Ch.03 Modeling in Time Domain

1/21/2016 System Dynamics and Control 3.01 Modeling in Time Domain 03 Modeling in Time Domain HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 3.03 Nguyen Tan Tien Modeling in Time Domain System Dynamics and Control 3.02 Modeling in Time Domain Chapter Objectives After completing this chapter, the student will be able to • Find a mathematical model, called a state-space representation, for a linear, time invariant system • Model electrical and mechanical systems in state space • Convert a transfer function to state space • Convert a state-space representation to a transfer function • Linearize a state-space representation 𝐶 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 3.04 Nguyen Tan Tien Modeling in Time Domain §1.Introduction Two approaches are available for the analysis and design of feedback control systems - The classical, or frequency-domain, technique Major disadvantage: can be applied only to linear, timeinvariant systems or systems that can be approximated as such Major advantage: rapidly provide stability and transient response information §1.Introduction - The modern, or time domain, state-space technique • A unified method for modeling, analyzing, and designing a wide range of systems • Can be used to represent nonlinear systems that have backlash, saturation, and dead zone • Can handle, conveniently, systems with nonzero initial conditions • Can be used to represent time-varying systems, (for example, missiles with varying fuel levels or lift in an aircraft flying through a wide range of altitudes) • Can be compactly represented in state space for multipleinput, multiple-output systems • Can be used to represent systems with a digital computer in the loop or to model systems for digital simulation HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 3.05 Nguyen Tan Tien Modeling in Time Domain System Dynamics and Control 3.06 Đ1.Introduction ã With a simulated system, system response can be obtained for changes in system parameters - an important design tool • The state space approach is also attractive because of the availability of numerous state-space software packages for the personal computer §2.Some Observations HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Nguyen Tan Tien Modeling in Time Domain Select the current 𝑖(𝑡) as a variable, write the loop equation 𝑑𝑖(𝑡) 𝐿 + 𝑅𝑖 𝑡 = 𝑣(𝑡) 𝑑𝑡 𝑑𝑖(𝑡) 𝑅 ⟹ = − 𝑖 𝑡 + 𝑣(𝑡) 𝑑𝑡 𝐿 Nguyen Tan Tien 1/21/2016 System Dynamics and Control 3.07 Modeling in Time Domain §2.Some Observations Select the current 𝑖(𝑡) as a variable, write the loop equation 𝑑𝑞(𝑡) 𝑖(𝑡) = 𝑑𝑡 𝑑𝑖(𝑡) 𝐿 + 𝑅𝑖 𝑡 + 𝑖𝑑𝑡 = 𝑣(𝑡) 𝑑𝑡 𝐶 ⟹ 𝑑𝑞 𝑡 = 𝑖(𝑡) 𝑑𝑡 𝑑𝑖(𝑡) 𝑅 = − 𝑞 𝑡 − 𝑖 𝑡 + 𝑣(𝑡) 𝑑𝑡 𝐿𝐶 𝐿 𝐿 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 3.09 Nguyen Tan Tien Modeling in Time Domain System Dynamics and Control 3.08 Modeling in Time Domain §2.Some Observations 𝑑𝑞 𝑡 = 𝑖(𝑡) 𝑑𝑡 𝑑𝑖(𝑡) 𝑅 =− 𝑞 𝑡 − 𝑖 𝑡 + 𝑣(𝑡) 𝑑𝑡 𝐿𝐶 𝐿 𝐿 The state equation can be written in vector-matrix form 𝒙 = 𝑨𝒙 + 𝑩𝑢 𝑞(𝑡) 𝑅 , 𝑩 = , 𝑢 = 𝑣(𝑡) ,𝑨 = where, 𝑥 = 𝑖(𝑡) − − 𝐿𝐶 𝐿 𝐿 The output equation can be written in vector-matrix form 𝑦 = 𝑪𝒙 + 𝑫𝑢 𝑞 𝑡 where, 𝑦 = 𝑣𝐿 𝑡 , 𝑪 = − , 𝐷 = 1, 𝑢 = 𝑣(𝑡) −𝑅 , 𝑥 = 𝑖 𝑡 𝐶 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 