1. B.Tech – II – ECE – II SEMESTER ACE ENGINEERING COLLEGE
CONTROL SYSTEMS
B.Tech – II – ECE – II SEMESTER
M.MYSIAH
Assistant Professor
ACE ENGINEERING COLLEGE 5

2. Course objectives
To understand the different ways of system representations such as Transfer function representation and state space representations and to assess the system dynamic response
To assess the system performance using time domain analysis
and methods for improving
To assess the system performance using frequency domain analysis and techniques for improving the performance
To design various controllers and compensators to improve system performance 5

3. Course Outcomes
After completion of this course the student is able to
Improve the system performance by selecting a suitable controller and/or a compensator for a specific application
Apply various time domain and frequency domain techniques to assess the system performance.
Apply various control strategies to different applications (example: Power systems, electrical drives etc…)
Test system Controllability and Observability using state space representation and applications of state space representation to various systems. 5

4. Syllabus UNIT – I Introduction:
Concepts of Control Systems- Open Loop and closed loop control systems and their differences.
Different examples of control systems.
Classification of control systems, Feed-Back Characteristics, Effects of feedback.
Mathematical models – Differential equations - Impulse Response and transfer functions - Translational and Rotational
mechanical systems.
Transfer Function Representation:
Transfer Function of DC Servo motor - AC Servo motor- Synchro transmitter and Receiver.
Block diagram representation of systems considering electrical systems as examples - Block diagram algebra .
Representation by Signal flow graph - Reduction using mason’s gain formula. 5

5. Time Response Analysis:
UNIT-II
Time Response Analysis:
Standard test signals - Time response of first order systems
Characteristic Equation of Feedback control systems, Transient response of second order systems.
Time domain specifications – Steady state response - Steady state errors and error constants.
Effects of proportional derivative, proportional integral systems.
UNIT – III
Stability Analysis:
The concept of stability - Routh stability criterion – qualitative stability and conditional stability.
Root Locus Technique:
The root locus concept - construction of root loci-effects of adding poles and zeros to G(s) H(s) on the root loci. 5

6. UNIT- I : Introduction to Control System
UNIT – III
Frequency Response Analysis:
Introduction, Frequency domain specifications.
Bode diagrams-Determination of Frequency domain specifications and transfer function from the Bode Diagram.
Phase margin and Gain margin-Stability Analysis from Bode Plots.
UNIT - IV
Stability Analysis In Frequency Domain:
Polar Plots.
Nyquist Plots and applications of Nyquist criterion to find the stability.
Effects of adding poles and zeros to G(s)H(s) on the shape of the Nyquist diagrams.
Classical Control Design Techniques:
Compensation techniques – Lag, Lead, and Lead-Lag Controllers design in frequency Domain, PID Controllers. 5

7. UNIT- I : Introduction to Control System
UNIT – V
State Space Analysis of Continuous Systems:
Concepts of state, state variables and state model.
Derivation of state models from block diagrams.
Diagonalization- Solving the Time invariant state Equations.
State Transition Matrix and its Properties.

8. REFERENCE BOOKS:
1. “I. J. Nagrath and M. Gopal”, “Control Systems Engineering”, New Age International (P) Limited, Publishers, 5th edition, 2009
2. “B. C. Kuo”, “Automatic Control Systems”, John wiley and sons, 8th edition, 2003.
3. “Katsuhiko Ogata”, “Modern Control Engineering”, Prentice Hall of India Pvt. Ltd., 3rd edition, 1998. 5

9. UNIT- I : Introduction to Control System
Introduction to Control systems 
Control System – Definition and Practical Examples
Classification of Control System : Open Loop and Closed Loop Systems – Definitions, Block diagrams, practical examples, and Comparison, Linear and Non-linear Control System, Time Varying and Time In-varying Systems
Servo System : Definition, Block Diagram, Classification (AC and DC Servo
System), Block diagram of DC Servo System.  
Laplace Transform and Transfer Function
 Laplace Transform : Signifiance in Control System
 Transfer Function : Definition, Derivation of transfer functions for Closed loop
Control System and Open Loop Control System, Differential Equations and
transfer functions of RC and RLC Circuit 
Block Diagram Algebra
 Order of a System : Definition, 0,1,2 order system Standard equation, Practical
Examples
 Block Diagram Reduction Technique: Need, Reduction Rules, Problems
M Mysaiah 5

10. Introduction to
Control
System 11

11. Explain different types of control system
Specific Objectives
Explain different types of control
Develop transfer functions
system
Differentiate between 1st&
2nd order of
system
Develop
system
and
solve
block
diagram of
control 12

12. UNIT- I : Introduction to Control System
Introduction to Control systems 
Control System – Definition and Practical Examples
Classification of Control System : Open Loop and Closed Loop Systems – Definitions, Block diagrams, practical examples, and Comparison, Linear and Non-linear Control System, Time Varying and Time In-varying Systems
Servo System : Definition, Block Diagram, Classification (AC and DC Servo
System), Block diagram of DC Servo System.  
Laplace Transform and Transfer Function
 Laplace Transform : Signifiance in Control System
 Transfer Function : Definition, Derivation of transfer functions for Closed loop
Control System and Open Loop Control System, Differential Equations and
transfer functions of RC and RLC Circuit 
Block Diagram Algebra
 Order of a System : Definition, 0,1,2 order system Standard equation, Practical
Examples
 Block Diagram Reduction Technique: Need, Reduction Rules, Problems 13

13. Input  The stimulus or excitation applied to a control system from an
external
source in
order to
produce
the
output is
called input 14

14. Output  The actual response obtained from a system is called output.
Input
Output
 The
actual
response
obtained
from a
system is
called output. 15

15. “System”  A system is an arrangement of or a combination of
Input
Output
 A system is an arrangement of or a combination of
different physical components connected or related in
such a manner so as to form an entire unit to attain a
certain objective. 16

16. Control  It means to regulate, direct or command a system so that the
desired
objective is
attained 17

17. Combining above
definitions
System +
Control =
Control
System 18

18. Control System  It is an arrangement of different physical elements
Input
Output
 It is an
arrangement of
different physical
elements
connected in such a
or command itself to
manner so as to
achieve a certain
regulate, direct
objective. 19

19. Difference between System and Control System System be desired)
Input
Input
Proper
Desired
System
Output
Output
(May or may not
be desired) 20

20. An example : Fan Difference between System and Control System (System)
230V/50Hz
Fan
(System)
Air
Flow
Input
Output
AC Supply 21

21. A Fan: Can't Say System Output)
 A Fan without blades cannot be a “SYSTEM”
Because it
i.e. airflow
cannot
provide a
desired/proper output
Input
Output
230V/50Hz
No Airflow
AC Supply
(No Proper/
Output)
Desired 22

