参考书籍：《自动控制原理》(第七版).胡寿松主编.
《自动控制原理PDF版下载》
齐次状态方程的解
状态方程为：
x
˙
(
t
)
=
A
x
(
t
)
(41)
\dot{x}(t)=Ax(t)\tag{41}
x˙(t)=Ax(t)(41)
式(41)称为齐次状态方程，通常采用幂级数法和拉普拉斯变换法求解；
幂级数法
设齐次状态方程的解是
t
t
t的向量幂级数，
x
(
t
)
=
b
0
+
b
1
t
+
b
2
t
2
+
⋯
+
b
k
t
k
+
⋯
+
(42)
x(t)=b_0+b_1t+b_2t^2+\dots+b_kt^k+\dots+\tag{42}
x(t)=b0+b1t+b2t2+⋯+bktk+⋯+(42)
式中，
x
,
b
0
,
b
1
,
…
,
b
k
,
…
x,b_0,b_1,\dots,b_k,\dots
x,b0,b1,…,bk,…都是
n
n
n维向量；
x
˙
(
t
)
=
b
1
+
2
b
2
t
+
⋯
+
k
b
k
t
k
−
1
+
⋯
=
A
(
b
0
+
b
1
t
+
b
2
t
2
+
⋯
+
b
k
t
k
+
…
)
(43)
\dot{x}(t)=b_1+2b_2t+\dots+kb_kt^{k-1}+\dots=A(b_0+b_1t+b_2t^2+\dots+b_kt^k+\dots)\tag{43}
x˙(t)=b1+2b2t+⋯+kbktk−1+⋯=A(b0+b1t+b2t2+⋯+bktk+…)(43)
可得：
x
(
t
)
=
(
I
+
A
t
+
1
2
A
2
t
2
+
⋯
+
1
k
!
A
k
t
k
+
⋯
+
)
x
(
0
)
(44)
x(t)=\left(I+At+\frac{1}{2}A^2t^2+\dots+\frac{1}{k!}A^kt^k+\dots+\right)x(0)\tag{44}
x(t)=(I+At+21A2t2+⋯+k!1Aktk+⋯+)x(0)(44)
定义：
e
A
t
=
I
+
A
t
+
1
2
A
2
t
2
+
⋯
+
1
k
!
A
k
t
k
+
⋯
+
=
∑
k
=
0
∞
1
k
!
A
k
t
k
(45)
e^{At}=I+At+\frac{1}{2}A^2t^2+\dots+\frac{1}{k!}A^kt^k+\dots+=\sum_{k=0}^\infty\frac{1}{k!}A^kt^k\tag{45}
eAt=I+At+21A2t2+⋯+k!1Aktk+⋯+=k=0∑∞k!1Aktk(45)
则有：
x
(
t
)
=
e
A
t
x
(
0
)
(46)
x(t)=e^{At}x(0)\tag{46}
x(t)=eAtx(0)(46)
标量微分方程
x
˙
=
a
x
\dot{x}=ax
x˙=ax的解为：
x
(
t
)
=
e
a
t
x
(
0
)
，
e
a
t
x(t)=e^{at}x(0)，e^{at}
x(t)=eatx(0)，eat称为指数函数，向量微分方程具有相似形式的解，故把
e
A
t
e^{At}
eAt称为矩阵指数函数，简称矩阵指数；由于
x
(
t
)
x(t)
x(t)由
x
(
0
)
x(0)
x(0)转移而来，对于线性定常系统，
e
A
t
e^{At}
eAt亦称状态转移矩阵，记为
Φ
(
t
)
\Phi(t)
Φ(t)，即：
Φ
(
t
)
=
e
A
t
(47)
\Phi(t)=e^{At}\tag{47}
Φ(t)=eAt(47)
拉普拉斯变换法
状态方程：
x
˙
(
t
)
=
A
x
(
t
)
\dot{x}(t)=Ax(t)
x˙(t)=Ax(t)
将上式进行拉氏变换，
s
X
(
s
)
=
A
X
(
s
)
+
x
(
0
)
(48)
sX(s)=AX(s)+x(0)\tag{48}
sX(s)=AX(s)+x(0)(48)
则有：
X
(
s
)
=
(
s
I
−
A
)
−
1
x
(
0
)
(49)
X(s)=(sI-A)^{-1}x(0)\tag{49}
X(s)=(sI−A)−1x(0)(49)
进行拉氏反变换：
x
(
t
)
=
L
−
1
[
(
s
I
−
A
)
−
1
]
x
(
0
)
(50)
x(t)=L^{-1}\left[(sI-A)^{-1}\right]x(0)\tag{50}
x(t)=L−1[(sI−A)−1]x(0)(50)
式(46)和式(50)比较：
e
A
t
=
L
−
1
[
(
s
I
−
A
)
−
1
]
e^{At}=L^{-1}\left[(sI-A)^{-1}\right]
eAt=L−1[(sI−A)−1]
状态转移矩阵的运算性质
状态转移矩阵
Φ
(
t
)
\Phi(t)
Φ(t)的幂级数展开式：
Φ
(
t
)
=
e
A
t
=
I
+
A
t
+
1
2
A
2
t
2
+
⋯
+
1
k
!
