此系列属于胡寿松《自动控制原理题海与考研指导》(第三版)习题精选,仅包含部分经典习题,需要完整版习题答案请自行查找,本系列属于知识点巩固部分,搭配如下几个系列进行学习,可用于期末考试和考研复习。
自动控制原理(第七版)知识提炼
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Example 9.6
已知系统传递函数:
- G ( s ) = ( s − 1 ) ( s + 2 ) ( s + 1 ) ( s − 2 ) ( s + 3 ) G(s)=\displaystyle\frac{(s-1)(s+2)}{(s+1)(s-2)(s+3)} G(s)=(s+1)(s−2)(s+3)(s−1)(s+2);
- G ( s ) = 2 s 3 + s 2 + 7 s s 4 + 3 s 3 + 5 s 2 + 4 s G(s)=\displaystyle\frac{2s^3+s^2+7s}{s^4+3s^3+5s^2+4s} G(s)=s4+3s3+5s2+4s2s3+s2+7s;
- G ( s ) = 3 s 3 + s 2 + s + 1 s 3 + 1 G(s)=\displaystyle\frac{3s^3+s^2+s+1}{s^3+1} G(s)=s3+13s3+s2+s+1;
- G ( s ) = 1 ( s + 3 ) 3 G(s)=\displaystyle\frac{1}{(s+3)^3} G(s)=(s+3)31;
写出系统可控标准型和可观测标准型最小实现。
解:
-
G ( s ) = ( s − 1 ) ( s + 2 ) ( s + 1 ) ( s − 2 ) ( s + 3 ) G(s)=\displaystyle\frac{(s-1)(s+2)}{(s+1)(s-2)(s+3)} G(s)=(s+1)(s−2)(s+3)(s−1)(s+2);
由于系统传递函数的分子、分母不存在零极点对消,故系统可控可观。
由于
G ( s ) = ( s − 1 ) ( s + 2 ) ( s + 1 ) ( s − 2 ) ( s + 3 ) = s 2 + s − 2 s 3 + 2 s 2 − 5 s − 6 G(s)=\displaystyle\frac{(s-1)(s+2)}{(s+1)(s-2)(s+3)}=\frac{s^2+s-2}{s^3+2s^2-5s-6} G(s)=(s+1)(s−2)(s+3)(s−1)(s+2)=s3+2s2−5s−6s2+s−2
则系统可控标准型最小实现为:
x ˙ c = [ 0 1 0 0 0 1 6 5 − 2 ] x c + [ 0 0 1 ] u , y = [ − 2 1 1 ] x c \dot{x}_c=\begin{bmatrix} 0 & 1 & 0\\ 0 & 0 & 1\\ 6 & 5 & -2 \end{bmatrix}x_c+\begin{bmatrix} 0\\0\\1 \end{bmatrix}u,y=\begin{bmatrix} -2 & 1 & 1 \end{bmatrix}x_c x˙c=⎣⎡00610501−2⎦⎤xc+⎣⎡001⎦⎤u,y=[−211]xc
可观测标准型最小实现为可控标准型最小实现的对偶形式:
x ˙ o = [ 0 0 6 1 0 5 0 1 − 2 ] x o + [ − 2 1 1 ] u , y = [ 0 0 1 ] x o \dot{x}_o=\begin{bmatrix} 0 & 0 & 6\\ 1 & 0 & 5\\ 0 & 1 & -2 \end{bmatrix}x_o+\begin{bmatrix} -2\\1\\1 \end{bmatrix}u,y=\begin{bmatrix} 0 & 0 & 1 \end{bmatrix}x_o x˙o=⎣⎡01000165−2⎦⎤xo+⎣⎡−211⎦⎤u,y=[001]xo -
G ( s ) = 2 s 3 + s 2 + 7 s s 4 + 3 s 3 + 5 s 2 + 4 s G(s)=\displaystyle\frac{2s^3+s^2+7s}{s^4+3s^3+5s^2+4s} G(s)=s4+3s3+5s2+4s2s3+s2+7s;
由于传递函数的分子、分母存在零极点对消,故系统不完全可控可观测,因此,最小实现的传递函数为:
G ( s ) = 2 s 2 + s + 7 s 3 + 3 s 2 + 5 s + 4 G(s)=\frac{2s^2+s+7}{s^3+3s^2+5s+4} G(s)=s3+3s2+5s+42s2+s+7
则系统的可控标准型最小实现为:
x ˙ c = [ 0 1 0 0 0 1 − 4 − 5 − 3 ] x c + [ 0 0 1 ] u , y = [ 7 1 2 ] x c \dot{x}_c=\begin{bmatrix} 0 & 1 & 0\\ 0 & 0 & 1\\ -4 & -5 & -3 \end{bmatrix}x_c+\begin{bmatrix} 0\\0\\1 \end{bmatrix}u,y=\begin{bmatrix} 7 & 1 & 2 \end{bmatrix}x_c x˙c=⎣⎡00−410−501−3⎦⎤xc+⎣⎡001⎦⎤u,y=[712]xc
可观测标准型最小实现为可控标准型最小实现的对偶形式:
x ˙ o = [ 0 0 − 4 1 0 − 5 0 1 − 3 ] x o + [ 7 1 2 ] u , y = [ 0 0 1 ] x o \dot{x}_o=\begin{bmatrix} 0 & 0 & -4\\ 1 & 0 & -5\\ 0 & 1 & -3 \end{bmatrix}x_o+\begin{bmatrix} 7\\ 1\\ 2 \end{bmatrix}u,y=\begin{bmatrix} 0 & 0 & 1 \end{bmatrix}x_o x˙o=⎣⎡010001−4−5−3⎦⎤xo+⎣⎡712⎦⎤u,y=[001]xo -
G ( s ) = 3 s 3 + s 2 + s + 1 s 3 + 1 G(s)=\displaystyle\frac{3s^3+s^2+s+1}{s^3+1} G(s)=s3+13s3+s2+s+1;
由于系统传递函数的分子、分母不存在零极点对消,故系统可控可观。
由于
G ( s ) = 3 + s 2 + s − 2 s 3 + 1 G(s)=3+\frac{s^2+s-2}{s^3+1} G(s)=3+s3+1s2+s−2
则系统的可控标准型最小实现为:
x ˙ c = [ 0 1 0 0 0 1 − 1 0 0 ] x c + [ 0 0 1 ] u , y = [ − 2 1 1 ] x c + 3 u \dot{x}_c=\begin{bmatrix} 0 & 1 & 0\\ 0 & 0 & 1\\ -1 & 0 & 0 \end{bmatrix}x_c+\begin{bmatrix} 0\\0\\1 \end{bmatrix}u,y=\begin{bmatrix} -2 & 1 & 1 \end{bmatrix}x_c+3u x˙c=⎣⎡00−1100010⎦⎤xc+⎣⎡001⎦⎤u,y=[−211]xc+3u
