UNIT I CONTROL SYSTEM (EC3351).pptx

COURSE OBJECTIVE
 To introduce the components and their representation
of control systems
 To learn various methods of analyzing
 1.Time response
 2.Frequency response
 3.Stability of the systems
 4.State variable analysis
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DrNSR/AP/ECE/PSNACET/EC3351

UNIT I-SYSTEM COMPONENTS AND THEIR
REPRESENTION
 Syllabus
 Control system; terminology and basic structure
 Feed forward and feedback control theory
 Electrical and mechanical transfer function models
 Block diagram models
 Signal flow graph models
 DC and AC servo systems
 Synchros
 Multivariable control system
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CONTROL SYSTEM; TERMINOLOGY AND
BASIC STRUCTURE
 Control system theory evolved in the following areas
 1.Engineering
 2.Economy
 3.Socialogy
 4.Biology
 5.Medicine
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BASIC STRUCTURE
 Automatic control has become an integral part of
modern manufacturing and industrial processes,
 1.Numerical control of machine tools in
manufacturing industries
 2.Controlling pressure
 3.Temperatre
 4.Humidity
 5.Viscosity
 6.Flow in process industry
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BASIC STRUCTURE
System
Group of components are connected in a sequence
to perform a specific function,
the group thus formed is called a system.
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BASIC STRUCTURE
 Control system
 In a system ,when the output quantity is
controlled by varying the input quantity, the
system is called control system
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OPEN LOOP SYSTEM
 Any physical system which does not automatically
correct the variation in its output is called an open
loop system
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CLOSED LOOP SYSTEM
 Control system in which the output has an effect
upon the input quantity in order to maintain the
desired output value are called closed loop system
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CLOSED LOOP SYSTEM- CAR DRIVING
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DIFFERENCE BETWEEN OPEN LOOP AND
CLOSED LOOP CONTROL SYSTEM
Open Loop Control System Closed-Loop Control System
 The controlled action is
free from the output
 Non feedback control
system
 The construction of this
system is simple
 The consistency is non-
reliable
 The output mainly
depends on the controlled
act of the system.
 Feedback control system
 The construction of this
system is complex
 The consistency is reliable
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FEED FORWARD AND FEEDBACK CONTROL
THEORY
FEEDBACK CONTROL WHEN WINTER GROWING UP
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FEED FORWARD AND FEEDBACK CONTROL
THEORY
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SAMPLE CONTROL SYSTEM
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FEEDFORWARD CONTROL
Control element responds to change in command
or
measured disturbance in a pre-defined way
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IMPLIMENTATION OF FEEDFORWARD
 Ideally consists of exact inverse model of the plant
 Can compensate for known plant dynamics, delays (before
you get errors)
 No sensors needed
 System response must be predictable
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LIMITATIONS OF FEEDFORWARD
 Effects of disturbance or command input must be
predictable
 May not generalize to other conditions
 Will not be accurate if the system changes
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FEEDBACK CONTROL
Plant- System to be controlled
Reference -Desired value of output (also ‘set point’)
Controller -Computes compensatory command to
the plant based on error
Sensor- (implied)
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FEATURES OF FEEDBACK
 Reactive / Error-driven
 Automatically compensates for disturbances (controller
acts on error)
 Automatically follows change in desired state (set point can
change)
 Can improve undesirable properties of system/plant
 Can be very simple
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COMBINING FEEDBACK AND FEEDFORWARD
 Feed forward component provides rapid response
 Feedback component fills in the rest of the
response accurately, compensating for errors in
the model
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ELECTRICAL AND MECHANICAL TRANSFER
FUNCTION MODELS
 The control systems can be represented with a set of
mathematical equations known as mathematical
model.
 These models are useful for analysis and design of
control systems.
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FUNCTION MODELS
 Analysis of control system means finding the output
when we know the input and mathematical model.
 Design of control system means finding the
mathematical model when we know the input and the
output

FUNCTION MODELS
 The following mathematical models are mostly used.
 1. Differential equation model
 2. Transfer function model
 3. State space model

DIFFERENTIAL EQUATION MODEL
Differential equation model is a time domain
mathematical model of control systems.

RLC series electrical system
Mesh equation for this circuit is

 Substitute, the current passing through capacitor
The above equation is a second order differential equation.

