线性时变系统状态方程的解
线性时变系统状态方程的解鲁鹏北京理工大学宇航学院2019.10.11文章目录线性时变系统状态方程的解线性时变系统齐次状态方程的解线性时变系统的状态转移矩阵线性时变系统非齐次状态方程的解参考文献随着学习的深入,线性定常系统(状态方程一般形式为x˙(t)=Ax(t)+Bu(t)\dot{\boldsymbol{x}}(t)=Ax(t)+Bu(t)x˙(t)=Ax(t)+Bu(t))已经无法满...
线性时变系统状态方程的解
鲁鹏
北京理工大学,宇航学院
2019.10.11
随着学习的深入,线性定常系统(状态方程一般形式为 x ˙ ( t ) = A x ( t ) + B u ( t ) \dot{\boldsymbol{x}}(t)=Ax(t)+Bu(t) x˙(t)=Ax(t)+Bu(t))已经无法满足我的需求了。线性时变系统(状态方程一般形式为 x ˙ ( t ) = A ( t ) x ( t ) + B ( t ) u ( t ) \dot{\boldsymbol{x}}(t)=A(t)x(t)+B(t)u(t) x˙(t)=A(t)x(t)+B(t)u(t))更具一般性。非线性时变系统也可以通过泰勒展开简化为线性时变系统[1]。本文将总结线性时变系统的状态方程的解,主要参考了东北大学课件[2]。
线性时变系统齐次状态方程的解
系统描述:
x ˙ ( t ) = A ( t ) x ( t ) (1) \dot{\boldsymbol{x}}(t) = A(t)\boldsymbol{x}(t) \tag{1} x˙(t)=A(t)x(t)(1)
初始时刻为 t 0 t_{0} t0, a i j ( t ) a_{ij}(t) aij(t)分段连续
微分方程(1)解为
x ( t ) = [ I + ∫ t 0 t A ( τ ) d τ + ∫ t 0 t A ( τ 1 ) ∫ t 0 t A ( τ 2 ) d τ 2 d τ 1 + ∫ t 0 t A ( τ 1 ) ∫ t 0 t A ( τ 2 ) ∫ t 0 t A ( τ 3 ) d τ 3 d τ 2 d τ 1 + ⋯ ] x ( t 0 ) \begin{gathered} \boldsymbol{x}(t) = \bigg[ I + \int^{t}_{t_{0}}A(\tau)d\tau + \int^{t}_{t_{0}}A(\tau_{1})\int^{t}_{t_{0}}A(\tau_{2})d\tau_{2}d\tau_{1} + \\\int^{t}_{t_{0}} A(\tau_{1}) \int^{t}_{t_{0}} A(\tau_{2}) \int^{t}_{t_{0}} A(\tau_{3}) d\tau_{3} d\tau_{2} d \tau_{1} + \cdots \bigg]\boldsymbol{x}(t_{0}) \end{gathered} x(t)=[I+∫t0tA(τ)dτ+∫t0tA(τ1)∫t0tA(τ2)dτ2dτ1+∫t0tA(τ1)∫t0tA(τ2)∫t0tA(τ3)dτ3dτ2dτ1+⋯]x(t0)
特殊情况:
A ( t ) [ ∫ t 0 t A ( τ ) d τ ] = [ ∫ t 0 t A ( τ ) d τ ] A ( t ) ⟺ x ( t ) = exp [ ∫ t 0 t A ( τ ) d τ ] x ( t 0 ) A(t)\left[\int^{t}_{t_{0}}A(\tau)d\tau \right] = \left[\int^{t}_{t_{0}}A(\tau)d\tau \right]A(t) \\ \iff \boldsymbol{x}(t) = \exp\left[\int^{t}_{t_{0}}A(\tau)d\tau \right]\boldsymbol{x}(t_{0}) A(t)[∫t0tA(τ)dτ]=[∫t0tA(τ)dτ]A(t)⟺x(t)=exp[∫t0tA(τ)dτ]x(t0)
线性时变系统的状态转移矩阵
对于连续线性时变系统
x ˙ ( t ) = A ( t ) x ( t ) + B ( t ) u ( t ) (2) \dot{\boldsymbol{x}}(t) = A(t)\boldsymbol{x}(t) + B(t)\boldsymbol{u}(t)\tag{2} x˙(t)=A(t)x(t)+B(t)u(t)(2)
初始时刻为 t 0 t_{0} t0, a i j ( t ) a_{ij}(t) aij(t), b i j ( t ) b_{ij}(t) bij(t)分段连续。
系统的状态转移矩阵是如下矩阵微分方程和初始条件的 n × n n\times n n×n阶解矩阵 Φ ( t , t 0 ) \Phi(t,t_{0}) Φ(t,t0),如果如下初值问题(Initial Value Problem, IVP)无法解析求解可以使用数值积分求解
Φ ˙ ( t , t 0 ) = A ( t ) Φ ( t , t 0 ) , Φ ( t 0 , t 0 ) = I n (3) \dot{\Phi}(t,t_{0}) = A(t)\Phi(t,t_{0}), \quad \Phi(t_{0},t_{0}) = I_{n} \tag{3} Φ˙(t,t0)=A(t)Φ(t,t0),Φ(t0,t0)=In(3)
线性时变系统非齐次状态方程的解
