Linearization

Linearization

A nonlinear state-space system
$\dot{x}(t)=f(x(t),u(t)),y(t)=h(x(t),u(t))$
can be approximated by a linear-time-invariant (LTI) model near an equilibrium point $(x_0,u_0)$ satisfying $f(x_0,u_0)=0$. Defining perturbations $\delta x=x-x_0$, $\delta u=u-u_0$, $\delta y=y-h(x_0,u_0)$, the first-order Taylor expansion gives
$\dot{\delta x}=A\delta x+B\delta u,\delta y=C\delta x+D\delta u,$
where the matrices are the Jacobians of $f$ and $h$ evaluated at the equilibrium:
$A={\frac{\partial f}{\partial x}|}_{(x_0,u_0)},B={\frac{\partial f}{\partial u}|}_{(x_0,u_0)},C={\frac{\partial h}{\partial x}|}_{(x_0,u_0)},D={\frac{\partial h}{\partial u}|}_{(x_0,u_0)}.$

The linearized model is valid only near $(x_0,u_0)$: it captures the first-order response to small perturbations and discards second-order and higher terms. The approximation is accurate as long as the state stays close to the equilibrium; far-from-equilibrium behavior requires the original nonlinear model or a different local linearization.

Linearization is the bridge that lets all of LTI theory — eigenvalues, pole placement, controllability — apply to real, nonlinear systems. For a robot, the standard move is: identify an operating configuration $({\theta }_0,{\dot{\theta }}_0=0)$ with gravity-compensating torque ${{\rm Y}}_0$, linearize the manipulator dynamics, and design a controller for small deviations.

Linearization Is About a Point, Not a Trajectory $\dot{\delta x}=A(t)\delta x+B(t)\delta u$) or uses feedback linearization to cancel the nonlinearity exactly. The simple "linearize about equilibrium" approach only works for set-point regulation.

Linearization about an equilibrium produces a single LTI model. For tracking a moving reference, one either re-linearizes continuously along the trajectory (yielding a time-varying linear model

Intuition

Linearization is a multivariable version of replacing a curve by its tangent line. At the equilibrium, $f$ and $h$ are zero (by construction) and vary smoothly with $x$ and $u$. The first-order Taylor expansion $f(x_0+\delta x,u_0+\delta u)\approx A\delta x+B\delta u$ captures how small deviations propagate — the curved nonlinear dynamics look straight if you zoom in enough.

The matrices $A$, $B$, $C$, $D$ are all Jacobians. This matches the convention of robot Jacobians, which are also Jacobians of a nonlinear map (the forward kinematics). In both cases, the Jacobian tells you how small changes in the input translate to small changes in the output. Control-theory Jacobians take partial derivatives with respect to state and input rather than configuration, but the calculus is identical.

The crucial point is that linearization is a local approximation. A simple pendulum linearized about its downward equilibrium behaves like a harmonic oscillator for small angles, but linearization about the upward (inverted) equilibrium gives a fundamentally different system — unstable, with eigenvalues on both sides of the imaginary axis. Which equilibrium you pick determines the LTI model you get.

Properties

Procedure

(Find the equilibrium) Solve $f(x_0,u_0)=0$ for a fixed-point $(x_0,u_0)$.

Equilibria are the states where, given a constant input, the system remains at rest. For the pendulum $\ddot{\theta }=-g/\ell \sin \theta +u$, the equilibria satisfying $\dot{\theta }=0$, $\ddot{\theta }=0$ are ${\theta }_0=0$ (down, $u_0=0$) and ${\theta }_0=\pi$ (up, $u_0=0$). A system may have multiple equilibria, each with its own LTI model.

(Compute the Jacobians) $A=\partial f/\partial x$, $B=\partial f/\partial u$, $C=\partial h/\partial x$, $D=\partial h/\partial u$, each evaluated at $(x_0,u_0)$.

The Jacobian $A$ captures how small state perturbations feed back into $\dot{x}$; $B$ captures how input perturbations drive $\dot{x}$. The output Jacobians $C$ and $D$ capture how $(x,u)$ perturbations affect the measured $y$. For physical systems $D=0$ almost always.

(Work in perturbation variables) $\delta x\triangleq x-x_0$, $\delta u\triangleq u-u_0$, $\delta y\triangleq y-y_0$.

The linearized system governs $\delta x$, not $x$ directly. Converting back to absolute variables requires adding $(x_0,u_0,y_0)$ to the linearized solution. This is straightforward but easy to forget when interpreting simulation results.

Properties of the Jacobian

(Linear map) The Jacobian of a linear function $f(x)=Mx$ is just $M$.

This is why LTI systems are their own linearization: $\partial (Ax)/\partial x=A$. For a linear input-affine system $\dot{x}=Ax+Bu$, linearization at any equilibrium gives back the same $(A,B)$.

