Hi @영탁 한
- Settling time less than 2 seconds
- Overshoot less than 5%
- Steady-stage error less than 1%
Like many collaborative robot arms and autonoumous vehicles, it is necessary to design the desired reference trajectory so that the control systems can follow. In this exercise, a reference model (transfer function) is designed to satisfy the design criteria.
From the reference model, one can obtain the desired closed-loop pole and the pole placement technique (solving algebraic equations) can be applied. The rest of the controller and observer design are procedural steps. So, the key is the reference model design!
% Parameters
Rs = 0.47;
Lq = 1.23e-3;
B = 0.0002;
Kb = 0.42;
Kt = 0.73;
J = 6.5e-4;
% State-space Matrices of the Uncompensated Plant
A = [-B/J Kt/J;
-Kb/Lq -Rs/Lq];
B = [0;
1/Lq];
C = [1 0];
D = 0;
% Reference trajectory design
wr = 3; % reference omega, ω
zr = 0.7; % reference zeta, ζ
nr = wr^2;
dr = [1 2*zr*wr wr^2]; % x" + 2·ζ·ω·x' + ω²·x
Gr = tf(nr, dr) % reference model
stepinfo(Gr)
% Controller design
cp = pole(Gr) % controller poles
K = place(A, B, cp) % controller gains
% Closed-loop control system without observer
sys = ss(A-B*K, B, C, D);
Nbar = 1/dcgain(sys); % Normalizer (to rescale the reference input)
cls = ss(A-B*K, B*Nbar, C, D)
stepinfo(cls)
% Observer design
or = 10*real(cp(1)); % make the observer response 10 times faster
op = [or+1 or-1] % observer poles (make op ≪ cp but no repeated poles)
L = place(A', C', op)' % observer gains
% Combine state-feedback controller and observer
Aco = [A-B*K B*K;
zeros(size(A)) A-L*C];
Bco = [B*Nbar;
zeros(size(B))];
Cco = [C zeros(size(C))];
Dco = 0;
% Closed-loop observer-based control system
clco = ss(Aco, Bco, Cco, Dco)
step(clco, 6), grid on
stepinfo(clco)
sse = fix(1 - dcgain(clco)) % steady-state error
As shown, all three design criteria are satisfied.
