CC 01: Block With Friction

June 1, 2025

Attempting to make a series of tutorials based on the Control Challenges by Janismac. Still working out on the presentation and integration.

By looking at the JS files¹² it’s possible to find the set of ODE used to model it.

$X = \begin{bmatrix} x & v & T \end{bmatrix}^T$ is the vector of the states.

The state space representation is:

\begin{cases} \dot{X} = \mathbb{A} X + \mathbb{B} U \\ Y = \mathbb{C} X + \mathbb{D} U \end{cases}

Then by converting in State Space form and using Julia:

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using DiscretePIDs, ControlSystems, Plots
using Plots
using LinearAlgebra

# System parameters
Ts = 0.02 # sampling time
Tf = 2; #final simulation time
g = 9.81 #gravity
α = 0.0 # slope
μ = 1.0 # friction coefficient
x_0 = -2.0 # starting position
dx_0 = 0.0 # starting velocity
τ = 20.0 # torque constant 

# State Space Matrix
A = [0 1 0
    0 -μ 1
    0 0 -τ
];
B = [0
    0
    τ];
C = [1 0 0
    0 1 0];




sys = ss(A, B, C, 0)      # Continuous

bodeplot(tf(sys))
current_poles = poles(sys)


ϵ = 0.01;
pp = 15;
p = -2 * [pp + ϵ pp - ϵ (pp / 4)];
observability(A, C)
controllability(A, B)


L = real(place(sys, p, :c));


# L = place(A', C', 5*p, opt=:c) # error, let's use kalman
R1 = diagm([0.01, 0.02, 0.03])
R2 = diagm([0.005, 0.002])
K = kalman(sys, R1, R2; direct=false)

cont = observer_controller(sys, L, K; direct=false)


closedLoop = feedback(sys, cont)

clpoles = poles(closedLoop)
setPlotScale("dB")

gangoffourplot(sys, cont; minimal=true)

bodeplot(closedLoop[1, 1], 0.1:40)

K = L[1]
Ti = 0;
Td = L[2] / L[1]


pid = DiscretePID(; K, Ts, Ti, Td)

sysreal = ss(A, B, [1 0 0], 0)

ctrl = function (x, t)
    y = (sysreal.C*x)[] # measurement
    d = 0 * [1.0]        # disturbance
    r = 2 * (t >= 0) # reference
    # u = pid(r, y) # control signal
    # u + d # Plant input is control signal + disturbance
    # u =1
    e = x - [r; 0; 0]
    e[3] = 0 # torque not observable, just ignore it in the final feedback
    u = -L * e + d
    u = [maximum([-20 minimum([20 u])])]
end
t = 0:Ts:Tf


res = lsim(sysreal, ctrl, t)

plot(res, plotu=true, plotx=true, ploty=false);
ylabel!("u", sp=1);
ylabel!("x", sp=2);
ylabel!("v", sp=3);
ylabel!("T", sp=4);

#  ylabel!(["u", "x","v","T"]);

# 
si = stepinfo(res);
plot(si);
title!("Step Response");

I think 1.32s is close to the optimal, I hope this doesn’t become a contest.

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