Physical Interaction via Dynamic Primitives

Abstract

Humans out-perform contemporary robots despite vastly slower ‘wetware’ (e.g. neurons) and ‘hardware’ (e.g. muscles). The basis of human sensory-motor performance appears to be quite different from that of robots. Human haptic perception is not compatible with Riemannian geometry, the foundation of classical mechanics and robot control. Instead, evidence suggests that human control is based on dynamic primitives, which enable highly dynamic behavior with minimal high-level supervision and intervention. Motion primitives include submovements (discrete actions) and oscillations (rhythmic behavior). Adding mechanical impedance as a class of dynamic primitives facilitates controlling physical interaction. Both motion and interaction primitives may be combined by re-purposing the classical equivalent electric circuit and extending it to a nonlinear equivalent network. It highlights the contrast between the dynamics of physical systems and the dynamics of computation and information processing. Choosing appropriate task-specific impedance may be cast as a stochastic optimization problem, though its solution remains challenging. The composability of dynamic primitives, including mechanical impedances, enables complex tasks, including multi-limb coordination, to be treated as a composite of simpler tasks, each represented by an equivalent network. The most useful form of nonlinear equivalent network requires the interactive dynamics to respond to deviations from the motion that would occur without interaction. That suggests some form of underlying geometric structure but which geometry is induced by a composition of motion and interactive dynamic primitives? Answering that question might pave the way to achieve superior robot control and seamless human-robot collaboration.

Submitted to: Laumond, J.-P. Geometric and Numerical Foundations of Movement.

Appendix: Choosing Impedance via Stochastic Optimization

Assume a manipulator and its control system are modeled as a mass $m_{m}$ moving in 1 degree of freedom, retarded by linear damping b and driven by a linear spring referenced to a ‘zero-force’ point $x_{0}$. It interacts with an object modeled as mass $m_{o}$ such that both move with common motion x. State-determined equations for the coupled system are

$$ \frac{d}{{dt}}\left[ {\begin{array}{*{20}c} x \\ v \\ \end{array} } \right] = \left[ {\begin{array}{*{20}c} 0 &{} 1 \\ { - k/m} &{} { - b/m} \\ \end{array} } \right] \left[ {\begin{array}{*{20}c} x \\ v \\ \end{array} } \right] + \left[ {\begin{array}{*{20}c} 0 \\ {k/m} \\ \end{array} } \right] x_{0} + \left[ {\begin{array}{*{20}c} 0 \\ {1/m} \\ \end{array} } \right] w_{e} $$

$$ f_{o} = \left[ {- \frac{{m_{o}}}{m}k} {-\frac{{m_{o}}}{m}} b \right] \left[ {\begin{array}{*{20}c} x \\ v \\ \end{array} } \right] + \left[ {\frac{{m_{o} }}{m}k} \right] x_{0} $$

where $m = m_{m} + m_{o}$, $f_{o}$ is the force exerted by the manipulator on the object, and $w_{e}$ denotes stochastic perturbation forces, modeled as zero-mean Gaussian white noise of strength S, i.e. $E\left\{ {w_{e} (t)} \right\} = 0 , E\left\{ {w_{e} \left( t \right) w_{e} \left( {t + \tau } \right) } \right\} =S\delta \left( \tau \right)$, where $E\left\{ \cdot \right\}$ is the expectation operator and $\delta \left( \cdot \right)$ denotes the unit impulse function. The objective function to be minimized is

$$ Q = E\left\{ {\frac{1}{{t_{{final}} }}\mathop \int \limits _{0}^{{t_{{final}} }} \left( {\frac{{f_{o}^{2} }}{{f_{{tol}}^{2} }} + \frac{{{{\Delta }}x^{2} }}{{{{\Delta }}x_{{tol}}^{2} }}} \right) dt} \right\} $$

