Control system Engineering Lecture notes

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CONTROL SYSTEM ENGINEERING-I Department of Electrical Engineering VEER SURENDRA SAI UNIVERSITY OF TECHNOLOGY, ODISHA, BURLA 1 Disclaimer This document does not claim any originality and cannot be used as a substitute for prescribed textbooks. The information presented here is merely a collection by the committee members for their respective teaching assignments. Various sources as mentioned at the end of the document as well as freely available material from internet were consulted for preparing this document. The ownership of the information lies with the respective authors or institutions. Further, this document is not intended to be used for commercial purpose and the committee members are not accountable for any issues, legal or otherwise, arising out of use of this document. The committee members make no representations or warranties with respect to the accuracy or completeness of the contents of this document and specifically disclaim any implied warranties of merchantability or fitness for a particular purpose. The committee members shall be liable for any loss of profit or any other commercial damages, including but not limited to special, incidental, consequential, or other damages. 5 CHAPTER1 1. Basic Concept of Control System Control Engineering is concerned with techniques that are used to solve the following six problems in the most efficient manner possible. (a)The identification problem :to measure the variables and convert data for analysis. (b)The representation problem:to describe a system by an analytical form or mathematical model (c)The solution problem:to determine the above system model response. (d)The stability problem:general qualitative analysis of the system (e)The design problem: modification of an existing system or develop a new one (f)The optimization problem: from a variety of design to choose the best. The two basic approaches to solve these six problems are conventional and modern approach. The electrical oriented conventional approach is based on complex function theory. The modern approach has mechanical orientation and based on the state variable theory. Therefore, control engineering is not limited to any engineering discipline but is equally applicable to aeronautical, chemical, mechanical, environmental, civil and electrical engineering. For example, a control system often includes electrical, mechanical and chemical components. Furthermore, as the understanding of the dynamics of business, social and political systems increases; the ability to control these systems will also increase. 1.1. Basic terminologies in control system System: A combination or arrangement of a number of different physical components to form a whole unit such that that combining unit performs to achieve a certain goal. Control: The action to command, direct or regulate a system. Plant or process: The part or component of a system that is required to be controlled. Input: It is the signal or excitation supplied to a control system. Output: It is the actual response obtained from the control system. Controller: The part or component of a system that controls the plant. Disturbances: The signal that has adverse effect on the performance of a control system. Control system: A system that can command, direct or regulate itself or another system to achieve a certain goal. Automation: The control of a process by automatic means Control System: An interconnection of components forming a system configuration that will provide a desired response. Actuator: It is the device that causes the process to provide the output. It is the device that provides the motive power to the process. 6 Design: The process of conceiving or inventing the forms, parts, and details of system to achieve a specified purpose. Simulation: A model of a system that is used to investigate the behavior of a system by utilizing actual input signals. Optimization: The adjustment of the parameters to achieve the most favorable or advantageous design. Feedback Signal: A measure of the output of the system used for feedback to control the system. Negative feedback: The output signal is feedback so that it subtracts from the input signal. Block diagrams: Unidirectional, operational blocks that represent the transfer functions of the elements of the system. Signal Flow Graph (SFG): A diagram that consists of nodes connected by several directed branches and that is a graphical representation of a set of linear relations. Specifications: Statements that explicitly state what the device or product is to be and to do. It is also defined as a set of prescribed performance criteria. Open-loop control system: A system that utilizes a device to control