msub> <mi>b</mi> <mi>T</mi> </msub> <msub> <mi>cos&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mfrac> <mrow> <mo>&amp;part;</mo> <msub> <mi>P</mi> <mrow> <mi>j</mi> <mi>i</mi> <mi>k</mi> </mrow> </msub> </mrow> <mrow> <mo>&amp;part;</mo> <msub> <mi>u</mi> <mi>j</mi> </msub> </mrow> </mfrac> <mo>=</mo> <mfrac> <mn>1</mn> <mi>K</mi> </mfrac> <msub> <mi>u</mi> <mi>i</mi> </msub> <msub> <mi>b</mi> <mi>T</mi> </msub> <msub> <mi>sin&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mfrac> <mrow> <mo>&amp;part;</mo> <msub> <mi>P</mi> <mrow> <mi>i</mi> <mi>j</mi> <mi>k</mi> </mrow> </msub> </mrow> <mrow> <mo>&amp;part;</mo> <msub> <mi>&amp;theta;</mi> <mi>j</mi> </msub> </mrow> </mfrac> <mo>=</mo> <mo>-</mo> <mfrac> <mn>1</mn> <mi>K</mi> </mfrac> <msub> <mi>u</mi> <mi>i</mi> </msub> <msub> <mi>u</mi> <mi>j</mi> </msub> <msub> <mi>b</mi> <mi>T</mi> </msub> <msub> <mi>cos&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> </mrow> </mtd> </mtr> </mtable> </mfenced> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>17</mn> <mo>)</mo> </mrow> </mrow> <mrow><mfenced open = "{" close = ""><mtable><mtr><mtd><mrow><mfrac><mrow><mo>&amp;part;</mo><msub><mi>P</mi><mrow><mi>j</mi><mi>i</mi><mi>k</mi></mrow></msub></mrow><mrow><mo>&amp;part;</mo><msub><mi>u</mi><mi>i</mi></msub></mrow></mfrac><mo>=</mo><mfrac><mn>1</mn><mi>K</mi></mfrac><msub><mi>u</mi><mi>j</mi></msub><msub><mi>b</mi><mi>T</mi></msub><msub><mi>sin&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub></mrow></mtd></mtr><mtr><mtd><mrow><mfrac><mrow><mo>&amp;part;</mo><msub><mi>P</mi><mrow><mi>j</mi><mi>i</mi><mi>k</mi></mrow></msub></mrow><mrow><mo>&amp;part;</mo><msub><mi>&amp;theta;</mi><mi>i</mi></msub></mrow></mfrac><mo>=</mo><mfrac><mn>1</mn><mi>K</mi></mfrac><msub><mi>u</mi><mi>i</mi></msub><msub><mi>u</mi><mi>j</mi></msub><msub><mi>b</mi><mi>T</mi></msub><msub><mi>cos&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub></mrow></mtd></mtr><mtr><mtd><mrow><mfrac><mrow><mo>&amp;part;</mo><msub><mi>P</mi><mrow><mi>j</mi><mi>i</mi><mi>k</mi></mrow></msub></mrow><mrow><mo>&amp;part;</mo><msub><mi>u</mi><mi>j</mi></msub></mrow></mfrac><mo>=</mo><mfrac><mn>1</mn><mi>K</mi></mfrac><msub><mi>u</mi><mi>i</mi></msub><msub><mi>b</mi><mi>T</mi></msub><msub><mi>sin&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub></mrow></mtd></mtr><mtr><mtd><mrow><mfrac><mrow><mo>&amp;part;</mo><msub><mi>P</mi><mrow><mi>i</mi><mi>j</mi><mi>k</mi></mrow></msub></mrow><mrow><mo>&amp;part