University of California at Berkeley Class Notes for ME134: Automatic Control Systems by Anouck Girard Masayoshi Tomizuka Roberto Horowitz ME134 Instructor: Professor Karl Hedrick 5104 Etcheverry Hall [email protected] Department of Mechanical Engineering University of California at Berkeley Fall 2008 © 2004-2008 A. Girard, M. Tomizuka and R. Horowitz 1 Introduction: Automatic Control Systems Recent trends in the development of modern civilization have been in the direction of greater control. With the advent of the steam engine, and the material improvements brought about by the industrial revolution, man has had available greater quantities of power for his use. To use this power effectively, he has been required to learn how to control and how to regulate it. As part of the control process, certain standards have to be established. The performance of the equipment is compared to these standards, and, according to the difference, appropriate action is taken to bring about a closer correspondence between the desired objectives and the actual performance. The need for good control is present in many phases of our existence. We will limit ourselves in this class to a study of problems pertaining to the field of engineering applications of automatic control. The problem is to determine the desired objectives, and the best ways of producing those objectives. open-loop Figure 1 is a block diagram showing an open-loop control system. The input is system sometimes called the reference, while the output is sometimes called the controlled variable. Disturbances can be present in the system. 2 Mathematical Modeling of Systems Concept Given a physical system, find a model (mathematical representation) that accurately predicts the behavior (the output for a given input) of the system. Key Points To analyze and design control systems, we need quantitative mathematical models of systems. The dynamic behavior of systems can be described using differential equations. If the equations can be linearized, then the Laplace transform can be used to simplify the solution. The input/output relationship for linear components and sub-systems can be described in the form of transfer functions. System Representation and Modeling In order to analyze (and subsequently control) most dynamic systems, it is essential to attain a reasonable understanding of how a system functions. To achieve this objective, we formulate mathematical models that help us describe the behavior of systems. Mathematical models generally serve two purposes: a. They are used in conjunction with analytical techniques to develop control schemes for the systems that they represent. 1 b. They are used as a design tool in computer simulation studies. In this context, the model is used as the control object to test and on which to evaluate possible control schemes. This procedure is generally more efficient, cheaper and less time consuming than to test the control schemes on the actual system. energetic Here we will lay the groundwork for the formulation of mathematical models for systems energetic (dynamic) systems. Of paramount importance in our methodology are the state of the concept of state of the system and the selection of a set of state variables that describe the system state state of the system and adequately serve our modeling objectives. variables State Variables and State Equations Consider a single input, single output (SISO) dynamic system, as shown schematically in dynamic system figure 1. Since the system is dynamic, the output y(t) at time t‡ 0 is a function of the past as well as the present inputs: u . Notice that u represents the entire history of the [0,t] [0,t] input variable u(t ) from t = 0 to t = t. It is obviously a problem to keep track of the entire system history in order to be able to predict the future system output. Thus, it is important to determine what the minimum amount of information needed, at time t, to predict the output of the system immediately after time t. u(t) y(t) Dynamic System input output Figure 1. SISO Dynamic System. The state variables are the minimum set of variables such that knowledge of these state variables at time k, together with the current (present) state of the system, is sufficient to variables: determine the future state and output of a system. minimum amount of information The state variables represent the minimum amount of information that needs to be needed to retained at any time