Control System

1. Mathematical Modeling of
Mechanical Systems
Control Systems – MCE
3406
Ashraf AlShalalfeh,
M.Sc.
ashalalfeh@hct.ac.ae

2. Lesson
Objectives:
 Find the transfer function for linear, time-invariant
translational mechanical systems. (SDOF)
 Find the transfer function for linear, time-invariant
translational mechanical systems. (MDOF)

3. Basic Types of Mechanical Systems
 Translational
 Linear Motion
 Rotational
 Rotational Motion

4. Translational
Spring
i)
Circuit Symbols
Translational Spring
A translational spring is a mechanical element that can be deformed by
an external force such that the deformation is directly proportional to
the force applied to it.
Translational Spring

5. Translational Spring
 If F is the applied force
 Then x1 is the deformation if x2=0
 Or (x1-x2) is the deformation.
 The Force Displacement equation is given as
 Where k is stiffness of spring expressed in N/m
2
x
1
x
)
( 2
1 x
x
k
F 

F
F

6. Translational Spring
 Given two springs with spring constant k1 and k2, obtain the
equivalent spring constant keq for the two springs connected
in:
(1) Parallel (2) Series

7. Translational Spring
(1) Parallel
F
x
k
x
k 
 2
1
F
x
k
k 
 )
( 2
1
F
x
keq 
• The two springs have same displacement therefore:
2
1 k
k
keq 

• If n springs are connected in parallel then:
n
eq k
k
k
k 


 
2
1

8. Translational Spring
(2) Series
F
x
k
x
k 
 2
2
1
1
• The force on two springs is same, F, however
displacements are different therefore:
1
1
k
F
x 
2
2
k
F
x 
• Since the total displacement is , and we have
2
1 x
x
x 
 x
k
F eq

2
1
2
1
k
F
k
F
k
F
x
x
x
eq






9. Translational Spring
• Then we can obtain
2
1
2
1
2
1
1
1
1
k
k
k
k
k
k
keq




2
1 k
F
k
F
k
F
eq


• If n springs are connected in series then:
n
n
eq
k
k
k
k
k
k
k






2
1
2
1

10. Translational Spring
Exercise: Obtain the equivalent stiffness for the following spring
networks.
3
k
i)
ii) 3
k

11. Translational Mass
Translational
Mass
ii)
• Translational Mass is an inertia
element.
• A mechanical system without mass
does not exist.
• If a force F is applied to a mass and it
is displaced to x meters then the
relation b/w force and displacements is
given by Newton’s law. M
)
(t
F
)
(t
x
x
M
F 



12. Translational Damper
Translational
Damper
iii)
• When the viscosity or drag is not
negligible in a system, we often model
them with the damping force.
• All the materials exhibit the property of
damping to some extent.
• If damping in the system is not enough
then extra elements (e.g. Dashpot) are
added to increase damping.

13. Common Uses of Dashpots
Door Stoppers
Vehicle Suspension
Bridge Suspension
Flyover Suspension

14. Translational Damper
x
C
F 

• Where C is damping coefficient (N/ms-1).
)
( 2
1 x
x
C
F 
 


15. Translational Damper
Translational Dampers in Series
2
1 C
C
Ceq 

2
1
2
1
C
C
C
C
Ceq


Translational Dampers in Parallel

16. Force-velocity, force-displacement, and
impedance relationships for springs, viscous
dampers, and mass
where, K, f v, and M are called spring constant, coefficient of viscous friction, and mass,

17. Analogies Between Electrical and
Mechanical Components
• Mechanical systems, like electrical networks, have three passive, linear
components.
• Two of them, the spring and the mass, are energy-storage elements;
• One of them, the viscous damper, dissipates energy.
• The two energy-storage elements are analogous to the two electrical energy-
storage elements, the inductor and capacitor.
• The energy dissipater is analogous to electrical resistance.
• The motion of translation is defined as a motion that takes place along a straight or
curved path. The variables that are used to describe translational motion are acceleration,

18. Newton’s Second Law
Newton's law of motion states that the algebraic sum of
external forces acting on a rigid body in a given
direction is equal to the product of the mass of the
body and its acceleration in the same direction. The
law can be expressed as
𝑭 = 𝑴𝒂

