facat/stateestimation-self-derivatives
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1.修复了变压器的公式

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Signed-off-by: facat <facat@facat.cn>
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facat
2013-08-31 11:48:03 +08:00
parent cff1dde6f4
commit 4b44a7684f
3 changed files with 8 additions and 8 deletions
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公式/公式.tex
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@@ -133,15 +133,13 @@ i----k:1----z----j \\
变压器有功输送功率
\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} \\
&=\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}]
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}&=-[V_1^2-V_1V_2cos(\theta_1 - \theta_2)]B_{ij}-V_1V_2sin (\theta_1 - \theta_2)G_{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}]
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