3.10 Nguyen Tan Tien Modeling in Time Domain §3.The General State-Space Representation Review - Linear combination: A linear combination of 𝑛 variables, 𝑥𝑖 , for 𝑖 = ÷ 𝑛, is given by the following sum, 𝑆 𝑆 = 𝐾𝑛 𝑥𝑛 + 𝐾𝑛−1 𝑥𝑛−1 + ⋯ + 𝐾1 𝑥1 , where 𝐾𝑖 is constant - Linear independence: A set of variables is said to be linearly independent if none of the variables can be written as a linear combination of the others - System variable: Any variable that responds to an input or initial conditions in a system - State variables: The smallest set of linearly independent system variables such that the values of the members of the set at time 𝑡0 along with known forcing functions completely determine the value of all system variables for all t ≥ 𝑡0 - State vector: A vector whose elements are the state variables §3.The General State-Space Representation - State space: The 𝑛 -dimensional space whose axes are the state variables In the figure • the state variables: 𝑣𝑅 and 𝑣𝐶 • the state trajectory can be thought of as being mapped out by the state vector, 𝑥(𝑡), for a range of 𝑡 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 3.11 Nguyen Tan Tien Modeling in Time Domain - State equations: A set of 𝑛 simultaneous, first-order differential equations with 𝑛 variables, where the 𝑛 variables to be solved are the state variables - Output equation: The algebraic equation that expresses the output variables of a system as linear combinations of the state variables and the inputs System Dynamics and Control 3.12 Nguyen Tan Tien Modeling in Time Domain §3.The General State-Space Representation - A system is represented in state space by the following equations 𝒙 = 𝑨𝒙 + 𝑩𝒖 𝒚 = 𝑪𝒙 + 𝑫𝒖 for 𝑡 ≥ 𝑡0 and initial conditions, 𝒙(𝑡0 ), where 𝒙 : state vector 𝒙 : derivative of the state vector with respect to time 𝒚 : output vector 𝒖 : input or control vector 𝑨 : system matrix 𝑩 : input matrix 𝑪 : output matrix 𝑫 : feedforward matrix §4.Applying the State-Space Representation In this section, we apply the state-space formulation to the representation of more complicated physical systems The first step in representing a system is to select the state vector, which must be chosen according to the following considerations 1.A minimum number of state variables must be selected as components of the state vector This minimum number of state variables is sufficient to describe completely the state of the system 2.The components of the state vector (that is, this minimum number of state variables) must be linearly independent HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Nguyen Tan Tien 1/21/2016 System Dynamics and Control 3.13 Modeling in Time Domain System Dynamics and Control 3.14 Modeling in Time Domain §4.Applying the State-Space Representation - Ex.3.1 Representing an Electrical Network Given the electrical network, find a state-space representation if the output is the current through the resistor Solution Step Label all of the branch currents in the network These include 𝑖𝐿 , 𝑖𝑅 , and 𝑖𝐶 Step Select the state variables by writing the derivative equation for all energy storage elements, 𝐿 and 𝐶 𝑑𝑣𝐶 (3.22) 𝐶 = 𝑖𝐶 𝑑𝑡 𝑑𝑖𝐿 (3.23) 𝐿 = 𝑣𝐿 𝑑𝑡 §4.Applying the State-Space Representation Step Apply network