Cs unit-i-ii

KARMAVEER BHAURAO PATIL
POLYTECHNIC,
SATARA
Rayat Shikshan Sanstha’s
Department Of Electronics And Telecommunication Engineering
Control System and PLC

6/30/2016 Amit Nevase 2
Control System and PLC
Amit Nevase
Lecturer,
Department of Electronics & Telecommunication Engineering,
Karmaveer Bhaurao Patil Polytechnic, Satara
EJ5G Subject Code: 17536 Third Year Entc

Objectives
The student will be able to:
 Understand classifications of control system.
 Understand Steady state, time response, and frequency
response analysis.
 Analyze the Stability of control system using RH criteria.
 Understand the fundamentals and diff. Hardware parts of
PLC.
 Draw ladder diagrams to program PLC

Teaching & Examination Scheme
 Two tests each of 25 marks to be conducted as per the schedule given by
MSBTE.
 Total of tests marks for all theory subjects are to be converted out of 50 and
to be entered in mark sheet under the head Sessional Work (SW).
Teaching Scheme Examination Scheme
TH TU PR
PAPER
HRS
TH PR OR TW TOTAL
03 -- 02 03 100 50# --- 25@ 175

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 (4 Marks)
 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 (8 Marks)
 Order of a System : Definition, 0,1,2 order system Standard equation, Practical
Examples
 Block Diagram Reduction Technique: Need, Reduction Rules, Problems

Module II – Time Response Analysis
 Time Domain Analysis (4 Marks)
 Transient and Steady State Response
 Standard Test Inputs : Step, Ramp, Parabolic and Impulse, Need, Significance
and corresponding Laplace Representation
 Poles and Zeros : Definition, S-plane representation
 First and Second order Control System (8 Marks)
 First Order Control System : Analysis for step Input, Concept of Time Constant
 Second Order Control System : Analysis for step input, Concept, Definition and
effect of damping
 Time Response Specifications (8 Marks)
 Time Response Specifications ( no derivations )
 Tp, Ts, Tr, Td, Mp, ess – problems on time response specifications
 Steady State Analysis – Type 0, 1, 2 system, steady state error constants,
problems

Module III – Stability
 Introduction to Stability (4 Marks)
 Definition of Stability, Analysis of stable, unstable, critically stable
and conditionally stable
 Relative Stability
 Root locations in S-plane for stable and unstable system
 Routh’s Stability Criterion (8 Marks)
 Routh’s Stability Criterion : Different cases and conditions
 Statement Method
 Numericals Problems

Module IV – Control Actions
 Process Control System (4 Marks)
 Process Control System – Block diagram, explanation of each block
 Control Actions (8 Marks)
 Discontinuous Mode : On-Off Controller, Equation, Neutral Zone
 Continuous modes: Proportional Controller (offset, proportional
band), Integral Controllers, Derivative Controllers – output
equations, corresponding Laplace transforms, Response of P, I, D
controllers
 Composite Controllers : PI, PD, PID Controllers – output equations,
response, comparison

Module V – PLC Fundamentals
 Introduction (4 Marks)
 Evolution of PLC in automation, need and benefits of PLC in
automation
 Block Diagram of PLC (12 Marks)
 Block diagram and description of different parts of PLC -
 CPU Function, Scanning cycle, speed of execution, Power supply
function,
 Memory – function , organization of ROM and RAM
 Input modules – function, different input devices used with PLC
and their uses
 Output modules – function, different output devices used with
PLC and their uses
 Fixed and Modular PLCs

Module VI – PLC Hardware and Programming
 PLC Hardware (8 Marks)
 Discrete Input Modules – Block diagram, typical wiring details, Specifications of
AC input modules and DC input modules. Sinking and sourcing concept in DC
input modules
 Discrete Output Modules – Block diagram, typical wiring details, Specifications
of AC output modules and DC output modules.
 Analog Input and output modules : Block diagram, typical wiring details and
specifications
 PLC Programming (16 Marks)
 I/O Addressing in PLC
 PLC Instruction Set : Relay instructions, timer instructions, counter instructions,
data handling instructions, logical and comparison instructions
 PLC programming examples based on above instruction using Ladder
programming

Module-I
Introduction to Control
System

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

Examples

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

Output
 The actual response obtained from a system is
called output.
OutputInput

“System”
 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.
SYSTEMInput Output

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

Combining above definitions
System + Control Control System=

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

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

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

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

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
230V/50Hz
AC Supply
Airflow
(Proper Output)
Input Output

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

Examples

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

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

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

OLCS Examples
 Automatic washing machine
– This machine runs
according to the pre-set time
irrespective of washing is
completed or not.

