?0?13??1?.????5.3 Consider the systemx(t)?Ax(t)?Bu(t),y(t)?Cx(t),withA?10?1B?0???????0???01?1??,
s2?s?1. Assume the system parameters C?[100]The system transfer function is32s?s?2s?2are all unknown.
?i?Verify or specify the conditions needed for the design of a model reference adaptive controller
in Section 5.5;
?ii?design a model reference adaptive controller for this system;and
?iii?simulate the adaptive control system with r(t) = 1, sin(t) and sin(t)+2sin(2t), respectively,
and plot both the tracking error and parameter errors. （1）解：参考模型自适应控制器需要三个条件：
a：Z(s)为稳定的多项式，可以保证y(t)跟踪ym(t)，也说明(A,B,C)是稳定的和可检测的。
b：Z(s)的维数m是已知的，这样可以为系统选择一个自适应参考模型。 c：zm是已知的，可以设计一个自适应参数规律。 b，c也是输出匹配控制和内部信号有界所需要的。 （2）参考模型：
ym(t)?Wm(s)r(t)，Wm(s)?T1，Pm的维数为n-m，且Pm?s?1。 Pm(s)T*T控制输出：u??(t)?(t),?(t)?[x(t),r(t)],?(t)?(K(t),kr(t))T,?*?(K*,kr)
辅助信号：
?(t)?Wm(s)?(t)，?(t)??T(t)?(t)?Wm(s)[?T?](t)
?(t)?e(t)??(t)?(t)，e(t)?y(t)?ym(t)
自适应律：
??(t)???sign(zm)?(t)?(t)；
1??T(t)?(t)??2(t)??(t)????(t)?(t)； T21??(t)?(t)??(t)（3）仿真如下：参数为??5I，?=1；
a：r(t)=1时，仿真结果如下所示：
b：r(t)=sin(t)时，仿真结果如下所示：
c：r(t)=sin(t)+2sin(2t)，仿真结果如下所示：
..3222.1、Analyze the stability of the system x1??x1?x2x1 ，x2??2x1x2?x2 (use a
2Lyapunov function candidate V(x)?2x1?x2)。
?.2x1?x1??x13?x2解：?
.?2x??2x?21x2?x2显然，x1?x2?0是系统的唯一的平衡态。
2又由于，V(x)?2x1?x2；
所以，
?.?2.??x13?x2x1??V(x)dx1?V(x)dx2??V(x)?V(x)??x1?42V(x)?????4x2x??(4x?2x)?2??12??.??12?x1dt?x2dt??x1?x2?????2x1x2?x2??x2?.由上可知，V(x)?0，等号仅在x?0时成立；当x?0时，V(x)?0，
且当x??时，V(x)??。
所以，系统的原点平衡状态是大范围渐进稳定的。
3.2As an application of the result in Problem 3.1, let us considerthe stability of the system
.x?A(t)x?g(x,t)， x(t0)?x0，where A(t)?Rn?n,g(x,t)?Rnare continuous. Show that
.if the equilibrium statexe?0of the unperturbed linear system x?A(t)xis uniformly
nasymptotically stable andg(x,t)??(t)x,?t?t0,?x?R
(which impliesg(0,t)?0,?t?t0) for some continuous function?(t)?0,?t?t0such that
?t2t1?(t)dt??(t2?t1)??0where?0?0,??0, then there exists a ?*?0such that for any
*.????0,??, the equilibrium statexe?0of the perturbed systemx?A(t)x?g(x,t)is
uniformly asymptotically stable. (Hint: use
tx(t)??(t,t0)???(t,?)g(x(?),?)d?and?(t,?)??e??(t??),
t0for some
??0,??0,?t???t0.)
.证明：由题可知，x?A(t)x，在xe?0是一致渐进稳定的
则存在一个与t0无关的T，使得，当x(t0)??0,t?t0?T时，有x(t)??(t,t0)x0??0，且lim?(t,t0)?0。
t??.当系统为x?A(t)x?g(x,t)时，系统的解如下： x(t)??(t,t0)x0???(t,?)g(x(?),?)d?
t0t所以，x(t)??(t,t0)x0??tt0?(t,?)g(x(?),?)d???(t,t0)x0?tt0?tt0?(t,?)g(x(?),?)d?
