1、最优控制习题及参考答案最优控制习题及参考答案习题 1 求通过 x(0) = 1 , x(1) = 2 ,使下列性能指标为极值的曲线:t fJ = (x 2 +1)dtt0解: 由已知条件知: t0 = 0 , t f = 1d由欧拉方程得: (2x ) = 0dtx = C1x = C1t + C2将 x(0) = 1,x(1) = 2 代入,有:C2 = 1,C1 = 1得极值轨线: x* (t) = t +1习题 2 求性能指标:J = 1(x 2 +1)dt0在边界条件 x(0) = 0 , x(1) 是自由情况下的极值曲线。解: 由上题得: x* (t) = C t + Cx* (t)
2、 由 x(0) = 0 得: C2 = 00L由x t=t f= 2x (t f ) = 2C1 t=t = 0 t0 1于是: x* (t) = 0【分析讨论】对于任意的 x(0) = x0 ,x(1) 自由。2 0 1 0有: C = x , C = 0 ,即: x* (t) = x其几何意义: x(1) 自由意味着终点在虚线上任意点。习题 3 已知系统的状态方程为:x 1 (t) = x2 (t) , x 2 (t) = u(t) 边界条件为: x1 (0) = x2 (0) = 1 , x1 (3) = x2 (3) = 0 ,3 1试求使性能指标 J = 0u2 (t)dt 2取极小
3、值的最优控制 u* (t) 以及最优轨线 x* (t) 。 x 解: 由已知条件知: f = 2 u Hamiton 函数: H = L + T fH = 1 u2 + x+ u = 0 = 由协态方程: 12 12 1 2 2 = C 得: 1 12 = C1t + C2 H由控制方程:u= u + 2 = 0得: u = 2 = C1t C2 由状态方程: x 2 = u = C1t C2得: x (t) = 1 C t2 C t + C 2 2由状态方程: x 1 = x21 2 3得: x (t) = 1 C t3 1 C t 2 + C t + C 1 6 12 2 3 410将 x
4、(0) = , x(3) = 0 代入,,1 10联立解得: C1 =由、式得:u* (t) = 10 t 29, C2 = 2 , C3 = C4 = 1 91x* (t) =5 t3 t 2 + t +127x* (t) = 5 t2 2t +12 9习题 4 已知系统状态方程及初始条件为x =u ,x(0) = 1试确定最优控制使下列性能指标取极小值。1J = 0解: H = x2e2t + u2e2t + ux = u列方程: = 2xe2t2e2t u + = 0(x2 + u2 )e2t dt由得, u代入得,x1 e2t 21 e2t = 2x 1 e2t e2t = +2将,代
5、入,并考虑到 u = x x 1 e2t (2xe2t ) + e2t (2e2t x ) 2整理可得: x + 2x x = 0特征方程: s2 + 2s 1 = 0s1 = 1+2,s2 = 1 21 2于是得: x* (t) = C es1t + C es2t) = u = * (t 2e2t 2e2t x * (t) = 2e2t(C1s1es1t + C s es2t )2 2由 x(0) = 1 ,得: C1 + C2 = 1 由 (t f ) = (1) = 0 得: C1s1e+ C2 s2e = 0 、联立,可得 C1、C2求导代回原方程可得 x* u*(略)习题 5 求使系
6、统: x 1 = x2 , x 2 = u由初始状态 x1 (0) = x2 (0) = 0出发,在 t f= 1 时转移到目标集 1x1 (1) + x2 (1) = 1,并使性能指标 J = 1u2 (t)dt2 0为最小值的最优控制 u* (t) 及相应的最优轨线 x* (t) 。解: 本题 f (i),L(i) 与习题 3 同,故 H (i) 相同方程同通解同1 = C1,2 = C1t + C2x = 1 C t3 1 C t2 + C t + C有: 1 6 1 2 2 3 4x = 1 C t 2 C t + C 2 21 2 3u = C1t C20x(0) = 0由 ,有:
7、C3 = C4 = 0 由 x1 (1) + x2 (1) = 1,有:1 C 1 C+ 1 C C = 16 1 2 22 1 22 C 3 C = 1 3 1 2 2 T由 (1) = + = 0 , = x1 + x2 1x x1有: (1) = = 0 (1) = (1)1 1 2于是: C1 = C1 + C22C1 = C2 3 6、联立,得: C1 =- 、C2 = -7 7于是: u* = 3 t + 67 7x * = 1 t3 + 3 t 21 14 7x * = 3 t 2 + 6 t2 14 7习题 6 已知一阶系统: x (t) = x(t) + u(t) , x(0
8、) = 3f()试确定最优控制 u* (t) ,使系统在 t = 2 时转移到 x(2) = 0 ,并使性能泛函2J = (1+ u2 )dt = min0f f()如果使系统转移到 x(t ) = 0 的终端时间 t 自由,问 u* (t) 应如何确定?解: H = 1+ u2 + u xx = x + u列方程: = 2u + = 01由协态方程得: = C et 1 t由控制方程: u= C1e t2 t f1 t代入状态方程: x = x C1e2= 2,x(2) = 0 x(t) = C2e 1 C et4 1 1 C = 3 2 4 1C e2 1 C e2 = 0 2 4 112
9、解得: C1 = 4 ,e 13e4C2 = 4e 1代入得: u* (t) = x(t f ) = 2,t f自由6 ete4 1 1 C = 3 2 4 1C et f 1 C et f = 0 2 1H (t f ) = 0解得: C1 =40 6 = 0.325u* (t) = 0.162et习题 7 设系统状态方程及初始条件为x (t) = u(t) , x(0) = 1试确定最优控制 u* (t) ,使性能指标1 t f 2f J = t +2 0u dt为极小,其中终端时间 t f 未定,x(t f ) = 0 。解: H = 1 u2 + u2由协态方程得: = 0 = C1