theory, such as Kirchhoff’s voltage and current laws, to obtain 𝑖𝐶 and 𝑣𝐿 in terms of the state variables, 𝑣𝐶 and 𝑖𝐿 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 3.15 Nguyen Tan Tien Modeling in Time Domain §4.Applying the State-Space Representation Step Substitute the results of Eqs (3.24) and (3.25) into Eqs (3.22) and (3.23) to obtain the following state equations 𝑑𝑣𝐶 1 =− 𝑣 + 𝑖 𝑑𝑡 𝑅𝐶 𝐶 𝐶 𝐿 𝑑𝑖𝐿 1 = − 𝑣𝐶 + 𝑣(𝑡) 𝑑𝑡 𝐿 𝐿 𝐶 𝑑𝑣𝐶 𝑑𝑡 = 𝑖𝐶 (3.22), 𝐿 𝑑𝑖𝐿 𝑑𝑡 = 𝑣𝐿 (3.23), 𝑖𝐶 = − 𝑅 𝑣𝐶 + 𝑖𝐿 (3.24), HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 3.17 𝑖𝐶 = −𝑖𝑅 + 𝑖𝐿 = − 𝑣𝐶 + 𝑖𝐿 (3.24) 𝑅 which yields 𝑖𝐶 in terms of the state variables, 𝑣𝐶 and 𝑖𝐿 Around the outer loop, 𝑣𝐿 = −𝑣𝐶 + 𝑣(𝑡) (3.25) which yields 𝑣𝐿 in terms of the state variable, 𝑣𝐶 , and the source, 𝑣(𝑡) System Dynamics and Control Nguyen Tan Tien Modeling in Time Domain Solution Step Label all of the branch currents in the network Step Select the state variables by listing the voltage-current relationships for all of the energy-storage elements 𝑑𝑖𝐿 (3.30.a) 𝐿 = 𝑣𝐿 𝑑𝑡 𝑑𝑣𝐶 (3.30.b) 𝐶 = 𝑖𝐶 𝑑𝑡 Select the state variables: 𝑥1 = 𝑖𝐿 and 𝑥2 = 𝑣𝐶 (3.31) Nguyen Tan Tien 3.16 Modeling in Time Domain 𝑣 (3.28) 𝑅 𝐶 The state-space representation is found by representing Eqs (3.27) and (3.28) in vector-matrix form 𝑣𝐶 −1/𝑅𝐶 1/𝐶 𝑣𝐶 = 𝑖𝐿 + 1/𝐿 𝑣(𝑡) 𝑖𝐿 −1/𝐿 𝑣𝐶 𝑖𝑅 = 1/𝑅 𝑖 𝐿 𝑖𝑅 = (3.27.b) 𝑣𝐿 = −𝑣𝐶 + 𝑣(𝑡) (3.25) Nguyen Tan Tien §4.Applying the State-Space Representation Step Find the output equation Since the output is 𝑖𝑅 (𝑡) (3.27.a) §4.Applying the State-Space Representation - Ex.3.2 Representing an Electrical Network with a Dependent Source Find the state and output equations for the electrical network if the output vector is 𝑦 = 𝑣𝑅2 𝑖𝑅2 𝑇 HCM City Univ of Technology, Faculty of Mechanical Engineering At Node 1, 𝑑𝑣𝐶 𝑑𝑡 1 = − 𝑅𝐶 𝑣𝐶 + 𝐶 𝑖𝐿 (3.27.a), 𝑑𝑖𝐿 𝑑𝑡 1 = − 𝐿 𝑣𝐶 + 𝐿 𝑣(𝑡) (3.27.b) HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control Nguyen Tan Tien 3.18 Modeling in Time Domain §4.Applying the State-Space Representation Step Using Kirchhoff’s voltage and current laws to find 𝑖𝐿 , 𝑣𝐶 in terms of the state variables Around the mesh containing 𝐿 and 𝐶 𝑣𝐿 = 𝑣𝐶 + 𝑣𝑅2 = 𝑣𝐶 + 𝑖𝑅2 𝑅2 At node 2, 𝑖𝑅2 = 𝑖𝐶 + 4𝑣𝐿 𝑣𝐿 = 𝑣𝐶 + (𝑖𝐶 + 4𝑣𝐿)𝑅2 = At node 𝑖𝐶 = 𝑖 − 𝑖𝑅1 − 𝑖𝐿 = 𝑖 − HCM City Univ of Technology, Faculty of Mechanical Engineering (𝑣 + 𝑖 𝑅 ) (3.35) − 4𝑅2 𝐶 𝐶 𝑣𝑅1 𝑣𝐿 − 𝑖𝐿 = 𝑖 − − 𝑖𝐿 (3.36) 𝑅1 𝑅1 Nguyen Tan Tien 1/21/2016 System Dynamics and Control 3.19 Modeling in Time Domain §4.Applying the State-Space Representation Rewriting Eqs (3.35) and (3.36) − 4𝑅2 𝑣𝐿 − 𝑅2 𝑖𝐶 = 𝑣𝐶 −(1/𝑅1 )𝑣𝐿 − 𝑖𝐶 = 𝑖𝐿 − 𝑖 Writing the result in vector-matrix form 𝑣𝐿 = 𝑅2 𝑖𝐿 − 𝑣𝐶 − 𝑅2 𝑖 (3.38) ∆ 1 𝑖𝐶 = − 4𝑅2 𝑖𝐿 + 𝑣𝐶 − − 4𝑅2 𝑖 (3.39) ∆ 𝑅1 where, 𝑅2 ∆= − − 4𝑅2 + 𝑅1 𝑅2/(𝐿∆) −1/(𝐿∆) 𝑖𝐿 −𝑅2/(𝐿∆) 𝑖𝐿 = + 𝑖 𝑣𝐶 − 4𝑅2 /(𝐶∆) 1/(𝑅1𝐶∆) 𝑣𝐶 − 4𝑅2 /(𝐶∆) 𝑣𝐿 = 1−4𝑅 (𝑣𝐶 + 𝑖𝐶 𝑅2 ) (3.35), Nguyen Tan Tien Modeling in Time Domain §4.Applying the State-Space Representation - Ex.3.3 Representing a Translational Mechanical System Find the state equations for the translational mechanical system Solution Find the Laplace-transformed equations of motion + 𝑀1 𝑠 + 𝐷𝑠 + 𝐾 𝑋1 − 𝐾𝑋2 = −𝐾𝑋1 + 𝑀2𝑠 + 𝐾 𝑋2 = 𝐹 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 3.23 Nguyen Tan Tien Modeling in Time Domain Modeling in Time Domain Since the specified output variables are 𝑣𝑅2 and 𝑖𝑅2 , note that around the mesh containing 𝐶, 𝐿, and 𝑅2 𝑣𝑅2 = −𝑣𝐶 + 𝑣𝐿 (3.42.a) 𝑖𝑅2 = 𝑖𝐶 + 4𝑣𝐿 (3.42.b) Substituting Eqs (3.38) and (3.39) into Eq.