1. 何为状态转移矩阵
一般地,对于一个线性定常系统,可以写成如下的柯西标准型形式
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\begin{cases} \dot x (t) = A(t) x(t) + B(t) u(t) \\ y(t) = C(t) x(t) + D(t) u(t) \end{cases}
{x˙(t)=A(t)x(t)+B(t)u(t)y(t)=C(t)x(t)+D(t)u(t)其中
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x(t), u(t),y(t)
x(t),u(t),y(t)分别为系统状态和、系统输入控制量和输出量,为向量。该方程组的解为
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(1)
x(t) = e^{At} x_0 = \left( I + At + \frac{A^2 t^2} {2!} + \cdots \right) x_0 = \Phi (t) x_0 \tag{1}
x(t)=eAtx0=(I+At+2!A2t2+⋯)x0=Φ(t)x0(1)其中
I
I
I为单位矩阵。
式(1)中的
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\Phi (t)
Φ(t)即为状态转移矩阵。
2. 状态转移矩阵计算举例
从式(1)可以看出,状态转移矩阵
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\Phi (t)=e^{At}
Φ(t)=eAt虽然具有指数形式,但由于
A
A
A是矩阵,因而其计算并不是简单地将
A
A
A中所有元素变为
e
e
e的指数就行的。相反,其计算需要具有式(1)的形式:
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(2)
\Phi (t) = e^{At} = I + At + \frac{A^2 t^2} {2!} + \frac{A^3 t^3} {3!} + \cdots + \frac{A^i t^i} {i!} + \cdots \tag{2}
Φ(t)=eAt=I+At+2!A2t2+3!A3t3+⋯+i!Aiti+⋯(2)以下列传递函数为例,进行状态转移矩阵的计算。
例:
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W(s) = \frac{s+1}{s^2 + 12s + 32}
W(s)=s2+12s+32s+1可以写出其对应的状态空间形式:
A
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−
12
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32
1
0
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,
A = \left[ \begin{matrix} -12 & -32 \\ 1 & 0 \end{matrix} \right],
A=[−121−320],
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1
0
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,
B = \left[ \begin{matrix} 1 \\ 0 \end{matrix} \right],
B=[10],
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1
1
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,
C = \left[ \begin{matrix} 1 & 1 \end{matrix} \right],
C=[11],
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D = 0,
D=0,则
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A^2 = \left[ \begin{matrix} 112 & 384 \\ -12 & -32 \end{matrix} \right], A^3 = \left[ \begin{matrix} -960 & -3584 \\ 112 & 384 \end{matrix} \right], \quad \cdots
A2=[112−12384−32],A3=[−960112−3584384],⋯代入式(2)有
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\begin{aligned} \Phi (t) &= e^{At} = I + At + \frac{A^2 t^2} {2!} + \frac{A^3 t^3} {3!} + \cdots + \frac{A^i t^i} {i!} + \cdots \\ &= \left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] + \left[ \begin{matrix} -12 & -32 \\ 1 & 0 \end{matrix} \right] t + \frac{1}{2} \left[ \begin{matrix} 112 & 384 \\ -12 & -32 \end{matrix} \right] t^2 + \frac{1}{6} \left[ \begin{matrix} -960 & -3584 \\ 112 & 384 \end{matrix} \right] t^3 + \cdots \\ &= \left[ \begin{matrix} 1-12t + 56t^2 - 160t^3 + \cdots & -32t + 192t^2 - \frac{1792}{3} t^3 + \cdots \\ t - 6t^2 + \frac{56}{3} t^3 + \cdots & 1 - 16t^2 + 64t^3 + \cdots \end{matrix} \right] \end{aligned}
Φ(t)=eAt=I+At+2!A2t2+3!A3t3+⋯+i!Aiti+⋯=[1001]+[−121−320]t+21[112−12384−32]t2+61[−960112−3584384]t3+⋯=[1−12t+56t2−160t3+⋯t−6t2+356t3+⋯−32t+192t2−31792t3+⋯1−16t2+64t3+⋯]
3. 离散系统的状态方程
对于离散系统
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\begin{cases} x (k+1) = A_d(k) x(k) + B_d(k) u(k) \\ y(k) = C_d(k) x(k) + D_d(k) u(k) \end{cases}