22. A Fan:
Can be a System
 A Fan with blades but without regulator can be a “SYSTEM”
Because it can provide a proper output
i.e. airflow
 But it
cannot be a
“Control
System”
Because it
cannot
provide desired
output
i.e.
controlled airflow
Input
Output
230V/50Hz
Airflow
AC Supply
(Proper Output) 23

23. A Fan: Can be a Control System Element
 A Fan with blades and with regulator can be a “CONTROL
SYSTEM” Because it can
i.e. Controlled airflow
provide a
Desired output.
Control
Element
Input
Output
Controlled Airflow
230V/50Hz
(Desired Output)
AC Supply 24

24. Module I – Introduction to Control System
Introduction to Control systems 
Control System – Definition and Practical Examples
Classification of Control System : Open Loop and Closed Loop Systems – Definitions, Block diagrams, practical examples, and Comparison, Linear and Non-linear Control System, Time Varying and Time In-varying Systems
Servo System : Definition, Block Diagram, Classification (AC and DC Servo
System), Block diagram of DC Servo System.  
Laplace Transform and Transfer Function
 Laplace Transform : Signifiance in Control System
 Transfer Function : Definition, Derivation of transfer functions for Closed loop
Control System and Open Loop Control System, Differential Equations and
transfer functions of RC and RLC Circuit 
Block Diagram Algebra
 Order of a System : Definition, 0,1,2 order system Standard equation, Practical
Examples
 Block Diagram Reduction Technique: Need, Reduction Rules, Problems 25

25. (Depending on control action)
Classification of Control System
Classification of Control System
(Depending on control action)
Closed
Loop
Control
Open Loop Control
System
System 26

26. Open Loop Control System Definition: “A system in which the controlof
which
the
control
action is
totally
as open
independent
loop system”
the
output of
the
system is
called
Controlled
o/p
c(t)
Controller
Process
Reference I/p
r(t)
u(t)
Fig. Block
Diagram of Open loop Control System 27

27. Open Loop Control Systems Examples
 Electric hand drier – Hot
air (output) comes out as
long as
you
keep
your
hand under the machine,
irrespective of how
your hand is dried.
much 28

28. Open Loop Control Systems Examples
 Automatic washing machine –
This
machine
runs
according to the pre-set time
irrespective of
washing is
completed or
not. 29

29. OLCS Examples  Bread toaster - This machine runs
per time
machine
adjusted
runs
irrespective of toasting
completed or not. is 30

30. OLCS Examples  Automatic tea/coffee Vending Machine – These machines
also
These
machines
function for
time only.
pre
adjusted 31

31. OLCS Examples  Light switch – lamps glow whenever light switch is on
required or not.
switch is on
irrespective of light is
 Volume on stereo system –
Volume is
adjusted
manually
irrespective of
output
volume
level. 32

32. Advantages of OLCS  Simple in construction and design.  Economical.
 Easy to maintain.
 Generally stable.
 Convenient to use
as output is difficult
to measure. 33

33. Disadvantages of OLCS  They are inaccurate  They are unreliable
 Any
change in
output
cannot be
corrected
automatically. 34

34. Closed Loop System Definition: “A system in which the control actionis
somehow
dependent
the
output is
called as
closed loop system” 35

35. Block Diagram of CLCS O/p Transducer Signal Signal Transducer I/p
Forward
Path
Command
Error
Signal
Manipulated
Controlled
O/p
I/p
Reference
Transducer
Controller
Signal
Plant
r(t)
e(t)
c(t)
m(t)
Reference
I/p
Feedback
Transducer
Feedback
Signal
b(t)
c(t)
Feedback
Path 36

36. CLCS Examples  Automatic Electric Iron- Heating elements are
controlled by
output
temperature of
the 37

37. Servo voltage stabilizer – Voltage controller operates system.
CLCS Examples
Servo voltage stabilizer –
Voltage
controller
operates
system.
depending
upon
output
voltage of
the 38

38. CLCS Examples
Perspiration 39

39. Advantages of CLCS Closed loop control systems are more
Closed loop control systems are more
presence of non-linearity.
accurate even in the 
Highly
accurate as
any
error
arising is
corrected
due to
presence of feedback signal.
Bandwidth range is large.
Facilitates automation. 
The
sensitivity of
system
may be
made
small to
make
system more stable.
This system is less affected by noise.  40

40. Disadvantages of CLCS  They are costlier. design.
 They are complicated to
design.
 Required more maintenance.
 Feedback leads to oscillatory response.
 Overall gain is reduced due to presence
of feedback.
 Stability is the major problem and more care is needed
to design a stable closed loop system. 41

41. Difference Between OLCS & CLCS
Open Loop Control System
Closed Loop Control System 1.
The
open loop
systems 1.
The
closed loop
systems
are simple & economical.
are complex and costlier 2.
They
power.
consume
less 2.
They
power.
consume
more 3.
The
easier OL
systems
are 3.
The
easy of CL to
systems
construct
are
not to of
construct
because
because
less
number
more
number
required. of
of components required.
components 4.
The
are
open
loop
systems 4.
The
are
closed
loop
systems
more
inaccurate
&
accurate
&
unreliable
reliable. 42

42. Difference Between OLCS & CLCS
Open Loop Control System
Closed Loop Control System 5.
Stability is in
not OL a
major
control
5. Stability is a major problem
in closed loop systems & more
problem
systems.
Generally OL
care is
needed to
design a
systems are stable.
6. Small bandwidth.
stable closed loop system.
6. Large bandwidth. 7.
Feedback
element is 7.
Feedback
element is
absent.
present. 8.
Output
measurement is 8.
Output
measurement is
not necessary.
Necessary. 43

43. Difference Between OLCS & CLCS
Open Loop Control System
Closed Loop Control System
9. The changes in the output due
to external disturbances are not corrected automatically. So they
9.The
due
changes
in the
output to
external
disturbances
are corrected automatically. So
are more sensitive
other disturbances. to
noise
and
they are less sensitive to
and other disturbances.
noise
10. Examples:
10. Examples:
Coffee Maker,
Guided Missile,
Automatic Toaster,
Temp control of oven,
Hand Drier.
Servo voltage stabilizer. 44

44. Introduction to Control System
Introduction to Control systems 
Control System – Definition and Practical Examples
Classification of Control System : Open Loop and Closed Loop Systems – Definitions, Block diagrams, practical examples, and Comparison, Linear and Non-linear Control System, Time Varying and Time In-varying Systems
Servo System : Definition, Block Diagram, Classification (AC and DC Servo
System), Block diagram of DC Servo System.  
Laplace Transform and Transfer Function
 Laplace Transform : Signifiance in Control System
 Transfer Function : Definition, Derivation of transfer functions for Closed loop
Control System and Open Loop Control System, Differential Equations and
transfer functions of RC and RLC Circuit 
Block Diagram Algebra
 Order of a System : Definition, 0,1,2 order system Standard equation, Practical
Examples
 Block Diagram Reduction Technique: Need, Reduction Rules, Problems 45

