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I have a nonlinear system (Ball & Beam) which is described by the following equations of motion:

$$ \ddot{y} + \frac{mg}{a} \sin(θ) -\frac{m}{a}y\dot{θ}^2 = 0 $$

$$ \ddot{θ} + \frac{2m}{b}y\dot{y}\dot{θ}+\frac{mg}{b}y\cos(θ) = u $$

where the letters stand for:

  • $y \rightarrow $ Ball Position
  • $ θ \rightarrow $ Angle of Beam
  • $ u \rightarrow $ External Torque Applied
  • $ b = my^2+J$
  • $a, m, g, J \rightarrow $ Positive Constants

Now, my final goal is to convert this nonlinear system into an equivalent nonlinear system of the Byrnes-Isidori normal form. The state-space representation ($\dot{x} = f(x)+g(x)u)$ of the system by considering the state variables $x_1 = y$, $x_2 = \dot{y}$, $x_3 = θ$ and $x_4 = \dot{θ}$ is:

\begin{gather*} \dot{x_1} = x_2 \\ \dot{x_2} = \frac{m}{a}x_1x_4^2-\frac{mg}{a}\sin(x_3) \\ \dot{x_3} = x_4 \\ \dot{x_4} = -\frac{2m}{b}x_1x_2x_4-\frac{mg}{b}x_1\cos(x_3)+\frac{1}{b}u \\ y = x_1 \rightarrow Output \end{gather*}

The relative degree of the system is $r=3 < n=4$ (resticted to the region where $D_o = \{x\epsilon R^4 \ | \ x_1 \neq 0 \ \& \ x_4 \neq 0\}$ ), so we have the internal states $μ_1$, $μ_2$ and $μ_3$, which are:

  • $μ_1 = y \Rightarrow \dot{μ_1} = μ_2$
  • $μ_2 = \dot{y} \Rightarrow \dot{μ_2} = μ_3 $
  • $μ_3 = \ddot{y} \Rightarrow \dot{μ_3} = \dddot{y} $

Now, in order to find the internal dynamics we need a function $ψ$ which validates the expression:

$$ \nabla{ψ}\cdot g(x) = 0 \Rightarrow \frac{\partial ψ}{\partial x_4}\cdot \frac{1}{mx_1^2+J} = 0 \Rightarrow $$

$$ \frac{\partial ψ}{\partial x_4} = 0, \ \text{since} \ \frac{1}{mx_1^2+J} > 0 $$

where $g = [0 \ \ 0 \ \ 0 \ \ \frac{1}{mx_1^2+J}]^T$. I chose the function to be $ψ(x) = x_1 + x_2 + x_3$. And now, the internal dynamics by differentiating this function is:

$$ \dot{ψ} = \dot{x_1}+\dot{x_2}+\dot{x_3} \Leftrightarrow \dot{ψ} = y + \dot{y} + x_4 $$

$$ \dot{ψ} = μ_1 + μ_2 + x_4 $$

Now, what's left to do is to express the term $x_4$ as a function of $ψ$ and $μ$. In order to do so, I followed the below procedure:

$$ μ_3 = \ddot{y} \Rightarrow μ_3 = \dot{x_2} = \frac{m}{a}x_1x_4^2-\frac{mg}{a}\sin(x_3) $$

The term $x_3 = ψ - x_1 - x_2 = ψ - μ_1 - μ_2$ and combibning these I obtained an expression for $x_4$:

$$ x_4 = [\frac{a}{m}\frac{μ_3}{μ_1}+g\frac{\sin(ψ-μ_1-μ_2)}{μ_1}]^{\frac{1}{2}} $$

My question is regarding the zero dynamics that can be found by $\dot{ψ} = w(0,ψ)$ meaning that I should set $μ_1=μ_2=μ_3=0$ while $μ_1 \neq 0$ (which also validates restriction for the relative degree to be well-defined). I am stuck at this point and considering what to do. I also considered to take a simpler $ψ$ function such as:

$$ ψ = x_1 + x_2 \Rightarrow \dot{ψ} = μ_2 + μ_3 $$

but this means that $\dot{ψ}$ only depends on $μ_1$ & $μ_2$ resulting the zero dynamics to be $\dot{ψ}(0,ψ) = 0$. Do I miss something regarding the procedure or maybe am I somewhere wrong ? Could use any help.

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  • $\begingroup$ Check Theorem 13.1 in H. K. Khalil, Nonlinear Systems: Pearson New International Edition, vol. Third editof Pearson New International Edition. Essex: Pearson, 2014. $\endgroup$ – useless-machine May 13 '20 at 16:57
  • $\begingroup$ I have only the old version and I assume the theorem is the one regarding the diffeomorphism $T(x)$. Well, haven't I followed the same procedure ? $\endgroup$ – Teo Protoulis May 13 '20 at 17:33
  • $\begingroup$ Yes and if you follow the proposed procedure you'll arrive at the desired. I did not check your answer. $\endgroup$ – useless-machine May 13 '20 at 18:23
  • $\begingroup$ You should pay attention to some terms. $\mu_1,\mu_2,$ and $\mu_3$ are not internal states. At least they are not related to the internal dynamics. They form the feedback-linearized part of the system and the $\Psi$ is your "internal" state. You should check if the nonlinear transformation $\chi=[\mu_1,\mu_2,\mu_3 \Psi]$ is, in fact, a diffeomorphism which you validate with the Jacobian that must have full rank (i.e. $n$). It should be sufficient to choose $\Psi=x_3$. If you find that the zero dynamics don't change, i.e. the rate of change is 0, you can apply the control law $\endgroup$ – link Nov 11 '20 at 12:54
  • $\begingroup$ Maybe you should take a look at this: ieeexplore.ieee.org/stamp/stamp.jsp?arnumber=119645 $\endgroup$ – link Nov 11 '20 at 12:54

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