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Root Locus from equation?

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Ali Almakhmari
Ali Almakhmari on 4 Nov 2023
Answered: Star Strider on 4 Nov 2023
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.

Accepted Answer

Star Strider
Star Strider on 4 Nov 2023
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)
eqn = function_handle with value:
@(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
The original ‘eqn’ would only work for fimplicit, for example —
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.
.

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