45. Classification of Control System
Non-linear
Control
Linear Control
System
System 46

46. When an input X1 produces an output Y1 & an input X2 produces an
Linear Control System
When an input X1 produces an output Y1 & an
input X2
produces an
output
Y2,
then
any
combination
 X1   X 2
should produce an
output
Y1  Y 2
. In such case system is linear.
Therefore,
linear
systems
are
those
where
the
principles of superposition
are obeyed.
and proportionality 47

47. Non-linear Control System
 Non-linear systems do not obey law of superposition.
 The stability of non-linear systems depends on root
location as well as initial conditions & type
of input.
 Non-linear systems
exhibit self
sustained
oscillations
of fixed frequency. 48

48. Difference Between Linear & Non-linear System1. 2.
Obey superposition.
Can be analyzed by test signals 1. 2.
Do not obey superposition
Cannot be analyzed by standard test signals
standard 3.
Stability
depends
only on 3.
Stability
locations,
depends on
root
root location
initial
conditions
&
type of input
Exhibits limit cycles 4. 5.
Do not exhibit limit
cycles 4. 5. Do
not
exhibit hysteresis/
Exhibits
resonance
hysteresis/
jump
jump resonance
Can be analyzed by Laplace transform, z- transform 6. 6.
Cannot be analyzed by
transform, z- transform
Laplace 49

49. Classification of Control System
Time
Invarying
System
Control
Time Varying
Control System 50

50. Time varying/In-varying Control System
 Systems whose parameters vary with
time
are
called
time varying control systems.
 When
parameters do
not
vary
with
time
are
called
Time
Invariant
control
systems. 51

51. The mass of missile/rocket reduces as fuel is burnt and hence the
Time varying/In-varying Control System
The
mass of
missile/rocket
reduces as
fuel is
burnt
and
hence
the
parameter
mass is
time
varying
and
the
control
system is
time
varying
type. 52

52. Module I – Introduction to Control System
Introduction to Control systems
(4 Marks) 
Control System – Definition and Practical Examples
Classification of Control System : Open Loop and Closed Loop Systems – Definitions, Block diagrams, practical examples, and Comparison, Linear and Non-linear Control System, Time Varying and Time In-varying Systems
Servo System : Definition, Block Diagram, Classification (AC and DC Servo
System), Block diagram of DC Servo System.  
Laplace Transform and Transfer Function
 Laplace Transform : Signifiance in Control System
(4 Marks)
 Transfer Function : Definition, Derivation of transfer functions for Closed loop
Control System and Open Loop Control System, Differential Equations and
transfer functions of RC and RLC Circuit 
Block Diagram Algebra
(8 Marks)
 Order of a System : Definition, 0,1,2 order system Standard equation, Practical
Examples
 Block Diagram Reduction Technique: Need, Reduction Rules, Problems 53

53. Servo System Definition: 1. Servo system is defined as automatic
Definition: 1.
Servo
system is
defined as
automatic
feedback
control
system
working
on error
signals
giving
the
output as
mechanical position, velocity or acceleration. 2.
Servo system is one type of feedback control
system in
&
which control variable
its time derivatives like
is the mechanical load position
velocity and acceleration. 54

54. General
block
diagram of
Servo
System 55

55. Difference between Servo System 1. Efficiency is low 1. Efficiency is
AC servo System
DC servo System 1. 2. 3.
Efficiency is low
Low power output 1. 2. 3.
Efficiency is
High power
high
output frequent It
requires
less It
requires
maintenance
Less problems
maintenance
More problems 4.
stability 4.
stability 5. 6.
Smooth operation 5. 6.
Noisy operation It
has
non-linear It
has
linear
characteristics
characteristics 56

56. DC
Servo
System 57

57. Working of Servo SystemDC
Servo
System
Working of Servo System 58

58. Module I – Introduction to Control System
Introduction to Control systems
(4 Marks) 
Control System – Definition and Practical Examples
Classification of Control System : Open Loop and Closed Loop Systems – Definitions, Block diagrams, practical examples, and Comparison, Linear and Non-linear Control System, Time Varying and Time In-varying Systems
Servo System : Definition, Block Diagram, Classification (AC and DC Servo
System), Block diagram of DC Servo System.  
Laplace Transform and Transfer Function
 Laplace Transform : Signifiance in Control System
(4 Marks)
 Transfer Function : Definition, Derivation of transfer functions for Closed loop
Control System and Open Loop Control System, Differential Equations and
transfer functions of RC and RLC Circuit 
Block Diagram Algebra
(8 Marks)
 Order of a System : Definition, 0,1,2 order system Standard equation, Practical
Examples
 Block Diagram Reduction Technique: Need, Reduction Rules, Problems
Amit Nevase 59

59. Module I – Introduction to Control System
Introduction to Control systems
(4 Marks) 
Control System – Definition and Practical Examples
Classification of Control System : Open Loop and Closed Loop Systems – Definitions, Block diagrams, practical examples, and Comparison, Linear and Non-linear Control System, Time Varying and Time In-varying Systems
Servo System : Definition, Block Diagram, Classification (AC and DC Servo
System), Block diagram of DC Servo System.  
Laplace Transform and Transfer Function
 Laplace Transform : Signifiance in Control System
(4 Marks)
 Transfer Function : Definition, Derivation of transfer functions for Closed loop
Control System and Open Loop Control System, Differential Equations and
transfer functions of RC and RLC Circuit 
Block Diagram Algebra
(8 Marks)
 Order of a System : Definition, 0,1,2 order system Standard equation, Practical
Examples
 Block Diagram Reduction Technique: Need, Reduction Rules, Problems 67

60. Transfer Function  The relationship between input & output of
a system is
given by the transfer function.
 Definition: The ratio of Laplace
transform of
the output to
the
Laplace transform of
the
input
under
the
assumption of
zero
initial
conditions is
defined as
“ Transfer Function” . 68

61. Transfer Function g(t) G(s) System System LT c(t) C(s) r(t) R(s)
For the system shown,
c(t)= output r(t)= input
L{c(t)}= C(s)
L{r(t)}= R(s) L{g(t)}= G(s)
g(t)= System
function
Therefore transfer function G(s)
for above
system
is given by,
C ( s )
Laplace of output
G(s)= =
Laplace of input
R ( s ) 69

62. Transfer Function of closed loop system E(s) + - Signal Signal T.F.=
Error
Signal
E(s)
Gain for
CL system is given by;
C(s)
R(s)
G(s) 
G(s)
C(s)
Output
E(s)
 G(s). E(s)      (3)
+ -
 C(s)
Input
Substitute value of E(s) from eq.
C(s)  G(s).(R(s)  B(s))
1 to 3
B(s)
H(s)
 C(s)  G(s). R(s)  G(s).B(s)      (4)
Feedback
Signal
Error signal is
Substitute value of B(s) from eq. 2 to
C(s)  G(s) R(s)  G(s). H(s). C(s) 4
given by;
E (s)  R(s)  B(s)      (1)
 R(s)  E (s)  B(s)
G(s). R(s) 
C(s)  G(s). H(s). C(s)
C(s)(1  G(s). H(s))
Gain of feedback
network is
given
by;
Transfer function is given by;
B(s)
C(s)
H (s) 
C (s)
G(s) 
T.F.=
R(s) 1  G(s). H(s)
0B1(6s)  H (s).C(s)      (2) 70