TRANSFER FUNCTION MODEL
 Transfer function model is an s-domain mathematical
model of control systems.

TRANSFER FUNCTION MODEL
 The Transfer function of a Linear Time Invariant
(LTI) system is defined as the ratio of Laplace
transform of output and Laplace transform of
input by assuming all the initial conditions are
zero.
transfer function model

TRANSFER FUNCTION OF RLC SERIES ELECTRICAL
SYSTEM
RLC series circuit
Transfer function model of
RLC series circuit
second order electrical system with a
block having the transfer function inside
it. And this block has an input Vi(s) & an
output Vo(s).

THE DIFFERENTIAL EQUATION MODELING OF
MECHANICAL SYSTEMS.
 There are two types of mechanical systems based
on the type of motion.
 1. Translational mechanical systems
 2.Rotational mechanical systems

MODELING OF TRANSLATIONAL MECHANICAL
SYSTEMS
 Translational mechanical systems move along a
straight line
 These systems mainly consist of three basic
elements.
 Mass,
 Spring
 Dashpot or damper.

MASS
BLOCK DIAGRAM OF MASS

MASS
 Mass is the property of a body,
 which stores kinetic energy.
 If a force is applied on a body having mass M, then
it is opposed by an opposing force due to mass.
 This opposing force is proportional to the
acceleration of the body.
 Assume elasticity and frictions are negligible.

 Where,
 F is the applied force
 Fm is the opposing force due to mass
 M is mass
 a is acceleration
 x is displacement

SPRING
BLOCK DIAGRAM OF SPRING

 Spring is an element,
 which stores potential energy.
 If a force is applied on spring K, then it is opposed
by an opposing force due to elasticity of spring.
 This opposing force is proportional to the
displacement of the spring.
 Assume mass and friction are negligible.

 Where,
 F is the applied force
 Fk is the opposing
force due to elasticity
of spring
 K is spring constant

DASHPOT
BLOCK DIAGRAM OF FRICTION

 If a force is applied on dashpot B,
 then it is opposed by an opposing force due to
friction of the dashpot.
 This opposing force is proportional to the velocity
of the body.
 Assume mass and elasticity are negligible.

 Where,
 Fb is the opposing
force due to friction of
dashpot
 B is the frictional
coefficient
 v is velocity

MODELING OF ROTATIONAL MECHANICAL
SYSTEMS
 Rotational mechanical systems move about a
fixed axis.
 These systems mainly consist of three basic
elements. Those are
 Moment of inertia,
 Torsional spring
 Dashpot.

Moment of Inertia
BLOCK DIAGRAM OF MOMENT OF INERTIA

 If a torque is applied to a rotational mechanical
system,
 then it is opposed by opposing torques due to
moment of inertia, elasticity and friction of the
system.
 Since the applied torque and the opposing
torques are in opposite directions,
 the algebraic sum of torques acting on the system
is zero.

Where,
 T is the applied torque
 Tj is the opposing torque
due to moment of
inertia
 J is moment of inertia
 α is angular acceleration
 θ is angular
displacement

Torsional spring

 Rotational mechanical system, torsional spring
stores potential energy.
 If a torque is applied on torsional spring K, then it
is opposed by an opposing torque due to the
elasticity of torsional spring.
 This opposing torque is proportional to the
angular displacement of the torsional spring.
 Assume that the moment of inertia and friction are
negligible.

 Where,
 T is the applied torque
 Tk is the opposing
torque due to elasticity
of torsional spring
 K is the torsional
spring constant
 θ is angular
displacement

Dashpot

 If a torque is applied on dashpot B, then it is
opposed by an opposing torque due to the
rotational friction of the dashpot.
 This opposing torque is proportional to the
angular velocity of the body.
 Assume the moment of inertia and elasticity are
negligible.

PROBLEMS
1.Write the differential equations governing the
mechanical system shown in fig, and determine the
transfer function (X(s)/F(s))

2.Write the differential equations governing the
mechanical system shown in the fig, and determine
the transfer function

Substituting X1(S) from node 1

DC AND AC SERVO SYSTEMS
 A DC servo system is constructed by using DC
motor which has armature and field winding
coupled with gearbox, controller and
Potentiometer,
 AC servo system is constructed using an Induction
motor consisting of rotor and stator with gear
system and Encoders.