设连续线性时变系统(2)的解为[3]
x ( t ) = Φ ( t , t 0 ) c ( t ) (4) \boldsymbol{x}(t)=\Phi(t,t_{0})\boldsymbol{c}(t) \tag{4} x(t)=Φ(t,t0)c(t)(4)
其中 c ( t ) \boldsymbol{c(}t) c(t)为待定量,类比高等数学中的常数易变法(A method of constant variation)理解。
x ˙ ( t ) = Φ ˙ ( t , t 0 ) c ( t ) + Φ ( t , t 0 ) c ˙ ( t ) = A ( t ) Φ ( t , t 0 ) c ( t ) + Φ ( t , t 0 ) c ˙ ( t ) (5) \begin{aligned} \dot{\boldsymbol{x}}(t) =& \dot{\Phi}(t,t_{0})\boldsymbol{c}(t) + \Phi(t,t_{0})\dot{\boldsymbol{c}}(t) \\ =& A(t)\Phi(t,t_{0})\boldsymbol{c}(t) + \Phi(t,t_{0})\dot{\boldsymbol{c}}(t) \end{aligned} \tag{5} x˙(t)==Φ˙(t,t0)c(t)+Φ(t,t0)c˙(t)A(t)Φ(t,t0)c(t)+Φ(t,t0)c˙(t)(5)
将等式(2)带入等式(5)可得
A ( t ) Φ ( t , t 0 ) c ( t ) + Φ ( t , t 0 ) c ˙ ( t ) = A ( t ) Φ ( t , t 0 ) c ( t ) + B ( t ) u ( t ) (6) A(t)\Phi(t,t_{0})\boldsymbol{c}(t) + \Phi(t,t_{0})\dot{\boldsymbol{c}}(t) = A(t)\Phi(t,t_{0})\boldsymbol{c}(t) + B(t)\boldsymbol{u}(t) \tag{6} A(t)Φ(t,t0)c(t)+Φ(t,t0)c˙(t)=A(t)Φ(t,t0)c(t)+B(t)u(t)(6)
c ˙ ( t ) = Φ − 1 ( t , t 0 ) B ( t ) u ( t ) (7) \dot{\boldsymbol{c}}(t) = \Phi^{-1}(t,t_{0})B(t)\boldsymbol{u}(t) \tag{7} c˙(t)=Φ−1(t,t0)B(t)u(t)(7)
等式(7)两端积分得:
c ( t ) − c ( t 0 ) = ∫ t 0 t Φ − 1 ( τ , t 0 ) B ( τ ) u ( τ ) d τ (8) \boldsymbol{c}(t) - \boldsymbol{c}(t_{0}) = \int^{t}_{t_{0}}\Phi^{-1}(\tau,t_{0})B(\tau)\boldsymbol{u}(\tau)d\tau \tag{8} c(t)−c(t0)=∫t0tΦ−1(τ,t0)B(τ)u(τ)dτ(8)
由等式(4)可知 t = t 0 t = t_{0} t=t0时有 c ( t 0 ) = x ( t 0 ) \boldsymbol{c}(t_{0}) = \boldsymbol{x}(t_{0}) c(t0)=x(t0),所以
c ( t ) = x ( t 0 ) + ∫ t 0 t Φ − 1 ( τ , t 0 ) B ( τ ) u ( τ ) d τ (9) \boldsymbol{c}(t) = \boldsymbol{x}(t_{0}) + \int^{t}_{t_{0}}\Phi^{-1}(\tau,t_{0})B(\tau)\boldsymbol{u}(\tau)d\tau \tag{9} c(t)=x(t0)+∫t0tΦ−1(τ,t0)B(τ)u(τ)dτ(9)
x ( t ) = Φ ( t , t 0 ) c ( t ) = ⋯ = Φ ( t , t 0 ) x ( t 0 ) + ∫ t 0 t Φ ( t , τ ) B ( τ ) u ( τ ) d τ (10) \begin{aligned} \boldsymbol{x}(t) =& \Phi(t,t_{0})\boldsymbol{c}(t) \\ =& \cdots\\ =& \Phi(t,t_{0})\boldsymbol{x}(t_{0}) + \int^{t}_{t_{0}}\Phi(t,\tau)B(\tau)\boldsymbol{u}(\tau)d\tau \end{aligned} \tag{10} x(t)===Φ(t,t0)c(t)⋯Φ(t,t0)x(t0)+∫t0tΦ(t,τ)B(τ)u(τ)dτ(10)
类比上述推导过程,非线性时变系统 x ˙ ( t ) = A ( t ) x ( t ) + B ( t ) u ( t ) + z ( t ) \dot{\boldsymbol{x}}(t) = A(t)\boldsymbol{x}(t) + B(t)\boldsymbol{u}(t) + \boldsymbol{z}(t) x˙(t)=A(t)x(t)+B(t)u(t)+z(t)的解很好得到。
参考文献
[3] 麻省理工学院公开课:微分方程
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