Ch.04 Time Response

2/3/2016 System Dynamics and Control 4.01 Time Response System Dynamics and Control 4.03 4.02 Time Response Learning Outcome After completing this chapter, the student will be able to • Use poles and zeros of transfer functions to determine the time response of a control system • Describe quantitatively the transient response of 1st-order systems • Write the general response of second-order systems given the pole location • Find the damping ratio and natural frequency of a second-order system • Find the settling time, peak time, percent overshoot, and rise time for an underdamped second-order system • Approximate higher-order systems and systems with zeros as first- or second-order systems • Describe the effects of nonlinearities on the system time response • Find the time response from the state-space representation 04 Time Response HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control Nguyen Tan Tien Time Response HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 4.04 Nguyen Tan Tien Time Response §1.Introduction - This chapter • Analyze of system transient response • Demonstrates applications of the system representation by evaluating the transient response from the system model • Evaluate the response of a subsystem prior to inserting it into the closed-loop system • Describe a valuable analysis and design tool, poles and zeros • Analyze the models to find the step response of first- and second-order systems §2.Poles, Zeros, and System Response - The output response of a system is the sum of two responses • The forced response • The natural response - Output response of a system depends on the positions of the poles and zeros and their relationships ⟹ study • the concept of poles and zeros • fundamental to the analysis and design of control systems • simplifies the evaluation of a system’s response HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 4.05 Nguyen Tan Tien Time Response §2.Poles, Zeros, and System Response - Transfer function System Dynamics and Control 4.06 Nguyen Tan Tien Time Response §2.Poles, Zeros, and System Response - Example of poles and zeros of a first order system 𝑌 𝑠 𝑏0 𝑠 𝑚 + 𝑏1 𝑠 𝑚−1 + ⋯ + 𝑏𝑚−1 𝑠 + 𝑏𝑚 = 𝑈 𝑠 𝑎0 𝑠 𝑛 + 𝑎1 𝑠 𝑛−1 + ⋯ + 𝑎𝑛−1 𝑠 + 𝑎𝑛 • The poles of a transfer function are roots of 𝑎0 𝑠 𝑛 + 𝑎1 𝑠 𝑛−1 + ⋯ + 𝑎𝑛−1 𝑠 + 𝑎𝑛 = • The zeros of a transfer function are roots of 𝑏0 𝑠 𝑚 + 𝑏1 𝑠 𝑚−1 + ⋯ + 𝑏𝑚−1 𝑠 + 𝑏𝑚 = 𝐺 𝑠 = HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien System 𝑠+2 𝑠+2 𝐴 𝐵 2/5 3/5 × = = + = + 𝑠 𝑠 + 𝑠(𝑠 + 5) 𝑠 𝑠+5 𝑠 𝑠+5 𝑠+2 𝑠+2 𝐴= = ,𝐵 = = (𝑠 + 5) 𝑠→0 𝑠 𝑠→−5 Response 𝑐 𝑡 = + 𝑒 −5𝑡 5 𝐶 𝑠 = HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 2/3/2016 System Dynamics and Control 4.07 Time Response §2.Poles, Zeros, and System Response System Dynamics and Control 4.09 Nguyen Tan Tien Time Response §2.Poles, Zeros, and System Response - Ex.4.1 Evaluating Response Using Poles Write the output, 𝑐(𝑡), in general terms, for the given system Specify the forced and natural parts of the solution Solution • Each system pole generates an exponential as part of the natural response • The input’s pole generates the forced response 𝐾1 𝐾2 𝐾3 𝐾4 + + + Thus 𝐶 𝑠 = 𝑠 𝑠+2 𝑠+4 𝑠+5 ⟹𝑐 𝑡 = System Dynamics and Control 4.11 𝑐(𝑡) 𝑡= 𝑎 𝑡= 𝑎 Nguyen Tan Tien Time Response 𝑡= 𝑎 4.10 Nguyen Tan Tien Time Response §3.First Order Systems - A first-order system without zeros can be described by the transfer function - If the input is step 𝑎 𝑎 1 × = = − 𝑠 𝑠 + 𝑎 𝑠(𝑠 + 𝑎) 𝑠 𝑠 + 𝑎 - Taking the inverse transform, the step response is given by 𝑐 𝑡 = 𝑐𝑓 𝑡 + 𝑐𝑛 𝑡 = − 𝑒 −𝑎𝑡 𝐶 𝑠 =𝑅 𝑠 𝐺 𝑠 = HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 4.12 𝑡=0 Nguyen Tan Tien Time Response = −𝑎 𝑎: the initial rate of change of the exponential at 𝑡 = - Transient response specifications 𝑎 : time constant 𝑇𝑟 : rise time 𝑇𝑠 : settling time = 0.63 First-order system response to a unit step - The time constant is the time it takes for the step response to rise to 63% of its final value 𝐶 𝑠 =𝑅 𝑠 𝐺 𝑠 = System Dynamics and Control §3.First Order Systems 𝑑 −𝑎𝑡 𝑒 = −𝑎𝑒 −𝑎𝑡 𝑑𝑡 𝑡=0 = 𝑒 −1 = 0.37 = − 𝑒 −𝑎𝑡 HCM City Univ of Technology, Faculty of Mechanical Engineering • system pole, −𝑎, generates the natural response 𝑐𝑛 𝑡 = −𝑒−𝑎𝑡 §3.First Order Systems 1.Time Constant 𝑇𝑐 - Time constant 𝑇𝑐 = 1/𝑎 𝑒 −𝑎𝑡 Time Response • input pole, 0, generates the forced response 𝑐𝑓 𝑡 = 𝐾1 + 𝐾2 𝑒 −2𝑡 + 𝐾3 𝑒 −4𝑡 + 𝐾4 𝑒 −5𝑡 HCM City Univ of Technology, Faculty of Mechanical Engineering 4.08 §2.Poles, Zeros, and System Response Conclusions • A pole of the input function generates the form of the forced response (that is, the pole at the origin generated a step function at the output) • A pole of the transfer function generates the form of the natural response (that is, the pole at generated 𝑒 5𝑡 ) • A pole on the real axis generates an exponential response of the form 𝑒 𝑎𝑡 , where 𝑎 is the pole location on the real axis Thus, the farther to the left a pole is on the negative real axis, the faster the exponential transient response will decay to zero • The zeros and poles generate the amplitudes for both the forced and natural responses (this can be seen from the calculation of 𝐴 and 𝐵) Evolution of a system response HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control - 𝑎 relates to the speed at which the system responds to a step input: the farther the pole from the imaginary axis, the faster the transient response - 𝑎 has the units 1/𝑠, or frequency ⟹ the exponential frequency 𝑎 𝑎 1 × = = − ⟹ 𝑐 𝑡 = 𝑐𝑓 𝑡 + 𝑐𝑛 𝑡 = − 𝑒 −𝑎𝑡 𝑠 𝑠 + 𝑎 𝑠(𝑠 + 𝑎) 𝑠 𝑠 + 𝑎 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 2/3/2016 System Dynamics and Control 4.13 Time Response System Dynamics and Control 4.14 Time Response §3.First Order Systems 2.Rise Time 𝑇𝑟 - The time for the waveform to go from 0.1 to 0.9 of its final value - Rise time is found by solving 𝑐 𝑡 = − 𝑒 −𝑎𝑡 𝑐 𝑡1 = 0.1 𝑙𝑛0.9 − 𝑒−𝑎𝑡1 = 0.1 ⟹ 𝑡1 = − 𝑎 𝑐 𝑡2 = 0.9 𝑙𝑛0.1 − 𝑒 −𝑎𝑡2 = 0.9 ⟹ 𝑡2 = − 𝑎 𝑙𝑛0.1 𝑙𝑛0.9 0.11 2.31 2.2 ⟹ 𝑡2 − 𝑡1 = − + =− + ⟹ 𝑇𝑟 = 𝑎 𝑎 𝑎 𝑎 𝑎 §3.First Order Systems 3.Settling Time 𝑇𝑠 - The time