Disturbance Rejection

Introduction

A control system must keep good performance (small errors) even when there is an external disturbance. This note shows how the partitioned controller handles a disturbance, and how an integral term removes the steady state error.

Effect of a disturbance

Add a disturbance force $d$ to the mass spring system:

$$m\ddot{x}+b\dot{x}+k{x}^3+d=u$$

Use the same controller:

$$u=m({\ddot{x}}_d+k_v\dot{e}+k_{p}e)+b\dot{x}+k{x}^3$$

The non-linear terms cancel, and the error equation becomes:

$$m(\ddot{e}+k_v\dot{e}+k_{p}e)=d\Rightarrow \ddot{e}+k_v\dot{e}+k_{p}e=\frac{d}{m}$$

Laplace transform:

$$\frac{E(s)}{D(s)}=\frac{1}{m(s^2+k_{v}s+k_p)}$$

Steady state error

For a constant disturbance $D(s)=d/s$, use the final value theorem:

$$e_{ss}=\underset{t\to \infty }{\lim }e(t)=\underset{s\to 0}{\lim }sE(s)=\underset{s\to 0}{\lim }\frac{d}{m(s^2+k_{v}s+k_p)}$$ $$\boxed{e_{ss}=\frac{d}{m{k}_p}}$$

A higher position gain $k_p$ gives a smaller steady state error, but the error does not go to zero.

Adding an integral term

To remove the steady state error, add an integral term to the servo law:

$$v={\ddot{x}}_d+k_v\dot{e}+k_{p}e+k_i\int edt$$

The error equation becomes:

$$\ddot{e}+k_v\dot{e}+k_{p}e+k_i\int edt=\frac{d}{m}$$ $$\frac{E(s)}{D(s)}=\frac{s}{m(s^3+k_v{s}^2+k_{p}s+k_i)}$$

For a constant disturbance:

$$e_{ss}=\underset{s\to 0}{\lim }\frac{ds}{m(s^3+k_v{s}^2+k_{p}s+k_i)}=0$$

The integral term removes the steady state error. But the system is now third order. See Third Order System for how to choose the gains.