可观测标准型最小实现为可控标准型最小实现的对偶形式:
x ˙ o = [ 0 0 − 1 1 0 0 0 1 0 ] x o + [ − 2 1 1 ] u , y = [ 0 0 1 ] x o + 3 u \dot{x}_o=\begin{bmatrix} 0 & 0 & -1\\ 1 & 0 & 0\\ 0 & 1 & 0 \end{bmatrix}x_o+\begin{bmatrix} -2\\ 1\\ 1 \end{bmatrix}u,y=\begin{bmatrix} 0 & 0 & 1 \end{bmatrix}x_o+3u x˙o=⎣⎡010001−100⎦⎤xo+⎣⎡−211⎦⎤u,y=[001]xo+3u -
G ( s ) = 1 ( s + 3 ) 3 G(s)=\displaystyle\frac{1}{(s+3)^3} G(s)=(s+3)31;
由于系统传递函数的分子、分母不存在零极点对消,故系统可控可观。
由于
G ( s ) = 1 s 3 + 9 s 2 + 27 s + 27 G(s)=\frac{1}{s^3+9s^2+27s+27} G(s)=s3+9s2+27s+271
则系统的可控标准型最小实现为:
x ˙ c = [ 0 1 0 0 0 1 − 27 − 27 − 9 ] x c + [ 0 0 1 ] u , y = [ 1 0 0 ] x c \dot{x}_c=\begin{bmatrix} 0 & 1 & 0\\ 0 & 0 & 1\\ -27 & -27 & -9 \end{bmatrix}x_c+\begin{bmatrix} 0\\0\\1 \end{bmatrix}u,y=\begin{bmatrix} 1 & 0 & 0 \end{bmatrix}x_c x˙c=⎣⎡00−2710−2701−9⎦⎤xc+⎣⎡001⎦⎤u,y=[100]xc
可观测标准型最小实现为可控标准型最小实现的对偶形式:
x ˙ o = [ 0 0 − 27 1 0 − 27 0 1 − 9 ] x o + [ 1 0 0 ] u , y = [ 0 0 1 ] x o \dot{x}_o=\begin{bmatrix} 0 & 0 & -27\\ 1 & 0 & -27\\ 0 & 1 & -9 \end{bmatrix}x_o+\begin{bmatrix} 1\\0\\0 \end{bmatrix}u,y=\begin{bmatrix} 0 & 0 & 1 \end{bmatrix}x_o x˙o=⎣⎡010001−27−27−9⎦⎤xo+⎣⎡100⎦⎤u,y=[001]xo
Example 9.7
已知系统的传递函数列向量:
- g ( s ) = [ 1 s + 1 1 s + 2 1 s + 3 ] T g(s)=\begin{bmatrix}\displaystyle\frac{1}{s+1} & \displaystyle\frac{1}{s+2} & \displaystyle\frac{1}{s+3}\end{bmatrix}^T g(s)=[s+11s+21s+31]T;
- g ( s ) = [ 1 s s 2 − 1 1 s − 1 ] T g(s)=\begin{bmatrix}1 & \displaystyle\frac{s}{s^2-1} & \displaystyle\frac{1}{s-1}\end{bmatrix}^T g(s)=[1s2−1ss−11]T;
- g ( s ) = [ 1 s 0 − 1 s 2 + 1 ] T g(s)=\begin{bmatrix}\displaystyle\frac{1}{s} & 0 & \displaystyle\frac{-1}{s^2+1}\end{bmatrix}^T g(s)=[s10s2+1−1]T;
- g ( s ) = [ 1 s 2 + 1 1 s 2 − 1 ] T g(s)=\begin{bmatrix}\displaystyle\frac{1}{s^2+1} & \displaystyle\frac{1}{s^2-1}\end{bmatrix}^T g(s)=[s2+11s2−11]T;
求系统可控标准型实现。
解:
-
g ( s ) = [ 1 s + 1 1 s + 2 1 s + 3 ] T g(s)=\begin{bmatrix}\displaystyle\frac{1}{s+1} & \displaystyle\frac{1}{s+2} & \displaystyle\frac{1}{s+3}\end{bmatrix}^T g(s)=[s+11s+21s+31]T;
引入中间变量 Z Z Z,使得:
g ( s ) = Y ( s ) Z ( s ) Z ( s ) U ( s ) = 1 ( s + 1 ) ( s + 2 ) ( s + 3 ) [ ( s + 2 ) ( s + 3 ) ( s + 1 ) ( s + 3 ) ( s + 1 ) ( s + 2 ) ] = 1 s 3 + 6 s 2 + 11 s + 6 [ s 2 + 5 s + 6 s 2 + 4 s + 3 s 2 + 3 s + 2 ] \begin{aligned} g(s)&=\frac{Y(s)Z(s)}{Z(s)U(s)}=\frac{1}{(s+1)(s+2)(s+3)}\begin{bmatrix} (s+2)(s+3)\\ (s+1)(s+3)\\ (s+1)(s+2) \end{bmatrix}\\\\ &=\frac{1}{s^3+6s^2+11s+6}\begin{bmatrix} s^2+5s+6\\ s^2+4s+3\\ s^2+3s+2 \end{bmatrix} \end{aligned} g(s)=Z(s)U(s)Y(s)Z(s)=(s+1)(s+2)(s+3)1⎣⎡(s+2)(s+3)(s+1)(s+3)(s+1)(s+2)⎦⎤=s3+6s2+11s+61⎣⎡s2+5s+6s2+4s+3s2+3s+2⎦⎤
令
Z ( s ) U ( s ) = 1 s 3 + 6 s 2 + 11 s + 6 , Y ( s ) Z ( s ) = [ s 2 + 5 s + 6 s 2 + 4 s + 3 s 2 + 3 s + 2 ] \frac{Z(s)}{U(s)}=\frac{1}{s^3+6s^2+11s+6},\frac{Y(s)}{Z(s)}=\begin{bmatrix} s^2+5s+6\\ s^2+4s+3\\ s^2+3s+2 \end{bmatrix} U(s)Z(s)=s3+6s2+11s+61,Z(s)Y(s)=⎣⎡s2+5s+6s2+4s+3s2+3s+2⎦⎤
选择 x 1 = z , x 2 = z ˙ , x 3 = z ¨ x_1=z,x_2=\dot{z},x_3=\ddot{z} x1=z,x2=z˙,x3=z¨,可得可控标准型实现为:
