diff --git a/.gitignore b/.gitignore
index 121711d..e0bad96 100644
--- a/.gitignore
+++ b/.gitignore
@@ -1 +1,5 @@
-/Powerflow
\ No newline at end of file
+/Powerflow
+/公式/*.aux
+/公式/*.log
+/公式/*.pdf
+/公式/*.synctex.gz(busy)
\ No newline at end of file
diff --git a/公式/公式.tex b/公式/公式.tex
index 38d99d8..7a861f7 100644
--- a/公式/公式.tex
+++ b/公式/公式.tex
@@ -9,146 +9,56 @@
 \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}
-%\begin{equation}
-%\begin{aligned}
-%\frac{\partial P_{ij}}{\partial V_i}
-%&=
-%-V_{ij}(g_{ij}cos\theta_{ij}+b_{ij}sin\theta_{ij})+
-%2(g_{ij}+g_{si})V_i
-%\end{aligned}
-%\end{equation}
 线路功率（不考虑接地导纳）
-\newline
 \begin{equation}
 \begin{aligned}
-\dot{I}_{12}&=(V_1 e^{j\theta_1} -  V_2 e^{j\theta_2} )
-(G+jB) \\
-&=[V_1 (cos\theta_1+jsin\theta_1) - V_2 (cos\theta_2+jsin\theta_2)](G+jB) \\
-&=(V_1 cos\theta_1+ jV_1 sin\theta_1- V_2 cos \theta_2 -j V_2 sin \theta_2)(G+jB) \\
-&=(V_1 cos \theta_1 -V_2 cos \theta_2)G + j(V_1 sin \theta_1- V_2 sin \theta_2)G
-\\&+j( V_1 cos \theta_1 -V_2 cos \theta_2)B + j(jV_1 sin \theta_1-jV_2 sin \theta_2)B \\
-&=G(V_1 cos \theta_1- V_2 cos \theta_2) - B(V_1 sin \theta_1 -V_2 sin \theta_2) \\
-&+jB(V_1 cos \theta_1 - V_2 cos \theta_2) + jG(V_1 sin \theta_1 - V_2 sin \theta_2) \\
-&= V_1( G cos \theta_1-B sin \theta_1) -V_2 (G cos \theta_2 - B sin \theta_2) \\
-&+jV_1(Bcos \theta_1+Gsin\theta_1)-jV_2(Bcos\theta_2+Gsin\theta_2)
+\dot{I}_{12}&=(V_1e^{j \theta_1} - V_2e^{j \theta_2})
+(G_{ij}+jB_{ij})
 \end{aligned}
 \end{equation}
-电流的共轭复数
+共轭后
 \begin{equation}
 \begin{aligned}
-\dot{I}^*_{12}&=V_1( G cos \theta_1-B sin \theta_1) -V_2 (G cos \theta_2 - B sin \theta_2) \\
-&-jV_1(Bcos \theta_1+Gsin\theta_1)+jV_2(Bcos\theta_2+Gsin\theta_2)
+\dot{I}^*_{12}&=(V_1e^{-j \theta_1} - V_2e^{-j \theta_2})
+(G_{ij}-jB_{ij})
 \end{aligned}
 \end{equation}
-线路复功率
 \begin{equation}
 \begin{aligned}
-\dot{S}_{12}&=(V_1 e^{j\theta_1} -  V_2 e^{j\theta_2} ) 
+\dot{S}^*_{12}&=V_1e^{j \theta_1}
 \dot{I}^*_{12} \\
-&=[V_1 (cos\theta_1+jsin\theta_1) - V_2 (cos\theta_2+jsin\theta_2)]\dot{I}^*_{12} \\
-&=[V_1 (cos\theta_1+jsin\theta_1) - V_2 (cos\theta_2+jsin\theta_2)] \\
-&[V_1( G cos \theta_1-B sin \theta_1) -V_2 (G cos \theta_2 - B sin \theta_2) 
--jV_1(Bcos \theta_1+Gsin\theta_1)+jV_2(Bcos\theta_2+Gsin\theta_2)]\\
-&=V_1( G cos \theta_1-B sin \theta_1) 
-[V_1( G cos \theta_1-B sin \theta_1) -V_2 (G cos \theta_2 - B sin \theta_2)]\\
-&-V_1( G cos \theta_1-B sin \theta_1)
-[jV_1(Bcos \theta_1+Gsin\theta_1)-jV_2(Bcos\theta_2+Gsin\theta_2)] \\
-&- V_2 (cos\theta_2+jsin\theta_2)
-[V_1( G cos \theta_1-B sin \theta_1) -V_2 (G cos \theta_2 - B sin \theta_2)]\\
-&V_2 (cos\theta_2+jsin\theta_2)
