COURSE FILE CONTENTS

Sl.No.	Topic		Page no.
1.	Department Vision and Mission		3
2.	Course Description		3
3.	Course Overview		3
4.	Course Pre-requisites		4
5.	Marks Distribution		4
6.	POs and PSOs		4 & 5
7.	Course outcomes (COs)		5
8.	CO mapping with POs and PSOs		5
9.	Syllabus, Textbooks and Reference Books		6
10.	Gaps in syllabus		6
11.	Course plan/Lesson Plan		8
12.	Lecture Notes
	*Unit-I*		10
	*Unit-II*		27
	*Unit-III*		46
	*Unit-IV*		61
	*Unit-V*		69
13.	Unit wise Question Bank
	a.	Short answer questions	81
	b.	Long answer questions
14.	Previous University Question papers		87
15.	Internal Question Papers with Key		95
16.	Unit wise Assignment Questions		106
17.	Content Beyond Syllabus		110
18.	Methodology used to identify Weak and bright students		110
	· Support extended to weak students
	· Efforts to engage bright students

CERTIFICATE

I, the undersigned, have completed the course allotted to me as shown below,

Sl. No.	Semester	Name of the Subject	Course ID	Total Units
1	V	Automatic Control Systems	PC503 EC	05

Date: Prepared by

Academic Year: 2020-21 1. A. NARMADA

Verifying authority:

1. Head of the Department: Dr. N. SRINIVASA RAO

PRINCIPAL

MATRUSRI ENGINEERING COLLEGE

Saidabad, Hyderabad-500 059.

(Approved by AICTE & Affiliated to Osmania University)

ELECTRONICS AND COMMUNICATION ENGINEERING

DEPARTMENT VISION

To become a reputed centre of learning in Electronics and Communication Engineering and transform the students into accomplished professionals.

DEPARTMENT MISSION

1. To provide the learning ambience to nurture the young minds with theoretical and practical knowledge to produce employable and competent engineers.

2. To provide a strong foundation in fundamentals of Electronics and Communication Engineering to make students explore advances in research for higher learning.

3. To inculcate awareness for societal needs, continuous learning and professional practices.

4. To imbibe team spirit and leadership qualities among students.

COURSE DESCRIPTION

Course Title	AUTOMATIC CONTROL SYSTEMS
Course Code	PC 503EC
Programme	BE-ECE
Semester	V
Course Type	Core
Regulation	AICTE
Course Structure	Theory			Practical
	Lectures	Tutorials	Credits	Laboratory	Credits
	3	1	4	-	-
Course Faculty	Mrs.A.NARMADA

I. COURSE OVERVIEW:

Control system engineering is applicable to electrical, mechanical, hydraulic and chemical processes as well as air craft control and design also. Control engineering plays a dominant role various Industrial system designs. This course covers basic concepts of control systems, time domain and frequency domain analysis, state variable design, stability of linear continuous time and discrete time systems.

II. COURSE PRE-REQUISITES:

Level	Course Code	Semester	Prerequisites	Credits
UG	BS102 MT	I SEM	MATHEMATICS - I	-

III. MARKS DISTRIBUTION:

Subject

SEE Examination

CIA

Examination

Total Marks

Automatic Control Systems

100

IV. PROGRAM OUTCOMES (POs):

The students will be able to:
PO1	Engineering knowledge: Apply the knowledge of mathematics, science, engineering fundamentals, and an engineering specialization to the solution of complex engineering problems.
PO2	Problem analysis: Identify, formulate, review research literature, and analyze complex engineering problems reaching substantiated conclusions using first principles of mathematics, natural sciences, and engineering sciences.
PO3	Design/development of solutions: Design solutions for complex engineering problems and design system components or processes that meet the specified needs with appropriate consideration for the public health and safety, and the cultural, societal, and environmental considerations.
PO4	Conduct investigations of complex problems: Use research-based knowledge and research methods including design of experiments, analysis and interpretation of data, and synthesis of the information to providevalid conclusions.
PO5	Modern tool usage: Create, select, and apply appropriate techniques, resources, and modern engineering and IT tools including prediction and modeling to complex engineering activities with an understanding of the limitations.
PO6	The engineer and society: Apply reasoning informed by the contextual knowledge to assess societal, health, safety, legal and cultural issues and the consequent responsibilities relevant to the professional engineering practice.
PO7	Environment and sustainability: Understand the impact of the professional engineering solutions in societal and environmental contexts, and demonstrate the knowledge of, and need for sustainable development.
PO8	Ethics: Apply ethical principles and commit to professional ethics and responsibilities and norms of the engineering practice.
PO9	Individual and team work: Function effectively as an individual, and as a member or leader in diverse teams, and in multidisciplinary settings.
PO10	Communication: Communicate effectively on complex engineering activities with the engineering community and with society at large, such as, being able to comprehend and write effective reports and design documentation, make effective presentations, and give and receive clear instructions.
PO11	Project management and finance: Demonstrate knowledge and understanding of the engineering and management principles and apply these to one’s own work, as a member and leader in a team, to manage projects and in multidisciplinary environments.
PO12	Life-long learning: Recognize the need for, and have the preparation and ability to engage in independent and life- long learning in the broadest context of technological change.

V. PROGRAM SPECIFIC OUTCOMES (PSOs):

The students will be able to:
PSO1	*Professional Competence:* Apply the knowledge of Electronics Communication Engineering principles in different domains like VLSI, Signal Processing, Communication, Embedded Systems.
PSO2	*Technical Skills:* Able to design and implement products using the state of art Hardware and Software tools and hence provide simple solutions to complex problems.

VI. COURSE OUTCOMES (COs):

The course should enable the students to:
CO1	Convert a given control system into equivalent block diagram and transfer function
CO2	Analyze system stability using time domain techniques
CO3	Analyze system stability using frequency domain techniques
CO4	Design a digital control system in the discrete time domain
CO5	Analyze a control system in the state space representation

VII. MAPPING COURSE OUTCOMES (COs) with POs and PSOs:

(3 = High; 2 = Medium; 1 = Low )

COs	POs												PSOs
COs	PO1	PO2	PO3	PO4	PO5	PO6	PO7	PO8	PO9	PO 10	PO 11	PO 12	PSO1	PSO2
PC503EC.1	2	2	2	1	-	-	-	-	-	-	-	2	2	2
PC503EC.2	2	2	2	1	-	-	-	-	-	-	-	2	2	2
PC503EC.3	2	2	2	1	-	-	-	-	-	-	-	2	2	2
PC503EC.4	2	2	2	2	-	1	-	-	-	-	-	2	2	2
PC503EC.5	2	2	2	2	-	1	-	-	-	-	-	2	2	2
PC504EC (AVG)	2	2	2	1.4	-	1	-	-	-	-	-	2	2	2

VIII. SYLLABUS :

UNIT I-Control System fundamentals and Components:		No. of Hrs
Classification of control systems including Open and Closed loop systems, Transfer function representation, Mathematical modeling of Mechanical systems and their conversion into electrical systems, Block diagram representation, Block diagram algebra and reduction and Signal flow graphs and Mason’s gain formula.		10
UNIT II-Time Response and Stability:
Transfer function and types of input. Transient response of second order system for step input. Time domain specifications Characteristic Equation of Feedback control systems Types of systems, static error coefficients, error series, Concept of Stability, Routh-Hurwitz criterion for stability, Root locus technique and its construction.		10
UNIT III-Frequency response plots and Compensation Techniques:
Bode plots, frequency domain specifications Gain and Phase margin. Principle of argument Nyquist plot and Nyquist criterion for stability. Cascade and feedback compensation. Phase lag, lead and lag-lead compensators PID controller.		10
UNIT IV- Discrete Control Systems:
Digital control, advantages and disadvantages, Digital control system architecture. The discrete transfer function sampled data system Transfer function of sample data systems. Analysis of Discrete data systems		6
UNIT V- State space representation:
Concept of state and state variables. State models of linear time invariant systems, State transition matrix, Solution of state equations. Controllability and Observability		10
TEXT BOOKS:
1.	Nagrath, I.J, and Gopal, M., “Control System Engineering”, 5/e, New Age Publishers, 2009
2.	NagoorKani.,” Control systems”, Second Edition, RBA Publications.
3.	Ogata, K., “Modern Control Engineering”, 5/e, PHI.
REFERENCES:
1	Ramesh Babu, “Digital Signal Processing”, 2/e,
2.	K.Deergha Rao, Swamy MNS, “Digital Signal Processing, Theory and Applications”, 1/e, Springer Publications, 2018

IX. GAPS IN THE SYLLABUS - TO MEET INDUSTRY / PROFESSION REQUIREMENTS:

S No

Description

Proposed Actions

Relevance With POs

Relevance With PSOs

X. COURSE PLAN/ LECTURE PLAN:

Lecture No.	Topics to be covered	PPT/BB/ OHP/ e-material	No. of Hrs	Relevant Cos	Text Book/Reference Book
1	Classification of control systems,	BB / e-material	1	CO1	NagoorKani.,” Control systems”, Second Edition, RBA Publications
2	Open and Closed loop systems,	BB / e-material	1	CO1	NagoorKani.,” Control systems”, Second Edition, RBA Publications
3	Mathematical modeling of mechanical systems and their conversion into electrical systems	BB / e-material	1	CO1	NagoorKani.,” Control systems”, Second Edition, RBA Publications
4	Problems	BB / e-material	1	CO1	NagoorKani.,” Control systems”, Second Edition, RBA Publications
5	Problems	BB / e-material	1	CO1	NagoorKani.,” Control systems”, Second Edition, RBA Publications
6	Block diagram reduction	BB / e-material	1	CO1	NagoorKani.,” Control systems”, Second Edition, RBA Publications
7	Problems	BB / e-material	1	CO1	NagoorKani.,” Control systems”, Second Edition, RBA Publications
8	Signal flow graphs	BB / e-material	1	CO1	NagoorKani.,” Control systems”, Second Edition, RBA Publications
9	Problems	BB	2	CO1	NagoorKani.,” Control systems”, Second Edition, RBA Publications
10	Transfer function and Impulse response, types of input.	BB / e-material	1	CO2	NagoorKani.,” Control systems”, Second Edition, RBA Publications
11	Transient response of second order system for step input	BB / e-material	1	CO2	NagoorKani.,” Control systems”, Second Edition, RBA Publications
12	Time domain specifications	BB / e-material	2	CO2	NagoorKani.,” Control systems”, Second Edition, RBA Publications
13	Types of systems, static error coefficients	BB / e-material	1	CO2	NagoorKani.,” Control systems”, Second Edition, RBA Publications
14	error series	BB / e-material	1	CO2	NagoorKani.,” Control systems”, Second Edition, RBA Publications
15	Routh - Hurwitz criterion for stability.	BB / e-material	1	CO2	NagoorKani.,” Control systems”, Second Edition, RBA Publications
16	Analysis of typical systems using root locus techniques.	BB / e-material	2	CO2	NagoorKani.,” Control systems”, Second Edition, RBA Publications
17	Effect of location of roots on system response.	BB / e-material	1	CO2	NagoorKani.,” Control systems”, Second Edition, RBA Publications
18	Bode plots	BB / e-material	2	CO3	NagoorKani.,” Control systems”, Second Edition, RBA Publications
19	Frequency domain specifications. Gain margin and Phase Margin.	BB / e-material	2	CO3	NagoorKani.,” Control systems”, Second Edition, RBA Publications
20	Principle of argument, Polar plot	BB / e-material	1	CO3	NagoorKani.,” Control systems”, Second Edition, RBA Publications
21	Nyquist plot and Nyquist criterion for stability.	BB / e-material	2	CO3	NagoorKani.,” Control systems”, Second Edition, RBA Publications
22	Cascade and feedback compensation using Bode plots.	BB / e-material	2	CO3	NagoorKani.,” Control systems”, Second Edition, RBA Publications
23	Phase lag, lead, lag-lead compensators. PID controller.	BB / e-material	1	CO3	NagoorKani.,” Control systems”, Second Edition, RBA Publications
24	Digital control, advantages and disadvantages	BB / e-material	1	CO4
25	Digital control system architecture	BB / e-material	1	CO4
26	The discrete transfer function of sampled data systems	BB / e-material	2	CO4
27	Stability of Discrete data systems	BB / e-material	2	CO4
28	Concept of state and state variables.	BB / e-material	1	CO5
29	State models of linear time invariant systems	BB / e-material	1	CO5
30	State transition matrix	BB / e-material	2	CO5
31	Solution of state equations	BB / e-material	2	CO5
32	Design of digital control systems using state-space concepts	BB / e-material	2	CO5
33	Controllability and Observability	BB / e-material	2	CO5

