[足式机器人]Part4 南科大高等机器人控制课 Ch01 Linear Differential Equations and Matrix Exponential

本文仅供学习使用
本文参考：
B站：CLEAR_LAB
课程主讲教师：
Prof. Wei Zhang

1. ODEs-Ordinary Differential Equations

Most engineering system(including most robotics systems) are modeled by Ordinary Differential (or Difference) Equations (ODEs)

1.1 Dynamics of 2R linkage

1.1.1 拉格朗日法

• 对于构件1而言：
  ${\vec{R}}_{{\mathrm{m}}_1}={\vec{R}}_B=l_1\cdot \cos {\theta }_1\hat{I}+l_1\cdot \sin {\theta }_1\hat{K}$
  ${\vec{V}}_{{\mathrm{m}}_1}={\dot{\vec{R}}}_{{\mathrm{m}}_1}=\left(-{\dot{\theta }}_1{l}_1\cdot \sin {\theta }_1\right)\hat{I}+\left({\dot{\theta }}_1{l}_1\cdot \cos {\theta }_1\right)\hat{K}$
  $K_1=\frac{1}{2}{m}_1{\left({\vec{V}}_{{\mathrm{m}}_1}\right)}^2=\frac{1}{2}{{\dot{\theta }}_1}^2{l_1}^2{m}_1,P_1=m_1\left(g\hat{K}\right)\cdot {\vec{R}}_{{\mathrm{m}}_1}=m_{1}g{l}_1\sin {\theta }_1$

• 对于构件2而言：
  ${\vec{R}}_{{\mathrm{m}}_2}={\vec{R}}_{\mathrm{C}}={\vec{R}}_{\mathrm{B}}+{\vec{R}}_{\mathrm{BC}}=\left[l_1\cdot \cos {\theta }_1+l_2\cdot \cos \left({\theta }_1+{\theta }_2\right)\right]\hat{I}+\left[l_1\cdot \sin {\theta }_1+l_2\cdot \sin \left({\theta }_1+{\theta }_2\right)\right]\hat{K}$
  ${\vec{V}}_{{\mathrm{m}}_2}={\dot{\vec{R}}}_{{\mathrm{m}}_2}=\left[-{\dot{\theta }}_1{l}_1\cdot \sin {\theta }_1-\left({\dot{\theta }}_1+{\dot{\theta }}_2\right)l_2\cdot \sin \left({\theta }_1+{\theta }_2\right)\right]\hat{I}+\left[{\dot{\theta }}_1{l}_1\cdot \cos {\theta }_1+\left({\dot{\theta }}_1+{\dot{\theta }}_2\right)l_2\cdot \cos \left({\theta }_1+{\theta }_2\right)\right]\hat{K}$
  $K_2=\frac{1}{2}{m}_2{\left({\vec{V}}_{{\mathrm{m}}_2}\right)}^2=\frac{1}{2}{m}_2\left[{{\dot{\theta }}_1}^2{l_1}^2+{\left({\dot{\theta }}_1+{\dot{\theta }}_2\right)}^2{l_2}^2+2{\dot{\theta }}_1\left({\dot{\theta }}_1+{\dot{\theta }}_2\right)l_1{l}_2\cos {\theta }_2\right]$
  $P_2=m_2\left(g\hat{K}\right)\cdot {\vec{R}}_{{\mathrm{m}}_2}=m_{1}g\left[l_1\sin {\theta }_1+l_2\cdot \sin \left({\theta }_1+{\theta }_2\right)\right]$

