Passive suspension state space equation This is the problem.
I have to choose state variables. It's a second order problem, so calling z the position of the car body, p the road profile, we could write:
$$u(t)=p(t)$$
$$y(t)=z(t)+Mg/k$$
$$M\ddot{z} = -Mg -k(z(t)-p(t)) - \beta(\dot{z}(t)-\dot{p}(t))$$
As state variables I thought:
$$x_1=z(t)$$
$$x_2=\dot{z}(t)$$
$$x_3=u(t)$$
$$x_4=\dot{u}(t)$$
The state equation would be:
$$\dot{x_1}(t)=x_2(t)$$
$$\dot{x_2}(t)=\frac{1}{M}(-Mg-k(z(t)-p(t)) - \beta(\dot{z}(t)-\dot{p}(t))$$
$$\dot{x_3}(t)=x_4(t)$$
$$\dot{x_4}(t)=????????$$
I badly chose the state variables I think... I did many attempts but with no result. Could you tell me how to go on?
Thanks a lot!
 A: You have got the wrong idea about the states of a dynamical system. The states are those independent variables that describe the whole dynamic of the system.
Generally a dynamical system is described by:
$$\dot{x}=f(x)+g(x,u)$$
where $x$ is the vector of states. In this example (as well as others), $u$ is the $\color{red}{\text{input}}$ of the system. So it is not part of the system and cannot be considered as one of the states. If a system is physically moving by an external force, then its position and velocity are changing according to Newton's law. So it is reasonable to take position and velocity as the two states of the system:
$$x_1=z,\;\;x_2=v_z=\dot{z}$$
Since $\sum F=Ma$ and the force applied by the spring is $k\Delta z$, choose the origin such that $\Delta z=z$ to get a more simplified equation:
$$Ma=M\ddot{z}=M\dot{x}_2=-Mg+k\,z-\beta\dot{z}+u=k x_1-\beta x_2+(u-Mg)$$
and be careful about the signs. Since this is the only motion of the system, it seems the system's dynamic is fully described. This finally leads to:
$$\pmatrix{\dot{x_1}\\\dot{x_2}}=\pmatrix{0&1\\\frac{k}{M}&-\frac{\beta}{M}}\pmatrix{x_1\\x_2}+\pmatrix{0\\\frac{u}{M}-g}$$
Edit:
It is worth mentioning that we generally prefer to choose the states in such a way that the fixed term is eliminated. Here for example, choose the origin where $kz_0=-Mg$, so that
$$k\Delta z-Mg=kz$$
and
$$\dot{x}=\pmatrix{0&1\\\frac{k}{M}&-\frac{\beta}{M}}x+\frac{1}{M}\pmatrix{0\\1}u$$