{x(k+1)=Ad(k)x(k)+Bd(k)u(k)y(k)=Cd(k)x(k)+Dd(k)u(k)其中
k
k
k指量化后的第
k
k
k时刻,即:若离散化的每个时间单位为
T
0
T_0
T0,则
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x(k+1)
x(k+1)即指在
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T
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(k+1)T_0
(k+1)T0时刻下的状态量。
在离散化的系统下,矩阵
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A_d, B_d, C_d, D_d
Ad,Bd,Cd,Dd与原连续系统下的
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A, B, C, D
A,B,C,D具有如下转换关系:
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(3)
A_d = \Phi \left( T_0 \right) = e^{A T_0 }, \quad B_d = \int _0 ^{T_0} \Phi (q) B dq, \quad C_d = C, \quad D_d = D \tag{3}
Ad=Φ(T0)=eAT0,Bd=∫0T0Φ(q)Bdq,Cd=C,Dd=D(3)文章来源:https://www.toymoban.com/news/detail-472358.html
4. 离散系统的状态方程计算举例
仍以如下连续系统传递函数为例:
W
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W(s) = \frac{s+1}{s^2 + 12s + 32}
W(s)=s2+12s+32s+1设量化时间单位为
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0
=
0.01
s
T_0 = 0.01s
T0=0.01s。连续系统下的矩阵
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D
A, B, C, D
A,B,C,D为
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A = \left[ \begin{matrix} -12 & -32 \\ 1 & 0 \end{matrix} \right], \quad B = \left[ \begin{matrix} 1 \\ 0 \end{matrix} \right], \quad C = \left[ \begin{matrix} 1 & 1 \end{matrix} \right], \quad D = 0,
A=[−121−320],B=[10],C=[11],D=0,则
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∣
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≈
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0.3014
0.0094
0.9985
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\begin{aligned} A_d &= \Phi \left( T_0 \right) = e^{A T_0} \\ &= I + AT_0 + \frac{A^2 T_0^2}{2!} + \frac{A^3 T_0^3}{3!} + \cdots \\ &= \left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] + \left[ \begin{matrix} -12 & -32 \\ 1 & 0 \end{matrix} \right] T_0 + \frac{1}{2} \left[ \begin{matrix} 112 & 384 \\ -12 & -32 \end{matrix} \right] T_0^2 + \frac{1}{6} \left[ \begin{matrix} -960 & -3584 \\ 112 & 384 \end{matrix} \right] T_0^3 + \cdots \\ &= \left[ \begin{matrix} 1-12T_0 + 56T_0^2 - 160T_0^3 + \cdots & -32T_0 + 192T_0^2 - \frac{1792}{2} T_0^3 + \cdots \\ T_0 - 6T_0^2 + \frac{56}{3} T_0^3 + \cdots & 1 - 16T_0^2 + 64T_0^3 + \cdots \end{matrix} \right] \Bigg\rvert _{T_0 = 0.01} \\ &\approx \left[ \begin{matrix} 0.8854 & -0.3014 \\ 0.0094 & 0.9985 \end{matrix} \right] \end{aligned}
Ad=Φ(T0)=eAT0=I+AT0+2!A2T02+3!A3T03+⋯=[1001]+[−121−320]T0+21[112−12384−32]T02+61[−960112−3584384]T03+⋯=[1−12T0+56T02−160T03+⋯T0−6T02+356T03+⋯−32T0+192T02−21792T03+⋯1−16T02+64T03+⋯]
T0=0.01≈[0.88540.0094−0.30140.9985]同时
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(4)
\begin{aligned} B_d &= \int _0 ^{T_0} \Phi (q) B dq = \int _0 ^{T_0} e^{Aq} B dq \\ &= \int _0 ^{T_0} \left( I + Aq + \frac{A^2 q^2}{2!} + \frac{A^3 q^3}{3!} + \cdots \right) B dq \\ &= \int _0 ^{T_0} B dq + \int _0 ^{T_0} AB q dq + \frac{1}{2} \int _0 ^{T_0} A^2 B q^2 dq + \frac{1}{6} \int _0 ^{T_0} A^3 B q^3 dq + \cdots \\ &= \int _0 ^{T_0} B dq + \int _0 ^{T_0} AB q dq + \frac{1}{2} \int _0 ^{T_0} A^2 B q^2 dq + \frac{1}{6} \int _0 ^{T_0} A^3 B q^3 dq + \cdots \\ &= B q \Big\rvert_0^{T_0} + AB \cdot \frac{1}{2} q^2 \bigg\rvert_0^{T_0} + \frac{1}{2} A^2B \cdot \frac{1}{3} q^3 \bigg\rvert_0^{T_0} + \frac{1}{6} A^3B \cdot \frac{1}{4} q^4 \bigg\rvert_0^{T_0} + \cdots \\ &= B T_0 + \frac{1}{2} AB T_0^2 + \frac{1}{6} A^2 B T_0^3+ \frac{1}{24} A^3 B T_0^4 + \cdots \tag{4} \end{aligned}