63.  The Laplace transform can be used independently on
Laplace Transform of Passive Element (R,L & C)
 The Laplace transform can be
used independently on
different circuit elements, and
then the circuit can be
solved entirely in the S Domain
(Which is much easier).
 Let's
take a
look at
some of
the
circuit
elements 71

64. Laplace Transform of R
 Resistors are time and frequency invariant. Therefore,
the
transform of a
resistor is
the
same as
the
resistance of the resistor.
L{Resistor}=R(s) 72

65.  i(t) dt Laplace Transform of C domain, we get the following: 1 1
Let us look at the relationship between voltage, current,
and capacitance, in
the time domain:
dv(t) dt
we get the following
i(t )  C
Solving for voltage,
integral: 
 i(t) dt 1
v(t)  C to
Then, transforming this equation
domain, we get the following:
1 1
into the Laplace
V (s) 
I (s)
C s 73

66. Laplace Transform of C Therefore, the transform for a capacitor with
Again, if we
solve for the
ratio
V(s)/I(s), we
get
the
following:
V (s) 1 
I (s) sC
Therefore,
the
transform
for a
capacitor
with
capacitance C is given by: 1
L{capacitor}  sC 74

67. Laplace Transform of L putting formula: this into the Laplace domain,
Let us look at the relationship between voltage, current,
and inductance, in the time domain:
di(t)
v(t)  L dt
putting
formula:
this
into
the Laplace
domain, we
get
the
V (s)  sLI(s)
for our ratio
V (s)
And solving
 sL
I(s) 75

68. Laplace Transform of L Therefore, the transform of an inductor with
inductance L is
given
by:
L{inductor}  sL 76

69.  i(t)dt  i(t)dt 1 ) Vi(s)  RI(s)  1 I (s)      (1)
Transfer Function of RC and RLC electrical circuits
Example:
Find
the
TF of given RC
network
i(t) i(t) C
Taking Laplace transform above equation 1 V
Vo(t)
Vo(s) 
I (s)      (2) sC
 I (s)  sC.Vo(s)      (3)
From equation 1,
Vi(s)  I (s)(R  1 )      (4)
Apply
KVL for input loop, 1 t
 i(t)dt
vi(t)
 Ri(t)  sC
From equation 3 and 4, C
Taking Laplace transform above
equation
Vi(s)  RI(s)  1 I (s)      (1)
1 )
Vi(s)  Vo(s).sC.(R  sC
Apply KVL for output loop, sC 1 t
 i(t)dt
vo(t)  C 77

70. 1 ) sCR ) 1 sCR  1 Vo(s) Vi(s) 1  sCR  1 Transfer Function=
G(s)= 
1 )
sC.(R  sC
Vo(s)
Vi(s) 1 
sCR ) 1
sC.( sC
Vo(s)
Vi(s) 1 
sCR  1 1
sCR  1
Vi(s)
Vo(s) 78

71. Vo(s)  1 I (s)      (2) V i(s) I(s) V
Transfer Function of RC and RLC electrical circuits
Example:
Find
the TF of
given
RLC
network L
i(t) i(t) C
Apply KVL for input loop, V
Vo(t) V 1
Vi(s)  RI(s)  sLI(s) 
I (s) sC
Taking
Laplace
transform
above network 1
Vi(s)  [R sL
]I (s)        (1) sC sL 1
i(s) I(s) sC
Apply KVL for output loop, V
Vo(s) V
Vo(s)  1 I (s)      (2) sC 79

72. s2 LC  sCR  1 sCR  s2 LC  1 I (s) sC Transfer Function=
From
equation 1 and 2, 1
I (s) sC
Vo(s)
Vi(s)
Transfer
Function= 
[R  sL 1 ] I(s) sC 1 sC 
1 ]
[R  sL sC 1 sC 
sCR  s 2 LC  1 sC 1 
sCR  s2 LC  1 1 
s2 LC  sCR  1 80

73. Module I – Introduction to Control System
Introduction to Control systems 
Control System – Definition and Practical Examples
Classification of Control System : Open Loop and Closed Loop Systems – Definitions, Block diagrams, practical examples, and Comparison, Linear and Non-linear Control System, Time Varying and Time In-varying Systems
Servo System : Definition, Block Diagram, Classification (AC and DC Servo
System), Block diagram of DC Servo System.  
Laplace Transform and Transfer Function
 Laplace Transform : Signifiance in Control System
 Transfer Function : Definition, Derivation of transfer functions for Closed loop
Control System and Open Loop Control System, Differential Equations and
transfer functions of RC and RLC Circuit 
Block Diagram Algebra
 Order of a System : Definition, 0,1,2 order system Standard equation, Practical
Examples
 Block Diagram Reduction Technique: Need, Reduction Rules, Problems 81

74. Order of System  The order of control
system is defined as the highest
power of s
present in of
denominator of
closed
loop
transfer
function
G(s)
unity
feedback
system. 82

75. Example1: Determine order of given system
s(s2) TF 
G(s)  s4
 7s3
 10s2
 5s
 5 83

76. Example1: Determine order of given system
s(s2)
TF  G(s)  s4
 7s3  10s2  5s  5
Answer:
The highest power of equation in
of given transfer function is ‘4’.
Hence the order of given system
denominator
is fourth 84

77. System Order and Proper System
 Highest power of s present in denominator of closed
loop transfer function is called as “Order of System”.
 A proper system is a system where the
degree of the
denominator is larger than or
equal to
the
degree of
the numerator polynomial. 85

78. Determine order of given system
Example 2 :
Determine order of given
system
(s5)(s2)
G(s) 
s(s 3)(s 4) 86

79. Example 2 : Determine order of given system
(s5)(s2)
G(s) 
s(s 3)(s 4)
Solution: To obtain highest power of
denominator,
Simplify
denominator polynomial.
s(s 3)(s 4)  0
s(s2  7 s 12)  0
s3  7 s2  12s  0
The highest power of equation in denominator
of given
transfer
function is ‘3’. Hence given system is “Third Order system”.
The degree of denominator is larger than the numerator hence
system is “Proper System” 87

80. Determine order of given system
Example 3 :
Determine order of given
system
K(s5)
G(s) 
s3 (7 s2  12s  5) 88

81. Example 3 : Determine order of given system
K(s5)
G(s) 
s3 (7 s2  12s  5)
Solution: To obtain highest power of
denominator,
Simplify
denominator polynomial.
s3 (7 s2  12s  5)  0
7 s5  12s4  5s3  0
The highest power of equation in denominator of given
transfer
function is ‘5’. Hence given system is “Fifth Order system”.
The degree of denominator is larger than the numerator hence
system is “Proper System” 89

