I have a third order process and I have to design a PID controller by cancelling out the two most dominant poles of the process. The process transfer function is:


The controller is of the form, where Kc is the controller gain: Controller

To cancel out the poles the controller becomes, this is multiplied by Kc:


Now I have to cascade the controller and the process and calculate the transfer function with feedback (negative feedback), the feedback gain is 4. The problem is that I'm supposed to calculate the value of Kc using the standard second order transfer function, but that is not what I get, because of the feedback gain being 4 and not 1.

Standard 2nd order tf


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    \$\begingroup\$ This can only be homework. Show your calculation of the open-loop gain with the plant, the controller, and the gain of 4 in the feedback. \$\endgroup\$ – TimWescott Dec 26 '18 at 23:16
  • \$\begingroup\$ @TimWescott I updated the question. My teacher says I have to take the value of KC using the standard tf however the third term in the denominator does not match the numerator \$\endgroup\$ – Pedro Dec 26 '18 at 23:36
  • \$\begingroup\$ OK. You simply can't do that. The standard transfer function, as you give it, has a DC gain of 1. A loop with a PID controller in it, and a plant with a DC gain \$\ne 0\$, has a DC gain equal to the reciprocal of the DC gain of the feedback (this is a good thing to work out on your own, BTW). So, no matter what, you're going to have a transfer function with a DC gain of 1/4. \$\endgroup\$ – TimWescott Dec 26 '18 at 23:52
  • \$\begingroup\$ Thank you, so I should do what I did first and assume the gain is 1 when calculating KC \$\endgroup\$ – Pedro Dec 27 '18 at 0:01
  • \$\begingroup\$ I hope that your instructor wants you to just pay attention to the denominator of the transfer function. If you assume that \$K_s = 1\$ then you'll surely get a value of \$K_v\$ that would be wrong in reality. But this is boiling down to second-guessing what your instructor really wants, and you're in a much better position to do that than me. \$\endgroup\$ – TimWescott Dec 27 '18 at 1:32

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