;</mo><msub><mi>&amp;theta;</mi><mi>j</mi></msub></mrow></mfrac><mo>=</mo><mo>-</mo><mfrac><mn>1</mn><mi>K</mi></mfrac><msub><mi>u</mi><mi>i</mi></msub><msub><mi>u</mi><mi>j</mi></msub><msub><mi>b</mi><mi>T</mi></msub><msub><mi>cos&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub></mrow></mtd></mtr></mtable></mfenced><mo>-</mo><mo>-</mo><mo>-</mo><mrow><mo>(</mo><mn>17</mn><mo>)</mo>mo></mrow></mrow> <mrow> <mfenced open = "{" close = ""> <mtable> <mtr> <mtd> <mrow> <mfrac> <mrow> <mo>&amp;part;</mo> <msub> <mi>Q</mi> <mrow> <mi>j</mi> <mi>i</mi> <mi>k</mi> </mrow> </msub> </mrow> <mrow> <mo>&amp;part;</mo> <msub> <mi>u</mi> <mi>i</mi> </msub> </mrow> </mfrac> <mo>=</mo> <mfrac> <mn>1</mn> <mi>K</mi> </mfrac> <msub> <mi>u</mi> <mi>j</mi> </msub> <msub> <mi>b</mi> <mi>T</mi> </msub> <msub> <mi>cos&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mfrac> <mrow> <mo>&amp;part;</mo> <msub> <mi>Q</mi> <mrow> <mi>j</mi> <mi>i</mi> <mi>k</mi> </mrow> </msub> </mrow> <mrow> <mo>&amp;part;</mo> <msub> <mi>&amp;theta;</mi> <mi>i</mi> </msub> </mrow> </mfrac> <mo>=</mo> <mo>-</mo> <mfrac> <mn>1</mn> <mi>K</mi> </mfrac> <msub> <mi>u</mi> <mi>i</mi> </msub> <msub> <mi>u</mi> <mi>j</mi> </msub> <msub> <mi>b</mi> <mi>T</mi> </msub> <msub> <mi>sin&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mfrac> <mrow> <mo>&amp;part;</mo> <msub> <mi>Q</mi> <mrow> <mi>j</mi> <mi>i</mi> <mi>k</mi> </mrow> </msub> </mrow> <mrow> <mo>&amp;part;</mo> <msub> <mi>u</mi> <mi>j</mi> </msub> </mrow> </mfrac> <mo>=</mo> <mo>-</mo> <mn>2</mn> <msub> <mi>b</mi> <mi>T</mi> </msub> <msub> <mi>u</mi> <mi>j</mi> </msub> <mo>+</mo> <mfrac> <mn>1</mn> <mi>K</mi> </mfrac> <msub> <mi>u</mi> <mi>i</mi> </msub> <msub> <mi>b</mi> <mi>T</mi> </msub> <msub> <mi>cos&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mfrac> <mrow> <mo>&amp;part;</mo> <msub> <mi>Q</mi> <mrow> <mi>j</mi> <mi>i</mi> <mi>k</mi> </mrow> </msub> </mrow> <mrow> <mo>&amp;part;</mo> <msub> <mi>&amp;theta;</mi> <mi>j</mi> </msub> </mrow> </mfrac> <mo>=</mo> <mfrac> <mn>1</mn> <mi>K</mi> </mfrac> <msub> <mi>u</mi> <mi>i</mi> </msub> <msub> <mi>u</mi> <mi>j</mi> </msub> <msub> <mi>b</mi> <mi>T</mi> </msub> <msub> <mi>sin&amp;theta;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> </mrow> </mtd> </mtr> </mtable> </mfenced> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>18</mn> <mo>)</mo> </mrow> </mrow> <mrow><mfenced open = "{" close = ""><mtable><mtr><mtd><mrow><mfrac><mrow><mo>&amp;part;</mo><msub><mi>Q</mi><mrow><mi>j</mi><mi>i</mi><mi>k</mi></mrow></msub></mrow><mrow><mo>&amp;part;</mo><msub><mi>u</mi><mi>i</mi></msub></mrow></mfrac><mo>=</mo><mfrac><mn>1</mn><mi>K</mi></mfrac><msub><mi>u</mi><mi>j</mi></msub><msub><mi>b</mi><mi>T</mi></msub><msub><mi>cos&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub></mrow></mtd></mtr><mtr><mtd><mrow><mfrac><mrow><mo>&amp;part;</mo><msub><mi>Q</mi><mrow><mi>j</mi><mi>i</mi><mi>k</mi></mrow></msub></mrow><mrow><mo>&amp;part;</mo><msub><mi>&amp;theta;</mi><mi>i</mi></msub></mrow></mfrac><mo>=</mo><mo>-</mo><mfrac><mn>1</mn><mi>K</mi></mfrac><msub><mi>u</mi><mi>i</mi></msub><msub><mi>u</mi><mi>j</mi></msub><msub><mi>b</mi><mi>T</mi></msub><msub><mi>sin&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub></mrow></mtd></mtr><mtr><mtd><mrow><mfrac><mrow><mo>&amp;part;</mo><msub><mi>Q</mi><mrow><mi>j</mi><mi>i</mi><mi>k</mi></mrow></msub></mrow><mrow><mo>&amp;part;</mo><msub><mi>u</mi><mi>j</mi></msub></mrow></mfrac><mo>=</mo><mo>-</mo><mn>2</mn><msub><mi>b</mi><mi>T</mi></msub><msub><mi>u</mi><mi>j</mi></msub><mo>+</mo><mfrac><mn>1</mn><mi>K</mi></mfrac><msub><mi>u</mi><mi>i</mi></msub><msub><mi>b</mi><mi>T</mi></msub><msub><mi>cos&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></mi>msub></mrow></mtd></mtr><mtr><mtd><mrow><mfrac><mrow><mo>&amp;part;</mo><msub><mi>Q</mi><mrow><mi>j</mi><mi>i</mi><mi>k</mi></mrow></msub></mrow><mrow><mo>&amp;part;</mo><msub><mi>&amp;theta;</mi><mi>j</mi></msub></mrow></mfrac><mo>=</mo><mfrac><mn>1</mn><mi>K</mi></mfrac><msub><mi>u</mi><mi>i</mi></msub><msub><mi>u</mi><mi>j</mi></msub><msub><mi>b</mi><mi>T</mi></msub><msub><mi>sin&amp;theta;</mi><mrow><mi>i</mi><mi>j</mi></mrow></msub></mrow></mtd></mtr></mtable></mfenced><mo>-</mo><mo>-</mo><mo>-</mo><mrow><mo>(</mo><mn>18</mn><mo>)</mo></mrow></mrow> 式中,Pjik、Qjik分别为变压器支路节点j侧的有功功率、无功功率;ui为节点i的电压幅值,uj为节点j的电压幅值;θi为节点i的电压相角,θj为节点j的电压相角,θij=θij,表示节点i和节点j的电压相角差;K为变压器非标准变比,j为标准侧,变比为1,i为非标准侧,变比为K;bT为变压器标准侧的电纳;In the formula, P jik and Q jik are the active power and reactive power of node j side of the transformer branch respectively; u i is the voltage amplitude of node i, u j is the voltage amplitude of node j; θ i is the voltage amplitude of node i Voltage phase angle, θ j is the voltage phase angle of node j, θ ij = θ i - θ j , indicating the voltage phase angle difference between node i and node j; K is the non-standard transformation ratio of the transformer, j is the standard side, and the