the process without using feedback. Thus the output has no effect upon the signal to the process. Closed-loop feedback control system: A system that uses a measurement of the output and compares it with the desired output. Regulator: The control system where the desired values of the controlled outputs are more or less fixed and the main problem is to reject disturbance effects. Servo system: The control system where the outputs are mechanical quantities like acceleration, velocity or position. Stability: It is a notion that describes whether the system will be able to follow the input command. In a non-rigorous sense, a system is said to be unstable if its output is out of control or increases without bound. Multivariable Control System: A system with more than one input variable or more than one output variable. Trade-off: The result of making a judgment about how much compromise must be made between conflicting criteria. 1.2. Classification 1.2.1. Natural control system and Man-made control system: Natural control system: It is a control system that is created by nature, i.e. solar system, digestive system of any animal, etc. Man-made control system: It is a control system that is created by humans, i.e. automobile, power plants etc. 1.2.2. Automatic control system and Combinational control system: 7 Automatic control system: It is a control system that is made by using basic theories from mathematics and engineering. This system mainly has sensors, actuators and responders. Combinational control system: It is a control system that is a combination of natural and man-made control systems, i.e. driving a car etc. 1.2.3. Time-variant control system and Time-invariant control system: Time-variant control system: It is a control system where any one or more parameters of the control system vary with time i.e. driving a vehicle. Time-invariant control system: It is a control system where none of its parameters vary with time i.e. control system made up of inductors, capacitors and resistors only. 1.2.4. Linear control system and Non-linear control system: Linear control system: It is a control system that satisfies properties of homogeneity and additive.  Homogeneous property: f x y f x f y   Additive property: f x f x  Non-linear control system: It is a control system that does not satisfy properties of 3 homogeneity and additive, i.e. f x x  1.2.5. Continuous-Time control system and Discrete-Time control system: Continuous-Time control system: It is a control system where performances of all of its parameters are function of time, i.e. armature type speed control of motor. Discrete -Time control system: It is a control system where performances of all of its parameters are function of discrete time i.e. microprocessor type speed control of motor. 1.2.6. Deterministic control system and Stochastic control system: Deterministic control system: It is a control system where its output is predictable or repetitive for certain input signal or disturbance signal. Stochastic control system:It is a control system where its output is unpredictable or non-repetitive for certain input signal or disturbance signal. 1.2.7. Lumped-parameter control system and Distributed-parameter control system: Lumped-parameter control system: It is a control system where its mathematical model is represented by ordinary differential equations. Distributed-parameter control system:It is a control system where its mathematical model is represented by an electrical network that is a combination of resistors, inductors and capacitors. 1.2.8. Single-input-single-output (SISO) control system and Multi-input-multi-output (MIMO) control system: SISO control system: It is a control system that has only one input and one output. MIMO control system:It is a control system that has only more than one input and more than one output. 1.2.9. Open-loop control system and Closed-loop control system: Open-loop control system: It is a control system where its control action only depends on input signal and does not depend on its output response. 8 Closed-loop control system:It is a control system where its control action depends on both of its input signal and output response. 