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.
