Electronic – Why linearise around an equilibrium point

control systemcontrol theorylinearitynon-linear

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.

Best Answer

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.