UNIT-I

CONTROL SYSTEM FUNDAMENTALS AND COMPONENTS

v Classification of control systems including Open and Closed loop systems

v Transfer function representation

v Mathematical modeling of Mechanical systems

v Mechanical system conversion into electrical systems

v Block diagram representation

v Block diagram algebra and reduction

v Signal flow graphs and Mason’s gain formula.

System: A system is a combination of different physical components which act together as an entire unit to achieve certain objective.

Control system: A control system is an arrangement of physical components connected in such a manner to regulate, direct or command itself or some other system.

Classification of Control systems:

Natural Control systems: The biological systems, systems inside the human body are of natural type.

Manmade Control systems: The systems which are manufactured by human beings are called Manmade Control systems.

Combinational Control systems:A system having combination of both natural and manmade is called combinational control system.

Time variant and Invariant Control systems: A Control system in which if the parameters are varying with respective to time then it is said to be a Time varying Control system and a system in which the parameters are independent of time, then it is said to be a Time invariant Control system.

Continuous time and Discrete time control systems: In continuous time control systems all the variables are functions of a continuous time variable and in Discrete Control systems all the variables of the systems are functions of a discrete time variable.

Transfer function of a system:

Transfer function of a system is given by the ratio of Laplace transform of the output to the Laplace transform of input. Mathematically it is represented by

Where Y(S) is the Laplace transform of the output and X(S) is the Laplace transform of input and H(S) is the Transfer function of the system.

Transfer function explains mathematical function of the parameters of system, performing on the applied input in order to produce the required output.

Mathematical model:

To study and examine a control system, it is necessary to have some type of equivalent representation of the system. Such representation can be obtained using mathematical equations, governing the behavior of the system.

The set of mathematical equations, describing the dynamic characteristics of a system is called mathematical modelling of the system.

Analysis of Mechanical systems:

In Mechanical systems motion can be of different types. They are Translational, Rotational or combination of both.

Translational motion:

If the motion takes place along a straight line, then it is called Translational motion.

The translational mechanical systems are characterized by displacement, linear velocity and linear acceleration. The following elements are involved in the analysis of Translational Mechanical systems.

(i) Mass (ii) Spring (iii) Friction

Mass:

This is the important property of the system itself which stores the kinetic energy of the translational motion. The displacement of the mass always takes place in the direction of the applied force results in inertial force.This force is always proportional to the acceleration produced in mass by the applied force.

The applied force f(t) produces displacement x(t), in the direction of the applied force f(t). Force required for the same is proportional to acceleration.

Taking Laplace Transform and neglecting initial conditions, we can write

Linear spring:

In actual mechanical system, there may be an actual spring or indication of spring action because of elastic cable or a belt. Spring has the property to store the potential energy. The force required to cause the displacement is proportional to the net displacement in the spring. All springs are nonlinear in nature, but for small deformations their behavior can be approximated as a linear one.

Consider a spring with negligible mass and spring constant k, connected to a rigid support. Force required to cause displacement in the spring is proportional to displacement.

Now, consider the spring connected between two moving elements having masses M₁ and M₂, where force is applied to Mass M₁.

Now, Mass M₁ will get displaced by x₁(t) but mass M₂ will get displayed by x₂(t) as spring of constant k stores some potential energy and will be the cause for change in displacement. Now consider the free body diagram of spring as shown in below figure.

Net displacement in the spring is x₁(t)-x₂(t).

Taking Laplace transform on both sides, we get

Friction:

Whenever there is a motion, there exists friction. Friction may be between moving element and fixed support or between two moving surfaces. The friction is shown by dashpot or a damper.

Consider a Mass M having friction with a support with a constant B represented by a Dashpot. Friction will oppose the motion of Mass M and opposing force is proportional to the velocity of Mass M.

Taking Laplace transform, neglecting initial conditionals,

If friction is between two moving surfaces, in such case the opposing force is given by

Rotational Motion:

Motion about a fixed axis is called Rotational motion. In such systems, Force gets replaced by Torque. Mass gets replaced by Inertia. Spring and Friction behaves in the same manner in the rotational systems.

To determine the transfer function of the mechanical systems:

Step1:In mechanical systems, the differential equations governing the systems are obtained by writing force balancing equations at node in the system. The nodes are meeting points of elements. Generally the nodes are mass elements in the system.

Step 2:The linear displacement of the masses are assumed as x₁, x₂, x₃…..and assign a displacement to each node.

Step 3: Draw the free body diagram of the system. The free body diagram is obtained by drawing each mass separately and then marking all the forces acting on that mass. Always opposite force acts in a direction opposite to applied force. The mass has to move in the direction of applied force. Hence the displacement, velocity and acceleration will be in the same direction of the applied force.

Step 4:For each free body diagram, write one differential equation by equating the sum of applied forces equal to the sum of opposing forces.

Step 5: Take the Laplace transform of differential equations to convert them into algebraic equations. Then rearrange the S-domain equations to eliminate the unwanted variables and obtain the ratio between output variable and input variable. This ratio is the Transfer function of the system.

Ex:Write the differential equation of the governing system and determine the transfer function of the below system.

Free body diagram of Mass M₁:

Free body diagram of Mass M₂:

From equations (1) and (2)

Analogous systems:

In between Mechanical systems and Electrical systems there exists a fixed analogy and there exists a similarity between their equilibrium equations. Due to this, it is possible to draw an equivalent system which will behave exactly similar to the given mechanical system, this is called electrical analogous of given mechanical system and vice versa.

There are two methods of obtaining electrical analogous networks, namely

1. F-V analogy

2. F-I analogy

F-V Analogy:

Translational	Rotational	Electrical system
Force, F	Torque, T	Voltage , V
Mass, M	Inertia, J	Inductance , L
Friction constant, B	Tortional Friction Constant, B	Resistor, R
Spring constant, K	Tortional Spring constant, K	1/C
Displacement, x		Charge, q
Velocity, V

F-I Analogy:

Translational	Rotational	Electrical system
Force, F	Torque, T	Current, I
Mass, M	Inertia, J	Capacitance, C
Friction constant, B	Tortional Friction Constant, B	Resistor, 1/R
Spring constant, K	Tortional Spring constant, K	1/L
Displacement, x
Velocity, V

Steps to solve problem on Analogous system:

Step 1: Identify all the displacements due to the applied force. The elements spring and Friction between two moving surfaces cause change in displacement.

Step 2: Draw the equivalent Mechanical system based on node basis. The elements under same displacement will get connected in parallel under that code. Each displacement is represented by separate node. Element causing change in displacement is always between the nodes.

Step 3: write the equilibrium equations. At each node algebraic sum of all the forces acting at the node is zero.

Step 4: In F-V analogy use the following replacements and rewrite the equations.

Step 5: In F-I analogy use the following replacements and rewrite the equations.

Ex:For the physical system shown below draw its equivalent systemand write equilibrium equations

When a Force f(t) is applied to Mass M, it will get displaced by an amount of x and spring is connected to a fixed support and friction B. Both K and B will be the under influence of x only.

F-V Method:

Use the analogous terms and express all the terms in terms of current.

In F-V analogy, the quantities which are under the same displacement in Mechanical system, carry the same current in electrical analogous circuit.

The equivalent analogous electric circuit is given below.

Analogous to K is a capacitor C but its value is proportional to 1/K hence it is indicated by 1/K in the bracket near C.

F-I Analogous circuit:

Use the analogous terms and express all the terms in terms of current.

In F-I analogy, the quantities which are under the same displacement in mechanical system, have the same voltage across them in analogous electrical system.

Analogous to K is an Inductor L while to B is a Resistor R. But their values are proportional to reciprocals of K and B respectively. This is indicated by writing 1/K and 1/B in the brackets near L and R respectively in the above figure.

Block diagram representation:

If a given system is complicated, it is very difficult to analyze it as a whole. With the help of transfer approach, we can find transfer function of each and every block of the complicated system. And by showing the connection between the elements, complete system can be splitted into different blocks and can be analyzed conveniently.

Block diagram is the pictorial presentation of the given system. In block diagram, the interconnection of system components to form a system can be conveniently shown by the blocks arranged in proper sequence.

To draw the block diagram of a system, each element of a practical system is represented by a block. The block is called functional block.

Any block diagram consists of the following five basic elements associated with it.

1. Blocks 2.Transfer functions of the elements inside the blocks 3.Summingpoints 4.Takeoff points

5. Arrows

Rules for Block Diagram Reduction:

Rule 1: Combining the blocks in cascade

Now the three blocks can be replaced by a single block whose transfer function is

Rule 2:Combining parallel blocks

Now, the three blocks can be replaced by single block whose transfer function will be equal to

Rule 3: Moving the branch point ahead of the block

Rule 4: Moving the branch point before the block

Rule 5: Removing minor feedback loop

Rule 6: Shifting takeoff point after summing point

Rule 7: Shifting take point before a summing point

Rule 8: Shifting take point after summing point

Rule 9: Converting a Non unity feedback into Unity feedback

Ex: Determine the transfer function of the system shown below.

