Waffles

EXAMPLE 3-7

Obtain the transfer functions X , ( s ) / U ( s ) and X 2 ( s ) / U ( s ) the mechanical system shown in of Figure 3-19. The equations of motion for the system shown in Figure 3-19 are

m1xl = -klxl - k2(xl - x2) - b(xl - x2) + u m2X2= -k3x2 - k2(x2- x l ) - b(x2 - x l )
Simplifying,we obtain

+ bxl + ( k , + k2)x, = bx, + k2x2+ u m 2 i 2 + bx2 + ( k z + k3)x2= b x , + k 2 x ,mlxl
Taking the Laplace transforms of these two equations, assuming zero initial conditions, we obtain

Solving Equation (3-45), for X 2 ( s )and substituting it into Equation (3-44) and simplifying, we get

[(m,s2 + bs

+ k , + k2)(m2s2+ bs + k, + k,)

- (bs

+ k2)2]~l(s)

from which we obtain

From Equations (3-45) and (3-46) we have

, Equations (3-46) and (3-47) are thetransfer functions X , ( s ) / U ( s )and X 2 ( s ) / U ( s )respectively.

Figure 3-19 Mechanical system.

EXAMPLE 3-8
I

An inverted pendulum mounted on a motor-driven cart is shown in Figure 3-20(a).This is a model of the attitude control of a space booster on takeoff. (The objective of the attitude control problem is to keep the space booster in a vertical position.) The inverted pendulumis unstable in that it may fall over any time in any direction unless a suitable control force is applied. Here we consider
Chapter 3 / Mathematical Modeling of Dynamic Systems

86

Figure 3-20 (a) Inverted pendulum system; (b) free-body diagram.

only a two-dimensional problem in which the pendulum moves only in the plane of the page.The control force u is applied to the cart.Assume thatthe center of gravity of the pendulum rod is at its geometric center. Obtain a mathematical model for the system. Define the angle of the rod from the vertical line as 0. Define also the (x, y) coordinates of the center of gravity of the pendulum rod as (x,, Yc).Then
x, = x

+ lsin0

Section 3-7

/ Mechanical Systems

To derive the equations of motion for the system, consider thefree-body diagram shown in Figure 3-2O(b). The rotational motion of the pendulum rod about its center of gravity can be described by

where I is the moment of inertia of the rod about its center of gravity. The horizontal motion of center of gravity of pendulum rod is given by d2 m-(x dt2

+ lsine)

=H

The vertical motion of center of gravity of pendulum rod is

The horizontal motion of cart isdescribed by

Since we must keep the inverted pendulum vertical, we can assume that % ( t ) and 0(t) are small quantities such that sin%= 0, cos e = 1,and %e2 0. Then, Equations (3-48) through (3-50) = can be linearized. The linearized equations are

10

=

vze - HI

(3-52) (3-53) (3-54)

m(x

+ le) = H

O=V-rng From Equations (3-51) and (3-53), we obtain

From Equations (3-52),(3-53), and (3-54), we have

10 = mgle
=

- HI

mgle - l ( m x

+ m10)

Equations (3-55) and (3-56) describe the motion of the inverted-pendulum-on-the-cart system. They constitute a mathematical model of the system.

EXAMPLE 3-9

Consider the inverted pendulum system shown in Figure 3-21. Since in this system the mass is concentrated at the top of the rod, the center of gravity isthe center of the pendulum ball. For this case, the moment of inertia of the pendulum about its center of gravity is small, and we assume I = 0 in Equation (3-56).Then the mathematical model for this system becomes as follows: (3-57) ( M + m ) i + ml9 = u Equations (3-57) and (3-58) can be modified to ~ 1 =0( M

+ m)ge - u

88

Chapter 3 / Mathematical Modeling of Dynamic Systems

+ x
'u

M

Figure 34!1 Inverted-pendulum system.
Equation (3-59) was obtained by eliminating x from Equations (3-57) and (3-58). Equation (3-60) was obtained by eliminating 9 from Equations (3-57) and (3-58). From Equation (3-59) we obtain the plant transfer function to be

@ --( s ) -U(s)

-

1 Mls2 - ( M t- m)g Ml ( s

+

JT) JT)
(s -

1

The inverted pendulum plant has one...