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

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

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

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

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

Closed Loop System
Definition:
“A system in which the control action is somehow
dependent on the output is called as closed loop system”

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

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

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

Perspiration
CLCS Examples

Advantages of CLCS
 Closed loop control systems are more accurate even in the
presence of non-linearity.
 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.

Disadvantages of CLCS
 They are costlier.
 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.

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

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

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

Examples

Linear Control
System
Non-linear Control
System

When an input X1 produces an output Y1 & an
input X2 produces an output Y2, then any
combination should produce an
output . In such case system is linear.
Therefore, linear systems are those where the
principles of superposition and proportionality
are obeyed.
Linear Control System
1 2X X 
1 2Y Y 

 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.
Non-linear Control System

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

Time Varying
Control System
Time Invarying 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.
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.
Time varying/In-varying Control System

Examples

 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 is the mechanical load position &
its time derivatives like velocity and acceleration.
Servo System

General block diagram of Servo System

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

DC Servo System

DC Servo System
Working of Servo System

Examples

The French Newton
Pierre-Simon Laplace
 Developed mathematics in
astronomy, physics, and statistics
 Began work in calculus which led
to the Laplace Transform
 Focused later on celestial
mechanics
 One of the first scientists to
suggest the existence of black
holes
Laplace Transform

 To evaluate the performance of an automatic control
system commonly used mathematical tool is “Laplace
Transform”
 Laplace transform converts the differential equation
into an algebraic equation in ‘s’.
 Laplace transform exist for almost all signals of
practical interest.
Laplace Transform

Why Laplace Transform?
Time domain
unknown f(t), d/dt, Diff Eqs
Frequency domain
unknown F(s), Alg Eqs
Laplace
Transformation
Solve
Algebraic
Equations
Frequency domain
known F(s)
Time domain
known f(t)
Solve
Differential
Equations
Inverse
Laplace
Transform

 Solution of intego differential equation of time
systems can be easily obtained.
 Initial conditions are automatically incorporated.
 Laplace transform provides an easy & effective
solution of many problems arising in automatic control
systems.
 Laplace transform allows the use of graphical
techniques, for predicting the system performance.
Advantages of Laplace Transform

The Laplace transform of a function, f(t), is defined as
Laplace Transform- Definition
where F(s) is the symbol for the Laplace transform, L is the Laplace
transform operator, and f(t) is some function of time, t.
Note: The L operator transforms a time domain function f(t)
into an s domain function, F(s). s is a complex variable:
s = a + bj,
   0
( ) ( ) (1-1)st
F s f t f t e dt
 
  L
1j B

Standard Laplace Transform
( )f t ( ) [ ( )]F s L f t
1 or ( )u t 1
s
t
e 
1
s 
sin t
2 2
s


cos t
2 2
s
s 
sint
e t

2 2
( )s

  
cost
e t

2 2
( )
s
s

 

 
t
2
1
s
n
t
1
!
n
n
s 
t n
e t
1
!
( )n
n
s  

( )t 1
*Use when roots are complex.

Inverse Laplace Transform
By definition, the inverse Laplace transform operator, L-1,
converts an s-domain function back to the corresponding
time domain function:
   1
f t F s
   L

Examples

 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”.
Transfer Function

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

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

 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
Laplace Transform of Passive Element (R,L & C)

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)

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

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

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

Laplace Transform of L
Therefore, the transform of an inductor with inductance
L is given by:
{inductor} sLL 

Transfer Function of RC and RLC electrical circuits
Example: Find the TF of given RC network
C
Vo(t)Vi(t) i(t)
Apply KVL for input loop,
0
1
(t) Ri(t) (t)
t
vi i dt
C
  