Sol:

The blocks G₂ and G₃ are in parallel. These two blocks can be replaced by a single block whose transfer function is given by G₂+G₃

The blocksG₁ and G₄ are in cascade/series. These two blocks can be replaced by a single block whose transfer function is G₁G₄.

The block with the transfer function G₁G₄ and H₁ form a feedback loop.

After removing the feedback loop

Signal Flow Graph Method:

The signal flow graph is used to represent the control system graphically and it was developed by S.J. Mason. Using Mason’s gain formula the overall gain of the system can be computed easily.

Properties of signal flowgraph:

1. The Signal flow graph is applicable only to LTI systems.

2. The signal in the system flows along the branches and along the arrowheads associated with the branches.

3. The signal gets multiplied by the branch gain or branch transmittance when it travels along it.

4. The value of the variable represented by any node is an algebraic sum of all the signals entering at the node.

5. The value of the variable represented by any node is available to all the branches leaving that node.

Terms used in SFG:

Source node: The node having only outgoing branches is known as source or input node.

Sink node: The node having only incoming branches is known as sink/output node.

Chain node: A node having incoming and outgoing node branches is known as chain node.

Forward path: A path from the input to output node is defined forward path.

Feedback loop:A path which originates from a particular node and terminating at the same node, travelling through at least one other node, without tracing any node twice is called feedback loop.

Self-loop: A feedback loop consisting of only node is called self-loop. A self-loop cannot appear while defining a forward path or feedback loop as node containing it gets traced twice which is not allowed.

Path gain: The product of branch gains while going through a forward path is known as path gain.

Dummy node: If there exists incoming and outgoing branches both at first and last node representing input and output variables, then as per definitions these cannot be called source and sink nodes. In such a case a separate input and output nodes can be created by adding branches with gain1. Such nodes are called dummy nodes.

Non touching loops: If there is no node common in between the two or more loops, such lops are said to be non-touching loops.

Loop gain: The product of all the gains of the branches forming a loop is called loop gain.

Mason’s gain formula:

K= number of forward paths

=gain of K^th forward path

System determinant = 1-[ ∑all individual feedback loop gains [including self loops]]+
[∑Gain Gain product of all possible combinations of two non-

touching loops] – [∑ Gain Gain product of all possible combinations

three non-touching loops] +…………….

Value of above by eliminating all loop gains and associated products which are touching to the K^th forward path.

Ex:Find the transfer function by using Mason’s Gain formula for the signal flow graph given below.

Two forward paths, K=2

Loops are

Out of these L₁ and L₂ is combination of non-touching loops

UNIT-2

Unit-II

Time response and stability

v Transfer function and types of input.

v Transient response of second order system for step input.

v Time domain specifications Characteristic Equation of Feedback control systems

v Types of systems, static error coefficients, error series,

v Concept of Stability, Routh-Hurwitz criterion for stability

v Root locus technique and its construction.

Time response: The response of the system, which is a function of time to the applied excitation is called Time response of a system.

Total response of the system can be divided into two parts.

1. Transient response 2. Steady state response

Transient response: The output variation during the time, it takes to achieve its final value is called as Transient response, To get the desired response, system must satisfactorilypass through transient period.

From the transient response we can get the following information about the system.

a. When the system has started showing its response to the applied excitation?

b. What is the rate of rise of output? From this, parameters of system can be designed which can withstand such rate of rise. It also gives the indication of speed of the system.

c. Whether the output is increasing exponentially or it is oscillating

d. If output is oscillating, whether it is overshooting its final value.

e. When it is settling down to its final value

Steady state response: It is the time response of the system which remains after complete transient response vanishes from the system output.

From the steady state response we can get the following information about the system.

a. How much away the system output is from its desired value which indicates error.

b. whether the error is constant or varying with time.

So, the entire information about system performance can be obtained from transient and steady state response.

Error:It is the difference between the actual output and the desired output.

Mathematically it is defined in Laplace domain as,

Steady state error in time domain is given by,

Therefore, Steady state error in S-domain is given by

Static error coefficients:

Consider a system having open loop transfer function G(S)H(S) and excited by

a) Reference input is Step of magnitude A:

R(S)=A/S

For a system selected, is a constant and is called Positional error coefficient of the system denoted as K_p.

And corresponding error is ,

So, whenever step input is selected as a reference input, positional error coefficient K_p will control the error in the system along with the magnitude of the input applied.

b) Reference input is ramp of magnitude A:

R(S)=A/S²

For a system selected, is a constant and is called Velocity error coefficient of the system denoted as K_v..

And corresponding error is ,

So, whenever step input is selected as a reference input, velocity error coefficient K_vwill control the error in the system along with the magnitude of the input applied.

c) Reference input is parabolic of Magnitude A:

R(S)=A/S³

For a system selected, is a constant and is called acceleration error coefficient of the system denoted as K_a.

And corresponding error is ,

So, whenever step input is selected as a reference input, acceleration error coefficient K_a will control the error in the system along with the magnitude of the input applied.

TYPE of a system:

Where K= Resultant system gain and j=TYPE of the system.

TYPE of the system means number of poles at origin of open loop TF G(s) H(S) of the system.

So j=0, TYPE zero system

J=2, TYPE two system

Analysis of Type 0, 1,2 systems:

TYPE 0:

For step input,

Now by increasing K,error can be reduced. But there is a limitation on the increase in value of K from stability point of view.

TYPE 1:

For step input,

TYPE 2:

For ramp input,

In general, for any TYPE of system more than zero, K_p will be infinite and error will be zero.

Ramp input:

TYPE 0:

For step input,

TYPE 0 systems will not follow ramp input of any magnitude and will give large error in the output which may damage the parameters of system.

TYPE 1:

For ramp input,

TYPE 1 systems follow the ramp type of input magnitude A with finite error.

TYPE 2:

For ramp input,

Hence all systems of type 2 and more than two follow ramp type of input with negligible small error.

Parabolic input:

TYPE 0:

For parabolic input,

TYPE 1:

For parabolic input,

For both TYPE 0 and 1 systems, error will be very large and uncontrollable if parabolic input is used. Hence parabolic input should not be used as reference input to excite TYPE 0 and 1 systems.

TYPE 2:

For parabolic input,

Hence TYPE 2 systems will follow parabolic input with finite error A/K which can be controlled by change in A or K or both.

Disadvantages of Static error coefficient Method:

1. Method cannot give the error if inputs are other than the three standard inputs.

2. Most of the times , method gives mathematical answer of the error as 0 or infiniteand hence does not provide precise value of the error.

3. This method does not provide variation of error with respect to time.

4. The method is applicable only to stable systems.

Generalised Error Coefficient method (Dynamic Error Coefficients):

Advantages:

i) It gives variation of error as a function of time.

Analysis of second order system

The closed loop transferfunction for a standard second order system is given by

Effect of on second order system performance:

Case 1:1< <∞

The roots are

i.e. real, unequal and negative, say –K₁and -K₂

Taking Inverse Laplace Transform,

where C_ss is the steady state output=A

The output is purely exponential. This means damping is so high that there are no oscillations in the output and purely exponential. Hence such systems are called Overdamped.

As value increases, output will take more time to reach its steady state and hence become sluggish and slow.

Case 2:

The roots are

i.e real, equal and negative.

Taking Inverse Laplace Transform,

This is purely exponential. But in comparison with Overdamped case, settling time required for this case is less and the system is called critically damped system. is the critical value of damping ratio because if it is decreased further, roots will become complex conjugates and this is the least value of damping ratio for which roots are real, negative and the output is exponential.

Case 3:

The roots are,

As the term is written as

Hence, roots are complex conjugate with negative real part.

Taking inverse Laplace Transform,

Where C_ss is the steady state output value = A

This response is oscillatory, with oscillating frequency but decreasing amplitude as it is associated with exponential term with negative index . Such oscillations are called damped oscillations and frequency of such oscillations is called damped frequency of oscillations which is nothing but

As damping is reduced, it is not sufficient to damp the oscillations completely hence oscillations are of damped type. As damping is not sufficient, systems are called underdamped systems.

Case 4:

The roots are

i.e. complex conjugates with zero real part . i.e. purely imaginary.

K” is a constant and C_ss steady state output value = A

The response is purely oscillatory, oscillating with constant frequency and amplitude. The frequency of such oscillations is the maximum frequency with which output can oscillate. As this frequency is under the condition i.e. no opposition condition, system oscillates freely and naturally. Hence this frequency is called natural frequency of oscillations denoted by rad/sec. the systems are classified as undamped systems.

Transient response specifications:

Delay time T_d:It is the time required for theresponse to reach 50% of the final value in the first attempt. It is given by

Rise time T_r: it is the time required for the response to rise from 10% to 90% of the final value for Overdamped systems and 0 to 100% of the final value for underdamped systems. The rise time is reciprocal of the slope of the response at the instant, the response is equal to 50% of the final value. It is given by

Where must be in radians.

Peak time T_p: It is the time required for the response to reach its peak value. It is also defined as the time at which response undergoes the first overshoot which is always peak overshoot.

Peak overshoot M_p: It is the largest error between reference input and output during the transient period. It is also defined as the amount by which output overshoots its reference steady state value during the first overshot.

Settling time T_s: This is the time required for response to decrease and stay within specified percentage of its final value (within tolerance band).

Time constant of a system =

T_s=4 Time constant

Routh’s stability criterion:

It is also called Routh’s array method or Routh-Hurwitz’s method. In this method the coefficients of characteristic equation are tabulated in a particular way.

Consider the general characteristic equation as,

Method of forming an array:

Coefficients for first two rows are written directly from characteristic equation.

From these two rows, next rows can be obtained as follows.

From 2^nd and 3^rd row, 4^th row can be obtained as

This process is to be continued till the coefficient for S⁰is obtained which will be a_n.From this array stability of a system can be predicted.

The necessary and sufficient condition for system to be stable is “All the terms in the first column of Routh’s array must have same sign. There should not be any sign changes in the first column of Routh’s array.

If there is any sign changes existing then,

a) System is unstable.

b) The number of sign changes equals the number of roots lying in the right half of the S-Plane.

Special case 1:

First element of any of the rows of Rouh’s array is zero and the same remaining row contains at least one non-zero element.

Following method is used to remove above said difficulty.

Substitute a small positive number in place of a zero occurred as a first element in a row. Completer the array with this number . Then examine the sign change by taking

To examine sign change,

Routh’s array is

As there are two sign changes, system is unstable.

Special case 2:

All the elements of a row are in a Routh’s array are zero.

Procedure to eliminate this difficulty:

i) Form an equation by using the coefficients of a row which is just above the row of zeros. Such an equation is called an Auxiliary equation denoted as A(S) and is given by

The coefficients of any row are corresponding to alternate powers of ‘s’ from 4 i.e. s², hence the term es² and so on.

ii) Take the derivative of an auxiliary equation with respect to ‘s’.

i.e.

iii) Replace row of zeros by the coefficients of

iv) Complete the array interms of these new coefficients.

