?c??112? N???cA??????125??cA2???????7411?? rank N=3，则系统能观所以此系统为能控并且能观系统?17??111???34?? 取T???4c2??21226?3???，则T?1c2??7?1??20?2????2?5????3144????002? 则A???10?5??1?，B?T?1?4?c2b??0?，c?cTc2??71323?????01???0??3-14求下列传递函数阵的最小实现。 （1） w?s??1?11?s?1??11??解： ??11???10?0?1，B0???11?，?Ac???0?1??B?10??11?c??，?01??Cc??，?11??Dc??00???00??系统能控不能观取R?1??11???1?1?0?01?，则?R0???01??所以A??R?1??10???11?0AR0???0?1?，?B??R10Bc???01??C??CcR0??10??，?10??D???00???00??所以最小实现为A?m?1，B?m??11?，C?m??1??，?1??D?m??00???00??验证：C?m?sI?A?m??1B?m?1?11?s?1??11???w?s?3-15设?1和?2是两个能控且能观的系统??01??0?1：A1??，??3?4??b1??，?1??C1??2?2：A2??2，b2?1，C2?1?1 因为 M???b 解：（2）?1和?2并联（1）?1和?2串联?c??210?????3?21? 因为N??cA????2??cA????654??Abrank M=3 则系统能控当?2得输出y2是?1的输入u1时则rank M=2<3，所以系统不完全能控。rank N=2<3 则系统不能观?001??01?6?A2b???????1?2?4??M???bAbW(s)?C(sI?A)?1B??3??2x3?2x1?x2当?1的输出y1是?2的输入u2时，x（2）试分析由?1和?2所组成的并联系统的能控性和能观性，并写出其传递函数。（1）试分析由?1和?2所组成的串联系统的能控性和能观性，并写出其传递函数；?010??0?????3?40?x??1?u，y??001?xx???????21?2???0???011??0?????3?41?x??0?u，y??210?xx???????00?2???1???010??0?????3?40?x??1?u，y??211?xx???????00?2???1??M???AAb?01?4???A2b????1?413???01?4??s?21?2(s?2)(s?3)(s?4)s?7s?12W(s)?C(sI?A)?1B?1s2?7s?12?01?4???Ab2????1?413???1?2?4?? ??x1x2解：（1）由已知得4-1判断下列二次型函数的符号性质：22（2）v(x)?x12?4x2?x3?2x1x2?6x2x3?2x1x322（1）Q(x)??x12?3x2?11x3?2x1x2?x2x3?2x1x3（2）由已知得?1??1?0，?2?因此Q(x)是负定的?Q(x)???x1?x2?x3?1x1?3x2?x32因为rank N=3，所以系统完全能观因为rank M=3，所以系统完全能控???11?x3??1?3??1??1??2??1???x1?1????x22??????x3???11???11?2?0，?3?11?3?11?3?1?12?x1??Q(x)??x1?x2?x3?x1?4x2?3x3?x1?3x2?x3??x2????x3???1?1?1??x1???x???x1x2x3???14?3???2????1?31????x3???x1?1???x1?x2?11x3??x22?????x3???c??211?????3?2?2?N??cA????2??cA????654???2??2?2?s?2?s?2????22?1???w?s??C?sI?A?B???s?1??s?2??s?3?现代控制理论第四章习题答案?11?1171????0242a11若 a120即：?P取Q?I，令P??11?P122a21a11?a222a12要求P正定，则要求因此Q(x)不是正定的其中detA?A?a11a22?a12a21因此a11?a22?0，且detA?0有解，且解具有负实部。即：a11?a22?0且a11a22?a12a214-2已知二阶系统的状态方程：试确定系统在平衡状态处大范围渐进稳定的条件。?2a11?a?12??0?I?A??02a21a11?a222a12?a11?x???a21??a11?a211?1?11?1?1?1?0，?2??3?0，?3??14?3??16?0?14?1?31方法（2）：系统的原点平衡状态xe?0为大范围渐近稳定，等价于ATP?PA??Q。解：方法（1）：要使系统在平衡状态处大范围渐进稳定，则要求满足A的特征值均具有负实部。0a21?4(a11?a22)(a11a22?a12a21)?0，则此方程组有唯一解。即2a22(a11?a22)2?(a12?a21)2?2?P??0?4(a11?a22)P12?，则带入ATP?PA??Q，得到?P22??a12??a22a12??xa22?22?A?a21?a22?(a12a22?a21a11)?1P????22A?a11?a122(a11?a22)A??(a12a22?a21a11)?22A?a21?a22?1?P?011??2(a11?a22)A??2?(a11?a22)??a11a22?a12a210??P??1?11??P???0?a21???12???2a22????P22?????1??????取P?I????（1）x??11????（2）x?x?1?1??J(x)?试确定平衡状态的稳定性。4-6设非线性系统状态方程为：解：若采用克拉索夫斯基法，则依题意有：?Q(x)?JT(x)?J(x)?11?x??2?3??1?2x2x?2V(x)?2x1x?1?2x2x?2V(x)?2x1x2??2x12?2x2?02??2x12?6x1x2?6x2x2??f(x)???2??a(1?x2)x2?x1?332??2(x1?x2)2?x2?022?2x1(?x1?x2)?2x2(?x1?x2)?2x1(?x1?2x2)?2x2(2x1?3x2)1??f(x)?0???1?a?4ax?3ax2??xT?22??1?x2x4-3试用lyapunov第二法确定下列系统原点的稳定性。2解：（1）系统唯一的平衡状态是xe?0。选取Lyapunov函数为V(x)?x12?x2?0，则V(x)是负定的。x??，有V(x)??。即系统在原点处大范围渐近稳定。2（2）系统唯一的平衡状态是xe?0。选取Lyapunov函数为V(x)?x12?x2?0，则V(x)是负定的。x??，有V(x)??。即系统在原点处大范围渐近稳定。?11?0??0?????2?2??1?a?4ax2?3ax2???1?a?4ax2?3ax2?0?0???2??0?2a?8ax2?6ax2??2??a(1?x2)2x2?x1,a?0x2很明显，Q(x)的符号无法确定，故改用李雅普诺夫第二法。选取Lyapunov函数为V(x)?x12?x2?0，则
VIP专享《现代控制理论》第3版（刘豹 - 唐万生）课后习题答案