Taking Laplace transform above equation
1
Vi(s) RI(s) (s) (1)I
sC
       
Apply KVL for output loop,
0
1
(t) (t)
t
vo i dt
C
 
Taking Laplace transform above equation
1
Vo(s) (s) (2)I
sC
      
(s) .Vo(s) (3)I sC       
From equation 1,
From equation 3 and 4,
1
Vi(s) (s)(R ) (4)I
sC
       
1
Vi(s) Vo(s). .(R )sC
sC
 

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

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

From equation 1 and 2, 1
( )
(s)
1(s)
[R sL ]I(s)
I s
Vo sC
Vi
sC

 
Transfer Function=
1
1
[R sL ]
sC
sC

 
2
1
1
sC
sCR s LC
sC

 
2
1
1sCR s LC

 
2
1
1s LC sCR

 6/30/2016 Amit Nevase 80

Examples

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

Example1: Determine order of given system
4 3 2
(s 2)
(s)
7 10 5 5
s
TF G
s s s s

 
   

Example1: Determine order of given system
4 3 2
(s 2)
(s)
7 10 5 5
s
TF G
s s s s

 
   
Answer: The highest power of equation in denominator
of given transfer function is ‘4’.
Hence the order of given system is fourth

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.

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

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”
(s 5)(s 2)
(s)
(s 3)(s 4)
G
s
 

 
Solution: To obtain highest power of denominator,
Simplify denominator polynomial.
(s 3)(s 4) 0s   
2
(s 7s 12) 0s   
3 2
s 7s 12 0s  

3 2
K(s 5)
(s)
(7s 12 5)
G
s s


 

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”
Solution: To obtain highest power of denominator,
Simplify denominator polynomial.
3 2
K(s 5)
(s)
(7s 12 5)
G
s s


 
3 2
(7s 12 5) 0s s  
5 4 3
7s 12 5 0s s  

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

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

Consider a unity feedback system with transfer function
Hence characteristics equation is given by,
or
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
1
(s)
1
G
T


1 0T 
0
1 0s T 
Zero (0) Order System

First Order System
In Characteristics equation is one, then it is called as
“First order System”
+
-
R(s) C(s)1
1 sCR

Hence the system given above is First order system.
Practical Example: RC circuits, thermal type systems
1
(s)
1
G
sCR


1 0sCR 
First Order System

Second Order System
In Characteristics equation is two, then it is called as
“Second order System”
+
-
R(s) C(s)
2
1
1s LC sCR 

Hence the system given above is Second order system.
Practical Example: RLC circuits, Robotic control system.
2
1
(s)
1
G
s LC sCR

 
2
1 0s LC sCR  
Second Order System

Examples

 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.
Need of Block Diagram Algebra

 To overcome this problem block diagram
representation method is used.
 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.
Need of Block Diagram Algebra

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

 Output: The value of the input is multiplied to the
value of block gain to get the output.
3sX(s) Y(s)
Output Y(s)= 3s. X(s)

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

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

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

Rule 1: For blocks in cascade
Gain of blocks connected in cascade gets
multiplied with each other.
Block Diagram Reduction Techniques
G1 G2R(s) R1(s) C(s)
R1(s)=G1R(s)
C(s) =G2R1(s)
=G1G2R(s)
C(s)= G1G2R(s)
G1G2R(s) C(s)

G1 G2 G3R(s) C(s)
Find Equivalent
G1G2G3R(s) C(s)

Find Equivalent
G1 G2 G3R(s) C(s)
R1(s)
G1G2G3
R(s) C(s) G1G2R(s) C(s)G3
R1(s)

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

Rule 3: Eliminate Feedback Loop
R(s)
C(s)G
H
+-
+
R(s) C(s)
1
G
GH
B(s)
E(s)
C(s)
(s) 1
G
R GH


In General

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

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

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) C(s)X
B1 B2
+ +
-
R(s) C(s)
B1B2
+ +X
-
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

Rule 5: Shift summing point before block
R(s) C(s)
X
+
G
C(s)=R(s)G+X
C(s)=G{R(s)+X/G}
=GR(s)+X
+
R(s) C(s)+
G
1/G
X
+

