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The system equations are:

$$ \ddot{\theta} = -\alpha \; |\theta| \; \theta + \sin(\theta) - \tau $$ $$ \tau(t) = 20 \; e^{-20t} \; v(t) $$

The system state space description is:

$$ \mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} \theta \\ \dot{\theta} \end{bmatrix} $$ $$ u = v $$ $$ y = \theta $$

The equilibrium values are $\alpha = -1,14$, $\theta = \frac{\pi}{4}$, $\dot{\theta} = 0$ and $v(t) = 0$.

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    – AJN
    Commented Jul 3, 2023 at 12:03
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    – mhdadk
    Commented Jul 6, 2023 at 1:38

1 Answer 1

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If we rewrite the system under the input affine form: $$\dot{x}=f(x)+g(t)v$$ i.e. $$\left[\begin{array}{c} \dot{x}_1 \\ \dot{x}_2\end{array}\right] = \left[\begin{array}{c} x_2 \\ -\alpha |x_1|x_1+\sin(x_1)\end{array}\right]+\left[\begin{array}{c} 0 \\ -20e^{-20t}\end{array}\right]v $$ The input-free system is at an equilibrium for all $x$ such that $f(x)=0$. In particular if $x_2=0$, $x_1=\pi/4$, and $\alpha=8 \sqrt{2}/\pi^2 \approx 1.146318...$, one can indeed check that $f(x)=0$.

The linearization of the input-free system is provided by the Jacobian of $f$ evaluated at the equilibrium $(x_1,x_2)=(\pi/4,0)$.

For all $x_1\neq 0$, the Jacobian is given by: $$\frac{\partial f}{\partial x}(x_1,x_2) = \left[\begin{array}{cc} 0 & 1 \\ -2\alpha |x_1|+\cos(x_1) & 0\end{array}\right]$$ Hence evaluated at $(x_1,x_2)=(\pi/4,0)$: $$\frac{\partial f}{\partial x}(\pi/4,0) = \left[\begin{array}{cc} 0 & 1 \\ (\sqrt{2}-\alpha \pi)/2 & 0\end{array}\right]$$

We obtain the following linearization: $$\left[\begin{array}{c} \dot{x}_1 \\ \dot{x}_2\end{array}\right] = \left[\begin{array}{cc} 0 & 1 \\ (\sqrt{2}-\alpha \pi)/2 & 0\end{array}\right]\left[\begin{array}{c} x_1 \\ x_2\end{array}\right]+\left[\begin{array}{c} 0 \\ -20e^{-20t}\end{array}\right]v $$

However, note that the input matrix is time-varying. This can easily be fixed by re-introducing $\tau$ with $$v(t)= \frac{e^{20t}}{20}\tau(t)$$

In the end:

$$\left[\begin{array}{c} \dot{x}_1 \\ \dot{x}_2\end{array}\right] = \left[\begin{array}{cc} 0 & 1 \\ (\sqrt{2}-\alpha \pi)/2 & 0\end{array}\right]\left[\begin{array}{c} x_1 \\ x_2\end{array}\right]+\left[\begin{array}{c} 0 \\ -1\end{array}\right]\tau $$

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