t in order to determine the future behavior of a system. Although the predict the number of state variables is unique (that is, it has to be the minimum and necessary output number of variables), for a given system, the choice of state variables is not. Choosing state variables and deriving models is in general a non-trivial problem. In many cases, there is no universal or general procedure. For engineering energetic systems, the choice of state variables still depends on the problem. However, there exists a fairly systematic procedure for selecting state variables and deriving models. The order of the system designates the number n of state variables needed (minimum and system order necessary) to describe the system. 2 The set of n differential or difference first-order equations that govern the relationship model = state equations + between the input to a system and the n state variables are the state equations. Together output with the output equation, they constitute the model of the system. equation Denoting the n state variables of a system as x, x, …, x and defining x& = dx dt, the 1 2 n state equations can be written as: x& (t)= f (x (t),x (t),...,x (t),u(t)) 1 1 1 2 n x& (t)= f (x (t),x (t),...,x (t),u(t)) 2 2 1 2 n ... x& (t)= f (x (t),x (t),...,x (t),u(t)) n n 1 2 n Note that the right-hand side of the state equations is only a function of the state variables and the input at time t. Define the nth order vectors: Ø x ø Ø f ø Œ 1œ Œ 1œ Œ x œ Œ f œ x = 2 and f = 2 Œ œ Œ œ ... ... Œ œ Œ œ º x ß º f ß n n We can write the system of n state equations in vector form as: x&(t) = f(x(t),u(t)) When the dynamic system is linear, we can write the above equation as follows: linear system x&(t) = A(t)x(t)+b(t)u(t) where A(t) is a n by n matrix and b(t) is an n by one vector. Ø b (t)ø Ø a (t) ... a (t)ø Œ 1 œ Œ 11 1n œ Œ b (t)œ A(t)= Œ ... ... œ and b(t) = 2 . Œ œ ... Œº a (t) ... a (t)œß Œ œ n1 nn º b (t)ß n If A and b are constants, then the system is an nth order Linear Time Invariant (LTI) linear time invariant system. (LTI) system x&(t) = Ax(t)+bu(t) The output equation is given by the algebraic equation: 3 y(t) = h(x(t),u(t)) When the system is LTI, the output equation can be written as: y(t)= cx(t)+du(t) [ ] where c = c c ... c is a one by n vector and d is a scalar. 1 2 n Mass Spring Damper Example Consider the mass-spring-damper example shown in figure 2, where x is the position of the mass, v = x& is the velocity of the mass, u is the input force applied to the mass, k is the coefficient of elasticity of the spring, and b is the viscous damping coefficient of the damper. x k u m b Figure 2. Mass-Spring-Damper System. Assuming that the system is linear and time invariant, a model of the system can be derived as follows: a. Define the state of the system to be the position and the velocity of the mass, r [ ] x = x v T. b. Use Newton’s laws to derive the equations of motion for the system: mv&= f + f +u(t), where f and f are the forces exerted by the spring and k v k v damper respectively. c. Use the constitutive relations for the spring and damper, f =- kx and f = - bv. k v d. Combine these equations and write them in matrix form: 4 d Ø x(t)ø Ø 0 1 ø Ø x(t)ø Ø 0 ø Œ œ = Œ œ Œ œ +Œ œ u(t) dt º v(t)ß º - k/m - b/mß º v(t)ß º 1/mß [ ]Ø x(t)ø y(t) = 1 0 Œ œ º v(t)ß Note that this is not the only state equation model that could have been used to describe the system. Difference Equations Example y(k) = u(k-2) The associated state equations would be: x (k) = u(k-1) 1 x (k) = x (k-1) 2 1 The output equation would be: y(k) = x (k-1) = u(k-2) 1 Re-writing: x (k+1) = u(k) 1 x (k+1) = x (k) 2 1 y(k) = x (k) 2 In matrix form: Ø x (k +1)ø Ø 0 0ø Ø x (k)ø Ø 1ø Œ 1 œ = Œ œ Œ 1 œ +Œ œ u(k) º x (k +1)ß º 1 0ß º x (k)ß º 0ß 2 2 [ ]Ø x (k)ø y(k)= 0 1Œ 1 œ º x (k)ß 2 (cid:222) x(k+1) = A.x(k) + b.u(k) y(k) = C.x(k) 5 Modeling of Dynamic Systems: General Approach A dynamic (or energetic) system is a collection of energy storage elements, power dissipative elements, power sources, transformers and transducers. For successful modeling of energetic systems, it is important to know the characteristics of each of these elements. In this class, we will model energetic systems as lumped parameter systems. That is, we will model the system such that each point in the system embodies the properties of the region immediately surrounding it. Lumped parameter systems are described by a finite number of state