Importance of an auxiliary equation:

Auxiliary equation is always the part of the original characteristic equation. This means the roots of the auxiliary equation are some of the roots of original characteristic equation.

Ex:For a system with characteristic equation

examine stability.

No sign change, hence no root is located in RHS of S-Plane. As a row of zeros occur, system may be marginally stable or unstable.

The roots of A’(s) = 0 are the roots of A(s) = 0.

As there are repeated roots on imaginary axis, system is unstable.

ROOT LOCUS:

The stability of the closed loop systems depends on the location of the roots of the characteristic equation i.e closed loop poles. Nature of the transient response is closely related to the location of the poles in the s-plane. It is advantageous to know how the closed loop poles move in the s-plane if some parameters of the system are varied. The knowledge of such movements of the closed loop poles with small changes in them parameters of the system helps in the design of any closed loop system.

Such movement of poles can be known by Rot locus method, introduced by W.R.Evans in 1948. This is graphical method, in which movement of poles in the s-plane is sketched when a particular parameter of system is varied from zero to infinity. For root locus method, gain is assumed to be a parameter which is to be varied from zero to infinity.

Basic concept of Root locus:

In general, the characteristic equation of a closed loop system is given by

1+G(S)H(S)=0

For root locus, the gain K is assumed to be a variable parameter and is a part of forward path of the closed loop system.

Consider the system shown in the above figure.

The characteristic equation becomes,

K is a variable parameter.

Rules for construction of root locus:

Rule 1: The root locus is symmetrical about the real axis.

Rule 2:Each branch of the root locus originates from an open loop pole corresponding to k=0 and terminates at either on a finite open loop zero (or open loop zero at infinity) corresponding to K=∞.

The number of branches of the root locus terminating on infinity is equal to (n-m), i.e, the number of open loop poles minus the number of finite zeros.

Rule 3:Segments of the real axis having an odd numbers of real axis open loop poles plus zeros to their right are parts of the root locus.

Rule 4:The (n-m) root locus branches that tend to infinity, do so along straight line asymptotes making angles with the real axis given by

Rule 5: the point of intersection of the asymptotes with the real axis at s=σ_Ais given by

Rule 6: The breakaway and break in points of the root locus are determined from the roots of the equation dK/dS =0.

Rule 7:The angle of departure from a complex open loop pole is given by

Where is the net angle contribution at the pole by all other open loop poles and zeros. Similarly the angle of arrival at complex open loop zero is given by

Where is the net angle contribution at the zero by all other open loop poles and zeros.

Rule 8:The intersection of root locus branches with the imaginary axis can be determined by use of Routh criterion, or by letting s=jω in the characteristic equation and equating the real part and imaginary part to zero, to solve for ω and k. the value of ω is the intersection point on imaginary axis and K is the value of gain at the intersection point.

Ex: For a unity feedback system, . Sketch the nature of the rootlocus. Comment on the stability of the system.

Step 1: P=3, Z=0, Number of branches, N=P=3. All branches will terminate at infinity. All the branches will start at open loop pole locations.

Step 2:Sketch the poles and zeros in the S-plane also identify the sections of real axis which lie in the root locus.

Step 3: Angle of Asymptotes

Three branches will approach to infinity, along the direction of the asymptotes.

Angles of Asymptotes,

Step 4: Centroid

Branches will approach to infinity along these lines which are asymptotes.

Step 5: Breakaway point

Characteristic equation is

Break away points =

As there is no root locus between -2 to -4, -3.15 cannot be a break point. It also can be confirmed by calculating ‘K’ for S= -3.15. it will be negative that confirms S=-3.15 is not a breakaway point.

For S = -3.15, K = -3.079. So, S = -0.845 is a breakaway point(K = 0.379)

Step 6: Intersection point with imaginary axis

The characteristic equation is

K_mar= 48 is the value which makes Row S¹as row of zeros.

Intersection of the Root locus with the imaginary axis at and corresponding value of K_mar=48.

For 0 < K < 48, all the roots will lie in the left half of the S- Plane. Hence the system is stable. For K= 48, a pair of dominant roots will lie in imaginary axis and remaining root will lie in the left half of the S-Plane. So, system is marginally stable oscillating at 2.82.rad/sec. For 48 < K < ∞, dominant roots will lie in the Right half of the S-Plane hence system is unstable.

UNIT-3

Unit-III

v Bode plots

v Frequency domain specifications Gain and Phase margin.

v Principle of argument Nyquist plot and Nyquist criterion for stability.

v Cascade and feedback compensation Phase lag, lead and lag-lead compensators

v PID controller.

Frequency domain specifications:

The basic objective of the control system design is to meet the performance specifications. These specifications are the constraints or limitations put on the mathematical functions describing the system characteristics.

Such frequency response specifications of a system are

Bandwidth, BW: It is defined as the range of frequencies over which the system will respond satisfactorily. It can also be defined as the range of frequencies in which the magnitude response is almost flat in nature.

Cut off frequency, f_c: The frequency at which the magnitude of the closed loop response is 3 dB down from its zero frequency value is called cut-off frequency.

Cut-off rate: The slope of the resultant magnitude curve near the cut-off frequency is called cut-off rate.

Resonant peak, M_p: It is the maximum value of magnitude of the closed loop frequency response Resonant Frequency: The frequency at which resonant peak occurs in closed loop frequency response is called resonant frequency.

Gain crossover frequency,ω_gc: The frequency at which magnitude of G(jω)H(jω) is unity is called gain crossover frequency.

Phase crossover frequency ω_pc: The frequency at which phase angle of G(jω)H(jω) is -180^o is called phase crossover frequency

Gain margin, GM: It is defined as the margin in gain allowable by which gain can be increased till system reaches on the verge of instability.

Phase margin, PM: It is the amount of additional phase lag that can be introduced in the system till system reaches on the verge of the instability.

Bode plot:

The bode plot is a frequency response plot of the transfer function of a system. A bode plot consists of two graphs. One is a plot of magnitude of a sinusoidal transfer function versus log

The other is a plot of the phase angle of a sinusoidal transfer function versus log .

The bode plot can be drawn for both open loop and closed loop transfer function. Usually it is drawn for open loop system. The main advantage of the bode plot is that multiplication of magnitudes can be converted into addition.

The basic factors that very frequently occur in a transfer function G(j ) are,

1. Constant gain ,

2. Integral factor,

3. Derivative factor,

4. First order factor in denominator,

5. First order factor in denominator,

6. Quadratic factor in denominator,

7. Quadratic factor in numerator,

Procedure for plotting the magnitude plot:

Step 1: Convert the given transfer function into bode form or time constant form. The bode form of the transfer function is

Step 2:

List the corner frequencies in the increasing order and prepare a table

Term

Corner frequency

rad/sec

Slope

db/dec

Change in slope db/dec

Step 3:

Choose an arbitrary frequency which is lesser than the lowest corner frequency. Calculate the db magnitude of K or K/(j ⁿ or K (j ⁿ at and at the lowest corner frequency.

Step 4:

Then calculate the gain at every corner frequency one by one by using the formula,

Step 5:

Chose an arbitary frequency which is greater than the highest corner frequency. Calculate the gain at by using the formula in step 4.

Step 6:

In a semilog graph sheet mark the required range of frequency on x-axis and the range of db magnitude on y-axis after choosing proper scale.

Step 7:

Mark all the points obtained in steps 3,4 and 5 on the graph and join the points by staright lines. Mark the slopes at every part of the graph.

Procedure for phase plot of bode plot:

The phase plot is an exact plot and no approximations are made while drawing the plot. Hence the exact phase angles of G(jω)are computed for various values of ω and tabulated.

Ex: Sketch the bode plot for the following transfer function and determine the system gain K for the gain crossover frequency to be 5 rad/sec.

Sol:

Let k=1,

Magnitude plot:

The corner frequencies are

The various terms of G(j ) are listed in below table in the increasing order of their corner frequency. Also the table shows the slope contributed by each term and the change in slope at the corner frequency.

Term	Corner frequency rad/sec	Slope db/dec	Change in slope db/dec
	-	+40
		-20	40-20=20
		-20	20-20=0

Choose a low frequency such that and choose a high frequency such that

Let

Let A=|G(jω)| in db

At =48 db

Phase plot:

Ω rad/sec
0.5	5.7	0.6	173.7
1	11.3	1.1	167.6
5	45	5.7	129.3
10	63.4	11.3	105.3
50	84.3	45	50.7
100	87.1	63.4	29.5

Calculation of K:

Given that the gain crossover frequency is 5 rad/sec. At ω=5 rad/sec the gain is 28 db. If gain crossover frequency is 5 rad/sec then at that frequency the db gain should be zero. Hence to every point of magnitude plot a db gain of -28 db should be added. The addition of -28db shifts the plot downwards. The corrected magnitude plot is obtained by shifting the plot with K=1 by 28 db onwards.The magnitude correction is independent of frequency. Hence the magnitude of -28 db is contributed by the term K. the value of K is calculated by equating 20logK to -28 db.

20 log K=-28 db

CONTROLLERS:

A controller is a device introduced in the system to modify the error signal and to produce a control signal. The controller modifies the transient response of the system. The controllers may be electrical, electronic, hydraulic or pneumatic, depending on the nature of the signal and the system.

Depending on the control actions provided the controllers can be classified as follows.

1. Proportional controller

2. Integral controller

3. Proportional + Integral controller

4. Proportional + Derivative Controller

5.Proportional + Integral + Derivative Controller

Proportional controller:

The proportional controller is a device that produces a control signal u(t) which is proportional to the input error signal, e(t)

In P- Controller,

Where Kp is proportional gain or constant.

The proportional controller improves the steady state tracking accuracy, disturbance signal rejection and the relative stability and also makes the system less sensitive to parameter variations. But increasing gain to very large values may lead to instability of the system.

Integral controller:

The integral controller is a device that produces a control signal u(t) which is proportional to the integral of the input error signal , e(t).

Where Ki = integral gain or constant

The Integral controller removes or reduces the steady state error. The drawback in integral controller is that it may lead to oscillatory response of increasing or decreasing amplitude which is undesirable and the system may become unstable.

Proportional + integral controller:

The proportional + Integral controller (PI-Controller) produces an output signal consisting of two terms – one is proportional to the error signal and the other proportional to the integral of error signal.

Where Kp = proportional gain

Ti = integral time

The advantages of both P-Controller and I-Controller are combined in PI controller. The proportional action increases the loop gain and makes the system less sensitive to variation of system parameters. The integral action eliminates or reduces the steady state error.

Proportional + Derivative Controller:

The proportional + Derivative controller (PD-Controller) produces an output signal consisting of two terms – one is proportional to the error signal and the other proportional to the Derivative of error signal.