Rule 6: Shift summing point after block
R(s) C(s)+
G
X
C(s)=G{R(s)+X}
=GR(s)+GX
C(s)=GR(s)+XG
=GR(s)+XG
+
R(s) C(s)
X
+
G
G
+

Rule 7: Shift a take off point before block
R(s) C(s)
G
C(s)=GR(s)
and
X=C(s)=GR(s)
C(s)=GR(s)
and
X=GR(s)
X
R(s) C(s)
X
G
G

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

 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

First Choice
First Preference: Rule 1 (For series)
Second Preference: Rule 2 (For parallel)
Third Preference: Rule 3 (For FB loop)

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

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

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

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

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

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

+
-
H2
G6
R(s) C(s)
G2(G3+G4+G5)
1
1 1 1
G
G H
Example 1 cont….

+
-
H2
G6R(s) C(s)
1 2(G3 G4 G5)
1 1 1
G G
G H
 

Example 1 cont….

G6R(s)
C(s)
1 2(G3 G4 G5)
1 1 1 1 2 2(G3 G4 G5)
G G
G H G G H
 
   
Example 1 cont….

R(s) C(s)1 2 6(G3 G4 G5)
1 1 1 1 2 2(G3 G4 G5)
G G G
G H G G H
 
   
Example 1 cont….

1 2 6(G3 G4 G5)
1 1 1 1 2 2(G3 G4 G5)
G G G
G H G G H
 
   
( )
( )
C s
R s

Example 1 cont….

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

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

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

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

G3+G4
H2
R(s) C(s)+
+
1 2
1 1 2 1
G G
G G H
Example 2 cont….

H2
R(s) C(s)+
+
1 2(G3 G 4)
1 1 2 1
G G
G G H


Example 2 cont….

R(s) C(s)
1 2(G3 G4)
1 1 2 1 1 2 3 2 1 2 4 2
G G
G G H G G G H G G G H

  
Example 2 cont….

( ) 1 2(G3 G4)
( ) 1 1 2 1 1 2 3 2 1 2 4 2
C s G G
R s G G H G G G H G G G H


  
Example 2 cont….

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

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

G3
H2
R(s) C(s)+ +
+
-
G1 G4
G5
2
1 2 1
G
G H
Example 3 cont….

H2
R(s) C(s)+ +
+
-
G4
G5
Apply Rule 2 Blocks in parallel
1 2 3
1 2 1
G G G
G H
Example 3 cont….

H2
R(s) C(s)+
-
G4
1 2 3
5
1 2 1
G G G
G
G H


Example 3 cont….

H2
R(s) C(s)+
-
1 2 3
G 4( 5 )
1 2 1
G G G
G
G H


Example 3 cont….

R(s) C(s)4 5 2 4 5 1 1 2 3 4
1 2 1 4 5 2 2 4 5 1 2 1 2 3 4 2
G G G G G H G G G G
G H G G H G G G H H G G G G H
 
   
Example 3 cont….

( ) 4 5 2 4 5 1 1 2 3 4
( ) 1 2 1 4 5 2 2 4 5 1 2 1 2 3 4 2
C s G G G G G H G G G G
R s G H G G H G G G H H G G G G H
 

   
Example 3 cont….

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

G1
H2
G2
H1
R(s) C(s)+ +
--
+
Example 4 cont….
-

G1
H1
R(s) C(s)+ + -
-
2
1 2 2
G
G H
Example 4 cont….

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 block reduce
by Rule 1, followed by Rule 3.
Which option we have to use????
2
1 2 2
G
G H

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

G1
H1
R(s) C(s)+ +
-
-
Apply Rule 3 Elimination feedback loop
2
1 2 2
G
G H
12
Example 4 cont….

R(s) C(s)+
-
Apply Rule 1 Bocks in series
2
1 2 2
G
G H
2
1
1 1 1
G
G H
Example 4 cont….

R(s) C(s)+
-
Now which Rule will be applied
-------It is blocks in parallel OR
-------It is feed back loop
2
1 2
1 1 1 2 2 1 2 1 2
G G
G H G H G G H H  
Example 4 cont….

