control engineering basics
TRANSCRIPT
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Vijay Gupta
MEC308 CONTROLS THEORY
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Introduction And
Applications
Lecture 1
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Introduction
Control SystemControl means to regulate, to direct or to
commandSystem means combination of devices and
components connected together by some form of regular interactions to act together and perform a certain objective.
Hence control system means a system that is controlled.
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Control Systems: some examples
(1)A bulb with
a switch
(2)A fan with a regulator
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Control Systems: some examples
(3) An air conditioner with a control to set the temperature of the room (4) A window to control the air flow in the room
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(5)Tap to control the flow
(6) A flush tank with the provision to control the level of water
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(7) A security guard at the gate controls the unauthorized entry
(8) A traffic police to control the traffic
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(9) A person riding bicycle
• Controls speed using brake and pedal• Controls direction using handle bar
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(10) James Watt’s Centrifugal Governor
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(11) Missile Direction Control system
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Types of Control Systems
• Open Loop Control System
• Feedback Control System OR Closed Loop Control System
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• Output has no effect on the control action• Output is neither measured nor fedback for
comparison with the input • Control system that operates on a time basis
are examples of Open Loop Control System. • Eg. traffic control by means of signals operated
on a time basis, automatic washing machine with a timer etc.
Open Loop Control System:
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• Maintains a prescribed relationship between the output and the reference input by comparing them.
• Output is measured and fedback for comparison with the input.
• Eg. room temperature control system, missile direction control system etc.
• Can be found in various non-engineering fields as well. Eg. Temperature and blood pressure control system of human body etc.
Feedback Control System OR Closed Loop Control System:
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Terminologies:
1) Controlled variable and Manipulated variable• Controlled variable: quantity or condition
that is measured and controlled
• Manipulated variable: quantity or condition that is varied by the controller so as to affect the value of the controlled variable
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Eg. In the speed control system of an engine with load variation, the controlled variable is the speed of the engine shaft and the manipulated variable is the displacement of the fuel supply valve.
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Terminologies:
2) Plant: a piece of equipment the purpose of which is to perform a particular operation. In control system, any physical object to be controlled is called a plant
Eg. The centrifugal governor controls the speed of the engine with load, since here the physical object to be controlled is an engine, hence engine is called plant.
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Terminologies:
3) Disturbance: a signal that tends to adversely affect the value of the output of a system
Eg. In the temperature control system of a room any external heat coming into the room from outside or leaving the room is an example of disturbance.
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Machine Tool Control, Boiler Control, Engine Governing
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Machine Tool Control• Motion of the cutting tool is controlled by a computer. This type of • control is termed NUMERICAL CONTROL.
• When there is a difference between the desired position r(t) and the • actual position y(t) of the cutting tool, the amplifier amplifies this • difference so that the output current of the amplifier is large enough to • activate the coil.
• The magnetic field produced around the coil creates a force on the • piston of the valve of the hydraulic servomotor, moving it to the left or • to the right.
• These small movements of the piston result in controlling the position of • the cutting tool in such a way that y(t)=r(t)
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Machine Tool Control for cutting(or shaping or engraving) metals
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Boiler-Generator Control
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Boiler- Generator Control
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• The steam produced by the boiler sets the shaft in rotation. As the shaft rotates, the generator produces electric power.
• This system has many inputs(water, air and liquid fuel) and one output (electric power).
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Lecture 3
Introduction to Mathematical Modeling
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Mathematical Model
• the most important part of the entire analysis of a control system is to derive a reasonable mathematical model
• It is defined as a set of equations that represents the dynamics of the system accurately or, at least, fairly well.
• dynamics of many system (mechanical, electrical , biological etc) may be described in terms of differential equations by using the physical laws governing them.
