线性系统理论课程笔记
对应教材内容第三章 线性系统的运动分析的内容
引言
运动分析的实质
\(\dot x = A(t)x + B(t)u,\quad x({t_0}) =
{x_0}\)
\(\dot x = Ax + Bu,\quad x(0) = {x_0},\quad t
\geqslant 0\)
从数学模型出发，定量地和精确地定出系统运动的变化规律，为系统的实际运动过程作出估计。
给定初始状态\(x_0\)和外输入作用\(u\)，求解出状态方程的解。
解的存在性和唯一性
①系统矩阵\(A(t)\)的各个元\(a_{ij}(t)\)在时间区间\([t_0, t_\alpha]\)上为绝对可积，即
\(\int_{t_0}^{t_\alpha}{|{a_{ij}}(t)|dt <
\infty ,\quad \quad i,j = 1,2, \cdots n}\)
②输入矩阵\(B(t)\)的各个元\(b_{ij}(t)\)在时间区间\([t_0, t_\alpha]\)上为平方可积，即
\(\int_{t_0}^{t_\alpha }{ {
{[{b_{ik}}(t)]}^2}dt < \infty ,\quad \quad i = 1,2, \cdots n,\quad
\;k = 1,2 \cdots p}\)
③输入\(u(t)\)的各个元\(u_k(t)\)在时间区间\([t_0, t_\alpha]\)上为平方可积，即
\(\int_{t_0}^{t_\alpha}{ { {[{u_k}(t)]}^2}dt
< \infty ,\quad \quad k = 1,2 \cdots p}\)
条件②③可一步合并为要求\(B(t)u(t)\)的各元在时间区间\([t_0, t_\alpha]\)上绝对可积，即
\(\int_{t_0}^{t_\alpha}{ {
{[{b_{ik}}(t){u_k}(t)]}^2}dt < \infty ,\quad \quad k = 1,2 \cdots
p}\)
零输入响应和零初始状态响应
线性系统满足叠加原理。在初始状态和输入向量作用下的运动，分解为两个单独的分运动。
\(x(t)=x_{ou}(t)+x_{ox}(t)\)
①自治系统（零输入响应）\(\dot x=Ax \quad
x({t_0}) = {x_0}\)
②受迫系统（零初始状态响应）\(\dot x = Ax + Bu
\quad x(t_0) = x_0\)
线性定常系统的运动分析
零输入响应
自治系统状态方程\(\dot x=Ax \quad x({t_0})
= {x_0}\)
\(sx-x_0=Ax \Rightarrow sx-Ax=x_0 \Rightarrow
(sI-A)x=x_0 \Rightarrow x=(sI-A)^{-1}x_0=e^{At}x_0\)
结论：系统自治状态方程的解\(x_{ou}(t)=e^{At}x_0 \quad (t \geqslant 0)\)
其中\(e^{At}\)为矩阵指数函数，可以展开为
\(e^{At}=I + At + \frac{1}{2!}{A^2}{t^2} +
\cdots = \sum\limits_{k=0}^\infty{\frac{1}{k!}{A^k}{t^k}}\)
渐近稳定的充要条件：
对于线性定常系统，零输入响应最终趋向于系统平衡状态\(x=0\)
当且仅当\(e^{At}\)最终趋于0，即\(\mathop {\lim }\limits_{t \to \infty } e^{At} =
0\)。
矩阵指数函数的性质
①\(\mathop {\lim }\limits_{t \to 0}
{e^{At}} = I\)
②\({e^{A(t + \tau )}} = {e^{At}} \cdot
{e^{A\tau }} = {e^{A\tau }} \cdot {e^{At}}\)
③\(e^{At}\)总是非奇异的，且其逆为：\({({e^{At}})^{ - 1}} = {e^{ - At}}\)
④设\(A\)和\(F\)为两个同维可交换方阵，即\(AF = FA\)，则有\({e^{(A + F)t}} = {e^{At}}{e^{Ft}} =
{e^{Ft}}{e^{At}}\)
⑤\(\frac{d}{dt}{e^{At}} = A{e^{At}} =
{e^{At}}A\)
⑥\(\frac{d}{dt}{e^{ - At}} = - A{e^{ - At}} =