What is servomotor
 Servo is an electromagnetic device uses a negative
feedback mechanism to converts an electric signal
into controlled motion.

DC AND AC SERVOSYSTEMS
 TYPES
 1.DC SERVO MOTORS
 2.AC SERVO MOTORS

DC SERVO MOTORS

 Servo motor is basically a closed loop feedback
system.
 A closed loop feedback system controls the
output of the system by varying input

CONSTRUCTION OF DC
SERVOMOTOR
 Stator Winding: This type of winding wound on the
stationary part of the motor. It is also known as field
winding of the motor.
 Rotor Winding: This type of winding wound on the
rotating part of the motor. It is also known as an
armature winding of the motor.

 Bearing: These are of two types,i.e, font bearing and
back bearing which are used for the movement of the
shaft.
 Shaft: The armature winding is coupled on the iron
rod is known as the shaft of the motor.
 Encoder: It has the approximate sensor which
determines the rotational speed of motor and
revolution per minute of the motor.

AC SERVOMOTOR

TORQUE –SPEED CHARACTERISTICS
WITH ROTOR RESISTNCE

TORQUE-SPEED CHARACTERISTICS

APPLICATIONS
 Robotics.
 Installed in cameras to adjust the focus for
obtaining better quality images
 Used to track the direction of the sun in solar
panels
 Textile industries in knitting the fabrics
 Aircraft for balancing.

SYNCHROS
 The Synchro is a type of transducer
 which transforms the angular position of the
shaft into an electric signal.
 It is used as an error detector and as a rotary
position sensor.

SYNCHROS
 The error occurs in the system because of the
misalignment of the shaft.
 The transmitter and the control transformer are
the two main parts of the synchro.

SYNCHROS SYSTEM TYPES
 1.Control Type Synchro.
 2.Torque Transmission Type Synchro.

CONTROL TYPE SYNCHRO
 The control type Synchro is used for driving the
large loads.

TORQUE TRANSMISSION TYPE SYNCHROS
 This type of synchros has small output torque,
 Used for running the very light load like a
pointer.

CONTROL TYPE SYNCHROS SYSTEM
 The controls synchros is used for error detection
in positional control systems. Their systems
consist two units. They are
 1.Synchro Transmitter
 2.Synchro receiver

Synchros Transmitter
 Their construction is similar to the three phase
alternator.
 The stator of the synchros is made of steel for
reducing the iron losses.
 The stator is slotted for housing the three phase
windings.
 The axis of the stator winding is kept 120º apart
from each other.

 The AC voltage is applied to the rotor of the
transmitter and it is expressed as

 The coils of the stator windings are connected in
star.
 The rotor of the synchros is a dumb bell in shape,
and a concentric coil is wound on it.
 The AC voltage is applied to the rotor with the
help of slip rings.

MULTIVARIABLE CONTROL SYSTEM
 Multivariable control system in which the
variable interacts strongly.
 This kind of system must have more than one
input and more than one output.
 A disturbance in any input causes a change of
response from at least one output.

 This kind of system have as many inputs and
outputs as needed to control the process.
 A system with an equal number of inputs and
outputs is said to be square.

APPLICATIONS
 A heated liquid tank where both the level and the
temperature shall be controlled.
 A distillation column where the top and bottom
concentration shall be controlled.
 A robot manipulator where the positions of the
manipulators (arms)shall be controlled.
 A chemical reactor where the concentration and
the temperature shall be controlled.

SIGNAL FLOW GRAPH MODELS
 Signal flow graph is used to represent the control
system graphically
 Signal flow graph is a diagram which represents a
set of simultaneous equations

SIGNAL FLOW GRAPH MODELS
 Signal flow graph of control system is further
simplification of block diagram of control system
 Blocks of transfer function, summing symbols
and take off points are eliminated by branches
and nodes.

Terms used in Signal Flow Graph
 Node
 Input node or source
 Output node or sink
 Mixed node
 Branch
 Transmittance
 Forward path
 Feedback loop

 Self-loop
 Path gain
 Loop gain
 Non-touching loops:

BLOCK DIAGRAM AND SIGNAL FLOW
GRAPH

Properties of signal flow graph
 The signal from a node to other flows through the
branch in the direction of arrowhead.
 The graphical method is valid only for linear time-
invariant systems.
 The signal flowing through a branch is multiplied by
the gain or transmittance of that branch. This product
is equivalent to the node where that branch is
terminating.

