Project - Position control of mass spring damper system

Transfer function:

Given,

mass m = 3.6 kg

Spring constant, k = 400 N/m

Damper coefficient, c = 100 Ns/m

Let, position is denoted as x(t) and faorce acting on mass is F(t)

Therefore,

m d^2x/dt^2 + c dx/dt + kx = F(t)

It can be written in laplace transform as:

m*s^2 X(s) + c*sX(s) + k*X(s) = F(s)

X(s)[m*s^2 + c*s + k] = F(s)

X(s)/F(s) = 1/(m*s^2 + c*s + k)

The above equation represents transfer function of wn^2/(S^2 + 2*zeta*wn*S + wn^2)

Therefore,

wn^2 = k/m = 400/3.6 = 111.1111

wn = 10.5409

Also, 2*zeta*wn = c/m = 0.25

So, zeta = 1.3176 % over damped system

We also know,

A = [0 1; -k/m 0]

By solving, A = [0 1; -111.1111 0]

B = [0; 1/m]

B = [0; 0.2778]

Poles design

p1 = -zeta*wn + wn*sqrt(zeta^2-1) = -4.8451

p2 = -zeta*wn - wn*sqrt(zeta^2-1) = -22.9327

p = [p1; p2]

k = place(A,B,p)

We get,

Kp = -0.0019

Kd = 100

Ki = 1/m = 0.2778

Simulink model

OUTPUT

Here, position is tracking the reference signal.