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I got a plant $G(s)=\left(0.13s+1\right)/s^2$ to design a compensator which provides below demands:

  • Settling time : max 2s
  • Overshoot : max %35
  • Gain margin : min 10 dB
  • Phase margin : min 30 deg
  • Controller effort (r to u) : max 0.9
  • Bandwith : min 10 rad/s

The best architecture so far was the one below but I couldn't reach the demands.

enter image description here

assigning $C1=0.03\cdot\left(\left(s+80\right)\left(s+10\right)\right)/\left(\left(s+0.12\right)\left(s+1\right)\right)$, $C2=17.5\text{m}$ and $H=1$ results as below:

enter image description here

Can anyone explain or guide a design approach on how to handle $\left(s+a\right)/s^2$ type plants when designing compensators or mention some tips/shortcuts for architecture selection? How do we select the order of the controller?

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  • $\begingroup$ Is this a homework question? Or like a job? Could you show what your compensator could accomplish in a plot? What it does manage, and also doesn‘t manage? What don‘t you understand when it comes to design and choosing an order? $\endgroup$
    – DrMrstheMonarch
    Commented Oct 23, 2019 at 8:07
  • $\begingroup$ it is a job like question and I editted the question to show how my compensator affects. As to order, I know that we should select lower order compensator as much as possible and there is no exact prescription for the type of compensator selection. what I want to learn is that is there a relation or at least an approach between plant order with compensator order when we start to design a compensator. for example my plant is second order and so should I start with 2th order compensator (like lead-lag) in order to save time by skipping working on 1th order compensators. $\endgroup$
    – lsn
    Commented Oct 23, 2019 at 9:09
  • $\begingroup$ Is there a reason why you choose that specific architecture, instead of for example a single feedback loop? $\endgroup$
    – fibonatic
    Commented Oct 23, 2019 at 9:20
  • $\begingroup$ this architecture performed better than the others. I tried the others also. $\endgroup$
    – lsn
    Commented Oct 23, 2019 at 9:40
  • $\begingroup$ Maybe this will help, $-\frac{8.20546 \left(1. s^2-3.84499 s+118.343\right)}{s}$ $\endgroup$
    – DrMrstheMonarch
    Commented Oct 23, 2019 at 21:50

1 Answer 1

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Consider the traditional control diagram below.

control-diagram

If you set $C=(2\zeta\omega_ns+\omega_n^2)/(0.13s+1)$, then you'll get the following closed-loop system transfer function: $$ T=\frac{GC}{1+GC}=\frac{2\zeta\omega_ns+\omega_n^2}{s^2+2\zeta\omega_ns+\omega_n^2}. $$

This can be achieved through zero-pole cancelation, which is doable since $G$ has a zero in LHP.

Finally, you can easily attain your design requirements with $\omega_n \approx 10\,\text{rad/s}$ and e.g. $\zeta=\sqrt{2}/2$.

The requirement on the maximum controller effort cannot be attained with a pure LTI controller though. You may thus consider LTV and/or nonlinear approaches to this end for controlling a double integrator (for example relying on optimal control).

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