Root Locus from equation?
Afficher commentaires plus anciens
I have this equation:
syms s u
eqn = 1472.*s.^4 - 99.2.*s.^3 -256.*s.^2.*u.^2 + 1393.*s.^2 + 2.4.*s.*u.^2 - 25.*s -24.*u.^2 + 150 == 0;
where u is speed, and s are the eigenvalues of the systems. I want to plot the root locus (real vs imaginary part of the eigenvalues) as a function of the speed but I am not sure where to start because the root locus command in MATLAB doesnt take something like this.
Réponse acceptée
Plus de réponses (1)
In order to use rlocus, we need to get the characteristic equation in a form of 1 + u^2*N(s)/D(s) = 0.
Start with the equation
syms s u
eqn = 1472.*s.^4 - 99.2.*s.^3 -256.*s.^2.*u.^2 + 1393.*s.^2 + 2.4.*s.*u.^2 - 25.*s -24.*u.^2 + 150 == 0;
Sub in u2 = u^2
syms u2
eqn2 = subs(eqn,u^2,u2)
Find the coefficients of u2 and 1
[C,T] = coeffs(lhs(eqn2),u2,'All')
Now eqn can be expressed as 1 + u2*N(s)/D(s) = 0, where
N(s) = C(1);
D(s) = C(2);
Convert to numeric
N = sym2poly(N(s));
D = sym2poly(D(s));
and plot the root locus. Here, the root locus "gain" is u^2
figure
rlocus(tf(N,D))
We can verify by first having rlocus return the roots of the eqn for each value of u^2
[r,u2] = rlocus(tf(N,D));
and then verify that eqn is satisfied for the first (or second/third/fourth) root returned for the values of u^2.
figure
plot(abs(double(subs(lhs(eqn),[s,u^2],{r(1,:).',u2.'}))))
Catégories
En savoir plus sur Classical Control Design dans Centre d'aide et File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!