• 对于该系统，则有：
  $L=K-P=\left(K_1+K_2\right)-\left(P_1+P_2\right)$
  $=\frac{1}{2}{{\dot{\theta }}_1}^2{l_1}^2{m}_1+\frac{1}{2}{m}_2\left[{{\dot{\theta }}_1}^2{l_1}^2+{\left({\dot{\theta }}_1+{\dot{\theta }}_2\right)}^2{l_2}^2+2{\dot{\theta }}_1\left({\dot{\theta }}_1+{\dot{\theta }}_2\right)l_1{l}_2\cos {\theta }_2\right]-m_{1}g{l}_1\sin {\theta }_1-m_{2}g\left[l_1\sin {\theta }_1+l_2\cdot \sin \left({\theta }_1+{\theta }_2\right)\right]$
  $=\frac{1}{2}\left(m_1{l_1}^2+m_2{l_1}^2+m_2{l_2}^2+2m_2{l}_1{l}_2\cos {\theta }_2\right){{\dot{\theta }}_1}^2+\frac{1}{2}{m}_2{{\dot{\theta }}_2}^2{l_2}^2+\left(m_2{l_2}^2+m_2{l}_1{l}_2\cos {\theta }_2\right){\dot{\theta }}_1{\dot{\theta }}_2-\left(m_{1}g{l}_1+m_{2}g{l}_1\right)\sin {\theta }_1-m_{2}g{l}_2\cdot \sin \left({\theta }_1+{\theta }_2\right)$
  $=\frac{1}{2}\left(m_1{l_1}^2+m_2{l_1}^2+m_2{l_2}^2\right){{\dot{\theta }}_1}^2+\frac{1}{2}{m}_2{l_2}^2{{\dot{\theta }}_2}^2+m_2{l}_1{l}_2\cos {\theta }_2{{\dot{\theta }}_1}^2+m_2{l_2}^2{\dot{\theta }}_1{\dot{\theta }}_2+m_2{l}_1{l}_2{\dot{\theta }}_1{\dot{\theta }}_2\cos {\theta }_2-\left(m_{1}g{l}_1+m_{2}g{l}_1\right)\sin {\theta }_1-m_{2}g{l}_2\cdot \sin \left({\theta }_1+{\theta }_2\right)$

• 求解力矩${\tau }_1$可得：
  $$\begin{gathered} {\tau }_1=\frac{\mathrm{d}}{\mathrm{dt}}\frac{\partial L}{\partial {\dot{\theta }}_1}-\frac{\partial L}{\partial {\theta }_1}=\frac{\mathrm{d}}{\mathrm{dt}}\left(\left(m_1{l_1}^2+m_2{l_1}^2+m_2{l_2}^2\right){\dot{\theta }}_1+2m_2{l}_1{l}_2{\dot{\theta }}_1\cos {\theta }_2+m_2{l_2}^2{\dot{\theta }}_2+m_2{l}_1{l}_2{\dot{\theta }}_2\cos {\theta }_2\right)+\left(m_{1}g{l}_1+m_{2}g{l}_1\right)\cos {\theta }_1+m_{2}g{l}_2\cdot \cos \left({\theta }_1+{\theta }_2\right) \\ =\left(m_1{l_1}^2+m_2{l_1}^2+m_2{l_2}^2\right){\ddot{\theta }}_1+2m_2{l}_1{l}_2{\ddot{\theta }}_1\cos {\theta }_2-2m_2{l}_1{l}_2{\dot{\theta }}_1{\dot{\theta }}_2\sin {\theta }_2+m_2{l_2}^2{\ddot{\theta }}_2+m_2{l}_1{l}_2{\ddot{\theta }}_2\cos {\theta }_2-m_2{l}_1{l}_2{{\dot{\theta }}_2}^2\sin {\theta }_2+\left(m_{1}g{l}_1+m_{2}g{l}_1\right)\cos {\theta }_1+m_{2}g{l}_2\cdot \cos \left({\theta }_1+{\theta }_2\right) \\ =\left(m_1{l_1}^2+m_2{l_1}^2+m_2{l_2}^2+2m_2{l}_1{l}_2\cos {\theta }_2\right){\ddot{\theta }}_1+\left(m_2{l_2}^2+m_2{l}_1{l}_2\cos {\theta }_2\right){\ddot{\theta }}_2+\left(-2m_2{l}_1{l}_2{\dot{\theta }}_2\sin {\theta }_2\right){\dot{\theta }}_1-m_2{l}_1{l}_2\sin {\theta }_2{{\dot{\theta }}_2}^2+\left(m_{1}g{l}_1+m_{2}g{l}_1\right)\cos {\theta }_1+m_{2}g{l}_2\cdot \cos \left({\theta }_1+{\theta }_2\right) \end{gathered}$$

