What does my Root Locus response mean? Why is it different?

2 views (last 30 days)
Joel Okanta
Joel Okanta on 31 Mar 2022
Edited: Paul on 1 Apr 2022
Right at the moment I am learning about Root Locus of the system and how they are normally used for damping/oscillating systems. I often see Magnetic Levitation Papers use these to determine the root locus, which I guess must be to determine the oscillations in their system. However, unlike their ones, mine is a bit different, only sitting at the real axis. Is this alright? I mean it does make sense since both my roots exist on the real plane, but I'm curious and want to know more as I cannot find any relevant info about MagLevs and root locuses anywhere else.
My system:
sys = tf([-19008],[960.704 0 -2376000])
rlocus(sys)
Results:
Relevant Papers
Both Taken from IEEE:
Links:
https://ieeexplore.ieee.org/document/5570565

Answers (1)

Paul
Paul on 31 Mar 2022
The root locus is much more general than as suggested in the Question. Given a system described by an open loop transfer fuction L(s), the root locus are all values of s that solve, i.e., are the roots of, the equation
D(s) = 1 + K*L(s) = 0
as the scalar K varies over the range -inf < K < inf. Often the root locus is plotted only for 0 < K < inf ( the default for rlocus() ), but don't forget that sometimes we need to consider K < 0.
That equation is critical becase the roots of D(s) are the closed loop poles of a system with open loop transfer function K*L(s), and the closed loop poles determine the modes of the system. So the root locus technique applies to much more than just dampled oscillator systems, as the example in the Question illustrates with the closed loop poles one the real axis.
  2 Comments
Paul
Paul on 1 Apr 2022
The system in the Question is
Y(s)/U(s) = P(s), where P(s) is:
P = tf([-19008],[960.704 0 -2376000]);
Is this system stable? No, as we can see because it has one pole in the right-half-plane (RHP)
pole(P)
ans = 2×1
49.7311 -49.7311
That means that some sort of feedback control will be needed to stabilize the system. If using proportional control, the root locus of H(s) shows how the closed loop poles migrate as a function of K using compensation of the form u(t) = K*(r(t) - y(t))
figure;
rlocus(P) % default assumption that K > 0
Because one branch of the root locus is always in the RHP, the root locus tells us that proportional control with K > 0 cannot stablize the system.
Plot the root locus for K < 0
figure
rlocus(P,-(0:1000))
This plot tells us that for low, negative K there will be one unstable real closed-loop pole and one stable real closed loop pole. As a critical value of the gain, we have two closed loop poles at the origin and as the gain becomes more negative we have a conjugate pair on the imaginary axis (beause the poles of P are mirror images). Between these two plots (which could have been combined into a single plot) we know that that proportional feedback cannot stabilize the system. Some sort of dynamic compensation is necessary of the form K(s) = k*G(s).
For example, let G(s) = (s + 20)/(s + 80)
G = tf([1 20],[1 80]);
Then we plot the root locus of G*P for k < 0
figure
rlocus(G*P,-1000:0);
The plot shows that for k < kcrit (which is the gain where the red branch crosses the imaginary axis and happens to be ~ -500), the closed loop system with compensation of the form U(s) = (R(s) - Y(s))*K(s), with K(s) = k*G(s), will be stable because all of the branches of the root locus are in the left half plane. Whether or not a particular value of k yields desirable closed loop pole locations is another question.

Sign in to comment.

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!