A
k
t
k
+
…
(51)
\Phi(t)=e^{At}=I+At+\frac{1}{2}A^2t^2+\dots+\frac{1}{k!}A^kt^k+\dots\tag{51}
Φ(t)=eAt=I+At+21A2t2+⋯+k!1Aktk+…(51)
Φ ( 0 ) = I \Phi(0)=I Φ(0)=I；
Φ ˙ ( t ) = A Φ ( t ) = Φ ( t ) A \dot{\Phi}(t)=A\Phi(t)=\Phi(t)A Φ˙(t)=AΦ(t)=Φ(t)A；
Φ ( t 1 ± t 2 ) = Φ ( t 1 ) Φ ( ± t 2 ) = Φ ( ± t 2 ) Φ ( t 1 ) \Phi(t_1±t_2)=\Phi(t_1)\Phi(±t_2)=\Phi(±t_2)\Phi(t_1) Φ(t1±t2)=Φ(t1)Φ(±t2)=Φ(±t2)Φ(t1)；
Φ − 1 ( t ) = Φ ( − t ) , Φ − 1 ( − t ) = Φ ( t ) \Phi^{-1}(t)=\Phi(-t),\Phi^{-1}(-t)=\Phi(t) Φ−1(t)=Φ(−t),Φ−1(−t)=Φ(t)；
x ( t 2 ) = Φ ( t 2 − t 1 ) x ( t 1 ) x(t_2)=\Phi(t_2-t_1)x(t_1) x(t2)=Φ(t2−t1)x(t1)；
Φ ( t 2 − t 0 ) = Φ ( t 2 − t 1 ) Φ ( t 1 − t 0 ) \Phi(t_2-t_0)=\Phi(t_2-t_1)\Phi(t_1-t_0) Φ(t2−t0)=Φ(t2−t1)Φ(t1−t0)；
[ Φ ( t ) ] k = Φ ( k t ) [\Phi(t)]^k=\Phi(kt) [Φ(t)]k=Φ(kt)；
若 Φ ( t ) \Phi(t) Φ(t)为 x ˙ ( t ) = A x ( t ) \dot{x}(t)=Ax(t) x˙(t)=Ax(t)的状态转移矩阵，则引入非奇异变换 x = P x ‾ x=P\overline{x} x=Px后的状态转移矩阵为： Φ ‾ ( t ) = P − 1 e A t P \overline{\Phi}(t)=P^{-1}e^{At}P Φ(t)=P−1eAtP；
两种常见的状态转移矩阵；设
A
=
d
i
a
g
[
λ
1
,
λ
2
,
…
,
λ
n
]
A=diag[\lambda_1,\lambda_2,\dots,\lambda_n]
A=diag[λ1,λ2,…,λn]，即
A
A
A为对角阵，且具有互异元素，则：
Φ
(
t
)
=
[
e
λ
1
t
e
λ
2
t
⋱
e
λ
n
t
]
(52)
\Phi(t)= [eλ1teλ2t⋱eλnt]\tag{52}
Φ(t)=⎣
⎡eλ1teλ2t⋱eλnt⎦
⎤(52)
设
A
A
A阵为
m
×
m
m\times{m}
m×m约当阵：
A
=
[
λ
1
λ
⋱
⋱
1
λ
]
(53)
A= [λ1λ⋱⋱1λ]\tag{53}
A=⎣
⎡λ1λ⋱⋱1λ⎦
⎤(53)
则：
Φ
(
t
)
=
[
e
λ
t
t
e
λ
t
t
2
2
e
λ
t
⋯
t
m
−
1
(
m
−
1
)
!
e
λ
t
0
e
λ
t
t
e
λ
t
⋯
t
m
−
2
(
m
−
2
)
!