x ˙ = [ 0 1 0 0 0 1 − 6 − 11 − 6 ] x + [ 0 0 1 ] u , y = [ 6 5 1 3 4 1 2 3 1 ] x \dot{x}=\begin{bmatrix} 0 & 1 & 0\\ 0 & 0 & 1\\ -6 & -11 & -6 \end{bmatrix}x+\begin{bmatrix} 0\\0\\1 \end{bmatrix}u,y=\begin{bmatrix} 6 & 5 & 1\\ 3 & 4 & 1\\ 2 & 3 & 1 \end{bmatrix}x x˙=⎣⎡00−610−1101−6⎦⎤x+⎣⎡001⎦⎤u,y=⎣⎡632543111⎦⎤x -
g ( s ) = [ 1 s s 2 − 1 1 s − 1 ] T g(s)=\begin{bmatrix}1 & \displaystyle\frac{s}{s^2-1} & \displaystyle\frac{1}{s-1}\end{bmatrix}^T g(s)=[1s2−1ss−11]T;
由于
g ( s ) = [ 1 s s 2 − 1 1 s − 1 ] = 1 s 2 − 1 [ 0 s s + 1 ] + [ 1 0 0 ] g(s)=\begin{bmatrix} 1\\ \displaystyle\frac{s}{s^2-1}\\ \displaystyle\frac{1}{s-1} \end{bmatrix}=\displaystyle\frac{1}{s^2-1}\begin{bmatrix} 0\\ s\\ s+1 \end{bmatrix}+\begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix} g(s)=⎣⎢⎢⎡1s2−1ss−11⎦⎥⎥⎤=s2−11⎣⎡0ss+1⎦⎤+⎣⎡100⎦⎤
利用传递函数直接分解法可得可控标准型实现为:
x ˙ = [ 0 1 1 0 ] x + [ 0 1 ] u , y = [ 0 0 0 1 1 1 ] x + [ 1 0 0 ] u \dot{x}=\begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix}x+\begin{bmatrix} 0\\1 \end{bmatrix}u,y=\begin{bmatrix} 0 & 0\\ 0 & 1\\ 1 & 1 \end{bmatrix}x+\begin{bmatrix} 1\\0\\0 \end{bmatrix}u x˙=[0110]x+[01]u,y=⎣⎡001011⎦⎤x+⎣⎡100⎦⎤u -
g ( s ) = [ 1 s 0 − 1 s 2 + 1 ] T g(s)=\begin{bmatrix}\displaystyle\frac{1}{s} & 0 & \displaystyle\frac{-1}{s^2+1}\end{bmatrix}^T g(s)=[s10s2+1−1]T;
由于
g ( s ) = [ 1 s 0 − 1 s 2 + 1 ] = 1 s 3 + s [ s 2 + 1 0 − s ] g(s)=\begin{bmatrix} \displaystyle\frac{1}{s}\\ 0\\ \displaystyle\frac{-1}{s^2+1} \end{bmatrix}=\frac{1}{s^3+s}\begin{bmatrix} s^2+1\\ 0\\ -s \end{bmatrix} g(s)=⎣⎢⎢⎡s10s2+1−1⎦⎥⎥⎤=s3+s1⎣⎡s2+10−s⎦⎤
利用传递函数直接分解法可得可控标准型实现为:
x ˙ = [ 0 1 0 0 0 1 0 − 1 0 ] x + [ 0 0 1 ] u , y = [ 1 0 1 0 0 0 0 − 1 0 ] x \dot{x}=\begin{bmatrix} 0 & 1 & 0\\ 0 & 0 & 1\\ 0 & -1 & 0 \end{bmatrix}x+\begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix}u,y=\begin{bmatrix} 1 & 0 & 1\\ 0 & 0 & 0\\ 0 & -1 & 0 \end{bmatrix}x x˙=⎣⎡00010−1010⎦⎤x+⎣⎡001⎦⎤u,y=⎣⎡10000−1100⎦⎤x -
g ( s ) = [ 1 s 2 + 1 1 s 2 − 1 ] T g(s)=\begin{bmatrix}\displaystyle\frac{1}{s^2+1} & \displaystyle\frac{1}{s^2-1}\end{bmatrix}^T g(s)=[s2+11s2−11]T;
由于
g ( s ) = [ 1 s 2 + 1 1 s 2 − 1 ] = 1 s 4 − 1 [ s 2 − 1 s 2 + 1 ] g(s)=\begin{bmatrix} \displaystyle\frac{1}{s^2+1}\\ \displaystyle\frac{1}{s^2-1} \end{bmatrix}=\frac{1}{s^4-1}\begin{bmatrix} s^2-1\\ s^2+1 \end{bmatrix} g(s)=⎣⎢⎡s2+11s2−11⎦⎥⎤=s4−11[s2−1s2+1]
利用传递函数直接分解法可得可控标准型实现为:
x ˙ = [ 0 1 0 0 0 0 1 0 0 0 0 1 1 0 0 0 ] x + [ 0 0 0 1 ] u , y = [ − 1 0 1 0 1 0 1 0 ] x \dot{x}=\begin{bmatrix} 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ 1 & 0 & 0 & 0 \end{bmatrix}x+\begin{bmatrix} 0\\ 0\\ 0\\ 1 \end{bmatrix}u,y=\begin{bmatrix} -1 & 0 & 1 & 0\\ 1 & 0 & 1 & 0 \end{bmatrix}x x˙=⎣⎢⎢⎡0001100001000010⎦⎥⎥⎤x+⎣⎢⎢⎡0001⎦⎥⎥⎤u,y=[−11001100]x
Example 9.8
已知系统的传递函数行向量:
- g ( s ) = [ 0 1 1 s + 1 1 s + 2 ] g(s)=\begin{bmatrix}0 & 1 & \displaystyle\frac{1}{s+1} & \displaystyle\frac{1}{s+2}\end{bmatrix} g(s)=[01s+11s+21];
- g ( s ) = [ 1 s + 1 1 s − 1 2 s s 2 − 1 ] g(s)=\begin{bmatrix}\displaystyle\frac{1}{s+1} & \displaystyle\frac{1}{s-1} & \displaystyle\frac{2s}{s^2-1}\end{bmatrix} g(s)=[s+11s−11s2−12s];
- g ( s ) = [ 3 2 s 2 + 1 s 2 s 2 + 1 ] g(s)=\begin{bmatrix}3& \displaystyle\frac{2}{s^2+1} & \displaystyle\frac{s^2}{s^2+1}\end{bmatrix} g(s)=[3s2+12s2+1s2];