for the response to reach, and stay within, 2% of its final value - Settling time is found by solving 𝑐 𝑡 = − 𝑒 −𝑎𝑡 𝑐 𝑇𝑠 = 0.98 𝑙𝑛0.02 − 𝑒−𝑎𝑇𝑠 = 0.98 ⟹ 𝑇𝑠 = − 𝑎 ⟹ 𝑇𝑠 = 𝑎 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 4.15 Nguyen Tan Tien Time Response §3.First Order Systems 4.First-Order Transfer Functions via Testing - Consider a simple first-order system with the step input System Dynamics and Control Nguyen Tan Tien 4.16 Time Response §3.First Order Systems Example using the plot of 𝐺 𝑠 = 𝑠+7 From the plot • Steady state value 𝑐𝑠𝑠 = 0.72 • Time constant 𝑇𝑐 = 0.13𝑠 Therefore 𝑇𝑐 = ⟹ 𝑎 = 1/0.13 = 7.7 𝑎 𝐾 = 𝑐𝑠𝑠 ⟹ 𝐾 = 𝑎𝑐𝑠𝑠 = 7.7 × 0.72 = 5.54 𝑎 𝐾 𝐾/𝑎 𝐾/𝑎 × = − 𝑠 𝑠+𝑎 𝑠 𝑠+𝑎 If we can identify 𝐾 and 𝑎 from laboratory testing, we can obtain the transfer function of the system 𝐶 𝑠 = 5.54 𝑠 + 7.7 Note: 1st order system has no overshoot and nonzero initial slope The transfer function of the plot HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 4.17 Nguyen Tan Tien Time Response 𝐺 𝑠 = HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 4.18 §3.First Order Systems Skill-Assessment Ex.4.2 Problem Find the time constant, 𝑇𝑐 , rising time, 𝑇𝑟 , and settling time, 𝑇𝑠 , of the system with the following TF 50 𝐺 𝑠 = 𝑠 + 50 Solution From the TF 𝑎 = 50 1 = 0.02𝑠 Time constant 𝑇𝑐 = = 𝑎 50 2.2 2.2 𝑇𝑟 = = = 0.044𝑠 Rising time 𝑎 50 4 𝑇𝑠 = = = 0.08𝑠 Settling time 𝑎 50 §4.Second Order Systems: Introduction - The general second order system HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Nguyen Tan Tien Time Response - Depending on component values, the second order system exhibits a wide range of responses such as • first-order system • damped oscillation • pure oscillation Nguyen Tan Tien 2/3/2016 System Dynamics and Control 4.19 Time Response §4.Second Order Systems: Introduction - Second-order systems, pole plots, and step responses • Overdamped 𝐶 𝑠 = = [r,p,k]=residue([9],[1,9,9,0]), 𝐶 𝑠 = 1/𝑠 → 𝑐(𝑡) = 1, 𝐶 𝑠 = 1/(𝑠 + 𝑎) → 𝑐(𝑡) = 𝑒 −𝑎𝑡 4.21 Time Response • Underdamped 𝑠(𝑠 + 1.146)(𝑠 + 7.854) ⟹ 𝑐 𝑡 = + 0.171𝑒 −7.854𝑡 − 0.171𝑒 −1.146𝑡 Poles: two real at −𝜎1 and −𝜎2 Natural response: 𝑐𝑛 𝑡 = 𝐾1 𝑒 −𝜎1 𝑡 + 𝐾2 𝑒 −𝜎2 𝑡 System Dynamics and Control 4.20 §4.Second Order Systems: Introduction 𝐶 𝑠 = HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control Nguyen Tan Tien Time Response §4.Second Order Systems: Introduction 𝑠(𝑠 + 2𝑠 + 9) 𝑠 𝑠 − −1 − 𝑗 𝑠 − −1 + 𝑗 ⟹ 𝑐 𝑡 = − 1.06𝑒 −𝑡 cos( 8𝑡 − 19.470 ) Poles: two complex at −𝜎𝑑 ± 𝑗𝜔𝑑 Natural response: 𝑐𝑛 𝑡 = 𝐴𝑒 −𝜎𝑑 𝑡 𝑐𝑜𝑠 𝜔𝑑 𝑡 − 𝜙 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 4.22 Nguyen Tan Tien Time Response Đ4.Second Order Systems: Introduction ã Undamped 𝑠 = − 𝑠(𝑠 + 9) 𝑠 𝑠 + 32 ⟹ 𝑐 𝑡 = − 𝑐𝑜𝑠3𝑡 Poles: two imaginary at ±𝑗𝜔1 Natural response: 𝑐𝑛 𝑡 = 𝐴𝑐𝑜𝑠 𝜔1 𝑡 − 𝜙 𝐶 𝑠 = 𝐶 𝑠 = 𝑠/(𝑠 + 𝑎 ) → 𝑐(𝑡) = 𝑐𝑜𝑠𝑎𝑡 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 4.23 Nguyen Tan Tien Time Response §4.Second Order Systems: Introduction HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 4.24 Nguyen Tan Tien Time Response Đ4.Second Order Systems: Introduction ã Step responses for second-order system damping cases • Critically damped 𝑠(𝑠 + 6𝑠 + 9) 1 = − − 𝑠 𝑠+3 𝑠+3 ⟹ 𝑐 𝑡 = − 3𝑡𝑒 −3𝑡 − 𝑒 −3𝑡 Poles: two real at −𝜎1 Natural response: 𝑐𝑛 𝑡 = 𝐾1 𝑒 −𝜎1 𝑡 + 𝐾2 𝑡𝑒 −𝜎1 𝑡 𝐶 𝑠 = 𝐶 𝑠 = 𝑎/(𝑠 + 𝑎)2 → 𝑐(𝑡) = 𝑎𝑡𝑒 −𝑎𝑡 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 2/3/2016 System Dynamics and Control 4.25 Time Response §4.Second Order Systems: Introduction Skill-Assessment Ex.4.3 Problem For each of the following transfer functions, write, by inspection, the general form of the step response 400 a 𝐺 𝑠 = 𝑠 + 12𝑠 + 400 900 b 𝐺 𝑠 = 𝑠 + 90𝑠 + 900 225 c 𝐺 𝑠 = 𝑠 + 30𝑠 + 225 625 d 𝐺 𝑠 = 𝑠 + 625 System Dynamics and Control 4.26 Time Response §4.Second Order Systems: Introduction 400 Solution a 𝐺 𝑠 = 𝑠 + 12𝑠 + 400 400 𝑠 − −6 − 𝑗19.08 𝑠 − −6 + 𝑗19.08 The system has two complex poles at −6 ± 𝑗19.08 The general form of the step response 𝑐 𝑡 = 𝐴 + 𝐵𝑒 −6𝑡 𝑐𝑜𝑠 19.08𝑡 + 𝜙 900 900 b 𝐺 𝑠 = = 𝑠 + 90𝑠 + 900 𝑠 + 78.54 𝑠 + 11.46 The system has two real poles at −78.54 and −11.46 The general form of the step response 𝑐 𝑡 = 𝐴 + 𝐵𝑒 −78.54𝑡 + 𝐶𝑒 −11.46𝑡 = 𝐺(𝑠) has poles at −𝜎𝑑 ± 𝑗𝜔𝑑 ⟹ natural response 𝑐𝑛 𝑡 = 𝐴𝑒 −𝜎𝑑 𝑡 𝑐𝑜𝑠 𝜔𝑑 𝑡 − 𝜙 𝐺(𝑠) has poles at −𝜎1 , −𝜎2 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 4.27 Nguyen Tan Tien Time Response §4.Second Order Systems: Introduction 225 400 c 𝐺 𝑠 = = 𝑠 + 30𝑠 + 225 𝑠 + 15 The system has two repeated real poles at −15 The general form of the step response 𝑐 𝑡 = 𝐴 + 𝐵𝑒 −15𝑡 + 𝐶𝑡𝑒 −15𝑡 625 625 d 𝐺 𝑠 = = 𝑠 + 625 𝑠 + 252 The system has two imaginary poles at ±𝑗25 The general form of the step response 𝑐 𝑡 = 𝐴 + 𝐵𝑐𝑜𝑠(25𝑡 + 𝜙) 𝐺(𝑠) has poles at −𝜎1 , −𝜎1 ⟹ natural response 𝑐𝑛 𝑡 = 𝐾1 𝑒 −𝜎1 𝑡 + 𝐾2 𝑡𝑒 −𝜎1 𝑡 𝐺(𝑠) has poles at ±𝑗𝜔1 ⟹ natural response 𝑐𝑛 𝑡 = 𝐴𝑐𝑜𝑠 𝜔1 𝑡 − 𝜙 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 4.29 Nguyen Tan Tien Time Response §5.The General Second Order Systems - Ex.4.5 Finding 𝜁 and 𝜔𝑛 For a Second-Order System Given the transfer function of Eq (4.23), find 𝜁and 𝜔𝑛 36 𝐺 𝑠 = (4.23) 𝑠 + 4.2𝑠 + 36 Solution From observation • The natural frequency of the given system 𝜔𝑛 = 𝑏 = 36 = • The damping ratio of the given system 𝑎 4.2 𝜁= = = 0.35 𝑏 2×6 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 4.28 Nguyen Tan Tien Time Response §5.The General Second Order Systems 1.Natural frequency 𝜔𝑛 - The natural frequency of a second-order system is the frequency of oscillation of the system without damping 2.Damping ratio 𝜁 - The natural frequency of a second-order system is the frequency of oscillation of the system without damping exponentialdeca𝑦frequency natural period 𝜁= = natural frequency 2𝜋 exponentialtimeconstant - Consider the general second order system 𝑏 𝜔𝑛2 𝐺 𝑠 = ≡ 𝑠 + 𝑎𝑠 + 𝑏 𝑠 + 2𝜁𝜔𝑛 𝑠 + 𝜔𝑛2 ⟹ 𝜔𝑛 = 𝑏 𝑎 𝜁= 𝑏 (4.18) (4.20) HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 4.30 Nguyen Tan Tien Time Response §5.The General Second Order Systems - The poles of the transfer function 𝜔𝑛2 𝐺 𝑠 = ⟹ 𝑝1,2 = −𝜁𝜔𝑛 ± 𝜔𝑛 𝜁2 − 𝑠 + 2𝜁𝜔𝑛 𝑠 + 𝜔𝑛2 ⟹ Second-order response as a function of damping ratio 𝜁=0 0