State-Space Representation of Mechanical Systems

Equations of Motion

Let us consider manipulator equations in the form $$M(q)\ddot{q}+c(q,\dot{q})-{\tau }_g(q)=B(q)u\text{\hspace{1em}(1)}$$ who's left-hand side may be for example derived from the system's kinetic $T(q,\dot{q})$ and potential $V(q)$ energy expressed in generalized coordinates by expanding Lagrange equations of the second kind $$\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\partial L}{\partial \dot{q}}\right)-\frac{\partial L}{\partial q}=\sum _i\frac{\partial r_i}{\partial q}{F}_i,$$ where $L=T-V$. First by separating the Lagrangian into its components $$\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\partial T}{\partial \dot{q}}-\cancel{\frac{\partial V}{\partial \dot{q}}}\right)-\frac{\partial T}{\partial q}+\frac{\partial V}{\partial q}=\sum _i\frac{\partial r_i}{\partial q}{F}_i$$ and then applying the change rule $$\frac{{\partial }^{2}T}{\partial {\dot{q}}^2}\ddot{q}+\frac{{\partial }^{2}T}{\partial \dot{q}\partial q}\dot{q}-\frac{\partial T}{\partial q}+\frac{\partial V}{\partial q}=\sum _i\frac{\partial r_i}{\partial q}{F}_i.$$ If the external forces are linearly dependent on inputs $u_i$, such that $F_i=f_i{u}_i$, the terms of (1) as derived by this approach are $$\begin{array}{rl}M(q)& =\frac{{\partial }^{2}T}{\partial {\dot{q}}^2}\\ c(q,\dot{q})& =\frac{{\partial }^{2}T}{\partial \dot{q}\partial q}\dot{q}-\frac{\partial T}{\partial q}\\ {\tau }_g(q)& =-\frac{\partial V}{\partial q}\\ B(q)& =\left[\begin{array}{ccc}\frac{\partial r_1}{\partial q}{f}_1& \dots & \frac{\partial r_m}{\partial q}{f}_m\end{array}\right].\end{array}$$

Continuous-time dynamics

Most commonly used state-space representation of a mechanical system in continuous time is attained by concatenating $q$ and $\dot{q}$ to form the system's state $$x=\left[\begin{array}{c}q\\ \dot{q}\end{array}\right],\dot{x}=f(x,u)=\left[\begin{array}{c}\dot{q}\\ M^{-1}(q)\left({\tau }_g(q)-c(q,\dot{q})+B(q)u\right)\end{array}\right].$$

Linearization

The system's dynamics can be linearized by performing a Taylor expansion around a point $\{x^{\ast },u^{\ast }\}$ which yields $$\dot{x}=f(x,u)\approx f(x^{\ast },u^{\ast })+\underset{A_C}{\underbrace{\frac{\partial f}{\partial x}(x^{\ast },u^{\ast })}}\underset{\bar{x}}{\underbrace{(x-x^{\ast })}}+\underset{B_C}{\underbrace{\frac{\partial f}{\partial u}(x^{\ast },u^{\ast })}}\underset{\bar{u}}{\underbrace{(u-u^{\ast })}},$$ where also $\dot{\bar{x}}=\frac{\mathrm{d}}{\mathrm{d}t}\left(x-x^{\ast }\right)=\dot{x}$.

If $\{x^{\ast },u^{\ast }\}$ is a fixed point, i.e. $f(x^{\ast },u^{\ast })=0$, first order partial derivatives of the system's dynamics simplify to $$\begin{array}{rl}\frac{\partial f}{\partial x}(x^{\ast },u^{\ast })& =\left[\begin{array}{cc}0& I\\ M^{-1}(q^{\ast })\left(\frac{\partial {\tau }_g}{\partial q}(q^{\ast })+\frac{\partial B}{\partial q}(q^{\ast })u^{\ast }\right)& 0\end{array}\right]\\ \frac{\partial f}{\partial u}(x^{\ast },u^{\ast })& =\left[\begin{array}{c}0\\ M^{-1}(q^{\ast })B(q^{\ast })\end{array}\right]\end{array}$$ as terms involving $\frac{\partial M^{-1}}{\partial q}$ drop out because ${\tau }_g-c+B{u}^{\ast }=0$ for all fixed points. Partial derivatives of $c$ also disappear as all of its terms contain second degree products of velocities and all velocities must be equal to zero at a fixed point.

Discretization via integration using the explicit Euler scheme

The most basic approach with which we may discretize continuous-time dynamics of a nonlinear system is by integrating the systems state with a timestep of $h$ using the explicit Euler scheme $$x_{k+1}=x_k+hf(x_k,u_k).$$

Linearization

Compared to more advanced integration methods, linearization of these discrete time dynamics around a point $\{x^{\ast },u^{\ast }\}$ is trivial: $$\begin{array}{rl}\underset{{\bar{x}}_{k+1}}{\underbrace{x_{k+1}-x^{\ast }}}& =x_k+hf(x_k,u_k)-x^{\ast }\\ & \approx hf(x^{\ast },u^{\ast })+\underset{A_D}{\underbrace{\left(I+h\frac{\partial f}{\partial x}(x^{\ast },u^{\ast })\right)}}\underset{{\bar{x}}_k}{\underbrace{(x_k-x^{\ast })}}+\underset{B_D}{\underbrace{h\frac{\partial f}{\partial u}(x^{\ast },u^{\ast })}}\underset{{\bar{u}}_k}{\underbrace{(u_k-u^{\ast })}}.\end{array}$$