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The length of the initial condition X0 must match the number of states.

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hello everyone i cant seem to figure out what the problem with my code. it keeps giving me an error regarding the initial state. any help would be appreciated.
thanks!
here is my code.
A= [0 1 0; 0 0 1; -52 -30 -4];
B = [0;0;1];
C = [20 1 0];
D = [0];
% compute open loop state space
sys_OL = ss(A,B,C,D);
% scaling the B matrix
w = -C/A*B;
% open loop state space with scaled input
sys_OL2 = ss(A, B/w, C, D);
p0 = 5;
Ts = 2;
term = pi^2 +log(p0/100)^2;
zeta = abs(log(p0/100))/sqrt(term);
wn = 4 /Ts/zeta;
beta = sqrt(1-zeta^2);
p3 = 10 * zeta *wn
eigDes = [-zeta*wn+j*wn*beta -zeta*wn-j*wn*beta -p3 -20+i]
% compute the close loop
Ae= [0 1 0 0; 0 0 1 0; -52 -30 -4 0; -20 -1 0 0];
Be = [0;0;1; 0];
Ce = [20 1 0 0];
De = [0];
Br = [0;0;0;1]
Ke = acker(Ae, Be,eigDes) % gain matrix
sys_Cl = ss(Ae - Be*Ke, Br ,Ce,De) %transfer function of close loop
t = 0:0.001:5;
u = ones(size(t));
x0 = [0.01; 0; 0; 0];
x0hat = [0; 0; 0; 0 ];
yOL = lsim(sys_OL2, u, t, x0);
yCl = lsim(sys_Cl, u, t, [x0; x0hat]);
  4 comentarios
Johannes James
Johannes James el 14 de Mzo. de 2021
It's very bad form to just say "i figured it out" and not exaplin what the solution was. Others coming here following search engine hits, looking for solutions, encounter your useless comment and absence of solution. If you bothered to post, concludet the matter with the solution!

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