where ${{\Delta }}x = x_{0} - x$ and $f_{{tol}}$ and ${{\Delta }}x_{{tol}}$ are tolerances on interface force and displacement. Due to the stochastic input, the state ($x\left( t\right) ,v\left( t \right)$) and output $f_{o} \left( t \right)$ are random variables. Assume $x_{0} \left( t \right) = {{\text {constant}}}= 0$ and find conditions for a steady-state solution (i.e. consider the limit as $t_{{final}} \rightarrow \infty$). The mean state and output variables propagate deterministically, i.e. $E\left\{ {x\left( t\right) } \right\} = E\left\{ {v\left( t \right) } \right\} = E\left\{ {f_{o} \left( t \right) } \right\} = 0$. Define the input covariance $W_{e} \left( t \right) = E\left\{ {\left( {w_{e}^{2} \left( t \right) } \right) } \right\} = S$ and the state covariance matrix

$$ {{\Sigma }}\left( t \right) = E\left\{ {\left[ {\begin{array}{*{20}c} {x\left( t \right) } \\ {v\left( t \right) } \\ \end{array} } \right] \left[ {\begin{array}{*{20}c} {x\left( t \right) } &{} {v\left( t \right) } \\ \end{array} } \right] } \right\} = E\left\{ {\left[ {\begin{array}{*{20}c} {x^{2} (t)} &{} {x\left( t \right) v\left( t \right) } \\ {x\left( t \right) v\left( t \right) } &{} {v^{2} (t)} \\ \end{array} } \right] } \right\} $$

For notational convenience, omit the explicit time dependence and use overbar notation ${{\Sigma }} = \left[ {\begin{array}{*{20}c} {\overline{{x^{2} }} } &{} {\overline{{xv}}} \\ {\overline{{xv}} } &{} {\overline{{v^{2} }} } \\ \end{array} } \right]$. Covariance propagation through a linear time-invariant system is described by the dynamic equation ${{\dot{\Sigma }}} = A{{\Sigma }} + {{\Sigma }}A^{T}+ BSB^{T}$ where A and B are system and input weighting matrices.

$$ \frac{d}{{dt}}\left[ {\begin{array}{*{20}c} {\overline{{x^{2} }} } &{} {\overline{{xv}} } \\ {\overline{{xv}} } &{} {\overline{{v^{2} }} } \\ \end{array} } \right] = \left[ {\begin{array}{*{20}c} {2\overline{{xv}} } &{} {\overline{{v^{2} }} - \overline{{x^{2} }} k/m - \overline{{xv}} b/m} \\ {\overline{{v^{2} }} - \overline{{x^{2} }} k/m - \overline{{xv}} b/m} &{} { - 2\overline{{xv}} k/m - 2\overline{{v^{2} }} b/m} \\ \end{array} } \right] + \left[ {\begin{array}{*{20}c} 0 &{} 0 \\ 0 &{} {1/m^{2} } \\ \end{array} } \right] S $$

Re-write as 3 coupled scalar differential equations

$$ \begin{array}{c} \frac{d}{{dt}}\overline{{x^{2} }} = 2\overline{{xv}}\\ \frac{d}{{dt}}\overline{{xv}} = \overline{{v^{2} }} - \overline{{x^{2} }} k/m - \overline{{xv}} b/m\\ \frac{d}{{dt}}\overline{{v^{2} }} = S/m^{2} - 2\overline{{xv}} k/m - 2\overline{{v^{2} }} b/m \end{array} $$

The scalar to be minimized is

$$ Q = \frac{1}{{t_{{final}} }}\mathop \int \limits _{0}^{{t_{{final}} }} \left( {\frac{{\overline{{f_{o}^{2} }} }}{{f_{{tol}}^{2} }} + \frac{{\overline{{{{\Delta }}x^{2} }} }}{{{{\Delta }}x_{{tol}}^{2} }}} \right) dt $$

$$f_{o}^{2} = \left( {\frac{{m_{o}^{2} }}{{m^{2} }}} \right) \left( {k^{2} x^{2} + 2kbxv + b^{2} v^{2} } \right) $$

Defining $q = ( {f_{{tol}} /{{\Delta }}x_{{tol}} })(m / m_o)$

$$ Q = \frac{1}{{f_{{tol}}^{2} }}\frac{{m_{o}^{2} }}{{m^{2} }}\frac{1}{{t_{{final}} }}\mathop \int \limits _{0}^{{t_{{final}} }} \left( {k^{2} \overline{{x^{2} }} + 2kb\overline{{xv}} + b^{2} \overline{{v^{2} }} + q^{2} \overline{{x^{2} }} } \right) dt $$