• Eg. For dynamics of mechanical system - Newton’s law For dynamics of electrical system - Kirchhoff’s law and so on
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Mechanical systems• are of two types, namely (i) translatory mechanical system,
and (ii) rotary mechanical system
• Both translatory and rotary mechanical system can be modelled by means of three ideal translatory and three ideal rotary elements as shown in table below:
S. No. Translatory elements Rotary elements*
1 Mass(m) Inertia(J)
2 Spring(k) Torsional spring(k)
3 Damper(b) Damper (b)
2
2
dtxdm F
kx F
dtdxb F
2
2
dtd J T
k T
dtdb T
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Eg. Consider the following translatory mechanical system:
• Here, k= spring constant or stiffness of the spring• b= viscous friction coefficient of the damper• x(t) = displacement of the mass, m• F = force
A mass-spring-dashpot system
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To obtain mathematical model of the given mechanical system:-
There are following two steps involved:STEP 1: To draw the free body diagram
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STEP 2: To apply Newton’s law of motion to the free body diagram
• Inertia is the resistance of any physical object to any change in its state of motion, including changes to its speed and direction. It is the tendency of objects to keep moving in a straight line at constant velocity
kxdtdxb
dtxdm F,or
dtxdm kx
dtdxb - F
2
2
2
2
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Eg. Consider the following rotary mechanical system:
• Here, k= torsional spring constant or stiffness of the shaft• b= viscous friction coefficient of the damper(viscous
fluid)• θ(t) = displacement of the inertia, J• T = torque
An inertia-torsional spring-dashpot system
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To obtain mathematical model of the given rotary mechanical system:-
Similarly, there are following two steps involved:STEP 1: To draw the free body diagram
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STEP 2: To apply Newton’s law of motion to the
free body diagram
kdtdb
dtd J T,or
dtd Jk
dtdb - T
2
2
2
2
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• any electrical system can be modelled by using three basic elements of electrical circuits, namely:
Electrical systems
S. No. eElectrical Elements vVoltage-current relation
1 Resistor(R) V=Ri2 Inductor(L)
3 Capacitor(C)dtdiL V
t
idtC1 V
Here inductor and capacitor are the storage elements exactly like mass and spring are storage elements in mechanical system.
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• Eg.(i) Consider the following L-R-C series circuit:
By applying Kirchhoff’s voltage law:
t
idtC1
dtdiLRi e
is the required mathematical model which is an ordinary differential equation
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• The mathematical model obtained can be written in the following way:
Putting and in the mathematical model obtained above we have:
t
idt qdtdqi
qC1
dtdqR
dtqdLe,or
qC1
dtdqR
dtdq
dtdL e
2
2
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• (2)Consider the following L-R-C parallel circuit:
By applying Kirchhoff’s current law:
dtdeC
Reedt
L1 i
t
is the required mathematical model which is an ordinary differential equation
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• The mathematical model obtained can be written in the following way:
Putting and in the mathematical model obtained above we have:dtde
t
edt
L1
dtd
R1
dtdCi 2
2
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Analogous System: • are systems whose differential equations are of
identical form.• Eg. If we compare the following differential equations
Mechanical System Electrical System
kxdtdxb
dtxdm F 2
2
qC1
dtdqR
dtqdLe 2
2
Clearly the two differential equations are of identical form, hence the two systems are analogous.
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• Here, Force(Torque) of the mechanical system is analogous to Voltage of the electrical system. Therefore this analogy is called
Mechanical System Electrical System
Translational Rotational
Force(F) Torque(T) Voltage(e)
Mass(m) Moment of Inertia(J) Inductance(L)
Viscous friction coefficient(b)
Viscous friction coefficient(b)
Resistance(R)
Spring stiffness(k) Torsional Spring stiffness(k)
1/Capacitance(1/C)
Displacement(x) Angular displacement(θ)
Charge(q)
Force(Torque)-Voltage analogy.
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• Similarly, if we compare the following differential equations:
Mechanical System Electrical System
kxdtdxb
dtxdm F 2
2
L1
dtd
R1
dtdCi 2
2
Clearly the two differential equations are of identical form, hence the two systems are analogous.