1.3. Open-loop control system and Closed-loop control system 1.3.1. Open-loop control system: It is a control system where its control action only depends on input signal and does not depend on its output response as shown in Fig.1.1. Fig.1.1. An open-loop system Examples: traffic signal, washing machine, bread toaster, etc. Advantages:  Simple design and easy to construct  Economical  Easy for maintenance  Highly stable operation Dis-advantages:  Not accurate and reliable when input or system parameters are variable in nature  Recalibration of the parameters are required time to time 1.3.2. Closed-loop control system: It is a control system where its control action depends on both of its input signal and output response as shown in Fig.1.2. Fig.1.2. A closed-loop system Examples: automatic electric iron, missile launcher, speed control of DC motor, etc. Advantages:  More accurate operation than that of open-loop control system  Can operate efficiently when input or system parameters are variable in nature  Less nonlinearity effect of these systems on output response  High bandwidth of operation  There is facility of automation  Time to time recalibration of the parameters are not required Dis-advantages:  Complex design and difficult to construct 9  Expensive than that of open-loop control system  Complicate for maintenance  Less stable operation than that of open-loop control system 1.3.3. Comparison between Open-loop and Closed-loop control systems: It is a control system where its control action depends on both of its input signal and output response. Sl. Open-loop control systems Closed-loop control systems No. 1 No feedback is given to the control system A feedback is given to the control system 2 Cannot be intelligent Intelligent controlling action Closed loop control introduces the There is no possibility of undesirable 3 possibility of undesirable system system oscillation(hunting) oscillation(hunting) The output will not very for a constant In the system the output may vary for a 4 input, provided the system parameters constant input, depending upon the remain unaltered feedback System output variation due to variation in System output variation due to variation in 5 parameters of the system is greater and the parameters of the system is less. output very in an uncontrolled way 6 Error detection is not present Error detection is present 7 Small bandwidth Large bandwidth 8 More stable Less stable or prone to instability 9 Affected by non-linearities Not affected by non-linearities 10 Very sensitive in nature Less sensitive to disturbances 11 Simple design Complex design 12 Cheap Costly 10 1.4. Servomechanism It is the feedback unit used in a control system. In this system, the control variable is a mechanical signal such as position, velocity or acceleration. Here, the output signal is directly fed to the comparator as the feedback signal, b(t) of the closed-loop control system. This type of system is used where both the command and output signals are mechanical in nature. A position control system as shown in Fig.1.3 is a simple example of this type mechanism. The block diagram of the servomechanism of an automatic steering system is shown in Fig.1.4. Fig.1.3. Schematic diagram of a servomechanism Fig.1.4. Block diagram of a servomechanism Examples:  Missile launcher  Machine tool position control  Power steering for an automobile  Roll stabilization in ships, etc. 1.5. Regulators It is also a feedback unit used in a control system like servomechanism. But, the output is kept constant at its desired value. The schematic diagram of a regulating 11 system is shown in Fig.1.5. Its corresponding simplified block diagram model is shown in Fig.1.6. Fig.1.5. Schematic diagram of a regulating system Fig.1.6. Block diagram of a regulating system Examples:  Temperature regulator  Speed governor  Frequency regulators, etc. 12 CHAPTER2 2. Control System Dynamics 2.1. Definition: It is the study of characteristics behaviour of dynamic system, i.e. (a) Differential equation i. First-order systems ii. Second-order systems (b) System transfer function: Laplace transform 2.2. Laplace Transform: Laplace transforms convert differential equations into algebraic equations. They are related to frequency response.  st L x t X s x(t)e dt (2.1)   0  st Lxt Xs x(t)e dt (2.2)  0 Time-domain 7 Sine sin ωt  Laplace domain 2 2 s No. Function x(t)= X(s)= ℒx(t) -1 ℒ X(s) 8 Cosine cos ωt s 2 2 -τs s 1 Delay δ(t-τ) e 9 Hyperbolic sinh αt 2 Unit impulse δ(t) 1  sine 2 2 s 1 3 Unit step u(t) s 10 Hyperbolic cosh αt s cosine 2 2 s 1 4 Ramp t 2 s t 11 Exponentiall  e sint 2 2 y decaying (s ) Exponential 1 -αt 5 e sine wave decay s t 12 Exponentiall s e cost  Exponential y decaying 2 2 t (s) 6 1 e approach cosine wave s(s) 2.3. Solution of system dynamics in Laplace form: Laplace transforms can be solved using partial fraction method. A system is usually represented by following dynamic equation. A s  N s (2.3)  B s  The factor of denominator, B(s) is represented by following forms, i. Unrepeated factors 13 ii. Repeated factors iii. Unrepeated complex factors (i) Unrepeated factors N(s) A B  (s a)(s b) s a s b (2.4) A(s b) B(s a)  (s a)(s b) By equating both sides, determine A and B. Example 2.1: Expand the following equation of Laplacetransform in terms of its partial fractionsand obtain its time-domain response. 