R(s) C(s)+
-
Let us rearrange the block diagram to understand
2
1 2
1 1 1 2 2 1 2 1 2
G G
G H G H G G H H  
Apply Rule 3 Elimination of feed back loop
Example 4 cont….

R(s) C(s)1 2
1 1 1 2 2 1 2 1 2 1 2
G G
G H G H G G H H G G   
Example 4 cont….

( ) 1 2
( ) 1 1 1 2 2 1 2 1 2 1 2
C s G G
R s G H G H G G H H G G

   
Example 4 cont….

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

G1
H1
R(s) C(s)+ +
-
G2 G3
H2
H3
-
Simplify, by splitting second
summing point as
said in note 1
Example 5
-

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

C(s)+
-
G2 G3
H2
H3
+
-R(s)
Apply rule 1 Blocks in series
1
1 1 1
G
G H
Example 5 cont….

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

C(s)
G3
H3
+
-R(s)
1 2
1 1 1 1 2 2
G G
G H G G H 
Example 5 cont….

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

C(s)R(s) 1 2 3
1 1 1 1 2 2 1 2 3 3
G G G
G H G G H G G G H  
Example 5 cont….

( ) 1 2 3
( ) 1 1 1 1 2 2 1 2 3 3
C s G G G
R s G H G G H G G G H

  
Example 5 cont….

H1
G2 G3
H2
R(s) C(s)+ +
- -
G1 G4
G5
+
+
Apply rule 8
Shift take off point beyond block
G3
Example 6

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

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

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

H2
R(s) C(s)+
-
G1
G4+(G5/
G3)
Apply rule 1
Blocks in series
2 3
1 2 3 1
G G
G G H
Example 6 cont….

H2
R(s) C(s)+
-
2 3 5
(G1)( )(G4 )
1 2 3 1 3
G G G
G G H G


Example 6 cont….

2 3 5
(G1)( )(G4 )
1 2 3 1 3
G G G
G G H G
 

1 2(G 4G3 G5)
1 2 3 1
G G
G G H



2 3 G4 3 5
(G1)( )( )
1 2 3 1 3
G G G G
G G H G



Example 6 cont….

H2
R(s) C(s)+
-
1 2(G 4G3 G5)
1 2 3 1
G G
G G H


Apply rule 3 Feedback loop
Example 6 cont….

R(s)
C(s)
1 2(G 4G3 G5)
1 2 3 1 1 2 2(G3G 4 G5)
G G
G G H G G H

  
Example 6 cont….

(S) 1 2(G 4G3 G5)
(S) 1 2 3 1 1 2 2(G3G 4 G5)
C G G
R G G H G G H


  
Example 6 cont….

H3
G2
H1
R(s)
C(s)
+ +
-
-
G1 G3
H2
-
+
Apply rule 8 Shift take off point after block G4
G4
Example 7

H3
G2
H1
R(s)
C(s)
+ +
-
-
G1 G3
H2
-
+
G4
1/G4
Example 7 cont….

H3
G2
H1
R(s)
C(s)
+ +
-
-
G1 G3G4
-
+
H2/
G4
Example 7 cont….

G2
H1
R(s)
C(s)
+ +
-
-
G1
H2/
G4
3 4
1 3 4 3
G G
G G H
Example 7 cont….

H1
R(s)
C(s)
+ +
-
-
G1
H2/
G4
2 3 4
1 3 4 3
G G G
G G H
Example 7 cont….

H1
R(s) C(s)+
-
G1
2 3 4
1 3 4 3 2 3 2
G G G
G G H G G H 
Example 7 cont….

H1
R(s) C(s)+
-
1 2 3 4
1 3 4 3 2 3 2
G G G G
G G H G G H 
Example 7 cont….

R(s) C(s)1 2 3 4
1 3 4 3 2 3 2 1 2 3 4 1
G G G G
G G H G G H G G G G H  
Example 7 cont….

(S) 1 2 3 4
(S) 1 3 4 3 2 3 2 1 2 3 4 1
C G G G G
R G G H G G H G G G G H

  
Example 7 cont….

