Performance and Stability

Introduction

After you choose a controller in Proportional, PD and PID Control, you must choose its gains. The performance of the system depends on factors such as maximum overshoot, rise time, steady state error and settling time.

Standard second order form

The characteristic equation of a second order system has the standard form:

$$s^2+2\zeta {\omega }_{n}s+{\omega }_n^2=0$$

• $\zeta$ is the damping ratio
• ${\omega }_n$ is the undamped natural frequency

For a general $s^2+bs+c=0$, compare terms: $b=2\zeta {\omega }_n$ and $c={\omega }_n^2$. Then:

$$b^2-4c=4{\omega }_n^2({\zeta }^2-1)$$

Condition	Damping ratio	Response
$b^2>4c$	$\zeta >1$	Overdamped, slow with no overshoot
$b^2=4c$	$\zeta =1$	Critically damped, fastest with no overshoot
$b^2<4c$	$\zeta <1$	Underdamped, oscillates and overshoots

These cases come from the solution of the second order differential equation $\ddot{y}+b\dot{y}+cy=r(t)$, whose characteristic equation is $s^2+bs+c$:

Design rules

1. No overshoot. For safety, the manipulator must not have an underdamped response. An overshoot can make the arm hit something, for example a wall near the target. So:

$$\zeta \ge 1$$

2. No resonance. Do not excite the structural oscillation and resonance of the joint. A rule of thumb is:

$${\omega }_n\le 0.5{\omega }_{res}$$

where ${\omega }_{res}$ is the lowest structural resonant frequency.

How it works

1. Write the closed loop transfer function
2. Get the characteristic equation from $1+G_c(s)G_p(s)=0$
3. Compare it with $s^2+2\zeta {\omega }_{n}s+{\omega }_n^2=0$
4. Use $\zeta \ge 1$ and ${\omega }_n\le 0.5{\omega }_{res}$ to find the gains

Example

A motor drives a single link robot with these parameters:

• $L_a=0.6\times 10^{-3}$ H, $R_a=4.50\Omega$, $n=1$
• $J_{eff}=32\times 10^{-7}$ kg m²
• $f_{eff}=32\times 10^{-7}$ Nm/(rad/s)
• $K_a=33\times 10^{-3}$ Nm/A
• $K_b=33\times 10^{-3}$ V/(rad/s)

Given ${\omega }_{res}=180$ rad/s, design a PD controller so that the system is critically damped and does not excite resonance.

Step 1: plant model

$$K=\frac{K_a}{R_a{f}_{eff}+K_a{K}_b}=\frac{33\times 10^{-3}}{4.5\times 32\times 10^{-7}+(33\times 10^{-3}{)}^2}=29.9$$ $$T_m=\frac{R_a{J}_{eff}}{R_a{f}_{eff}+K_a{K}_b}=13.05\times 10^{-3}$$ $$G_p(s)=\frac{29.9}{s(13.05\times 10^{-3}s+1)}=\frac{2291.2}{s(s+76.6)}$$

Step 2: closed loop with PD

$$G(s)=\frac{2291.2(K_p+s{K}_v)}{s(s+76.6)}$$ $$\frac{{\theta }_L(s)}{{\theta }_d(s)}=\frac{2291.2(K_p+s{K}_v)}{s^2+(76.6+2291.2K_v)s+2291.2K_p}$$

Step 3: target values

• critically damped, so $\zeta =1$
• ${\omega }_n\le 0.5\times 180=90$ rad/s, so take ${\omega }_n=90$ rad/s

$$s^2+2\zeta {\omega }_{n}s+{\omega }_n^2=s^2+180s+8100$$

Step 4: compare terms

$$76.6+2291.2K_v=180\Rightarrow K_v=0.0451$$ $$2291.2K_p=8100\Rightarrow K_p=3.535$$

From one joint to a robot

A single joint is a single input, single output (SISO) system. A robot is a multi input, multi output (MIMO) system. With classical control, you use inverse kinematics to get the desired angle of each joint, and each joint has its own PID loop.

Question

The slide asks: can you use the same PID control law for all joints? The slide does not give the answer. Each joint has a different load and inertia, so each joint has a different model $G_p(s)$, and each joint needs its own gains.