10、由控制方程: u + = 0 u = C1 由状态方程: x = u = C1 x(t) = C1t + C2 由始端: x(0) = 1 C2 = 1由末端: x(t f ) = 0 C1t f +1 = 0 考虑到: H (t f ) = t t = 1 f f1 2有: u + u = 121 C 2 C 2 = 1 C 2 = 22 1 1 1C1 = 2 当 C1 =2 时,代入有: t f= 1 = 1C1 2当 C1 = 2 时,代入有: t f= 1 = 1 ,不合题意,故有 C = 2C1 2最优控制u* = 2习题 8 设系统状态方程及初始条件为x 1 (t) = x2 (
11、t) , x1 (0) = 2性能指标为x 2 (t) = u(t) ,J = 1 t f u2 dtx2 (0) = 12 0要求达到 x(t f ) = 0 ,试求:(1) t f= 5 时的最优控制 u* (t) ;f(2) t 自由时的最优控制 u* (t) ; 解:本题 f (i ),L(i ),H (i ) 与前同,故有1 = C12 = C1t + C2x = 1 C t3 1 C t2 + C t + C 1 12 2 3 4x = 1 C t 2 C t + C 2 21 2 3u = C1t C220C4 = 2C3 = 1125 25 由 x(0) = x(5) = 0
12、,得: C1 C2 + 5C3 + C4 = 01 6 2 25C 5C + C = 0 1 2 3 2联立得: C1 = 0.432,C2 = 1.28 , u*= 0.432t 1.28 t f 自由C = 1 4C3 = 21 C t3 1 C t 2 + C t+ C = 0 1 f2 2 f 3 f 4 1 C t2 C t+ C = 0 2 1 f2 f 3H (t f ) = 0联立有: C 2t 2 2C t+ 2 = 0 , 无论 C 为何值, t 均无实解。2 f 2 f 2 f习题 9 给定二阶系统x (t) = x (t) + 1 , x (0) = 11 2 4 1
13、41x 2 (t) = u(t) ,1x2 (0) = 4控制约束为 u(t) ,要求最优控制 u* (t) ,使系统在 t = t2 f并使时转移到 x(t f ) = 0 ,其中 t f 自由。t fJ = u2 (t)dt = min0解: H = u2 + x+ 1 + u1 2 4 1 2 1 1 2 2 2本题属最小能量问题,因此:u* (t) = 1 1 2 1 1 2 = 0 = C由协态方程: 1 1 12 1 2 1 2 = = C t + C2 是 t 的直线函数。当 u* (t) = 1 = 1 C t 1 C时(试取)2 2 2 1 2 2x (t) = 1 C t2
14、 1 C t + C2 4 12 2 3x (t) =1 C t3 1 C t2 + 1 t + C t + C1 12 14 2 4 3 41由始端条件 C3 = C4 =4由末端条件 1 Ct 3 1 C t2 + 1 t+ 1 = 012 1 f4 2 f2 f 41 Ct 2 1 C t+ 1 = 04 1 f2 2 f 4另: H (t f ) = 01联立解得: C1 = ,C2 = 0,t = 39 f于是, 1 t 2 = 1时,t 02 = 9 2= 1时,t = 9在 t 从 0 3 段, 2 1满足条件。故, u*= 1 = 1 t2 2 1810 1 2 3 4 t习题
15、 10 设二阶系统x 1 (t) = x1 (t) + u(t) , x1 (0) = 1x 2 (t) = x1 (t) ,x2 (0) = 0控制约束为 u(t) 1 ,当系统终端自由时,求最优控制 u* (t) ,使性能指标J = 2x1 (1) + x2 (1)取极小值,并求最优轨线 x* (t) 。解:由题意, fx1 + u = , = x+ x ,L = 0 , H = u x+ x 1 21 1 1 2 1 x1 1由控制方程可得: u* = +11 0 = = C et + C由协态方程可得: 1 1 2 1 2 1 2 = 0 2 2 = C1由 (t) = = C = 1
16、,C= e1x(t f )1 1 2 = et 1 +1 在t 0的范围内 1 1 1故: u* = 1t 0,12 = 1若需计算 最优轨线 ,只需把 u* = 1 代入状态 方程,可 得: x * (t) = 2et 1 1x * (t) = 2et t + 2 2习题 11 设系统状态方程为x 1 (t) = x2 (t) , x1 (0) = x10性能指标为 J = 12 0x 2 (t) = u(t) ,(4x2 + u2 )dtx2 (0) = x20试用调节器方法确定最优控制 u* (t) 。0 1解: 由已知条件得: A = 0 00, B = ,14 0Q = 0 0, R
17、 = 11 0B AB = 0 1 , 可控 最优解存在考虑到Q = 4 0 = 2 2 0 = DT D ,故0 0 0 D = 2 0 D 2 0 = DA 0 2闭环系统渐近稳定由 Riccati 方程 AT P + PA PBR1BT P + Q = 0 ,有0 0 P1P2 + P1P2 0 1 P1P2 0 0 1 P1P2 + 4 0 = 0 1 0 P2P3 P2P3 0 0P2P3 1P2P3 0 02 2P 2 + 4 = 0 P= 2(取 + 2舍 2)展开得: P1 P2 P3 = 0 P1 = 4(由正定舍 4)2P P2 = 0 P2 = 2P P = 2 2 3 3 2 34 2故 P = 2 2于是, u* = R1BT Px = 2x 2x1 21 2即: u* (t) = 2x (t) 2x (t) Welcome ToDownload !欢迎您的下载,资料仅供参考!
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