5第五节组合系统的状态空间描述

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6
= G2 ( s ) ⋅ G1 ( s )
∴ G ( s ) = ∏ Gi ( s )
i= N
1
Saturday, June 25, 2011
[结论]:当两系统串联时,组合系统的传递函数阵等于后一子系 统的传递函数阵乘以前一子系统的传递函数阵。 反馈连接: Y1 U1 U
Y
,关系: 1 = U − Y2 U
[
]
对应的传递函数阵为:
G ( s ) = C1
[
G 化简后得: ( s ) = G1 ( s )
[I + G (s)G (s)] 或: ( s ) = [I + G ( s )G ( s ) ] G ( s ) G
−1 2 1 −1 1 2 1
A1 0 ( sI − B2 C1
]
− B1 C2 −1 B1 ) A2 0
Saturday, June 25, 2011
8
传递函数的推导:
U
U1
Y1
N1
U2
Y
,关系: 1 = U − Y2 U
− Y2
Y = Y1 = U 2
N2
Y ( s ) = Y1 ( s ) = G1 ( s )[U ( s ) − G2 ( s )Y ( s )] = G1 ( s )U ( s ) − G1 ( s )G2 ( s )Y ( s )
]
0 −1 B1 A1 G ( s ) = D2 C1 C2 ( sI − ) B D + D2 D1 B2 C1 A2 2 1 (化简略) 2 ( sI − A2 ) −1 B2 + D2 ⋅ C1 ( sI − A1 ) −1 B1 + D1 C
传递函数阵为:
整理后得: [ I + G1 ( s )G2 ( s )]Y ( s ) = G1 ( s )U ( s )
Y = [ I + G1 ( s )G2 ( s )]−1 G1 ( s )U ( s ) 同样可得到: Y = G1 ( s)[ I + G2 ( s)G21 ( s)]−1U ( s)
Saturday, June 25, 2011
− Y2
N1
U2
Y = Y1 = U 2
N2
为简化起见,令 D1 = D2 = O
& 有: 1 = A1 X 1 + B1U1 = A1 X 1 + B1 (U − Y2 ) = A1 X 1 + B1U − B1 C2 X 2 X & X 2 = A2 X 2 + B2 U 2 = A2 X 2 + B2 C1 X 1 Y = Y1 = C1 X 1
9
小结
子系统的串联 子系统的并联 子系统的反馈联接
Saturday, June 25, 2011
10
Saturday, June 25, 2011
7
写成矩阵形式:
& − B1 C2 X 1 B1 X 1 A1 & = X + U A2 2 0 X 2 B2 C1 X1 Y = C1 0 X2
第五节 组合系统的状态空间描述
Saturday, June 25, 2011
1
系统的组合有并联、串联和反馈等三种形式。 设子系统Ⅰ的动态方程为:
& X 1 = A1 X 1 + B1U1 N1 : Y1 = C1 X 1 + D1U1
子系统Ⅱ的动态方程为:
& X 2 = A2 X 2 + B2 U 2 N2 : Y2 = C2 X 2 + D2 U 2
写成矩阵形式: & X 1 A1 0 X 1 B1 X1 & = X + B U = A X + BU X 2 0 A2 2 1 2 X1 X1 Y = C1 C 2 X + (D1 + D2 )U = C X + DU 2 2 传递函数阵为:
Saturday, June 25, 2011
2
两个子系统并联时: U1
U
U2
N1 N2
+
+
Y
,关系:
U1 = U 2 = U Y = Y1 + Y2
则有:& = A X + B U X1 1 1 1
& X 2 = A2 X 2 + B2 U Y = Y1 + Y2 = C1 X 1 + D1U + C2 X 2 + D2 U
Saturday, June 25, 2011
3
G ( s ) = C ( sI − A) −1 B + D (sI − A )−1 0 −1 1 ∴ sI − A为分块对角阵,由性质知: − A) = (sI −1 (sI − A2 ) 0 sI − A1 −1 B1 0 ∴G(s) = C1 C2 ⋅ + D1 + D2 −1 ⋅ sI − A2 B2 0
N1
Y1 = U 2
N1
Y2 = Y
,关系: 1 = U , Y1 = U 2 , Y2 = Y U
有:& 1 = A1 X 1 + B1U X
& X 2 = A2 X 2 + B2 Y1 = A2 X 2 + B2 C1 X 1 + B2 D1U Y = Y2 = C2 X 2 + D2 Y1 = C2 X 2 + D2 C1 X 1 + D2 D1U
5
Saturday, June 25, 2011
写成矩阵形式:
& X 1 A1 & = X 2 B2 C1 Y = D2 C1
0 X 1 B1 X + B D U A2 2 2 1
[
X1 C2 + D2 ⋅ D1U X2
[
]
[
( ]
)
(
)ห้องสมุดไป่ตู้
= C(sI − A1 )−1 B1 + D1 + C(sI − A2 )−1 B2 + D2
Saturday, June 25, 2011
4
[结论]:当两系统并联时,组合系统的传递函数阵等于各子系统 N 传递函数阵之和, 。
G(s) = ∑Gi (s)
i=1
子系统串联:
U = U1
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