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I was given the following equations and asked to design a controller for $u=-Kx$ using a pole placement method with the closed loop system having the damping and settling time $T_s$ given.

Am I supposed to use the $\dot x_1$ and $\dot x_2$ equations or are they totally unrelevant?

Also, which pole placement method is easier for the case, root locus or Bode and Nyquist plots?

$$\begin{gather} \dot x_1 = x_2 \\ \dot x_2 = -x_1 +\dfrac{1}{6}x_1^5-x_2+u \\ u = -Kx \quad \zeta=1.02 \quad T_s = 0.40 \end{gather}$$

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    $\begingroup$ Neither of your proposed methods will be applicable, because your system is not linear (because of $x_1^5$). $\endgroup$
    – fibonatic
    Nov 23, 2016 at 0:42
  • $\begingroup$ so, the process would be to linealize and then design the controller? $\endgroup$
    – spe4ker
    Nov 23, 2016 at 14:02
  • $\begingroup$ Yes. Also are you sure that $\dot{x}_1=\dot{x}_2$, or should it be $\dot{x}_1=x_2$? $\endgroup$
    – fibonatic
    Nov 23, 2016 at 15:58
  • $\begingroup$ I'm pretty sure it is just $x_2$. $\endgroup$
    – Suba Thomas
    Nov 29, 2016 at 15:44

1 Answer 1

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Step 1

The first thing to do is to determine the desired poles.

The natural frequency can be computed using $\omega =\frac{4}{\zeta T_s}$. This is a rule-of-thumb calculation for underdamped systems. The system here is slightly overdamped and is nonlinear as well. If the desired settling time is not obtained in the end, we have to come back and increase the constant 4 slightly. The design procedure is typically iterative. So we start with $\omega =9.80392$

Then the characteristic equation can be computed as $s^2+2 \zeta s \omega +\omega ^2$, which after substituting values gives $s^2+20 s+96.1169$ and has roots $-11.9706$ and $-8.02944$

Step 2

Put the system in a linear from $$\dot{x}=\text{A}.x+\text{B}.v$$ where $$x=\left( \begin{array}{cc} x_1 & x_2 \\ \end{array} \right)^T \ \ \ v=u+\frac{x_1^5}{6}$$ $$A=\left( \begin{array}{cc} 0 & 1 \\ -1 & -1 \\ \end{array} \right) \ \ \ \ \ B=\left( \begin{array}{c} 0 \\ 1 \\ \end{array} \right)$$

Step 3 Do the pole-placement design which gives $v=-k.x$ using Ackerman's formula. $$k=\left( \begin{array}{cc} 0 & 1 \\ \end{array} \right). \left( \begin{array}{cc} \text{B} & \text{A}.\text{B} \\ \end{array} \right)^{-1}. (\text{A}^2+20 \text{A}+96.1169I)$$

Substituting values, we get $$k=\left( \begin{array}{cc} 95.1169 & 19 \\ \end{array} \right)$$

Step 4

Do the back transformation to get the value for $u$. $$u+\frac{x_1^5}{6}=-95.1169 x_1-19 x_2$$ $$u=-95.1169 x_1-19 x_2-\frac{x_1^5}{6}$$

Step 5

Verification. We have to see if the design has met the requirements. (These simulations were done in Mathematica. The above calculations could also have been done there. I went through them manually above to explain things.) From the plot we see that the settling time constraint has been satisfied.

enter image description here

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  • $\begingroup$ Very detailed explanation. I understand that the main point was to start from the assumption of a dominant second order behaviour and finding the closed loop poles of s2+2ζsω+ω2=0. If I don´t have access to mathematica or matlab could I calculate the settling time with e^ζsωts by just looking at the standard graph? $\endgroup$
    – spe4ker
    Dec 1, 2016 at 4:30
  • $\begingroup$ Additionaly, if I wanted to prove the stability of the system via lyapunov function method, how can I approximate an initial equation? $\endgroup$
    – spe4ker
    Dec 1, 2016 at 4:38
  • $\begingroup$ 1. There is actually no assumption or approximation involved in the design because the system is second-order (and additionally we did not throw out any of the nonlinear dynamics). 2. If we look at the controlled system we see that the nonlinearity $\frac{x_1^5}{6}$ is small, so it could be neglected and the settling time can again be approximated from the linear system as $\frac{4}{\zeta \omega }$. This is an approximation. For a more conclusive verification you will have to resort to simulation as I have shown. $\endgroup$
    – Suba Thomas
    Dec 1, 2016 at 14:11
  • $\begingroup$ The equations $\dot{x}=A.x+B.v$, $v=-K.x$ are equivalent to the final system. You can do the Lyapunov stability analysis for this as you would do for any such linear system. $\endgroup$
    – Suba Thomas
    Dec 1, 2016 at 14:18

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