- {e^{ - At}}A\)
⑦\({({e^{At}})^m} = {e^{A(mt)}},\quad m =
0,1,2, \cdots\)
矩阵指数函数的求解
定义法
\(e^{At}=I + At + \frac{1}{2!}{A^2}{t^2} + \cdots = \sum\limits_{k=0}^\infty{\frac{1}{k!}{A^k}{t^k}}\)
特征值法
若\(\bar A =
diag(\begin{array}{*{20}{c}}{\lambda _1}&{\lambda _2}& \cdots
&{ {\lambda _n})} \end{array}\)，则有\({e^{\bar At}} =
diag(\begin{array}{*{20}{c}}{e^{\lambda _1 t}}&{e^{\lambda _2
t}}& \cdots &{e^{\lambda _n t})}\end{array}\)
且\(\bar A = {P^{ - 1}}AP \quad A = P\bar
A{P^{ - 1}}\)，则有\({e^{At}} =
P{e^{\bar At}}{P^{ - 1}}\)
【例1】\(A = \left[
{\begin{array}{*{20}{c}}0&1&0 \\ 0&0&1 \\
-6&-11&-6 \end{array}} \right]\)
由\(|\lambda
I-A|=\lambda^3+6\lambda^2+11\lambda+6=(\lambda+1)(\lambda+2)(\lambda+3)=0
\quad \Rightarrow \quad
\lambda_1=1,\lambda_2=2,\lambda_3=3\)
由等式\(\lambda_i
v_i=Av_i\)，解方程组\(|\lambda_i
I-A|v_i=0\)，得到特征向量
\[{\begin{array}{*{20}{c}} {v_1=\left[
{\begin{array}{*{20}{c}} 1\\-1\\1 \end{array}}\right]}&{v_2=\left[
{\begin{array}{*{20}{c}} 1\\-2\\4 \end{array}}\right]}&{v_3=\left[
{\begin{array}{*{20}{c}} 1\\-3\\9 \end{array}}\right]}
\end{array}}\] 得到\(P =[v_1,v_2,v_3] =
\left[ {\begin{array}{*{20}{c}}1&1&1 \\ -1&-2&-3 \\
1&4&9\end{array}} \right]\)
该式构成范德蒙德行列式，其行列式值\(|P|=[(-2)-(-1)][(-3)-(-2)][(-3)-(-1)]=-2\)
求其逆矩阵\(P^{-1} =
\frac{P^*}{|P|}=\frac{1}{2}\left[ {\begin{array}{*{20}{c}}6&5&1
\\ -6&-8&-2 \\ 2&3&1\end{array}} \right]\)
则有\(\bar A = {P^{ - 1}}AP = \left[
{\begin{array}{*{20}{c}}{ - 1}&0&0 \\ 0&{ - 2}&0 \\
0&0&{ - 3} \end{array}} \right]\)
由\(\bar A = {P^{ - 1}}AP \quad \Rightarrow
\quad A = P\bar A{P^{ - 1}}\)计算出
\({e^{At}} = P{e^{\bar At}}{P^{ - 1}} =
P\left[ {\begin{array}{*{20}{c}} e^{-t}&0&0 \\
0&e^{-2t}&0 \\ 0&0&e^{-3t} \end{array}} \right]{P^{ -
1}}\)
\(= \left[ {\begin{array}{*{20}{c}}{3{e^{ -
t}} - 3{e^{ - 2t}} + {e^{ - 3t}}}&{\frac{5}{2}{e^{ - t}} - 4{e^{ -
2t}} + \frac{3}{2}{e^{ - 3t}}}&{\frac{1}{2}{e^{ - t}} - {e^{ - 2t}}
+ \frac{1}{2}{e^{ - 3t}}} \\ { - 3{e^{ - t}} + 6{e^{ - 2t}} - 3{e^{ -