求系统可观测标准型实现。
解:
-
g ( s ) = [ 0 1 1 s + 1 1 s + 2 ] g(s)=\begin{bmatrix}0 & 1 & \displaystyle\frac{1}{s+1} & \displaystyle\frac{1}{s+2}\end{bmatrix} g(s)=[01s+11s+21];
由于
g ( s ) = Y ( s ) U ( s ) = 1 ( s + 1 ) ( s + 2 ) [ 0 0 s + 2 s + 1 ] + [ 0 1 0 0 ] g(s)=\frac{Y(s)}{U(s)}=\frac{1}{(s+1)(s+2)}\begin{bmatrix} 0 & 0 & s+2 & s+1 \end{bmatrix}+\begin{bmatrix} 0 & 1 & 0 & 0 \end{bmatrix} g(s)=U(s)Y(s)=(s+1)(s+2)1[00s+2s+1]+[0100]
此为四输入-单输出系统。由于线性系统满足叠加原理,故系统微分方程可写为:
y ¨ + 3 y ˙ + 2 y = u ˙ 3 + 2 u 3 + u ˙ 4 + u 4 \ddot{y}+3\dot{y}+2y=\dot{u}_3+2u_3+\dot{u}_4+u_4 y¨+3y˙+2y=u˙3+2u3+u˙4+u4
令 x 1 = y ˙ + 3 y − u 3 − u 4 , x 2 = y x_1=\dot{y}+3y-u_3-u_4,x_2=y x1=y˙+3y−u3−u4,x2=y,则有
{ x ˙ 1 = − 2 x 2 + 2 u 3 + u 4 x ˙ 2 = x 1 − 3 x 2 + u 3 + u 4 y = x 2 \begin{cases} &\dot{x}_1=-2x_2+2u_3+u_4\\\\ &\dot{x}_2=x_1-3x_2+u_3+u_4\\\\ &y=x_2 \end{cases} ⎩⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎧x˙1=−2x2+2u3+u4x˙2=x1−3x2+u3+u4y=x2
即系统可观测标准型实现的向量-矩阵形式为:
x ˙ = [ 0 − 2 1 − 3 ] x + [ 0 0 2 1 0 0 1 1 ] u , y = [ 0 1 ] x + [ 0 1 0 0 ] u \dot{x}=\begin{bmatrix} 0 & -2\\ 1 & -3 \end{bmatrix}x+\begin{bmatrix} 0 & 0 & 2 & 1\\ 0 & 0 & 1 & 1 \end{bmatrix}u,y=\begin{bmatrix} 0 & 1 \end{bmatrix}x+\begin{bmatrix} 0 & 1 & 0 & 0 \end{bmatrix}u x˙=[01−2−3]x+[00002111]u,y=[01]x+[0100]u -
g ( s ) = [ 1 s + 1 1 s − 1 2 s s 2 − 1 ] g(s)=\begin{bmatrix}\displaystyle\frac{1}{s+1} & \displaystyle\frac{1}{s-1} & \displaystyle\frac{2s}{s^2-1}\end{bmatrix} g(s)=[s+11s−11s2−12s];
系统的传递函数向量:
g ( s ) = Y ( s ) U ( s ) = 1 s 2 − 1 [ s − 1 s 2 + s 2 s ] g(s)=\frac{Y(s)}{U(s)}=\frac{1}{s^2-1}\begin{bmatrix} s-1 & s^2+s & 2s \end{bmatrix} g(s)=U(s)Y(s)=s2−11[s−1s2+s2s]
此为三输入-单输出系统。由于线性系统满足叠加原理,故系统微分方程可写为:
y ¨ − y = u ˙ 1 − u 1 + u ¨ 2 + u ˙ 2 + 2 u ˙ 3 \ddot{y}-y=\dot{u}_1-u_1+\ddot{u}_2+\dot{u}_2+2\dot{u}_3 y¨−y=u˙1−u1+u¨2+u˙2+2u˙3
令 x 1 = y ˙ − u 1 − u ˙ 2 − u 2 − 2 u 3 , x 2 = y − u 2 x_1=\dot{y}-u_1-\dot{u}_2-u_2-2u_3,x_2=y-u_2 x1=y˙−u1−u˙2−u2−2u3,x2=y−u2,则有:
{ x ˙ 1 = x 2 − u 1 + u 2 x ˙ 2 = x 1 + u 1 + u 2 + 2 u 3 y = x 2 + u 2 \begin{cases} &\dot{x}_1=x_2-u_1+u_2\\\\ &\dot{x}_2=x_1+u_1+u_2+2u_3\\\\ &y=x_2+u_2 \end{cases} ⎩⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎧x˙1=x2−u1+u2x˙2=x1+u1+u2+2u3y=x2+u2
即系统可观测标准型实现的向量-矩阵形式为:
x ˙ = [ 0 1 1 0 ] x + [ − 1 1 0 1 1 2 ] u , y = [ 0 1 ] x + [ 0 1 0 ] u \dot{x}=\begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix}x+\begin{bmatrix} -1 & 1 & 0\\ 1 & 1 & 2 \end{bmatrix}u,y=\begin{bmatrix} 0 & 1 \end{bmatrix}x+\begin{bmatrix} 0 & 1 & 0 \end{bmatrix}u x˙=[0110]x+[−111102]u,y=[01]x+[010]u -
g ( s ) = [ 3 2 s 2 + 1 s 2 s 2 + 1 ] g(s)=\begin{bmatrix}3& \displaystyle\frac{2}{s^2+1} & \displaystyle\frac{s^2}{s^2+1}\end{bmatrix} g(s)=[3s2+12s2+1s2];
系统的传递函数向量:
g ( s ) = Y ( s ) U ( s ) = 1 s 2 + 1 [ 3 s 2 + 3 2 s 2 ] g(s)=\frac{Y(s)}{U(s)}=\frac{1}{s^2+1}\begin{bmatrix} 3s^2+3 & 2 & s^2 \end{bmatrix} g(s)=U(s)Y(s)=s2+11[3s2+32s2]
此为三输入-单输出系统。由于线性系统满足叠加原理,故系统微分方程可写为:
y ¨ + y = 3 u ¨ 1 + 3 u 1 + 2 u 2 + u ¨ 3 \ddot{y}+y=3\ddot{u}_1+3u_1+2u_2+\ddot{u}_3 y¨+y=3u¨1+3u1+2u2+u¨3