Where Kp = Proportional gain

Td = Derivative time

The derivative control acts on rate of change of error and not on the actual error signal. The derivative control action is effective only during transient periods and so it does not produce corrective measures for any constant error.

Hence derivative controller is never used alone, but it is employed in association with proportional and integral controllers. Its disadvantage is that it amplifies the noise signals.

Proportional + Integral +Derivative controller:

The proportional + Integral + Derivative controller (PID-Controller) produces an output signal consisting of three terms – one is proportional to the error signal, another one is proportional to the integral of the error signal and the third one proportional to the Derivative of error signal.

The proportional controller stabilizes the gain but produces a steady state error. The integral controller reduces or eliminates the steady state error. The derivative controller reduces the rate of change of error.

Polar plot:

In polar plot , the magnitude of G(j )H( ) is plotted against phase angle of G(j )H( ) for various values of .

= Phase


0

-	-	-
-	-	-
-	-	-

This is the data required for the polar plot. For each value of M and corresponding to particular frequency decides a point as per the polar coordinate system. i.e for , and So it decides a point having polar coordinates .this is the tip of the phasor of magnitude plotted at an angle of

Fig:Polar plot

So, polar plot starts at a point representing magnitude and phase angle for While it terminates at appoint representing magnitude and phase angle for

Stability determination from polar plot:

1. If

2. If

3. If

Nyquist plot

The concept of Nyquist plot is based on polar plot which can be conveniently applied to the stability analysis of any kind of system.

Nyquist stability criterion:

If the G(s)H(s) contour in the G(S)H(S) plane corresponding to Nyquist contour in the s-plane encircles the point -1+j0 in the anticlockwise direction as many times as the number of right half s-plane poles of G(S)H(S), then the closed loop system is stable.

In examining the stability of linear control systems using the Nyquist stability criterion, we come across the following three situations.

1.No encirclement of -1+j0 point:This implies that the system is stable if there are no poles of G(S)H(S) in the right of s-plane. If there are poles on right half s-plane, then the system is unstable.

2. Anticlockwise encirclement of -1+j0 point: in this case the system is stable if the number of anticlockwise encirclement is same as the number of poles of G(S)H(S) in the right half s-plane. If the number of anticlockwise encirclements is not equal to the number of poles on right half s-plane then the system is unstable.

Procedure for investigating the stability using Nyquist criterion:

1. Choose a Nyquist contour, which encloses the right half of the S-plane except the singular points. The Nyquist contour encloses all the right half S-plane poles and zeros of G(S)H(S).

2. The Nyquist contour should be mapped in the G(S)H(S) plane using the function to determine the encirclement -1+j0 point in the G(S)H(S) plane.The Nyquist contour can be divided into four sections. The mapping of four sections in the G(S)H(S) plane can be carried section wise and then combined together to get entire G(S)H(S) contour.

3. In section C₁, the value of varies from 0 to The locus of G(j ) H(j ) as varied from 0 to will be the G(S)H(S) contour in G(S)H(S) plane corresponding to section C₁ in s-plane. This locus is the polar plot of G(j ) H(j ).

4. The section C₂ of Nyquist contour has a semi-circle of infinite radius. Therefore every point on section C₂ has infinite magnitude but the argument varies from Hence the mapping of section C₂ from s-plane to G(S)H(S) plane can be obtained by letting in G(S)H(S) and varying from

5. In section C₃, the value of varies from The mapping of section C₃is obtained by letting s=j in G(S)H(S) and varying from . The locus is inverse polar plot of G(j )H(j ). The inverse polar plot is given by the mirror image of polar plot with respect to real axis.

6. The section C₄ of Nyquist contour has a semicircle of radius of zero. Therefore every point on semicircle has zero magnitude but the argument varies from

Hence the mapping of section C₄ from s-plane to G(S)H(S) plane can be obtained by letting in G(S)H(S) and varying from

Compensators:

The compensator is a physical device. It may be an electrical network, mechanical unit, hydraulic or combinations of various types of devices. The commonly ised various electrical compensating networks are

1. Lead compensator

2. Lag compensator

3. Lag-Lead compensator

When as sinusoidal input is applied to network and it produces a sinusoidal state output having a phase lead with respect to input then the network is called Phase lead compensator.if the steady state output has phase lag then the network is called phase lag compensator. In the Lag-Lead network both phase lag and phase lead occurs but in the different frequency regions. The phase lag occurs in the low frequency region while the phase lead occurs in the high frequency region.

Lead Compensator:

Consider an electrical network which is a lead compensatingnetwork as shown in below figure.

Applying KCL to the output node,

I1+I₂=I

Taking Laplace Transform of the above equation,

[

This is generally expressed as,

Where T= and < 1

The lead compensator has zero at s= -1/T and pole at s = -1/

As 0 < , the zero is always located to the right of the pole. The pole zero plot is shown below.

The maximum lead angle provided by the lead compensator is

and the frequency at which maximum phase lead occurs is given by

From the above equation, we can say that is the geometric mean of two corner frequencies.

Effects of Lead compensation:

The various effects of a lead compensation are,

1. The lead compensator adds a dominant zero and a pole. This increases the damping of the closed loop system.

2. The increased damping means less overshoot, less rise time and less settling time. Thus there is improvement in the transient response.

3. It improves the phase margin of the closed loop system.

4. It increases bandwidth of the closed loop system. More the bandwidth, faster is the response.

5. The steady state error does not get affected.

Limitations of Lead Compensator:

1. Lead compensation requires additional increase in gain to offset the attenuation inherent in the lead network. Larger gain requirement means, larger space, more elements, greater weight and higher cost.

2. More bandwidth is sometimes not desirable. This is because the noise entering the system at the input may become objectionable.

Lag Compensator:

Applying KVL to the loop,

Now the output equation is,

This is generally expressed as,

Where T=R₂C,

The Lag compensator has zero at and a pole at . As the pole is always located to the right of the zero.

The frequency at which the phase lag is at its maximum is given by

The two corner frequencies of lag compensator are,

From the above equation, we can say that is the geometric mean of two corner frequencies.

Effects and Limitations of Lag Compensator:

The various effects and limitations of Lag Compensator are,

1. Lag compensator allows high gain at low frequencies thus it basically a low pass filter. Hence it improves the steady state performance.

2. In lag compensation, the attenuation characteristics is used for the compensation. The phase lag characteristics is of no use in the compensation.

3. The attenuation due to lag compensator shifts the gain crossover frequency to a lower frequency point. Thus the bandwidth of the system gets reduced.

4. Reduced bandwidth means slower response. Thus rise time and settling time are usually longer. The transient response lasts for long time.

5. Lag Compensator approximately acts as proportional plus integral controller and thus tends to make system less stable.

Lag-Lead Compensator:

A combination of Lad and Lead compensators is nothing but a Lag-Lead Compensator.

Now,

The output equation is,

Taking Laplace transform on both sides,

After solving the above two equations in Laplace domain, we get

Where T₁=R₁C₁, T₂=R₂C₂

Where

The phase lead portion involving T₁, adds phase angle while the phase lag portion involving T₂ provide attenuation near and above the gain crossover frequency.

The poles are while the zeros are located at

Effects of Lag-Lead Compensator:

Lag-Lead Compensator is used when both fast response and good static accuracy are desired. Use of Lag-Lead Compensator increases the low frequency gain which improves the steady state. While at the same time it increases the bandwidth of the system, making the system response very fast.

In general, the phase lead portion of this compensator is used to achieve large bandwidth and hence shorter rise time and settling time. While the phase lag portion provides the major damping of the system.

UNIT-IV

Contents:

v Digital control, advantages and disadvantages

v Digital control system architecture.

v The discrete transfer function sampled data system

v Transfer function of sample data systems.

v Analysis of Discrete data systems

When the signal or information at any point in a system is in the form of discrete pulses, then the system is called discrete data system. In control engineering the discrete data system is popularly known as sampled data system.

The control system becomes a sampled data system in any one of the following situations.

1. When a digital computer or microprocessor or digital device is employed as a part of the control loop.

2. When the control components are used on time sharing basis.

3. When the control signals are transmitted by pulse modulation.

4. When the output or input of a component in the system is a digital or discrete signal.

The controllers are provided in control systems to modify the error signal for better control action. If the controllers are constructed using analog elements, then they are called analog controllers and their input and output are analog signals.

A digital controller can be employed to implement complex or time shared control functions. The digital controller can be a special purpose computer or a general purpose computer or it is constructed using non-programmable devices.

A sampled data controller system using digital controller is shown in below figure.

e(t)-Error signal(Analog) f(kT)-Digital error signal

u(t)-Control signal (Analog) g(kT)-Digital control signal

the input and output signal in a digital computer will be digital signals, but the error signal (input to the controller) to be modified by the controller and the control signal (output of the controller) to drive the plant are analog in nature. Hence a sampler and an analog-to-digital converter (ADC) are provided at the computer input. A digital to Analog Converter (DAC) and a hold circuit are provided at the computer output.

The sampler converts the continuous time error signal into a sequence of pulses and ADC produces a binary code (binary number) for each sample. These codes are the input data to the Digital computer which process the binary codes and produces another stream of binary codes as output. The DAC and hold circuit converts the output binary codes to continuous time signal (Analog signal) called control signal. This output control signal is used to drive the plant.

Advantages of Digital Controllers:

1. The digital controllers can perform large and complex computation with any desired degree of accuracy at very high speed. In analog controllers, the cost of controllers increases rapidly with the increase in complexity of computation and desired accuracy.

2. The digital controllers are easily programmable and so they are more versatile.

3. Digital controllers have better resolution.

Advantages of Sampled Data Control Systems:

1. The sampled data systems are highly accurate, fast and flexible.

2. Use of time sharing concept of Digital computer results in economical cost and space.

3. Digital transducers used in the system have better resolution.\

4. The digital components used in the system are less affected by noise, non linearities and transmission errors of noisy channel.

5. The sampled data system requires low power instruments which can be built to have high sensitivity.

6. Digital coded signals can be stored, transmitted, retransmitted, detected, analyzed or processed as desired.

7. The system performance can be modified by compensation techniques.

Analysis of Sampler and Zero – Order Hold:

Consider a pulse sampler with zero-order hold (ZOH) is shown in below figure. Let the output of sampler be a pulse train of pulse width For each input pulse, the ZOH produces a pulse of duration T, where T is the sampling period.

Fig: Pulse sampler with ZOH

Fig: Equivalent representation of pulsesampler with ZOH

The Transfer function of the sample and hold circuit is given by

Frequency response characteristics of zero order holding device:

Analysis of system with impulse sampling:

Consider a linear continuous time system fed from an impulse sampler as shown in below fig. Let H(S) be the transfer function of the system in s-domain. In such a system we are intersected in reading the output at sampling instants. This can be achieved by means of a mathematical sampler or read-out sampler.