2s Y (s) (s1)(s 2) Solution: The following equation in Laplacetransform is expandedwith its partial fractions as follows. 2s A B  (s1)(s 2) (s1) (s 2) 2s A(s 2) B(s1)  (s1)(s 2) (s1)(s 2) By equating both sides, A and B are determined as . Therefore, A2, B 4 2 4 Y(s) (s1) (s 2) Taking Laplace inverse of above equation, t2t y(t)2e 4e (ii) Unrepeated factors N (s) A B A B(s a)  (2.5) 2 2 2 (s a) (s a) (s a) (s a) By equating both sides, determine A and B. Example 2.2: Expand the following equation of Laplacetransform in terms of its partial fractionsand obtain its time-domain response. 2s Y(s) 2 (s1) (s 2) Solution: The following equation in Laplacetransform is expandedwith its partial fractions as follows. 2s A B C  2 2 (s1) (s 2) (s1) (s1) (s 2) By equating both sides, A and B are determined as . Therefore, A2, B 4 2 4 4 Y(s) 2 (s1) (s1) (s 2) Taking Laplace inverse of above equation, tt2t y(t)2te 4e 4e 14 (iii) Complex factors: They contain conjugate pairs in the denominator. N(s) As B  (2.6) 2 2 (s a)(s a) (s) By equating both sides, determine A and B. Example 2.3: Expand the following equation of Laplacetransform in terms of its partial fractionsand obtain its time-domain response. 2s1 Y(s) (s1 j)(s1 j) Solution: The following equation in Laplacetransform is expandedwith its partial fractions as follows. 2s 1 Y(s) 2 2 (s1)1 (s1)1 Taking Laplace inverse of above equation, tt y(t) 2e cost e sint 2.4. Initial value theorem: (2.7) y(t)  lim sY(s)  lim t0 s Example 2.4: Determine the initial value of the time-domain response of the following equation using the initial-value theorem. 2s1 Y(s) (s1 j)(s1 j) Solution: Solution of above equation, tt y(t) 2e cost e sint Applying initial value theorem, s(2s1)  2 lim (s1 j)(s1 j) s 2.5. Final value theorem: (2.8) y(t)  lim sY(s) lim t s0 Example 2.5: Determine the initial value of the time-domain response of the following equation using the initial-value theorem. 2s Y(s) 2 (s1) (s 2) Solution: Solution of above equation, 15 tt2t y(t)2te 4e 4e Applying final value theorem, s(2s1)  2 lim (s1 j)(s1 j) s 16 CHAPTER3 3. Transfer Function 3.1. Definition: It is the ratio of Laplace transform of output signal to Laplace transform of input signal assuming all the initial conditions to be zero, i.e. Let, there is a given system with input r(t) and output c(t) as shown in Fig.3.1 (a), then its Laplace domain is shown in Fig.3.1 (b). Here, input and output are R(s) and C(s) respectively. (a) (b) (c) Fig.3.1. (a) A system in time domain, (b) a system in frequency domainand (c) transfer function with differential operator G(s) is the transfer function of the system. It can be mathematically represented as follows. Cs Gs Equation Section (Next)(3.1) R s  zero initial condition Example 3.1: Determine the transfer function of the system shown inFig.3.2. Fig.3.2. a system in time domain Solution: Fig.3.1 is redrawn in frequency domain as shown in Fig.3.2. Fig.3.2. a system in frequency domain 17 Applying KVL to loop-1 of the Fig.3.2  1 Vs R Ls Is (3.2) i Cs  Applying KVL to loop-2 of the Fig.3.2 1  (3.3) Vs Is o Cs  From eq (2.12), 1  I s V s / CsV s (3.4)  o o Cs  Now, using eq (2.13) in eq (2.10),  1 Vs R Ls CsVs i o Cs  (3.5) V s  1 1 o  2 1 V s  LCs RCs1 i R Ls Cs   Cs Then transfer function of the given system is 1 G s (3.6)  2 LCs RCs1 3.2. General Form of Transfer Function m s z   i K s z s z ... s z  1 2 m i1 G s K (3.7)  n s p s p ... s p  1 2 n s z   j i1 Where, z , z ...z are called zeros and p , p ...p are called poles. 1 2 m 1 2 n Number of poles n will always be greater than the number of zeros m Example 3.2: Obtain the pole-zero map of the following transfer function. (s 2)(s 2 j4)(s 2 j4) G(s) (s 3)(s 4)(s 5)(s1 j5)(s1 j5) Solution: The following equation in Laplacetransform is expandedwith its partial fractions as follows. Zeros Poles s=2 s=3 s=-2-j4 s=4 s=-2+j4 s=5 18 s=-1-j5 s=-1+j5 Fig.3.3. pole-zero map 3.3. Properties of Transfer function:  Zero initial condition  It is same as Laplace transform of its impulse response d  Replacing ‘s’ by in the transfer function, the differential equation can be obtained dt  Poles and zeros can be obtained from the transfer function  Stability can be known  Can be applicable to linear system only 3.4. Advantages of Transfer function:  It is a mathematical model and gain of the system d  Replacing ‘s’ by in the transfer function, the differential equation can be obtained dt  Poles and zeros can be obtained from the transfer function  Stability can be known  Impulse response can be found 3.5. Disadvantages of Transfer function:  Applicable only to linear system  Not applicable if initial condition cannot be neglected  It gives no information about the actual structure of a physical system 19 CHAPTER4 4. Description of physical system 4.1. Components of a mechanical system: Mechanical systems are of two types, i.e. (i) translational mechanical system and (ii) rotational mechanical system. 