control system PPT

1. Control system
Dr. Devaraj Somasundaram
Department of ECE,
CMR institute of Technology,
Bangalore

2. Module 1
• Introduction to Control Systems:
Types of Control Systems, Effect of Feedback System’s, Differential
equation of Physical Systems –Mechanical Systems, Electrical Systems,
Electromechanical systems, Analogous Systems.

3. open loop control systems & closed loop control system

5. Types of Feedback

6. Effect of Feedback on Overall Gain
• The overall gain of negative feedback closed loop control system is
the ratio of 'G' and (1+GH). So, the overall gain may increase or
decrease depending on the value of (1+GH).
• If the value of (1+GH) is less than 1, then the overall gain increases. In
this case, 'GH' value is negative because the gain of the feedback path
is negative.
• If the value of (1+GH) is greater than 1, then the overall gain
decreases. In this case, 'GH' value is positive because the gain of the
feedback path is positive

7. Effect of Feedback on Sensitivity
• Sensitivity of the overall gain of negative feedback closed loop control
system (T) to the variation in open loop gain (G).
• If the value of (1+GH) is less than 1, then sensitivity increases. In this case,
'GH' value is negative because the gain of feedback path is negative.
• If the value of (1+GH) is greater than 1, then sensitivity decreases. In this
case, 'GH' value is positive because the gain of feedback path is positive.

8. Effect of Feedback on Stability
• A system is said to be stable, if its output is under control. Otherwise,
it is said to be unstable.
• if the denominator value is zero (i.e., GH = -1), then the output of the
control system will be infinite. So, the control system becomes
unstable.

9. Effect of Feedback on Noise

10. mathematical models
• Differential equation model
• Transfer function model
• State space model

11. Example:

12. Basic Types of Mechanical Systems
• Translational
• Linear Motion
• Rotational
• Rotational Motion
12

13. Translational Mechanical Systems
Part-I
13

14. Basic Elements of Translational Mechanical Systems
14
Translational Mass Translational Damper Translational Spring

15. Modeling of Rotational Mechanical Systems

16. 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
16

17. Translational Spring
• If F is the applied force
• Then is the deformation if
• Or is the deformation.
• The equation of motion is given as
• Where is stiffness of spring expressed in N/m
2
x
1
x
0
2 
x
1
x
)
( 2
1 x
x 
)
( 2
1 x
x
k
F 

k
F
F
17

18. 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 


18

19. 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.
19

20. Common Uses of Dashpots
Door Stoppers
Vehicle Suspension
Bridge Suspension
Flyover Suspension
20

21. Translational Damper
x
C
F 

• Where C is damping coefficient (N/ms-1).
)
( 2
1 x
x
C
F 
 

21

22. Modeling a Simple Translational System
• Example-1: Consider a simple horizontal spring-mass system on a
frictionless surface, as shown in figure below.
or
22
kx
x
m 



0

 kx
x
m


23. Example-2
• Consider the following system (friction is negligible)
23
• 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

24. Example-2
24
• 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

25. 25
)
(
)
(
)
( 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-2

26. 26
• The pole-zero map of the system is
2
001
0
2


s
s
F
s
X .
)
(
)
(
Example-2
-1 -0.5 0 0.5 1
-40
-30
-20
-10
0
10
20
30
40
Pole-Zero Map
Real Axis
Imaginary
Axis

27. Example-3
• Consider the following system
27
• Free Body Diagram
k
F
x
M
C
M
F
k
f
M
f
C
f
C
M
k f
f
f
F 



28. Example-3
28
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
)
(
)
(

29. Example-3
29
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 .
)
(
)
(
-1 -0.5 0 0.5 1
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
Pole-Zero Map
Real Axis
Imaginary
Axis

30. Example-4
• Consider the following system
30
• Free Body Diagram (same as example-3)
M
F
k
f
M
f
B
f
B
M
k f
f
f
F 


k
Bs
Ms
s
F
s
X


 2
1
)
(
)
(

31. Example-5
• Consider the following system
31
• Mechanical Network
k
F
2
x
M
1
x B
↑ M
k
B
F
1
x 2
x

32. Example-5
32
• Mechanical Network
↑ M
k
B
F
1
x 2
x
)
( 2
1 x
x
k
F 

At node 1
x
At node 2
x
2
2
1
2
0 x
B
x
M
x
x
k 

 


 )
(

33. Example-6
• Find the transfer function X2(s)/F(s) of the following system.
1
M 2
M
k
B
33

34. Example-7
34
k
)
(t
f
2
x
1
M
4
B
3
B
2
M
1
x
1
B 2
B
↑ M1
k 1
B
)
(t
f
1
x 2
x
3
B
2
B M2
4
B

35. Example-8
• Find the transfer function of the mechanical translational
system given in Figure-1.
35
Free Body Diagram
Figure-1
M
)
(t
f
k
f
M
f
B
f
B
M
k f
f
f
t
f 


)
(
k
Bs
Ms
s
F
s
X


 2
1
)
(
)
(

36. Example-9
36
• Restaurant plate dispenser

37. Example-10
37
• Find the transfer function X2(s)/F(s) of the following system.
Free Body Diagram
M1
1
k
f
1
M
f
B
f
M2
)
(t
F
1
k
f
2
M
f
B
f
2
k
f
2
k
B
M
k
k f
f
f
f
t
F 


 2
2
1
)
(
B
M
k f
f
f 

 1
1
0

38. Example-11
38
1
k
)
(t
u
3
x
1
M
4
B
3
B
2
M
2
x
2
B 5
B
2
k 3
k
1
x
1
B

39. Example-12: Automobile Suspension
39

40. Automobile Suspension
40

41. Automobile Suspension
41
)
.
(
)
(
)
( 1
0 eq




 i
o
i
o
o x
x
k
x
x
b
x
m 



2
eq.
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
)
(
)
(

42. Example-13: Train Suspension
42
Car Body
Bogie-2
Bogie
Frame
Bogie-1
Wheelsets
Primary
Suspension
Secondary
Suspension

