Linearisation

        The linear life lifts all spirits

The general state space representation doesn't have so many assumptions, which is great, however, it's also not very useful for analysis. Now we will assume something that very often isn't true, but most of control science is built around this wonderful assumption: linearity. There is an easy solution for systems that aren't linear, linearisation and forget anything ever happened!

Linearisation isn't super difficult: it's just a first order Taylor expansion of $f$ around an equilibrium point. So $$f(x,u) \approx f(\bar x, \bar u) + \underbrace{\left.\frac{\partial f(x,u)}{\partial x}\right|_{(\bar x, \bar u)}}_A \underbrace{(x-\bar x)}_x + \underbrace{\left.\frac{\partial f(x,u)}{\partial u}\right|_{(\bar x, \bar u)}}_B \underbrace{(u-\bar u)}_u$$ $$h(x,u) \approx h(\bar x, \bar u) + \underbrace{\left.\frac{\partial h(x,u)}{\partial x}\right|_{(\bar x, \bar u)}}_C \underbrace{(x-\bar x)}_x + \underbrace{\left.\frac{\partial h(x,u)}{\partial u}\right|_{(\bar x, \bar u)}}_D \underbrace{(u-\bar u)}_u.$$ If we are extremely intelligent, we can translate $f$ beforehand such that $f(\bar x, \bar u)=0$, perform the underbraced substitutions, and commit some proper notation heresy to end up with the prettiest equation of all: the linear state space $$ \dot x = A x + B u $$ $$ y = C x + D u .$$

$\frac{\partial f(x,u)}{\partial x}$ is called the Jacobian of $f$. A nice memory support is to remember that the Jacobian is multiplied with the state from the right such that each state derivative is multiplied with the same state, so the states vary horizontally and the function elements vertically.