Linearization

In the main text, we discussed linearization of relatively simple dynamical systems, such as a pendulum or a moving car. In this appendix, we provide a more comprehensive and formal treatment of linearization for general nonlinear systems. For the curious and the brave!

General nonlinear dynamical systems¶

A general dynamical system might consist of multiple coupled differential equations involving input signals $u_1,\dots,u_m$ and other signals $x_1,\dots,x_n$. So it will look something like:

What about higher derivatives? We can always deal with those by introducing additional variables. For example, consider the equations of motion for a pendulum:

We can let $x_1 = \theta$ and $x_2 = \dot\theta$ and $u_1 = T$. Then our equations become:

We can go a step further, and define $\x = [x_1,\dots,x_n]$ and $\u = [u_1,\dots,u_m]$, and also let $\f$ be the vector-valued function $[f_1,\dots,f_n]$. Then Eq. (1) collapses neatly to a single vector differential equation that can represent pretty much anything:

If we use the constant input $\u(t) = \u_0$ and initialize our system at $\x(0) = \x_0$, we will have $\dot x=0$, so $\x$ will not change and the system will remain at $\x(t) = \x_0$ for all $t\geq 0$. This is why we call $(\x_0,\u_0)$ an equilibrium point.

A system will typically have many possible equilibrium points. For example, consider the pendulum of Eq. (2). We can form many possible equilibria:

• Pick $\theta_0=0$ and $T_0=0$. The pendulum is pointing down and we apply no torque.

• Pick $\theta_0=\pi$ and $T_0=0$. The pendulum is pointing up and we apply no torque.

• Pick $\theta_0=\tfrac{\pi}{2}$ and $T_0=mg\ell$. The pendulum is perfectly balanced at 90°.

• Pick any $\theta_0$ and use $T_0=mg\ell\sin\theta_0$.

Linearizing a dynamical system¶

We can linearize a dynamical system about an equilibrium point in a similar way to how we linearized a multivariable function about a reference point. Start with a general dynamical system (4), pick an equilibrium point $(\x_0,\u_0)$, and apply the linearization of Eq. (5) to the right-hand side. We have one approximation for each $i=1,\dots,n$:

We can write this as one neat vector equation by defining the Jacobian matrix of partial derivatives with respect to $\x$ and $\u$

Then the linearization can be written compactly as

Now define the deviations from the reference as we did before, but now these are going to be small signals rather than being small numbers:

Since $\x_0$ is a constant, we have $\bdelta\dot\x = \dot\x$. Also, we chose $(\x_0,\u_0)$ to be an equilibrium point, so $\f(\x_0,\u_0)=\bm{0}$. Therefore, we can write (7) even more compactly as:

Since the Jacobians $\J_{\f}^{\x}(\x_0,\u_0)$ and $\J_{\f}^{\u}(\x_0,\u_0)$ are evaluated at the (constant) equilibrium point $(\x_0,\u_0)$, they are just matrices of numbers. In other words, The approximation of Eq. (9) is an LTI system!

Example: pendulum on a cart¶

Consider an inverted pendulum on a cart. We will not derive the equations of motion here since our focus is on linearization; if you’re interested, there are several nice derivations available online (here and here, for example).

The equations of motion are:

We would like to linearize these equations about the equilibrium point where the pendulum is perfectly upright ($\theta_0=0$) and the cart is stationary ($\dot x_0=0$, $\dot\theta_0=0$), with no external force applied ($f_0=0$).

We begin by putting our equations into the standard form $\dot\x = \f(\x,\u)$. We choose our state variables as

Substituting these definitions into the equations of motion and rearranging, we obtain:

Written in matrix form to expose the state derivatives, we obtain:

Inverting the matrix on the left and multiplying both sides by the inverse, we can isolate the derivative terms $\dot x_i$.

Multliplying out the right-hand side, we obtain the nonlinear equations of motion in the standard form of (4):

Let’s compute the Jacobian matrices $\J_{\f}^{\x}$ and $\J_{\f}^{\u}$ evaluated at the equilibrium point $(\x_0,\u_0) = 0$. We will show one sample calculation. To compute the $(2,3)$ entry of $\J_{\f}^{\x}$, we need to compute $\frac{\partial f_2}{\partial x_3}$ and evaluate it at the equilibrium:

Proceeding in a similar manner for all entries, we obtain the following Jacobian matrices at the equilibrium point:

Finally, substituting these Jacobians into Eq. (9), we obtain the linearized equations of motion for the inverted pendulum on a cart about the upright equilibrium:

If we want, we can return this to our non-vector coordinates:

Linearizing about a trajectory¶

So far, we only discussed linearization about an equilibrium point, which leads to an LTI approximation. It is also possible to linearize about a time-varying trajectory $(\x_0(t),\u_0(t))$. This is useful for control systems where we want to follow a desired trajectory rather than stay at a fixed point.

In this case, we must ensure that the trajectory is feasible, meaning that it satisfies the system dynamics:

With this in place, we can linearize the right-hand side of (4) about the time-varying trajectory $(\x_0(t),\u_0(t))$ using the same approach as before. We obtain:

Before we had $\bdelta\x(t) = \x(t)-\x_0$ (constant reference), and therefore $\bdelta\dot{\x}(t) = \dot\x(t)$, but now $\x_0$ is time-varying, so we have to be careful when taking derivatives. We now have: $\bdelta\x(t) = \x(t)-\x_0(t)$ (time-varying reference), so $\bdelta\dot{\x}(t) = \dot\x(t) - \dot\x_0(t)$. Substituting this into (22) and using (21), we obtain:

Although this looks very similar to (9), there is a key difference: we are now evaluating the Jacobians at the time-varying trajectory $(\x_0(t),\u_0(t))$ rather than at the fixed equilibrium point $(\x_0,\u_0)$. So the Jacobians $\J_{\f}^{\x}(\x_0(t),\u_0(t))$ and $\J_{\f}^{\u}(\x_0(t),\u_0(t))$ will be time-varying matrices, and the resulting linearized system will be a linear time-varying (LTV) system rather than an LTI system.