加入有功潮流的雅克比矩阵

Signed-off-by: dmy@lab <dmy@lab.lab>

公式/公式.tex

@@ -4,10 +4,12 @@
 \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}
@@ -201,7 +203,214 @@ Q_{ij}&=-\frac{V_1^2}{k^2}B_{ij}-\frac{V_1}{k} V_2[sin(\theta_1 - \theta_2)G_{ij
 \end{aligned}
 \end{equation}
 
-ps.已检验过线路的公式。
+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}
+
+
+
+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