\documentclass[10pt,a4paper,final]{report} \usepackage[utf8]{inputenc} \usepackage{amsmath} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{fontspec}%使用xetex \usepackage[top=1in, bottom=1in, left=1.0in, right=1.0in]{geometry} \setmainfont[BoldFont=黑体]{宋体} % 使用系统默认字体 \XeTeXlinebreaklocale "zh" % 针对中文进行断行 \XeTeXlinebreakskip = 0pt plus 1pt minus 0.1pt % 给予TeX断行一定自由度 \linespread{1.5} % 1.5倍行距 \begin{document} 线路功率（不考虑接地导纳） \begin{equation} \begin{aligned} \dot{I}_{12}&=(V_1e^{j \theta_1} - V_2e^{j \theta_2}) (G_{ij}+jB_{ij}) \end{aligned} \end{equation} 共轭后 \begin{equation} \begin{aligned} \dot{I}^*_{12}&=(V_1e^{-j \theta_1} - V_2e^{-j \theta_2}) (G_{ij}-jB_{ij}) \end{aligned} \end{equation} \begin{equation} \begin{aligned} \dot{S}_{12}&=V_1e^{j \theta_1} \dot{I}^*_{12} \\ &=V_1e^{j \theta_1} (V_1e^{-j \theta_1} - V_2e^{-j \theta_2}) (G_{ij}-jB_{ij}) \\ &=(V_1e^{j \theta_1}V_1e^{-j \theta_1} -V_1e^{j \theta_1}V_2e^{-j \theta_2} ) (G_{ij}-jB_{ij}) \\ &=[V_1^2-V_1V_2e^{j (\theta_1 - \theta_2) } ] (G_{ij}-jB_{ij}) \\ &=\{V_1^2-V_1V_2[cos(\theta_1 - \theta_2)+jsin (\theta_1 - \theta_2) ]\} (G_{ij}-jB_{ij}) \\ &=[V_1^2-V_1V_2cos(\theta_1 - \theta_2)]G_{ij} \\ &+[V_1^2-V_1V_2cos(\theta_1 - \theta_2)](-jB_{ij}) \\ &-jV_1V_2sin (\theta_1 - \theta_2)G_{ij} \\ &-jV_1V_2sin (\theta_1 - \theta_2)(-jB_{ij}) \\ &=[V_1^2-V_1V_2cos(\theta_1 - \theta_2)]G_{ij}-V_1V_2sin (\theta_1 - \theta_2)B_{ij} \\ &-j[V_1^2-V_1V_2cos(\theta_1 - \theta_2)]B_{ij}-jV_1V_2sin (\theta_1 - \theta_2)G_{ij} \end{aligned} \end{equation} 有功传输功率 \begin{equation} \begin{aligned} P_{ij}&=[V_1^2-V_1V_2cos(\theta_1 - \theta_2)]G_{ij}-V_1V_2sin (\theta_1 - \theta_2)B_{ij} \\ &=V_1^2G_{ij}-V_1V_2[cos(\theta_1 - \theta_2)G_{ij}+sin (\theta_1 - \theta_2)B_{ij}] \end{aligned} \end{equation} 无功传输功率 \begin{equation} \begin{aligned} Q_{ij}&=-[V_1^2-V_1V_2cos(\theta_1 - \theta_2)]B_{ij}-V_1V_2sin (\theta_1 - \theta_2)G_{ij} \\ &=-V_1^2B_{ij}-V_1V_2[sin(\theta_1 - \theta_2)G_{ij}-cos (\theta_1 - \theta_2)B_{ij}] \end{aligned} \end{equation} 线路有功功率Jacobi \begin{equation} \begin{aligned} \frac{\partial P_{ij}}{\partial V_1}= 2V_1G_{ij}-V_2[cos(\theta_1 - \theta_2)G_{ij}+sin (\theta_1 - \theta_2)B_{ij}] \end{aligned} \end{equation} \begin{equation} \begin{aligned} \frac{\partial P_{12}}{\partial V_2}= -V_1[cos(\theta_1 - \theta_2)G_{ij}+sin (\theta_1 - \theta_2)B_{ij}] \end{aligned} \end{equation} \begin{equation} \begin{aligned} \frac{\partial P_{12}}{\partial \theta_1}&= -V_1V_2[-sin(\theta_1 - \theta_2)G_{ij}+cos (\theta_1 - \theta_2)B_{ij}] \\ &=V_1V_2[sin(\theta_1 - \theta_2)G_{ij}-cos (\theta_1 - \theta_2)B_{ij}] \end{aligned} \end{equation} \begin{equation} \begin{aligned} \frac{\partial P_{12}}{\partial \theta_2}&= -V_1V_2[sin(\theta_1 - \theta_2)G_{ij}-cos (\theta_1 - \theta_2)B_{ij}] \\ \end{aligned} \end{equation} 线路无功功率Jacobi \begin{equation} \begin{aligned} \frac{\partial Q_{12}}{\partial V_1}&= -2V_1B_{12}-V_2[sin(\theta_1 - \theta_2)G_{ij}-cos (\theta_1 - \theta_2)B_{ij}] \end{aligned} \end{equation} \begin{equation} \begin{aligned} \frac{\partial Q_{12}}{\partial V_2}&= -V_1[sin(\theta_1 - \theta_2)G_{ij}-cos (\theta_1 - \theta_2)B_{ij}] \end{aligned} \end{equation} \begin{equation} \begin{aligned} \frac{\partial Q_{12}}{\partial \theta_1}&= -V_1V_2[cos(\theta_1 - \theta_2)G_{ij}+sin (\theta_1 - \theta_2)B_{ij}] \end{aligned} \end{equation} \begin{equation} \begin{aligned} \frac{\partial Q_{12}}{\partial \theta_2}&= -V_1V_2[-cos(\theta_1 - \theta_2)G_{ij}-sin (\theta_1 - \theta_2)B_{ij}] \\ &=V_1V_2[cos(\theta_1 - \theta_2)G_{ij}+sin (\theta_1 - \theta_2)B_{ij}] \end{aligned} \end{equation} 变压器支路功率 \newline 变压器模型为 \newline \begin{center} ----k:1----z----高压侧 \\ or \\ i----k:1----z----j \\ \end{center} 变压器支路功率 \newline 变压器有功输送功率 \begin{equation} \begin{aligned} P_{ij}&=\frac{V_1^2}{k^2} G_{ij}-\frac{V_1}{k} V_2[cos(\theta_1 - \theta_2)G_{ij}+sin (\theta_1 - \theta_2)B_{ij}] \end{aligned} \end{equation} 变压器无功输送功率 \begin{equation} \begin{aligned} Q_{ij}&=-\frac{V_1^2}{k^2}B_{ij}-\frac{V_1}{k} V_2[sin(\theta_1 - \theta_2)G_{ij}-cos (\theta_1 - \theta_2)B_{ij}] \end{aligned} \end{equation} 变压器输送有功功率Jacobi \begin{equation} \begin{aligned} \frac{\partial P_{ij}}{\partial V_1}= 2\frac{V_1}{k^2}G_{ij}-\frac{V_2}{k}[cos(\theta_1 - \theta_2)G_{ij}+sin (\theta_1 - \theta_2)B_{ij}] \end{aligned} \end{equation} \begin{equation} \begin{aligned} \frac{\partial P_{12}}{\partial V_2}= -\frac{V_1}{k}[cos(\theta_1 - \theta_2)G_{ij}+sin (\theta_1 - \theta_2)B_{ij}] \end{aligned} \end{equation} \begin{equation} \begin{aligned} \frac{\partial P_{12}}{\partial \theta_1} &=\frac{V_1}{k}V_2[sin(\theta_1 - \theta_2)G_{ij}-cos (\theta_1 - \theta_2)B_{ij}] \end{aligned} \end{equation} \begin{equation} \begin{aligned} \frac{\partial P_{12}}{\partial \theta_2}&= -\frac{V_1}{k}V_2[sin(\theta_1 - \theta_2)G_{ij}-cos (\theta_1 - \theta_2)B_{ij}] \\ \end{aligned} \end{equation} 变压器输送无功功率Jacobi \begin{equation} \begin{aligned} \frac{\partial Q_{12}}{\partial V_1}&= -2\frac{V_1}{k^2}B_{12}-\frac{V_2}{k}[sin(\theta_1 - \theta_2)G_{ij}-cos (\theta_1 - \theta_2)B_{ij}] \end{aligned} \end{equation} \begin{equation} \begin{aligned} \frac{\partial Q_{12}}{\partial V_2}&= -\frac{V_1}{k}[sin(\theta_1 - \theta_2)G_{ij}-cos (\theta_1 - \theta_2)B_{ij}] \end{aligned} \end{equation} \begin{equation} \begin{aligned} \frac{\partial Q_{12}}{\partial \theta_1}&= -\frac{V_1}{k}V_2[cos(\theta_1 - \theta_2)G_{ij}+sin (\theta_1 - \theta_2)B_{ij}] \end{aligned} \end{equation} \begin{equation} \begin{aligned} \frac{\partial Q_{12}}{\partial \theta_2} &=\frac{V_1}{k}V_2[cos(\theta_1 - \theta_2)G_{ij}+sin (\theta_1 - \theta_2)B_{ij}] \end{aligned} \end{equation} 为了推潮流公式，先从简单的开始推起。 \begin{equation} \Delta P =diag ( \left[ \begin{array}{c} V_1\\ V_2 \end{array} \right] ) \left[ \begin{array}{cc} a & b \\ c & d \end{array} \right] \left[ \begin{array}{c} V_1\\ V_2 \end{array} \right] \end{equation} \begin{equation} \Delta P = \left[ \begin{array}{c} aV_1^2+bV_1V_2 \\ cV_1V_2+dV_2^2 \end{array} \right] \end{equation} 所以 \begin{equation} \begin{array}{cc} \dfrac{\Delta P}{\partial V}&= \left[ \begin{array}{cc} 2aV_1+bV_2 & bV_1\\ cV_2 & cV_1+2dV_2 \end{array} \right] \\ &=diag( \left[ \begin{array}{c} V_1\\ V_2 \end{array} \right] ) \left[ \begin{array}{cc} a & b\\ c & d \end{array} \right] +diag( \left[ \begin{array}{cc} a & b\\ c & d \end{array} \right] \left[ \begin{array}{c} V_1\\ V_2 \end{array} \right] ) \end{array} \end{equation} 再看 \begin{equation} \Delta P =diag ( \left[ \begin{array}{c} V_1\\ V_2 \end{array} \right] ) \left[ \begin{array}{cc} cos(t_1-t_1) & cos(t_1-t_2) \\ cos(t_2-t_1) & cos(t_2-t_2) \end{array} \right] \left[ \begin{array}{c} V_1\\ V_2 \end{array} \right] \end{equation} \begin{equation} \Delta P= diag( \left[ \begin{array}{c} V_1 \\ V_2 \end{array} \right] ) \left[ \begin{array}{c} V_1cos(t_1-t_1)+V_2cos(t_1-t_2) \\ V_1cos(t_2-t_1)+V_2cos(t_2-t_2) \\ \end{array} \right] \end{equation} \begin{equation} \frac{\Delta P}{\partial t}= diag( \left[ \begin{array}{c} V_1 \\ V_2 \end{array} \right] ) \left[ \begin{array}{cc} -V_2sin(t_1-t_2) & V_2sin(t_1-t_2) \\ V_1sin(t_2-t_1) & -V_1sin(t_2-t_1) \\ \end{array} \right] \end{equation} \begin{equation} \frac{\Delta P}{\partial t}= \begin{array}{c} diag( \left[ \begin{array}{c} V_1 \\ V_2 \end{array} \right] ) \\ \left[ \begin{array}{cc} V_1sin(t_1-t_1)-V_2sin(t_1-t_2)-V_1sin(t_1-t_1) & V_2sin(t_1-t_2) \\ V_1sin(t_2-t_1) & V_2sin(t_2-t_2)-V_2sin(t_2-t_2)-V_1sin(t_2-t_1) \end{array} \right] \end{array} \end{equation} \begin{equation} \frac{\Delta P}{\partial t}= diag( \left[ \begin{array}{c} V_1 \\ V_2 \end{array} \right] ) \left\{ \left[ \begin{array}{cc} sin(t_1-t_1) & sin(t_1-t_2) \\ sin(t_2-t_1) & sin(t_2-t_2) \end{array} \right] diag( \left[ \begin{array}{c} V_1 \\ V_2 \end{array} \right] ) - diag( \left[ \begin{array}{cc} sin(t_1-t_1) & sin(t_1-t_2) \\ sin(t_2-t_1) & sin(t_2-t_2) \end{array} \right] \left[ \begin{array}{c} V_1\\ V_2 \end{array} \right] ) \right\} \end{equation} 接下来推导潮流无功公式 \begin{equation} \Delta Q =diag ( \left[ \begin{array}{c} V_1\\ V_2 \end{array} \right] ) \left[ \begin{array}{cc} sin(t_1-t_1) & sin(t_1-t_2) \\ sin(t_2-t_1) & sin(t_2-t_2) \end{array} \right] \left[ \begin{array}{c} V_1\\ V_2 \end{array} \right] \end{equation} \begin{equation} \Delta Q= diag( \left[ \begin{array}{c} V_1 \\ V_2 \end{array} \right] ) \left[ \begin{array}{c} V_1sin(t_1-t_1)+V_2sin(t_1-t_2) \\ V_1sin(t_2-t_1)+V_2sin(t_2-t_2) \\ \end{array} \right] \end{equation} \begin{equation} \dfrac{\partial \Delta Q}{\partial t}= diag( \left[ \begin{array}{c} V_1 \\ V_2 \end{array} \right] ) \left[ \begin{array}{cc} V_2cos(t_1-t_2) & -V_2cos(t_1-t_2) \\ -V_1cos(t_2-t_1) & V_1cos(t_2-t_1) \end{array} \right] \end{equation} \begin{equation} \dfrac{\partial \Delta Q}{\partial t}= \begin{array}{c} diag( \left[ \begin{array}{c} V_1 \\ V_2 \end{array} \right] )\\ \left[ \begin{array}{cc} -V_1cos(t_1-t_1) & -V_2cos(t_1-t_2) \\ -V_1cos(t_2-t_1) & -V_2cos(t_2-t_2) \end{array} \right] + \left[ \begin{array}{cc} V_1cos(t_1-t_1)+V_2cos(t_1-t_2) & 0 \\ 0 & V_1cos(t_2-t_1)+V_2cos(t_2-t_2) \end{array} \right] \end{array} \end{equation} \begin{equation} \dfrac{\partial \Delta Q}{\partial t}= diag( \left[ \begin{array}{c} V_1 \\ V_2 \end{array} \right] ) \left\{ - \left[ \begin{array}{cc} cos(t_1-t_1) & cos(t_1-t_2) \\ cos(t_2-t_1) & cos(t_2-t_2) \end{array} \right] diag( \left[ \begin{array}{c} V_1 \\ V_2 \end{array} \right] ) +diag( \left[ \begin{array}{cc} cos(t_1-t_1) & cos(t_1-t_2) \\ cos(t_2-t_1) & cos(t_2-t_2) \end{array} \right] \left[ \begin{array}{c} V_1 \\ V_2 \end{array} \right] ) \right\} \end{equation} 潮流方程有功的公式为 \begin{equation} \Delta P=diag(V)[Y.*cos(\theta e^T -e \theta^T-\alpha)]V+P_{D}-P_{G}=0 \end{equation} \begin{equation} \frac{\partial P}{\partial V}= \end{equation} ps.已检验过线路以及变压器支路的公式。 \par 以上公式已经可以完成状态估计，若要实现更好的收敛性，需要利用二阶导数。 \par 线路支路功率二阶导数 \begin{equation} \begin{aligned} \frac{\partial^2 P_{12}}{\partial V_1^2}&= \frac{-2}{k^2}B_{12}\\ &=\frac{-2B_{12}}{k^2} \end{aligned} \end{equation} \begin{equation} \begin{aligned} \frac{\partial^2 P_{12}}{\partial V_1 \partial V_2 }&= \frac{-1}{k}[sin(\theta_1 - \theta_2)G_{ij}-cos (\theta_1 - \theta_2)B_{ij}] \\ &= \frac{-[sin(\theta_1 - \theta_2)G_{ij}-cos (\theta_1 - \theta_2)B_{ij}]}{k} \end{aligned} \end{equation} \begin{equation} \begin{aligned} \frac{\partial^2 P_{12}}{\partial V_2^2}&=0 \end{aligned} \end{equation} \begin{equation} \begin{aligned} \frac{\partial P_{12}}{\partial \theta_1^2}&= V_1V_2[cos(\theta_1 - \theta_2)G_{ij}+sin (\theta_1 - \theta_2)B_{ij}] \end{aligned} \end{equation} \begin{equation} \begin{aligned} \frac{\partial P_{12}}{\partial \theta_1 \partial \theta_2}&= V_1V_2[-cos(\theta_1 - \theta_2)G_{ij}-sin (\theta_1 - \theta_2)B_{ij}] \\ &= -V_1V_2[cos(\theta_1 - \theta_2)G_{ij}+sin (\theta_1 - \theta_2)B_{ij}] \end{aligned} \end{equation} \begin{equation} \begin{aligned} \frac{\partial P_{12}}{\partial \theta_2^2}&= -V_1V_2[-cos(\theta_1 - \theta_2)G_{ij}-sin (\theta_1 - \theta_2)B_{ij}] \\ &= V_1V_2[cos(\theta_1 - \theta_2)G_{ij}+sin (\theta_1 - \theta_2)B_{ij}] \end{aligned} \end{equation} \end{document}