Would you, please, help to linearize this system state equations?

Question

State equation: $$\dot{x}_1 = x_2$$ $$\dot{x}_2 = -\frac{k}{M}\exp(-x_1)x_1-\frac{h}{M}x_2+\frac{1}{M}u$$

Output equation: $$y= x_1$$ Would you, please, help to linearize this system state equations?

Answer 1

In order to linearize a system of the form $$\dot{\boldsymbol{x}}=\boldsymbol{f}(\boldsymbol{x},\boldsymbol{u})$$ $$\boldsymbol{y}=\boldsymbol{h}(\boldsymbol{x},\boldsymbol{u})$$

at the reference state $\boldsymbol{x}_R$ and the reference input $\boldsymbol{u}_R$ you need to evaluate the following expressions $$\Delta\dot{\boldsymbol{x}}=\dfrac{\partial\boldsymbol{f}}{\partial\boldsymbol{x}}\bigg|_{\boldsymbol{x}_R,\boldsymbol{u}_R} \Delta\boldsymbol{x}+\dfrac{\partial\boldsymbol{f}}{\partial\boldsymbol{u}}\bigg|_{\boldsymbol{x}_R,\boldsymbol{u}_R} \Delta\boldsymbol{u}$$ $$\Delta\boldsymbol{y}=\dfrac{\partial\boldsymbol{h}}{\partial\boldsymbol{x}}\bigg|_{\boldsymbol{x}_R,\boldsymbol{u}_R} \Delta\boldsymbol{x}+\dfrac{\partial\boldsymbol{h}}{\partial\boldsymbol{u}}\bigg|_{\boldsymbol{x}_R,\boldsymbol{u}_R} \Delta\boldsymbol{u}.$$

The following abbreviations were used $\Delta \boldsymbol{x}=\boldsymbol{x}-\boldsymbol{x}_R$ and $\Delta \boldsymbol{u}=\boldsymbol{u}-\boldsymbol{u}_R$.

The gradients

$$\dfrac{\partial\boldsymbol{f}}{\partial\boldsymbol{x}},\dfrac{\partial\boldsymbol{f}}{\partial\boldsymbol{u}}, \dfrac{\partial\boldsymbol{h}}{\partial\boldsymbol{x}} \text{ and } \dfrac{\partial\boldsymbol{h}}{\partial\boldsymbol{x}} $$

are all evaluated at ${\boldsymbol{x}_R,\boldsymbol{u}_R}$.

For your case, this is very easy. Your first state equation is already linear so is your output equation. You only need to linearize the second state equation. We obtain:

$$\Delta \dot{x}_1=\Delta x_2$$ $$\Delta \dot{x}_2 = -\frac{k}{M}\left[-\exp(-x_1)x_1+\exp(-x_1)\right]\bigg|_{x_{R1}}\Delta x_1-\frac hM \Delta x_2 + \frac{1}{M}\Delta u$$

$$\Delta y = \Delta x_1.$$

Answer 2

$$\left(\array{\dot{z_1} \\ \dot{z_2}}\right)=\left[\array{0 & 1\\ \frac{-k}{M}e^{-x_1}(1-x_1) & \frac{-h}{M}}\right] \left(\array{z_1 \\ z_2}\right) +\left(\array{0\\ \frac{u}{M}}\right)$$

$$y=(\array{1 & 0})\left(\array{z_1 \\ z_2}\right)+\left(\array{x_1 \\ 0}\right)$$

I have used z variables instead of x variables to state the linearized equation in terms of the basis vectors of the linear space (the tangent space) at $x$. Looking at the answer should tell you how we get to that answer. If something is not clear, you can leave me a message in the comments and I'll reply.