82. Types of System  Zero (0) Order System  First Order System
(depending
on highest
power of
denominator)
 Zero (0) Order System
 First Order System
 Second Order System 90

83. Zero (0) Order System 1 1  T Definition: If highest power
of complex variable ‘s’ present
in Characteristics equation is
zero,
then it is
called as
“Zero order
System” 1
1  T
R(s) +
C(s) - 91

84. Zero (0) Order System 1  s0T  0
Consider a unity feedback system with transfer
function 1
G(s) 
1  T
Hence characteristics equation
1  T  0
is given by, or
1  s0T  0
Here the highest power of s is equal to 0,
Hence the system given above is zero order
system.
Practical Example: Amplifier type control system 92

85. First Order System
Definition: If highest power of complex variable ‘s’ present
In Characteristics equation is
one,
then it is
called as
“First order
System” 1
1  sCR +
C(s)
R(s) - 93

86. First Order System 1  sCR  0 1 G(s)  1  sCR Consider a
unity feedback system with transfer
function 1
G(s) 
1  sCR
Hence characteristics equation is given by,
1  sCR  0
Here the highest power of s is equal to 1,
Hence the system given above is
First order
system.
Practical Example: RC circuits,
thermal type systems 94

87. Second Order System
Definition: If highest power of complex variable ‘s’ present
In Characteristics equation is
two,
then it is
called as
“Second
order
System” 1
s2 LC  sCR  1
R(s) +
C(s) - 95

88. Second Order System s2 LC  sCR  1
Consider a unity feedback system with transfer
function 1
G(s) 
s2 LC  sCR  1
Hence characteristics equation is given by,
s2 LC  sCR  1  0
Here the highest power of s is equal to 2, Hence the system given above is Second order
system.
Practical Example: RLC circuits, Robotic control system. 96

89. Module I – Introduction to Control System
Introduction to Control systems
(4 Marks) 
Control System – Definition and Practical Examples
Classification of Control System : Open Loop and Closed Loop Systems – Definitions, Block diagrams, practical examples, and Comparison, Linear and Non-linear Control System, Time Varying and Time In-varying Systems
Servo System : Definition, Block Diagram, Classification (AC and DC Servo
System), Block diagram of DC Servo System.  
Laplace Transform and Transfer Function
 Laplace Transform : Signifiance in Control System
(4 Marks)
 Transfer Function : Definition, Derivation of transfer functions for Closed loop
Control System and Open Loop Control System, Differential Equations and
transfer functions of RC and RLC Circuit 
Block Diagram Algebra
(8 Marks)
 Order of a System : Definition, 0,1,2 order system Standard equation, Practical
Examples
 Block Diagram Reduction Technique: Need, Reduction Rules, Problems 97

90. Need of Block Diagram Algebra
 If the system is simple & has limited parameters then it is
easy to
analyze
such
systems
using the methods
discussed earlier i.e. transfer function, if the system is
complicated and also have number of
parameters
then
it is very difficult to analyze it. 98

91. Need of Block Diagram Algebra
 To
overcome
this
problem
is used.
block
diagram
representation method
 It is a
simple
way to
represent
any
practically
complicated
system. In
this
each
component of
the
system is represented by a separate block known as
functional block.
 These blocks are interconnected in a proper sequence. 99

92. Block Diagram Fundamentals
 Block Diagram: It is shorthand, pictorial representation
of the cause and effect relationship
between
input
and
output of a
physical
system.
BLOCK
Input
Output
100

93. Block Diagram Fundamentals
 Output:
The value of the input is multiplied to
the
value of
block
gain to
get
the
output. 3s
X(s)
Y(s)
Output
Y(s)= 3s. X(s)
101

94. Block Diagram Fundamentals
 Summing Point: Two or
more
signals
can be
added/
substracted
at summing point. y +
output x +
Output
=x+y-z - z
102

95. Block Diagram Fundamentals
 Take off Point:
The output signal can be
applied to
two or
more
points
from a
take off
point. Z Z Z
Take off point Z
103

96.  Feedback Path: Block Diagram Fundamentals of signal is from
 Forward Path:
The
direction of flow of
signal
is from
input
to output
Forward
Path
C(s)
R(s) + G1 G2 - H1
Feedback Path
The direction of flow
 Feedback Path:
of signal is
from
output to input
104

97. Gain of blocks connected in cascade gets multiplied with each other.
Block Diagram Reduction Techniques
Rule 1: For blocks in cascade
Gain of blocks connected in
cascade
gets
multiplied with each
other.
R(s)
C(s)
R(s) G1
R1(s)
G1G2
C(s) G2
R1(s)=G1R(s)
C(s)=
G1G2R(s)
C(s) =G2R1(s)
=G1G2R(s)
105

98. G1 G2 G3
R(s)
C(s)
Find
Equivalent
G1G2G3
R(s)
C(s)
106

99. G1 G2 G3 G1G2G3 C(s) R(s) R1(s) Find Equivalent R(s) C(s) R(s) C(s)
107

100. Block Diagram Reduction Techniques
Rule 2: For blocks in Parallel
Gain of blocks connected
algebraically. in
parallel gets
added G1
R1(s) +
R(s)
R(s) -
C(s)
G1-G2+G3
C(s) G2
R2(s) + G3
R3(s)
C(s)=
(G1-G2+G3) R(s)
C(s)= R1(s)-R2(s)+R3(s)
= G1R(s)-G2R(s)+G3R(s)
C(s)=(G1-G2+G3) R(s)
108
M Mysaiah

101. Block Diagram Reduction Techniques
Rule 3: Eliminate Feedback Loop
R(s)
E(s) G
C(s)
+ -
R(s) G
1  GH
C(s) +
B(s) H
C(s)
R(s) G 
In General
1  GH
109

102. + - G H C (s) G   R(s) 1  GH C(s)  G.E(s) For Negative Feedback
From Shown Figure,
E(s)  R(s)  B(s)
and
C(s)  G.E(s)
 G[R(s)  B(s)]
 GR(s)  GB(s)
R(s)
E(s) G
C(s)
+ -
But
B(s)  H.C(s)
C(s)  G.R(s)  G .H.C(s)
B(s) H
C(s)  G . H  GR(s)
C(s){1  G.H}  G.R(s)
C (s) G  
For Negative
Feedback
R(s)
1  GH
110

103. G H C(s) G   R(s) 1  GH C(s)  G.E(s) + For Positive Feedback
From Shown Figure,
E(s)  R(s)  B(s)
and
C(s)  G.E(s)
 G[R(s)  B(s)]
 GR(s)  GB(s)
B(s)  H.C(s)
R(s)
E(s) G
C(s) + +
But
B(s) H
C(s)  G.R(s)  G.H.C(s)
C(s)  G .H  GR(s)
C(s){1  G.H}  G.R(s)
C(s) G  
For Positive
Feedback
R(s)
1  GH
111

