# What is the appropriate type of controller for this system?

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:

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 $$K_p=6$$:

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?

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:

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:

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.

• 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? – snitchben Apr 11 '19 at 14:09
• 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. – am304 Apr 11 '19 at 15:32
• 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. – snitchben Apr 12 '19 at 2:05