Fig: Linear continuous time system with impulse sampled input

In sampled data control systems, the z-domain transfer function can be obtained by taking the z-transform of H(S).

H(z) = Z{H(s)}

Procedure to find the z-Transfer function from s-domain transfer function:

1. Determine h(t) from H(s), where h(t)=L^-1[H(s)]

2. Determine the discrete sequence h(kT) by replacing t by kT in h(t).

3. Take z-transform of h(kT), which is z-transfer function of the system (i.e., H(Z)=z{h(kT)}).

Ex:

Sol:

h(t) = ate^-at

h(kT) =a kTe^-akT

H(z) = akT z{ke^-akT}

Analysis of sampled data control systems using Z Transforms:

The following points are used to determine the output in z-domain and hence the z-Transfer function of the sampled data control systems.

1. The pulse sampling is approximated as impulse sampling.

2. The ZOH is replaced by a block with transfer function, G₀(s) = (1-e^-sT)/s.

3. When the input to a block is impulse sampled signal then the z-Transform of the output of the block can be obtained from the z-transform of the input and the z-Transform of the s-domain transfer function of the block. In determining the output of a block one may come across the following cases.

Case (i): The impulse sampler is located at the input of a block as shown in below figure.

In this case, C(z) = G(z) R(z)

Here, G(z) = Z{G(s)} ; R(z) = Z{R’(s)} and R’(s)=L[r’(t)]

Case (ii):The impulse sampler is located at the input of s-domain cascaded blocks as shown in below figure.

In this case, C(z) = Z{G₁(s) G₂(s)}R(z) = G₁G₂(z) R(z)

Case (iii):The impulse sampler is located at the input of each block as shown in below figure.

In this case, C(z) = G₁(z) G₂(z) R(z)

Here, G₁(z)=Z{G₁(s)} and G₂(z) = Z{G₂(s)}

Case (iv): The impulse sampler is located at the input of ZOH in cascade with G(s) as shown in below figure.

In this case, C(z) = Z{G₀(s) G(s) R(z)} = (1-Z^-1) Z{G(s)/s}R(z)

The z–domain and s-domain relationship:

z = e^sT

Stability analysis of sampled data control systems:

The sampled data control system is stable if all the poles of the z-transfer function of the system lies inside the unit circle in z-plane. The poles of the transfer function are given by the roots of the characteristic equation. Hence the system stability can be determined from the roots of the characteristic equation.

The following methods are available for the stability analysis of sampled data control systems using the characteristic equation.

1. Jury’s stability test

2. Bilinear Transformation

Jury’s stability test:

The Jury’s stability test is used to determine whether the roots of the characteristic polynomial lie within a unit circle or not. The Jury’s test consists of two parts. One test for necessary condition and another test for sufficient condition for stability.

Let F(z) be the n^th order characteristic polynomial of a sampled data control system.

Where are constant coefficients.

The necessary condition to be satisfied for the stability of the system with characteristic polynomial, F(z) are

F(1) > 0 and (-1)ⁿ> 0

Method for testing sufficiency:

To test the sufficiency,prepare a table as shown below using the coefficients of the characteristic polynomial F(z). the table consists of (2n-3) rows, where n is the order of the characteristic equation.

Row	Z⁰	Z¹	Z²	…	Z^n-k	…	Z^n-2	Z^n-1	Zⁿ
1	a₀	a₁	a₂		a_n-k		a_n-2	a_n-1	a_n
2	a_n	a_n-1	a_n-2		a_k		a₂	a₁	a₀
3	b₀	b₁	b₂				b_n-2	b_n-1
4	b_n-1	b_n-2	b_n-3				b₁	b₀
5	c₀	c₁	c₂				c_n-2
6	c_n-2	c_n-3	c_n-4				c₀
. . .	. . .	. . .
2n-5	s₀	s₁	s₂	s₃
2n-4	s₃	s₂	s₁	s₀
2n-3	r₀	r₁	r₂

and so on

The first column elements of the table are used to check the following (n-1) conditions. These (n-1) conditions are the sufficient conditions for stability of the system.

Stability analysis using Bilinear Transformation:

The bilinear transformation maps the interior of unit circle in the z-plane into the left half of the r-plane. In this transformation,

The transformation is performed by substituting in the characteristic equation of the system.Find the roots of the new characteristic equation. If all the rots are lying in the left half of the r-plane, then the system is said to be stable. The routh stability criterion can be applied to the new characteristic equation to determine if all its roots are lie in the left half of the r-plane.

Ex:Check for the stability of the sampled data control systems represented by the characteristic equation

Sol:

F(z) =

F(1) = 5

` (-1)ⁿ F(-1) = (-1)² [5(-1)²-2(-1)+2]=9

Since F(1) > 0 and (-1)ⁿ F(-1) > 0; the necessary conditions for stability are satisfied.

Check for sufficient condition:

The sufficient condition for stability can be checked by constructing a table consisting of (2n-3) rows as shown below.

Here n=2, therefore (2n-3) = 1 and so the table consists of only one row.

Row	Z⁰	Z¹	Z²
1	2	-2	5

The necessary condition to be satisfied is |a₀| < |a₂|

Here |a₀|=2 and |a₂|=5 and so the condition |a₀| < |a₂|is satisfied.

The necessary and sufficient condition for stability are satisfied. hence the system is stable.

UNIT V

Contents

v Concept of state and state variables

v State models of linear time invariant systems

v State transition matrix

v Solution of state equations

v Controllability and Observability

The limitations of Conventional approach are

1. The method is not applicable for Multiple Input Multi Output systems.

2. It gives analysis of systems for specific types of inputs like step, ramp etc.

3. It is only applicable for Linear Time Variant Systems.

Hence it is necessary to use a method of analyzing systems which overcomes most of the above said difficulties. The modern method, uses the concept of total internal state of the system considering all initial conditions. This technique is called state space analysis.

Advantages of state space approach:

1. This method takes into the effect of initial conditions.

2. It can be applied to Nonlinear as well as Time varying systems.

3. It can be applied to Multiple Input and Multiple output systems.

4. Any type of the input can be considered for designing the system.

5. The state variables need not be the physical quantities of the system.

Definitions:

State: The state of a dynamic system is defined as a minimal set of variables such that the knowledge of these variables at t=t₀ together with the knowledge of inputs for completely determines the behavior of the system for

State variables:The variables involved in determining the state of a dynamic system X(t) are called State variables.

State Vector:The ‘n’ state variables necessary to describe the complete behavior of the system can be considered as ‘n’ components of a vector x(t) called the state Vector at time ‘t’.

State Space:The space whose coordinate axes are nothing but the n state variables with time as the implicit variable is called the state space.

State Trajectory:It is the locus of the tips of the state vectors, with time as the implicit variable.

State model of single input single output system:

State equation:

Output equation:

Matrix A is called Evolution Matrix

B is called Control Matrix

C is called Observation Matrix

D is called Transmission Matrix.

In electrical networks currents through Inductors and Voltage across the Capacitors are considered as state variables.

State variable representation using Physical variables:

The state variables are minimum number of variables which are associated with all the initial conditions of the system. To obtain the state model for a given system, it is necessary to select the state variables. Many a times, the various physical quantities of system itself are selected as the state variables.

For the electrical systems, the currents through various inductors and the voltage across the capacitors are selected to be the state variables. Then by any method of network analysis, the equations must be written in terms of the selected state variables, their derivatives and the inputs. The equations must be rearranged in the standard form so as to obtain the required state model.

Ex: Obtain the state model of the given electrical system.

There are two energy storage elements L and C. so, the two state variables are current through inductor i(t) and and voltage across capacitor i.e. v₀(t).

X₁(t) = i(t) and X₂(t) = v₀(t)

And U(t) = v_i(t) = input variable

Applying KVL to the loop,

Arranging it for di(t) / dt,

While

Advantages:

The advantages of using physical variables as the state variables are,

1. The physical variables which are selected as the state variables are the physical quantities and can be determined.

2. As state variables can be physically measured, the feedback may consist the information about state variables in addition to the output variables. Thus design with state feedback is possible.

3. Once the state equations are resolved and solution is obtained, directly the behavior of various physical variables with time is available.

State diagram of standard state model:

State space representation using Phase Variables:

The phase variables are those state variables which are obtained by assuming one of the system variable as a state variable and other state variables are the derivatives of the selected system variable. Such set of phase variables is easily obtained if the differential equation of the system is known or the transfer function is available.

Ex: Construct the state model using phase variables if the system is described by the differential equation, Draw the state diagram.

Sol:

Choose output Y(t) as the state variable X₁(t) and successive derivatives of it give us remaining state variables. As order of the equation is 3, only 3 state variables are allowed.

And

Thus

To obtain substitute state variablesobtained in the differential equation.

Therefore,

The output is, Y(t)=X₁(t)

Y(t)= C X(t)+D U(t)

Where

This is the required state model using phase variables.

The state diagram is shown in below figure.

State model from Transfer function:

Consider a system characterized by the differential equation containing derivatives of the input variable U(t) as,

In such a case, it is advantageous to obtain the transfer function, assuming zero initial conditions. Taking Laplace Transform of both sides of the above equation and neglecting initial conditions we get,

By using direct decomposition of transfer function, we can obtain state model from the transfer function,

UsingDirect Decomposition approach:

This is also called direct programming. in this method, the denominator of transfer function is rearranged in a specific form. To understand the rearrangement, consider an element with the transfer function . From block diagram algebra, the transfer function of minor feedback loop is for negative feedback.

Let

Where

The feedback is negative and the transfer function can be simulated as shown in the below figure with a minor feedback loop.

Now, if such a loop is added in the forward path of another such loop then we get the block diagram as shown in the below diagram.

The transfer function now becomes,

Where X=(s+a)

If a number of such loops are added in the forward path, then assign output of each integrator as the state variable, state model in the phase variable form can be obtained.

Ex: Obtain the state model by direct decomposition method of a system whose transfer function is

To simulate numerator, shift take-off point for 6s and shift twice for . Assign output of each integrator as the state variable. Now complete the state diagram.

While output,

Where,

Advantages:

The various advantages of phase variables i.e. direct programming method are,

1. Easy to implement

2. The phase variables need not be physical variables hence mathematically powerful to obtain state model.

3. It is easy to establish the link between the transfer function design and time domain design using phase variables.

Transfer function from State Model:

Characteristic Equation:

Solution of State Equation:

And the state transition matrix is

Properties of State Transition Matrix:

Laplace Transform method of finding State Transition Matrix:

Applying Laplace Transform to the above equation

Solution of state equation by Laplace Transform Method:

The Laplace transform method converts integro-differential equation to simple algebraic equations. Due to this important property, it is very convenient to use Laplace transform method to obtain the solution of state equation.

Consider the non-homogeneous state equation as

Taking Laplace Transform on both sides,

Multiplying both sides by

Ex: A linear time invariant system is characterized by the homogeneous stare equation:

Compute the solution of homogeneous equation, assume the initial state vector:

Sol:

From the given model,

This is the required solution.

