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\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}
考虑接地支路后(利用Ali Abur 书上的公式)
\newline
有功传输功率
\begin{equation}
\begin{aligned}
P_{ij}&=V_1^2(G_{ij}+G_{is})-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(B_{ij}+B_{is})-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_1(G_{ij}+G_{is})-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_1(B_{12}+B_{is})-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}

ps.已检验过线路的公式。
\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
有功部分
\newline
海森矩阵对电压
\begin{equation}
\begin{aligned}
\frac{\partial^2 P_{12}}{\partial V_1^2}&=
\frac{2}{k^2}(G_{12}+G_{1s})\\
&=\frac{2 (G_{12}+G_{1s} ) } {k^2}
\end{aligned}
\end{equation}

\begin{equation}
\begin{aligned}
\frac{\partial^2 P_{12}}{\partial V_1 \partial V_2 }&=
\frac{-1}{k}[cos(\theta_1 - \theta_2)G_{ij}+sin (\theta_1 - \theta_2)B_{ij}] \\
&=
\frac{-[cos(\theta_1 - \theta_2)G_{ij}+sin (\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}
海森矩阵对相角
\begin{equation}
\begin{aligned}
\frac{\partial P_{12}}{\partial \theta_1^2}&=
\frac{V_1}{k} V_2[cos(\theta_1 - \theta_2)G_{ij}+sin (\theta_1 - \theta_2)B_{ij}]
\end{aligned}
\end{equation}

\begin{equation}
\begin{aligned}
\frac{\partial P_{12}}{\partial \theta_1 \partial \theta_2}&=
\frac{V_1}{k}  V_2[-cos(\theta_1 - \theta_2)G_{ij}-sin (\theta_1 - \theta_2)B_{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}

\begin{equation}
\begin{aligned}
\frac{\partial P_{12}}{\partial \theta_2^2}&=
- \frac{V_1}{k} V_2[-cos(\theta_1 - \theta_2)G_{ij}-sin (\theta_1 - \theta_2)B_{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}
海森矩阵对电压相角

\begin{equation}
\begin{aligned}
\frac{\partial P_{12}}{\partial V_1 \partial \theta_1}&=
-\frac{V_2}{k}[-sin(\theta_1 - \theta_2)G_{ij}+cos (\theta_1 - \theta_2)B_{ij}] \\
&=\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 P_{12}}{\partial V_1 \partial \theta_2}&=
-\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 P_{12}}{\partial V_2 \partial \theta_1}&=
-\frac{V_1}{k}[-sin(\theta_1 - \theta_2)G_{ij}+cos (\theta_1 - \theta_2)B_{ij}] \\
&=\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 P_{12}}{\partial V_2 \partial \theta_2}&=
-\frac{V_1}{k}[sin(\theta_1 - \theta_2)G_{ij}-cos (\theta_1 - \theta_2)B_{ij}]
\end{aligned}
\end{equation}

无功部分
\newline
海森矩阵对电压
\begin{equation}
\begin{aligned}
\frac{\partial^2 Q_{12}}{\partial V_1 \partial V_1}&=
-2\frac{( B_{12}+B_{1s})}{k^2}
\end{aligned}
\end{equation}

\begin{equation}
\begin{aligned}
\frac{\partial^2 Q_{12}}{\partial V_1 \partial V_2}&=
-\frac{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^2 Q_{12}}{\partial V_2^2}&=0
\end{aligned}
\end{equation}
海森矩阵对电压
\begin{equation}
\begin{aligned}
\frac{\partial^2 Q_{12}}{\partial \theta_1^2}&=
-\frac{V_1}{k}V_2[-sin(\theta_1 - \theta_2)G_{ij}+cos (\theta_1 - \theta_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}

\begin{equation}
\begin{aligned}
\frac{\partial^2 Q_{12}}{\partial \theta_1 \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}

\begin{equation}
\begin{aligned}
\frac{\partial^2 Q_{12}}{\partial \theta_2^2}
&=\frac{V_1}{k}V_2[sin(\theta_1 - \theta_2)G_{ij}-cos (\theta_1 - \theta_2)B_{ij}]
\end{aligned}
\end{equation}
海森矩阵对电压相角
\begin{equation}
\begin{aligned}
\frac{\partial Q_{12}}{\partial V_1 \partial \theta_1}&=
-\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 Q_{12}}{\partial V_1 \partial \theta_2}&=
-\frac{V_2}{k}[-cos(\theta_1 - \theta_2)G_{ij}-sin (\theta_1 - \theta_2)B_{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 Q_{12}}{\partial V_2 \partial \theta_1}&=
-\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 Q_{12}}{\partial V_2 \partial \theta_2}&=
-\frac{V_1}{k}[-cos(\theta_1 - \theta_2)G_{ij}-sin (\theta_1 - \theta_2)B_{ij}] \\
&=\frac{V_1}{k}[cos(\theta_1 - \theta_2)G_{ij}+sin (\theta_1 - \theta_2)B_{ij}]
\end{aligned}
\end{equation}

以上公式对线路和没有计及接地支路的变压器适用，只是线路中变比$k$为1.

\end{document}