4.1.1. Translational mechanical system There are three basic elements in a translational mechanical system, i.e. (a) mass, (b) spring and (c) damper. (a) Mass: A mass is denoted by M. If a force f is applied on it and it displays 2 d x distance x, then as shown in Fig.4.1. f M 2 dt Fig.4.1. Force applied on a mass with displacement in one direction If a force f is applied on a massM and it displays distance x in the direction of f and 1 2 2  d x d x 1 2 f M distance x in the opposite direction, then as shown in Fig.4.2. 2 2 2 dt dt  X X 2 1 f M Fig.4.2. Force applied on a mass with displacement two directions (b) Spring: A spring is denoted by K. If a force f is applied on it and it displays distance x, then f Kx as shown in Fig.4.3. Fig.4.3. Force applied on a spring with displacement in one direction If a force f is applied on a springK and it displays distance x in the direction of f and 1 distance x in the opposite direction, then f K x x as shown in Fig.4.4. 2 1 220 Fig.4.4. Force applied on a spring with displacement in two directions (c) Damper: A damper is denoted by D. If a force f is applied on it and it displays dx f D distance x, then as shown in Fig.4.5. dt Fig.4.5. Force applied on a damper with displacement in one direction If a force f is applied on a damperD and it displays distance x in the direction of f and 1  dx dx 1 2 distance x in the opposite direction, then f D as shown in Fig.4.6. 2  dt dt  Fig.4.6. Force applied on a damper with displacement in two directions 4.1.2. Rotational mechanical system There are three basic elements in a Rotational mechanical system, i.e. (a) inertia, (b) spring and (c) damper. (a) Inertia: A body with aninertia is denoted by J. If a torqueT is applied on it and it 2 d displays distanceӨ, then . If a torqueT is applied on a body with inertia T J 2 dt J and it displays distance Ө in the direction of T and distance Ө in the opposite 1 2 2 2  d d 1 2 direction, then T J .  2 2 dt dt  (b) Spring: A spring is denoted by K. If a torqueT is applied on it and it displays distanceӨ, then T K . If a torqueT is applied on a body with inertia J and it displays distance Ө in the direction of T and distance Ө in the opposite 1 2 direction, then T K .  1 2 (c) Damper: A damper is denoted by D. If a torqueT is applied on it and it displays d distanceӨ, then T D . If a torqueT is applied on a body with inertia J and it dt21 displays distance Ө in the direction of T and distance Ө in the opposite 1 2 d d  1 2 direction, then T D .  dt dt  4.2. Components of an electrical system: There are three basic elements in an electrical system, i.e. (a) resistor (R), (b) inductor(L) and (c) capacitor (C). Electrical systems are of two types, i.e. (i) voltage source electrical system and (ii) current source electrical system. 4.2.1. Voltage source electrical system: If i is the current through a resistor(Fig.4.7) and v is the voltage drop in it, then v Ri . If i is the current through an inductor (Fig.4.7) and v is the voltage developed in it, di v L then . dt If i is the current through a capacitor(Fig.4.7) and v is the voltage developed in it, 1 then v idt .  C Fig.4.7. Current and voltage shown in resistor, inductor and capacitor 4.2.2. Current source electrical system: v If i is the current through a resistor and v is the voltage drop in it, then i . R If i is the current through an inductor and v is the voltage developed in it, then 1 i vdt .  L If i is the current through a capacitor and v is the voltage developed in it, then dv i C . dt 4.2.3. Work out problems: Q.4.1. Find system transfer function betweenvoltage drop across the capacitanceand input voltage in the followingRC circuit as shown in Fig.4.8. Fig.4.8. 22 Solution Voltage across resistance, e (t) i(t)R R 1 Voltage across capacitance, e (t) i(t)dt C  C 1 Total voltage drop, e e e i(t)R i(t)dt i R C  C  1 Laplace transform of above equation, E (s) I(s) R i Cs  System transfer function betweenvoltage drop across the capacitanceand input E (s) 1 1 C voltage,  E (s) RCs1 s1 i where, RC is the time-constant Q.4.2. Find system transfer function betweenfunction between the inductance currentto the source currentin the followingRL circuit as shown in Fig.4.9. Fig.4.9. e(t) Voltage across the Resistance, e(t) i R i R R R di 1 L Voltage across the Inductance, e(t) L i e(t)dt L  dt L e(t) 1 Total current, i i i e(t)dt a R L  R L Laplace transform of the current source, 1 1 E  I (s) E(s) and I (s) a L R Ls Ls  Transfer function between the inductance current to the source current, I (s) 1 1 L  L I (s) s1 a s1 R

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