Why linearise around an equilibrium point?

Question

In the control theory dealt in class I was taught to always linearise a nonlinear system around an equilibrium point, i.e., where $$\dot x = 0$$

However, linearization is a Taylor series expansion, so why the equilibrium point in particular? What happens if you don't use it?

I know that if the system is at the equilibrium states then the system is 'unperturbed' and will continue to stay in unperturbed state unless given an external disturbance / excitation / input. If you give an example please use non-zero equilibrium states.

Answer 1

When using Taylor series, there are two fundamental choices to be made:

1. A point (\$x_0\$) in which to evaluate the derivatives and obtain the series coefficients
2. The coefficient order to which the series should be truncated.

The truncation of the Taylor series causes the linearized representation to be not exact, in other words, a linearization error is incurred. This linearization error is zero at point \$x_0\$. The further away you take the system from \$x_0\$, the greater this error is in magnitude.

Why equilibrium point in particular? What happens if you don't use it?

The equilibrium point is frequently a sensible choice for \$x_0\$ when it is inferred that the system oscillates or gravitates around this point. Depending on application, this may change. If I were to word a statement on how to choose a center point \$x_0\$ for Taylor series, it would be something like this:

"Systems should be linearized around a point \$x_0\$, in such way that the choice of \$x_0\$ minimizes linearization error for all expected system trajectories. If the system oscillates or gravitates around an equilibrium point, then this is a sensible choice for \$x_0\$."

A choice of \$x_0\$ that does not follow such guidelines would lead to a linearized model that does not represent, within acceptable tolerances, the non-linear system.

Answer 2

Remember at the equilibrium points, the dynamical system doesn't move (i.e. the velocity is zero). It stays at this state forever. This makes sense since the velocity is zero (i.e. \$\dot{x}=0\$). Also remember, the trajectory of the system will be constant at these equilibrium points. Moreover, the equilibrium points could be any point, however, we can perform some transformation to convert the non-zero equilibrium point to the equilibrium point at the origin to simplify the analysis and apply theorems directly because a lot of theorems talk about the stability of the origin. Finally, once we know the trajectory is constant at these equilibria, starting from an arbitrary initial value or if it is disturbed, does it converge or diverge to/from an equilibrium point. Understanding this information allows us to check the stability without solving the differential equation.

Now why we use the linearization? Remember the difference between linear and nonlinear systems. Linear systems are very well understood and it is indeed a mature field. You can tell if the system is stable or unstable just by looking at the eigenvalues of the system's matrix or equivalently at the poles of the characteristics equation. Unfortunately, nonlinear systems can't be check by a glance. Each nonlinear system must be investigated independently. So instead of going through this hassle, we linearize the system and check its stability and we are done. Unfortunately, if at least one of the eigenvalues has a zero real part, then linearization may mislead you about the stability and it could provide you with incorrect answer. You may consider the linearization as a dirty/lazy or even clever approach but the price is it doesn't handle all scenarios.