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I have to design a PI-controller with zero order hold for the plant $G_s(s) = \frac{0.24}{(s+4)(s+12)}$ to meet the specifications of $\omega_n=5\,s^{-1}$ and $\zeta = 0.6$.

I choose a $T=0.12\,s$ which leads to a approx. 13 samples per oscillation (and $T \leq 0.125 T_d$)and applied the Z.O.H. to $H_o(s)G_s(s)=G(s)\approx \frac{1}{1+0.5 T s}\frac{0.24}{(s+4)(s+12)}$.

With the transfer function of the PI-controller: $G_D(s) = K_c\frac{a + s}{s}$ I get the characteristic equation with $H_c(s) = \frac{G(s)}{1+G_D(s)G(s)}$ to $\frac{4s}{s^3 + 32.667 s^2 (314.672 + 4 K_c)s + 800.016 + K_c a 4}$.

With the desired system of: $\frac{1}{(s^2 + 2 \omega_n \zeta s + \omega_n^2)(s + \alpha)}= \frac{1}{s^3 + 32.6s^2 + 184.6 s + 665}$ where I choose $\alpha=26.6$ with $\alpha \approx 4 \ldots 5 \omega_n$ (hence, the $s^2$-term is equal).

Now I did the comparison of coefficents:

$314.672 + 4 K_c = 184.6 \rightarrow K_c=-32.518$

$800.916 + K_c a 4 = 665 \rightarrow a = 1.038 $

Theoretically this should be now a proper $K_c$ and $a$. But simulation shows undesired behavior (step response goes to infinity). Should I introduce another pole? Is the general approach right?

(Remark: I edited the problem considerable with help from Design of a PI controller.)

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  • $\begingroup$ Why do you have to design a closed loop system with a given natural frequency and damping factor? Since usually you have bandwidth and phase margin specifications. Also why do you use $\frac{G}{1+CG}$ as closed loop and not $\frac{CG}{1+CG}$ (which I think is much more intuitive way of controlling)? $\endgroup$
    – fibonatic
    Commented May 12, 2016 at 23:27
  • $\begingroup$ It is for an exercise example so my guess the lecturer wanted to spare (us) the calculation of $\omega_n$ and $\zeta$. For the second question, I don't know which way is more common, and was using the common description of our course. $\endgroup$
    – Lageos
    Commented May 13, 2016 at 7:07
  • $\begingroup$ In that case I would assume that the dominant poles should have those $\omega_n$ and $\zeta$. You do not need to bother about zero-pole-cancellation in the open loop, just make sure that the two other poles die out faster. $\endgroup$
    – fibonatic
    Commented May 13, 2016 at 7:24
  • $\begingroup$ So the general approach - the values for $K_c$ / $a$ - are right? I don't see my mistake. $\endgroup$
    – Lageos
    Commented May 13, 2016 at 7:50
  • $\begingroup$ You should always check whether your results are correct when you get negative gains. $\endgroup$
    – fibonatic
    Commented May 13, 2016 at 8:25

1 Answer 1

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I think you've made a mistake when setting up your characteristic equation. In MATLAB I get the following for your system transfer function:

enter image description here

for your system transfer function.

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