Motion Control

We would partition the controller into two parts, model-based portion and servo portion.

The model-based portion makes use of feedback to reduce the system such that it appears like a unit mass.

Control-law decomposition

Given model

$$m\ddot{x}+b\dot{x}+kx=f,$$

choose $f=\alpha f^{\prime }+\beta$ such that

$$\begin{array}{}\text{(1)}& \ddot{x}=f^{\prime }\text{open-loop}\end{array}$$

so

$$\alpha =m,\beta =b\dot{x}+kx.$$

we design a control law to compute $f^{\prime }$, so

$$f^{\prime }=-k_v\dot{x}-k_{p}x$$

the system yields

$$\ddot{x}+k_v\dot{x}+k_{p}x=0.$$

Trajectory-following control

Given a planned trajectory $x_d(t)$ is smooth, which means our trajectory generator could give access to $x_d,{\dot{x}}_d,{\ddot{x}}_d$ all the time.

Trajectory following

Define error $e=x_d-x$, then a servo-control law that cause trajectory following

$$f^{\prime }={\ddot{x}}_d+k_v\dot{e}+k_{p}e.$$

combined with equation $\text{1}$, we have

$$\begin{array}{rl}{\ddot{x}}_d+k_v\dot{e}+k_{p}e& =\ddot{x}\\ \text{(2)}& \Rightarrow \ddot{e}+k_v\dot{e}+k_{p}e& =0.\end{array}$$

which is also called error space.

Disturbance Rejection

Consider a noise $f_{dist}$, then equation $\text{2}$ becomes

$$\ddot{e}+k_v\dot{e}+k_{p}e=f_{dist}.$$

If $\underset{t}{max}{f}_{dist}(t)<a$, then the consequent error $e$ is a also bounded.

Modelling & Control of a single joint

We first develop a simplified model of a single rotary joint.

Modelling

For direct current (DC) motor, we have

$${\tau }_m=k_m{i}_a,V=k_e{\dot{\theta }}_m.$$

where the first one is a driven torque of the motor, and the second one is the generated voltage of rotation.

In the circuit of the armature, we have

$$l_a{\dot{i}}_a+r_a{i}_a=V_a-V=V_a-k_e{\dot{\theta }}_m.$$

The gear ratio $\eta$ causes an increase in the torque seen at the load and a reduction in the speed of the load, given by

$$\begin{array}{}\text{(3)}& \tau =\eta {\tau }_m,\dot{\theta }=\frac{\dot{\theta }}{\eta }.\end{array}$$

• Write a dynamic equaiton of the rotor

$${\tau }_m=I_m{\ddot{\theta }}_m+b_m{\dot{\theta }}_m+(1/\eta )(I\ddot{\theta }+b\dot{\theta }).$$

where $I_m$, $I$ are the inertias of the motor rotor and of the load, respectively, and $b_m$, $b$ are viscous friction coefficients for the rotor and load bearings, respectively.

Using equation $\text{3}$, we write the above equation in terms of motor variables ${\theta }_m$ or joint/load variables $\theta$

$$\begin{array}{rl}{\tau }_m& =(I_m+\frac{I}{{\eta }^2}){\ddot{\theta }}_m+(b_m+\frac{b}{{\eta }^2}){\dot{\theta }}_m\\ \tau & =(I+I_m{\eta }^2)\ddot{\theta }+(b+b_m{\eta }^2)\dot{\theta }.\end{array}$$

where $I+I_m{\eta }^2$ is called the effective inertia seen at the output of the gearing, $b+b_m{\eta }^2$ is called the effective damping.