e
λ
t
⋮
⋮
⋮
⋮
0
0
0
⋯
t
e
λ
t
0
0
0
⋯
e
λ
t
]
(54)
\Phi(t)= [eλtteλtt22eλt⋯tm−1(m−1)!eλt0eλtteλt⋯tm−2(m−2)!eλt⋮⋮⋮⋮000⋯teλt000⋯eλt]\tag{54}
Φ(t)=⎣
⎡eλt0⋮00teλteλt⋮002t2eλtteλt⋮00⋯⋯⋯⋯(m−1)!tm−1eλt(m−2)!tm−2eλt⋮teλteλt⎦
⎤(54)
实例分析：
Example4： 设状态方程为：
[
x
˙
1
(
t
)
x
˙
2
(
t
)
]
=
[
0
1
−
2
−
3
]
[
x
1
(
t
)
x
2
(
t
)
]
[˙x1(t)˙x2(t)]= [01−2−3] [x1(t)x2(t)]
[x˙1(t)x˙2(t)]=[0−21−3][x1(t)x2(t)]
求状态方程的解.
解：
用拉氏变换求解：
s
I
−
A
=
[
s
0
0
s
]
−
[
0
1
−
2
−
3
]
=
[
s
−
1
2
s
+
3
]
sI-A= [s00s]- [01−2−3]= [s−12s+3]
sI−A=[s00s]−[0−21−3]=[s2−1s+3]
( s I − A ) − 1 = a d j ( s I − A ) ∣ s I − A ∣ = 1 ( s + 1 ) ( s + 2 ) [ s + 3 1 − 2 s ] = [ 2 s + 1 − 1 s + 2 1 s + 1 − 1 s + 2 − 2 s + 1 + 2 s + 2 − 1 s + 1 + 2 s + 2 ] (sI−A)−1=adj(sI−A)|sI−A|=1(s+1)(s+2)[s+31−2s]=[2s+1−1s+21s+1−1s+2−2s+1+2s+2−1s+1+2s+2] (sI−A)−1=∣sI−A∣adj(sI−A)=(s+1)(s+2)1[s+3−21s]=⎣ ⎡s+12−s+21s+1−2+s+22s+11−s+21s+1−1+s+22⎦ ⎤
可得：
Φ
(
t
)
=
L
−
1
[
(
s
I
−
A
)
−
1
]
=
[
2
e
−
t
−
e
−
2
t
e
−
t
−
e
−
2
t
−
2
e
−
t
+
2
e
−
2
t
−
e
−
t
+
2
e
−
2
t
]
\Phi(t)=L^{-1}\left[(sI-A)^{-1}\right]= [2e−t−e−2te−t−e−2t−2e−t+2e−2t−e−t+2e−2t]
Φ(t)=L−1[(sI−A)−1]=[2e−t−e−2t−2e−t+2e−2te−t−e−2t−e−t+2e−2t]
状态方程的解为：
[
x
1
(
t
)
x
2
(
t
)
]
=
Φ
(
t
)
[
x
1
(
0
)
x
2
(
0
)
]
=
[
2
e
−
t
−
e
−
2
t
e
−
t
−
e
−
2
t
−
2
e
−
t
+
2
e
−
2
t
−
e
−
t
+
2
e
−
2
t
]
[
x
1
(
0
)
x
2
(
0
)
]
[x1(t)x2(t)]= \Phi(t)[x1(0)x2(0)]= [2e−t−e−2te−t−e−2t−2e−t+2e−2t−e−t+2e−2t] [x1(0)x2(0)]
[x1(t)x2(t)]=Φ(t)[x1(0)x2(0)]=[2e−t−e−2t−2e−t+2e−2te−t−e−2t−e−t+2e−2t][x1(0)x2(0)]
非齐次状态方程的解
非齐次状态方程如下：
x
˙
(
t
)
=
A
x
(
t
)
+
B
u
(
t
)
(55)
\dot{x}(t)=Ax(t)+Bu(t)\tag{55}
x˙(t)=Ax(t)+Bu(t)(55)
方程的解为：
x
(
t
)
=
Φ
(
t
)
x
(
0
)
+
∫
0
t
Φ
(
t
−
τ
)
B
u
(
τ
)
d
τ
(56)
x(t)=\Phi(t)x(0)+\int_0^t\Phi(t-\tau)Bu(\tau)d\tau\tag{56}
x(t)=Φ(t)x(0)+∫0tΦ(t−τ)Bu(τ)dτ(56)
式中第一项是对初始状态的响应，第二项是对输入作用的响应；
亦可表示为：
x
(
t
)
=
Φ
(
t
)
x
(
0
)
+
∫
0
t
Φ
(
τ
)
B
u
(
t
−
τ
)
d
τ
(57)
x(t)=\Phi(t)x(0)+\int_{0}^t\Phi(\tau)Bu(t-\tau)d\tau\tag{57}
x(t)=Φ(t)x(0)+∫0tΦ(τ)Bu(t−τ)dτ(57)
