Why does my closed loop pole-zero map appear to indicate a critically damped system but a step input results in an under-damped response?

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Joseph Donnelly
Joseph Donnelly on 17 Dec 2021
Commented: Jon on 17 Dec 2021
**EDIT:forgot to show plant
As the question says my pole-zero map for a closed loop system with negative unity feedback appears to indicate that I should expect an critically damped response but when I subject my system to a unit step it is oscillatory. Can anyone offer an explanation or point me in the direction of releveant literature? Thanks
I have a model plant, G, with a time delay (dead time)** of -1 seconds which is defined as:
G = (2*exp(-1*s))/(3*s+1);
**Just as a side note, taking away the dead time has the same final result.
Plotting plant:
plot(t,y), grid;
I have the following Ziegler-nichols tuning parameters:
Kp = 1.8;
Ti = 3;
I have made the following PI controller for the plant:
controller = pidstd(Kp,Ti)
I have calculated the open and closed loop system as follows:
gol = controller*G;
gcl = feedback(gol,1);
The PZ map for the closed loop system is:
pzmap(gcl), grid
Based on the PZ map where there is a pole-zero cancellation at -0.333, leaving a single real pole at -1.2 and a damping of 1, I am expecting a critically damped reponse with no oscillations.
However when I plot a step response I get the following:
step(gcl,40),grid

Accepted Answer

Jon
Jon on 17 Dec 2021
Edited: Jon on 17 Dec 2021
If you remove the time delay the response has no oscillations as you expected. I know you said "**Just as a side note, taking away the dead time has the same final result.", but I think perhaps you didn't evaluate this correctly. Here I show the system without time delay and no oscillation:
G = 2/(3*s+1);
Kp = 1.8
Ti = 3
controller = pidstd(Kp,Ti)
gol = controller*G
gcl = feedback(gol,1)
step(gcl,40),grid
  4 Comments
Jon
Jon on 17 Dec 2021
If you are already somewhat familiar with the Nyquist Plot, and Nyquist Stability Criteria, I think this is a good way to see the effects of time delay on stability. The time delay keeps adding more and more phase lag with increasing frequency, which in turn changes the encirclements of the -1 point. I just skimmed this, but it looked like it was a good quick reference on the topic (see the later sections) http://www.cds.caltech.edu/~murray/courses/cds110/fa15/pdf/L7-3_delay+nyquist.pdf

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