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• Here, Force(Torque) of the mechanical system is analogous to Current of the electrical system. Therefore this analogy is called Force(Torque)-Current analogy.
Mechanical System Electrical System
Translational Rotational
Force(F) Torque(T) Current(i)
Mass(m) Moment of Inertia(J) Capacitance(C)
Viscous friction coefficient(b)
Viscous friction coefficient(b)
1/Resistance(1/R)
Spring stiffness(k) Torsional Spring stiffness(k)
1/Inductance(1/L)
Displacement(x) Angular displacement(θ)
Magnetic flux(Ф)
Force(Torque)-Current analogy:
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Mathematical ModelingTransfer Functions
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Transfer Function• Of a linear time-invariant system is defined to be the ratio of the • Laplace transform of the output variable to Laplace transform of • the input variable under the assumption that all initial conditions • are zero.• Transfer Function, G(s) = • Where • = Laplace transformation of output variable
• Eg. Consider the following translational mechanical system:)s(R)s(C
= Laplace transformation of input variable
We have the following mathematical model for this system:
kxdtdxb
dtxdm F 2
2
)s(C)s(R
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To find the Transfer Function:• Taking the Laplace transformation of each term of this equation(assuming
zero initial condition) we have:
kx
dtdxb
dtxdm L FL 2
2
xkLdtdxbL
dtxdLm FL 2
2
)s(kX)s(Xbs)s(Xms F(s) 2
kbsms1
)s(FX(s)
2
Here, and outputL X(s) inputL)s(F
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By definition, Transfer function = inputLoutputL
Hence, Transfer function is:
G(s)= )s(FX(s)
kbsms1
2
ORDER OF THE SYSTEM: The highest power of the complex variable s in the denominator of the transfer function determines the order of the system
NOTE: In the above transfer function since the highest power of the complex variable (s) in the denominator is 2, therefore the above mechanical system is a SECOND ORDER SYSTEM .
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The corresponding block diagram representation of the mechanical system is given as:
)s(F G(s)= kbsms
1 2
X(s)
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• Consider the following electrical system:
The mathematical model can be written in the following way:
qC1
dtdqR
dtqdLe 2
2
G(s)= )s(E
Q(s))C/1(RsLs
1 2
Hence, Transfer function is:
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The corresponding block diagram representation of the electrical system is given as:
G(s) )C/1(RsLs1
2 Q(s))s(E
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• To find the transfer function of the following electrical system:
The mathematical model can be written in the following way:
L1
dtd
R1
dtdCi 2
2
Hence, Transfer function is:
G(s)= )s(I
(s))L/1(s)(1/RCs
1 2
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The corresponding block diagram representation is given as:
)s(I G(s) )L/1(s)(1/RCs
1 2
(s)
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• Transfer Function of a Closed Loop Control System:
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Comments on Transfer Function:• Only applicable to systems described by linear, time-invariant, differential equation.
•Is an operational method of expressing the differential equation that relates the output variable to the input variable.
• Is a property of a system itself, independent of the magnitude and nature of the input.
• Gives full description of the dynamic characteristics of the system.
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BLOCK DIAGRAM
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Control System consists of a number of components
The pictorial representation of the functions performed by each of these components and of the flow of signals is called a Block Diagram.
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Speed Control System of a Bicycle
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Block Diagram Reduction Techniques
Mr. Varun Panwar
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• We often represent control systems using block diagrams. A block diagram consists of blocks that represent transfer functions of the different variables of interest.
• If a block diagram has many blocks, not all of which are in cascade, then it is useful to have rules for rearranging the diagram such that you end up with only one block.