??(t,t0)x0??tt0?(t,?)g(x(?),?)d???(t,t0)x0???(t,?)g(x(?),?)d?
??(t??)由于，?(t,?)??e，g(x,t)??(t)x，
?t2t1?(t)dt??(t2?t1)??0
所以，x(t)??e??(t?t0)x0???e??(t??)?(?)x(?)d?
t0t又由于x(t)??(t,t0)x0??0 所以，x(t)??0(1???(?)d?)??(1??(t?t)??t000t0)，则系统是稳定的。
*0,?而，???，当??0时，令???0(1??0)，则x(t)??，?与t0无关，并且??.lim?(t,t0)?0，所以系统x?A(t)x?g(x,t)是一致渐进稳定的。
t??4.2 Consider the system?s?p0?[y](t)?z0u(t)withy(0)?0,z0?0andp0?0
?i?Design a least squares estimator for the parametersz0,p0; ?ii?design a gradient estimator for the parametersz0,p0;
?iii?show that for u(t)?sin(t),?t?0the estimated parameters converge to the true ones as t
goes to infinity for any initial estimates (hint: use the definition of persistently exciting signals to show that in this case the signal?(t)is persistently exciting);
?iv?simulate both the gradient and least squares estimators withz0?1,p0?2for(a)u(t)?1
and(b)u(t)?sin(t), and plot the parameter errors for?(0)?(1,1);
T?v?comment on the simulation results.
解：（1）、由题可知，系统为(s?p0)y(t)?z0u(t)，取为?(s)?s??0 所以，?*?(z0,?0?p0)T，?(t)?(~T11u(t),y(t))T，?(s)?(s)~?(t)??(t)?(t)?y(t)??(t)?(t),?(t)??(t)??*
.P(t)?(t)?(t)?(t)其最小二乘法估计器为：?(t)??(t)??,?(t0)??0,t?t0 2m(t)~T.~T.P(t)?(t)?(t)TP(t)其中P(t)??,P(t0)?P0?P0T,t?t0 2m(t)且m2(t)?1???(t)T?(t)???(t)TP(t)?(t),??0,??0 （2）、由（1）可知?*?(z0,?0?p0)T，?(t)?(~T~11u(t),y(t))T ?(s)?(s)?(t)??(t)?(t)?y(t)??(t)?(t),?(t)??(t)??*，
.??(t)?(t)?(t)其梯度估计器为，?(t)??(t)??,?(t0)??0,t?t0，且2m(t)~T.~Tm2(t)?1???(t)T?(t),??0，其中?是矩阵增益常数。
（3）、因为u(t)?sin(t)有?1?1,?2??1两个频率值，而本系统中n?m?1?2，所以?(x)是持续激励的信号。并且?(t)?L?，所以?(t)??(t)m(t)也是持续激励的信号。对于梯度算法
?(t)或者最小二乘算法，都有limt???2?0，即limt???(t)??*。
（4）、梯度估计器：??1,?0?1,??15I
（a）、z0 = 1, p0 = 2 ,u(t) = 1，梯度估计器的最终的结果曲线如图1所示。
图1 u=1时梯度估计器的误差曲线
（b）、z0 = 1, p0 = 2 ,u(t) = sin(t)，梯度估计器的最终的结果曲线如图2所示。
图2 u=sin(t)时梯度估计器的误差曲线
最小二乘估计器：??1,??1,?0?1,??15I
a）、z0 = 1, p0 = 2 ,u(t) = 1，最小二乘估计器的最终的结果曲线如图3所示。
图3 u=sin(t)时最小二乘估计器曲线图
（b）、z0 = 1, p0 = 2 ,u(t) = sin(t)，最小二乘估计器的最终的结果曲线如图4所示。
图4 u=sin(t)时最小二乘估计器曲线图
（5）通过MATLAB实验仿真得到上述几个图形，当u=1时，对比图1与图3发现两个估计器的效果类似，最小二乘估计器的曲线有小量的波动，二者趋向于0的时间都一样；当u=sin(t)时，对比图2和图4发现最小二乘估计器的效果比较好，曲线波动少，误差趋向0的速度明显比梯度估计器的快。
6.1 Consider the
.system(s?p1s?p0)[y](t)?kp(s?z0)[u](t),y(0)?y(0)?0wherez0?0, p1,p0,kpare
2..unknown but sign(kp) is known, and y(t) is measured but noty(t), and choose the reference
model(s?1)[ym](t)?r(t),ym(0)?0:
?i?Design an adaptive control scheme including controller structure and adaptive law for the
unknown system, and prove that the tracking errory(t)?ym(t)converges to zero ast??;
?ii?simulate the adaptive control system withp1??2,p0?2,kp??1,z0?1for
r(t)?10sin(t),r(t)?10sin(t)?13sin(3.3t), and plot the tracking error for each case;
?iii?comment on the simulation results.