令 x 1 = y ˙ − 3 u ˙ 1 − u ˙ 3 , x 2 = y − 3 u 1 − u 3 x_1=\dot{y}-3\dot{u}_1-\dot{u}_3,x_2=y-3u_1-u_3 x1=y˙−3u˙1−u˙3,x2=y−3u1−u3,则有:
{ x ˙ 1 = − x 2 + 2 u 2 − u 3 x ˙ 2 = x 1 y = x 2 + 3 u 1 + u 3 \begin{cases} &\dot{x}_1=-x_2+2u_2-u_3\\\\ &\dot{x}_2=x_1\\\\ &y=x_2+3u_1+u_3 \end{cases} ⎩⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎧x˙1=−x2+2u2−u3x˙2=x1y=x2+3u1+u3
即系统可观测标准型实现的向量-矩阵形式为:
x ˙ = [ 0 − 1 1 0 ] x + [ 0 2 − 1 0 0 0 ] u , y = [ 0 1 ] x + [ 3 0 1 ] u \dot{x}=\begin{bmatrix} 0 & -1\\ 1 & 0 \end{bmatrix}x+\begin{bmatrix} 0 & 2 & -1\\ 0 & 0 & 0 \end{bmatrix}u,y=\begin{bmatrix} 0 & 1 \end{bmatrix}x+\begin{bmatrix} 3 & 0 & 1 \end{bmatrix}u x˙=[01−10]x+[0020−10]u,y=[01]x+[301]u
Example 9.9
已知系统传递函数:
- G ( s ) = s 2 + s − 6 s 4 + 10 s 3 + 35 s 2 + 50 s + 24 G(s)=\displaystyle\frac{s^2+s-6}{s^4+10s^3+35s^2+50s+24} G(s)=s4+10s3+35s2+50s+24s2+s−6;
- G ( s ) = 4 s 2 + 17 s + 16 s 3 + 7 s 2 + 16 s + 12 G(s)=\displaystyle\frac{4s^2+17s+16}{s^3+7s^2+16s+12} G(s)=s3+7s2+16s+124s2+17s+16;
- G ( s ) = 2 s 2 + 5 s + 1 s 3 + 6 s 2 + 12 s + 8 G(s)=\displaystyle\frac{2s^2+5s+1}{s^3+6s^2+12s+8} G(s)=s3+6s2+12s+82s2+5s+1;
- G ( s ) = ( s + 1 ) 3 s 3 G(s)=\displaystyle\frac{(s+1)^3}{s^3} G(s)=s3(s+1)3;
求系统的约当标准型或对角线标准型实现。
解:
-
G ( s ) = s 2 + s − 6 s 4 + 10 s 3 + 35 s 2 + 50 s + 24 G(s)=\displaystyle\frac{s^2+s-6}{s^4+10s^3+35s^2+50s+24} G(s)=s4+10s3+35s2+50s+24s2+s−6;
将系统传递函数分解为部分分式形式:
G ( s ) = Y ( s ) U ( s ) = ( s + 3 ) ( s − 2 ) ( s + 1 ) ( s + 2 ) ( s + 3 ) ( s + 4 ) = − 1 s + 1 + 2 s + 2 + − 1 s + 4 G(s)=\frac{Y(s)}{U(s)}=\frac{(s+3)(s-2)}{(s+1)(s+2)(s+3)(s+4)}=\frac{-1}{s+1}+\frac{2}{s+2}+\frac{-1}{s+4} G(s)=U(s)Y(s)=(s+1)(s+2)(s+3)(s+4)(s+3)(s−2)=s+1−1+s+22+s+4−1
令
{ X 1 ( s ) = 1 s + 1 U ( s ) X 2 ( s ) = 1 s + 2 U ( s ) X 3 ( s ) = 1 s + 4 U ( s ) \begin{cases} &X_1(s)=\displaystyle\frac{1}{s+1}U(s)\\\\ &X_2(s)=\displaystyle\frac{1}{s+2}U(s)\\\\ &X_3(s)=\displaystyle\frac{1}{s+4}U(s) \end{cases} ⎩⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎧X1(s)=s+11U(s)X2(s)=s+21U(s)X3(s)=s+41U(s)
即
{ x ˙ 1 = − x 1 + u x ˙ 2 = − 2 x 2 + u x ˙ 3 = − 4 x 3 + u \begin{cases} &\dot{x}_1=-x_1+u\\\\ &\dot{x}_2=-2x_2+u\\\\ &\dot{x}_3=-4x_3+u \end{cases} ⎩⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎧x˙1=−x1+ux˙2=−2x2+ux˙3=−4x3+u
向量-矩阵形式,可得系统对角线标准型最小实现为:
x ˙ = [ − 1 0 0 0 − 2 0 0 0 − 4 ] x + [ 1 1 1 ] u , y = [ − 1 2 − 1 ] x \dot{x}=\begin{bmatrix} -1 & 0 & 0\\ 0 & -2 & 0\\ 0 & 0 & -4 \end{bmatrix}x+\begin{bmatrix} 1\\ 1\\ 1 \end{bmatrix}u,y=\begin{bmatrix} -1 & 2 & -1 \end{bmatrix}x x˙=⎣⎡−1000−2000−4⎦⎤x+⎣⎡111⎦⎤u,y=[−12−1]x -
G ( s ) = 4 s 2 + 17 s + 16 s 3 + 7 s 2 + 16 s + 12 G(s)=\displaystyle\frac{4s^2+17s+16}{s^3+7s^2+16s+12} G(s)=s3+7s2+16s+124s2+17s+16;
将系统传递函数分解为部分分式形式:
G ( s ) = Y ( s ) U ( s ) = 4 s 2 + 17 s + 16 ( s + 3 ) ( s + 2 ) 2 = − 2 ( s + 2 ) 2 + 3 s + 2 + 1 s + 3 G(s)=\frac{Y(s)}{U(s)}=\frac{4s^2+17s+16}{(s+3)(s+2)^2}=\frac{-2}{(s+2)^2}+\frac{3}{s+2}+\frac{1}{s+3} G(s)=U(s)Y(s)=(s+3)(s+2)24s2+17s+16=(s+2)2−2+s+23+s+31
令