43. Example: Train Suspension
43

44. Rotational Mechanical Systems
Part-I
44

45. Basic Elements of Rotational Mechanical Systems
Rotational Spring
)
( 2
1 
 
 k
T
2

1

45

46. Basic Elements of Rotational Mechanical Systems
Rotational Damper
2

1

)
( 2
1 
 
 
 C
T
T
C
46

47. Basic Elements of Rotational Mechanical Systems
Moment of Inertia


J
T 

T
J
47

48. Example-1
1

T 1
J
1
k
1
B
2
k
2
J
2

3

↑ J1
1
k
T
1
 3

1
B
J2
2

2
k
48

49. Example-2
↑ J1
1
k
1
B
T
1
 3

2
B
3
B J2
4
B
2

1

T 1
J
1
k
3
B
2
B
4
B
1
B
2
J
2

3

49

50. Example-3
1

T
1
J
1
k
2
B 2
J
2

2
k
50

51. Example-4
51

54. Mechanical Linkages
Part-III
54

55. Gear
• Gear is a toothed machine part, such
as a wheel or cylinder, that meshes
with another toothed part to transmit
motion or to change speed or
direction.
55

56. Fundamental Properties
• The two gears turn in opposite directions: one clockwise and
the other counterclockwise.
• Two gears revolve at different speeds when number of teeth on
each gear are different.
56

57. Gearing Up and Down
• Gearing up is able to convert torque to
velocity.
• The more velocity gained, the more torque
sacrifice.
• The ratio is exactly the same: if you get three
times your original angular velocity, you
reduce the resulting torque to one third.
• This conversion is symmetric: we can also
convert velocity to torque at the same ratio.
• The price of the conversion is power loss due
to friction.
57

58. Why Gearing is necessary?
58
• A typical DC motor operates at speeds that are far too
high to be useful, and at torques that are far too low.
• Gear reduction is the standard method by which a
motor is made useful.

59. Gear Trains
59

60. Gear Ratio
• You can calculate the gear ratio by using
the number of teeth of the driver divided
by the number of teeth of the follower.
• We gear up when we increase velocity
and decrease torque.
Ratio: 3:1
• We gear down when we increase torque
and reduce velocity.
Ratio: 1:3
Gear Ratio = # teeth input gear / # teeth output gear
= torque in / torque out = speed out / speed in
Follower
Driver
60

61. Example of Gear Trains
• A most commonly used example of gear trains is the gears of
an automobile.
61

62. Mathematical Modeling of Gear Trains
• Gears increase or reduce angular velocity (while
simultaneously decreasing or increasing torque, such
that energy is conserved).
62
2
2
1
1 
 N
N 
1
N Number of Teeth of Driving Gear
1
 Angular Movement of Driving Gear
2
N Number of Teeth of Following Gear
2
 Angular Movement of Following Gear
Energy of Driving Gear = Energy of Following Gear

63. Mathematical Modeling of Gear Trains
• In the system below, a torque, τa, is applied to gear 1 (with
number of teeth N1, moment of inertia J1 and a rotational friction
B1).
• It, in turn, is connected to gear 2 (with number of teeth N2,
moment of inertia J2 and a rotational friction B2).
• The angle θ1 is defined positive clockwise, θ2 is defined positive
clockwise. The torque acts in the direction of θ1.
• Assume that TL is the load torque applied by the load connected
to Gear-2.
63
B1
B2
N1
N2

64. Mathematical Modeling of Gear Trains
• For Gear-1
• For Gear-2
• Since
• therefore
64
B1
B2
N1
N2
2
2
1
1 
 N
N 
1
1
1
1
1 T
B
J
a 

 

 

 Eq (1)
L
T
B
J
T 

 2
2
2
2
2 
 

 Eq (2)
1
2
1
2 

N
N
 Eq (3)

65. Mathematical Modeling of Gear Trains
• Gear Ratio is calculated as
• Put this value in eq (1)
• Put T2 from eq (2)
• Substitute θ2 from eq (3)
65
B1
B2
N1
N2
2
2
1
1
1
2
1
2
T
N
N
T
N
N
T
T



2
2
1
1
1
1
1 T
N
N
B
J
a 

 

 


)
( L
a T
B
J
N
N
B
J 



 2
2
2
2
2
1
1
1
1
1 



 





)
( L
a T
N
N
N
N
B
N
N
J
N
N
B
J
2
1
2
2
1
2
1
2
1
2
2
1
1
1
1
1 



 



 






66. Mathematical Modeling of Gear Trains
• After simplification
66
)
( L
a T
N
N
N
N
B
N
N
J
N
N
B
J
2
1
2
2
1
2
1
2
1
2
2
1
1
1
1
1 



 



 





L
a T
N
N
B
N
N
B
J
N
N
J
2
1
1
2
2
2
1
1
1
1
2
2
2
1
1
1 



















 



 





L
a T
N
N
B
N
N
B
J
N
N
J
2
1
1
2
2
2
1
1
1
2
2
2
1
1 



































 

 


2
2
2
1
1 J
N
N
J
Jeq 








 2
2
2
1
1 B
N
N
B
Beq 









L
eq
eq
a T
N
N
B
J
2
1
1
1 

 

 



67. Mathematical Modeling of Gear Trains
• For three gears connected together
67
3
2
4
3
2
2
1
2
2
2
1
1 J
N
N
N
N
J
N
N
J
Jeq 


























3
2
4
3
2
2
1
2
2
2
1
1 B
N
N
N
N
B
N
N
B
Beq 

