• 求解力矩${\tau }_2$可得：
  $$\begin{gathered} {\tau }_2=\frac{\mathrm{d}}{\mathrm{dt}}\frac{\partial L}{\partial {\dot{\theta }}_2}-\frac{\partial L}{\partial {\theta }_2}=\frac{\mathrm{d}}{\mathrm{dt}}\left(m_2{l_2}^2{\dot{\theta }}_2+m_2{l_2}^2{\dot{\theta }}_1+m_2{l}_1{l}_2{\dot{\theta }}_1\cos {\theta }_2\right)+m_2{l}_1{l}_2\sin {\theta }_2{{\dot{\theta }}_1}^2+m_2{l}_1{l}_2{\dot{\theta }}_1{\dot{\theta }}_2\sin {\theta }_2+m_{2}g{l}_2\cdot \cos \left({\theta }_1+{\theta }_2\right) \\ =m_2{l_2}^2{\ddot{\theta }}_2+m_2{l_2}^2{\ddot{\theta }}_1+m_2{l}_1{l}_2\cos {\theta }_2{\ddot{\theta }}_1+m_2{l}_1{l}_2\sin {\theta }_2{{\dot{\theta }}_1}^2+m_{2}g{l}_2\cdot \cos \left({\theta }_1+{\theta }_2\right) \\ =\left(m_2{l_2}^2+m_2{l}_1{l}_2\cos {\theta }_2\right){\ddot{\theta }}_1+m_2{l_2}^2{\ddot{\theta }}_2+m_2{l}_1{l}_2\sin {\theta }_2{{\dot{\theta }}_1}^2+m_{2}g{l}_2\cdot \cos \left({\theta }_1+{\theta }_2\right) \end{gathered}$$

• 整理可得：
  $$\left[\begin{array}{c}{\tau }_1\\ {\tau }_2\end{array}\right]=\left[\begin{array}{cc}\left(m_1+m_2\right){l_1}^2+m_2{l_2}^2+2m_2{l}_1{l}_2\cos {\theta }_2& m_2{l_2}^2+m_2{l}_1{l}_2\cos {\theta }_2\\ m_2{l_2}^2+m_2{l}_1{l}_2\cos {\theta }_2& m_2{l_2}^2\end{array}\right]\left[\begin{array}{c}{\ddot{\theta }}_1\\ {\ddot{\theta }}_2\end{array}\right]+\left[\begin{array}{c}-2m_2{l}_1{l}_2{\dot{\theta }}_2\sin {\theta }_2{\dot{\theta }}_1-m_2{l}_1{l}_2\sin {\theta }_2{{\dot{\theta }}_2}^2\\ m_2{l}_1{l}_2\sin {\theta }_2{{\dot{\theta }}_1}^2\end{array}\right]+\left[\begin{array}{c}\left(m_{1}g{l}_1+m_{2}g{l}_1\right)\cos {\theta }_1+m_{2}g{l}_2\cdot \cos \left({\theta }_1+{\theta }_2\right)\\ m_{2}g{l}_2\cdot \cos \left({\theta }_1+{\theta }_2\right)\end{array}\right]$$
  即：$\tau =M\left(\theta \right)\ddot{\theta }+c\left(\theta ,\dot{\theta }\right)+g\left(\theta \right)=M\left(\theta \right)\ddot{\theta }+h\left(\theta ,\dot{\theta }\right)$

Screw theory, exponential coordinate, and Product of Exponential (PoE) are based on the (linear) differential equation view of robot kinematics.

1.1.2 else method?

1.2 Linear Differential Equations (Autonomous/ɔː’tɒnəməs/)

• Linear Differential Equation：ODEs that are linear wrt variables

eg1:
$\{\begin{array}{l}{\dot{x}}_1\left(t\right)+x_2\left(t\right)=0\\ {\dot{x}}_2\left(t\right)+x_1\left(t\right)+x_2\left(t\right)=0\end{array}$
$$\begin{gathered} x\left(t\right)=\left[\begin{array}{c}{x}_1\left(t\right)\\ x_2\left(t\right)\end{array}\right] \\ \dot{x}\left(t\right)=\left[\begin{array}{c}{\dot{x}}_1\left(t\right)\\ {\dot{x}}_2\left(t\right)\end{array}\right]=\left[\begin{array}{cc}0& -1\\ -1& -1\end{array}\right]\left[\begin{array}{c}{x}_1\left(t\right)\\ x_2\left(t\right)\end{array}\right]=Ax\left(t\right) \end{gathered}$$

