Assume x is available for state feedback. Design LQR control law by letting R=1 and choosing Q so that all the elements of the feedback gain K have absolute value less than 50.

Solution 

A= [■(0&1&0@0&0&1@-1&4&-3)]  ,     B= [■(0@0@2)]        C=[■(0&1&0@0&0&1)]

Pick Q [■(2000&0@0&2000)]  ,       R=1

K=lqr(A<B,C^'*Q*C,R);   K=[0.0000   47.2523   43.7714]

eig(A-B*K)={-89.5321,-0.0112,-0.9996)

MATLAB Simulink Model

Output response 

2). 2) Assume that only the output is available. Design an observer so that the poles of the observer are -5+5j,-5-j5,-10. Choose the observer gain so that all the elements have absolute value less than 80. Form a closed-loop system along with the LQR controller part 1). Plot y_1 (t) and y_2 (t) in same figure for t∈[0,20]

Solution 

Using approach of full dimensional estimators 

G_(n-q+1)^0= [■(C@CA)]= [■(■(0&1&0@0&0&1)@■(0&0&1@-1&4&-3))]

G_(n-q+1)^0=ρ(G_(n-q+1)^0 )=3 ,observable 

Given the poles of observer, -5+5j,-5-j5,-10 choose 

F= [■(-5&-5&0@5&-5&0@5&5&-10)]  ,L_0= [■(2&3@3&2@2&1)]

G_(n-ρ+1)^c= [L_0   FL_o ]= [■(■(2&3@3&2@2&1)&■(-25&-25@-5&5@-20&-10))]

G_(n-ρ+1)^c=3 →(F,L_o )Controllable 

T=lyap(-F,A,-L_0*C); L=inv(T)*L_o; 

L= [■(19.0148&-25.5544@15.1669&7.2402@-4.1955&1.8331)]

eig(A-L*C)= {-5.0000+5.0000i,-5.0000-5.0000i,-10.0000}

Simulink 

Output response 

Problem 3

A Cart with an inverted pendulum 

The control problems are; 

Stabilization: designed a feedback law u=Fx such that x(t)→0for x(0)close to origin 

A). Find the matrices A, B for state space equation 

(M+m) Y ̈+mI(θ ̈=u                       (1) ) ̈

Y ̈=1θ ̈=gθ                                   (2)

Solving equations 1 and 2 then we get; 

Y ̈=(-mg/M)θ+(1/M)u                (a)

θ ̈=(g/l=mg/Ml  )  θ-(1/Ml)u            (b)

y=x_1,y`=x_2,θ=x_3,θ`=x_4                                     

State space equations: 

x`=Ax+Bu,y(t)=Cx+Du

Solving the equation (a), and (b) in the state space equations, we get; 

A=[■(■(0&1@0&0)&■(0&0@-1.96&0)@■(0&0@0&0)&■(0&1@58.5&0))]

B= [■(■(0&0.2)&■(0&-1))]`

B). Design a feedback law u=F1x so that A+BF1 has eigenvalues at -3±j3;-6 and -8. Build a simulink model for the closed-loop linear system. Plot the response under initial condition x(0)=[-1.5,0,1,3].

Solution 

F= [■(■(-1&1)&■(0&0  ; ■(■(-1&-1)&■(0&0)))) ; ■(■(0&0)&■(-2.5&0)) ; ■(■(0&0)&■(0&-5))]

F =

   -1.0000    1.0000         0         0

   -1.0000   -1.0000         0         0

         0         0    2.5000         0

         0         0         0   -5.0000

K_o= [■(■(1&0)&■(-1&1))]

K_0= ■(■(1&0)&■(-1&1))

a=K_0*F 

a=■(■(-1&1)&■(2.5&-5))

b=a*F 

b=■(■(0&-2)&■(-6.25&25))

c=b*F 

c=■(■(2&2)&■(15.625&-125.000))

G=[K_O;a;b;c]

G= ■(■(-1&0@-1&1)&■(-1&1@2.5&-5)@■(0&-2@2&2)&■(-6.25&25@15.62&-125.00))

rank(G)= 4

A= [0 1 0 0; 0 0 -1.96 0 ;0 0 0 1;0 0 58.8 0]

A =

0    1.0000         0         0

0         0   -1.9600         0

0         0         0    1.0000

0         0   58.8000         0

B= [0; 0.2;0;-1]

B =

         0

    0.2000

         0

   -1.0000

T= lyap(A,-F,-B*K_0)

T =

   0.0006   -0.0834    0.0260   -0.0057

    0.0828    0.0839    0.0651    0.0284

   -0.0170    0.0006    0.0190   -0.0296

    0.0164   -0.0176    0.0476    0.1479

K= K_0 *inv (T) 

K =

    2.5510    2.0408  -52.7898   -4.0918

eig(A-B*K) 

ans =

  -5.0000 + 0.0000i

  -1.0000 + 1.0000i

  -1.0000 - 1.0000i

   2.5000 + 0.0000i

Design a feedback law 

Results 

Simulink model for the closed-loop linear system

3). Build a Simulink model for the original nonlinear system, verify that stabilization is achieved by u=F1x when x(0) is close to the origin. Find the maximal _o  so that the nonlinear system can be stabilized from x(0)=(0,0,_0,0).

Solutions 

%Simulink model for nonlinear system, 

function dx=ff(x) 

g= 9.8;

l=0.2;

M=5;

m=0.5;

x1= x(1);x2=x(2);

x3=x(3);x4=x(4);

u=x(6);

dx2=((m*l*x4^2*sin(x3)+u)-(m*g*cos(x3)*sin(x3)))/(M+m-m*cos(x3)^2);

dx4=((g*sin(x3)+(M+m))-(cos(x3)*(u+m*1*x4^2*sin(x3))))/(1*(M+mm*cos(x3)^2));

dx=[dx2;dx4];

end

4). . For x(0)=(0,0,compare the response y(t) and (t) for the linearized system and the nonlinear system under the same feedback u=F1x.

Solution 

Graph: