似乎公式没推错,只是牛顿法对初值敏感,带入潮流的电压和相角迭代几次就可以收敛。
Signed-off-by: dmy <dugg@21cn.com>
This commit is contained in:
71
公式/公式.tex
71
公式/公式.tex
@@ -221,6 +221,8 @@ ps.已检验过线路的公式。
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\par
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线路支路功率二阶导数
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有功部分
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\newline
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海森矩阵对电压
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\begin{equation}
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\begin{aligned}
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\frac{\partial^2 P_{12}}{\partial V_1^2}&=
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@@ -243,7 +245,7 @@ ps.已检验过线路的公式。
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\frac{\partial^2 P_{12}}{\partial V_2^2}&=0
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\end{aligned}
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\end{equation}
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海森矩阵对相角
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\begin{equation}
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\begin{aligned}
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\frac{\partial P_{12}}{\partial \theta_1^2}&=
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@@ -268,7 +270,41 @@ ps.已检验过线路的公式。
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\frac{V_1}{k} V_2[cos(\theta_1 - \theta_2)G_{ij}+sin (\theta_1 - \theta_2)B_{ij}]
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\end{aligned}
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\end{equation}
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海森矩阵对电压相角
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\begin{equation}
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\begin{aligned}
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\frac{\partial P_{12}}{\partial V_1 \partial \theta_1}&=
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-\frac{V_2}{k}[-sin(\theta_1 - \theta_2)G_{ij}+cos (\theta_1 - \theta_2)B_{ij}] \\
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&=\frac{V_2}{k}[sin(\theta_1 - \theta_2)G_{ij}-cos (\theta_1 - \theta_2)B_{ij}]
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\end{aligned}
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\end{equation}
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\begin{equation}
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\begin{aligned}
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\frac{\partial P_{12}}{\partial V_1 \partial \theta_2}&=
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-\frac{V_2}{k}[sin(\theta_1 - \theta_2)G_{ij}-cos (\theta_1 - \theta_2)B_{ij}]
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\end{aligned}
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\end{equation}
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\begin{equation}
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\begin{aligned}
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\frac{\partial P_{12}}{\partial V_2 \partial \theta_1}&=
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-\frac{V_1}{k}[-sin(\theta_1 - \theta_2)G_{ij}+cos (\theta_1 - \theta_2)B_{ij}] \\
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&=\frac{V_1}{k}[sin(\theta_1 - \theta_2)G_{ij}-cos (\theta_1 - \theta_2)B_{ij}]
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\end{aligned}
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\end{equation}
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\begin{equation}
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\begin{aligned}
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\frac{\partial P_{12}}{\partial V_2 \partial \theta_2}&=
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-\frac{V_1}{k}[sin(\theta_1 - \theta_2)G_{ij}-cos (\theta_1 - \theta_2)B_{ij}]
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\end{aligned}
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\end{equation}
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无功部分
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\newline
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海森矩阵对电压
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\begin{equation}
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\begin{aligned}
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\frac{\partial^2 Q_{12}}{\partial V_1 \partial V_1}&=
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@@ -288,7 +324,7 @@ ps.已检验过线路的公式。
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\frac{\partial^2 Q_{12}}{\partial V_2^2}&=0
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\end{aligned}
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\end{equation}
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海森矩阵对电压
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\begin{equation}
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\begin{aligned}
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\frac{\partial^2 Q_{12}}{\partial \theta_1^2}&=
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@@ -310,6 +346,37 @@ ps.已检验过线路的公式。
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&=\frac{V_1}{k}V_2[sin(\theta_1 - \theta_2)G_{ij}-cos (\theta_1 - \theta_2)B_{ij}]
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\end{aligned}
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\end{equation}
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海森矩阵对电压相角
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\begin{equation}
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\begin{aligned}
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\frac{\partial Q_{12}}{\partial V_1 \partial \theta_1}&=
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-\frac{V_2}{k}[cos(\theta_1 - \theta_2)G_{ij}+sin (\theta_1 - \theta_2)B_{ij}]
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\end{aligned}
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\end{equation}
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\begin{equation}
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\begin{aligned}
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\frac{\partial Q_{12}}{\partial V_1 \partial \theta_2}&=
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-\frac{V_2}{k}[-cos(\theta_1 - \theta_2)G_{ij}-sin (\theta_1 - \theta_2)B_{ij}] \\
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&=\frac{V_2}{k}[cos(\theta_1 - \theta_2)G_{ij}+sin (\theta_1 - \theta_2)B_{ij}]
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\end{aligned}
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\end{equation}
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\begin{equation}
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\begin{aligned}
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\frac{\partial Q_{12}}{\partial V_2 \partial \theta_1}&=
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-\frac{V_1}{k}[cos(\theta_1 - \theta_2)G_{ij}+sin (\theta_1 - \theta_2)B_{ij}]
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\end{aligned}
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\end{equation}
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\begin{equation}
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\begin{aligned}
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\frac{\partial Q_{12}}{\partial V_2 \partial \theta_2}&=
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-\frac{V_1}{k}[-cos(\theta_1 - \theta_2)G_{ij}-sin (\theta_1 - \theta_2)B_{ij}] \\
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&=\frac{V_1}{k}[cos(\theta_1 - \theta_2)G_{ij}+sin (\theta_1 - \theta_2)B_{ij}]
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\end{aligned}
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\end{equation}
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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.
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\end{document}
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12
公式/内点法公式.tex
12
公式/内点法公式.tex
@@ -15,7 +15,7 @@ f(x)=[z-h(x)]^T W [z-h(x)]
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\end{equation}
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最优条件等价为
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\begin{equation}
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\bigtriangledown f(x)=J^T W [z-h(x)]=0
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\bigtriangledown f(x)=J^T W [h(x)-z]=0
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\end{equation}
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其中$J$是$h(x)$的$Jacobi$矩阵
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\newline
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@@ -55,6 +55,16 @@ f(x)=[z-h(x)]^T W [z-h(x)]
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J ^T W J + \tilde{Q}
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\end{equation}
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其中
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$J$具有以下形式
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\begin{equation}
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J=\left[ %左括号
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\begin{array}{cc} %该矩阵一共3列,每一列都居中放置
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\frac{\partial f_1}{ \partial x_1} & \frac{\partial f_1}{ \partial x_2}\\ %第一行元素
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\frac{\partial f_2}{ \partial x_1} & \frac{\partial f_2}{ \partial x_2}\\ %第二行元素
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\end{array}
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\right] %右括号
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\end{equation}
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\begin{equation}
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J ^T W J=
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(\omega_1 \frac{ \partial f_1 }{\partial x_1} \frac{ \partial f_1 }{\partial x_1}
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