transformation ratio is 1, i is the non-standard side, and the transformation ratio is K; b T is the susceptance of the standard side of the transformer; 对于节点电压,基于公式(19)形成其雅克比矩阵元素;For node voltages, form its Jacobian matrix elements based on formula (19); <mrow> <mfenced open = "{" close = ""> <mtable> <mtr> <mtd> <mrow> <mfrac> <mrow> <mo>&amp;part;</mo> <msub> <mi>U</mi> <mi>i</mi> </msub> </mrow> <mrow> <mo>&amp;part;</mo> <msub> <mi>u</mi> <mi>i</mi> </msub> </mrow> </mfrac> <mo>=</mo> <mn>1</mn> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mfrac> <mrow> <mo>&amp;part;</mo> <msub> <mi>U</mi> <mi>i</mi> </msub> </mrow> <mrow> <mo>&amp;part;</mo> <msub> <mi>&amp;theta;</mi> <mi>i</mi> </msub> </mrow> </mfrac> <mo>=</mo> <mn>0</mn> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mfrac> <mrow> <mo>&amp;part;</mo> <msub> <mi>U</mi> <mi>i</mi> </msub> </mrow> <mrow> <mo>&amp;part;</mo> <msub> <mi>u</mi> <mi>j</mi> </msub> </mrow> </mfrac> <mo>=</mo> <mn>0</mn> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mfrac> <mrow> <mo>&amp;part;</mo> <msub> <mi>U</mi> <mi>i</mi> </msub> </mrow> <mrow> <mo>&amp;part;</mo> <msub> <mi>&amp;theta;</mi> <mi>j</mi> </msub> </mrow> </mfrac> <mo>=</mo> <mn>0</mn> </mrow> </mtd> </mtr> </mtable> </mfenced> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>19</mn> <mo>)</mo> </mrow> </mrow> <mrow><mfenced open = "{" close = ""><mtable><mtr><mtd><mrow><mfrac><mrow><mo>&amp;part;</mo><msub><mi>U</mi><mi>i</mi></msub></mrow><mrow><mo>&amp;part;</mo><msub><mi>u</mi><mi>i</mi></msub></mrow></mfrac><mo>=</mo><mn>1</mn></mrow></mtd></mtr><mtr><mtd><mrow><mfrac><mrow><mo>&amp;part;</mo><msub><mi>U</mi><mi>i</mi></msub></mrow><mrow><mo>&amp;part;</mo><msub><mi>&amp;theta;</mi><mi>i</mi></msub></mrow></mfrac><mo>=</mo><mn>0</mn></mrow></mtd></mtr><mtr><mtd><mrow><mfrac><mrow><mo>&amp;part;</mo><msub><mi>U</mi><mi>i</mi></msub></mrow><mrow><mo>&amp;part;</mo><msub><mi>u</mi><mi>j</mi></msub></mrow></mfrac><mo>=</mo><mn>0</mn></mrow></mtd></mtr><mtr><mtd><mrow><mfrac><mrow><mo>&amp;part;</mo><msub><mi>U</mi><mi>i</mi></msub></mrow><mrow><mo>&amp;part;</mo><msub><mi>&amp;theta;</mi><mi>j</mi></msub></mrow></mfrac><mo>=</mo><mn>0</mn></mrow></mtd></mtr></mtable></mfenced><mo>-</mo><mo>-</mo><mo>-</mo><mrow><mo>(</mo><mn>19</mn><mo>)</mo></mrow></mrow> 式中,Ui、ui均为节点i的电压幅值,uj为节点j的电压幅值,θi为节点i的电压相角,θj为节点j的电压相角;In the formula, U i and u i are the voltage amplitude of node i, u j is the voltage amplitude of node j, θ i is the voltage phase angle of node i, and θ j is the voltage phase angle of node j; 零注入等式约束的雅克比矩阵为其中c(x)是零注入节点的量测方程,其表示节点的注入有功功率和无功功率值;x是n维的状态量,n是状态估计中实际的状态变量个数;The Jacobian matrix of the zero-injection equality constraint is Where c(x) is the measurement equation of the zero injection node, which represents the injected active power and reactive power value of the node; x is the n-dimensional state quantity, and n is the actual number of state variables in the state estimation; III)计算权函数III) Calculate the weight function 基于公式(20)计算权函数W的对角阵元素,计算公式为:Based on the formula (20) to calculate the diagonal matrix elements of the weight function W, the calculation formula is: <mrow> <msup> <msub> <mi>w</mi> <mi>i</mi> </msub> <mo>*</mo> </msup> <mo>=</mo> <mfenced open = "{" close = ""> <mtable> <mtr> <mtd> <mn>1</mn> </mtd> <mtd> <mrow> <mo>|</mo> <msub> <mi>r</mi> <mi>i</mi> </msub> <mo>|</mo> <mo>&amp;le;</mo> <msub> <mi>r</mi> <mi>min</mi> </msub> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <msup> <msub> <mi>R</mi> <mi>i</mi> </msub> <mrow> <mo>-</mo> <mn>1</mn> </mrow> </msup> <mfrac> <mrow> <mo>|</mo> <mn>1</mn> <mo>/</mo> <msub> <mi>r</mi> <mi>i</mi> </msub> <mo>|</mo> </mrow> <mrow> <mo>|</mo> <mn>1</mn> <mo>/</mo> <msub> <mi>r</mi> <mi>min</mi> </msub> <mo>|</mo> </mrow> </mfrac> <mo>=</mo> <msup> <msub> <mi>R</mi> <mi>i</mi> </msub> <mrow> <mo>-</mo> <mn>1</mn> </mrow> </msup> <mfrac> <msub> <mi>r</mi> <mi>min</mi> </msub> <mrow> <mo>|</mo> <msub> <mi>r</mi> <mi>i</mi> </msub> <mo>|</mo> </mrow> </mfrac> </mrow> </mtd> <mtd> <mrow> <mo>|</mo> <msub> <mi>r</mi> <mi>i</mi> </msub> <mo>|</mo> <mo>&gt;</mo> <msub> <mi>r</mi> <mi>min</mi> </msub> </mrow> </mtd> </mtr> </mtable> </mfenced> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>20</mn> <mo>)</mo> </mrow> </mrow> <mrow><msup><msub><mi>w</mi><mi>i</mi></msub><mo>*</mo></msup><mo>=</mo>< mfenced open = "{" close = ""><mtable><mtr><mtd><mn>1</mn></mtd><mtd><mrow><mo>|</mo><msub><mi>r</mi><mi>i</mi></msub><mo>|</mo><mo>&amp;le;</mo><msub><mi>r</mi><mi>min</mi></msub></mrow></mtd></mtr><mtr><mtd><mrow><msup><msub><mi>R</mi><mi>i</mi></msub><mrow><mo>-</mo><mn>1</mn></mrow></msup><mfrac><mrow><mo>|</mo><mn>1</mn><mo>/</mo><msub><mi>r</mi><mi>i</mi></msub><mo>|</mo></mrow><mrow><mo>|</mo><mn>1</mn><mo>/</mo><msub><mi>r</mi><mi>min</mi></msub><mo>|</mo></mrow></mfrac><mo>=</mo><msup><msub><mi>R</mi><mi>i</mi></msub><mrow><mo>-</mo><mn>1</mn></mrow></msup><mfrac><msub><mi>r</mi><mi>min</mi></msub><mrow><mo>|</mo><msub><mi>r</mi><mi>i</mi></msub><mo>|</mo></mrow></mfrac></mrow></mtd><mtd><mrow><mo>|</mo><msub><mi>r</mi><mi>i</mi></msub><mo>|</mo><mo>&gt;</mo><msub><mi>r</mi><mi>min</mi></msub></mrow></mtd></mtr></mtable></mfenced><mo>-</mo><mo>-</mo><mo>-</mo><mrow><mo>(</mo><mn>20</mn><mo>)</mo></mrow></mrow> 