-[jV_1(Bcos \theta_1+Gsin\theta_1)-jV_2(Bcos\theta_2+Gsin\theta_2)]
-\end{aligned}
-\end{equation}
-另一种推导方式
-\begin{equation}
-\begin{aligned}
-\dot{I}_{12}=(V_1 e^{j\theta_1} -  V_2 e^{j\theta_2} )
-(G+jB)
-\end{aligned}
-\end{equation}
-电流的共轭复数
-\begin{equation}
-\begin{aligned}
-\dot{I}^*_{12}=(V_1 e^{-j\theta_1} -  V_2 e^{-j\theta_2} )
-(G-jB)
-\end{aligned}
-\end{equation}
-线路复功率
-\begin{equation}
-\begin{aligned}
-\dot{S}_{12}&=(V_1 e^{j\theta_1} -  V_2 e^{j\theta_2} )\dot{I}^*_{12} \\
-&=(V_1 e^{j\theta_1} -  V_2 e^{j\theta_2} )
-(V_1 e^{-j\theta_1} -  V_2 e^{-j\theta_2} )
-(G-jB) \\
-&=(V_1 e^{j\theta_1}V_1 e^{-j\theta_1} - V_1 e^{j\theta_1}V_2 e^{-j\theta_2} 
--  V_2 e^{j\theta_2}V_1 e^{-j\theta_1}+  V_2 e^{j\theta_2}V_2 e^{-j\theta_2}
+&=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-jB) \\
-&=(V_1^2+V_2^2-V_1V_2e^{j(\theta_1-\theta_2)} -V_1V_2e^{j(\theta_2-\theta_1)} )
-(G-jB) \\
-&=[V_1^2+V_2^2
--V_1V_2(e^{j(\theta_1-\theta_2)}+e^{j(\theta_2-\theta_1)})
-](G-jB) \\
-&=\{V_1^2+V_2^2-V_1V_2[cos(\theta_1-\theta_2)+jsin(\theta_1-\theta_2)
-+cos(\theta_2-\theta_1)+jsin(\theta_2-\theta_1)]\}(G-jB) \\
-&=\{V_1^2+V_2^2-V_1V_2[cos(\theta_1-\theta_2)
-+cos(\theta_1-\theta_2)]\}(G-jB) \\
-&=[V_1^2+V_2^2-2V_1V_2cos(\theta_1-\theta_2)](G-jB) \\
-&=(V_1^2+V_2^2)G-j(V_1^2+V_2^2)B-j2V_1V_2cos(\theta_1-\theta_2)G-j2V_1V_2cos(\theta_1-\theta_2)(-jB) \\
-&=(V_1^2+V_2^2)G-2V_1V_2cos(\theta_1-\theta_2)B
--j(V_1^2+V_2^2)B-j2V_1V_2cos(\theta_1-\theta_2)G
+(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_{12}=(V_1^2+V_2^2)G+2V_1V_2sin(\theta_1-\theta_2)B
+\dot{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_{12}=-(V_1^2+V_2^2)B+2V_1V_2sin(\theta_1-\theta_2)G
-\end{aligned}
-\end{equation}
-考虑接地支路,接地电流为
-\begin{equation}
-\label{接地电流}
-\dot{I}_{1}=V_1e^{j\theta_1}(G_d+jB_d)
-\end{equation}
-(\ref{接地电流})共轭为
-\begin{equation}
-\dot{I}_{1}^{*}=V_1e^{-j\theta_1}(G_d-jB_d)
-\end{equation}
-%Sd
-\begin{equation}
-\begin{aligned}
-\dot{S}_{1}&=V_1e^{j\theta_1}\dot{I}_{1}^{*} \\
-&=V_1e^{j\theta_1} 
-V_1e^{-j\theta_1}(G_d-jB_d) \\
-&=V_1^2(G_d-jB_d)
-\end{aligned}
-\end{equation}
-因而
-\begin{equation}
-\begin{aligned}
-P_{1}&=V_1^2 G_d \\
-\end{aligned}
-\end{equation}
-\begin{equation}
-\begin{aligned}
-Q_{1}&=-V_1^2 B_d \\
-\end{aligned}
-\end{equation}
-考虑节点支路后的注入电流
-\begin{equation}
-\begin{aligned}
-\tilde{P}_{12}&=V_1^2 G_d 
-+(V_1^2+V_2^2)G+2V_1V_2sin(\theta_1-\theta_2)B
-\end{aligned}
-\end{equation}
-\begin{equation}
-\begin{aligned}
-\tilde{Q}_{12}&=-V_1^2 B_d 
--(V_1^2+V_2^2)B+2V_1V_2sin(\theta_1-\theta_2)G
+\dot{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}
 \end{document}