104. Block Diagram Reduction Techniques
Rule 4: Associative Law for Summing Points
The order of summing points
can be
changed if
two or more
summing
points
are
in series + +
R(s) X
C(s)
R(s) + +
C(s) X - - B1 B2 B2 B1
X=R(s)-B1
C(s)=X-B2
C(s)=R(s)-B1-B2
X=R(s)-B2
C(s)=X-B1
C(s)=R(s)-B2-B1
112

105. Block Diagram Reduction Techniques
Rule 5: Shift summing point before block G +
R(s)
C(s)
R(s) + G
C(s) + +
1/G X X
C(s)=R(s)G+X
C(s)=G{R(s)+X/G}
=GR(s)+X
113

106. Block Diagram Reduction Techniques
Rule 6: Shift summing point after block
R(s) + G
C(s) G +
R(s)
C(s) + + G X X
C(s)=GR(s)+XG
=GR(s)+XG
C(s)=G{R(s)+X}
=GR(s)+GX
114

107. Block Diagram Reduction Techniques
Rule 7: Shift a take off point before block
R(s) G
C(s)
R(s) G
C(s) G X X
C(s)=GR(s)
and
X=C(s)=GR(s)
C(s)=GR(s)
and
X=GR(s)
115

108. Block Diagram Reduction Techniques
Rule 8: Shift a take off point after block
C(s)
R(s) G
C(s)
R(s) G
1/G X X
C(s)=GR(s)
and
X=C(s).{1/G}
=GR(s).{1/G}
= R(s)
C(s)=GR(s)
and
X=R(s)
116

109. Block Diagram Reduction Techniques
 While
solving
block diagram
for
getting
single block
equivalent,
the
said rules
need to be
applied.
After
each
simplification a decision needs to be
taken.
For
each
decision we
suggest
preferences as
117

110. First Choice First Preference: Rule 1 (For series) Second Preference:
Block Diagram Reduction Techniques
First Choice
First Preference:
Rule 1 (For
series)
Second Preference:
Rule 2
(For
parallel)
Third Preference:
Rule 3
(For
FB loop)
118

111. Block Diagram Reduction Techniques
Second Choice
(Equal Preference)
Rule 4 Adjusting summing order
Rule 5/6
Shifting
summing point before/after
block
Rule7/8
Shifting
take
off point
before/after
block
119

112. Example 1 G4 + +
R(s) + G1 G2 G3 +
C(s) G6 - - + H1 G5 H2
120

113. immediate series blocks. can be applied to G4, G3, G5 in parallel to
Rule 1
cannot be
used as
there
are no
immediate series blocks.
Hence Rule 2
can be applied to G4, G3, G5 in
parallel to
get an
equivalent of
G3+G4+G5
121

114. Example 1 cont…. Blocks Apply Rule 2 in Parallel + + R(s) + + C(s) - -G4 + +
R(s) + G1 G2 G3 +
C(s) G6 - - + H1 G5 H2
122

115. Example 1 cont…. Apply Rule 1 Blocks in series R(s) + + C(s) - - G1 G2
G3+G4+G5 G6 - - H1 H2
123

116. Example 1 cont…. Apply Rule 3 Elimination of feedback loop R(s) + +G1
C(s)
G2(G3+G4+G5) G6 - - H1 H2
124

117. Example 1 cont…. G1 1  G1H1 Apply Rule 1 Blocks in series R(s) + C(s)
G2(G3+G4+G5) G6 - H2
125

118. Example 1 cont…. G1G2(G 3 G 4 G 5) 1  G1H1 Apply Rule 3
Elimination of
feedback
loop
G1G2(G 3 G 4 G 5)
1  G1H1
R(s) +
C(s) G6 - H2
126

119. Example 1 cont…. G1G2(G 3  G 4  G 5)
Apply
Rule 1
Blocks in
series
G1G2(G 3  G 4  G 5)
1  G1H1  G1G2H 2(G 3  G 4  G 5)
C(s) G6
R(s)
127

120. Example 1 cont…. G1G2G6(G 3  G 4  G 5)
1  G1H1  G1G2H 2(G 3  G 4  G 5)
R(s)
C(s)
128

121. Example 1 cont…. G1G2G6(G 3  G 4  G 5)
C (s) 
R(s)
1  G1H1  G1G2H 2(G 3  G 4  G 5)
129

122. Example 2 G4 +
R(s) + + +
C(s) G1 G2 G3 + - H1 H2
130

123. Example 2 cont…. Apply Rule 1 Blocks in series + R(s) + + + C(s) + -G4
Apply Rule 1
Blocks
in series +
R(s) + + +
C(s) G1 G2 G3 + - H1 H2
131
M Mysaiah

124. Example 2 cont…. Apply Rule 2 Blocks in parallel + R(s) + + + C(s) + -G4
Apply Rule 2
Blocks
in parallel +
R(s) + + +
C(s)
G1G2 G3 + - H1 H2
132

125. Example 2 cont…. Elimination of Apply Rule 3 feedback loop R(s) + +
C(s)
G1G2
G3+G4 + - H1 H2
Amit Nevase
133

126. Example 2 cont…. G1G2 1  G1G2H1 Apply Rule 2 Blocks in series R(s) +
C(s)
G3+G4 + H2
134

127. Example 2 cont…. G1G2(G 3  G 4) 1  G1G2H1 Apply Rule 3 Eliminationof
feedback
loop
G1G2(G 3  G 4)
1  G1G2H1
R(s) +
C(s) + H2
135

128. Example 2 cont…. G1G2(G 3  G 4) 1  G1G2H1  G1G2G3H 2  G1G2G4H 2
R(s)
C(s)
136

129. Example 2 cont…. C (s) G1G 2(G 3 G 4)  R(s) 1  G1G2H1  G1G2G3H 2
137

130. Example 3 G5 +
R(s) + + +
C(s) G1 G2 G3 G4 - - H1 H2
138

131. Example 3 cont…. Apply Rule 3 Elimination of feedback loop + R(s) + +G5 +
R(s) + + +
C(s) G1 G2 G3 G4 - - H1 H2
139

132. Example 3 cont…. Apply Rule 1 Blocks in series + R(s) + + C(s) - G 2
1  G 2H1
R(s) + +
C(s) G1 G3 G4 - H2
140

133. Example 3 cont…. G1G 2G3 1  G 2H1 Apply Rule 2 Blocks in parallel +
R(s) + +
C(s) G4 - H2
141

134. Example 3 cont…. G5  G1G 2G3 1  G2H1 Apply Rule 1 Blocks in series
R(s) +
C(s) G4
1 
G2H1 - H2
142

135. Example 3 cont…. G 4(G5  G1G2G3 ) 1  G2H1 R(s) + Apply Rule 3
Elimination of
feedback
loop
G 4(G5  G1G2G3 )
1  G2H1
R(s) +
C(s) - H2
143

