Root Locus from equation?
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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.
采纳的回答
It might if you aske it to and give it a causal system it can work with —
s = tf('s');
eqn = @(u) 1 / (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)
u = 1;
figure
rlocus(eqn(u))
grid
u = 10;
figure
rlocus(eqn(u))
grid
eqn = @(s,u) 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;
figure
fimplicit(eqn, [[-1 1]*10 [-1 1]*10])
grid
I would not consider that to be a root locus plot, however that is likely the only way to work with it as originally stated.
.
更多回答（1 个）
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.'}))))
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