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I am given this transfer function for a system: $T\left(s\right)=\frac{9.81}{s^4+0.5s^3+11.2815s^2+4.905s}$

Now, I want to design a controller that will make the closed loop system have an overshoot due to a step input equal to $5\%$. A controller with a proportional gain $K_p=1$ gives me this step response of the closed-loop system on MATLAB:

pcontrol

As you can see, the overshoot with this proportional gain is way bigger than $5\%$. As K increases, the response becomes weirder.

Here, I have tried $K_p=6$:

pcontrol

PI, PD or PID controllers give a similar response to the figure above. So, it seems the best I can do is a controller with $K_p=1$, which is not acceptable in my case.

What would be the appropriate type of controller for this system to get a small overshoot?

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I used optiPID in Octave where the plant P(s) is controlled by a PID controller with second-order roll-off:

                 1                1
C(s) = Kp (1 + ---- + Td s) -------------
               Ti s         (tau s + 1)^2

in the usual negative feedback structure

         L(s)       P(s) C(s)
T(s) = -------- = -------------
       1 + L(s)   1 + P(s) C(s)

With your transfer function, and by and tweaking the weighing factors to put more weight on minimising overshoot, I got the following values (the ones of interest here are the optimised values):

optiPID: Astrom/Hagglund PID controller parameters:
kp_AH =  0.032753
Ti_AH =  0.63360
Td_AH =  0.16028
warning: optiPID: optimization starts, please be patient ...
warning: called from
    optiPID at line 215 column 1
Max no. of function evaluations exceeded...quitting
optiPID: optimized PID controller parameters:
kp_opt =  0.017183
Ti_opt =  0.14305
Td_opt =  0.50410
optiPID: closed-loop stability check:
st_AH = 1
st_opt = 1
optiPID: gain margin gamma [-] and phase margin phi [deg]:
gamma_AH =  2.2796
phi_AH =  80.408
gamma_opt =  4.3438
phi_opt =  64.139

The step response looks like this:

enter image description here

which gives and overshoot of approx. 5.7%. The amount of overshoot is actually governed by the integral gain, not the proportional one (which only influences the speed of response) so I have played around with it to reduce the overshoot further and using Ti_opt = 0.2, I get the following step response:

enter image description here

Which gives only about 1.6% overshoot.

The roll-off parameter tau is calculated as Td/10, which gives the following transfer function for the PID controller:

>> C_opt

Transfer function 'C_opt' from input 'u1' to output ...

      0.001732 s^2 + 0.003437 s + 0.01718
 y1:  -----------------------------------
      0.0005082 s^3 + 0.02016 s^2 + 0.2 s

Continuous-time model.

If you're not bothered about the response time, the Astroem/Haegglund PID (with kp_AH, Ti_AH and Td_AH) has essentially zero overshoot:

 >> C_AH

Transfer function 'C_AH' from input 'u1' to output ...

        0.003326 s^2 + 0.02075 s + 0.03275
 y1:  --------------------------------------
      0.0001628 s^3 + 0.02031 s^2 + 0.6336 s

Continuous-time model.
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  • $\begingroup$ Thank you for your answer @am304. Based on the root locus of my transfer function and the overshoot criteria, first I found the proportional gain to be $K_p=671$. However, when closing the loop, the step response would be not how I expected, similar to the second figure in my question. In the problems I had done previously, the overshoot criteria would be satisfied immediately after finding a desired point on the Root Locus. Thus, I played with the control system toolbox in MATLAB and it seems that for $K_p<1$ I get a very small overshoot. How do you explain that theoretically? $\endgroup$ – snitchben Apr 11 at 14:09
  • $\begingroup$ Using the root locus, I found that Kp = 0.075 (approximately) is the maximum value you can have before poles move into the complex right-hand plane, indicating that the system becomes unstable beyond that. You must have made a mistake. $\endgroup$ – am304 Apr 11 at 15:32
  • $\begingroup$ You are right, I had made a mistake. However, I played a bit with the control system toolbox and I managed to achieve the smallest settling time for a proportional gain $K_p=0.08$. How is this possible? Normally, settling time would decrease the more we go towards the left, but here it is decreasing for a certain area of root locus when moving to the right of it. $\endgroup$ – snitchben Apr 12 at 2:05

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