Controllability and Observability:

A system is said to be completely state controllable if it is possible to transfer the system state from any initial state X(t₀) to any other desired state X(t_f) in a specified finite time interval t_f by a control vector U(t).

Kalman’s test for Controllability:

Consider n^thorder multiple input linear time invariant system represented by its state equation as,

Where A has an order n×n matrix

And U(t) is m×1 vector.

X(t) is n×1 state vector.

The necessary and sufficient condition for the system to be completely state controllable if the rank of the composite matrix Q_cis ‘n’.

The composite matrix Q_cis given by,

Q_c=

Ex:

Consider the system with state equation

Evaluate the state controllability by kalman’s test.

Sol:

Q_c= [B AB A²B]……..n=3

is non-singular.

Hence the rank of Q_C is 3 which is ‘n’.

Thus the system is completely state controllable.

Observability:

A system is said to be completely observable, if every state X(t₀) can be completely identified by measurements of the outputs Y(t) over finite time interval.

Kalman’s test for Observability:

Consider n^thorder multiple output linear time invariant system, represented by its state equation as,

andY(t) = C X(t)

The system is completely observable if and only if the rank of the composite matrix Q₀ is ‘n’.

The composite matrix Q₀ is given by,

Q₀ = [C^T A^T C^T………..(A^T)^n-1 C^T]

Ex: Evaluate the observability of the system with

observable.

Sol:

Consider the determinant,

Hence a non-zero determinant existing in Q₀ is having order less than 3.

Therefore, rank 0f Q₀≠ n

Hence the system is not completely .

QUESTION BANK

· Short answer questions.

· Long answer questions.

UNIT -1
PART-A: SHORT ANSWER QUESTIONS
S.NO	QUESTION	Blooms Taxonomy Level	Course Outcome
1	Write the merits and demerits of open loop and closed loop systems	L1	CO1
3	Differentiate between Force-voltage and Force-current analogy	L1	CO1
4	Why negative feedback is preferred in control systems	L1	CO1
5	What are the characteristics of Servo motors	L1	CO1
6	What is a synchro? Write the applications of synchros	L1	CO1
7	Explain force-voltage analogy	L1	CO1
8	Write Mason’s gain formula	L1	CO1
9	.What are the basic properties of signal flow graphs	L1	CO1
10	Write short notes on AC Servomotor	L1	CO1
PART-B: LONG ANSWER QUESTIONS
S.NO	QUESTION	Blooms Taxonomy Level	Course Outcome
1	For the given below system write the equilibrium equations and draw the F-V and F-I analogous circuits.	L3	CO1
2	For the given rotational system, obtain the electrical analogous system.	L3	CO1
3	Find the transfer function of the given figure using block diagram reduction technique.	L3	CO1
4	Find the transfer function of the given figure using Signal flow graph method.	L3	CO1
UNIT –II
PART-A: SHORT ANSWER QUESTIONS
S.NO	QUESTION	Blooms Taxonomy Level	Course Outcome
1	State the limitations of Routh-Hurwitz criterion.	L1	CO2
2	Define the time domain specifications for a second order system.	L1	CO2
3	Using Routh’s array determine the stability of the system with the characteristic equation s4+2s³+18s²+4s+3=0	L2	CO2
4	Find the ststic error constants for a unity feedback system given by G(S)=20(S+3)/S(s+5)(S+6)	L2	CO2
5	Sketch the response of a second order underdamped system	L2	CO2
6	Define the specifications of 2^nd order system	L1	CO2
7	Explain the error series	L2	CO2
8	What are the advantages of the Root Locus	L2	CO2
9	What are the advantages of and drawbacks of Routh stabi;lity criterion	L1	CO2
10	A unity feedback system has a open loop transfer function of G(s) = . Determine the steady state error for unit ramp input.	L2	CO2
11	The step response of a system is (1-10e^-t)u(t),find the transfer function of a system.	L2	CO2
12	Closed loop transfer function of a unity feedback system is given by 1/(s²+s+1).Find the velocity error coefficient of the system.	L2	CO2
13	Given r(t)=(1-t²)3u(t).Find steady state error	L2	CO2
PART-B: LONG ANSWER QUESTIONS
S.NO	QUESTION	Blooms Taxonomy Level	Course Outcome
1	A unity feedback system is characterized by the open loop transfer function G(S)=1/s (0.5s+1)(0.2s+1). Determine the steady state errors for unit-step, unit-ramp and unit acceleration input.	L3	CO2
2	A unity feedback system has open loop transfer function G(S)=K/s(s²+8s+32).Sketch the root locus.	L3	CO2
3	A unity feedback system is characterized by an open loop transfer function G(S)=K/s(s+10). Determine the gain K so that the system will have a damping ratio of 0.5. For this value of K determine the settling time, peak overshoot and time to peak overshoot for unit step input.	L3	CO2
4	OLTF of a certain feedback control system is K/s(s+4) (s²+4s+5). Draw the root locus diagram of the system	L3	CO2
7	State and explain the Routh stability criterion.	L3	CO2
8	Determine the stability of the system represented by the characteristic equation s⁵+S⁴+6s³+12s²+18s+6=0	L3	CO2
9	A unity feedback control system has an open loop transfer function G(S) = (1+0.4S)/s(s+0.6). Obtain the unit step response of the system.	L3	CO2
UNIT –III
PART-A: SHORT ANSWER QUESTIONS
S.NO	QUESTION	Blooms Taxonomy Level	Course Outcome
1	What are the limitations of bode plots?	L1	CO3
2	Define gain margin and phase margin	L1	CO3
3	State the Nyquist stability criterion	L1	CO3
4	Explain PID controller	L1	CO3
5	Explain PI,PD controller	L1	CO3
6	How the roots of the characteristic equations are related to stability	L2	CO3
7	Sketch the bode plot of 6/s(s+6)	L2	CO3
8	What is lead compensator?	L1	CO3
9	What is Bode plot? Write its advantages.	L1	CO3
10	Draw the Bode plot of a Lag network	L1	CO3
11	Explain PI,PD controller	L1	CO3
12	How the roots of the characteristic equations are related to stability	L1	CO3
13	What are the limitations of bode plots?	L1	CO3
14	Define gain margin and phase margin	L1	CO3
15	State the Nyquist stability criterion	L1	CO3
PART-B: LONG ANSWER QUESTIONS
S.NO	QUESTION	Blooms Taxonomy Level	Course Outcome
1	Derive the transfer function of Lag-Lead network and what are the effects of Lag-Lead network	L2	CO3
2	Explain PI,PID controllers	L2	CO3
3	For the system with TF G(S) =400(S+2)/2S(S+5)(S+10).Draw the bode plot and obtain the gain crossover frequency and phase crossover frequency.	L3	CO3
4	For the system with TF G(S) =5(2S+1)/(4S+1)(0.25S+1).Draw the bode plot and obtain the gain crossover frequency and phase crossover frequency.	L3	CO3
5	Construct the complete Nyquist plot for a unity feedback control system whose open loop transfer function is G(S)H(S=K/s(s2+2s+2).Find maximum value of K for which the system is stable	L3	CO3
UNIT –IV
PART-A: SHORT ANSWER QUESTIONS
S.NO	QUESTION	Blooms Taxonomy Level	Course Outcome
1	What are the advantages of digital control systems over analog control systems	L1	CO4
2	What is a Digital controller	L1	CO4
3	What is zero order hold circuit	L1	CO4
4	What is the equivalent representation of pulse sampler with ZOH	L2	CO4
5	Sketch the frequency response of ZOH device	L2	CO4
6	What are the methods available for the stability analysis of sampled data control system?	L1	CO4
7	What are the necessary conditions to be satisfied for the stability of sampled data control system?	L1	CO4
8	What is bilinear transformation?	L1	CO4
PART-B: LONG ANSWER QUESTIONS
S.NO	QUESTION	Blooms Taxonomy Level	Course Outcome
1	For the sampled data control system shown in below figure, find the response to unit step input G(s) = 1/(s+1).	L3	CO4
2	Check the stability of the sampled data control systems represented by the characteristic equation z³-0.2z²-0.25z+0.05=0 using Jury’s stability criterion.	L3	CO4
3	Check the stability of the sampled data control systems represented by the characteristic equation z⁴-1.7z³+1.04z²-0.268+0.024=0 using bilinear transformation method.	L3	CO4
UNIT -V
PART-A: SHORT ANSWER QUESTIONS
S.NO	QUESTION	Blooms Taxonomy Level	Course Outcome
1	What are the advantages of a state variable analysis	L1	CO5
2	What are the drawbacks in transfer function model analysis	L1	CO5
3	What is state and state variable	L1	CO5
4	What is a state vector	L1	CO5
5	What are phase variables	L1	CO5
6	What is state transition matrix	L1	CO5
8	Write the properties of state transition matrix	L1	CO5
9	Write the solution of homogeneous state equation	L1	CO5
10	Write the solution of non-homogeneous state equation		CO5
11	Define Controllability and Observability		CO5
PART-B: LONG ANSWER QUESTIONS
1	Construct a state model for a system characterized by the differential equation,	L3	CO5
2	Obtain the state model of the system whose transfer function is given as,	L3	CO5
3	Consider the Matrix A. Compute	L3	CO5
4	A linear time invariant system is characterized by homogeneous state equation Compute the solution of the homogeneous equation, assuming the initial state vector =	L3	CO5

PREVIOUS UNIVERSITY QUESTION PAPERS

INTERNAL QUESTION PAPERS WITH KEY

MATRUSRI ENGINEERING COLLEGE

SAIDABAD, HYDERABAD – 500 059

DEPARTMENT OF ELECTRONICS &COMMUNICATION ENGINEERING

Automatic Control Systems Scheme and Key for Internal-I

Academic Year: 2017-18 IIIYear I SEM

1.Classify control systems (L-1,CO-1)

Classification-2M

2.What are the different time domain specifications of second order systems (L-1,CO-2)

Spcifications-2M

3.Explain Routh’s criterion (L-2,CO-2)

Explanation-2M

Part-B 7x2=14 Marks

4.Find the stability of the system described by the characteristic equation

by making use of Routh-Hurwitz criterion. (L- 3,CO-2)

Statement for the stabiliy-2M

Solution using Routh’s criterion-5M

5 .Find the transfer function of a system using block diagram technique (L-3,CO-1)

Rules-3M

Solution-4M

6.Find transfer function of a system using signal flow graph method (L-3,CO-1)