3t}}}&{ - \frac{5}{2}{e^{ - t}} + 8{e^{ - 2t}} - \frac{9}{2}{e^{ -
3t}}}&{ - \frac{1}{2}{e^{ - t}} + 2{e^{ - 2t}} - \frac{3}{2}{e^{ -
3t}}} \\ {3{e^{ - t}} - 12{e^{ - 2t}} + 9{e^{ -
3t}}}&{\frac{5}{2}{e^{ - t}} - 16{e^{ - 2t}} +
\frac27{2}{e^{ - 3t}}}&{\frac{1}{2}{e^{ - t}} -
4{e^{ - 2t}} + \frac{9}{2}{e^{ - 3t}}} \end{array}} \right]\)
注：其中\(P^*\)为\(P\)的伴随矩阵
伴随矩阵公式\(A^*=\left[
{\begin{array}{*{20}{c}}A_{11}&A_{21}&A_{31} \\
A_{12}&A_{22}&A_{32} \\ A_{13}&A_{23}&A_{33}\end{array}}
\right]\)
其中\(A_{ij}=(-1)^{i+j}M_{ij}\)为元素\(a_{ij}\)的代数余子式。
约当阵的情况：
若\(\bar A = J =
diag(\begin{array}{*{20}{c}}{J_1}&{J_2}& \cdots &{J_l})
\end{array}\)
其中\({J_i} = \left[
{\begin{array}{*{20}{c}}{\lambda _i}&1&0& \cdots &0 \\
0&{\lambda _i}&1& \cdots &0 \\ 0&0&{\lambda
_i}& \cdots &0 \\ {}&{}& \cdots &{}&{} \\
0&0&0& \cdots &{\lambda _i} \end{array}}
\right]\)
则\({e^{\bar At}} =
diag(\begin{array}{*{20}{c}}e^{J_1}t&e^{J_2}t& \cdots
&e^{J_l}t) \end{array}\)
其中\({e^{ {J_i}t}} = \left[
{\begin{array}{*{20}{c}}{ {e^{ {\lambda _i}t}}}&{t{e^{
{\lambda_i}t}}}&{\frac{1}{2!} {t^2}{e^{ {\lambda_i}t}}}&\cdots
&{} \\ 0&{ {e^{ {\lambda _i}t}}}&{t{e^{ {\lambda
_i}t}}}& \cdots &{} \\ 0&0&{ {e^{ {\lambda _i}t}}}&
\cdots &{} \\ {}&{}& \cdots &{}&{} \\
0&0&0& \cdots &{ {e^{ {\lambda _i}t}}} \end{array}}
\right]\)
有限项展开法
设\(\lambda_1,\lambda_2,\cdots,\lambda_n\)为\(A\)的\(n\)个互异特征值
\({e^{At}} = {\alpha _0}I + {\alpha _1}A +
{\alpha _2}{A^2} + \cdots {\alpha _{n - 1}}{A^{n - 1}}\)
\(\left\{ \begin{gathered}{e^{ {\lambda _1}t}}
= {\alpha _0} + {\alpha _1}{\lambda _1} + {\alpha _2}{\lambda _1}^2 +
\cdots {\alpha _{n - 1}}{\lambda _1}^{n - 1} \hfill \\{e^{ {\lambda
_2}t}} = {\alpha _0} + {\alpha _1}{\lambda _2} + {\alpha _2}{\lambda
_2}^2 + \cdots {\alpha _{n - 1}}{\lambda _2}^{n - 1} \hfill \\ \cdots
\hfill \\{e^{ {\lambda _n}t}} = {\alpha _0} + {\alpha _1}{\lambda _n} +
{\alpha _2}{\lambda _n}^2 + \cdots {\alpha _{n - 1}}{\lambda _n}^{n - 1}
