Modern Control Engineering

Section 3–2 / Mathematical Modeling of Mechanical Systems 71

Equation (3–20) was obtained by eliminating from Equations (3–18) and (3–19). Equation

(3–21) was obtained by eliminating from Equations (3–18) and (3–19). From Equation (3–20)

we obtain the plant transfer function to be

The inverted-pendulum plant has one pole on the negative real axis and

another on the positive real axis Hence, the plant is open-loop unstable.

Define state variables x 1 , x 2 , x 3 ,andx 4 by

Note that angle uindicates the rotation of the pendulum rod about point P, and xis the location

of the cart. If we consider uandxas the outputs of the system, then

(Notice that both uandxare easily measurable quantities.) Then, from the definition of the state

variables and Equations (3–20) and (3–21), we obtain

x# 4 =-

m

M

gx 1 +

1

M

u

x# 3 =x 4

x# 2 =

M+m

Ml

gx 1 -

1

Ml

u

x# 1 =x 2

y=B

y 1

y 2

R =B

u

x

R = B

x 1

x 3

R

x 4 =x#

x 3 =x

x 2 =u

x 1 =u

Cs=A 1 M+m 1 MlB 1 gD.

Cs=-A 1 M+m 1 MlB 1 gD

=

1

Mlas+

A

M+m

Ml

gbas-

A

M+m

Ml

gb

Q (s)

• U(s)

=

1

Mls^2 - (M+m)g

u

$ x

$

0

M

P

z

u

mg

m

 sin u

x

x

 cos u



u

Figure 3–6
Inverted-pendulum
system.