Construct the ‘control Hamiltonian’

$$\begin{array}{c} H = \left( {k^{2} + q^{2} } \right) \overline{{x^{2} }} + 2kb\overline{{xv}} + b^{2} \overline{{v^{2} }}\\ + \lambda _{1} 2\overline{{xv}}\\ + \lambda _{2} \left( {\overline{{v^{2} }} - \overline{{x^{2} }} k/m - \overline{{xv}} b/m} \right) \\ + \lambda _{3} \left( {S/m^{2} - 2\overline{{xv}} k/m - 2\overline{{v^{2} }} b/m} \right) \end{array}$$

where $\lambda _{i}$ denote Lagrange multipliers. Minimize with respect to k and b.

$$\begin{aligned}&\frac{{\partial H}}{{\partial k}} = 2k\overline{{x^{2} }} + 2b\overline{{xv}} - \lambda _{2} \overline{{x^{2} }} /m - \lambda _{3} 2\overline{{xv}} /m = 0\\&\frac{{\partial H}}{{\partial b}} = 2k\overline{{xv}} + 2b\overline{{v^{2} }} - \lambda _{2} \overline{{xv}} /m - \lambda _{3} 2\overline{{v^{2} }} /m = 0\\ \end{aligned}$$

The Lagrange multipliers are defined by ‘co-state’ equations.

$$\begin{aligned}&\frac{{\partial H}}{{\partial \overline{{x^{2} }} }} = - \dot{\lambda }_{1} = k^{2} + q^{2} - \lambda _{2} k/m\\ \frac{{\partial H}}{{\partial \overline{{xv}} }}&= - \dot{\lambda }_{2} = 2kb + 2\lambda _{1} - \lambda _{2} b/m - \lambda _{3} 2k/m\\ \frac{{\partial H}}{{\partial \overline{{v^{2} }} }}&= - \dot{\lambda }_{3} = b^{2} + \lambda _{2} - \lambda _{3} 2b/m \end{aligned}$$

Assume steady state exists and set all rates of change to zero.

$$\begin{array}{c} 2\overline{{xv}} = 0\\ \overline{{v^{2} }} - \overline{{x^{2} }} k/m - \overline{{xv}} b/m = 0\\ S/m^{2} - 2\overline{{xv}} k/m - 2\overline{{v^{2} }} b/m = 0\\ k^{2} + q^{2} - \lambda _{2} k/m = 0\\ 2kb + 2\lambda _{1} - \lambda _{2} b/m - \lambda _{3} 2k/m = 0\\ b^{2} + \lambda _{2} - \lambda _{3} 2b/m = 0 \end{array}$$

A little manipulation shows that

$$\begin{aligned}&\overline{{xv}} = 0\\ \overline{{v^{2} }}&= \overline{{x^{2} }} k/m\\ \overline{{x^{2} }}&= S/2kb\\ \overline{{v^{2} }}&= S/2bm \end{aligned}$$

The first co-state equation yields

$$k^{2} + q^{2} = \lambda _{2} k/m$$

$$\lambda _{2} = km + q^{2} m/k$$

The optimal stiffness is defined by

$$2k_{{opt}} \overline{{x^{2} }} = \lambda _{2} \overline{{x^{2} }} /m$$

$$k_{{opt}} = q = \frac{{f_{{tol}} }}{{{{\Delta }}x_{{tol}} }}\frac{m}{{m_{o} }}$$

Note that this manipulation requires $\overline{{x^{2} }} \ne 0$ and hence $S \ne 0$. However, $k_{{opt}}$ is independent of noise strength S.

The third co-state equation yields

$$b^{2} + \lambda _{2} = \lambda _{3} 2b/m$$

$$\lambda _{3} = \left( {b^{2} + \lambda _{2} } \right) m/2b = \frac{{bm}}{2} + \frac{{km}}{{2b}} + \frac{{q^{2} m^{2} }}{{2kb}}$$

The optimal damping is defined by

$$2b_{{opt}} \overline{{v^{2} }} = \lambda _{3} 2\overline{{v^{2} }} /m$$

$$b_{{opt}} = \sqrt{2k_{{opt}} m}$$

This manipulation requires $\overline{{v^{2} }} \ne 0$ and hence $S\ne 0$. However, $b_{{opt}}$ is independent of noise strength S.