Question

My question is Problem 11.4-8

11.4-8. Consider the third-order continuous-time LTI system * = Ax + Bu y = Cx 102 with A = To 0 Lo 2 0 -8 07 3 , B = -6] 0 , and C = [1 0 0]. Using Q = [800] 0 6 0, LO 0 4 R = 1.5 (a) First design a LQ controller for this continuous time-system using the MATLAB function iqr. Let the optimal controller gain vector be K. Simulate the closed-loop system x = (A – BK) with X(0) = (2 0 -2]. Sample the closed-loop output response y(t) with sampling time T = 0.2 s. (b) Next, discretize the continuous-time system directly with sampling time T = 0.2 s and design a LQ controller for this discrete-time LTI system with the same Q and R as in part (a). Compare the closed-loop output response with the sampled continuous-time output obtained in part (a). (c) Increase Q and R by a factor of 10 and repeat both the continuous-time and discrete-time designs, Describe how the respective output responses change.

Thanks for your help!

Answer 1

(a)

clear;
close all;
clc;

A = [0 2 0;
0 0 3;
0 -8 -6];

B = [0;0;1];

C = [1 0 0];

D = 0;

Q = [8 0 0;
0 6 0;
0 0 4];
R = 1.5;

% gain matrix using lqr
[K1,S1,P1] = lqr(A,B,Q,R);

% closed-loop step response with the generated gain matrix K1.
% continuous model
sys1 = ss(A-B*K1,B,C,D)
x0 = [2;0;-2]; % initial state
tspan = 0:0.2:10;
initial(sys1,x0,tspan)
[y1,t1,x1]= initial(sys1,x0,tspan);

(b)

Response to Initial Conditions Amplitude 8 10 Time (seconds)

clear;
close all;
clc;

A = [0 2 0;
0 0 3;
0 -8 -6];

B = [0;0;1];

C = [1 0 0];

D = 0;

mul = 1;

Q = mul*[8 0 0;
0 6 0;
0 0 4];
R = mul*1.5;

% gain matrix using lqr
[K1,S1,P1] = lqr(A,B,Q,R);

% closed-loop step response with the generated gain matrix K1.

% discrete model
Ts = 0.2;
sys2 = ss(A-B*K1,B,C,D,Ts)
x0 = [2;0;-2];
tspan = 0:0.2:0.6;
figure()
initial(sys2,x0,tspan)
[y2,t2,x2]=initial(sys2,x0,tspan);

Comparison: The response in the discrete case diverges unlike the continuous case where the response converges to 0. The comparison plot is given below.

Response to Initial Conditions Amplitude 0.1 0.2 0.4 0.5 0.6 0.3 Time (seconds)

(c) The following lines are replaced.

10

mul = 10;

Q = mul*[8 0 0;
0 6 0;
0 0 4];
R = mul*1.5;

No noticeable change is observed in the responses.

Response to Initial Conditions Amplitude 0.1 0.2 0.4 0.5 0.6 0.3 Time (seconds)