(Chain rule) Composition composes Jacobians: if $f=g\circ h$, then $Df=Dg\cdot Dh$.

This is the multivariable chain rule. For cascaded systems (a motor driving a gearbox driving a link), the overall linearization is the product of per-stage Jacobians.

(Symbolic tools) SymPy, MATLAB Symbolic Math Toolbox, Mathematica, and Maple compute Jacobians symbolically.

For anything beyond a 2- or 3-DoF system, computing $A=\partial f/\partial x$ by hand is error-prone. Symbolic computation is the standard practice; the result is then evaluated numerically at the equilibrium point.

Validity

(Local approximation) The linear model agrees with the nonlinear one to first order in $(\delta x,\delta u)$; errors grow quadratically with distance from equilibrium.

If the nonlinear $f$ is Taylor-expanded as $f(x_0+\delta x)=A\delta x+\frac{1}{2}\delta x^{T}H\delta x+\cdots$, the Hessian term $H$ bounds how fast the linear model degrades. Rule of thumb: within $\sim 10°$ of a joint equilibrium the linear model is usually adequate; beyond $\sim 30°$ you probably need the nonlinear model.

(Stability preservation) If the linearization has all eigenvalues in the open LHP, the nonlinear system is locally asymptotically stable at the equilibrium (Lyapunov’s first method).

Linearization provides a rigorous stability test: negative real parts imply local stability of the nonlinear system. Eigenvalues with zero real parts are inconclusive — the nonlinear terms determine stability there.

Examples

Linearizing a Pendulum

The pendulum $\ddot{\theta }=-(g/\ell )\sin \theta +(1/m{\ell }^2)\tau$ has state $x=(\theta ,\dot{\theta })$. Linearize about the downward equilibrium $({\theta }_0,{\dot{\theta }}_0,{\tau }_0)=(0,0,0)$.

Solution $f(x,u)=\left[\begin{array}{c}{x}_2\\ -(g/\ell )\sin x_1+u/(m{\ell }^2)\end{array}\right]$.

$A=\partial f/\partial x=\left[\begin{array}{cc}0& 1\\ -(g/\ell )\cos x_1& 0\end{array}\right]$. At $x_0=0$: $A=\left[\begin{array}{cc}0& 1\\ -g/\ell & 0\end{array}\right]$.

$B=\partial f/\partial u=\left[\begin{array}{c}0\\ 1/(m{\ell }^2)\end{array}\right]$.

Eigenvalues of $A$: $\lambda =\pm i\sqrt{g/\ell }$ — purely imaginary, so small oscillations about the downward equilibrium are undamped sinusoids at the pendulum’s natural frequency. This matches the expected behavior of a frictionless pendulum.

Inverted Pendulum

Repeat the linearization at the upward equilibrium ${\theta }_0=\pi$.

Solution $x_1=\pi$: $\cos \pi =-1$, so $A=\left[\begin{array}{cc}0& 1\\ g/\ell & 0\end{array}\right]$.

At

Eigenvalues: $\lambda =\pm \sqrt{g/\ell }$ — one positive, one negative. The system is an unstable saddle: small angular deviations grow exponentially (the pendulum falls). Stabilizing the inverted equilibrium requires feedback — see 7.6. State Feedback.

Linearizing Manipulator Dynamics

Linearize the 1-DoF revolute link $I\ddot{\theta }=\tau -\frac{1}{2}mg\ell \sin \theta$ about equilibrium $({\theta }_0,0,{\tau }_0)$.

Solution ${\tau }_0=\frac{1}{2}mg\ell \sin {\theta }_0$ (gravity compensation).

At equilibrium:

State $x=(\theta ,\dot{\theta })$, input $u=\tau$. $f=(x_2,(u-\frac{1}{2}mg\ell \sin x_1)/I)$.

$A=\left[\begin{array}{cc}0& 1\\ -(mg\ell /2I)\cos {\theta }_0& 0\end{array}\right],B=\left[\begin{array}{c}0\\ 1/I\end{array}\right]$.

For ${\theta }_0\in (-\pi /2,\pi /2)$ (arm pointing downward from horizontal), $\cos {\theta }_0>0$ so the $A_{21}$ entry is negative: stable oscillatory dynamics. For ${\theta }_0=\pi /2$ (arm horizontal), $A_{21}=0$ — marginally stable, unstable under nonlinear terms. For ${\theta }_0\in (\pi /2,3\pi /2)$ (arm pointing up), $\cos {\theta }_0<0$, $A_{21}>0$: unstable equilibrium.

See Also

• 7.1. State-Space Representation — the nonlinear form being linearized
• 7.3. Solving LTI Systems — how to simulate the resulting linear model
• 7.5. Stability — eigenvalue-based stability criterion, applied to the linearized $A$
• 5.2. Robot Jacobians — a related Jacobian: of forward kinematics rather than dynamics