Modelling Inverted Pendulum
Inverted Pendulum Derivation with Lagrange
Cart Kinetic Energy
Pendulum Kinetic Energy
Potential Energy
Lagrange's Equation of Motion
Respective to
1
Respective to
Final Answer
State Sapce
2
Linearization
M
m
b
I
g
l
q
p
=
=
=
=
=
=
=
=
0.5;
0.2;
0.1;
0.006;
9.8;
0.3;
(M+m)*(I+m*l^2)-(m*l)^2;
I*(M+m)+M*m*l^2; %denominator for the A and B matrices
A = [
0, 1, 0, 0;
0, 0, 1/(M+m)*m*l, 0;
0, 0, 0, 1;
0, 1/(m*l^2+I)*m*l, 1/(m*l^2+I)*m*g*l*(M+m), 0
];
B = [
0;
(I+m*l^2)/p;
0;
m*l/p];
C = [1 0 0 0;
3
0 0 1 0];
D = [0;
0];
states = {'x' 'x_dot' 'phi' 'phi_dot'};
inputs = {'u'};
outputs = {'x'; 'phi'};
sys_ss = ss(A,B,C,D,'statename',states,'inputname',inputs,'outputname',outputs)
sys_ss =
A =
x
0
0
0
0
x
x_dot
phi
phi_dot
x_dot
1
0
0
2.5
phi
0
0.08571
0
17.15
phi_dot
0
0
1
0
B =
x
x_dot
phi
phi_dot
u
0
1.818
0
4.545
C =
x
1
0
x
phi
x_dot
0
0
phi
0
1
phi_dot
0
0
D =
x
phi
u
0
0
Continuous-time state-space model.
sys_tf = tf(sys_ss)
sys_tf =
From input "u" to output...
1.818 s^2 - 6.782e-14 s - 30.79
x: -----------------------------------------s^4 + 8.882e-16 s^3 - 17.15 s^2 - 0.2143 s
phi:
4.545 s + 4.545
-------------------------------------s^3 + 8.882e-16 s^2 - 17.15 s - 0.2143
Continuous-time transfer function.
step(sys_ss);
4
5