(3.42) 𝑣𝑅2 𝑅2/∆ −(1 + 1/∆) 𝑖𝐿 −𝑅2/∆ + 𝑖 𝑖𝑅2 = 1/∆ −1/∆ − 4𝑅1 /(𝑅1∆) 𝑣𝐶 𝑣𝐿 = ∆ 𝑅2 𝑖𝐿 − 𝑣𝐶 − 𝑅2 𝑖 (3.38), 3.21 3.20 §4.Applying the State-Space Representation Step Derive the output equation 𝑣 𝑖𝐶 = 𝑖 − 𝑅𝐿 − 𝑖𝐿 (3.36) HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control System Dynamics and Control 1 𝑖𝐶 = ∆ − 4𝑅2 𝑖𝐿 + 𝑅 𝑣𝐶 − − 4𝑅2 𝑖 (3.39) HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 3.22 Nguyen Tan Tien Modeling in Time Domain §4.Applying the State-Space Representation Take the inverse Laplace transform assuming zero initial conditions 𝑑2𝑥1 𝑑𝑥1 𝑀1𝑠2 + 𝐷𝑠 + 𝐾 𝑋1 − 𝐾𝑋2 = ⟹ 𝑀1 + 𝐷 + 𝐾(𝑥1 − 𝑥2) = (3.44) 𝑑𝑡 𝑑𝑡 𝑑 𝑥 −𝐾𝑋1 + 𝑀2𝑠2 + 𝐾 𝑋2 = 𝐹 ⟹ −𝐾𝑥1 + 𝑀2 + 𝐾𝑥2 = 𝑓(𝑡) (3.45) 𝑑𝑡 𝑑𝑥1 𝑑𝑥2 𝑑 𝑥1 𝑑𝑣1 𝑑 𝑥2 𝑑𝑣2 Let ≡ 𝑣1 , ≡ 𝑣2 ⟹ = , = 𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑑𝑥1 State equations = +𝑣1 𝑑𝑡 𝑑𝑣1 𝐾 𝐷 𝐾 =− 𝑥 − 𝑣 + 𝑥 𝑑𝑡 𝑀1 𝑀1 𝑀1 𝑑𝑥2 = +𝑣2 𝑑𝑡 𝑑𝑣2 𝐾 𝐾 =+ 𝑥 − 𝑥 + 𝑓(𝑡) 𝑑𝑡 𝑀2 𝑀2 𝑀2 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 3.24 Nguyen Tan Tien Modeling in Time Domain §4.Applying the State-Space Representation 𝑑𝑥1 = +𝑣1 𝑑𝑡 𝑑𝑣1 𝐾 𝐷 𝐾 =− 𝑥 − 𝑣 + 𝑥 𝑑𝑡 𝑀1 𝑀1 𝑀1 𝑑𝑥2 = +𝑣2 𝑑𝑡 𝑑𝑣2 𝐾 𝐾 =+ 𝑥 − 𝑥 + 𝑓(𝑡) 𝑑𝑡 𝑀2 𝑀2 𝑀2 In vector matrix form 0 𝐾 𝐷 𝐾 𝑥1 − − 0 𝑣1 𝑀1 𝑀1 𝑀1 = + 𝑓(𝑡) 𝑥2 0 1 𝐾 𝐾 𝑣2 − 𝑀2 𝑀2 𝑀2 §4.Applying the State-Space Representation Note The equations of motion (3.44) and (3.45) can be derived directly from the figure using Newton’s laws of motion HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 𝑑 𝑥1 𝑑𝑥1 +𝐷 + 𝐾(𝑥1 − 𝑥2 ) = 𝑑𝑡 𝑑𝑡 𝑑 𝑥2 𝑀2 + 𝐾(𝑥2 − 𝑥1 ) = 𝑓(𝑡) 𝑑𝑡 𝑑 𝑥1 𝑑𝑥1 ⟹ 𝑀1 + 𝐷 + 𝐾(𝑥1 − 𝑥2 ) = 𝑑𝑡 𝑑𝑡 𝑑 𝑥2 −𝐾𝑥1 + 𝑀2 + 𝐾𝑥2 = 𝑓(𝑡) 𝑑𝑡 𝑀1 (3.44) (3.45) Nguyen Tan Tien 1/21/2016 System Dynamics and Control 3.25 Modeling in Time Domain System Dynamics and Control 3.26 Modeling in Time Domain §4.Applying the State-Space Representation Skill-Assessment Ex.3.1 Problem Find the state-space representation of the electrical network with the output is 𝑣𝑜 (𝑡) Solution Identifying appropriate variables on the circuit yields Writing the derivative relations 𝑑𝑣𝐶1 𝐶1 = 𝑖𝐶1 𝑑𝑡 𝑑𝑖𝐿 𝐿 = 𝑣𝐿 𝑑𝑡 𝑑𝑣𝐶2 𝐶2 = 𝑖𝐶2 𝑑𝑡 §4.Applying the State-Space Representation Using Kirchhoff’s current and voltage laws HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 3.27 Nguyen Tan Tien Modeling in Time Domain 𝑖𝐶1 = 𝑖𝐿 + 𝑖𝑅 = 𝑖𝐿 + (𝑣 − 𝑣𝐶2 ) 𝑅 𝐿 𝑣𝐿 = −𝑣𝐶1 + 𝑣𝑖 𝑖𝐶2 = 𝑖𝑅 = (𝑣𝐿 − 𝑣𝐶2 ) 𝑅 Substituting and rearranging 𝑑𝑣𝐶1 1 1 =− 𝑣 + 𝑖 − 𝑣 + 𝑣 𝑑𝑡 𝑅𝐶1 𝐶1 𝐶1 𝐿 𝑅𝐶1 𝐶2 𝑅𝐶1 𝑖 𝑑𝑖𝐿 1 = − 𝑣𝐶1 + 𝑣𝑖 𝑑𝑡 𝐿 𝐿 𝑑𝑣𝐶2 1 =− 𝑣𝐶1 − 𝑣 + 𝑣 𝑑𝑡 𝑅𝐶2 𝑅𝐶2 𝐶2 𝑅𝐶2 𝑖 The output 𝑣𝑜 = 𝑣𝐶2 System Dynamics and Control 3.28 Nguyen Tan Tien Modeling in Time Domain §4.Applying the State-Space Representation 𝑑𝑣𝐶1 1 1 =− 𝑣 + 𝑖 − 𝑣 + 𝑣 𝑑𝑡 𝑅𝐶1 𝐶1 𝐶1 𝐿 𝑅𝐶1 𝐶2 𝑅𝐶1 𝑖 𝑑𝑖𝐿 1 = − 𝑣𝐶1 + 𝑣𝑖 𝑑𝑡 𝐿 𝐿 𝑑𝑣𝐶2 1 =− 𝑣𝐶1 − 𝑣 + 𝑣 𝑑𝑡 𝑅𝐶2 𝑅𝐶2 𝐶2 𝑅𝐶2 𝑖 The output 𝑣𝑜 = 𝑣𝐶2 , the equations in vector-matrix form 1 1 − − 𝑅𝐶1 𝐶1 𝑅𝐶1 𝑅𝐶1 𝑣𝐶1 1 𝒙 = 𝑖𝐿 = − + 𝑣𝑖 0 𝐿 𝐿 𝑣𝐶2 1 − − 𝑅𝐶2 𝑅𝐶2 𝑅𝐶2 𝑦= 0 1𝒙 §4.Applying the State-Space Representation Skill-Assessment Ex.3.2 Problem Represent the translational mechanical system in statespace, where 𝑥3 (𝑡) is the output HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 3.29 Nguyen Tan Tien Modeling in Time Domain §4.Applying the State-Space Representation Defining state variables 𝑧1 = 𝑥1 , 𝑧2 = 𝑥1 , 𝑧3 = 𝑥2 , 𝑧4 = 𝑥2 , 𝑧5 = 𝑥3 , 𝑧6 = 𝑥3 Write the state equations 𝑧1 = 𝑧2 𝑧2 = 𝑥1 = −𝑥1 − 𝑥1 + 𝑥2 + 𝑓 = −𝑧2 − 𝑧1 + 𝑧4 + 𝑓 𝑧3 = 𝑥2 = 𝑧4 𝑧4 = 𝑥2 = 𝑥1 − 𝑥2 − 𝑥2 + 𝑥3 = 𝑧2 − 𝑧4 − 𝑧3 + 𝑧5 𝑧5 = 𝑥3 = 𝑧6 𝑧6 = 𝑥3 = −𝑥3 − 𝑥3 + 𝑥2 = −𝑧6 − 𝑧5 + 𝑧3 𝑥1 = −𝑥1 − 𝑥1 + 𝑥2 + 𝑓, 𝑥2 = +𝑥1 − 𝑥2 − 𝑥2 + 𝑥3 , HCM City Univ of Technology, Faculty of Mechanical Engineering Solution Writing the equations of motion 𝑠 + 𝑠 + 𝑋1 − 𝑠𝑋2 = 𝐹 −𝑠𝑋1 + 𝑠 + 𝑠 + 𝑋2 − 𝑋3 = −𝑋2 + 𝑠 + 𝑠 + 𝑋3 = Taking the inverse Laplace transform and simplifying 𝑥1 = −𝑥1 − 𝑥1 + 𝑥2 + 𝑓 𝑥2 = +𝑥1 − 𝑥2 − 𝑥2 + 𝑥3 𝑥3 = −𝑥3 − 𝑥3 + 𝑥2 System Dynamics and Control 3.30 Nguyen Tan Tien Modeling in Time Domain §4.Applying the State-Space Representation 𝑧1 = 𝑧2 𝑧2 = −𝑧2 − 𝑧1 + 𝑧4 + 𝑓 𝑧3 = 𝑧4 𝑧4 = 𝑧2 − 𝑧4 − 𝑧3 + 𝑧5 𝑧5 = 𝑧6 𝑧6 = −𝑧6 − 𝑧5 + 𝑧3 The output 𝑦 = 𝑧5, the equations in vector-matrix form 0 0 −1 −1 0 0 0 𝒛= 𝒛+ 𝑓(𝑡) −1 −1 0 0 0 0 −1 −1 𝑦= 0 0 0𝒛 𝑥3 = −𝑥3 − 𝑥3 + 𝑥2 Nguyen Tan Tien HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 1/21/2016 System Dynamics and Control 3.31 Modeling in Time Domain §5.Converting a Transfer Function to State-Space Consider the differential equation 𝑑𝑦 𝑑𝑛 𝑦 𝑑 𝑛−1 𝑦 + 𝑎𝑛−1 + ⋯ + 𝑎1 + 𝑎0 𝑦 = 𝑏0 𝑢 𝑛 𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑛−1 𝑦 : output 𝑢 : input 𝑎𝑖 ’s, 𝑏0 : constant Choose the output, 𝑦, and its derivatives as the state variables 𝑑𝑦 𝑑2𝑦 𝑑 𝑛−1 𝑦 𝑥1 = 𝑦, 𝑥2 = , 𝑥3 = , ⋯ , 𝑥𝑛 = 𝑛−1 𝑑𝑡 𝑑𝑡 𝑑𝑡 Differentiating both sides yields 𝑑𝑦 𝑑 𝑦 𝑑 𝑦 𝑑𝑛 𝑦 𝑥1 = , 𝑥2 = , 𝑥3 = , ⋯ , 𝑥𝑛 = 𝑛 𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑑𝑡 Differentiating both sides yields 𝑥1 = 𝑥2 , 𝑥2 = 𝑥3 , 𝑥3 = 𝑥4 , ⋯ , 𝑥𝑛−1 = 𝑥𝑛 𝑥𝑛 = −𝑎0 𝑥1 − 𝑎1 𝑥2 ⋯ − 𝑎𝑛−1 𝑥𝑛 + 𝑏0 𝑢 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 3.33 Nguyen Tan Tien Modeling in Time Domain System Dynamics and Control 3.32 Modeling in Time Domain §5.Converting a Transfer Function to State-Space The phase-variable form of the state equations 𝑥1 𝑥1 ⋯ 0 𝑥2 𝑥2 0 ⋯ 0 𝑥3 0 ⋯ 0 𝑥3 = + 𝑢 ⋮ ⋮ ⋮ ⋮ ⋱ ⋮ ⋮ ⋮ ⋮ 𝑥𝑛−1 0 ⋯ 𝑥𝑛−1 −𝑎1 −𝑎2 −𝑎3 ⋯ −𝑎𝑛−2 −𝑎𝑛−1 𝑥𝑛 𝑏0 𝑥𝑛 𝑥1 𝑥2 𝑥3 𝑦= 0 ⋯ 0 ⋮ 𝑥𝑛−1 𝑥𝑛 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 3.34 Nguyen Tan Tien Modeling in Time Domain §5.Converting a Transfer Function to State-Space - Ex.3.4 Converting a TF with Constant Term in Numerator Find the state-space representation in phase-variable form for the TF Solution Step Find the associated differential equation 𝐶(𝑠) 24 = 𝑅(𝑠) 𝑠 + 9𝑠 + 26𝑠 + 24 ⟹ 𝑠 + 9𝑠 + 26𝑠 + 24 𝐶 𝑠 = 24𝑅(𝑠) Take the inverse Laplace transform, assuming zero initial conditions 𝑐 + 9𝑐 + 26𝑐 + 24𝑐 = 24𝑟 §5.Converting a Transfer Function to State-Space Step Select the state variables Choosing the state variables 𝑥1 = 𝑐, 𝑥2 = 𝑐, 𝑥3 = 𝑐 𝑥1 = 𝑥2 𝑥2 = 𝑥3 𝑥3 = −24𝑥1 − 26𝑥2 − 9𝑥3 + 24𝑟 𝑦 = 𝑐 = 𝑥1 In vector-matrix form 𝑥1 𝑥1 𝑥2 = 0 𝑥2 + 𝑟 𝑥3 −24 −26 −9 𝑥3 24 𝑥1 𝑦 = 0 𝑥2 𝑥3 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering 𝑐 + 9𝑐 + 26𝑐 + 24𝑐 = 24𝑟 System Dynamics and Control 3.35 Nguyen Tan Tien Modeling in Time Domain §5.Converting a Transfer Function to State-Space An equivalent block diagram of