线性系统的能控性

一、引言

{ 能控性:分析 u ( t ) 对状态 x ( t ) 的控制能力 能观性:输出 y ( t ) 对状态 x ( t ) 的反映能力 \begin{cases} \color{blue}能控性:分析 \pmb{u}(t) 对状态 \pmb{x}(t) 的控制能力\\ \color{blue}能观性:输出\pmb{y}(t)对状态\pmb{x}(t)的反映能力 \end{cases} { 能控性:分析u(t)对状态x(t)的控制能力能观性:输出y(t)对状态x(t)的反映能力

二、能控性

线性连续定常系统: x ˙ = A x + B u \color{red}\pmb{\dot{x}}=\pmb{Ax}+\pmb{Bu} x˙=Ax+Bu

如果存在一个分段连续的输入 u ( t ) \pmb{u}(t) u(t),能在有限时间区间 [ t 0 , t f ] [t_0,t_f] [t0,tf] 内,使系统由某一初始状态 x ( t 0 ) \pmb{x}(t_0) x(t0),转移到指定的任一终端状态 x ( t f ) \pmb{x}(t_f) x(tf),则称此状态是能控的。若系统的所有状态都是能控的, 则称此系统是状态完全能控的 \color{red}则称此系统是状态完全能控的 则称此系统是状态完全能控的,或简称系统是能控的。


线性定常离散系统: x ( k + 1 ) = G x ( k ) + H u ( k ) \color{red}\pmb{x}(k+1)=\pmb{Gx}(k)+\pmb{Hu}(k) x(k+1)=Gx(k)+Hu(k)
初始状态为 x ( 0 ) \pmb{x}(0) x(0),利用递推法

k = 0 k=0 k=0 x ( 1 ) = G x ( 0 ) + H u ( 0 ) \pmb{x}(1)=\pmb{Gx}(0)+\pmb{Hu}(0) x(1)=Gx(0)+Hu(0)
k = 1 k=1 k=1 x ( 2 ) = G x ( 1 ) + H u ( 1 ) = G 2 x ( 0 ) + G H u ( 0 ) + H u ( 1 ) \pmb{x}(2)=\pmb{Gx}(1)+\pmb{Hu}(1)=\pmb{G^2x}(0)+\pmb{GHu}(0)+\pmb{Hu}(1) x(2)=Gx(1)+Hu(1)=G2x(0)+GHu(0)+Hu(1)
k = 2 k=2 k=2 x ( 3 ) = G x ( 2 ) + H u ( 2 ) = G 3 x ( 0 ) + G 2 H u ( 0 ) + G H u ( 1 ) + H u ( 2 ) \pmb{x}(3)=\pmb{Gx}(2)+\pmb{Hu}(2)=\pmb{G^3x}(0)+\pmb{G^2Hu}(0)+\pmb{GHu}(1)+\pmb{Hu}(2) x(3)=Gx(2)+Hu(2)=G3x(0)+G2Hu(0)+GHu(1)+Hu(2)
⋮ \vdots ⋮ \vdots
k = n − 1 k=n-1 k=n1 x ( n ) = G n x ( 0 ) + G n − 1 H u ( 0 ) + G n − 2 H u ( 1 ) + ⋯ + G H u ( n − 2 ) + H u ( n − 1 ) \pmb{x}(n)=\pmb{G^nx}(0)+\pmb{G^{n-1}Hu}(0)+\pmb{G^{n-2}Hu}(1)+\cdots+\pmb{GHu}(n-2)+\pmb{Hu}(n-1) x(n)=Gnx(0)+Gn1Hu(0)+Gn2Hu(1)++GHu(n2)+Hu(n1)

若系统是能控的,则应在 k = n − 1 k=n-1 k=n1 时,从上式解得 u ( 0 ) , u ( 1 ) , ⋯   , u ( n − 1 ) \pmb{u}(0),\pmb{u}(1),\cdots,\pmb{u}(n-1) u(0),u(1),,u(n1),使 x ( n ) = 0 \pmb{x}(n)=0 x(n)=0,从而有

G n − 1 H u ( 0 ) + G n − 2 H u ( 1 ) + ⋯ + G H u ( n − 2 ) + H u ( n − 1 ) = − G n x ( 0 ) \pmb{G^{n-1}Hu}(0)+\pmb{G^{n-2}Hu}(1)+\cdots+\pmb{GHu}(n-2)+\pmb{Hu}(n-1)=-\pmb{G^nx}(0) Gn1Hu(0)+Gn2Hu(1)++GHu(n2)+Hu(n1)=Gnx(0)

( G n − 1 H , G n − 2 H , ⋯   , G H , H ) ( u ( 0 ) u ( 1 ) ⋮ u ( n − 2 ) u ( n − 1 ) ) = − G n x ( 0 ) \left( \begin{matrix} \pmb{G^{n-1}H}, & \pmb{G^{n-2}H}, & \cdots, & \pmb{GH}, & \pmb{H}\\ \end{matrix} \right)\left( \begin{matrix} \pmb{u}(0) \\ \pmb{u}(1) \\ \vdots \\ \pmb{u}(n-2) \\ \pmb{u}(n-1)\\ \end{matrix} \right)=-\pmb{G^nx}(0) (Gn1H,Gn2H,,GH,H) u(0)u(1)u(n2)u(n1) =Gnx(0)

故方程组有解得充分必要条件是能控性矩阵 M = ( H , G H , ⋯   , G n − 2 H , G n − 1 H ) \color{red}\pmb{M}=\left( \begin{matrix} \pmb{H}, & \pmb{GH}, & \cdots, & \pmb{G^{n-2}H}, & \pmb{G^{n-1}H}\\ \end{matrix} \right) M=(H,GH,,Gn2H,Gn1H) 的秩等于 n n n

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转载自blog.csdn.net/m0_72748751/article/details/132830099