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I am trying to design a PID controller, to get the response I want to, so I am beginning to think that my approach with tuning the values might be incorrect.

I have to Design a controller for the system G(S) such, that the step response to close loop transfer function has an overshoot of 0% and a settling time of less than 1 sec.

$$G(s) = \frac{1}{10s^2+5s+10}$$

the overshoot has to be 0% which means that we interested in a critical damped system ζ = 1 since the settling time has to be lower than 1 sec, I can deduce that ωn has to be less than 4. which means that, my poles shall be on the real axis and less than -4.

I chose I want my poles to be at -5 and -7, and using solve I am able to see that a PD controller consisting of d = 115 and p = 340 would do the job. But the step response shows me that there is about 12.5 percent overshoot, but the settling time matches. The close loop transfer function I end up with

What mathematically do: I write up the mathematical equation for the close loop transfer each each consisting of P,PI,PD,and PID, and solve the denominator for it's pole.

So if I want the poles to lie at s = -5 and s = -7 I would for P controller solve this equation

s1:= -5
s2:= -7

Solve [10s1^2 +5s1 + c+kp ==0 && 10s2^2 +5s2 + c+kp ==0 , p] for a PI controller.

Solve [s1 (a*s1^2 + b*s1 + c) + i + p*s1 == 0 && 
       s2 (a*s2^2 + b*s2 + c) + i + p*s2 == 0 , {p, i}  ]

And for a PID controller:

Solve[s1 (a*s1^2 + b*s1 + c) d*s1^2 + p*s1 + i == 0 && 
      s2 (a*s2^2 + b*s2 + c) d*s2^2 + p*s2 + i == 0, {p, i, d}]

Where a, b, c is parameter corresponding a second order equation $$as^2 + bs + c$$

$$G(s)=\frac{115s+340}{10s^2+120s+350}$$

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  • \$\begingroup\$ Are you wanting the overall closed-loop response to be as in your 1st equation? If not, what does the 1st equation represent? \$\endgroup\$
    – Andy aka
    Commented May 18, 2014 at 11:24
  • \$\begingroup\$ The close loop transfer function contains both the PD controller and the system which it is applied to. I just want to make an controller which make the response of the system due to a step a critical damped system. \$\endgroup\$ Commented May 18, 2014 at 11:41
  • \$\begingroup\$ 0% overshoot? Quick operation? With a super simple device? Please invite me when getting the Nobel award! \$\endgroup\$
    – Vorac
    Commented Aug 5, 2020 at 12:18

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