variables. Example of a Lumped System: Mass-Spring System x k m Figure 3. Mass-Spring System. mx&&+kx= 0 The properties of the surrounding region are lumped (concentrated) at each point. The state equations describing lumped parameter systems can be written as finite order ordinary differential equations. Many energetic systems cannot be modeled as lumped parameter systems. These systems must be modeled as distributed parameter systems. Example of a Distributed System: The Cantilever Beam u(x,t): deflection x Figure 4. Cantilever beam. 6 ¶ 2u(x,t) 1 ¶ 2u(x,t) - = 0 ¶ x2 a2 ¶ t2 Each little “chunk” of mass acts as an elastic segment. The properties of mass and elasticity cannot be separated from each other. The state equations that describe distributed parameter systems are partial differential equations. In the next pages, we will present a unified approach for modeling mechanical, electrical, fluid and/or thermal lumped parameter systems. The first step in modeling energetic systems is to break the system up into elements, then find the basic relations that describe the individual elements that form the system. Physical laws are generally used to obtain these relations. Mechanical systems: Newton’s Laws Isaac Newton, a 17th-18th century English physicist and mathematician (also the guy who had an apple hit his head supposedly), stated three laws that basically describe the statics and dynamics of objects. Statics is the study of forces on an object at rest. Dynamics is the study of how forces affect the motion of a body. Newton's First Law of Motion (Law of Inertia): Every body continues in its state of rest or of uniform speed in a straight line unless it is compelled to change that state by forces acting on it. Newton's Second Law of Motion: The acceleration of an object is directly proportional to the net force acting on it and is inversely proportional to its mass. The direction of the acceleration is in the direction of the applied net force. Newton's Third Law of Motion (Law of Action-Reaction): Whenever one object exerts a force on a second object, the second exerts an equal and opposite force on the first. The Law of the Conservation of Momentum: The total momentum of an isolated system of bodies remains constant. Electric Systems: Kirchhoff’s Rules There are only two relatively simple rules known as Kirchhoff's rules (developed by Gustav Robert Kirchhoff (1824-1887)). • Point Rule: The algebraic sum of the currents toward any branch point is zero. 7 (cid:229) I = 0 • Loop Rule: The algebraic sum of the potential differences in any loop, including those associated with EMFs and those of resistive elements, must equal zero. (cid:229) (cid:229) x - IR= 0 Here is a good set of guidelines when you are working with these laws: 1. Choose any closed loop in the network and designate a direction to traverse the loop when you apply the loop rule (If you have a diagram, draw it on the diagram). 2. Go around the loop in that direction, adding EMF's and potential differences. An EMF is positive if you go from (-) to (+) (which is in the direction of the field E in the source) and negative when you go from (+) to (-). A -IR term is counted negative if the resistor is traversed in the same direction as the assumed current, positive if backwards. 3. Equate the sum of found in the previous step to zero. 4. If you need to, choose another loop to obtain a different relationship. The number of loop equations you have will be one less than the number of loops you have. You will be setting up a system of equations in order to find the current in each loop. 5. The last equation should be a branch equation, where I + I = I (substitute the 1 2 3 currents for whichever branch you choose). 6. Solve the systems of equations. If one of the currents you solve is negative, it simply means that the current goes in the direction opposite the way you assumed. Thermal Systems: The Laws of Thermodynamics Thermodynamics is the study of the inter-relation between heat, work and internal energy of a system. The British scientist and author C.P. Snow had an excellent way of remembering the three laws: 1. You cannot win (that is, you cannot get something for nothing, because matter and energy are conserved). 2. You cannot break even (you cannot return to the same energy state, because there is always an increase in disorder; entropy always increases). 3. You cannot get out of the game (because absolute zero is unattainable). 8