eg2:
$\{\begin{array}{l}\ddot{y}\left(t\right)+z\left(t\right)=0\\ \dot{z}\left(t\right)+y\left(t\right)=0\end{array}$
$$\begin{gathered} \to x_1\left(t\right)=y\left(t\right),x_2\left(t\right)=\dot{y}\left(t\right),x_3\left(t\right)=z\left(t\right) \\ \dot{x}\left(t\right)=\left[\begin{array}{c}{\dot{x}}_1\left(t\right)\\ {\dot{x}}_2\left(t\right)\\ {\dot{x}}_3\left(t\right)\end{array}\right]=\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& -1\\ -1& 0& 0\end{array}\right]\left[\begin{array}{c}{x}_1\left(t\right)\\ x_2\left(t\right)\\ x_3\left(t\right)\end{array}\right]=Ax\left(t\right) \end{gathered}$$

• State-space form(1-st-order ODE with vector variables)
  Linear $\dot{x}=Ax$——vector field

1.3 General Linear Control Systems

• General(Autonomous) Dynamical Systems: $\dot{x}\left(t\right)=f\left(x\left(t\right)\right)$
  $x\left(t\right)\in {\mathbb{R}}^n$——state vector, $f:{\mathbb{R}}^n\to {\mathbb{R}}^n$——vector field
  “Autonomous” means ‘f’ does not depend on non-‘x’ variables——$\dot{x}=Ax+b$

• Non-autonomous: $\dot{x}\left(t\right)=f\left(x\left(t\right),t\right)$
  captures all non-‘x’ dependence——$\dot{x}=Ax+2t,\dot{x}=Ax+b\left(t\right)$

• Control System: $\dot{x}\left(t\right)=f\left(x\left(t\right),u\left(t\right)\right)$
  vector field $f:{\mathbb{R}}^n\to {\mathbb{R}}^m$ depends on external variable $u\left(t\right)\in {\mathbb{R}}^m$——$\dot{x}=Ax+\sin \left(u\right)=f\left(x,u\right)$

• General Linear Control Systems:
  $\{\begin{array}{l}\dot{x}\left(t\right)=Ax\left(t\right)+Bu\left(t\right)\\ y\left(t\right)=Cx\left(t\right)+Du\left(t\right)\end{array}$, with $x\left(0\right)=x_0$
  $x\in {\mathbb{R}}^n$——system state, $u\in {\mathbb{R}}^m$——control input, $y\in {\mathbb{R}}^p$——system output

$y\left(t\right)=Cx\left(t\right)+Du\left(t\right)$——static relation(无微分项)

1.4 Existence and Uniqueness of ODE Solutions

• Function: $g:{\mathbb{R}}^n\to {\mathbb{R}}^p$ is called Lipschitz over domain $\mathcal{D}\subseteq {\mathbb{R}}^n$ if $\exists L<\infty$
  $$\Vert g\left(x\right)-g\left(x\prime \right)\Vert ⩽L\Vert x-x\prime \Vert ,\forall x,x\prime \in \mathcal{D}$$
  smooth, continuous, piecewise(分段) continuous——$g\left(x\right)=\sqrt{\left|x\right|}$

直觉上，利普希茨连续函数限制了函数改变的速度，符合利普希茨条件的函数的斜率，必小于一个称为利普希茨常数的实数（该常数依函数而定）——学习一下

• Theorem [Existense & Uniqueness] Nonlinear ODE
  $$\dot{x}\left(t\right)=f\left(x\left(t\right),t\right),\mathrm{I}.\mathrm{C}.x\left(t_0\right)=x_0$$
  has a unique solution if $f\left(x,t\right)$ is Lipschitz in $x$ and piecewise continuous in $t$

$\mathrm{I}.\mathrm{C}.$——initial condition

上述的唯一解（unique solution）指的是给定初值问题（Initial Value Problem, IVP），存在且只存在一个解决方案。具体来说，对于给定的初始条件 ，存在且只存在一个函数x(t)，在区间内满足微分方程和初始条件。
证明这一唯一解的存在性和唯一性通常基于庞加莱－林德洛夫（Peano-Lindelöf）定理或者皮卡－林德洛夫（Picard-Lindelöf）定理，这两者都属于常微分方程理论的基本结果。