136. Example 3 cont…. G4G5  G2G4G5H1  G1G2G3G4
1  G2H1  G4G5H 2  G2G4G5H1H 2  G1G2G3G4H 2
R(s)
C(s)
144

137. Example 3 cont…. C (s) G4G5 G2G4G5H1 G1G2G3G4  R(s )
1  G2H1  G4G5H 2  G2G4G5H1H 2
 G1G2G3G4H 2
145

138. Example 4 -
R(s) + + +
C(s) G1 G2 - - H1 H2
146

139. Example 4 cont…. Apply Rule 3 Elimination of feedback loop - R(s) + +
C(s) G1 G2 - - H1 H2
147

140. Example 4
cont…. G2
1  G2H 2 -
R(s) + +
C(s) G1 - H1
148

141. Now Rule 1, 2 or 3 cannot be used directly.
There are possible ways of going ahead. a.
Use Rule 4 & interchange order of summing
so that Rule 3 can be used on G.H1 loop. b.
Shift take off point after
G 2
block
reduce
1  G 2 H 2
by Rule 1, followed by Rule 3.
Which option we have to use????
149

142. Example 4 cont…. G 2 1  G2H 2 Apply Rule 4 Exchange summing order 1 2-
R(s) + +
C(s) G1 - H1
150

143. Example 4 cont…. 2 G 2 1  G2H 2 Apply Rule 3 Elimination feedback
loop
G 2
1  G2H 2 1 2 + -
R(s) +
C(s) G1 - H1
151

144. Example 4 cont…. 2 G2 1  G2H 2 G1 1  G1H1 Apply Rule 1 Bocks in
series G1
1  G1H1 G2
1  G2H 2 2 + -
R(s)
C(s)
152

145. Example 4 cont…. 2 + - R(s) C(s) Now which Rule will be applied
G1G2
1  G1H1  G2H 2  G1G2H1H 2 2 + -
R(s)
C(s)
Now which
Rule
will be applied
blocks in parallel feed back loop It is OR
153

146. Example 4 cont…. 2 Let us rearrange the block diagram to understand
Apply Rule 3
Elimination of
feed
back
loop
G1G2
1  G1H1  G2H 2  G1G2H1H 2 2 +
R(s)
C(s) -
154

147. Example 4 cont…. G1G2 1  G1H1  G2H 2  G1G2H1H 2  G1G2 R(s) C(s)
155

148. Example 4 cont…. R(s) C (s) G1G 2  1  G1H1  G2H 2  G1G2H1H 2 
156

149. Note 1: According to Rule 4
 By corollary, one can split a summing point to
two summing point
and
sum in
any
order B B +
R(s) + G
C(s) + + +
R(s) G
C(s) - - H H
157

150. Example 5 Simplify, by splitting second summing point as said in noteH1
said in
note 1 -
R(s) + +
C(s) G1 G2 G3 - - H2 H3
158

151. Example 5 cont…. Apply rule 3 Elimination of feedback loop + - + +H1 +
R(s) - + +
C(s) G1 G2 G3 - - H2 H3
159

152. Example 5 cont…. G1 1  G1H1 Apply rule 1 Blocks in series + + C(s) -
R(s) - - H2 H3
160

153. Example 5 cont…. G1G 2 1  G1H1 Apply rule 3 Elimination of feedback
loop
G1G 2
1  G1H1 + + G3
C(s)
R(s) - - H2 H3
161

154. Example 5 cont…. G1G 2 1  G1H1  G1G2H 2 Apply rule 1 Blocks in
series
G1G 2
1  G1H1  G1G2H 2 + G3
C(s)
R(s) - H3
162

155. Example 5 cont…. Apply rule 3 Elimination of feedback loop G1G 2G3
1  G1H1  G1G 2H 2 +
R(s)
C(s) - H3
163

156. Example 5
cont….
G1G2G3
1  G1H1  G1G2H 2  G1G2G3H 3
R(s)
C(s)
164

157. Example 5 cont…. C (s) G1G2G3  R(s) 1  G1H1  G1G2H 2  G1G2G3H 3
165

158. Example 6 G3 Apply rule 8 Shift take off point beyond block + R(s) + +
C(s) G1 G2 G3 G4 - - H1 H2
166

159. Example 6 cont…. Apply rule 1 Blocks in series + R(s) + + + C(s) - -
1/ G3 G5 +
R(s) + + +
C(s) G1 G2 G3 G4 - - H1 H2
167

160. Example 6 cont…. Apply rule 2 Blocks in parallel + R(s) + + + C(s) - -
G5/ G3 +
R(s) + + +
C(s) G1
G2G3 G4 - - H1 H2
168

161. Example 6 cont…. Apply rule 3 Feedback loop R(s) + + C(s) - -
G4+(G5/ G3)
R(s) + +
C(s) G1
G2G3 - - H1 H2
169

162. Example 6 cont…. Apply rule 1 Blocks in series R(s) + C(s) - G2G3
1  G2G3H1
G4+(G5/ G3)
R(s) +
C(s) G1 - H2
170

163. Example 6 cont…. R(s) + C(s) - (G1)( G2G3 )(G 4  G5) 1  G2G3H1 G3 H2
171

164. Example 6 cont…. (G1)( G2G3 )( G 4G3 G5)  (G1)( G2G3 )(G 4  G5) G3
1  G2G3H1
(G1)( G2G3 )( G 4G3 G5) 
1  G2G3H1 G3
G1G2(G 4 G 3 G 5) 
1  G2G3H1
172

165. Example 6 cont…. G1G2(G 4 G 3  G 5) 1  G2G3H1 Apply rule 3 Feedback
loop
G1G2(G 4 G 3  G 5)
1  G2G3H1
C(s)
R(s) + - H2
173

166. Example 6 cont…. G1G2(G 4 G 3  G 5) 1  G2G3H1  G1G2H 2(G 3 G 4 
C(s)
R(s)
1 
G2G3H1 
G1G2H 2(G 3 G 4 
G 5)
174

167. Example 6 cont…. C (S) G1G2(G 4 G 3 G 5)  R(S) 1  G2G3H1 
G1G2H 2(G 3 G 4 
G 5)
175

168. Example 7 Apply rule 8 Shift take off point after block G4 - R(s) + +
C(s) - H3 H1
176

169. Example 7 cont…. Apply rule 1 Blocks in series - R(s) + + + - - 1/G4H2
1/G4 -
R(s) + + + G1 G2 G3 G4 -
C(s) - H3 H1
177

170. Example 7 cont…. G4 Apply rule 3 Feedback loop - R(s) + + + - - H2/
G3G4 -
C(s) - H3 H1
178

171. Example 7 cont…. G4 Apply rule 1 Blocks in series - R(s) + + - H2/
G3G 4
1  G3G4H 3
R(s) + + G1 G2
C(s) - H1
179

172. Example 7 cont…. Apply rule 3 Feedback loop - R(s) + + - H2/ G4 C(s)
G 2G3G 4
1  G3G 4 H 3
R(s) + + G1
C(s) - H1
180