式中,rmin为小残差的检测门槛值,取0.002;wi *为量测i的权函数;为量测i的固定权重;ri为量测i的残差;In the formula, r min is the detection threshold of small residual error, which is 0.002; w i * is the weight function of measurement i; is the fixed weight of measurement i; r i is the residual of measurement i; (5)状态变量更新和收敛性判断(5) State variable update and convergence judgment i)状态变量更新i) State variable update 第(4)步完成后,根据公式(21)计算状态变量的修正量Δx(time),然后更新状态变量,得到状态变量新值x(time+1)=x(time)+Δx(time),time=time+1;After step (4) is completed, calculate the correction amount Δx (time) of the state variable according to the formula (21), and then update the state variable to obtain the new value of the state variable x (time+1) = x (time) + Δx (time) , time=time+1; <mrow> <mfenced open = "[" close = "]"> <mtable> <mtr> <mtd> <mrow> <msup> <mi>&amp;Delta;x</mi> <mrow> <mo>(</mo> <mi>t</mi> <mi>i</mi> <mi>m</mi> <mi>e</mi> <mo>)</mo> </mrow> </msup> </mrow> </mtd> </mtr> <mtr> <mtd> <msup> <mi>&amp;lambda;</mi> <mrow> <mo>(</mo> <mi>t</mi> <mi>i</mi> <mi>m</mi> <mi>e</mi> <mo>)</mo> </mrow> </msup> </mtd> </mtr> </mtable> </mfenced> <mo>=</mo> <msup> <mfenced open = "[" close = "]"> <mtable> <mtr> <mtd> <mrow> <msup> <mi>H</mi> <mi>T</mi> </msup> <mi>W</mi> <mi>H</mi> </mrow> </mtd> <mtd> <msup> <mi>C</mi> <mi>T</mi> </msup> </mtd> </mtr> <mtr> <mtd> <mi>C</mi> </mtd> <mtd> <mn>0</mn> </mtd> </mtr> </mtable> </mfenced> <mrow> <mo>-</mo> <mn>1</mn> </mrow> </msup> <mfenced open = "[" close = "]"> <mtable> <mtr> <mtd> <msup> <mi>H</mi> <mi>T</mi> </msup> <mi>W</mi> <mo>(</mo> <mi>z</mi> <mo>-</mo> <mi>h</mi> <mrow> <mo>(</mo> <msup> <mi>x</mi> <mrow> <mo>(</mo> <mi>t</mi> <mi>i</mi> <mi>m</mi> <mi>e</mi> <mo>)</mo> </mrow> </msup> <mo>)</mo> </mrow> <mo>)</mo> </mtd> </mtr> <mtr> <mtd> <mo>-</mo> <mi>c</mi> <mo>(</mo> <msup> <mi>x</mi> <mrow> <mo>(</mo> <mi>t</mi> <mi>i</mi> <mi>m</mi> <mi>e</mi> <mo>)</mo> </mrow> </msup> <mo>)</mo> </mtd> </mtr> </mtable> </mfenced> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>21</mn> <mo>)</mo> </mrow> </mrow> <mrow><mfenced open = "[" close = "]"><mtable><mtr><mtd><mrow><msup><mi>&amp;Delta;x</mi><mrow><mo>(</mo><mi>t</mi><mi>i</mi><mi>m</mi><mi>e</mi><mo>)</mo></mrow></mo>msup></mrow></mtd></mtr><mtr><mtd><msup><mi>&amp;lambda;</mi><mrow><mo>(</mo><mi>t</mi><mi>i</mi><mi>m</mi><mi>e</mi><mo>)</mo></mrow></msup></mtd></mtr></mtable></mfenced><mo>=</mo><msup><mfenced open = "[" close = "]"><mtable><mtr><mtd><mrow><msup><mi>H</mi><mi>T</mi></msup><mi>W</mi><mi>H</mi></mrow></mtd><mtd><msup><mi>C</mi><mi>T</mi></msup></mtd></mtr><mtr><mtd><mi>C</mi></mtd><mtd><mn>0</mn></mtd></mtr></mtable></mfenced><mrow><mo>-</mo><mn>1</mn></mrow></msup><mfenced open = "[" close = "]"><mtable><mtr><mtd><msup><mi>H</mi><mi>T</mi></msup><mi>W</mi><mo>(</mo><mi>z</mi><mo>-</mo><mi>h</mi><mrow><mo>(</mo><msup><mi>x</mi><mrow><mo>(</mo><mi>t</mi><mi>i</mi><mi>m</mi><mi>e</mi><mo>)</mo></mrow></msup><mo>)</mo></mrow><mo>)</mo></mtd></mtr><mtr><mtd><mo>-</mo><mi>c</mi><mo>(</mo><msup><mi>x</mi><mrow><mo>(</mo><mi>t</mi><mi>i</mi><mi>m</mi><mi>e</mi><mo>)</mo></mrow></msup><mo>)</mo></mtd></mtr></mtable></mfenced><mo>-</mo><mo>-</mo><mo>-</mo><mrow><mo>(</mo><mn>21</mn><mo>)</mo></mrow></mrow> 式中,time为计算迭代次数;x(time)为第time次迭代时的状态量;W为权函数对角阵,其对角元素等于权函数,即Wii=wi *为量测量的雅克比矩阵,HT为其转置;c(x(time))为迭代值是x(time)时的零注入等式约束,为零注入等式约束的雅克比矩阵,CT为其转置;z-h(x(time))表示迭代值为x(time)时的残差;λ(time)为第time次迭代时的拉格朗日乘子向量;In the formula, time is the number of calculation iterations; x (time) is the state quantity during the time iteration; W is the weight function diagonal matrix, and its diagonal elements are equal to the weight function, that is, W ii =w i * ; is the measured Jacobian matrix, H T is its transpose; c(x (time) ) is the zero-injection equality constraint when the iteration value is x (time) , is the Jacobian matrix of equality constraints injected into zero, C T is its transpose; zh(x (time) ) indicates the residual error when the iteration value is x (time) ; Grangian multiplier vector; ii)收敛性判断ii) Judgment of convergence 当状态变量的修正量Δx(time)满足max(|Δx(time)|)<ε,则结束迭代计算,输出结果;当max(|Δx(time)|)≥ε且迭代次数time≥Tmax,则停止迭代,输出“不收敛!”;When the correction amount Δx (time) of the state variable satisfies max(|Δx (time) |)<ε, the iterative calculation is ended and the result is output; when max(|Δx (time) |)≥ε and the number of iterations time≥Tmax, Then stop the iteration and output "no convergence!"; 当max(|Δx(time)|)≥ε且迭代次数time<Tmax,使迭代次数time增加1,返回第(3)步,进行重新迭代计算。When max(|Δx (time) |)≥ε and the number of iterations time<Tmax, increase the number of iterations time by 1, and return to step (3) for re-iteration calculation.