\hfill \\ \end{gathered} \right.\)
若\(\lambda_i\)为\(l\)重特征值，则相应的\(l\)个方程为
\(\left\{ \begin{gathered} {e^{ {\lambda
_i}t}} = {\alpha _0} + {\alpha _1}{\lambda _i} + {\alpha _2}{\lambda
_i}^2 + \cdots + {\alpha _{n - 1}}{\lambda _i}^{n - 1} \hfill \\ \frac{
{\text{d}}}{ { {\text{d}}{\lambda _i}}}{e^{ {\lambda _i}t}} = {\alpha
_1} + 2{\alpha _2}{\lambda _i} + \cdots + (n - 1){\alpha _{n -
1}}{\lambda _i}^{n - 2} \hfill \\ \cdots \hfill \\ \frac{ { {
{\text{d}}^{(l - 1)}}}}{ { {\text{d}}{\lambda _i}^{(l - 1)}}}{e^{
{\lambda _i}t}} = {\alpha _{l - 1}}(l - 1)! + {\alpha _l}l!{\lambda _i}
+ \cdots + {\alpha _{n - 1}}\frac{ {(n - 1)!}}{ {(n - l)!}}{\lambda
_i}^{n - l} \hfill \\ \end{gathered} \right.\)
【例2】\(A = \left[ {\begin{array}{*{20}{c}}
0&0&-2 \\ 0&1&0 \\ 1&0&3
\end{array}}\right]\)
\(|\lambda I - A| = 0 \Rightarrow {(\lambda -
1)^2}(\lambda - 2) = 0\)
\(\Rightarrow {\lambda_1,2} = 1,{\lambda_3} =
2\)
令\({e^{At}} = {\alpha _0}I + {\alpha _1}A +
{\alpha _2}{A^2}\)
列出方程组\(\left\{ \begin{gathered}{e^t} =
{\alpha _0} + {\alpha _1} + {\alpha _2} \hfill \\t{e^t} = {\alpha _1} +
2{\alpha _2} \hfill \\{e^{2t}} = {\alpha _0} + 2{\alpha _1} + 4{\alpha
_2} \hfill \\ \end{gathered} \right.\)
解得\(\left\{ \begin{gathered}{\alpha _0} = -
2t{e^t} + {e^{2t}} \hfill \\{\alpha _1} = 3t{e^t} + 2{e^t} - 2{e^{2t}}
\hfill \\{\alpha _2} = - t{e^t} - {e^t} + {e^{2t}} \hfill \\
\end{gathered} \right.\)
进而得到\({e^{At}} = {\alpha _0}I + {\alpha
_1}A + {\alpha _2}{A^2}\)
\(= ( - 2t{e^t} + {e^{2t}})I + (3t{e^t} +
2{e^t} - 2{e^{2t}})A + ( - t{e^t} - {e^t} +
{e^{2t}}){A^2}\)
\(= \left[ {\begin{array}{*{20}{c}}{2{e^t} -
{e^{2t}}}&0&{2{e^t} - 2{e^{2t}}} \\ 0&{ {e^t}}&0 \\ { -
{e^t} + {e^{2t}}}&0&{ - {e^t} + 2{e^{2t}}} \end{array}}
\right]\)
预解矩阵法
\({e^{At}} = {L^{ - 1}}[{\left( {sI - A}
\right)^{ - 1}}]\)
【例3】\(A = \left[ {\begin{array}{*{20}{c}}
0&1 \\ -2&-3 \end{array}}\right]\)
\((sI-A)^{-1}=\left[ {\begin{array}{*{20}{c}}