实例分析：
Example5： 系统状态方程为：
[
x
˙
1
x
˙
2
]
=
[
0
1
−
2
−
3
]
[
x
1
x
2
]
+
[
0
1
]
u
[˙x1˙x2]= [01−2−3] [x1x2]+ [01]u
[x˙1x˙2]=[0−21−3][x1x2]+[01]u
且
x
(
0
)
=
[
x
1
(
0
)
x
2
(
0
)
]
T
x(0)=[x1(0)x2(0)]^T
x(0)=[x1(0)x2(0)]T；求在
u
(
t
)
=
1
(
t
)
u(t)=1(t)
u(t)=1(t)作用下状态方程的解；
解：
由于：
u
(
t
)
=
1
,
u
(
t
−
τ
)
=
1
u(t)=1,u(t-\tau)=1
u(t)=1,u(t−τ)=1，可得：
x
(
t
)
=
Φ
(
t
)
x
(
0
)
+
∫
0
t
Φ
(
t
)
B
d
τ
x(t)=\Phi(t)x(0)+\int_{0}^t\Phi(t)Bd\tau
x(t)=Φ(t)x(0)+∫0tΦ(t)Bdτ
s I − A = [ s 0 0 s ] − [ 0 1 − 2 − 3 ] = [ s − 1 2 s + 3 ] sI-A= [s00s]- [01−2−3]= [s−12s+3] sI−A=[s00s]−[0−21−3]=[s2−1s+3]
( s I − A ) − 1 = a d j ( s I − A ) ∣ s I − A ∣ = 1 ( s + 1 ) ( s + 2 ) [ s + 3 1 − 2 s ] = [ 2 s + 1 − 1 s + 2 1 s + 1 − 1 s + 2 − 2 s + 1 + 2 s + 2 − 1 s + 1 + 2 s + 2 ] (sI−A)−1=adj(sI−A)|sI−A|=1(s+1)(s+2)[s+31−2s]=[2s+1−1s+21s+1−1s+2−2s+1+2s+2−1s+1+2s+2] (sI−A)−1=∣sI−A∣adj(sI−A)=(s+1)(s+2)1[s+3−21s]=⎣ ⎡s+12−s+21s+1−2+s+22s+11−s+21s+1−1+s+22⎦ ⎤
Φ ( t ) = L − 1 [ ( s I − A ) − 1 ] = [ 2 e − t − e − 2 t e − t − e − 2 t − 2 e − t + 2 e − 2 t − e − t + 2 e − 2 t ] \Phi(t)=L^{-1}\left[(sI-A)^{-1}\right]= [2e−t−e−2te−t−e−2t−2e−t+2e−2t−e−t+2e−2t] Φ(t)=L−1[(sI−A)−1]=[2e−t−e−2t−2e−t+2e−2te−t−e−2t−e−t+2e−2t]
∫ 0 t Φ ( τ ) B d τ = ∫ 0 t [ e − τ − e − 2 τ − e − τ + 2 e − 2 τ ] d τ = [ − e − τ + 1 2 e − 2 τ e − τ − e − 2 τ ] ∣ 0 t = [ − e − t + 1 2 e − 2 t + 1 2 e − t − e − 2 t ] \int_0^t\Phi(\tau)Bd\tau=\int_0^t [e−τ−e−2τ−e−τ+2e−2τ]d\tau= \left.[−e−τ+12e−2τe−τ−e−2τ]\right|_0^t= [−e−t+12e−2t+12e−t−e−2t] ∫0tΦ(τ)Bdτ=∫0t[e−τ−e−2τ−e−τ+2e−2τ]dτ=[−e−τ+21e−2τe−τ−e−2τ]∣ ∣0t=[−e−t+21e−2t+21e−t−e−2t]
因此：
x
(
t
)
=
[
x
1
(
t
)
x
2
(
t
)
]
=
[
2
e
−
t
−
e
−
2
t
e
−
t
−
e
−
2
t
−
2
e
−
t
+
2
e
−
2
t
−
e
−
t
+
2
e
−
2
t
]
[
x
1
(
0
)
x
2
(
0
)
]
+
[
−
e
−
t
+
1
2
e
−
2
t
+
1
2
e
−
t
−
e
−
2
t
]
x(t)= [x1(t)x2(t)]= [2e−t−e−2te−t−e−2t−2e−t+2e−2t−e−t+2e−2t] [x1(0)x2(0)]+ [−e−t+12e−2t+12e−t−e−2t]
x(t)=[x1(t)x2(t)]=[2e−t−e−2t−2e−t+2e−2te−t−e−2t−e−t+2e−2t][x1(0)x2(0)]+[−e−t+21e−2t+21e−t−e−2t]