G1
H2
G2
G3
R(s) C(s)+ +
- -
G4+
+
H1
-
Simplify, by splitting 3rd summing point as given in Note 1
1 2
3
Example 8

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

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

G1
H2
G2
G3/
G2
R(s) C(s)+ +
- -
+
+
Apply Rule 1
Blocks in series
4
1 4 1
G
G H
Example 8 cont….

G1G2
H2
G3/
G2
R(s) C(s)+ +
- -
+
+
Now which rule we have to use?
4
1 4 1
G
G H
Example 8 cont….

G1G2
H2
G3/
G2
R(s) C(s)+ +
- -
+
+
Apply Rule 2
Blocks in parallel
4
1 4 1
G
G H
1
Example 8 cont….

G1G2
H2
R(s) C(s)+ +
- -
4
1 4 1
G
G H
3
1
2
G
G

Example 8 cont….

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

R(s) C(s)+
-
(G 3 G 2) 4
G 2(1 4 1)
G
G H


1 2
1 1 2 2
G G
G G H
Example 8 cont….

R(s) C(s)+
-
Apply Rule 3 Elimination of Feedback loop
G1 4(G3 G 2)
(1 G1G 2H 2)(1 4 1)
G
G H

 
Example 8 cont….

R(s) C(s)G1 4(G3 G2)
1 G4H1 1 2 2 1 2 4 1 2 1 4(G2 G3)
G
G G H G G G H H G G

    
Example 8 cont….

(s) G1 4(G3 G2)
(s) 1 G4H1 1 2 2 1 2 4 1 2 1 4(G2 G3)
C G
R G G H G G G H H G G


    
Example 8 cont….

H2
G2
H3
R(s) C(s)+ +
- -
G1 G3
-
+
Apply rule 2 Blocks in Parallel
G4
G5 H1
+
+
Example 9

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

H3
R(s) C(s)+
-
G1+G4+G5
Apply rule 1 Blocks in Series
2
1 2 1
G
G H
3
1 3 2
G
G H
Example 9 cont….

H3
R(s) C(s)+
-
Apply rule 3 Elimination of Feedback loop
2 3(G1 G 4 G 5)
(1 2 1)(1 G 3H 2)
G G
G H
 
 
Example 9 cont….

R(s) C(s)2 3(G1 G4 G5)
1 2 1 3 2 2 3 1 2 2 3 3(G1 G4 G5)
G G
G H G H G G H H G G H
 
     
Example 9 cont….

(s) 2 3(G1 G4 G5)
(s) 1 2 1 3 2 2 3 1 2 2 3 3(G1 G4 G5)
C G G
R G H G H G G H H G G H
 

     
Example 9 cont….

H1
R(s)
C(s)
G1 +
-
+
+
G2 G3
-
+
H3
-
Example 10

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

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

H1
R(s)
C(s)
G1 +
-
+
+
1+G3
H3
-
2
1 2
G
G
1
1 3G
Example 10 cont….

H1
R(s)
C(s)
G1 +
-
+
+
H3
-
2(1 G3)
1 2
G
G


1
1 3G
Example 10 cont….

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

R(s)
C(s)
Apply rule 3 Elimination of Feedback loop
G1 +
-
-
2(1 G3)
1 2
G
G


1(H2 H2G3 1)
1 3
H
G
 

Example 10 cont….

R(s)
C(s)
G1
2(1 G3)
1 2 2 1(1 H 2 H 2G3)
G
G G H

   
Example 10 cont….

R(s) C(s)
1 2(1 G3)
1 2 2 1(1 H 2 H 2G3)
G G
G G H

   
Example 10 cont….

(s) 1 2(1 G3)
(s) 1 2 2 1(1 H 2 H 2G3)
C G G
R G G H


   
Example 10 cont….

References
Control System Engineering
- J. J. Nagrath, M. Gopal
Feedback Control System
- R. A. Barapate
Modern Control
Engineering
- K. Ogata

Online Tutorials
 http://www.electrical4u.com
/control-engineering-
historical-review-and-types-
of-control-engineering/
 http://www.academia.edu/6
729369/EC2255-
_Control_System_Notes_solv
ed_problems_

Thank You
Amit Nevase

Cs unit-i-ii

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