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《线性系统理论》作业参考答案
1-8 证1: AB BA
Be
At
B
k 0
1k!
At
kk
k!
k 0
1
BAt
kk
k!
k 0
1
ABt
kk
(
k 0
1k!
At)B eB
kkAt
e AtBeAtx e AteAtBx Bx (t,t0) e x
B(t t0)
而 AB BA e 故
e
At
A B
e e e e
ABBA
e
(A B)(t t0)
e
At0
e e
At
e
A(t t0)
e
B(t t0)
e
At0
e
At
ee
At
At0
e
B(t t0)
e
At0
e
At0
e
B(t t0)
e
At0
e
At0
e
At0
e
B(t t0)B(t t0)
(t,t0)
证毕。
证2:取 ex T(t)x x e 对 e
At
At At
x
两边求导得
AeAtx eAtx AeAt e At eAt e AtBeAtx A BeAtx A B (A B) (A B) 其状态转移矩阵为 (t,t) e(A B)(t t) 即0
根据相似矩阵性质得 (t,t0) T
1
(t) (t,t0)T(t0) e
At
e
(A B)(t t0)
e
At0
1-9 解:由题可得A,B,C的值,则
s
1 300
1(s 1)(s 3)
1
s 10
(s 1)(s 2)(s 3)
1
(s 1)(s 2)
1
s 2
s
(sI A)
1
传递函数阵为
2
s 4s 2 1(s 1)(s 2)(s 3)G(s) C(sI A)B
1
s 2
(s 1)(s 2)(s 3)
1
s 2
2
脉冲响应阵为
1 3(t ) 1 (t ) 2(t )
e 2e e 1
G(t ) L[G(s)] 22
2(t )
e
e
(t )
2ee
2(t )
e
3(t )
2(t )
1-10 方法一:最小公分母为g(s) s(s 1),
2
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