1. Obtain mathematical models ( ) of the mechanical systems shown

1. Obtain mathematical models (
) of the mechanical systems shown
below.
Solution a) Here, the input is u(t) and the output is the displacement x as shown in the figure. the rollers under the mass means there is no friction. Converting the above equation to Laplace domain. Taking X(s) out of the brackets: Then the transfer function (T.F) equal to: b) The solution will be the same procedure, but the spring factor will be: Then the T.F would be: Note: The second derivative of x could be presented as: ” 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 automobile suspension system.The motion of this system consists of a translational motion of the center 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 vertical 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)
Xi(s)
=
bs + k
ms + bs + k
2
m
k
b
xo
Center of mass
Auto body
Figure 3–20
(a) Automobile
suspension system;
(b) simplified
suspension system.
86
P
xi
(a)
(b)
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 D X(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
Am1 s2 + bs + k1 + k2 B
m2 s2 + bs + k2
Y(s) = Abs + k2 BY(s) + k1 U(s)
bs + k2
which yields
Y(s)
U(s)
=
k1 Abs + k2 B
4
3
m1 m2 s + Am1 + m2 Bbs + Ck1 m2 + Am1 + m2 Bk2 Ds2 + k1 bs + k1 k2
y
m2
k2
b
m1
x
k1
u
Figure 3–21
Suspension system.
Example Problems and Solutions
87
2. Obtain the transfer function
/
of the electrical circuit shown
in.
The equations for the given circuit are as follow:
The Laplace transforms of these three equations, with zero initial conditions,
are
(1)
(2)
(3)
From Equation (2) we obtain.
or
(4)
Or
Substituting equation (4) into equation (1), we get
Or
(5)
From Equation (3) and (4), we get
(6)
From equation (5) and (6), we obtain
3.
Consider the system shown in Figure below. An armature-controlled dc
servomotor drives a load consisting of the moment of inertia
.The
torque developed by the motor is T. The moment of inertia of the motor
rotor is Jm. The angular displacements of the motor rotor and the load
element are
, respectively. The gear ratio is /
Obtain
the transfer function
/
.
Solution
Define the current in the armature circuit to be
then, we have
‫اﻟﻘوة اﻟداﻓﻌﺔ اﻟﻛﮭرﺑﺎﺋﯾﺔ‬
‫اﻟﻣﺗوﻟدة ﻧﺗﯾﺟﺔ ﺗﯾﺎر‬
‫اﻟﺣﻣل وﺗﻛون داﺋﻣﺎ‬
‫ﻣﻌﺎﻛﺳﺔ‬ Or
Taking Laplace transforms
(1)
Where Kb is the back emf constant of the motor. We also have
‫اﻟزﺧم‬
‫اﻟزاوي‬
‫اﻟﻌزم اﻟﻛﮭرﺑﺎﺋﻲ اﻟﻧﺎﺗﺞ‬
‫ﻣن ﻣرور اﻟﺗﯾﺎر ﻓﻲ‬
‫اﻟﻣﺣرك‬
(2)
‫اﻟﻌزم اﻟدوران اﻟﺻﺎﻓﻲ‬
‫اﻟذي ﯾﺳﻠط ﻋﻠﻰ اﻟﺣﻣل‬ ‫ﻋزم‬
‫اﻟﺣﻣل‬
Where K is the motor torque constant and ia is the armature current.
Equation (2) can be rewritten as.
(
Where Or
Taking Laplace transforms
(3)
By substituting equation (3) into equation (1), we obtain.
Or
Hence

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