现代控制理论状态转移矩阵.pptx

1、现代控制理论 Modern Control Theory (5),俞 立,浙江工业大学 信息工程学院,2.2 状态转移矩阵,x&(t) = Ax(t) 的状态轨线 x(t) e,= A(tt ),x,0,0,x = x = A(t t ) x,(t ) e,x,1,0,1,1,0,x(t ),x(t ),0,1,x = x = A(t t ) x,(t ) e,x(t ),2,0,2,2,2,0,O,t0,t1,t2,t,定义： (t, t ) = eA(tt ) = (t t ),= x(t) (t t )x,0,0,0,0,0,问题：状态转移矩阵的性质； 状态转移矩阵的计算。,状态转移矩阵

2、：将初始 状态转移到终值状态,1,1,状态转移矩阵的性质, = At = I + A + A2 2 +L+ An n +L,(t) e,t,t,t,2!,n!,&,(0) I,=,= (t) A(t),1.,2. 对任意的t和s， + =,(t s) (t)(s),(t) (s) e e =,A,t,As,1,1,= I + A + A2 2 +L + A + A2 2 +L,(,t,t,)(I s,s,),2!,2!, 1 = I + A + + A2 ,1 2! , 1 3!,1,1,1 3! ,2 + A3 ,3 +,2 +,2 +,3 +L,(t s),t ts s,2 + +,t,t

3、 s ts,s, 2!,2!,2!,1,1,= I + A + + A2 + 2 + A3 + 3 +L,(t s),(t s),(t s),2!,3!,= A(t+s) = +,e,(t s),3. 状态转移矩阵是可逆的，且 1 = , (t) ( t), = At A(t) = = = I,(t) ( t) e e,(t t) (0),根据 + =,(t s) (t)(s),x,x(t ),x(t ),0,1,x = A(t t ) x = x,(t ) e,(t ) (t t ) (t ),2,1,2,1,2,1,1,x(t ),2,x = A(t t ) x = x,(t ) e,(t

4、 ) (t t ) (t ),1,0,1,0,1,0,0,x(t ) = (t t )x(t ),2,2,1,1,O,t0,t1,t2,t,= (t t )(t t )x(t ),2,1,1,0,0,= (t t + t t )x(t ),2,1,1,0,0,= (t t )x(t ),2,0,0,从初始状态x(t )转移到状态x(t )和先从初始状态x(t )转,0,2,0,移到状态x(t ) ，然后再以x(t )作为初始状态转移到状态,1,1,x(t )的效果是相同的。,2,状态转移矩阵包含了自由运动的全部信息。 例 已知系统的状态转移矩阵, ,2t, ,2t,2e e,t ,e e,t

5、,(t) =,t,2t,t,2t, 2e + 2e, e + 2e,求系统的状态矩阵 A,& = 利用 (t) A (t) = A0 = (0) e I, , , 2e t + 2e 2t e t + 2e 2t,& (t) =, ,t,2t,t,2t,2e 4e,e 4e, 0 1 ,A = A(0) = & (0) =, 2 3,状态转移矩阵的计算,1,1,方法1：直接计算法 方法2：线性变换法 （1）对角矩阵,= At = + +,A t L 2 2 + +,A t L n n +,(t) e I At,2!,n!,2 0 ,A =, ,0 3,2,n,1 0 2 0 ,2 0 ,2 0

6、 1,1,eAt =,+,t,+,2 + + t L,n + t L, , ,0 1 0 3 2! 0 3,n! 0 3, ,1,1, ,+ +,2 t,2 2 +L +,n n +L 2 t,0,1 2t,2!,n!,= ,1,1,0,1 ( 3)t,+ + 2 2 +L + n n + L,( 3) t,( 3) t,2!,n!, ,e2t 0 e3t,0,=, ,（2）若矩阵 A 是一个可对角化的矩阵， 即存在非奇异矩阵T，使得：,0 ,1, ,2,TAT D 1 = =,O,0,n,1,A = T DT ，由于,1,2,1,1,1,1,1 2,(T DT) = (T DT)(T DT)

7、 =T D(TT )DT =T D T,1,3 =,1,2,1,=,1,2,1,= 1,2,1,= 1,3,(T DT) (T DT) (T DT) (T D T)(T DT) T D (TT )DT T D T,M,1,n,1,n,(T DT) = T D T,故,1,1,1,At = T DTt = +,I (T DT)t,1,+,(T DT) t L (T DT) t L n!,1,2 2 + +,1,n n +,e e,2!,1,1,1,1,1,2,2,1,n,n,= T IT + T DTt + T D Tt +L + T D Tt + L,2!,n!,1,1, ,= T1 I +

