Modern Control Engineering

68 Chapter 3 / Mathematical Modeling of Mechanical Systems and Electrical Systems

EXAMPLE 3–4 Obtain the transfer functions and of the mechanical system shown in

Figure 3–4.

The equations of motion for the system shown in Figure 3–4 are

Simplifying, we obtain

Taking the Laplace transforms of these two equations, assuming zero initial conditions, we obtain

(3–5)

(3–6)

Solving Equation (3–6) for and substituting it into Equation (3–5) and simplifying, we get

from which we obtain

(3–7)

From Equations (3–6) and (3–7) we have

(3–8)

Equations (3–7) and (3–8) are the transfer functions and respectively.

EXAMPLE 3–5 An inverted pendulum mounted on a motor-driven cart is shown in Figure 3–5(a). This is a model

of the attitude control of a space booster on takeoff. (The objective of the attitude control prob-

lem is to keep the space booster in a vertical position.) The inverted pendulum is unstable in that

it may fall over any time in any direction unless a suitable control force is applied. Here we consider

X 1 (s)U(s) X 2 (s)U(s),

X 2 (s)

U(s)

=

bs+k 2

Am 1 s^2 +bs+k 1 +k 2 BAm 2 s^2 +bs+k 2 +k 3 B-Abs+k 2 B^2

X 1 (s)

U(s)

=

m 2 s^2 +bs+k 2 +k 3

Am 1 s^2 +bs+k 1 +k 2 BAm 2 s^2 +bs+k 2 +k 3 B-Abs+k 2 B^2

=Am 2 s^2 +bs+k 2 +k 3 BU(s)

CAm 1 s^2 +bs+k 1 +k 2 BAm 2 s^2 +bs+k 2 +k 3 B-Abs+k 2 B^2 DX 1 (s)

X 2 (s)

Cm 2 s^2 +bs+Ak 2 +k 3 BDX 2 (s)=Abs+k 2 BX 1 (s)

Cm 1 s^2 +bs+Ak 1 +k 2 BDX 1 (s)=Abs+k 2 BX 2 (s)+U(s)

m 2 x$ 2 +bx# 2 +Ak 2 +k 3 Bx 2 =bx# 1 +k 2 x 1

m 1 x$ 1 +bx# 1 +Ak 1 +k 2 Bx 1 =bx# 2 +k 2 x 2 +u

m 2 x$ 2 =-k 3 x 2 - k 2 Ax 2 - x 1 B-bAx# 2 - x# 1 B

m 1 x$ 1 =-k 1 x 1 - k 2 Ax 1 - x 2 B-bAx# 1 - x# 2 B+u

X 1 (s)U(s) X 2 (s)U(s)

m 1 m 2

k 2

x 1

k 1 k 3

b

u x 2

Figure 3–4

Mechanical system.

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