19. Steps to Obtain the Transfer Function of
Mechanical System
• The mechanical system requires just one differential equation, called the equation
of motion, to describe it.
• Assume a positive direction of motion, for example, to the right.
• This assumed positive direction of motion is similar to assuming a current
direction in an electrical loop.
• First Step, draw a free-body diagram, placing on the body all forces that act on
the body either in the direction of motion or opposite to it.
• Second Step, use Newton’s law to form a differential equation of motion by
summing the forces and setting the sum equal to zero.
• Third Step, assuming zero initial conditions, we take the Laplace transform of the
differential equation, separate the variables, and arrive at the transfer function.

21. Example-1(a): Find the transfer function of the
mechanical translational system given in the Figure.
Free Body Diagram
Figure
M
)
(t
f
k
f
M
f
B
f
B
M
k f
f
f
t
f 


)
(

22. Example-1(b): Find the transfer function, X(s)/F(s), of the
system.
Free Body Diagram (FBD)
• First step is to draw the free-body diagram.
• Place on the mass all forces felt by the mass.
• We assume the mass is traveling toward the right. Thus, only the applied force
points to the right; all other forces impede the motion and act to oppose it. Hence,
the spring, viscous damper, and the force due to acceleration point to the left.
• Second step is to write the differential equation of motion using Newton’s law to
sum to zero all of the forces shown on the mass.

23. Example-1(b): Continue.
• Third step is to take the Laplace transform, assuming zero initial conditions,
• Finally, solving for the transfer function yields
Block Diagram

24. Impedance Approach to Obtain the Transfer
Function of Mechanical System
For the spring,
For the viscous damper,
and for the mass,
• Taking the Laplace transform of the force-displacement terms of mechanical
components , we get
• We can define impedance for mechanical components as

25. Example-2: Solve example-1 using the Impedance
Approach.
Laplace Transformed FBD
• Summing the forces in the Laplace Transformed FBD, we get
• Which is in the form of

26. Example-3: Consider a simple horizontal spring-mass
system on a frictionless surface, as shown in figure
below.
 The differential equation of the above system is
or
kx
x
m 



0

 kx
x
m


27. Example-4: Find the transfer function, X(s)/F(s), of the system.
Consider the system friction is negligible.
• Free Body Diagram
M
F
k
f
M
f
k
F
x
M
• Where and are force applied by the spring and inertial
force respectively.
k
f M
f

28. Example-4: continue
• Then the differential equation of the system is:
kx
x
M
F 
 

• Taking the Laplace Transform of both sides and ignoring initial conditions
we get
M
F
k
f
M
f
M
k f
f
F 

)
(
)
(
)
( s
kX
s
X
Ms
s
F 
 2

29. )
(
)
(
)
( s
kX
s
X
Ms
s
F 
 2
• The transfer function of the system is
k
Ms
s
F
s
X

 2
1
)
(
)
(
• if
1
2000
1000



Nm
k
kg
M
2
001
0
2


s
s
F
s
X .
)
(
)
(
Example-4: continue.

30. • The pole-zero map of the system is
2
001
0
2


s
s
F
s
X .
)
(
)
(
Example-4: continue.

31. Example-5: Find the transfer function, X(s)/F(s), of the
following system, where the system friction is not
negligible.
• Free Body Diagram
k
F
x
M
C
M
F
k
f
M
f
C
f
C
M
k f
f
f
F 



32. Example-5: continue.
Differential equation of the system is:
kx
x
C
x
M
F 

 


Taking the Laplace Transform of both sides and ignoring
Initial conditions we get
)
(
)
(
)
(
)
( s
kX
s
CsX
s
X
Ms
s
F 

 2
k
Cs
Ms
s
F
s
X


 2
1
)
(
)
(

33. k
Cs
Ms
s
F
s
X


 2
1
)
(
)
(
• if
1
1
1000
2000
1000





ms
N
C
Nm
k
kg
M
/
1000
001
0
2



s
s
s
F
s
X .
)
(
)
(
Example-5: continue.
-1 -0.5 0 0.5 1
-40
-30
-20
-10
0
10
20
30
40
Pole-Zero Map
Real Axis
Imaginary
Axis

