Linearized system: Value of the δu(t)

Question

To obtain the response of the nonlinear system for any input, Jacobian linearization procedure can be applied. According to procedure, we have to change the state variables by using values of equilibrium points. \begin{align} \delta x(t)&=x(t)-x_0\\ \delta y(t)&=y(t)-y_0\\ \delta u(t)&=u(t)-u_0\\ \end{align}

According to MATLAB documentation, $y_0$ is the value of $y$ at the operating point and $x_0$ is the state vector at the operating point. In here i have a question. What is the value of the $u_0$?

My purpose is only simulate the nonlinear model, there is no controller in here. In my opinion, i should assume an acceptable $u_0$ according to system input. For example, if $u$ is a unit step function, what may be the value of $u_0$ for succesful simulation?

Note: I linearize my system around equilibrium points. There might be no equilibrium points or they may be unstable, i will investigate the details of them later.

Thanks,

Answer 1

The constant $u_0$ is the input signal value around which you perform linearization. Suppose that the nonlinear system is given by $$\dot{x}(t) = f(x(t),u(t)).$$ For the linearization around the equilibrium $(x_0, u_0)$ it holds $f(x_0,u_0)=0$. In other words, $u_0$ is such an input signal that makes the state $x_0$ an equilibrium of the autonomous system $\dot{x}(t)=f(x(t),u_0)$.

If $x_0$ is already an equilibrium, then $u_0=0$.