SFG from system equations

SFG from Block Diagram

Mason’s Gain Formula
 Suppose there are ‘N’ forward paths in a signal
flow graph. The gain between the input and the
output nodes of a signal flow graph is nothing but
the transfer function of the system.
 It can be calculated by using Mason’s gain
formula.

Block diagram
Transfer Function: Ratio between transformation of output to the transformation of
input when all the initial conditions are zero.
A Block diagram is basically modelling of any simple or complex system.
It Consists of multiple Blocks connected together to represent a system to explain
how it is functioning
)
(s
R
2
G 3
G
1
G
4
G
1
H
2
H
)
(s
Y
)
(s
G
)
(s
C
)
(s
R G(s)=C(s)/R(s)
Simple system:
Complex System:

• It is normally required to reduce multiple
blocks into single block or for convenient
understanding it may sometimes required to
rearrange the blocks from its original order.
• For the calculation of Transfer function its
required to be reduced.
NEED FOR BLOCK DIAGRAM
REDUCTION

Block Diagram Reduction techniques
2
G
1
G 2
1G
G
2. Moving a summing point behind a block
G G
G
1
G
2
G
2
1 G
G 
1. Combining blocks which are in cascade or in parallel

5. Moving a pickoff point ahead of a block
G G
G G
G
1
G
3. Moving a summing point ahead of a block
G G
G
1
4. Moving a pickoff point behind a block

6. Eliminating a feedback loop
G
H
GH
G

1
7. Swapping with two adjacent summing points
A B A
B
G
1

H
G
G

1

1−Check for the blocks connected in series and
simplify. RULE NO-1
2−Check for the blocks connected in parallel
and simplify. RULE NO-1
3−Check for the blocks connected in feedback
loop and simplify. RULE NO-6
4− If there is difficulty with take-off point while
simplifying, shift it towards right. RULE NO-4,5
5−If there is difficulty with summing point
while simplifying, shift it towards left. RULE NO-2,3
6-Repeat the above steps till you get the
simplified form, i.e., single block.

Example 1
Find the transfer function of the following block diagrams
2
G 3
G
1
G
4
G
1
H
2
H
)
(s
Y
)
(s
R
(a)

1. Apply the rule that Moving pickoff/takeoff point ahead of block
2. Eliminate loop I & simplify as
3
2
4 G
G
G 
B
1
G
2
H
)
(s
Y
4
G
2
G
1
H
A
B
3
G
2
G
)
(s
R
I
Solution:
2
G

3. Moving pickoff point B behind block
3
2
4 G
G
G 
1
G
B
)
(s
R
2
1G
H 2
H
)
(s
Y
)
/(
1 3
2
4 G
G
G 
II
1
G
B
)
(s
R C
3
2
4 G
G
G 
2
H
)
(s
Y
2
1G
H
4
G
2
G A
3
G 3
2
4 G
G
G 

4. Eliminate loop III
)
(s
R
)
(
1
)
(
3
2
4
2
1
2
1
3
2
4
1
G
G
G
H
H
G
G
G
G
G
G



 )
(s
Y
)
(
)
(
1
)
(
)
(
)
(
)
(
3
2
4
1
3
2
4
2
1
2
1
3
2
4
1
G
G
G
G
G
G
G
H
H
G
G
G
G
G
G
s
R
s
Y
s
T








)
(s
R
1
G
C
3
2
4
1
2
G
G
G
H
G

)
(s
Y
3
2
4 G
G
G 
2
H
C
)
(
1 3
2
4
2
3
2
4
G
G
G
H
G
G
G



Using rule 6

2
G
1
G
1
H 2
H
)
(s
R )
(s
Y
3
H
(b) Find the transfer function of the following block diagram

Solution:
1. Eliminate loop I
2. Moving pickoff point A behind block
2
2
2
1 H
G
G