Bd=∫0T0Φ(q)Bdq=∫0T0eAqBdq=∫0T0(I+Aq+2!A2q2+3!A3q3+⋯)Bdq=∫0T0Bdq+∫0T0ABqdq+21∫0T0A2Bq2dq+61∫0T0A3Bq3dq+⋯=∫0T0Bdq+∫0T0ABqdq+21∫0T0A2Bq2dq+61∫0T0A3Bq3dq+⋯=Bq
0T0+AB⋅21q2
0T0+21A2B⋅31q3
0T0+61A3B⋅41q4
0T0+⋯=BT0+21ABT02+61A2BT03+241A3BT04+⋯(4)下面计算
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B
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AB, A^2 B, A^3 B
AB,A2B,A3B。由于
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B
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[
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A = \left[ \begin{matrix} -12 & -32 \\ 1 & 0 \end{matrix} \right], \quad B = \left[ \begin{matrix} 1 \\ 0 \end{matrix} \right]
A=[−121−320],B=[10]故
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−
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1
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,
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B
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[
112
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−
960
112
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AB = \left[ \begin{matrix} -12 \\ 1 \end{matrix} \right], \quad A^2 B = \left[ \begin{matrix} 112 \\ -12 \end{matrix} \right], \quad A^3 B = \left[ \begin{matrix} -960 \\ 112 \end{matrix} \right]
AB=[−121],A2B=[112−12],A3B=[−960112]则式(4)为
B
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B
T
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1
2
A
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1
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[
112
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12
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T
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1
24
[
−
960
112
]
T
0
4
∣
T
0
=
0.01
=
[
0.00942
4.8047
×
1
0
−
5
]
\begin{aligned} B_d &= B T_0 + \frac{1}{2} AB T_0^2 + \frac{1}{6} A^2 B T_0^3+ \frac{1}{24} A^3 B T_0^4 + \cdots \\ &= \left[ \begin{matrix} 1 \\ 0 \end{matrix} \right] T_0 + \frac{1}{2} \left[ \begin{matrix} -12 \\ 1 \end{matrix} \right] T_0^2 + \frac{1}{6} \left[ \begin{matrix} 112 \\ -12 \end{matrix} \right] T_0^3 + \frac{1}{24} \left[ \begin{matrix} -960 \\ 112 \end{matrix} \right] T_0^4 \Bigg\rvert_{T_0 = 0.01} \\ &= \left[ \begin{matrix} 0.00942 \\ 4.8047\times 10^{-5} \end{matrix} \right] \end{aligned}
Bd=BT0+21ABT02+61A2BT03+241A3BT04+⋯=[10]T0+21[−121]T02+61[112−12]T03+241[−960112]T04
T0=0.01=[0.009424.8047×10−5]进一步地,
C
d
=
C
=
[
1
1
]
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D
d
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C_d = C = \left[ \begin{matrix} 1 & 1 \end{matrix} \right], \quad D_d = D = 0
Cd=C=[11],Dd=D=0文章来源地址https://www.toymoban.com/news/detail-472358.html
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