{ X 1 ( s ) = 1 ( s + 2 ) 2 U ( s ) = 1 s + 2 X 2 ( s ) X 2 ( s ) = 1 s + 2 U ( s ) X 3 ( s ) = 1 s + 3 U ( s ) \begin{cases} &X_1(s)=\displaystyle\frac{1}{(s+2)^2}U(s)=\displaystyle\frac{1}{s+2}X_2(s)\\\\ &X_2(s)=\displaystyle\frac{1}{s+2}U(s)\\\\ &X_3(s)=\displaystyle\frac{1}{s+3}U(s) \end{cases} ⎩⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎧X1(s)=(s+2)21U(s)=s+21X2(s)X2(s)=s+21U(s)X3(s)=s+31U(s)
则
{ x ˙ 1 = − 2 x 1 + x 2 x ˙ 2 = − 2 x 2 + u x ˙ 3 = − 3 x 3 + u \begin{cases} &\dot{x}_1=-2x_1+x_2\\\\ &\dot{x}_2=-2x_2+u\\\\ &\dot{x}_3=-3x_3+u \end{cases} ⎩⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎧x˙1=−2x1+x2x˙2=−2x2+ux˙3=−3x3+u
向量-矩阵形式,可得系统约当标准型实现为:
x ˙ = [ − 2 1 0 0 − 2 0 0 0 − 3 ] x + [ 0 1 1 ] u , y = [ − 2 3 1 ] x \dot{x}=\begin{bmatrix} -2 & 1 & 0\\ 0 & -2 & 0\\ 0 & 0 & -3 \end{bmatrix}x+\begin{bmatrix} 0\\ 1\\ 1 \end{bmatrix}u,y=\begin{bmatrix} -2 & 3 & 1 \end{bmatrix}x x˙=⎣⎡−2001−2000−3⎦⎤x+⎣⎡011⎦⎤u,y=[−231]x -
G ( s ) = 2 s 2 + 5 s + 1 s 3 + 6 s 2 + 12 s + 8 G(s)=\displaystyle\frac{2s^2+5s+1}{s^3+6s^2+12s+8} G(s)=s3+6s2+12s+82s2+5s+1;
将系统传递函数分解为部分分式形式:
G ( s ) = Y ( s ) U ( s ) = 2 s 2 + 5 s + 1 ( s + 2 ) 3 = − 1 ( s + 2 ) 3 + − 3 ( s + 2 ) 2 + 2 s + 2 G(s)=\frac{Y(s)}{U(s)}=\frac{2s^2+5s+1}{(s+2)^3}=\frac{-1}{(s+2)^3}+\frac{-3}{(s+2)^2}+\frac{2}{s+2} G(s)=U(s)Y(s)=(s+2)32s2+5s+1=(s+2)3−1+(s+2)2−3+s+22
令
{ X 1 ( s ) = 1 ( s + 2 ) 3 U ( s ) = 1 s + 2 X 2 ( s ) X 2 ( s ) = 1 ( s + 2 ) 2 U ( s ) = 1 s + 2 X 3 ( s ) X 3 ( s ) = 1 s + 2 U ( s ) \begin{cases} &X_1(s)=\displaystyle\frac{1}{(s+2)^3}U(s)=\frac{1}{s+2}X_2(s)\\\\ &X_2(s)=\displaystyle\frac{1}{(s+2)^2}U(s)=\frac{1}{s+2}X_3(s)\\\\ &X_3(s)=\displaystyle\frac{1}{s+2}U(s) \end{cases} ⎩⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎧X1(s)=(s+2)31U(s)=s+21X2(s)X2(s)=(s+2)21U(s)=s+21X3(s)X3(s)=s+21U(s)
则
{ x ˙ 1 = − 2 x 1 + x 2 x ˙ 2 = − 2 x 2 + x 3 x ˙ 3 = − 2 x 3 + u \begin{cases} &\dot{x}_1=-2x_1+x_2\\\\ &\dot{x}_2=-2x_2+x_3\\\\ &\dot{x}_3=-2x_3+u \end{cases} ⎩⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎧x˙1=−2x1+x2x˙2=−2x2+x3x˙3=−2x3+u
向量-矩阵形式,可得系统的约当标准型实现为:
x ˙ = [ − 2 1 0 0 − 2 1 0 0 − 2 ] x + [ 0 0 1 ] u , y = [ − 1 − 3 2 ] x \dot{x}=\begin{bmatrix} -2 & 1 & 0\\ 0 & -2 & 1\\ 0 & 0 & -2 \end{bmatrix}x+\begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix}u,y=\begin{bmatrix} -1 & -3 & 2 \end{bmatrix}x x˙=⎣⎡−2001−2001−2⎦⎤x+⎣⎡001⎦⎤u,y=[−1−32]x -
G ( s ) = ( s + 1 ) 3 s 3 G(s)=\displaystyle\frac{(s+1)^3}{s^3} G(s)=s3(s+1)3;
将系统的传递函数分解为部分分式形式:
G ( s ) = Y ( s ) U ( s ) = ( s + 1 ) 3 s 3 = 1 + 3 s 2 + 3 s + 1 s 3 = 1 + 1 s 3 + 3 s 2 + 3 s G(s)=\frac{Y(s)}{U(s)}=\frac{(s+1)^3}{s^3}=1+\frac{3s^2+3s+1}{s^3}=1+\frac{1}{s^3}+\frac{3}{s^2}+\frac{3}{s} G(s)=U(s)Y(s)=s3(s+1)3=1+s33s2+3s+1=1+s31+s23+s3
令
{ X 1 ( s ) = 1 s 3 U ( s ) = 1 s X 2 ( s ) X 2 ( s ) = 1 s 2 U ( s ) = 1 s X 3 ( s ) X 3 ( s ) = 1 s U ( s ) \begin{cases} &X_1(s)=\displaystyle\frac{1}{s^3}U(s)=\frac{1}{s}X_2(s)\\\\ &X_2(s)=\displaystyle\frac{1}{s^2}U(s)=\frac{1}{s}X_3(s)\\\\ &X_3(s)=\displaystyle\frac{1}{s}U(s) \end{cases} ⎩⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎧X1(s)=s31U(s)=s1X2(s)X2(s)=s21U(s)=s1X3(s)X3(s)=s1U(s)
则
