Modern Control Engineering

Section 10–4 / Design of Servo Systems 747

x.=Ax+Bu y=Cx

k 1

 kI

k 2

k 3

k 4

ru

x

j j y

.

+– +–

Figure 10–9
Inverted-pendulum
control system. (Type
1 servo system when
the plant has no
integrator.)

in Figure 10–9. We assume that the pendulum angle uand the angular velocity are small, so that

and We also assume that the numerical values for M, m,andlare

given as

Earlier in Example 3–6 we derived the equations for the inverted-pendulum system shown in

Figure 3–6, which is the same as that in Figure 10–8. Referring to Figure 3–6, we started with the

force-balance and torque-balance equations and ended up with Equations (3–20) and (3–21) to

model the inverted-pendulum system. Referring to Equations (3–20) and (3–21), the equations for

the inverted-pendulum control system shown in Figure 10–8 are

(10–43)

(10–44)

When the given numerical values are substituted, Equations (10–43) and (10–44) become

(10–45)

(10–46)

Let us define the state variables x 1 ,x 2 ,x 3 ,andx 4 as

Then, referring to Equations (10–45) and (10–46) and Figure 10–9 and considering the cart position

xas the output of the system, we obtain the equations for the system as follows:

(10–47)

(10–48)

(10–49)

j (10–50)

=r-y=r-Cx

u =-Kx+kI j

y =Cx

x

=Ax+Bu

x 4 =x

x 3 =x

x 2 =u

x 1 =u

x

$

=0.5u-0.4905u

u

$

=20.601u-u

Mx

$

=u-mgu

Mlu

$

=(M+m)gu-u

M=2 kg, m=0.1 kg, l=0.5 m

uu

# 2

sinuu,cosu1, 0.

u