173. Example 7 cont…. G2G3G4 1  G3G4H 3  G2G3H 2 Apply rule 1 Blocks in
series
G2G3G4
1  G3G4H 3  G2G3H 2
C(s)
R(s) + G1 - H1
181

174. Example 7 cont…. G1G2G3G4 1  G3G4H 3  G2G3H 2 Apply rule 3 Feedback
loop
G1G2G3G4
1  G3G4H 3  G2G3H 2
R(s) +
C(s) - H1
182

175. Example 7 cont…. G1G2G3G4 1  G3G4H 3  G2G3H 2  G1G2G3G4H1 R(s) C(s)
183

176. Example 7 cont…. C (S) G1G 2G3G 4  R(S) 1  G3G4H 3  G2G3H 2 
G1G2G3G4H1
184

177. Example 8 3rd Simplify, by splitting summing point as given in Note 12 +
R(s) + + +
C(s) G1 G2 G4 - - - H2 H1
185

178. Example 8 cont…. Apply Rule 3 Elimination of Feedback loop + + R(s) +G3 + +
R(s) + + +
C(s) G1 G2 G4 - - - H2 H1
186

179. Example 8 cont…. Apply Rule 8 Shift take off point after block + R(s)G3 +
G 4
1  G 4H1
R(s) + + +
C(s) G1 G2 - - H2
187

180. Example 8 cont…. G2 Apply Rule 1 Blocks in series + R(s) + + + C(s) -
1  G 4H1
R(s) + + +
C(s) G1 G2 - - H2
188

181. Example 8 cont…. G2 Now which rule we have to use? + R(s) + + + C(s) -
1  G 4H1
R(s) + + +
C(s)
G1G2 - - H2
189

182. Example 8 cont…. G2 Apply Rule 2 Blocks in parallel + R(s) + + + C(s)
1  G 4H1
R(s) + + +
C(s)
G1G2 1 - - H2
190

183. Example 8 cont…. G3  1 G 2 Apply Rule 1 Blocks in series R(s) + +
1  G 4H1
R(s) + +
C(s)
G1G2 - - H2
191

184. Example 8 cont…. (G 3 G 2)G 4 G 2(1  G 4H1) Apply Rule 3
Elimination of
Feedback
Loop
(G 3 G 2)G 4
G 2(1  G 4H1)
R(s) + +
C(s)
G1G2 - - H2
192

185. Example 8 cont…. G1G2 (G 3 G 2)G 4 G 2(1  G 4H1) 1  G1G2H 2
Apply Rule 1
Blocks in
series
G1G2
1  G1G2H 2
(G 3 G 2)G 4
G 2(1  G 4H1)
R(s) +
C(s) -
193

186. Example 8 cont…. G 1G 4(G 3 G 2) (1  G1G 2 H 2)(1  G4H1)
Apply Rule 3
Elimination of
Feedback
loop
G 1G 4(G 3 G 2)
(1  G1G 2 H 2)(1  G4H1)
R(s) +
C(s) -
194

187. 1  G 4 H1  G1G2H 2  G1G2G4H1H 2  G1G4(G 2  G 3)
Example 8
cont….
G1G4(G 3  G 2)
1  G 4 H1  G1G2H 2  G1G2G4H1H 2  G1G4(G 2  G 3)
R(s)
C(s)
195

188. Example 8 cont…. C (s) G 1G4(G 3 G 2)  R(s)
1  G 4 H1  G1G2H 2  G1G2G4H1H 2  G1G4(G 2  G 3)
196

189. Example 9 Apply rule 2 Blocks in Parallel + R(s) + + + - - - C(s) G4H1 H2 H3
197

190. Example 9 cont…. Apply rule 3 Elimination of Feedback Loop R(s) + + +
G1+G4+ G5 G2 G3 - -
C(s) - H1 H2 H3
198

191. Example 9 cont…. G 2 G3 1  G 2H1 1  G3H 2 Apply rule 1 Blocks in
Series
G 2
1  G 2H1 G3
1  G3H 2
R(s) +
G1+G4+G5
C(s) - H3
199

192. Example 9 cont…. G 2G3(G 1 G 4 G 5) (1  G 2H1)(1  G 3 H 2)
Apply rule 3
Elimination of
Feedback
loop
G 2G3(G 1 G 4 G 5) (1 
G 2H1)(1  G 3 H 2)
R(s) +
C(s) - H3
200

193. 1  G2H1  G3H 2  G2G3H1H 2  G2G3H 3(G1  G 4  G 5)
Example 9
cont….
G2G3(G1  G 4  G 5)
1  G2H1  G3H 2  G2G3H1H 2  G2G3H 3(G1  G 4  G 5)
R(s)
C(s)
201

194. Example 9 cont…. C (s) G2G3(G1 G 4 G 5)  R(s)
1  G2H1  G3H 2  G2G3H1H 2  G2G3H 3(G1  G 4  G 5)
202

195. Example 10 Apply rule 2 Blocks in Parallel R(s) + + - - - + + C(s) G1H1 H3 +
203

196. Example 10 cont…. Apply rule 3 Elimination of Feedback Loop R(s) + - -G1 G2
1+G3 -
C(s) - + H1 H3 +
204

197. Example 10 cont…. Apply rule 8 Shift take off point after block R(s) +
G 2
1  G 2
R(s) + G1
1+G3 -
C(s) - + H1 H3 +
205

198. Example 10 cont…. Apply rule 1 Blocks in series R(s) + - - + + G 2
C(s) - 1
1  G3 + H1 H3 +
206

199. Example 10 cont…. Apply rule 2 Blocks in Parallel R(s) + - - + +
G 2(1 G 3)
1  G2
R(s) + G1 -
C(s) - 1
1  G3 + H1 H3 +
207

200. Example 10 cont…. H 2  1 Apply rule 1 Blocks in Series R(s) + - -
G 2(1 G 3)
1  G2
R(s) + G1 -
C(s) -
H 2  1 H1
1  G3
208

201. Example 10 cont…. Apply rule 3 Elimination of Feedback loop R(s) + - -
G 2(1 G 3)
1  G2
R(s) + G1 -
C(s) -
H1(H 2  H 2 G 3  1)
1  G3
209

202. Example 10 cont…. Apply rule 1 Blocks in series R(s) G 2(1 G 3)
1  G2  G2H1(1  H 2  H 2 G 3)
R(s) G1
C(s)
210

203. Example 10 cont…. R(s) G1G2(1  G 3) C(s)
1  G2  G2H1(1  H 2  H 2 G 3)
C(s)
R(s)
211

204. Example 10 cont…. C (s) G1G 2(1 G 3)  R(s) 1  G2  G2H1(1  H 2 
212