{ x ˙ 1 = x 2 x ˙ 2 = x 3 x ˙ 3 = u \begin{cases} &\dot{x}_1=x_2\\\\ &\dot{x}_2=x_3\\\\ &\dot{x}_3=u \end{cases} ⎩⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎧x˙1=x2x˙2=x3x˙3=u
向量-矩阵形式,可得系统的约当标准型实现为:
x ˙ = [ 0 1 0 0 0 1 0 0 0 ] x + [ 0 0 1 ] u , y = [ 1 3 3 ] x + u \dot{x}=\begin{bmatrix} 0 & 1 & 0\\ 0 & 0 & 1\\ 0 & 0 & 0 \end{bmatrix}x+\begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix}u,y=\begin{bmatrix} 1 & 3 & 3 \end{bmatrix}x+u x˙=⎣⎡000100010⎦⎤x+⎣⎡001⎦⎤u,y=[133]x+u
Example 9.10
已知系统状态矩阵为: A = [ 0 1 0 0 0 1 2 3 0 ] A=\begin{bmatrix}0&1&0\\0&0&1\\2&3&0\end{bmatrix} A=⎣⎡002103010⎦⎤,用下列方法求系统的状态转移矩阵 Φ ( t ) \Phi(t) Φ(t)。
- 拉普拉斯变换法;
- 线性变换法;
- 凯莱-哈密顿定理(待定系数法);
解:
-
拉普拉斯变换法;
因为
( s I − A ) − 1 = [ s − 1 0 0 s − 1 − 2 − 3 s ] − 1 = 1 ( s + 1 ) 2 ( s − 2 ) [ s 2 − 3 s 1 − 2 s 2 s 2 s − 3 s − 2 s 2 ] (sI-A)^{-1}=\begin{bmatrix} s & -1 & 0\\ 0 & s & -1\\ -2 & -3 & s \end{bmatrix}^{-1}=\frac{1}{(s+1)^2(s-2)}\begin{bmatrix} s^2-3 & s & 1\\ -2 & s^2 & s\\ 2s & -3s-2 & s^2 \end{bmatrix} (sI−A)−1=⎣⎡s0−2−1s−30−1s⎦⎤−1=(s+1)2(s−2)1⎣⎡s2−3−22sss2−3s−21ss2⎦⎤
所以
Φ ( t ) = L − 1 [ ( s I − A ) − 1 ] = 1 9 [ ( 8 − 6 t ) e − t + e 2 t ( − 2 + 3 t ) e − t + 2 e 2 t − ( 1 + 3 t ) e − t + e 2 t − ( 2 + 6 t ) e − t + 2 e 2 t ( 5 − 3 t ) e − t + 4 e 2 t ( − 2 + 3 t ) e − t + 2 e 2 t ( − 4 + 6 t ) e − t + 4 e 2 t ( − 8 + 3 t ) e − t + 8 e 2 t ( 5 − 3 t ) e − t + 4 e 2 t ] \begin{aligned} \Phi(t)&=L^{-1}[(sI-A)^{-1}]\\\\ &=\frac{1}{9}\begin{bmatrix} (8-6t){\rm e}^{-t}+{\rm e}^{2t} & (-2+3t){\rm e}^{-t}+2{\rm e}^{2t} & -(1+3t){\rm e}^{-t}+{\rm e}^{2t}\\\\ -(2+6t){\rm e}^{-t}+2{\rm e}^{2t} & (5-3t){\rm e}^{-t}+4{\rm e}^{2t} & (-2+3t){\rm e}^{-t}+2{\rm e}^{2t}\\\\ (-4+6t){\rm e}^{-t}+4{\rm e}^{2t} & (-8+3t){\rm e}^{-t}+8{\rm e}^{2t} & (5-3t){\rm e}^{-t}+4{\rm e}^{2t} \end{bmatrix} \end{aligned} Φ(t)=L−1[(sI−A)−1]=91⎣⎢⎢⎢⎢⎡(8−6t)e−t+e2t−(2+6t)e−t+2e2t(−4+6t)e−t+4e2t(−2+3t)e−t+2e2t(5−3t)e−t+4e2t(−8+3t)e−t+8e2t−(1+3t)e−t+e2t(−2+3t)e−t+2e2t(5−3t)e−t+4e2t⎦⎥⎥⎥⎥⎤ -
线性变换法;
由于 A A A为友矩阵,故系统特征多项式为:
f ( λ ) = λ 3 − 3 λ − 2 = ( λ + 1 ) 2 ( λ − 2 ) f(\lambda)=\lambda^3-3\lambda-2=(\lambda+1)^2(\lambda-2) f(λ)=λ3−3λ−2=(λ+1)2(λ−2)
解得特征根为:
λ 1 = λ 2 = − 1 , λ 3 = 2 \lambda_1=\lambda_2=-1,\lambda_3=2 λ1=λ2=−1,λ3=2
因此:
P = [ 1 0 1 λ 1 1 λ 3 λ 1 2 2 λ 1 λ 3 2 ] = [ 1 0 1 − 1 1 2 1 − 2 4 ] , P − 1 = 1 9 [ 8 − 2 − 1 6 3 − 3 1 2 1 ] Λ = P − 1 A P = [ − 1 1 0 0 − 1 0 0 0 2 ] , e A t = [ e − t t e − t 0 0 e − t 0 0 0 e 2 t ] \begin{aligned} &P=\begin{bmatrix} 1 & 0 & 1\\ \lambda_1 & 1 &\lambda_3\\ \lambda_1^2 & 2\lambda_1 & \lambda_3^2 \end{bmatrix}=\begin{bmatrix} 1 & 0 & 1\\ -1 & 1 & 2\\ 1 & -2 & 4 \end{bmatrix},P^{-1}=\frac{1}{9}\begin{bmatrix} 8 & -2 & -1\\ 6 & 3 & -3\\ 1 & 2 & 1 \end{bmatrix}\\\\ &\Lambda=P^{-1}AP=\begin{bmatrix} -1 & 1 & 0\\ 0 & -1 & 0\\ 0 & 0 & 2 \end{bmatrix},{\rm e}^{At}=\begin{bmatrix} {\rm e}^{-t} & t{\rm e}^{-t} & 0\\ 0 & {\rm e}^{-t} & 0\\ 0 & 0 & {\rm e}^{2t} \end{bmatrix} \end{aligned} P=⎣⎡1λ1λ12012λ11λ3λ32⎦⎤=⎣⎡1−1101−2124⎦⎤,P−1=91⎣⎡861−232−1−31⎦⎤Λ=P−1AP=⎣⎡−1001−10002⎦⎤,eAt=⎣⎡e−t00te−te−t000e2t⎦⎤
从而可得系统的状态转移矩阵为:
