EXAMPLE PROBLEMS AND SOLUTIONS

A–3–1. Figure 3–20(a) shows a schematic diagram of an automobile suspension system. As the car moves
along the road, the vertical displacements at the tires act as the motion excitation to the auto-
mobile suspension system.The motion of this system consists of a translational motion of the cen-
ter of mass and a rotational motion about the center of mass. Mathematical modeling of the
complete system is quite complicated.
A very simplified version of the suspension system is shown in Figure 3–20(b). Assuming that
the motion xi at point P is the input to the system and the vertical motion xo of the body is the
output, obtain the transfer function Xo(s)兾Xi(s). (Consider the motion of the body only in the ver-
tical direction.) Displacement xo is measured from the equilibrium position in the absence of
input xi .
Solution. The equation of motion for the system shown in Figure 3–20(b) is

mxo + bAxo - xi B + kAxo - xi B = 0

$ # #

or
$ # #
mxo + bxo + kxo = bxi + kxi

Taking the Laplace transform of this last equation, assuming zero initial conditions, we obtain

Ams2 + bs + kBXo(s) = (bs + k)Xi(s)

Hence the transfer function Xo(s)/Xi(s) is given by

Xo(s) bs + k
=
Xi(s) ms2 + bs + k

k b xo

Center of mass

Auto body

P
Figure 3–20
(a) Automobile xi
suspension system;
(b) simplified
suspension system. (a) (b)

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

 A–3–2. Obtain the transfer function Y(s)/U(s) of the system shown in Figure 3–21. The input u is a
displacement input. (Like the system of Problem A–3–1, this is also a simplified version of an
automobile or motorcycle suspension system.)
Solution. Assume that displacements x and y are measured from respective steady-state
positions in the absence of the input u. Applying the Newton’s second law to this system, we
obtain
$ # #
m1 x = k2(y - x) + b(y - x) + k1(u - x)
$ # #
m2 y = -k2(y - x) - b(y - x)

Hence, we have

m1 x + bx + Ak1 + k2 Bx = by + k2 y + k1 u
$ # #

$ # #
m2 y + by + k2 y = bx + k2 x

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

C m1 s2 + bs + Ak1 + k2 B DX(s) = Abs + k2 BY(s) + k1 U(s)

Cm2 s2 + bs + k2 DY(s) = Abs + k2 BX(s)

Eliminating X(s) from the last two equations, we have

m2 s2 + bs + k2
Am1 s2 + bs + k1 + k2 B Y(s) = Abs + k2 BY(s) + k1 U(s)
bs + k2

which yields

Y(s) k1 Abs + k2 B
m1 m2 s + Am1 + m2 Bbs + Ck1 m2 + Am1 + m2 Bk2 Ds2 + k1 bs + k1 k2
= 4 3
U(s)

y
m2

k2 b

x
m1

k1
u

Figure 3–21
Suspension system.

Example Problems and Solutions 87

 y1 y2

b
k
m1 m2 u

Figure 3–22
Mechanical system.

A–3–3. Obtain a state-space representation of the system shown in Figure 3–22.

Solution. The system equations are

m1 y1 + by1 + kAy1 - y2 B = 0
$ #

m2 y2 + kAy2 - y1 B = u
$

The output variables for this system are y1 and y2 . Define state variables as
x1 = y1
#
x2 = y 1
x3 = y2
#
x4 = y 2
Then we obtain the following equations:
#
x 1 = x2

C-by1 - kAy1 - y2 B D = -
# 1 # k b k
x2 = x - x + x
m1 m1 1 m1 2 m1 3
#
x 3 = x4

C-kAy2 - y1 B + uD =
# 1 k k 1
x4 = x - x + u
m2 m2 1 m2 3 m2

Hence, the state equation is

0 1 0 0
# 0
x1 k b k x1
# - - 0 0
x m1 m1 m1 x
D # 2T = F V D 2T + E 0 U u
x3 0 0 0 1 x3
# 1
x4 k k x4
0 - 0 m 2
m2 m2
and the output equation is
x1
y 1 0 0 0 x
B 1R = B R D 2T
y2 0 0 1 0 x3
x4

A–3–4. Obtain the transfer function Xo(s)/Xi(s) of the mechanical system shown in Figure 3–23(a). Also
obtain the transfer function Eo(s)/Ei(s) of the electrical system shown in Figure 3–23(b). Show that
these transfer functions of the two systems are of identical form and thus they are analogous systems.

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