EC2255- Control System Notes( solved problems) PDF

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EC2255- Solved Problems in Control System IV Semester ECE Control Systems PHYSICAL SYSTEMS: INTRODUCTION:  First step towards analysis of a control system is preparation of a mathematical model which is linear over a satisfactory range operating conditions giving us the property of linearity and su...

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EC2255- Solved Problems in Control System

IV Semester ECE

Control Systems PHYSICAL SYSTEMS: INTRODUCTION:         



First step towards analysis of a control system is preparation of a mathematical model which is linear over a satisfactory range operating conditions giving us the property of linearity and superposition. A model may be defined as a representation of the essential aspects of the system which presents knowledge of the system in a usable manner. To be useful, the model must not so complicated that it cannot be understood and thereby be unsuitable for analysis. The components of a control system are diverse in nature and may include electrical, mechanical, thermal and fluidic devices. At the same time, it must not so complicated that it cannot be understood and thereby be unsuitable for analysis. At the same time, it must not be oversimplified and trivial. While dealing with control systems, we shall be concerned mostly with dynamic systems. The behavior of such systems is described in the form of differential equations. Although these will normally be nonlinear, it is customary to linearize them about an operating point to obtain linear differential equations. The components of a control system are diverse in nature and may include electrical, mechanical, thermal and fluidic devices. The differential equations relating the input and output quantities for these devices are obtained using the basic law of physics. These include balancing forces, energy and mass. In practice, some simplifying assumptions are often made to obtain linear differential equations. With constant coefficients, although in most cases exact analysis would lead to nonlinear partial differential equations. For most physical devices one may classify the variables as either THROUGH or ACROSS variables, in the sense that the former refer to a point while the latter are measured between two points. The input , output relations of various physical components of a system is governed by differential equation. The mathematical model of a control system constitutes a set of differential equation.

Solved by A.Devasena., Associate Professor., Dept/ECE

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EC2255- Solved Problems in Control System

SYSTEM Electrical Mechanical (translational) Mechanical (rotational) Thermal Fluidic

THROUGH VARIABLE Current I Force, F Torque, T Rate of flow of heat energy,q Volumetric rate of fluid flow,Q

IV Semester ECE

ACROSS VARIABLE Potential difference or voltage V. Relative velocity,v Relative angular velocity ,ω Difference in temp. T Difference in pressure,P

A mathematical model will be linear if the differential equations describing the system have constant coefficients. If the coefficients of the differential equations describing the system are constants, then the model is linear time invariant. If the coefficients of the differential equations governing the system are functions of time, then the model is linear time varying. The differential equations of a linear time invariant system can be reshaped into different form for the convenience of analysis. One such model for single input and single output system analysis is transfer function of the system. The transfer function of a system analysis is transfer function of the system. The transfer function of a system is defined as the ratio of laplace transfer function of the system. The transfer function of a defined as the ratio of laplace transfer function of a system is defined as the ratio of laplace transform of output to the laplace transform of input with zero initial conditions.

An equation describing a physical has integrals and differentials. The response can be obtained by solving such equations. The steps involved in obtaining the transfer function are:  

Write differential equation of the system. Replace terms involving d/dt by s and integral of dt by 1/s.

Advantages of open loop system. The advantages of open loop system are 1. Such systems are simple in construction. 2. Very much convenient when input is difficult to measure. 3. Such systems are easy for maintenance point of view. 4. Generally these are not troubled with problems of stability. 5. Such systems are simple to design and hence economical. Solved by A.Devasena., Associate Professor., Dept/ECE

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EC2255- Solved Problems in Control System

IV Semester ECE

Disadvantages of open loop system. The disadvantages of open loop system are 1. Such systems are inaccurate and unreliable because accuracy of such systems are totally dependent on the accurate precalibration of the controller. 2. Such systems give inaccurate results if there are variation in the external environment i.e. systems cannot sense environmental changes. 3. Similarly they cannot sense internal disturbances in the system, after the controller stage.