s&-1 \\ 2&s+3 \end{array}}\right]^{-1}=\left[
{\begin{array}{*{20}{c}} \frac{s+3}{(s+1)(s+2)}&\frac{1}{(s+1)(s+2)}
\\ \frac{-2}{(s+1)(s+2)}&\frac{s}{(s+1)(s+2)}
\end{array}}\right]=\left[ {\begin{array}{*{20}{c}}
\frac{2}{s+1}-\frac{1}{s+2}&\frac{1}{s+1}-\frac{1}{s+2} \\
-\frac{2}{s+1}+\frac{2}{s+2}&-\frac{1}{s+1}+\frac{2}{s+2}
\end{array}}\right]\)
\(L^{-1}[(sI-A)^{-1}]=\left[
{\begin{array}{*{20}{c}} 2e^{-t}-e^{-2t}&e^{-t}-e^{-2t} \\
-2e^{-t}+2e^{-2t}&-e^{-t}+2e^{-2t} \end{array}}\right]\)
零初始状态响应
线性定常系统的零初始状态响应\(x_{ox}(t)\)具有以下表达式：
\(x_{ox}(t)=\int_0^t { {e^{A(t - \tau
)}}Bu(\tau ){\text{d}}\tau}\)
【例4】\(\dot x = \left[
{\begin{array}{*{20}{c}} 0&1 \\ -2&-3 \end{array}}\right]x +
\left[ {\begin{array}{*{20}{c}} 0 \\ 1 \end{array}}\right]u,\quad
t\geqslant 0\)
其中，初始状态\(x_1(0)=x_2(0)=0\)，输入\(u(t)=I(t)\)
由【例3】知\(e^{A(t-\tau)}=\left[
{\begin{array}{*{20}{c}}
2e^{-(t-\tau)}-e^{-2(t-\tau)}&e^{-(t-\tau)}-e^{-2(t-\tau)} \\
-2e^{-(t-\tau)}+2e^{-2(t-\tau)}&-e^{-(t-\tau)}+2e^{-2(t-\tau)}
\end{array}}\right]\) \(\Rightarrow
x_{ox}(t)=\int_0^t { {e^{A(t - \tau )}}Bu(\tau ){\text{d}}\tau}=\int_0^t
{\left[ {\begin{array}{*{20}{c}} e^{-(t-\tau)}-e^{-2(t-\tau)} \\
-e^{-(t-\tau)}+2e^{-2(t-\tau)} \end{array}}\right]{\text{d}}\tau}=\left[
{\begin{array}{*{20}{c}} \frac{1}{2}-e^{-t}+\frac{1}{2}e^{-2t} \\
e^{-t}-e^{-2t} \end{array}}\right]\)
系统状态运动规律的基本表达式
同时作用初始状态和输入的状态方程的解
对初始时刻\(t_0=0\)，有表达式\(x(t) = {e^{At}}{x_0} + \int_0^t {e^{A(t - \tau
)}Bu(\tau ){\text{d}}\tau }, \quad t \geqslant 0\)
对初始时刻\(t_0 \ne 0\)，有表达式\(x(t) = {e^{A(t - {t_0})}}{x_0} + \int_{t_0}^t
{e^{A(t - \tau )}Bu(\tau ){\text{d}}\tau ,\quad t \geqslant
{t_0}}\)
连续时间线性时不变系统的状态转移矩阵
定义：矩阵方程\(\dot \Phi (t - {t_0})=A\Phi
(t - {t_0}),\quad \Phi (0)=I\)的解\(\Phi (t - {t_0})\)称为状态转移矩阵。
\(t_0=0\)时，\(\Phi (t) = {e^{At}}\quad \quad t \geqslant
0\)
\(t_0 \ne 0\)时，\(\Phi (t - {t_0}) = {e^{A(t - {t_0})}},\quad t
\geqslant {t_0}\)
基于状态转移矩阵的系统响应表达式：
\({x_{ou}}(t) = \Phi (t -
