UNIT 1.pptx control system modeling in control

EE3503 – CONTROL SYSTEMS
UNIT - I

CONTROL SYSTEM; TERMINOLOGY AND
BASIC STRUCTURE
2
System
Group of components are connected in a
sequence to perform a specific function,
the group thus formed is called a system.

CONTROL SYSTEM; TERMINOLOGY AND
BASIC STRUCTURE
3
⚫Control
system
⚫In a system ,when
the
controlled by varying
the
outpu
t
input
quantity
is
quantity,
the
system is called control
system

OPEN LOOP SYSTEM
⚫Any physical system which does not
automatically correct the variation in its
output is called an open loop system
4

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
5

CLOSED LOOP SYSTEM- CAR DRIVING
7

DIFFERENCE BETWEEN OPEN LOOP AND
CLOSED LOOP CONTROL SYSTEM
8
Open 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
Closed-Loop Control
System
⚫ 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

FEEDFORWARD CONTROL
Control element responds to change in
command or
measured disturbance in a pre-defined way
10

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
11

LIMITATIONS OF FEEDFORWARD
12
⚫Effects of disturbance or command input
must be predictable
⚫May not generalize to other conditions
⚫Will not be accurate if the system changes

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)
13

FEATURES OF FEEDBACK
⚫Reactive / Error-driven
⚫Automatically compensatesfor 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 14

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
15

ELECTRICAL AND MECHANICAL TRANSFER
FUNCTION MODELS
16
⚫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.

FUNCTION MODELS
17
⚫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
9/29/2022 DrNSR/AP/ECE/PSNACET/
EC3351
18
⚫The following mathematical models are mostly
used.
⚫1. Differential equation model
⚫2. Transfer function model
⚫3. State space model

DIFFERENTIAL EQUATION MODEL
19
Differential equation model is a time
domain mathematical model of control systems.

RLC series electrical
system
Mesh equation for this circuit
is
20

⚫Substitute, the current passing through
capacitor
The above equation is a second order differential
equation.
21

TRANSFER FUNCTION MODEL
22
⚫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
23

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).
24

THE DIFFERENTIAL EQUATION MODELING OF
MECHANICAL SYSTEMS.
25
⚫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
26
⚫Translational mechanical systems move along
a straight line
consist of three
basic
⚫These systems
mainly
elements.
⚫Mass,
⚫Spring
⚫Dashpot or damper.

MASS
28
⚫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
30

SPRING
BLOCK DIAGRAM OF
SPRING
31

⚫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.
32
to
the
⚫This opposing force is
proportional 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
⚫ x is displacement
33

DASHPOT
BLOCK DIAGRAM OF
FRICTION
34

⚫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.
35

⚫Where,
⚫Fb is the opposing
force due to friction
of dashpot
⚫B is the
frictional
coefficient
⚫v is velocity
⚫ x is
displacement
36

MODELING OF ROTATIONAL MECHANICAL
SYSTEMS
37
⚫Rotational mechanical systems move about
a fixed axis.
consist of three
basic
⚫These systems
mainly elements.
Those are
⚫Moment of inertia,
⚫Torsional spring
⚫Dashpot.

Moment of Inertia
BLOCK DIAGRAM OF MOMENT OF
INERTIA
38

⚫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.
39

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
displacemen
t
40

⚫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.
42

⚫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 43

⚫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.
45

PROBLEMS
1.Write the differential equations governing
the mechanical system shown in fig, and
determine the transfer function (X(s)/F(s))
46

2.Write the differential equations governing
the mechanical system shown in the fig, and
determine the transfer function
51

Substituting X1(S) from
node 1
60

MULTIVARIABLE CONTROL SYSTEM
62
⚫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 disturbancein any input causes a change
of response from at least one output.

63

64
⚫This kind of system have as many inputs
and outputs as needed to control the process.
⚫Asystem with an equal number of inputs
and outputs is said to be square.

APPLICATIONS
65
⚫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.
⚫Arobot manipulator where the
positions of the manipulators
(arms)shall be controlled.
⚫Achemical 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
66

SIGNAL FLOW GRAPH MODELS
67
⚫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
68
⚫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:
69

BLOCK DIAGRAM AND SIGNAL FLOW
GRAPH
70

Properties of signal flow graph
71
⚫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.

