Question

3. Consider a system with the following state equation h(t)] [0 0 21 [X1(t) [x1(t) y(t) [0.1 0 0.1x2(t) X3(t) The unit step response is required to have a settling time of less than 2 seconds and a percent overshoot of less than 5%. In addition a zero steady-state error is needed. The goal is to design the state feedback control law in the form of u(t) Kx(t) + Gr(t) (a) Find the desired regon of the S-plane for two dominant closed loop eigenvalues to ensure the settling time and overshoot requirements are satisfied. Pick the desired location of dominant eigenvalues as well as the third eigenvalue. (b) Use MATLAB to find the feedback gain matrix K that places the closed-loop eigenvalues at desired locations selected in part (a). (use place or acker commands) (c) Find the value of G that satisfies the steady-state error requirement. (d) With r(t) as the input, use MATLAB to plot the unit step response of the closed-loop system and verify that all requirements are satisfied.

Answer 1

Dear user,

Solution (a)

Given from the requirements calculate zeta and natural frequency .

This comes out to be

zets =-0.6901 and wn= 2/zeta from setting time condition .

Using matlab , we will find K and G value.

Please find the code attached

matlab Code:

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

clc;

clear all;

close all;

%%

A=[0 0 2;0 1 1;-2 -1 3];

B=[1 1 1]';

C=[0.1 0 0.1];

%% Given overshoot less than 5% and settling time less than 2 sec

zeta=0.6901; % Value got from given 5% Overshoot

wn=2/zeta;

poles=[-zeta*wn+i*wn*sqrt(1-zeta^2);

-zeta*wn-i*wn*sqrt(1-zeta^2);

-50]; % Choose the third pole to be 10 times of dominant pair of poles

K=acker(A,B,poles);

G=-inv((C*inv(A-B*K))*B);

system=ss(A-B*K,B*G,C,0);

system1=ss(A-B*K,B,C,0);

[y,t]=step(system);

[y1,t1]=step(system1);

plot(t,y);hold on;grid on

plot(t1,y1);

legend('With G','Without G');

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

Output:

With G Without G 6 4 X: 3.174 0 -2 6 2.5 3.5 0 0.5 1.5

As you can see with out G , we have steady state error but requirements are met.

But as you keep R we have no control on overshoot but settling time condition is met and steady state error is zero.

Thus we have met our requirements except peakovershoot .