• Corollary/kə’rɒlərɪ/推论：Linear system $\dot{x}\left(t\right)=Ax\left(t\right)+Bu\left(t\right)$ has a unique solution for any piecewise continuous input $u\left(t\right)$ + $\mathrm{I}.\mathrm{C}.$
  $$\begin{gathered} f\left(x,t\right)=Ax+Bu\left(t\right) \\ \Rightarrow \Vert f\left(x,t\right)-f\left(x\prime ,t\right)\Vert =\Vert Ax-Ax\prime \Vert ⩽\Vert A\Vert \Vert x-x\prime \Vert \end{gathered}$$
  且 $x$ is fixed

Suppose $A$ becomes time-varying $A\left(t\right)$, can you derive conditions to ensure existence and uniqueness of $\dot{x}\left(t\right)=A\left(t\right)x\left(t\right)+Bu\left(t\right)$
矩阵函数 A(t) 应在所关心的时间区间内连续。
矩阵函数 A(t) 应在所关心的时间区间内连续。存在常数 M，使得 (t) 的范数在整个时间区间内都受到 M 的限制
输入函数 u(t) 应在时间区间上是分段连续的。

2. Matrix Exponential

2.1 How to Solve Linear Differential Equations?

General linear ODE: $\dot{x}\left(t\right)=Ax\left(t\right)+d\left(t\right)=Ax+Bu\left(t\right)$

• The key is to derive solutions to the autonomous linear case: $\dot{x}\left(t\right)=Ax\left(t\right)$, with $x\left(t\right)\in {\mathbb{R}}^n$, $A\in {\mathbb{R}}^{n\times n}$ and initial condition (IC) $x\left(0\right)=x_0$
• By existence and unqueness theorem, the ODE $\dot{x}=Ax$ admits a unique solution.
• It turns out that the solution can be found analytically via the Matrix Exponential

2.2 What is the “Euler’s Number” e ?

Consider a scalar linear system : $z\left(t\right)\in {\mathbb{R}}^n$ and $a\in \mathbb{R}$ is a constant, $\dot{z}\left(t\right)=az\left(t\right)$ with IC $z\left(0\right)=z_0$

• The above ODE has a unique solution: $z\left(t\right)=z_0{e}^{at}$
• What is the number “e”?

1. Euler’s Number
2. Defined as the number such that: $\left(e^x\right)\prime =e^x$——$e=\underset{h\to 0}{\lim }{\left(h+1\right)}^{\frac{1}{h}}$

2.3 What is the “Euler’s Number” e ?

For real variable $x\in \mathbb{R}$, Taylor series expansion for $e^x$ around $x=0$:
$$e^x=\sum _{k=0}^{\infty }\frac{z^k}{k!}=1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+\cdots$$
This can be extended to complex variables:
$$e^z=\sum _{k=0}^{\infty }\frac{z^k}{k!}=1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+\cdots$$, this power series is well defined for all $z\in \mathbb{C}$
In particular, we have: $e^{j\theta }={\sum }_{k=0}^{\infty }\frac{x^k}{k!}=1+j\theta -\frac{{\theta }^2}{2!}-j\frac{{\theta }^3}{3!}+\cdots$

Comparing with Taylor expansions for $\cos \theta$ and $\sin \theta$ leads to the Euler’s Formula $\sin \theta =\theta -\frac{{\theta }^3}{3!}+\frac{{\theta }^5}{5!}-\frac{{\theta }^7}{7!}+\cdots ,\cos \theta =1-\frac{{\theta }^2}{2!}+\frac{{\theta }^4}{4!}+\cdots$
Euler’s Formula$$e^{j\theta }=\cos \theta +j\sin \theta$$

2.4 Matrix Exponential Definition

• Similar to the ral and complex cases, we can define the so-called matrix exponential/ˌekspə'nenʃ(ə)l/
  $$e^A=\sum _{k=0}^{\infty }\frac{A^k}{k!}=I+A+\frac{A^2}{2!}+\frac{A^3}{3!}+\cdots$$

eg: $A=\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]$, $A^2=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$
$e^A={\sum }_{k=0}^{\infty }\frac{A^k}{k!}=I+A+\frac{A^2}{2!}+\frac{A^3}{3!}+\cdots =I+A=\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]$