Advantages of closed loop system. The advantages of closed loop system are: 1.accuracy of such system is always very high because controller modifies and manipulates the actuating signal such that error in the system will be zero. 2. Such systems senses environmental changes, as well as internal disturbances and accordingly modifies the error. 1. In such system, there is reduced effect of non-linearites and distortions. Comparison between open loop system and closed loop system OPEN LOOP SYSTEM Output measurement is not required for operation of the system. Highly affected by non-lineariteis Highly sensitive to the disturbances and environmental changes Feedback element and error detector are absent Generally stable in nature

CLOSED LOOP SYSTEM Output measurement is necessary. Reduced effect of nonlinearities. Less sensitive to disturbances and environmental changes. Feedback element and error detector are absent Stability is the major consideration while designing.

Linear system. A system is said to be linear if it obeys the principle of superposition and homogeneity. The principle of superposition states that the response of the system to a weighted sum of the responses of the system to each individual input signals. The system is said to be linear, if it satisfies the following two properties:  Adaptive property that is for any x and y belonging to the domain of the function f, we have F(x+y) = f(x) +f(y) Solved by A.Devasena., Associate Professor., Dept/ECE

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EC2255- Solved Problems in Control System

IV Semester ECE



Homogeneous property that is for any x belonging to the domain of the function f and for any scalar constant α We have  F(αx) = αf(x) These two properties together constitute a principle of superposition. Most of the systems are non-linear in nature because of different non-linearities such as saturation, friction, dead zone etc. present in the system.

Definition of Transfer Function Transfer function of a given system is defined as the ratio of the laplace transform of output variable to laplace transform of input variables at zero input conditions.

Properties of transfer functions. The properties of transfer function are as follows:  The transfer function of a system is the laplace transform of its impulse response. I.e. if the input to a system with transfer function P(s) is an impulse and all initial conditions are zero, the transform of the output is P(s).  The roots of the denominator are the system poles and the roots of the numerator are system zeros. The system stability can be described in terms of the location of the roots of the transfer function. Advantages of transfer function.  It helps in the stability analysis of the system.  It helps in determining the important information about the system Poles, zeros, characteristic equation etc.  Once transfer function is known, output response for any type of reference input can be calculated.  The system differential equation can be easily obtained by replacing variable ‘s’ by d/dt. Disadvantages of transfer function. The disadvantages of transfer function approach are:  Only applicable to linear time invariant systems.  It does not provide any information concerning the physical structure of the system. From transfer function, physical nature of the system, whether it is electrical, mechanical, thermal or hydraulic cannot be judged. Solved by A.Devasena., Associate Professor., Dept/ECE

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EC2255- Solved Problems in Control System 

IV Semester ECE

Effects arising due to initial conditions are totally neglected. Hence initial conditions loose importance.

Important features of feedback.  Reduced effects of non-linearities and distortion.  Increased accuracy.  Reduced sensitivity of the ratio of the output to input to variations in system characteristics.  Tendency toward oscillation or instability.

The basic elements used for modeling mechanical translational system: The model of mechanical translational system can be obtained by using three basic elements mass spring and dashpot. Translational system Consider a mechanical system in which motion is taking place along a straight line. Such systems are of translational type. These systems are characterized by displacement, linear velocity and linear acceleration. Definition of torque. This is the motion about a fixed axis. In such systems, the force gets replaced by a moment about the fixed axis. I.e. {force x distance from fixed axis} which is called torque. Definition of friction. Whenever there is a motion, there exists a friction. Friction may be between moving element and fixed support or between two moving surfaces. Friction is also nonlinear in nature. The types of friction. Friction can be divided into three types. They are  Viscous friction.  Static friction.  Coulomb friction. The two types of analogies for mechanical system are force-voltage analogy and force-current analogy. Analogous systems Systems whose differential equations are identical form are called analogous systems.

Solved by A.Devasena., Associate Professor., Dept/ECE

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EC2255- Solved Problems in Control System

IV Semester ECE

Two types of variables in physical system. The two variables in physical systems are through variables and across variable. Through variables refer to a point, the across variable is measured between two points. The other names for force voltage analogy and force current analogy are Force voltage analogy- Loop analysis. Force current analogy- nodal analysis.