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Families Citing this family (8)

* Cited by examiner, † Cited by third party
Publication number Priority date Publication date Assignee Title
CN106208055B (en) * 2016-07-18 2019-03-29 西南交通大学 A kind of harmonic source identification method based on binary system population optimization and compressed sensing
CN107016489A (en) * 2017-03-09 2017-08-04 中国电力科学研究院 A kind of electric power system robust state estimation method and device
CN108460547A (en) * 2018-05-16 2018-08-28 山东大学 Cogeneration system combined heat and power method for estimating state and system
EP3579366A1 (en) * 2018-06-08 2019-12-11 Siemens Aktiengesellschaft Method and apparatus for control of an electric grid
CN109255171B (en) * 2018-08-29 2023-09-05 深圳十沣科技有限公司 Method for automatically judging convergence of numerical simulation calculation
CN110943445A (en) * 2018-09-21 2020-03-31 中国电力科学研究院有限公司 A method and device for estimating robustness state of flexible, direct and AC hybrid power grid
CN110297887B (en) * 2019-06-26 2021-07-27 山东大学 Personalized dialogue system and method for service robot based on cloud platform
CN112383048B (en) * 2020-10-20 2024-07-09 重庆大学 Medium-voltage feeder line power grid three-phase state estimation method considering measurement characteristics of public special distribution transformer

Citations (10)

* Cited by examiner, † Cited by third party
Publication number Priority date Publication date Assignee Title
CN102708187A (en) * 2012-05-14 2012-10-03 成都信息工程学院 Reverse index mixed compression and decompression method based on Hbase database
CN102831315A (en) * 2012-08-23 2012-12-19 清华大学 Accurate linearization method of measurement equation for electric power system state estimation
CN103344844A (en) * 2013-05-31 2013-10-09 国家电网公司 Grid security warning terminal and warning method thereof
CN103593566A (en) * 2013-11-14 2014-02-19 华北电力大学 Mixing-quadratic-programming-form electrical power system synthesis state estimation method
CN104102836A (en) * 2014-07-14 2014-10-15 国家电网公司 Method for quickly estimating robust state of power system
CN104166060A (en) * 2014-08-15 2014-11-26 国家电网公司 Robustness state estimation method with large-scale wind power connection considered
CN104184144A (en) * 2014-09-05 2014-12-03 国家电网公司 Robust state estimation method used for multi-voltage-class power grid model
CN104467601A (en) * 2014-12-19 2015-03-25 东南大学 Synchronous generator parameter identification and control method based on WAMS
CN104821577A (en) * 2015-04-18 2015-08-05 安庆师范学院 Three-phase four-wire system distribution network robust estimation method based on intelligent electric meter measurement
CN104899435A (en) * 2015-05-25 2015-09-09 清华大学 Power system dynamic state estimation method considering zero-injection constraint

Patent Citations (10)

* Cited by examiner, † Cited by third party
Publication number Priority date Publication date Assignee Title
CN102708187A (en) * 2012-05-14 2012-10-03 成都信息工程学院 Reverse index mixed compression and decompression method based on Hbase database
CN102831315A (en) * 2012-08-23 2012-12-19 清华大学 Accurate linearization method of measurement equation for electric power system state estimation
CN103344844A (en) * 2013-05-31 2013-10-09 国家电网公司 Grid security warning terminal and warning method thereof
CN103593566A (en) * 2013-11-14 2014-02-19 华北电力大学 Mixing-quadratic-programming-form electrical power system synthesis state estimation method
CN104102836A (en) * 2014-07-14 2014-10-15 国家电网公司 Method for quickly estimating robust state of power system
CN104166060A (en) * 2014-08-15 2014-11-26 国家电网公司 Robustness state estimation method with large-scale wind power connection considered
CN104184144A (en) * 2014-09-05 2014-12-03 国家电网公司 Robust state estimation method used for multi-voltage-class power grid model
CN104467601A (en) * 2014-12-19 2015-03-25 东南大学 Synchronous generator parameter identification and control method based on WAMS
CN104821577A (en) * 2015-04-18 2015-08-05 安庆师范学院 Three-phase four-wire system distribution network robust estimation method based on intelligent electric meter measurement
CN104899435A (en) * 2015-05-25 2015-09-09 清华大学 Power system dynamic state estimation method considering zero-injection constraint

Non-Patent Citations (3)

* Cited by examiner, † Cited by third party
Title
基于指数权函数的抗差状态估计算法;蔡凝露等;《中国电力》;20130430;第46卷(第4期);全文 *
基于权函数的电网参数分区辨识与估计方法;颜伟等;《电力系统自动化》;20110310;第35卷(第5期);全文 *
考虑大规模风电接入的快速抗差状态估计研究;李静等;《电力系统保护与控制》;20141116;第42卷(第22期);全文 *

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