• This power series is well defined for any finite square martix

import numpy as np
import math
import scipy.linalg as la
import matplotlib.pyplot as pltA = np.mat('1,2;3,4')def matrixExp(A):  # this is not a numerically stable way to compute matrix exponenteA = ny.eye(len(A))for i in rang(15):eA = eA + la.fractional_martix_power(A,i+1)/math.factorial(i+1)return eAprint(la.expm(A))
print(matrixExp(A))

• Some Important Properties of Matrix Exponential
  $A{e}^A=e^{A}A$
  but $A{e}^B=e^{B}A,AB\ne BA$
  $e^A{e}^B=e^{A+B},AB=BA$
  $A=PD{P}^{-1},e^A=P{e}^D{P}^{-1}$
  for every $t,\tau \in \mathbb{R}$, $e^{At}{e}^{A\tau }=e^{A\left(t+\tau \right)}$
  ${\left(e^A\right)}^{-1}=e^{-A}$

3. Solution to Linear Differential Equations

3.1 Autonomous Linear Systems

$\dot{x}\left(t\right)=Ax\left(t\right)$ , with initial condition $x\left(0\right)=x_0$, $x\left(t\right)\in \mathbb{R},A\in {\mathbb{R}}^{n\times n}$ is constant matrix, $x_0\in {\mathbb{R}}^n$ is given
With the definition of matrix exponential, we can show that the solution is given by : $x\left(t\right)=e^{At}{x}_0$

• Computation of Matrix Exponential

1. Directly from definition: $e^{At}={\sum }_{k=0}^{\infty }\frac{A{t}^k}{k!}$——It’s hard to compute. But for special case, this series have analytical form(见2.4)

2. For diagonalizable martix: (见矩阵的对角化，相似矩阵，)
   $A=\left[\begin{array}{cc}1.5& -0.5\\ -0.5& 1.5\end{array}\right]=\left[\begin{array}{cc}1& 1\\ 1& -1\end{array}\right]\left[\begin{array}{cc}1& 0\\ 0& 2\end{array}\right]\left[\begin{array}{cc}0.5& 0.5\\ 0.5& -0.5\end{array}\right]$
   $\Rightarrow e^{At}=\left[\begin{array}{cc}1& 1\\ 1& -1\end{array}\right]\left[\begin{array}{cc}{e}^t& 0\\ 0& e^{2t}\end{array}\right]\left[\begin{array}{cc}0.5& 0.5\\ 0.5& -0.5\end{array}\right]$

3. Using Laplace transform:$\dot{x}=Ax,x\left(0\right)=x_0\in {\mathbb{R}}^n$
   $x\left(t\right)\leftrightarrow X\left(s\right)\in {\mathbb{R}}^n,X\left(s\right)=\int x\left(t\right)e^{-st}\mathrm{d}t$
   $\dot{x}\left(t\right)\leftrightarrow sX\left(s\right)-x_0=AX\left(s\right)\Rightarrow \left(sI-A\right)X\left(s\right)=x_0\Rightarrow X\left(s\right)={\left(sI-A\right)}^{-1}{x}_0$
   $x\left(t\right)={\mathcal{L}}^{-1}\left[{\left(sI-A\right)}^{-1}{x}_0\right]\Rightarrow e^{At}={\mathcal{L}}^{-1}\left[{\left(sI-A\right)}^{-1}{x}_0\right]$

3.2 Solution to General Linear Systems

$\{\begin{array}{l}\dot{x}\left(t\right)=Ax\left(t\right)+Bu\left(t\right)\\ y\left(t\right)=Cx\left(t\right)+Du\left(t\right)\end{array}$, with IC $x\left(0\right)=x_0$
$x\in {\mathbb{R}}^n$——system state, $u\in {\mathbb{R}}^m$——control input, $y\in {\mathbb{R}}^p$——system output, A, B, C, D are constant matrices with appropriate dimensions

• The solution is given by $\{\begin{array}{l}x\left(t\right)=e^{At}{x}_0+{\int }_0^t{e}^{A\left(t-\tau \right)}Bu\left(\tau \right)\mathrm{d}\tau \\ y\left(t\right)=C{e}^{At}{x}_0+C{\int }_0^t{e}^{A\left(t-\tau \right)}Bu\left(\tau \right)\mathrm{d}\tau +Du\left(t\right)\end{array}$

• 证明见：