8、D + D2 2 +L + Dn n +L T,t,t,t, ,2!,n!,1 Dt = T e T, , , t, ,e,0,1,0,1, ,2, t 2,e,A = T1DT = T1,T,= T 1,T,O,O, t,0, 0,e ,n,n,例 求以下线性时不变系统的状态转移矩阵, 0 1,x&(t) =,x(t), 2 3, = = ,1,2,解 状态矩阵的特征值是 故可对角化。,1,2,1 0, 2 1,D = TAT 1 =,T =,0 2,1 1, t, ,e,0,e T At = 1,T,因此，,0 e2t, , , 1 1,e,t,0, 2 1,= =,1 2 , 2t,1

9、1 , 0 e , ,t,2t,t,2t ,2e e,e e,2e 2e,t +,2t t +,e 2e,2t,（3）矩阵A 可等价变换为约当块。, 1,0 ,1, 1 1,TAT = J =,1, ,1,O,O 1, , ,0, 1,0 0 1,0 ,1, , + ,1,0 1 0 O O 1, ,1 2,1,J = ,1,2,n1 n2,1 t,t L (n 1)!t, ,O, , , ,1,0, 1,0,0,1,t,O,(n 2)!t,e,Jt =, t 1,e ,k,0 1,0 ,O,M,0 1, ,t, = 0,0 O,0,1,O 1, , ,0,0,方法3：拉普拉斯变换法,At =

10、 I A ,) ,有用的公式：,1,1,e L (s,例 求以下系统的状态转移矩阵：,x 0 1 x ,&,1 =,1, , , 2 3,x& ,x, ,2,2,1, s 0,0 1 , ,由于 故,(s,I A 1 =,) , , ,0 s,2 3, , 2,1 +,1,1 , ,1,s 1 ,+,+ s 1 s 2,+,s 1 s 2, ,=, =,2 s + 3,2,2,1,2,+,+,s +1 s + 2,s +1 s + 2,(t) = e t = L (sI A) ,A,1,1, ,2t, ,2e e,t ,e e,t 2t,=, 2et + 2e2t et + 2e2t,方法4：

11、凯莱哈密尔顿方法,凯莱哈密尔顿定理：对给定的n维矩阵A，特征多项式,det(,I A = n + n1 + L+,),a,a0,n1,则,n +,n1 + + =,A a A L a I 0,n1,0,A = a I a AL a A,n,n1,由以上定理可得,0,1,n1,n+1 = ,a A a A L a A,2 ,n,A,0,1,n1,n+1,A 表示成 A的低于n次幂矩阵的线性组合。,n1,n2,n,n+1,均可用 A , A , , A, I 线性表示。矩阵指数 A , A ,L L,函数eAt 的无穷级数表示式可转化为有限项的和,eAt =0 (t)I (t)A (t)A L,+

12、,+,2 + +,(t)An1,1,2,n1, L ,(t),问题：如何确定系数函数 (t), (t), ,0,1,n1,eAt = 0 (t)I (t)A (t)A L,+,+,2 + +,(t)An1,1,2 an1 2 2,n1,矩阵A的特征多项式： +,n,n,1 + + =,L a 0,0,1,1,t,n n,e =1+ t + t +L + t + L,2,n,矩阵A的任意特征值 满足特征方程，故,n = n 1 L ,a,a a,n1,1,0,+,n n 1 L, , 可以表示成以下项的线性组合： , , , ,1,n 1 n 2,L ,et,n1 n2, ,L, ,1,也可以表

13、示成以下有限项的线性组合：,且具有类似的系数：, (t), (t),L, (t),0,1,n1,因此,t,n1,e = (t) + (t) +L + (t),0,1,n1,eAt = 0 (t)I (t)A (t)A L,+,+,2 + +,(t)An1,1,2,n1, (t), (t), L, (t) 是标量函数。,0,1,n1, L ,对n个特征值 , , , n,1,2, t,n1 1,e = (t) + (t) +L + (t),1,0,1,1,n1, t,n1 2,e = (t) + (t) +L + (t),2,0,1,2,n1,M, t,n1 n,e = (t) + (t) +L

14、 + (t),n,0,1,n,n1, L ,得到了关于未知函数 (t), (t), , (t) 一组方程。,0,1,n1, (t), (t),L, (t) 的确定问题,0,1,n1,转化为一个线性方程组的求解问题。, , e,1 L n (t) ,2 1,1,t,1,0,1,1, , , , ,1 L n, ( ), t,2 2,1,t,e, ,2,1,=,2,2, ,M,M M M O M,M, ,范德蒙行列式,1 L n, ( ), t n,2 n,1,t, ,e , ,n1,n,n,可解的条件是系数行列式不为零。, , , L, 互不相同时，范德蒙行列式不为零。,1,2,n,以上线性方程组有惟一解。, L ,通过求解线性方程组，得到系数函数 (t), (t), , (t),0,1,n1,例 应用凯莱-哈密尔顿方法求以下系统的状态转移矩阵, 0 1,x&(t) =,x(t), 2 3,解 表示式：,eAt = 0 (t)I (t)A,+,1,系统矩阵的特征值是1, 2，范德蒙行列式不为零。, = ,(t) 2e e,t,2t, (t),1,e e,t = ,(t),