{t_0}){x_0}\)
\({x_{ox}}(t) = \int_{t_0}^t {\Phi (t - \tau
)Bu(\tau ){\text{d}}\tau } \quad t \geqslant {t_0}\)
\(x(t) = \Phi (t - {t_0}){x_0} + \int_{t_0}^t
{\Phi (t - \tau )Bu(\tau ){\text{d}}\tau }\)
状态转移矩阵的特性：
①\(\Phi (0) = I\)
②\({\Phi ^{ - 1}}(t) = \Phi ( - t) \quad {\Phi
^{ - 1}}(t - {t_0}) = \Phi ({t_0} - t)\)
③\(\Phi ({t_2} - {t_1})\Phi ({t_1} - {t_0}) =
\Phi ({t_2} - {t_0})\)
④\(\Phi ({t_2} + {t_1}) = \Phi ({t_2})\Phi
({t_1}) = \Phi ({t_1})\Phi ({t_2})\)
⑤\(\Phi (mt) = {\left[ {\Phi (t)}
\right]^m}\)
⑥\(\frac{\text{d}}{ {\text{d}}t}\Phi (t -
{t_0}) = A\Phi (t - {t_0}) = \Phi (t - {t_0})A\)
⑦\(\frac{\text{d}}{ {\text{d}t}}{\Phi ^{ -
1}}(t - {t_0}) = - A\Phi ({t_0} - t) = - \Phi ({t_0} - t)A\)
连续时间线性时不变系统的脉冲响应矩阵
注：自学内容，仅供参考。
定义：对于输入维数为\(p\)，输出维数为\(q\)的连续时间线性时不变系统，脉冲响应矩阵定义为零初始状态下以脉冲响应\(h_{ij}(t-\tau)\)为元构成的一个\(q \times p\)输出响应矩阵。
其中\(h_{ij}(t-\tau)\)为第\(j\)个输入端在时刻\(\tau\)加以单位脉冲\(\delta(t-\tau)\)而所有其他输入为零时，在第\(i\)个输出端的脉冲响应 \[H(t - \tau ) = \left[
{\begin{array}{*{20}{c}}{h_{11}(t - \tau )}&{h_{12}(t - \tau )}&
\cdots &{h_{1p}(t - \tau )} \\ {h_{21}(t - \tau )}&{h_{22}(t -
\tau )}& \cdots &{h_{2p}(t - \tau )} \\ \cdots & \cdots
& \cdots & \cdots \\ {h_{q1}(t - \tau )}&{h_{q2}(t - \tau
)}& \cdots &{h_{qp}(t - \tau )} \end{array}} \right]\]
且\(H(t - \tau ) = 0\)，\(\forall \tau\)和\(\forall t \prec \tau\)
输出响应：
对\(p\)维输入，\(q\)维输出连续时间线性时不变系统，假设初始状态为零，则系统在任意输入\(u\)作用下基于脉冲响应矩阵的输出响应\(y(t)\)为
\(y(t) = \int_{t_0}^t {H(t - \tau )u(\tau
){\text{d}}\tau } = \int_{t_0}^t {H(\tau )u(t - \tau ){\text{d}}\tau
\quad t \geqslant {t_0}}\)
脉冲响应矩阵\(H(t)\)和传递函数矩阵\(G(s)\)之间的关系：
\(G(s) = L\left[ {H(t)} \right],H(t) = {L^{ -
1}}\left[ {G(s)} \right]\)
\(G(s) = C{(sI - A)^{ - 1}}B + D\)
对连续时间线性时不变系统\((A, B, C,
D)\)，设初始状态为零，则系统的脉冲响应矩阵为
\(H(t - \tau ) = C{e^{A(t - \tau )}}B +
D\delta (t - \tau )\)
\(= C\Phi (t - \tau )B + D\delta (t - \tau
)\)
①两个代数等价的连续时间线性时不变系统具有相同的脉冲响应矩阵。
②两个代数等价的连续时间线性时不变系统具有相同的“输出零状态响应”和“输出零输入响应”。