（1） 自适应控制方案
由题可知：Pm?s?1，且n?1
*T*T?*?(?1*T?2?20?3*)T，?(t)?(?1T(t)?2(t)?20(t)?3(t))T T?(t)?(?1T(t)?2(t)y(t)r(t))T ~*?(t)??(t)??*,e(t)?y(t)?ym(t)
?1*T*T**，?1a(s)p(s)?(?2a(s)??20?(s))kpZ(s)??(s)(P(s)?kp?3Z(s)P?3*?kpm(s))
令a(s)?1,?(s)?s?1
~T 所以e?(t)??e(t)?kp?(t)?(t)
控制律：??(t)??sign(kp)??(t)e(t),????0 证明：令ep(t)?(e(t),?(t))T 选择一个正定函数如下：
~T~~TTV(ep)?e?kp???
V???2ame(t)?2e(t)kp??(t)?2kp?(t)??(t)
~2?12~T~T?1~?又由于?(t)???(t)??sign(kp)??(t)e(t),????0，带入V(ep)?e?kp???1?中可得，V???2ame(t)?0
因此，e(t)?L,?(t)?L,y(t)?L,?2(t)?L
?????T2~T~2siP(s)si由于P(s)y(t)?kpZ(s)u(t)，可以得到u(t)?y(t)
?(s)kpZ(s)?(s)P(s)si而是稳定的，y(t)?L?，则u(t)?L?,?1(t)?L?，e?(t)?L?,e(t)?L2 kpZ(s)?(s)因此，lime(t)?0
t??（2） 仿真如下：参数??5I,p1??2,p0?2,kp??1,z0?1，a(s)?1,?(s)?s?1
*T*T***?1利用?1 a(s)p(s)?(?2a(s)??20?(s))kpZ(s)??(s)(P(s)?kp?3Z(s)Pm(s))，?3?kp*T**可得到，?1*T=0，?2=-5，?20=4，?3=-1
a：r(t)=10sin(t),仿真结果如下：
b：r(t)=10sin(t)+13sin(3.3t)，仿真结果如下：
.6.2 Consider the system(s?p1s?p0)[y](t)?kp[u](t),y(0)?y(0)?0where y(t) is
2.measured but noty(t), and choose the reference model
..(s?3s?2)[ym](t)?r(t),ym(0)?ym(0)?0
2.?i?Design an adaptive control scheme including controller structure and adaptive law for the
system withp1,p0,kpunknown but sign(kp) known;
?ii?design an adaptive control scheme including controller structure and adaptive law for the
system with p1,p0unknown butkp=1known;
?iii?simulate the adaptive system of part?i?with
p1??2,p0?3,kp?1for r(t) = 10sin(t), r(t)
= 10sin(t)+13sin(3.3t), and plot the tracking errors and controller parameters; and
?iv?comment on the simulation results and compare the controller parameters with the matching
parameters calculated from Problem 1.2.