BLOCK DIAGRAM ALGEBRA; INTRODUCTION: In block diagram, the system consists of so many components. These components are linked together to perform a particular function. Each component can be represented with the help of individual block. NEED FOR BLOCK DIAGRAM REDUCTION: Block diagrams of some of the systems turn out to be complex, such that the evaluation of their performance required simplification (or reduction) of block diagrams which is carried out by block diagram rearrangements. DEFINITION: A block diagram of a system is a pictorial representation of the functions performed by each component of the system and shows the flow of signals. ADVANTAGES OF BLOCK DIAGRAM:  Very simple to construct the block diagram for complicated systems.  Individual as well as overall performance of the system can be studied by using transfer functions shown in the block diagram.  Overall closed loop transfer function can be easily calculated using block diagram rules.  The function of the individual element can be visualized from the block diagram. DISADVANTAGES OF BLOCK DIAGRAM: The disadvantages of block diagram are:  Block diagram does not include any information about the physical construction of the system.  Source of energy is generally not shown in the block diagram, so block diagram for a given system is not unique. The basic components of block diagram are block, branches, summing point, arrows. Solved by A.Devasena., Associate Professor., Dept/ECE

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EC2255- Solved Problems in Control System

IV Semester ECE

BLOCK: It indicates the function of particular system. R(s) is the reference or controlling variable. G(s) is the transfer function of the particular system.C(s) is output or controlled variable.

R(S)

C(S) G(S)

SUMMING POINT: It is used to add or subtract one or more signals. + indicates the add or subtract function. + indicates the signal is added to reference signal. -indicates the signal is subtracted to reference signal. It is called negative feedback. The signal which is added or subtracted to the reference signal is called feedback signal. Summing point R(S)

C(S) +

TAKE OF POINT Some of the signal at the top are bypass and it is given to other block for further performance. Take off point R(S)

C(S)

G(s)

The steps to reduce the block diagram.  Reduce the series blocks.  Reduce the parallel blocks. Solved by A.Devasena., Associate Professor., Dept/ECE

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EC2255- Solved Problems in Control System   

IV Semester ECE

Reduce minor feedback loops. As for as possible shift summing point to the left and take-off point to the right. Repeat the above steps till canonical form is obtained.

BLOCK DIAGRAM FOR A CANONICAL SYSTEM: R(s) = reference or controlling variable. C(S) = Output or controlled variable. E(S) = Error actuating signal. B(S) = Feedback signal H(S) = Feedback element. G(S) = transfer function for the system. G(S) = C(S) E(S) Closed loop transfer function is C(S) . R(S) C(S) = G(S) E(S). Consider a negative feedback signal is applied Actuating signal E(S) = R(S) – B(S)

(1)

(2)

Feedback signal C(S)H(S) = B(S) Put equation (3) in eq.(2) E(S) = R(S)- C(S)H(S) Put eq (4) in eq.(1) C(S) = G(S){R(S) – C(S)H(S)} C(S)[ 1+ G(S) H(S)] = G(S)R(S)

(3) (4) (5) (6)

(transfer function for negative feedback signal C(S) = G(S)______ R(S) 1+G(S) R(S) (transfer function for positive feedback signal.) C(S) = G(S)______ R(S) 1-G(S) R(S)

Rules for reduction of block diagram Rule 1: If the blocks are in cascade then Solved by A.Devasena., Associate Professor., Dept/ECE

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EC2255- Solved Problems in Control System

IV Semester ECE

R(S)

R(S)

C(S )

G2

G1

C(S)

=

G1G2

Rule 2: if the blocks are in parallel then, the blocks are added or subtracted depending on the summing point signal.

G2

R(S)

G1+G2

+

C(S)

=

G1 R(S)

+ +

C(s)

Rule 3: Moving the take-off point after the block

R(S)

G

R(S)

C (S)

G

C(S)

= 1/G

Rule 4: moving the take-off point ahead of the block

G G G Solved by A.Devasena., Associate Professor., Dept/ECE

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EC2255- Solved Problems in Control System

IV Semester ECE

Rule 5: Moving summing point after the block, then

Rule 6: Moving the summing point ahead (before) the blockff

Rule 7: Eliminating feedback loop, then

Solved by A.Devasena., Associate Professor., Dept/ECE

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EC2255- Solved Problems in Control System