34. Example-6: Find the transfer function, X(s)/F(s), of the
following system.
• Free Body Diagram
M F
k
f M
f
B
f
B
M
k f
f
f
F 


k
Bs
Ms
s
F
s
X


 2
1
)
(
)
(

35. Example-8: Find the transfer function Xo(s)/Xi(s) of
the automobile suspension system.

36. Example-8: continue.

37. Example-13: continue.
0
)
(
)
( 



 i
o
i
o
o x
x
k
x
x
b
x
m 



i
i
o
o
o kx
x
b
kx
x
b
x
m 


 



Taking Laplace Transform of the equation (2)
)
(
)
(
)
(
)
(
)
( s
kX
s
bsX
s
kX
s
bsX
s
X
ms i
i
o
o
o 



2
k
bs
ms
k
bs
s
X
s
X
i
o



 2
)
(
)
(
The transfer function of the system is

38. Example-15: Find the transfer function, X2(s)/F(s), of the
system.
• The system has two degrees of freedom, since each mass can be moved in the
horizontal direction while the other is held still.
• Thus, two simultaneous equations of motion will be required to describe the system.
• The two equations come from free-body diagrams of each mass.
• Superposition is used to draw the free body diagrams.
• For example, the forces on M1 are due to (1) its own motion and (2) the motion of
M2 transmitted toM1 through the system.
• We will consider these two sources separately.

39. Combine (a) & (b)
All forces on M1
Example-15:
Continue.
Case-I: Forces on M1
Figure-1.
Figure-1:
a. Forces on M1 due only to motion of M1;
b. Forces on M1 due only to motion of M2;
c. All forces on M1.

40. Example-15: Continue.
Combine (a) & (b)
All forces on M1
Figure-1:
a. Forces on M1 due only to motion of M1;
b. Forces on M1 due only to motion of M2;
c. All forces on M1.
• The Laplace transform of the equations of motion can be written from Figure-1 (c)
as;
(1)
Case-I: Forces on M1
• If we hold M2 still and move M1 to the right, we see the forces shown in Figure-1(a).
• If we holdM1 still and moveM2 to the right, we see the forces shown in Figure 1(b).
• The total force on M1 is the superposition, or sum of the forces, as shown in Figure-
1(c).

41. Example-15:
Continue.
Combine (a) & (b)
All forces on M2
Case-II: Forces on
M2
Figure-2.
Figure-2:
a. Forces on M2 due only to motion of M2;
b. Forces on M2 due only to motion of M1;
c. All forces on M2.

42. Example-15: Continue.
Combine (a) & (b)
All forces on M2
Figure-2:
a. Forces on M2 due only to motion of M2;
b. Forces on M2 due only to motion of M1;
c. All forces on M2.
• The Laplace transform of the equations of motion can be written from Figure-2 (c)
as;
(2)
Case-II: Forces on
M2
• If we hold M1 still and move M2 to the right, we see the forces shown in Figure-2(a).
• If we move M1 to the right and hold M2 still, we see the forces shown in Figure-2(b).
• For each case we evaluate the forces on M2.
• The total force on M2 is the superposition, or sum of the forces, as shown in Figure-
2(c).

43. • From equation (1) and (2), the transfer function, X2(s)/F(s), is
• Where,
Example-15: Continue.
(1)
(2)
Block Diagram

44. Example-15: Continue.

45. Example-16: Write, but do not solve, the equations of motion for the
mechanical network shown below.
• The system has three degrees of freedom, since each of the three masses can be
moved independently while the others are held still.
• M1 has two springs, two viscous dampers, and mass associated with its motion.
• There is one spring between M1 and M2 and one viscous damper between M1 and
M3.

47. Example-10: Find the transfer function Y(s)/F(s) of
the restaurant plate dispenser system.

48. Example-14: Find the transfer function Y(s)/U(s) of the
train suspension system.
Car Body
Bogie-2
Bogie
Frame
Bogie-1
Wheelsets
Primary
Suspension
Secondary
Suspension

49. Example-14: continue.

50. Electric Circuit Analogs
• An electric circuit that is analogous to a system from another discipline is called
an electric circuit analog.
• The mechanical systems with which we worked can be represented by
equivalent electric circuits.
• Analogs can be obtained by comparing the equations of motion of a
mechanical system, with either electrical mesh or nodal equations.
• When compared with mesh equations, the resulting electrical circuit is called a
series analog.
• When compared with nodal equations, the resulting electrical circuit is called a
parallel analog.