1
G
1
H
)
(s
R )
(s
Y
3
H
B
A
2
2
2
1 H
G
G

2
2
2
1
G
H
G

1
G
1
H
)
(s
R )
(s
Y
3
H
2
G
2
H
B
A
II
I
2
2
2
1 H
G
G

Not a feedback loop
)
1
(
2
2
2
1
3
G
H
G
H
H



3. Eliminate loop II
)
(s
R )
(s
Y
2
2
2
1
1 H
G
G
G

2
2
2
1
3
)
1
(
G
H
G
H
H


2
1
2
1
1
1
3
2
1
2
2
2
1
1
)
(
)
(
)
(
H
H
G
G
H
G
H
G
G
H
G
G
G
s
R
s
Y
s
T






Using rule 6

2
G 4
G
1
G
4
H
2
H
3
H
)
(s
Y
)
(s
R
3
G
1
H
(c) Find the transfer function of the following block diagrams

Solution:
2
G 4
G
1
G
4
H
)
(s
Y
3
G
1
H
2
H
)
(s
R
A B
3
H
4
1
G
4
1
G
I
4
G
4
3
G
H
4
2
G
H

Solution:
2
G
1
G
4
H
)
(s
Y
4
3G
G
1
H
2
H
)
(s
R
B
3
H
4
1
G
4
1
G
I
4
G
4
3
G
H
4
2
G
H

Solution:
2
G
1
G
)
(s
Y
1
H
2
H
)
(s
R
B
3
H
4
1
G
4
1
G
I
4
G
4
3
G
H
4
2
G
H
2
4
4
3
4
3
1 H
G
G
G
G


2. Eliminate loop I and Simplify
II
III
4
4
3
4
3
2
1 H
G
G
G
G
G

1
G
)
(s
Y
1
H
B
4
2
G
H
)
(s
R
4
3
G
H
II
3
3
2
4
4
3
4
3
2
1 H
G
G
H
G
G
G
G
G


III
4
1
4
2
G
H
G
H 
Not feedback
feedback

)
(s
R )
(s
Y
4
1
4
2
G
H
G
H 
3
3
2
4
4
3
4
3
2
1
1 H
G
G
H
G
G
G
G
G
G


3. Eliminate loop II & III
1
4
3
2
1
2
3
2
1
4
4
3
3
3
2
4
3
2
1
1
)
(
)
(
)
(
H
G
G
G
G
H
G
G
G
H
G
G
H
G
G
G
G
G
G
s
R
s
Y
s
T






Using rule 6

3
G
1
G
1
H
2
H
)
(s
R )
(s
Y
4
G
2
G A
B
(d)
Find the transfer function of the following block diagram

Solution:
3
G
I
1
H
3
1
G
)
(s
Y
1
G
1
H
2
H
)
(s
R
4
G
2
G
A B
3
1
G
3
G

2. Eliminate loop I & Simplify
3
G
1
H
2
G B
3
1
G
2
H
3
2G
G B
2
3
1
H
G
H

1
G
)
(s
R )
(s
Y
4
G
3
1
G
H
2
3
2
1
2
3
2
1 H
G
G
H
G
G
G


II

)
(s
R )
(s
Y
1
2
1
2
3
2
1
2
3
2
1
1 H
G
G
H
G
G
H
G
G
G
G



3. Eliminate loop II
1
2
1
2
3
2
1
2
3
2
1
4
1
)
(
)
(
)
(
H
G
G
H
G
G
H
G
G
G
G
G
s
R
s
Y
s
T






4
G

Find the transfer function of the following block diagram

Step 1 − Use Rule 1 for blocks G1and G2. Use Rule 2 for
blocks G3 and G4. The modified block diagram is shown in the
following figure.

Step 2 − Use Rule 3 for blocks G1G2 and H1. Use Rule 4 for
shifting take-off point after the block G5.
The modified block diagram is shown in the following figure.

Step 3 − Use Rule 1 for blocks (G3+G4) and G5.
The modified block diagram is shown in the
following figure.

Step 4 − Use Rule 3 for blocks (G3+G4)G5 and H3.
The modified block diagram is shown in the following
figure.

Step 5 − Use Rule 1 for blocks connected in
series. The modified block diagram is shown
in the following figure.

Step 6 − Use Rule 3 for blocks connected in
feedback loop. The modified block diagram is
shown in the following figure. This is the
simplified block diagram.

UNIT I CONTROL SYSTEM (EC3351).pptx

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UNIT I CONTROL SYSTEM (EC3351).pptx