Rules-3M

Solution-4M

MATRUSRI ENGINEERING COLLEGE

SAIDABAD, HYDERABAD – 500 059

DEPARTMENT OF ELECTRONICS &COMMUNICATION ENGINEERING

Automatic Control Systems Scheme and Key for Internal-II

Academic Year: 2017-18 IIIYear I SEM

1. Whatis lead compensator and when it is preferred (L-1,CO-3)

Definition-1M

Application-1M

2. Explain Nyquist stability criterion (L-2,CO-3)

Statement-1M

3. Explain Controllability and observability (L-2,CO-5 )

Controllability-1M

Observabilit -1M

4. Develop the bode plot of a given system G(S)=

Determine GM, PM, wgc and wpc (L- 3,CO-3)

Slope at corner frequencies-1M

Formulae to calculte gain-2M

Gain margin-2M

Phase margin-2M

5. For the sampled data system shown in below (L-3, CO-4)

(a) Find the discrete transfer fuction

Procedure-3.5M

(b) Unit step response

Procedure-3.5M

6. An LTI system is characterized by the homogeneous state equation (L-3, CO-5)

Compute the solution of the equation, assuming initial state vector as

state space representation-2M

Procedure-5M

MATRUSRI ENGINEERING COLLEGE

SAIDABAD, HYDERABAD – 500 059

DEPARTMENT OF ELECTRONICS &COMMUNICATION ENGINEERING

Automatic Control Systems Scheme and Key for Internal-I

Academic Year: 2015-16 IIIYear I SEM

1. What are the characteristics of negative feedback ?

Characteristics-2M

2. Find the static error constants K_p, K_v& K_a for a unity feedback system given by

G(s) = 20(s+3)/ s(s+5)(s+6). Find K_p, K_v& K_a

_Formulae-1M

_Solution-1M

3. State the Mason’s Gain Formula.

Formula-2M

4. Obtain the transfer function of the mechanical system shown in Fig. 1.

G₁ 111111111111111233333333333R5555

G₂ 111111111111111233333333333R5555

R(s) 111111111111111233333333333R5555

H₂ 111111111111111233333333333R5555

H₁ 111111111111111233333333333R5555

C(s) 111111111111111233333333333R5555

Fig. 1

Rules-3M

Procedure-4M

5. A second-order system is represented by a transfer function

, where, Q_o(s) is the proportional output and T is the input torque. A step input 10N-m is applied to the system and test results are given below.

a) Peak overshoot M_p = 6%

Formula-1M

Solution-1M

b) Peak time t_p = 1 sec

Formula-1M

Solution-1M

c) the steady-state output of the system is 0.5 radian. Determine the values of J, K and F.

Formula-1M

Solution-2M

5.Obtain the transfer function of the system of Fig. 2.

Fig. 2

C(s) 111111111111111233333333333R5555

R(s) 111111111111111233333333333R5555

G₁ 111111111111111233333333333R5555

G₂ 111111111111111233333333333R5555

G₃ 111111111111111233333333333R5555

G₄ 111111111111111233333333333R5555

G₅ 111111111111111233333333333R5555

G₆ 111111111111111233333333333R5555

G₇ 111111111111111233333333333R5555

G₉ 111111111111111233333333333R5555

G₁₀ 111111111111111233333333333R5555

G₈ 111111111111111233333333333R5555

Formula-1M

Identyfying Individual loops-2M

Solution-4M

MATRUSRI ENGINEERING COLLEGE

SAIDABAD, HYDERABAD – 500 059

DEPARTMENT OF ELECTRONICS &COMMUNICATION ENGINEERING

Automatic Control Systems Scheme and Key for Internal-II

Academic Year: 2016-17 IIIYear I SEM

1. Derive the transfer function of ZOH (L-1,CO-4)

2. An LTI system is described by the following state equation. Compute on stability of the system(L-3,CO-5)

3. For the system described in Q.2 check the controllability. (L-3,CO-5)

Part-B 7x2=14 Marks

4. For the sampled data system of given fig, find the response to unitstep input given G(S)=1/(S+1) (L- 3,CO-4)

(i) If D(Z)=1 and

(ii) If D(Z)=Z-0.367

5 .(i) Explain architecture of digital control system (L-2,CO-4)

(iii) List out any two advantages and disadvantages of Digital control sysytem (L-2,CO-4)

6.A system is described by the following state equation where u is unit step function

(L-3,CO-5)

The output equation is given by

(i) Calculate state transition matrix.

(ii) Calculate X(t)

(iii) Calculate Y(t)

MATRUSRI ENGINEERING COLLEGE

SAIDABAD, HYDERABAD – 500 059

DEPARTMENT OF ELECTRONICS &COMMUNICATION ENGINEERING

Automatic Control Systems Scheme and Key for Internal-I

Academic Year: 2016-17 IIIYear I SEM

1. Desceribe two block diagram reduction rules with example (L-1,CO-1)

Rules-1M

Examples-1M

2. Explain & derive the error constsants Kp,Kv,Ka for type1 system. Given the open loop transfer function of a unity feedback control system as under G(S) = 100/S(0.1S+1). Find Kp,Kv,Ka (L-2,CO-2)

Formulae-1M

Solution-1M

3. Define type and order of a system with examples.(L-1,CO-2

Type number definition-0.5M

Order number definition-0.5M

Examples-1M

4. Obtain the transfer function of the mechanical system shown in fig1. (L-3,CO-1)

Force balance equations-2M

Solution-5M

5 . A second order system is represented by a transfer function

= , where Qo(S) is the proportional output and T is the input torque. A step input 10N-m is applied to the system and test results are given below.

a) Peak overshoot=6%

Formula-1M

Solution-1M

b) peak time is 1 sec

Formula-1M

Solution-1M

c) the steady state output of the system is 0.5 radian. Determine the values of J,K and F. (L-3,CO-2)

Formula-1M

Solution-2M

6.Obtain the transfer function of the system of fig2 (L-3,CO-1)

Formula-1M

Identyfying Individual loops-2M

Solution-4M

MATRUSRI ENGINEERING COLLEGE

SAIDABAD, HYDERABAD – 500 059

DEPARTMENT OF ELECTRONICS &COMMUNICATION ENGINEERING

Automatic Control Systems Scheme and Key for Internal-II

Academic Year: 2015-16 IIIYear I SEM

1. What is the effect of Root locus by adding poles and Zeros(L-2,CO-2)

Explanation-2M

2. Explain PID controller (L-2,CO-3)

Definition-1M

Properties-1M

3. Define state-transition matrix and write its properties(L-1,CO-5)

Definitiopn-1M

Properties-1M

4. Sketch the root locus for the given G(S)H(S)=k/s(s+2)(s²+2s+2) (L-3,CO-2)

Rules-3M

Solution-4M

5 .For the sampled data system shown in Fig 1, find out the response to unit step input with G(S) = 1/(S+1) (L-3,CO-4)

Procedure-2M

Solution-5M

6.For the system with transfer function G(S) = S(1+2S)/(1+4S)(1+0.25S). Draw the Bode plot (L-3,CO-3)

Rules-3M

Solution-5M

ASSIGNMENT QUESTIONS

UNIT-1

1.Write the differential equations governing the mechanical system shown in below figure. Draw the F-V and F-I analogous circuits.

2.Write the differential equations governing the mechanical system shown in below figure. Draw the Torque-Voltage and Torque - Current analogous circuits.

3.For the system represented by the block diagram shown in below figure, evaluate the closed loop transfer function whrn the input is applied at (i) station I (ii)station-II

4. For the signal flow graph given below, find the transfer function using mason’s gain formula.

UNIT 2

1.The unity feedback system is characterized by an open loop transfer function G(s)=k/s(s+10). Determine the gain k, so that the system will have a damping ratio of 0.5 for this value of k. determine settling time, peak overshoot and time to peak overshoot for a unit step input.

2.A unity feedback control system is has an open loop transfer function G(s)=10/s(s+2). Find the rise time, percentage overshoot, peak time and settling time for a step input of 12 units.

3.for a unity feedback controls system the open loop transfer function G(s)=10(s+2) / s²(s+1). Find (a)position, velocity and acceleration error constants (b) the steady state error when the input is

4.A unity feedback control system is characterized by the following open loop transfer function G(s)=(0.4s+1)/s(s+0.6). determine its transient response for unit step input.

5.For a unity feedback system having

i) Find the factor by which k should be multiplied by to increase from 0.15 to 0.6.

ii) Find the factor which time constant should be multiplied to reduce from 0.8 to 0.4.

6.Sketch the root locus for the unity feedback system whose open loop transfer function is

7.Construct the Routh’s array and determine the stability of the system whose characteristic equation is .

UNIT III

1. Plot the Bode diagram for the transfer function and obtain the gain and phase crossover frequencies.

2.For the following transfer function draw bode plot and obtain gain crossover frequency.

3.Construct the Nyquist plot for a system whose open loop transfer function is given by

. Find the range of k for stability.

4.What are the various types of controllers and explain them in brief.

5.What is Lead compensator? Obtain its transfer function.

UNIT IV

1.Solve the difference equation c(k+2)+3c(k+1)+2c(k)=u(k). Given that c(0)=1, c(1)=-3, c(k)=0 for k<0.

2.Determine the z-domain transfer function for the following s-domain transfer function.

(a) H(s)= (b)=

3.Find C(z) / R(z) for following closed loop sampled data control system. Assume all the samplers of impulse type.

4.Check for stability of the sampled data control systems represented by the characteristic equation

using Jury’s stability test.

5.Check for stability of the sampled data control systems represented by the characteristic equation

using Bilinear transformation method.

UNIT IV

1.A linear time-invariant system is characterized by homogeneous state equation.

Compute the solution of the homogeneous equation, assuming the initial state vector =

2.Obtain the state model of the system whose transfer function is given as,

3.Construct the state model for a system characterized by the differential equation,

4.Compute the solution of following state equations.

(a)

(b)

CONTENT BEYOND SYLLABUS

Sl.No.	TOPIC	Mode of Teaching	Relevance with POs and PSOs
1.	Control systems using Matlab	BB

TOPIC Description:

Control systems refer to a very wide area, covering many disciplines and phenomena. Control systems theory is a wide area covering a range of physical phenomena. Control systems are systems that are designed to operate under strict specifications, to satisfy certain aims, like safety regulations in the industry, optimal production of goods, disturbance rejection in vehicles, smooth movement and placement of objects in warehousing, regulation of drug administration in medical operations, level control in chemical processes and many more.

This topic provides an introduction to the fundamental principles of control system’s analysis and design through the programming environment of Matlab and Simulink. Analysis of transfer function models is carried out though multiple examples in Matlab and Simulink, analyzing the dynamics of 1st and 2nd order systems, the role of the poles and zeros in the system’s dynamic response, the effects of delay and the possibility to approximate higher order systems by lower order ones

METHODOLOGY USED TO IDENTIFY WEAK AND BRIGHT STUDENTS:

SUPPORT TO WEAK STUDENTS:

EFFORTS TO ENGAGE BRIGHT STUDENTS:

END

A Narmada

Wednesday, November 11, 2020

TEXT BOOKS:

Fig. 1

No comments:

Post a Comment

Search This Blog