facat/stateestimation-self-derivatives-measurementdistribution
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1.推导的潮流公式是对的，已经验证过

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2.已经做成了只有负荷功率以及电压幅值量测的WLS方法。

3.如果ieee输电网算例不收敛，是因为不满足客观性（要考虑0注入约束）
4.ieee4-DN.dat是在原有ieee4节点算例上修改，改成所有节点都有负荷的形式。

Signed-off-by: dmy@lab <dmy@lab.lab>
This commit is contained in:
dmy@lab
2015-03-28 21:57:04 +08:00
parent df1943a96d
commit 537b5e4699
2 changed files with 162 additions and 10 deletions
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129
公式/公式.tex
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@@ -399,6 +399,135 @@ V_2
\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