Φ ( t ) = P e Λ t P − 1 = 1 9 [ ( 8 − 6 t ) e − t + e 2 t ( − 2 + 3 t ) e − t + 2 e 2 t − ( 1 + 3 t ) e − t + e 2 t − ( 2 + 6 t ) e − t + 2 e 2 t ( 5 − 3 t ) e − t + 4 e 2 t ( − 2 + 3 t ) e − t + 2 e 2 t ( − 4 + 6 t ) e − t + 4 e 2 t ( − 8 + 3 t ) e − t + 8 e 2 t ( 5 − 3 t ) e − t + 4 e 2 t ] \begin{aligned} \Phi(t)&=P{\rm e}^{\Lambda{t}}P^{-1}\\\\ &=\frac{1}{9}\begin{bmatrix} (8-6t){\rm e}^{-t}+{\rm e}^{2t} & (-2+3t){\rm e}^{-t}+2{\rm e}^{2t} & -(1+3t){\rm e}^{-t}+{\rm e}^{2t}\\\\ -(2+6t){\rm e}^{-t}+2{\rm e}^{2t} & (5-3t){\rm e}^{-t}+4{\rm e}^{2t} & (-2+3t){\rm e}^{-t}+2{\rm e}^{2t}\\\\ (-4+6t){\rm e}^{-t}+4{\rm e}^{2t} & (-8+3t){\rm e}^{-t}+8{\rm e}^{2t} & (5-3t){\rm e}^{-t}+4{\rm e}^{2t} \end{bmatrix} \end{aligned} Φ(t)=PeΛtP−1=91⎣⎢⎢⎢⎢⎡(8−6t)e−t+e2t−(2+6t)e−t+2e2t(−4+6t)e−t+4e2t(−2+3t)e−t+2e2t(5−3t)e−t+4e2t(−8+3t)e−t+8e2t−(1+3t)e−t+e2t(−2+3t)e−t+2e2t(5−3t)e−t+4e2t⎦⎥⎥⎥⎥⎤ -
凯莱-哈密顿定理(待定系数法);
根据凯莱-哈密顿定理可知:
e A t = ∑ k = 0 n − 1 α k ( t ) A k = α 0 ( t ) I + α 1 ( t ) A + α 2 ( t ) A 2 {\rm e}^{At}=\sum_{k=0}^{n-1}\alpha_k(t)A^k=\alpha_0(t)I+\alpha_1(t)A+\alpha_2(t)A^2 eAt=k=0∑n−1αk(t)Ak=α0(t)I+α1(t)A+α2(t)A2
由于 A A A的特征值为: λ 1 = λ 2 = − 1 , λ 3 = 2 \lambda_1=\lambda_2=-1,\lambda_3=2 λ1=λ2=−1,λ3=2,因此:
[ α 0 ( t ) α 1 ( t ) α 2 ( t ) ] = [ 1 λ 1 λ 1 2 0 1 2 λ 1 1 λ 3 λ 3 2 ] − 1 [ e λ 1 t t e λ 1 t e λ 3 t ] = [ 1 − 1 1 0 1 − 2 1 2 4 ] − 1 [ e λ 1 t t e λ 1 t e λ 3 t ] = 1 9 [ e − t + 6 t e − t + e 2 t − 2 e − t + 3 t e − t + 2 e 2 t − e − t − 3 t e − t + e 2 t ] \begin{bmatrix} \alpha_0(t)\\ \alpha_1(t)\\ \alpha_2(t) \end{bmatrix}=\begin{bmatrix} 1 & \lambda_1 & \lambda_1^2\\ 0 & 1 & 2\lambda_1\\ 1 & \lambda_3 & \lambda_3^2 \end{bmatrix}^{-1}\begin{bmatrix} {\rm e}^{\lambda_1t}\\ t{\rm e}^{\lambda_1t}\\ {\rm e}^{\lambda_3t} \end{bmatrix}=\begin{bmatrix} 1 & -1 & 1\\ 0 & 1 & -2\\ 1 & 2 & 4 \end{bmatrix}^{-1}\begin{bmatrix} {\rm e}^{\lambda_1t}\\ t{\rm e}^{\lambda_1t}\\ {\rm e}^{\lambda_3t} \end{bmatrix}=\frac{1}{9}\begin{bmatrix} {\rm e}^{-t}+6t{\rm e}^{-t}+{\rm e}^{2t}\\ -2{\rm e}^{-t}+3t{\rm e}^{-t}+2{\rm e}^{2t}\\ -{\rm e}^{-t}-3t{\rm e}^{-t}+{\rm e}^{2t} \end{bmatrix} ⎣⎡α0(t)α1(t)α2(t)⎦⎤=⎣⎡101λ11λ3λ122λ1λ32⎦⎤−1⎣⎡eλ1tteλ1teλ3t⎦⎤=⎣⎡101−1121−24⎦⎤−1⎣⎡eλ1tteλ1teλ3t⎦⎤=91⎣⎡e−t+6te−t+e2t−2e−t+3te−t+2e2t−e−t−3te−t+e2t⎦⎤
将上式代入可得:
Φ ( t ) = α 0 ( t ) I + α 1 A + α 2 A 2 = 1 9 [ ( 8 − 6 t ) e − t + e 2 t ( − 2 + 3 t ) e − t + 2 e 2 t − ( 1 + 3 t ) e − t + e 2 t − ( 2 + 6 t ) e − t + 2 e 2 t ( 5 − 3 t ) e − t + 4 e 2 t ( − 2 + 3 t ) e − t + 2 e 2 t ( − 4 + 6 t ) e − t + 4 e 2 t ( − 8 + 3 t ) e − t + 8 e 2 t ( 5 − 3 t ) e − t + 4 e 2 t ] \begin{aligned} \Phi(t)&=\alpha_0(t)I+\alpha_1A+\alpha_2A^2\\\\ &=\frac{1}{9}\begin{bmatrix} (8-6t){\rm e}^{-t}+{\rm e}^{2t} & (-2+3t){\rm e}^{-t}+2{\rm e}^{2t} & -(1+3t){\rm e}^{-t}+{\rm e}^{2t}\\\\ -(2+6t){\rm e}^{-t}+2{\rm e}^{2t} & (5-3t){\rm e}^{-t}+4{\rm e}^{2t} & (-2+3t){\rm e}^{-t}+2{\rm e}^{2t}\\\\ (-4+6t){\rm e}^{-t}+4{\rm e}^{2t} & (-8+3t){\rm e}^{-t}+8{\rm e}^{2t} & (5-3t){\rm e}^{-t}+4{\rm e}^{2t} \end{bmatrix} \end{aligned} Φ(t)=α0(t)I+α1A+α2A2=91⎣⎢⎢⎢⎢⎡(8−6t)e−t+e2t−(2+6t)e−t+2e2t(−4+6t)e−t+4e2t(−2+3t)e−t+2e2t(5−3t)e−t+4e2t(−8+3t)e−t+8e2t−(1+3t)e−t+e2t(−2+3t)e−t+2e2t(5−3t)e−t+4e2t⎦⎥⎥⎥⎥⎤