the system 𝑥1 𝑥1 𝑥2 = 0 𝑥2 + 𝑟 𝑥3 −24 −26 −9 𝑥3 24 𝑥1 𝑦 = 0 𝑥2 𝑥3 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 3.36 Nguyen Tan Tien Modeling in Time Domain §5.Converting a Transfer Function to State-Space Run ch3p1 through ch3p4 in Appendix B Learn how to use MATLAB to • represent the system matrix A, the input matrix B, and the output matrix C • convert a transfer function to the state-space representation in phase-variable form • solve Ex.3.4 Nguyen Tan Tien HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 1/21/2016 System Dynamics and Control 3.37 Modeling in Time Domain §5.Converting a Transfer Function to State-Space - If a TF has a polynomial in 𝑠 in the numerator that is of order less than the polynomial in the denominator the numerator and denominator can be handled separately: separate the transfer function into two cascaded TFs HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 3.39 Nguyen Tan Tien Modeling in Time Domain §5.Converting a Transfer Function to State-Space - Ex.3.5 Converting a TF with Polynomial in Numerator Find the state-space representation of the TF Solution Step Separate the system into two cascaded blocks Step Find the state equations for the block containing the denominator 𝑥1 𝑥1 𝑥2 = 0 𝑥2 + 𝑟 𝑥3 −24 −26 −9 𝑥3 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 3.41 Nguyen Tan Tien Modeling in Time Domain System Dynamics and Control 3.38 Modeling in Time Domain §5.Converting a Transfer Function to State-Space • The first TF with just the denominator is converted to the phase-variable representation in state space, phase variable 𝑥1 is the output, and the rest of the phase variables are the internal variables of the first block • The second TF with just the numerator yields 𝑌 𝑠 = 𝐶 𝑠 = (𝑏2 𝑠 + 𝑏1 𝑠 + 𝑏0 )𝑋1 (𝑠) Taking the inverse Laplace transform with zeros initial conditions 𝑑 𝑥1 𝑑𝑥1 𝑦 𝑡 = 𝑏2 + 𝑏1 + 𝑏0 𝑥1 𝑑𝑡 𝑑𝑡 = 𝑏0 𝑥1 + 𝑏1 𝑥2 + 𝑏2 𝑥3 Hence, the second block simply forms a specified linear combination of the state variables developed in the first block HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 3.40 Nguyen Tan Tien Modeling in Time Domain §5.Converting a Transfer Function to State-Space Step Introduce the effect of the block with the numerator The second block states that 𝐶 𝑠 = 𝑏2 𝑠 + 𝑏1 𝑠 + 𝑏0 𝑋1 𝑠 = 𝑠 + 7𝑠 + 𝑋1 𝑠 Taking the inverse Laplace transform with zero initial conditions 𝑐 = 𝑥1 + 7𝑥1 + 2𝑥1 = 𝑥3 + 7𝑥2 + 2𝑥1 The output equation 𝑥1 𝑦 = 𝑏0 𝑏1 𝑏2 𝑥2 𝑥3 𝑥1 = 𝑥2 𝑥3 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 3.42 Nguyen Tan Tien Modeling in Time Domain §5.Converting a Transfer Function to State-Space An equivalent block diagram of the system 𝑥1 𝑥1 𝑥1 𝑥2 = 𝑦 = 𝑥2 𝑥2 + 𝑟, 𝑥3 𝑥3 −24 −26 −9 𝑥3 §5.Converting a Transfer Function to State-Space Skill-Assessment Ex.3.3 Problem Find the state equations and output equation for the phase-variable representation of the TF 2𝑠 + 𝐺 𝑠 = 𝑠 + 7𝑠 + Solution First TF 𝑋(𝑠) = ⟹ 𝑠 + 7𝑠 + 𝑋 𝑠 = 𝑅(𝑠) 𝑅(𝑠) 𝑠 + 7𝑠 + Taking inverse Laplace transform with zero initial conditions 𝑥 + 7𝑥 + 9𝑥 = 𝑟 Defining the state variables as 𝑥1 = 𝑥, 𝑥2 = 𝑥 𝑥1 = 𝑥2 𝑥2 = 𝑥 = −7𝑥 − 𝑥 + 𝑟 = −9𝑥1 − 7𝑥2 + 𝑟 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Nguyen Tan Tien 1/21/2016 System Dynamics