Mason’s Gain Formula
76
⚫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
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
Transfer Function: Ratio between transformation of output to the
transformation of input when all the initial conditions are zero.
G2 G3
G1
G4
H1
H2
Y
(s)
G(s)
C(s)
R(s) G(s)=C(s)/R(s)
Simple
system:
Complex
System:
R(s)

• 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
G2
G1 G1G2
2. Moving a summing point behind a
block
G G
G
G1
G2
G1  G2
1. Combining blocks which are in cascade or in
parallel

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

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

1−Check for the blocks connected in series and
RULE NO-1
simplify.
2−Check for the blocks connected in parallel
feedback
and simplify. RULE NO-1
3−Check for the blocks connected
in 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
G2 G3
G1
G4
H1
H2
Y
(s)
R(s)
(a)

2. Eliminate loop I & simplify
as
G4  G2G3
B
G1
H2
Y
(s)
G4
G2
H1
A
B
3
G
G2
R(s)
I
Solution:
1. Apply the rule that Moving pickoff/takeoff point ahead of
block
G2

3. Moving pickoff point B behind
block
G4  G2G3
G1
B
R(s)
H1G2
H2
Y
(s)
1/(G4  G2G3 )
II
G1
B
R(s) C
G4  G2G3
H2
Y
(s)
H1G2
G4
A
G2
G3
G4  G2G3

4. Eliminate loop
III
R(s) G1 (G4  G2G3 )
1 G1G2 H1  H2 (G4  G2G3 )
Y
(s)
G1 (G4  G2G3 )
1 G1G2 H1  H2 (G4  G2G3 )  G1 (G4  G2G3 )
T (s) 
Y (s)

R(s)
R(s)
G1
C
G2 H1
G4  G2G3
Y
(s)
C
G4  G2G3
H2
G4  G2G3
1 H2 (G4  G2G3 )
Using rule
6

G2
1
G
H1
H2
R(s) Y
(s)
H3
(b) Find the transfer function of the following block
diagram

Solution
:
1. Eliminate loop
I
2. Moving pickoff point A behind
block
G2
1 G2 H 2
G1
H1
R(s) Y
(s)
H3
B
A G2
1 G2 H 2
1 G2 H 2
G2
G1
H1
R(s) Y
(s)
H3
B
A
II
I
G
2
H2
G
2
1 G2 H2
Not a feedback
loop
2
3
G
1 G2 H 2
)
H  H1 (

3. Eliminate loop
II
R(s) Y
(s)
G1G2
1 G2 H 2
2
3
H 
H1 (1 G2 H 2 )
G
G1G2
1 G2 H2  G1G2 H3  G1H1  G1G2 H1H2
T (s) 
Y (s)

R(s)
Using rule
6

G2 G4
G1
H4
H2
H3
Y
(s)
R(s)
G3
H1
(c) Find the transfer function of the following block
diagrams

Solution
:
G2 G4
G1
H4
Y
(s)
G3
H1
R(s)
A B
1
H
3
G 4
1
H
2
G 4
I
block
G4
H 3
G4
H 2
G4

Solution
:
G2
G1
H4
Y
(s)
4
3
G G
H1
R(s)
B
1
H
3
G 4
1
H
2
G 4
I
block
G4
H 3
G4
H 2
G4

Solution
:
G2
G1
Y
(s)
H1
R(s)
B
1
H
3
G 4
1
H
2
G 4
I
block
G4
H 3
G4
H 2
G4
G3 G4
1  G3 G4 H 4

2. Eliminate loop I and
Simplify
II
III
G2G3G4
1 G3G4 H 4
G1
Y
(s)
H1
B
H 2
G4
R(s)
H 3
G4
II
G2G3G4
1 G3G4 H 4  G2 G3 H3
III
G4
H 2  G4 H1
Not
feedback
feedbac
k

R(s) Y
(s)
H 2  G4 H1
G4
G1G2G3G4
1 G3G4 H 4  G2 G3 H3
3. Eliminate loop II &
III
G1G2G3G4
1 G2G3 H3  G3G4 H4  G1G2G3 H2  G1G2G3G4 H1
T (s) 
Y (s)

R(s)
Using rule
6

3
G
G1
H1
H2
R(s) Y
(s)
G4
G2
A
B
(d)
diagram

Solution
:
block
G3
I
H1
3
1
G
Y
(s)
G1
H1
H2
R(s)
G4
G2
A B
1
G3
G3

2. Eliminate loop I &
Simplify
G3
H1
G2
B
1
G3
H2
G2G3
B
2
3
H1
 H
G
G1
R(s) Y
(s)
G4
H1
G3
G2G3
1 G2 H1  G2G3 H 2
II

R(s) Y
(s)
G1G2G3
1 G2 H1  G2 G3 H 2  G1G2 H1
3. Eliminate loop
II
4
G1G2G
3
R(s
)
1 G2 H1  G2G3 H2  G1G2 H1
T (s) 
Y (s)
 G
G4

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 1.pptx control system modeling in control

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UNIT 1.pptx control system modeling in control