# How do you linearize this nonlinear system?

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 $$v(t) = 0$$.

• Welcome to Engineering SE!. Please visit the tour page if you have not already done so. Please improve your question by editing it. Please mention what you have tried so far and where you are stuck.
– AJN
Commented Jul 3, 2023 at 12:03
• What have you tried so far? Showing some effort on your part will likely yield answers from others Commented Jul 6, 2023 at 1:38

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:

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