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For example, we would want to transform the following diagram
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Block Diagram Transformations1.Combining blocks in cascade(series)
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2 (Blocks in Parallel)
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3. Moving a summingpoint behind a block
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4. Moving a pickoff point ahead of a block
![Page 74: Control engineering basics](https://reader036.vdocuments.net/reader036/viewer/2022062400/58a1b0291a28ab537c8b4635/html5/thumbnails/74.jpg)
Vijay Gupta
5. Moving a pickoffpoint behind a block
![Page 75: Control engineering basics](https://reader036.vdocuments.net/reader036/viewer/2022062400/58a1b0291a28ab537c8b4635/html5/thumbnails/75.jpg)
Vijay Gupta
6. Moving a summingpoint ahead of a block
![Page 76: Control engineering basics](https://reader036.vdocuments.net/reader036/viewer/2022062400/58a1b0291a28ab537c8b4635/html5/thumbnails/76.jpg)
Vijay Gupta
7. Eliminating afeedback loop
![Page 77: Control engineering basics](https://reader036.vdocuments.net/reader036/viewer/2022062400/58a1b0291a28ab537c8b4635/html5/thumbnails/77.jpg)
Vijay Gupta
8. replacing summing points
![Page 78: Control engineering basics](https://reader036.vdocuments.net/reader036/viewer/2022062400/58a1b0291a28ab537c8b4635/html5/thumbnails/78.jpg)
Vijay Gupta
9. Combining summing points
![Page 79: Control engineering basics](https://reader036.vdocuments.net/reader036/viewer/2022062400/58a1b0291a28ab537c8b4635/html5/thumbnails/79.jpg)
Vijay Gupta
block diagram: reduction example
R_+
_+1G 2G 3G
1H
2H
+ +
C
![Page 80: Control engineering basics](https://reader036.vdocuments.net/reader036/viewer/2022062400/58a1b0291a28ab537c8b4635/html5/thumbnails/80.jpg)
Vijay Gupta
block diagram: reduction example
R_+
_+
1G 2G 3G
1H
1
2
GH
+ +
C
![Page 81: Control engineering basics](https://reader036.vdocuments.net/reader036/viewer/2022062400/58a1b0291a28ab537c8b4635/html5/thumbnails/81.jpg)
Vijay Gupta
block diagram: reduction example
R_+
_+
21GG 3G
1H
1
2
GH
+ +
C
![Page 82: Control engineering basics](https://reader036.vdocuments.net/reader036/viewer/2022062400/58a1b0291a28ab537c8b4635/html5/thumbnails/82.jpg)
Vijay Gupta
block diagram: reduction example
R_+
_+ 21GG 3G
1H
1
2
GH
+ +
C
![Page 83: Control engineering basics](https://reader036.vdocuments.net/reader036/viewer/2022062400/58a1b0291a28ab537c8b4635/html5/thumbnails/83.jpg)
Vijay Gupta
block diagram: reduction example
R_+
_+ 21GG 3G
1H
1
2
GH
+ +
C
![Page 84: Control engineering basics](https://reader036.vdocuments.net/reader036/viewer/2022062400/58a1b0291a28ab537c8b4635/html5/thumbnails/84.jpg)
Vijay Gupta
block diagram: reduction example
R_+
_+
121
21
1 HGGGG
3G
1
2
GH
C
![Page 85: Control engineering basics](https://reader036.vdocuments.net/reader036/viewer/2022062400/58a1b0291a28ab537c8b4635/html5/thumbnails/85.jpg)
Vijay Gupta
block diagram: reduction example
R_+
_+
121
321
1 HGGGGG
1
2
GH
C
![Page 86: Control engineering basics](https://reader036.vdocuments.net/reader036/viewer/2022062400/58a1b0291a28ab537c8b4635/html5/thumbnails/86.jpg)
Vijay Gupta
block diagram: reduction example
R_+
232121
321
1 HGGHGGGGG
C
![Page 87: Control engineering basics](https://reader036.vdocuments.net/reader036/viewer/2022062400/58a1b0291a28ab537c8b4635/html5/thumbnails/87.jpg)
Vijay Gupta
block diagram: reduction example
R
321232121
321
1 GGGHGGHGGGGG
C