好东西
(A,B)能控 B
[
AB"An 1B行满秩 X=0.
]
充分性：反设(A,B)不能控，对(A,B)进行能控性分解.
A11 1 A=TAT= 0 A12 X1 B1 1 1
,B=TB= 0 ,X=TXT= XA22 3X2
. X4
X ,X =X A B =0 AX=XA,XB=0 A B X
(
B n 1B "A =0 X=0,X=0. A13
A12 0
A22 0X2 A11 X4 0
X2 0
= X4 0A12 0 = A22 0
A11X2+A12X4
A22X4 X2A22
X4A22
)
A11
X = A 0
0 = A X 0
令X4= I, 则A22X4=X4A22,A11X2 X2A22= A12.
X ,X =X A B=0成立. 产生矛盾. 因此(A,B)能控. A
12. 解：x(t)=ex(0)+
At
www.
∞
k
11 1 0 0 0
B= 1 ，AB= 22 1 1 = 1 =B,",AkB=B (A,B)不完全能控.
42 1 1 1 1
0 0
1
x0= 2 ∈Xc 2 = ∫e AtBu(t)dt有解u(t).
2 2
∞
k
0
t ( At)( 1)kk( t) At t
eB=∑=∑ABt=∑B=eB= e ,
kkk!!!k=0k=0k=0 e t
∞
k
0 0 t 1
t
则 2 = ∫ e u(t)dt. 取u(t)= 2e即可.
2 e t
khd
∫
t0
课
= 由λi(A11)≠λi(A22)知上述方程有唯一解.因此存在X 0
eA(t τ)Bu(τ)dτ.
aw.
网
0
后
答
案
com
X2 ≠0使得 I
(A11,B1)完全能控 B1
[
A11B1"A11B1行满秩.
n 1
]