and Control 3.43 Modeling in Time Domain §5.Converting a Transfer Function to State-Space Second TF The output equation 𝑐 = 2𝑥 + 𝑥 = 𝑥1 + 2𝑥2 Putting all equation in vector-matrix form 𝒙= 𝒙+ 𝑟 −9 −7 𝑐= 2𝒙 𝑥2 = −9𝑥1 − 7𝑥2 + 𝑟, 3.43 Modeling in Time Domain §6.Converting from State Space to a Transfer Function - Given the state and output equations 𝒙 = 𝑨𝒙 + 𝑩𝒖 𝒚 = 𝑪𝒙 + 𝑫𝒖 - Take the Laplace transform assuming zero initial conditions 𝑠𝑿 𝑠 = 𝑨𝑿 𝑠 + 𝑩𝑼(𝑠) 𝒀 𝑠 = 𝑪𝑿 𝑠 + 𝑫𝑼(𝑠) - After some arrangement 𝑿 𝑠 = (𝑠𝑰 − 𝑨)−1 𝑩𝑼(𝑠) 𝒀 𝑠 = 𝑪 𝑠𝑰 − 𝑨 −1 𝑩 + 𝑫 𝑼(𝑠) The matrix 𝑪 𝑠𝑰 − 𝑨 −1 𝑩𝑼 𝑠 + 𝑫: the transfer function matrix - If 𝑼 𝑠 = 𝑈(𝑠) and 𝒀 𝑠 = 𝑌(𝑠) are scalars, the TF 𝑌(𝑠) 𝑇 𝑠 = = 𝑪 𝑠𝑰 − 𝑨 −1 𝑩 + 𝑫 (3.73) 𝑈(𝑠) 𝑐 = 𝑥1 + 2𝑥2 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control System Dynamics and Control 3.45 Nguyen Tan Tien Modeling in Time Domain HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 3.46 Nguyen Tan Tien Modeling in Time Domain §6.Converting from State Space to a Transfer Function - Ex.3.6 State-Space Representation to Transfer Function Find the TF, 𝑇 𝑠 = 𝑌(𝑠)/𝑈(𝑠), for given the system 10 𝒙= 𝑦= 0𝒙 𝒙 + 𝑢, −1 −2 −3 Solution Find (𝑠𝑰 − 𝑨)−1 𝑠 −1 𝑠 0 𝑠𝑰 − 𝑨 = 𝑠 − −1 = 𝑠 𝑠+3 0 𝑠 −1 −2 −3 𝑠 + 3𝑠 + 𝑠+3 −1 𝑠 + 3𝑠 𝑠 −𝑠 −(2𝑠 + 1) 𝑠 adj(𝑠𝑰 − 𝑨) (𝑠𝑰 − 𝑨)−1 = = det(𝑠𝑰 − 𝑨) 𝑠 + 3𝑠 + 2𝑠 + §6.Converting from State Space to a Transfer Function 𝑠 + 3𝑠 + 𝑠+3 −1 𝑠 + 3𝑠 𝑠 −𝑠 −(2𝑠 + 1) 𝑠 adj(𝑠𝑰 − 𝑨) −1 (𝑠𝑰 − 𝑨) = = det(𝑠𝑰 − 𝑨) 𝑠 + 3𝑠 + 2𝑠 + Then 𝑇 𝑠 = 𝑪 𝑠𝑰 − 𝑨 −1 𝑩 + 𝑫 𝑠2 + 3𝑠 + 𝑠+3 −1 𝑠2 + 3𝑠 𝑠 −𝑠 −(2𝑠 + 1) 𝑠2 10 = 0 + 𝑠3 + 3𝑠2 + 2𝑠 + 10(𝑠 + 3𝑠 + 2) = 𝑠 + 3𝑠2 + 2𝑠 + HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 3.47 Nguyen Tan Tien Modeling in Time Domain 𝒙= 0 −1 10 𝒙 + 𝑢, −2 −3 System Dynamics and Control 𝑦= 0𝒙 3.48 Nguyen Tan Tien Modeling in Time Domain §6.Converting from State Space to a Transfer Function Run ch3p5 in Appendix B Learn how to use MATLAB to • convert a state-space representation to a transfer function ã solve Ex.3.6 Đ6.Converting from State Space to a Transfer Function Run ch3sp1 in Appendix F Learn how to use the Symbolic Math Toolbox to • write matrices and vectors • solve Ex.3.6 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Nguyen Tan Tien 1/21/2016 System Dynamics and Control 3.49 Modeling in Time Domain System Dynamics and Control 3.50 Modeling in Time Domain §6.Converting from State Space to a Transfer Function Skill-Assessment Ex.3.4 Problem Convert the state and output equations to a TF −4 −1.5 𝒙= 𝒙+ 𝑢 𝑡 , 𝑦 = 1.5 0.625 𝒙 (3.78) Solution −4 −1.5 𝑠 + 1.5 𝑠𝑰 − 𝑨 = 𝑠 − = −4 𝑠 𝑠 −1.5 adj(𝑠𝑰 − 𝑨) 𝑠+4 (𝑠𝑰 − 𝑨)−1 = = det(𝑠𝑰 − 𝑨) 𝑠 + 4𝑠 + 𝑠 −1.5 𝑠+4 𝐺 𝑠 = 𝑪 𝑠𝑰 − 𝑨 −1 𝑩 = 1.5 0.625 𝑠 + 4𝑠 + 3𝑠 + = 𝑠 + 4𝑠 + §6.Converting from State Space to a Transfer Function HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 3.51 Nguyen Tan Tien Modeling in Time Domain §7.Linearization −4 −1.5 𝒙+ 𝑢 𝑡 𝑦 = 1.5 0.625 𝒙 𝒙= Matlab Result 3s+5 s^2 + s + Continuous-time transfer function System Dynamics and Control Walking robots, such as Hannibal shown here, can be used to explore hostile environments and rough terrain, such as that found on other planets or inside volcanoes System Dynamics and Control 3.53 Nguyen Tan Tien Modeling in Time Domain §7.Linearization 𝑑 𝜃 𝑀𝑔𝐿 + 𝑠𝑖𝑛𝜃 = 𝑇 𝑑𝑡 2 Letting 𝑥1 = 𝜃, 𝑥2 = 𝑑𝜃/𝑑𝑡, the state equation 𝑥1 = 𝑥2 (3.80.a) 𝑀𝑔𝐿 𝑇 (3.80.b) 𝑥2 = − 𝑠𝑖𝑛𝑥1 + 2𝐽 𝐽 The nonlinear Eq (3.80) represent a valid and complete model of the pendulum in state space even under nonzero initial conditions and even if parameters are time varying To apply classical techniques and convert these state equations to a transfer function ⟹ The nonlinear must be linearized 𝐽 Nguyen Tan Tien Nguyen Tan Tien 3.52 Modeling in Time Domain Representing a Nonlinear System First represent the simple pendulum in state space: 𝑀𝑔 is the weight, 𝑇 is an applied torque in the 𝜃 direction, and 𝐿 is the length of the pendulum Assume the mass is evenly distributed, with the center of mass at 𝐿/2 Then linearize the state equations about the pendulum’s equilibrium point-the vertical position with zero angular velocity Solution Drawing the free body diagram Summing the torques 𝑑 𝜃 𝑀𝑔𝐿 𝐽 2+ 𝑠𝑖𝑛𝜃 = 𝑇 𝑑𝑡 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control Nguyen Tan Tien 3.54 Modeling in Time Domain §7.Linearization Linearize the equation about the equilibrium point, 𝑥1 = 𝜃 = 0, 𝑥2 = 𝑑𝜃/𝑑𝑡 = Let 𝑥1 and 𝑥2 be perturbed about the equilibrium point, or 𝑥1 = + 𝛿𝑥1 𝑥2 = + 𝛿𝑥2 Using Eqs (2.182) 𝑑(𝑠𝑖𝑛𝑥1 ) 𝑠𝑖𝑛𝑥1 − 𝑠𝑖𝑛0 = 𝛿𝑥 = 𝛿𝑥1 ⟹ 𝑠𝑖𝑛𝑥1 = 𝛿𝑥1 𝑑𝑥1 𝑥 =0 1 The state equations now become 𝛿𝑥1 = 𝛿𝑥2 𝑀𝑔𝐿 𝑇 𝛿𝑥2 = − 𝛿𝑥1 + 2𝐽 𝐽 𝑓 𝑥 − 𝑓(𝑥0 ) ≈ HCM City Univ of Technology, Faculty of Mechanical Engineering A=[-4 -1.5; 0]; B=[2 0]'; C=[1.5 0.625]; D=0; T=ss(A,B,C,D); T=tf(T) T= §7.Linearization - Ex.3.7 HCM City Univ of Technology, Faculty of Mechanical Engineering (3.78) 𝑑𝑓 𝑑𝑥 𝑥=𝑥0 (𝑥 − 𝑥0 ) (2.182), 𝑥1 = 𝑥2 (3.80.a), 𝑥2 = − HCM City Univ of Technology, Faculty of Mechanical Engineering 𝑀𝑔𝐿 2𝐽 𝑇 𝑠𝑖𝑛𝑥1 + (3.80.b) 𝐽 Nguyen Tan Tien 1/21/2016 System Dynamics and Control 3.55 Modeling in Time Domain System Dynamics and Control 3.56 Modeling in Time Domain §7.Linearization Skill-Assessment Ex.3.5 Problem Represent the translational mechanical system in state space about the equilibrium displacement The spring is nonlinear, 𝑓𝑠 𝑡 = 2𝑥𝑠2 (𝑡) The applied force is 𝑓 𝑡 = 10 + 𝛿𝑓(𝑡), where 𝛿𝑓(𝑡) is a small force about the 10𝑁 constant value Assume the output to be the displacement of the mass, 𝑥(𝑡) Solution The equation of motion 𝑑2𝑥 + 2𝑥 = 10 + 𝛿𝑓(𝑡) (1) 𝑑𝑡 Letting 𝑥 = 𝑥0 + 𝛿𝑥 𝑑 (𝑥0 + 𝛿𝑥) (2) + 2(𝑥0 + 𝛿𝑥)2 = 10 + 𝛿𝑓(𝑡) 𝑑𝑡 §7.Linearization Linearize 𝑥 at 𝑥0 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 3.57 Nguyen Tan Tien (𝑥0 + 𝛿𝑥)2 −𝑥02 = 𝑑 𝑥2 𝑑𝑥 𝛿𝑥 = 2𝑥0 𝛿𝑥 𝑥0 ⟹ (𝑥0 + 𝛿𝑥)2 = 𝑥02 + 2𝑥0 𝛿𝑥 (3) Substituting Eq.(3) into Eq.(1) 𝑑 𝛿𝑥 + 4𝑥0 𝛿𝑥 = −2𝑥02 + 10 + 𝛿𝑓(𝑡) (4) 𝑑𝑡 The force of the spring at equilibrium 𝐹 = 10 = 2𝑥02 ⟹ 𝑥0 = Substituting this value of 𝑥0 into Eq.(4) 𝑑 𝛿𝑥 + 5𝛿𝑥 = 𝛿𝑓(𝑡) 𝑑𝑡 𝑓 𝑥 − 𝑓(𝑥0 ) ≈ 𝑑𝑓 𝑑𝑥 𝑥=𝑥0 (𝑥 − 𝑥0 ) (2.182), 𝑑2 𝑥 𝑑𝑡 + 2𝑥 = 10 + 𝛿𝑓(𝑡) (1) Nguyen Tan Tien Modeling in Time Domain §7.Linearization 𝑑 𝛿𝑥 + 5𝛿𝑥 = 𝛿𝑓(𝑡) 𝑑𝑡 Selecting the state variables 𝑥1 = 𝛿𝑥, 𝑥2 = 𝛿𝑥 The state and output equations 𝑥1 = 𝑥2 𝑥2 = 𝛿𝑥 = −4 5𝑥1 + 𝛿𝑓 𝑡 𝑦 = 𝑥1 Converting to vector-matrix form 𝒙= 𝒙+ 𝛿𝑓(𝑡) −4 𝑦= 0𝒙 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 10 ... Control 3.23 Nguyen Tan Tien Modeling in Time Domain Modeling in Time Domain Since the specified output variables are