51. Series Analog
Equation of motion of the
above translational
mechanical system is;
Kirchhoff’s mesh
equation for the above
simple series RLC
network is;
For a direct analogy b/w Eq
(1) & (2), convert
displacement to velocity by
divide and multiply the left-
hand side of Eq (1) by s,
yielding;
(1) (2) (3)
Comparing Eqs. (2) & (3), we recognize the sum
of impedances & draw the circuit shown in
Figure (c). The conversions are summarized in
Figure (d).

52. Converting a Mechanical System to a Series Analog
Example-17: Draw a series analog for the mechanical
system.
(1)
(2)
• Eqs (1) & (2) are analogous t0 electrical mesh equations after conversion to
velocity. Thus,
• The equations of motion in the Laplace transform domain are;
(3)
(4)

53. Example-17: Continue.
• Coefficients represent sums of electrical impedance.
• Mechanical impedances associated withM1 form the first mesh,
• whereas impedances between the two masses are common to the two loops.
• Impedances associated with M2 form the second mesh.
• The result is shown in Figure below, where v1(t) and v2(t) are the velocities of
M1 and M2, respectively.
(3)
(4)

54. Skill-Assessment
Exercise
PROBLEM: Find the transfer function, G)s) =X2(s)/F(s), for the
translational
mechanical system shown in Figure

55. Answer Skill-Assessment Exercise

56. Fluid Modeling:
This part consider fluid systems comprising tanks
and pipes and shows that they have a 1st order
system.

57.  Fluid flow is driven by differential pressure
 A common source of fluid pressure is fluid depth,
the deeper the fluid the higher the associated
pressure.
 Pressure elevation equation is:
Pressure in
tanks:
𝑷 = 𝝆𝒈𝒉

58. HYDRAULIC BUILDING BLOCKS
8/27/2022
BLOCK PHYSICAL REPRESENTATION
HYDRAULIC RESISTANCE Opposition to flow
HYDRAULIC CAPACITANCE Store as potential energy
HYDRAULIC INERTANCE Accelerate a fluid by pressure
difference

59. HYDRAULIC RESISTANCE
8/27/2022
P1 − 𝑃2 = q . R
q =
𝑃1 − 𝑃2
R
𝑃𝑅 =
1
𝑅
(𝑃1 − 𝑃2 )2
P1 & P2 – pressure @ 1 & 2
q – volume flow rate
R – hydraulic resistance
PR – power dissipiated
Orifices, valves, nozzles
and friction in pipes can be
modeled as fluid resistors

60. HYDRAULIC CAPACITANCE
8/27/2022
𝑞1 − 𝑞2 = C .
𝑑𝑃
𝑑𝑡
𝑃1 − 𝑃2 =
1
𝐶
. 𝑞1 − 𝑞2 . 𝑑𝑡
𝐸𝐶 =
1
2
𝐶 . (𝑃1 − 𝑃2 )2
q1 & q2 – volume flow rate @ inlet & outlet
P1 & P2– pressure @ 1 & 2
C – hydraulic capacitance
EC – energy stored
C =
𝐴
ρ.𝑔 Refer
Notes for
Derivation
Hydraulic cylinder
chambers,
tanks, and accumulators
are
examples of fluid
capacitors

61. Example #1:
Derive the transfer function: G(s)=
𝑄2
𝑄1

62. Example #1: Solution
Derive the transfer function:
Solution:
𝑞1 − 𝑞2 = C .
𝑑𝑃
𝑑𝑡
(1)
Hydraulic Capacitance (C) is related to the inlet
and outlet flow rates by:
C =
𝐴
ρ.𝑔
Where;
Also, hydraulic Resistance (R) is related to the
pressure at the top surface and bottom of tank
by:
𝑃2 − 𝑃1 = 𝑝 = 𝑅𝑞2 (2)
G(s)=
𝑄2
𝑄1

