Transfer Function Model of a Single Joint

Introduction

To design a linear controller in Motion Control Schemes, you need a transfer function for one joint. A common actuator in industrial robots is the armature controlled DC motor. The motor shaft turns the link through a gear train.

Mechanical part

The gear train links the motor angle ${\theta }_m$ to the load (joint) angle ${\theta }_L$ with gear ratio $n$:

$${\theta }_L(t)=n{\theta }_m(t)$$

The torque at the motor shaft is the torque for the motor plus the load torque referred to the motor:

$$\tau (t)={\tau }_m(t)+n{\tau }_L(t)$$ $${\tau }_L(t)=J_L{\ddot{\theta }}_L(t)+f_L{\dot{\theta }}_L(t)=J_{L}n{\ddot{\theta }}_m(t)+f_{L}n{\dot{\theta }}_m(t)$$ $${\tau }_m(t)=J_m{\ddot{\theta }}_m(t)+f_m{\dot{\theta }}_m(t)$$

where:

• $J_L$, $f_L$ are the moment of inertia and viscous friction of the load
• $J_m$, $f_m$ are the moment of inertia and viscous friction of the motor

Combine these:

$$\tau (t)=J_{eff}{\ddot{\theta }}_m(t)+f_{eff}{\dot{\theta }}_m(t)$$ $$J_{eff}=J_m+J_L{n}^2{f}_{eff}=f_m+f_L{n}^2$$

Laplace transform:

$$T(s)=s^2{J}_{eff}{\theta }_m(s)+s{f}_{eff}{\theta }_m(s)$$

Electrical part

Kirchhoff's voltage law on the armature circuit:

$$V_a(t)=R_a{i}_a(t)+L_a\frac{d{i}_a(t)}{dt}+e_b(t)$$

The back emf is proportional to the motor speed, $e_b(t)=K_b{\dot{\theta }}_m(t)$. So:

$$I_a(s)=\frac{V_a(s)-s{K}_b{\theta }_m(s)}{R_a+s{L}_a}$$

The motor torque is proportional to the armature current, $\tau (t)=K_a{i}_a(t)$, where $K_a$ is the motor torque constant.

Transfer function

Put the electrical and mechanical parts together:

$$\frac{{\theta }_m(s)}{V_a(s)}=\frac{K_a}{s[s^2{J}_{eff}{L}_a+(L_a{f}_{eff}+R_a{J}_{eff})s+R_a{f}_{eff}+K_a{K}_b]}$$

This is a third order system, which is hard to work with. The electrical time constant is much smaller than the mechanical time constant, so you can neglect $L_a$:

$$\boxed{\frac{{\theta }_m(s)}{V_a(s)}=\frac{K_a}{s(R_a{J}_{eff}s+R_a{f}_{eff}+K_a{K}_b)}=\frac{K}{s(T_{m}s+1)}}$$

• motor gain constant: $K=\frac{K_a}{R_a{f}_{eff}+K_a{K}_b}$
• motor time constant: $T_m=\frac{R_a{J}_{eff}}{R_a{f}_{eff}+K_a{K}_b}$

In terms of the joint angle ${\theta }_L=n{\theta }_m$:

$$\boxed{G_p(s)=\frac{{\theta }_L(s)}{V_a(s)}=\frac{n{K}_a}{s(R_a{J}_{eff}s+R_a{f}_{eff}+K_a{K}_b)}=\frac{nK}{s(T_{m}s+1)}}$$

In the later slides, $J_{eff}$ and $f_{eff}$ are written as $J_e$ and $f_e$:

$$G_p(s)=\frac{n{K}_a}{s(R_a{J}_{e}s+R_a{f}_e+K_a{K}_b)}$$

Note

Slides 25 and 26 write $K_b$ in the numerator. The correct constant is $K_a$, because the motor torque is $\tau =K_a{i}_a$. Slide 27 and the later slides use $K_a$.