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采用拉格朗日松弛法, 通过引入拉格朗日乘子参数$ \alpha \geq 0 $和$ \beta \geq 0 $以
及松弛参数$ \mu $和$ {\boldsymbol{v}} $, 将条件极值问题转化为如下不带约束的极值问
题:
$$ {J^{*} = \min\limits_{u_{i}(k), \alpha, \beta , \mu , v} J^{\prime}} $$
其中, $J' = J+\alpha[u_{i}(k)-m+\mu^{2}]+\beta[-u_{i}(k)- m+ v^{2}]$.
定理 1. 饱和约束一步最优控制律为:
$$ {u_{i}(k) = u_{i}^{\prime}(k)-\frac{\alpha-\beta}{2 D^{2}}} $$
其中, $D=F b_{i}+Q_{0}$, $ u_{i}^{\prime}(k) $通过下式计算:
$$ \begin{split} D{{u'}_i}\left( k \right) =\;& R\left( {{z^{ - 1}}} \right)\frac{1}{{y_i^*\left( k \right)}} - \bar Q({z^{ - 1}}){u_i}\left( {k - 1} \right)-
\\ & G\left( {{z^{ - 1}}} \right){x_i}\left( k \right) - \left[ {F + S\left( {{z^{ - 1}}} \right)} \right]{a_i} \end{split} $$
当$ |u_{i}^{\prime}(k)|< m $时, $\alpha=0$, $\beta =0$; 当$ u_{i}^{\prime}(k)\leq -m
$时, $\alpha = 0$, $\beta=2 D^{2}[-m-u_{i}^{\prime}(k)]$; 当$ u_{i}^{\prime}(k) \geq m
$时, $\alpha=2D^{2}[- m+u_{i}^{\prime}(k)]$, $\beta=0$.
证明. 引入 Diophantine 方程:
$$ {P\left(z^{-1}\right) = A\left(z^{-1}\right) F+z^{-1} G\left(z^{-1}\right)} $$
其中, $ F $为常数, $ G\left(z^{-1}\right) $为 1 阶多项式. 用$ F $乘以式(4)等号左右两
边, 并利用式(10), 可以得到:
$$ \begin{split}P\left(z^{-1}\right) &x_{i}(k+1) = G\left(z^{-1}\right) x_{i}(k)+ \\ &F b_{i} u_{i}(k)+F a_{i}\end{split} $$
于是由式(7)和式(11)得$ J^{\prime} $对$ u_{i}(k) $的偏导为:
$$ \begin{split} &\frac{{\partial {J^\prime }}}{{\partial {u_i}(k)}} = 2\Bigg[ {P\left( {{z^{ - 1}}} \right){x_i}(k + 1) - R\left( {{z^{ - 1}}}
\right)\frac{1}{{y_i^*(k)}} + } \\ & \quad {Q\left( {{z^{ - 1}}} \right){u_i}(k) + S\left( {{z^{ - 1}}} \right){a_i}} \Bigg]\left( {F{b_i} + {Q_0}}
\right) + \alpha - \beta \end{split} $$
令$ D = F b_{i}+Q_{0} $, 并将式(11)代入式(12), 则
$$ \begin{split} \frac{\partial {{J}^{'}}}{\partial {{u}_{i}}\left( k \right)} = \;&2D\bigg[D{{u}_{i}}\left( k \right)+\bar{Q}\left( {{z}^{-1}}
\right){{u}_{i}}\left( k-1 \right)+ \\ &G\left( {{z}^{-1}} \right){{x}_{i}}\left( k \right)-R\left( {{z}^{-1}} \right)\frac{1}{y_{i}^{*}\left( k \right)}+\\
&\left[ S\left( {{z}^{-1}} \right)+F \right]{{a}_{i}}+\frac{\alpha -\beta }{2D}\bigg] \end{split}\tag{13a} $$
$ J^{\prime} $对$ \alpha $、$ \beta $、$ \mu $和$ v $的偏导分别为:
$$ {\frac{\partial J^{\prime}}{\partial \alpha} = u_{i}(k)-m+\mu^{2}}\tag{13b} $$
$$ {\frac{\partial J^{\prime}}{\partial \beta} = -u_{i}(k)-m+v^{2}}\tag{13c} $$
$$ {\frac{\partial J^{\prime}}{\partial \mu} = 2 \alpha \mu}\tag{13d} $$
$$ {\frac{\partial J^{\prime}}{\partial v} = 2 \beta v}\tag{13e} $$