IV Semester ECE

1. Using block diagram reduction techniques find C(s) / R(s) as in fig. Fig1.2.1.1 R(s)

G1 -

C(s)

-

I H1

II H2 Fig1.2.1.1

Step 1: eliminating feedback loop I R(s)

C(s)

G1

-

1+G1H1

II H1

Step2: eliminating feedback loop II

G(s) = G1 / 1 +G1H1 H(s) = H2 G1 G1 / 1 +G1H1 C(s) / R(s) =

= 1 +G1H1 + G1H2

1 + (G1 / 1 +G1H1 )(H2)

Solved by A.Devasena., Associate Professor., Dept/ECE

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EC2255- Solved Problems in Control System

IV Semester ECE

G1 R(s)

C(s) 1 + G1H1+G1H2 Fig1.2.1.2 Answer.

2. Using block diagram reduction technique find closed loop transfer function C(s) / R(s) shown in fig 1.2.2.1

G3 + C(S)

R(S) G1 -

G4

G2

H1

H2 Fig 1.2.2.1

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EC2255- Solved Problems in Control System

IV Semester ECE

Step 1: Combine the blocks G1 &G2 which are in cascade and combine the blocks G2 &G3 which are in parallel as shown in fig 1.2.2.2.

R(S)

C(S) G1 G4 -

-

G2 +G3 I

H1 II H2 Fig 1.2.2.2

Step 2: Eliminate feedback loopI and combine the blocks (G1G4 / 1+ G1G4H1) & (G2 + G3) which are in parallel as shown in fig. Fig 1.2.2.3 C(S)

G1G4 G2+G3

R(S) 1+ G1G4H1

II H2

Solved by A.Devasena., Associate Professor., Dept/ECE

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EC2255- Solved Problems in Control System

IV Semester ECE

Step 3: eliminate feedback loop II

G1G4 G2 +G3 C(s)

1 +G1G4H1

= R(s) 1+

G1G4

G2 +G3 1 + G1G4H1

G1G4

G2 +G3

= 1 + G1G4H1

+

Solved by A.Devasena., Associate Professor., Dept/ECE

G1G4 (G2 +G3)

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EC2255- Solved Problems in Control System

IV Semester ECE

3.Find the transfer function C(s) / R(s) for the block diagram shown below as shown in (fig 1.2.3.1) (A.U.2004)

R(s)

C(s) G1(s) -

G2(s) I

II

(fig 1.2.3.1)

Step 1: Eliminating feedback loopI (as in fig.1.2.3.2) R(s)

C(s) G1(s)

G2(s)__ 1 + G2 (s)

II

(fig 1.2.3.2)

Solved by A.Devasena., Associate Professor., Dept/ECE

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EC2255- Solved Problems in Control System

IV Semester ECE

Step 2: Combining blocks G1(s) & {G2(s) / 1+ G2(s)} which are in cascade (as in fig 1.2.3.3) C(s)

R(s)

G1(s)G2(s)__ 1 + G2 (s) II

Fig 1.2.3.3

Step 3: Eliminating feedback loop II (as in fig 1.2.3.4) G1(S) G2(S) 1 + G2(S) C(s) R(S)

= 1 +

G1(S) G2(S) 1 + G2(S)

G1(S) G2(S) C(s) R(S)

R(S)

= (1 + G2(S)) + G1(S) G2(S)

C(S) G1(S) G2(S) (1 + G2(S)) + G1(S) G2(S) Answer.

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EC2255- Solved Problems in Control System

IV Semester ECE

4. Determine the transfer function C(s) / R(s) of the system shown in fig.1.2.4.1 by block diagram reduction method. (AU: Dec.2003)

H3 T1

R(s) S1 +

+

G1

-

+

C(s)

S3

G2

-

G3

S2

G4 T2

-

H1

H2 Fig.1.2.4.1.

Step:1. Shifting the summing point S2 before the block G1 and shifting the take off point T2 after the block G4

H

1 / G1

1 / G4

3

R(s) S1 +

+

-

+ G2

G1 -

C(s)

S3 G3

G4

-

S2 H1

H2 Fig.1.2.4.2.

Solved by A.Devasena., Associate Professor., Dept/ECE

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