Tuning a PID Controller (or why linear systems theory is incredible)

Setup

Consider a double integrator:

x˙1=x2mx˙2=u\begin{align*} \dot x_1 &= x_2\\ m\dot x_2 &= u \end{align*}

where the object has a mass mm, and we apply the control input uu.

Suppose a PD controller is used to control this system:

u=kx(xref−x1)+kv(vref−x2)u = k_x (x_{ref} - x_1) + k_v (v_{ref} - x_2)

Assuming vref=0v_{ref} = 0, the closed-loop system has a transfer function

X1Xref=H(s)=kxms2+kvs+kx\frac{X_1}{X_{ref}} = H(s) = \frac{k_x}{m s^2 + k_v s + k_x}

This makes it a second order linear system, of the form

H(s)=ωn2s2+2ξωns+ωn2H(s) = \frac{\omega_n^2}{s^2 + 2 \xi \omega_n s + \omega_n^2}

using the map:

ωn=kxm,ξ=kv21mkx\begin{align*} \omega_n = \sqrt{\frac{k_x}{m}}, \quad && \xi = \frac{k_v}{2} \sqrt{\frac{1}{m k_x}} \end{align*}

Analysis

We can now analyze a few things about this system:

  • the closed loop response (i.e. solution to x1(t)x_1(t) such that x1(0)=0,x2(0)=0,xref(t)=1x_1(0) = 0, x_2(0) = 0, x_{ref}(t) = 1) is
x1(t)=1−e−σt(cos⁡(ωdt)+σωdsin⁡(ωdt))x_1(t) = 1 - e^{-\sigma t} \left( \cos(\omega_d t) + \frac{\sigma}{\omega_d} \sin(\omega_d t) \right)

where σ=ξωn\sigma = \xi \omega_n and ωd=ωn1−ξ2\omega_d = \omega_n \sqrt{1 - \xi^2}.

  • the rise time (defined as the first time that x1(t)=1x_1(t) = 1):
tr=2tan⁡−1(1+ξ1−ξ)ωn1−ξ2t_r = \frac{2 \tan^{-1}{\left( \sqrt{ \frac{1 + \xi}{1 - \xi} } \right)}} {\omega_n \sqrt{1 - \xi^2}}
  • the overshoot (defined as the maximum value reached minus one):
M=exp⁡(−πξ1−ξ2)M = \exp{\left(- \frac{\pi \xi}{\sqrt{1 - \xi^2}}\right)}

There are other properties you can analyse, see Ref 1.

Design

Notice that there are two design parameters in the controller: kx,kvk_x, k_v.

Therefore, if the overshoot and the rise time are specified, the controller gains kx,andkvk_x, and k_v can be determined, as follows:

1) Using the overshoot formula, determine ξ\xi:

ξ=−log⁡Mπ2+(log⁡M)2\xi = \frac{- \log M}{\sqrt{\pi^2 + (\log M)^2}}

2) Using the rise time, choose ωn\omega_n:

ωn=2trtan⁡−1(1+ξ1−ξ)1−ξ2\omega_n = \frac{2}{t_r} \frac{\tan ^{-1}\left(\sqrt{\frac{1 + \xi}{1-\xi }}\right)}{\sqrt{1-\xi ^2} }

3) Using ωn\omega_n, determine kxk_x:

kx=mωn2k_x = m \omega_n^2

4) Using ξ\xi, determine kvk_v:

kv=4ξ2mkxk_v = 4 \xi^2 \sqrt{m k_x}

Examples

Matlab Code:

%% inputs
m = 1.0;   % mass of system
M  = 0.10; % overshoot
tr = 0.4;  % rise time

%% outputs
xi = -(log(M)/sqrt(pi^2 + (log(M))^2))
omega_n = (2* atan(sqrt((1 + xi)/(1 - xi))))/(sqrt(1 - xi^2) * tr )
kx = m * omega_n^2
kv = 4 * xi^2 * sqrt(m * kx)

For m=1.0m = 1.0 kg:

kxk_x:

  100 ms 200 ms 500 ms 1 s 2 s
1% 2.04E+03 5.09E+02 8.14E+01 2.04E+01 5.09E+00
5% 1.04E+03 2.60E+02 4.15E+01 1.04E+01 2.60E+00
10% 7.46E+02 1.87E+02 2.98E+01 7.46E+00 1.87E+00
20% 5.28E+02 1.32E+02 2.11E+01 5.28E+00 1.32E+00
50% 3.35E+02 8.38E+01 1.34E+01 3.35E+00 8.38E-01

kvk_v:

  100 ms 200 ms 500 ms 1 s 2 s
1% 1.23E+02 6.16E+01 2.46E+01 1.23E+01 6.16E+00
5% 6.14E+01 3.07E+01 1.23E+01 6.14E+00 3.07E+00
10% 3.82E+01 1.91E+01 7.64E+00 3.82E+00 1.91E+00
20% 1.91E+01 9.55E+00 3.82E+00 1.91E+00 9.55E-01
50% 3.40E+00 1.70E+00 6.80E-01 3.40E-01 1.70E-01

Here are a few starting points for PID tuning:

MM trt_r kxk_x kvk_v
0.25 0.5 18.8 2.83
0.25 1.0 4.714 1.42
0.125 0.5 26.7 6.29
0.125 1.0 6.68 3.14

As a rough gauge of the control input required, the peak control input is likely to be at the very start, where the control input is u=kx(xref−x)+kv(vref−v)=kxu = k_x (x_{ref} - x) + k_v (v_{ref} - v) = k_x. So by looking at the kxk_x term, we can get a sense of the order of magnitude of the control input for a step input.

References:

  1. https://courses.engr.illinois.edu/ece486/fa2019/handbook/lec06.html