《现代控制理论》课程教学资源——现控课件_Ch1-2 Obtaining State Space Description from I_O Description

将（1.46-1.47）写成向量方程的形式可得系统的状态空间描述0010X：0βixB0000X2β;00000X3(1.48)u十....·.0000βn-11Xn-1本βnxn-an-an-1-an-2-a2-aj.···XiX2X3(1.49)+[β] u0y=1Xn-lxn

将（1.46-1.47）写成向量方程的形式可得系统的状态空间描述： u x x x x x x a a a a a x x x x n n n n n n n n n                     +                                         − − − − − =                     − − − − −      1 3 2 1 1 3 2 1 1 2 2 1 1 3 2 1 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0                        u x x x x x y n n 0 1 3 2 1 1 0 0 0 0 +                      = −   (1.49) (1.48)

例1.4写出下列微分方程的状态空间描述：+18i+192i+640y=160u+640u解：由微分方程可知：n=3a = 18,a, =192,a, = 640,b = b, = 0,b, = 160,b, =640由结论2，可得：[β= b。= 0β =b, -a,β= 0βz = bz -α,β, -aαzβ。= bz = 160β, =bs -α,β2 -a,β, -a,β= 640-18×160 =-2240将得到的β,代入(1.43)式可得状态变量表达式：xi =y-βux=-βi-βuxs =j-βi-β,ü-β,u

例1.4 写出下列微分方程的状态空间描述：  y  +18 y  +192y  + 640y = 160u  + 640u 解：由微分方程可知： n=3 1 2 3 0 1 2 3 a a a b b b b = = = = = = = 18, 192, 640, 0, 160, 640 由结论2，可得：        = − − − = −  = − = − − = = = − = = = 640 18 160 2240 160 0 0 3 3 1 2 2 1 3 0 2 2 1 1 2 0 2 1 1 1 0 0 0           b a a a b a a b b a b 将得到的 i 代入(1.43)式可得状态变量表达式：      = − − − = − − = − x y u u u x y u u x y u 3 0 1 2 2 0 1 1 0           

最后，定义状态向量X=[xxx,,可得微分方程的状态空间描述为：β00X=00X+β21u[β,]-ara2a30001001160X+=u-640-192-18-2240y=[1 0 o]X +[β]u=[1 0 o]x

最后, 定义状态向量 ,可得微分方程的状态空间 描述为：   T 1 2 3 X = x x x u u a a a  − +  − − − =   +  − − − = 2240 1600 640 192 18 0 0 1 0 1 0 0 0 1 0 1 0 321 3 2 1 X X X   y = 1 0 0X +  0 u = 1 0 0X

E.g. 1.5 Find a state space description for the system describebythefollowingdifferential equation+16+194j+640y=4u+160u+720uSolution : For the system, n=3a = 16,az = 194,a, = 640,b = 4,b, = 0,b, = 160,b, = 720The parameters β, can be chosen as :[β = b, = 4β, = b, -a,β。= 0-16×4 =-64βz = b -α,β, -azβ=160 -16×(-64)-194×4 = 408β, = b, -a,β2 -a,β, -a,β。= 720-16× 408 -194×(-64)-640×4 = 4048The state variables can be defined by :xi = y-βouxz=j-βi-βux=j-βi-βi-β,u

E.g. 1.5 Find a state space description for the system described by the following differential equation y y y y u u u + + + = + + 16 194 640 4 160 720 Solution ： For the system, n=3 1 2 3 0 1 2 3 a a a b b b b = = = = = = = 16, 194, 640, 4, 0, 160, 720        = − − − = −  −  − −  = = − − = −  − −  = = − = −  = − = = 720 16 408 194 ( 64) 640 4 4048 160 16 ( 64) 194 4 408 0 16 4 64 4 3 3 1 2 2 1 3 0 2 2 1 1 2 0 1 1 1 0 0 0           b a a a b a a b a b The state variables can be defined by ：      = − − − = − − = − x y u u u x y u u x y u 3 0 1 2 2 0 1 1 0            The parameters can be chosen as i ：

So, the state space description of system can be obtained asβ,00X=00X+β21u[β,]a3a2-a-001- 6400X +4081uI-1944048-640-16y=[1 0 o]x +[β.]u=[1 0 o]x + 4uWhere X-[xiX2x3]

So, the state space description of system can be obtained as   T 1 2 3 X = x x x u u a a a           − +           − − − =           +           − − − = 4048 408 64 640 194 16 0 0 1 0 1 0 0 0 1 0 1 0 3 2 1 3 2 1 X X X     y = 1 0 0X +  0 u = 1 0 0X + 4u Where