63. Example #1: Solution (Cont.)
Derive the transfer function:
Solution:
𝑑𝑝
𝑑𝑡
= 𝑅 ×
𝑑𝑞2
𝑑𝑡
(2)’
Deriving eqn. (2) w.r.t. time:
Now, substitute eqn. (2) into eqn. (1):
Taking the Laplace transform for both sides for eqn. (1)’:
𝑞1 = 𝑞2 +
𝐴𝑅
𝜌𝑔
×
𝑑𝑞2
𝑑𝑡
(1)’
G(s)=
𝑄2
𝑄1
𝑄1 𝑠 = 𝑄2 𝑠 +
𝐴𝑅
𝜌𝑔
× 𝑄2 𝑠 𝑥 𝑠

64. Example #1: Solution (Cont.)
Derive the transfer function:
Solution:
Taking 𝑄2 𝑠 as a common facto and re-arrange
the eqn.:
G(s)=
𝑄2
𝑄1
𝑄2 𝑠
𝑄1 𝑠
=
1
(
𝐴𝑅
𝜌𝑔
× 𝑠 ) + 1

65. Example #2:
Derive the transfer function: G(s)=
𝐻
𝑄1

66. Example #2: Solution
Derive the transfer function:
Solution:
𝑞1 − 𝑞2 = C .
𝑑𝑃
𝑑𝑡
(1)
Hydraulic Capacitance (C) is related to the inlet
and outlet flow rates by:
C =
𝐴
ρ.𝑔
Where;
Also, hydraulic Resistance (R) is related to the
pressure at the top surface and bottom of tank
by:
𝑃2 − 𝑃1 = 𝑝 = 𝜌𝑔ℎ (2)
G(s)=
𝐻
𝑄1

67. Example #1: Solution (Cont.)
Derive the transfer function:
Solution:
𝑑𝑝
𝑑𝑡
= 𝜌𝑔 ×
𝑑ℎ
𝑑𝑡
(2)’
Deriving eqn. (2) w.r.t. time:
Now, substitute eqn. (2)’ into eqn. (1):
But;
𝑞1 = 𝑞2 + 𝐴 ×
𝑑ℎ
𝑑𝑡
(1)’
𝜌𝑔ℎ = 𝑞2. R (3)
G(s)=
𝐻
𝑄1

68. Example #1: Solution (Cont.)
Derive the transfer function:
Solution:
Substitute eqn. (3) into eqn. (1)’:
G(s)=
𝐻
𝑄1
𝑞1 =
𝜌𝑔ℎ
𝑅
+ 𝐴 ×
𝑑ℎ
𝑑𝑡
(1)’
Re-writing eqn. (1)’:
𝐴𝑅
𝜌𝑔ℎ
×
𝑑ℎ
𝑑𝑡
+ ℎ =
𝑅
𝜌𝑔
𝑞1 (1)’’

69. Example #1: Solution (Cont.)
Derive the transfer function:
Solution:
Substitute eqn. (3) into eqn. (1)’:
G(s)=
𝐻
𝑄1
𝑞1 =
𝜌𝑔ℎ
𝑅
+ 𝐴 ×
𝑑ℎ
𝑑𝑡
(1)’
Re-writing eqn. (1)’:
𝐴𝑅
𝜌𝑔ℎ
×
𝑑ℎ
𝑑𝑡
+ ℎ =
𝑅
𝜌𝑔
𝑞1 (1)’’

70. Example #1: Solution (Cont.)
Derive the transfer function:
Solution:
G(s)=
𝐻
𝑄1
Taking the Laplace transform for both
sides for eqn. (1)’’:
𝐴𝑅
𝜌𝑔ℎ
× 𝐻 𝑠 . 𝑠 + 𝐻(𝑠) =
𝑅
𝜌𝑔
𝑄1(𝑠) (1)’’
Taking 𝐻 𝑠 as a common facto and re-arrange
the eqn.:
𝐻 𝑠
𝑄1 𝑠
=
𝐴𝑅
𝜌𝑔
(
𝐴𝑅
𝜌𝑔
× 𝑠 ) + 1
OR
𝐻 𝑠
𝑄1 𝑠
=
1
( 1 × 𝑠 ) +
𝜌𝑔
𝐴𝑅

71. 8/27/2022